Solution Manual

Notes to instructorsIntroduction

The following ideas and information are provided to assist the instructor in the design and implementationof the course. Traditionally this course is taught at Washington State University and the University of Idaho as athree-credit semester course which means 3 hours of lecture per week for 15 weeks. Basically the rst 11 chaptersand Chapter 13 (Flow Measurements) are covered in Mechanical Engineering. Chapters 12 (Compressible Flow)and Chapter 14 (Turbomachinery) may be covered depending on the time available and exposure to compressibleow in other courses (Thermodynamics). Open channel ow (Chapter 15) is generally not covered in MechanicalEngineering. When the text is used in Civil Engineering, Chapters 1-11 and 13 are nominally covered and Chapters14 and 15 may be included if time permits and exposure to open channel ow may not be available in other courses.The book can be used for 10-week quarter courses by selecting the chapters, or parts of the chapters, most appropriatefor the course.

Author Contact

Every e!ort has been made to insure that the solution manual is error free. If errors are found (and theywill be!) please contact Professors Crowe or Elger.

Donald Elger Clayton CroweMechanical Engineering Dept School of Mechanical Eng. & Matl. ScienceUniversity of Idaho Washington State UniversityMoscow, ID 83844-0902 Pullman, WA 99164-2920Phone (208) 885-7889 Phone (509) 335-3214Fax (208) 885-9031 Fax (509) 335-4662e-mail: [email protected] e-mail: [email protected]

Design and Computer Problems

Design problems (marked in the text in blue) are those problems that require engineering practices suchas estimation, making asummptions and considering realistic materials and components. These problems provide aplatform for student discussion and group activity. One approach is to divide the class into small groups of three orfour and have these groups work on the design problems together. Each group can then report on their design tothe rest of the class. The role of the professor is to help the student learn the practices of the design review—that is,teach the student to ask in-depth questions and teach them how to develop meaningful and in-depth answers. Thisdialogue stimulates interest and class discussion. Solutions to most design problems are included in the solutionmanual.

Computer-oriented problems (marked in the text is blue) are those problems may best be solved usingsoftware such as spreadsheets, TK Solver or MathCad. The choice is left to the student. The answer book alsoincludes the results for the computer-oriented problems.

1

PROBLEM 2.1

Situation: An engineer needs density for an experiment with a glider.Local temperature = 74.3 !F = 296!7K!Local pressure = 27.3 in.-Hg = 92!45 kPa!

Find: (a) Calculate density using local conditions.(b) Compare calculated density with the value from Table A.2, and make a recom-mendation.

Properties: From Table A.2, "air = 287 Jkg·K , # = 1!22 kg$m

3!

APPROACH

Apply the ideal gas law for local conditions.

ANALYSIS

a.) Ideal gas law

# =%

"&

=92' 450N$m2

(287 kg$m3) (296!7K)

= 1!086 kg/m3

# = 1!09 kg/m3 (local conditions)

b.) Table value. From Table A.2

# = 1!22 kg/m3 (table value)

COMMENTS

1. The density di!erence (local conditions versus table value) is about 12%. Mostof this di!erence is due to the e!ect of elevation on atmospheric pressure.

2. Answer ! Recommendation—use the local value of density because the e!ectsof elevation are signicant.

1

PROBLEM 2.2

Situation: Carbon dioxide is at 300 kPa and 60oC.

Find: Density and specic weight of CO2!

Properties: From Table A.2, "CO2 = 189 J/kg·K.

APPROACH

First, apply the ideal gas law to nd density. Then, calculate specic weight using( = #)!

ANALYSIS

Ideal gas law

#CO2 =*

"&

=300' 000

189(60 + 273)

= 4!767 kg/m3

Specic weight( = #)

Thus

(CO2 = #CO2 × )= 4!767× 9!81= 46.764 N/m3

2

PROBLEM 2.3

Situation: Methane is at 500 kPa and 60oC.

Find: Density and specic weight.

Properties: From Table A.2, "Methane = 518 Jkg·K .

APPROACH


ANALYSIS

Ideal gas law

#He =*

"&

=500' 000

518(60 + 273)

= 2.89 kg/m3

Specic weight( = #)

Thus

(He = #He × )= 2!89× 9!81= 28.4 N/m3

3

PROBLEM 2.4

Situation: Natural gas (10 !C) is stored in a spherical tank. Atmospheric pressure is100 kPa.Initial tank pressure is 100 kPa-gage. Final tank pressure is 200 kPa-gage.Temperature is constant at 10 !C!

Find: Ratio of nal mass to initial mass in the tank.

APPROACH

Use the ideal gas law to develop a formula for the ratio of nal mass to initial mass.

ANALYSIS

Mass+ = #, (1)

Ideal gas law

# =%

"&(2)

Combine Eqs. (1) and (2)

+ = #,"= (%$"& ),"

Volume and gas temperature are constant so

+2

+1=%2%1

and

+2

+1=

300 kPa200 kPa

= 1.5

4

PROBLEM 2.5

Situation: Water and air are at & = 100o- and % = 5 atm.

Find: Ratio of density of water to density of air.

Properties: From Table A.2, "air = 287 J/kg·K. From Table A.5, #water = 958 kg/m3!

APPROACH

Apply the ideal gas to air. Look up the density of water in Table A.5.

ANALYSIS

Ideal gas law

#air =%

"&

=506' 600

287(100 + 273)

= 4!73 kg/m3

For water#water = 958 kg/m

3

Ratio

#water#air

=958

4!73

= 202

5

PROBLEM 2.6

Situation: Oxygen (% = 400 psia, & = 70 !F)lls a tank. Tank volume = 10 ft3! Tankweight =100 lbf.

Find: Weight (tank plus oxygen).

Properties: From Table A.2, "O2 = 1555 ft·lbf/(slug ·o ") !

APPROACH

Apply the ideal gas law to nd density of oxygen. Then nd the weight of the oxygenusing specic weight (() and add this to the weight of the tank.

ANALYSIS

Ideal gas law

%abs. = 400 psia× 144 psf/psi = 57' 600 psf& = 460 + 70 = 530!"

# =%

"&

=57' 600

1555× 530= 0!0699 slugs/ft3

Specic weight (oxygen)

( = #)

= 0!0699× 32!2= 2!25 lbf/ft3

Weight of lled tank

.oxygen = 2!25 lbf/ft3 × 10 ft3

= 22!5 lbf

.total = .oxygen +.tank

= 22!5 lbf + 100 lbf

.total = 122.5 lbf

COMMENTS

For compressed gas in a tank, pressures are often very high and the ideal gas assump-tion is invalid. For this problem the pressure is about 27 atmospheres—it is a goodidea to check a Thermodynamics reference to analyze whether or not real gas e!ectsare signicant.

6

PROBLEM 2.7

Situation: Air is at an absolute pressure of % = 600 kPa and a temperature of& = 50oC.

Find: (a) Specic weight, and (b) density

Properties: From Table A.2, " = 287 Jkg·K !

APPROACH


ANALYSIS

Ideal gas law

#air =*

"&

=600' 000

287(50 + 273)

= 6.47 kg/m3

Specic weight

(air = #air × )= 6!47× 9!81= 63.5 N/m3

7

PROBLEM 2.8

Situation: Consider a mass of air with a volume of 1 cubic mile.

Find: Mass of air in a volume of 1 mi3. Express the answer using units of slugs andkg.

Properties: From Table A.2, #air = 0!00237 slugs/ft3!

Assumptions: The density of air is the value at sea level for standard conditions.

ANALYSIS

Units of slugs

+ = #,

= 0!00237 slugft3× (5280)3 ft3

+ = 3!49× 108 slugs

Units of kg

+ =¡3!49× 108 slug

¢×µ14!59

kg

slug

¶

+ = 5!09× 109 kg

COMMENTS

The mass will probably be somewhat less than this because density decreases withaltitude.

8

PROBLEM 2.9

Situation: This problem involves the e!ects of temperature on the properties of air.The application is a bicyclist.

Find: a.) Plot air density versus temperature for a range of -10oC to 50oC.b.) Plot tire pressure versus temperature for the same temperature range.

Properties: From Table A.2, "air = 287 J/kg/K.

Assumptions: For part b, assume that the bike tire was initially inated to %tire = 450kPa, abs at & = 20oC.

APPROACH

Apply the ideal gas law.

ANALYSIS

Ideal gas law

# =%

"&=

101000

287× (273 + & )

Temperature (o C)

-20 -10 0 10 20 30 40 50 601.05

1.10

1.15

1.20

1.25

1.30

1.35

1.40

Den

sity

(kg/

m)3

with density constant

% = %!&

&!

9

Temperature, oC

-20 -10 0 10 20 30 40 50 60380

400

420

440

460

480

500

520

Tire

pre

ssur

e, k

Pa

10

PROBLEM 2.10

Situation: A design team needs to know how much CO2 is needed to inate a rubberraft.Raft is shown in the sketch below.Ination pressure is 3 psi above local atmospheric pressure. Thus, ination pressureis 17.7 psi = 122 kPa.

Find: (a)Estimate the volume of the raft.(b) Calculate the mass of CO2 in grams to inate the raft.

Properties: From Table A.2, RCO2 = 189 J/kgK.

Assumptions: 1.) Assume that the CO2 in the raft is at 62 !F = 290K!2.) Assume that the volume of the raft can be approximated by a cylinder of diameter0.45 m and a length of 16 m (8 meters for the length of the sides and 8 meters forthe lengths of the ends plus center tubes).

APPROACH

Mass is related to volume by / = ##Volume. Density can be found using the idealgas law.

ANALYSIS

Volume contained in the tubes.

!,— =012

4× 2

=

µ0 × 0!452

4× 16

¶m3

= 2!54m3

!,— = 2!54m3

Ideal gas law

# =%

"&

=122' 000N$m2

(189 J$ kg · K) (290K)= 2!226 kg/m3

11

Mass of CO2

/ = #×Volume=

¡2!226 kg/m3

¢×¡2!54m3

¢

= 5!66 kg

/ = 5!66 kg

COMMENTS

The nal mass (5.66 kg = 12.5 lbm) is large. This would require a large and poten-tially expensive CO2 tank. Thus, this design idea may be impractical for a productthat is driven by cost.

12

PROBLEM 2.11

Situation: The application is a helium lled balloon of radius 3 = 1!3m!% = 0!89 bar = 89 kPa!& = 22 !C = 295!2K!

Find: Weight of helium inside balloon.

Properties: From Table A.2, RHe = 2077 J/kg·K.

APPROACH

Weight is given by . = /)! Mass is related to volume by / = ##Volume. Densitycan be found using the ideal gas law.

ANALYSIS

Volume in a sphere

Volume =4

3033

=4

301!33m3

= 9!203m3

Ideal gas law

# =%

"&

=89' 000N$m2

(2077 J$ kg · K) (295!2K)= 0!145 kg/m3

Weight of helium

. = #×Volume× )=

¡0!145 kg/m3

¢×¡9!203m3

¢×¡9!81m$ s2

¢

= 13!10N

Weight = 13.1 N

13

PROBLEM 2.12

Situation: In the wine and beer industries, fermentation involves glucose (-641256)being converted to ethyl alcohol (-43-4254) plus carbon dioxide gas that escapesfrom the vat.

-641256 $ 2(-43-4254) + 2(-52)

The initial specic gravity is 1.08.Specic gravity of alcohol is 0.80.Saturated solution (water + sugar) has a specic gravity of 1.59.

Find: (a.) Final specic gravity of the wine.(b.) Percent alcohol content by volume after fermentation.

Assumptions: All of the sugar is converted to alcohol.

APPROACH

Imagine that the initial mixture is pure water plus saturated sugar solution and thenuse this visualization to nd the mass of sugar that is initially present (per unitof volume). Next, apply conservation of mass to nd the mass of alcohol that isproduced (per unit of volume). Then, solve for the problem unknowns.

ANALYSIS

The initial density of the mixture is

#"#$ =#%,% + #&,&

,!

where #% and #& are the densities of water and sugar solution (saturated), ,! is theinitial volume of the mixture, and ,& is the volume of sugar solution. The totalvolume of the mixture is the volume of the pure water plus the volume of saturatedsolution

,% + ,& = ,!

The specic gravity is initially 1.08. Thus

6# =#"#$#%

= (1",&,!) +

#&#%

,&,!

1!08 = (1",&,!) + 1!59

,&,!

,&,!

= 0!136

Thus, the mass of sugar per unit volume of mixture

/&

,!= 1!59× 0!136

= 0!216 kg/m3

14

The molecular weight of glucose is 180 and ethyl alcohol 46. Thus 1 kg of glucoseconverts to 0.51 kg of alcohol so the nal density of alcohol is

/'

,!= 0!216× 0!51

= 0!110 kg/m3

The density of the nal mixture based on the initial volume is

/(

,!= (1" 0!136) + 0!110

= 0!974 kg/m3

The nal volume is altered because of conversion

,(,!

=/%

#%,!+/'

#',!

=,%,!+0!51/&

#',!

=,%,!+0!51#&#'

,&,!

= 0!864 +0!51× 1!59

0!8× 0!136

= 1!002

The nal density is

/(

,(=

/(

,!×,!,(

= 0!974×1

1!002= 0!972 kg/m3

The nal specic gravity is6( = 0!972

The alcohol content by volume

,',(

=/'

#',(

=/'

,!

1

#'

,!,(

= 0!110×1

0!8×

1

1!002= 0!137

Thus,Percent alcohol by volume = 13.7%

15

PROBLEM 2.13

Situation: This problem involves the viscosity and density of air and water.

Find: (a)Change in viscosity and density of water for a temperature change of 10!Cto 70!C.(b)Change in viscosity and density of air for a temperature change of 10!C to 70!C.

APPROACH

For water, use data from Table A.5. For air, use data from Table A.3

ANALYSIS

Water

770 = 4!04× 10"4N·s/m2710 = 1!31× 10"3N·s/m2

!7=-9. 06×10"4 8 · 9$/2

#70 = 978 kg/m3

#10 = 1000 kg/m3

!#=-22 kg/m3

Air

770 = 2!04× 10"5 N · s/m2

710 = 1!76× 10"5 N · s/m2

!7 = 2! 8× 10"6 8 ·9$/2

#70 = 1!03 kg/m3

#10 = 1!25 kg/m3

!# = "0!22 kg/m3

16

PROBLEM 2.14

Situation: Air at 10oC and 60oC.

Find: Change in kinematic viscosity from 10oC to 60oC.

Properties: From table A.3, :60 = 1!89× 10"5 m2/s, :10 = 1!41× 10"5 m2/s.

APPROACH

Use properties found in table A.3.

ANALYSIS

!;air,10#60 = (1!89" 1!41)× 10"5 = 4.8×10"6 m2/s

17

PROBLEM 2.15

Situation: This problem involves viscosity of SAE 10W-30 oil, kerosene and water.

Find: Dynamic and kinematic viscosity of each uid at 38!C.

APPROACH

Use property data found in Table A.4, Fig. A.2 and Table A.5.

ANALYSIS

Oil (SAE 10W-30) kerosene water

7(N · s/m2) 6.7×10"2 1.4×10"3 (Fig. A-2) 6.8×10"4

#(kg/m3) 880 993

:(m2/s) 7.6×10"5 1.7×10"6 (Fig. A-2) 6.8×10"7

18

PROBLEM 2.16

Situation: Air and water at 20!C.

Find: (a)Ratio of dynamic viscosity of air to that of water.(b)Ratio of kinematic viscosity of air to that of water.

Properties: From Table A.3, 7air,20!) = 1!81× 10"5 N·s/m2; : = 1!51× 10"5 m2/sFrom Table A.5, 7water,20!) = 1!00× 10"3 N·s/m2; : = 1!00× 10"6 m2/s

ANALYSIS

7air$7water =1!81× 10"5N · s$m2

1!00× 10"3N · s$m2= 1.81×10"2

:air$:water =1!51× 10"5m2$ s1!00× 10"6m2$ s

= 15.1

19

PROBLEM 2.17 Computer Problem - no solution is provided.

20

PROBLEM 2.18

Situation: Sutherland’s equation and the ideal gas law describe behaviors of commongases.

Find: Develop an expression for the kinematic viscosity ratio :$:!, where : is attemperature & and pressure %!

Assumptions: Assume a gas is at temperature &! and pressure %!, where the subscript”o” denes the reference state.

APPROACH

Combine the ideal gas law and Sutherland’s equation.

ANALYSIS

The ratio of kinematic viscosities is

:

:!=

7

7!

#!#=

µ&

&!

¶3*2&! + 6

& + 6

%!%

&

&!

++!=,!

,

³--!

´5*2-!+.-+.

21

PROBLEM 2.19

Situation: The viscosity of air is 7air (15o-) = 1!78× 10"5 N·s/m2!

Find: Dynamic viscosity 7 of air at 200 !C using Sutherland’s equation.

Properties: From Table A.2, 6 = 111<.

ANALYSIS

Sutherland’s equation

7

7!=

µ&

&!

¶3*2&! + 6

& + 6

=

µ473

288

¶3*2288 + 111

473 + 111= 1!438

Thus

7 = 1!4387!= 1!438×

¡1!78× 10"5N · s$m2

¢

7 = 2!56× 10"5 N·s/m2

22

PROBLEM 2.20

Situation: Kinematic viscosity of methane at 15!C and 1 atm is 1!59× 10"5m2$ s!

Find: Kinematic viscosity of methane at 200!C and 2 atm.

Properties: From Table A.2, 6 = 198 K.

APPROACH

Apply the ideal gas law and Sutherland’s equation.

ANALYSIS

: =7

#:

:!=

7

7!

#!#

Ideal-gas law:

:!=7

7!

%!%

&

&!


:

:!=%!%

µ&

&!

¶5*2&! + 6

& + 6

so

:

:!=

1

2

µ473

288

¶5*2288 + 198

473 + 198= 1!252

and

: = 1!252× 1!59× 10"5 m2/s

= 1!99× 10"5m2$ s

23

PROBLEM 2.21

Situation: Nitrogen at 59!F has a dynamic viscosity of 3!59× 10"7 lbf · s$ ft2!

Find: 7 at 200oF using Sutherland’s equation.

Properties: From Table A.2, 6 =192!R.

ANALYSIS


7

7!=

µ&

&!

¶3*2&! + 6

& + 6

=

µ660

519

¶3*2519 + 192

660 + 192= 1!197

7 = 1!197×µ3!59× 10"7

lbf · sft2

¶

= 4! 297× 10"7

7 = 4!30× 10"7 lbf-s/ft2

24

PROBLEM 2.22

Situation: Helium at 59!F has a kinematic viscosity of 1!22× 10"3 ft2$ s!

Find: Kinematic viscosity at 30oF and 1.5 atm using Sutherland’s equation.

Properties: From Table A.2, 6 =143!R.

APPROACH

Combine the ideal gas law and Sutherland’s equation.

ANALYSIS

:

:!=

%!%

µ&

&!

¶5*2&! + 6

& + 6

=1!5

1

µ490

519

¶5*2519 + 143

490 + 143= 1!359

: = 1!359×µ1!22× 10"3

ft2

s

¶

= 1! 658× 10"3ft2

s

: = 1!66× 10"3 ft2$ s

25

PROBLEM 2.23

Situation: Information about propane is provided in the problem statement.

Find: Sutherland’s constant.

ANALYSIS


6

&!=

//!

¡-!-

¢1*2 " 1

1" //!

¡-!-

¢3*2

Also

7

7!= 1!72

&!&

=373

673

Thus

6

&!= 0!964

6 = 360 K

26

PROBLEM 2.24

Situation: Information about ammonia is provided in the problem statement.

Find: Sutherland’s constant.

ANALYSIS


6

&!=

//!

¡-!-

¢1*2 " 1

1" //!

¡-!-

¢3*2 (1)

Calculations

7

7!=

3!46× 10"7

2!07× 10"7= 1!671 (a)

&!&

=528

852= 0!6197 (b)

Substitute (a) and (b) into Eq. (1)

6

&!= 1!71

6 = 903 oR

27

PROBLEM 2.25

Situation: Information about SAE 10W30 motor oil is provided in the problem state-ment.

Find: The viscosity of motor oil at 60 !C' 7(60oC), using the equation 7 = -=0*- .

APPROACH

Use algebra and known values of viscosity (7) to solve for the constant b. Then,solve for the unknown value of viscosity.

ANALYSIS

Viscosity variation of a liquid can be expressed as 7 = -=0*- ! Thus, evaluate 7 attemperatures & and &! and take the ratio:

7

7!= exp

·>(1

&"1

&!)

¸

Take the logarithm and solve for >!

> =ln (7$7!)

( 1-" 1

-!)

Data

7$7! = 0!011$0!067 = 0!164

& = 372

&! = 311

Solve for >> = 3430 (K)

Viscosity ratio at 60oC

7

7!= exp[3430(

1

333"

1

311)

= 0!4833

7 = 0!4833× 0!067

= 0!032 N · s$m2

28

PROBLEM 2.26

Situation: Information about grade 100 aviation oil is provided in the problem state-ment

Find: 7(150oF), using the equation 7 = -=0*- .

APPROACH

Use algebra and known values of viscosity (7) to solve for the constant b. Then,solve for the unknown value of viscosity.

ANALYSIS

Viscosity variation of a liquid can be expressed as 7 = -=0*- ! Thus, evaluate 7 attemperatures & and &! and take the ratio:

7

7!= exp

·>(1

&"1

&!)

¸

Take the logarithm and solve for >

> =ln (7$7!)

( 1-" 1

-!)

Data

7

7!=

0!39× 10"3

4!43× 10"3= 0!08804

& = 670

&! = 560

Solve for >> = 8293 (!R)

Viscosity ratio at 150!F

7

7!= exp[8293(

1

610"

1

560)

= 0!299

7 = 0!299×µ4!43× 10"3

lbf · sft2

¶

= 1!32× 10"3 lbf-s/ft2

29

PROBLEM 2.27

Situation: This problem involves the creation of a computer program to nd Suther-land’s constant and application to CO2!

Find: Develop a computer program and carry out the activities described in thetextbook.

ANALYSIS

Sutherland’s constant

6

273=

//!

¡273-

¢1*2 " 1

1" //!

¡273-

¢3*2 (1)

Program Eq. (1), process data and take the average

6 = 127 K

Dene error

error = 100×

¯¯¯

//!" /

/!|1'21

//!

¯¯¯

The results are

T(K) 260 270 280 290 300 350 500 1000 1500//!|1'21 .9606 .991 1.021 1.050 1.079 1.217 1.582 2.489 3.168

error(%) .013 .039 .084 .118 .108 .366 .486 1.17 3.56

COMMENTS

The error is less than 0.5% for temperatures up to 500 K. The error is greater than3.5% for temperatures above 1500K.

30

PROBLEM 2.28

Situation: Oil (SAE 10W30) lls the space between two plates. Plate spacing is!? = 1$8 = 0!125 in!Lower plate is at rest. Upper plate is moving with a speed @ = 25 ft$ s.

Find: Shear stress.

Properties: Oil (SAE 10W30 @ 150 !F) from Figure A.2: 7 = 5!2× 10"4 lbf·s$ft2!

Assumptions: 1.) Assume oil is a Newtonian uid. 2.) Assume Couette ow (linearvelocity prole).

ANALYSIS

Rate of strain

A@

A?=

!@

!?

=25 ft$ s

(0!125$12) ft

= 2400 s"1

Newton’s law of viscosity

B = 7

µA@

A?

¶

=

µ5!2× 10"4

lbf · sft2

¶×µ2400

1

s

¶

= 1! 248lbf

ft2

B = 1!25 lbf$ ft2

31

PROBLEM 2.29

Situation: Air and water at 40 !C and absolute pressure of 170 kPa

Find: Kinematic and dynamic viscosities of air and water.

Properties: Air data from Table A.3, 7air = 1!91× 10"5 N·s/m2Water data from Table A.5, 7water = 6!53× 10"4 N·s/m2, #water = 992 kg/m3.

APPROACH

Apply the ideal gas law to nd density. Find kinematic viscosity as the ratio ofdynamic and absolute viscosity.

ANALYSIS

A.) AirIdeal gas law

#air =%

"&

=170' 000

287× 313!2= 1!89 kg/m3

7air = 1!91× 10"5 N· sm2

: =7

#

=1!91× 10"5

1!89

:air = 10!1× 10"6m2$ s

B.) water7water = 6!53× 10"5 N·s/m2

: =7

#

: =6!53× 10"4

992

:water = 6!58× 10"7 m2/s

32

PROBLEM 2.30

Situation: Water ows near a wall. The velocity distribution is

@(?) = C³?>

´1*6

where C = 10m$ s, > = 2mm and ? is the distance from the wall in units of mm.

Find: Shear stress in the water at ? = 1 mm.

Properties: Table A.5 (water at 20 !C): 7 = 1!00× 10"3N · s$m2.

ANALYSIS

Rate of strain (algebraic equation)

A@

A?=

A

A?

·C³?>

´1*6¸

=C

>1*61

6?5*6

=C

6>

µ>

?

¶5*6

Rate of strain (at ? = 1mm)

A@

A?=

C

6>

µ>

?

¶5*6

=10m$ s

6× 0!002m

µ2mm

1mm

¶5*6

= 1485 s"1

Shear Stress

B 3=1mm = 7A@

A?

=

µ1!00× 10"3

N · sm2

¶¡1485 s"1

¢

= 1!485Pa

B (? = 1mm) = 1!49Pa

33

PROBLEM 2.31

Situation: Information is provided in problem statement.

Find: Shear stress at walls.

ANALYSIS

Velocity distribution@ = 100?(0!1" ?) = 10? " 100?2

Rate of strain

A@$A? = 10" 200?(A@$A?)3=0 = 10 s"2 (A@$A?)3=041 = "10 s"1

Shear stress

B 0 = 7A@

A?= (8× 10"5)× 10 = 8× 10"4 lbf/ft2

B 041 = 8× 10"4 lbf/ft2

Plot

0.00

0.02

0.04

0.06

0.08

0.10

Dis

tanc

e

Velocity

34

PROBLEM 2.32

Situation: Information is provided in problem statement.

Find: (a) Maximum and minimum shear stress.(b) Maximum shear stress at wall.

ANALYSIS

B = 7A,$A?

Bmax % 7(!,$!?) next to wall

Bmax = (10"3N · s/m2)((1 m/s)$0!001 m) = 1.0 N/m2

The minimum shear stress will be zero, midway between the two walls, where thevelocity gradient is zero.

35

PROBLEM 2.33

Situation: Glycerin is owing in between two stationary plates. The plate spacing isD = 5cm!The velocity distribution is

@ = "1

27

A%

AE

¡D? " ?2

¢

where the pressure gradient is A%$AE = "1!6 kN$m3Pressure gradient

Find:a.) Velocity and shear stress at12 mm from wall (i.e. at ? = 12mm).b.) Velocity and shear stress at the wall (i.e. at ? = 0mm).

Properties: Glycerin at 20 !C from Table A.4: 7 = 1!41N · s$m2.

APPROACH

Find velocity by direct substitution into the specied velocity distribution. Findshear stress using B = 7 (A@$A?), where the rate-of-strain (i.e. the derivative A@$A?)is found by di!erentiating the velocity distribution.

ANALYSIS

a.) Velocity (at ? = 12mm)

@ = "1

27

A%

AE

¡D? " ?2

¢

= "1

2 (1!41N · s$m2)¡"1600N$m3

¢ ¡(0!05m) (0!012m)" (0!012m)2

¢

= 0!258 7m

s

@ (? = 12mm) = 0!259m$ s

Rate of strain (general expression)

A@

A?=

A

A?

µ"1

27

A%

AE

¡D? " ?2

¢¶

=

µ"1

27

¶µA%

AE

¶A

A?

¡D? " ?2

¢

=

µ"1

27

¶µA%

AE

¶(D " 2?)

36


A@

A?=

µ"1

27

¶µA%

AE

¶(D " 2?)

=

µ"

1

2 (1!41N · s$m2)

¶µ"1600

N

m3

¶(0!05m" 2× 0!012m)

= 14!75 s"1

Shear stress

B = 7A@

A?

=

µ1!41

N · sm2

¶¡14!75 s"1

¢

= 20! 798Pa

B (? = 12mm) = 20!8Pa

b.) Velocity (at ? = 0mm)

@ = "1

27

A%

AE

¡D? " ?2

¢

= "1

2 (1!41N · s$m2)¡"1600N$m3

¢ ¡(0!05m) (0m)" (0m)2

¢

= 0!00m

s

@ (? = 0mm) = 0m$ s


A@

A?=

µ"1

27

¶µA%

AE

¶(D " 2?)

=

µ"

1

2 (1!41N · s$m2)

¶µ"1600

N

m3

¶(0!05m" 2× 0m)

= 28!37 s"1

Shear stress (at ? = 0mm)

B = 7A@

A?

=

µ1!41

N · sm2

¶¡28!37 s"1

¢

= 40!00Pa

B (? = 0mm) = 40!0Pa

COMMENTS

37

1. As expected, the velocity at the wall (i.e. at ? = 0) is zero due to the no slipcondition.

2. As expected, the shear stress at the wall is larger than the shear stress awayfrom the wall. This is because shear stress is maximum at the wall and zeroalong the centerline (i.e. at ? = D$2).

38

PROBLEM 2.34

Situation: Laminar ow occurs between two parallel plates–details are provided inthe problem statement.

Find: Is the maximum shear greater at the moving plate or the stationary plate?

ANALYSIS

B = 7A@$A?

7A@$A? = "7(1$27)(A%$A9)(4 " 2?) + @57$4Evaluate B at ? = 4 :

B6 = "(1$2)(A%$A9)(4 " 24) + @57$4= (1$2)(A%$A9)4 + @57$4

Evaluate B at ? = 0

B 0 = "(1$2)(A%$A9)4 + @57$4

Observation of the velocity gradient lets one conclude that the pressure gradient A%$A9is negative. Also @5 is negative. Therefore |B7| F |B 0| ! The maximum shear stressoccurs at ? = 4!

Maximum shear stress occur along the moving plate where ? = 4 .

39

PROBLEM 2.35


Find: Position (?) of zero shear stress.

ANALYSIS

B = 7A@$A?

= "7(1$27)(A%$A9)(4 " 2?) + @57$4= "(1$2)(A%$A9)(4 " 2?) + @57$4

Set B = 0 and solve for ?

0 = "(1$2)(A%$A9)(4 " 2?) + @57$4

? = (4$2)" (7@5$(4A%$A9))

40

PROBLEM 2.36


Find: Derive an expression for plate speed (@5) to make the shear stress zero at ? = 0!

ANALYSIS

From solution to 2.34

B = 7A@$A? = 0 at ? = 0

A@$A? = "(1$27)(A%$A9)(4 " 2?) + @5$4Then, at ? = 0 : A@$A? = 0 = "(1$27)(A%$A9)4 + @5$4

Solve for @5 : @5 = (1$27)(A%$A9)42

Note : because A%$A9 G 0' @5 G 0!

41

PROBLEM 2.37

Situation: A damping device is described in the problem statement.

Find: Torque on shaft.

Properties: From Table A.4, 7(38oC)=3!6× 10"2 N·s/m2!

ANALYSIS

Rd! !R

"R

Rsin

A& = 3AH

A& = 3BAI

where B = 7(A,$A?) = 7(!,$!")

= 7(J" sin K$!")

= 3!6× 10"2 N · s/m2)(10× 20$60) rad/s(0!05 m sin K$10"3 m)= 1!885 sin K N/m2

AI = 20" sin K"AK

= 20"2 sin K"AK

= 20"2 sin KAK

3 = " sin K

Then

A& = " sin K(1!885 sin K)(20"2 sin KAK)

A& = 11!84"3 sin3 KAK

& = 11!84"38Z

0

sin3 KAK

= 11!84(0!05)3["(1$3) cos K(sin2 K + 2)]80= 11!84(0!05)3["(1$3)("1)(2)" ("1$3)(1)(2)]

Torque =1!97× 10"3N · m

42

PROBLEM 2.38

Situation: Oxygen at 50 !F and 100 !F.

Find: Ratio of viscosities: /100/50.

ANALYSIS

Because the viscosity of gases increases with temperature 7100$750 F 1. Correctchoice is (c) .

43

PROBLEM 2.39

Situation: This problem involves a cylinder falling inside a pipe that is lled with oil.

Find: Speed at which the cylinder slides down the pipe.

Properties: SAE 20W oil from Figure A.2: 7(10oC) = 0.35 N·s/m2!

ANALYSIS

B = 7A,$A?

.$(0AL) = 7,fall$[(1 " A)$2],fall = . (1 " A)$(20AL7),fall = 20(0!5× 10"3)$(20 × 0!1× 0!2× 3!5× 10"1)

= 0.23 m/s

44

PROBLEM 2.40

Situation: This problem involves a cylinder falling inside a pipe–details are providedin problem statement.

Find: Weight of cylinder.

Properties: From Figure A.2, 7(10oC)=0.35 N·s/m2!

ANALYSIS

Newton’s second law". + HB = /C

". + 0AL7,$[(1 " A)$2] = (.$)) C". + (0 × 0!1× 0!2× 3!5× 10"1, )$(0!5× 10"3$2) =.C$9!81

Substituting , = 0!5 m/s and C = 14 m/s2 and solving yields . = 18!1N

45

PROBLEM 2.41

Situation: A disk is rotated very close to a solid boundary–details are provided inproblem statement.

Find: (a) Ratio of shear stress at 3 = 2 cm to shear stress at 3 = 3 cm.(b) Speed of oil at contact with disk surface.(c) Shear stress at disk surface.

Assumptions: Linear velocity distribution: A,$A? = ,$? = J3$?!

ANALYSIS

B = 7A,$A? = 7J3$?

B 2$B 3 = (7× 1× 2$?)$(7× 1× 3$?) = 2$3 = 0.667

, = J3 = 2× 0!03 = 0.06 m/s

B = 7A,$A? = 0!01× 0!06$0!002 = 0.30 N/m2

46

PROBLEM 2.42

Situation: A disk is rotated very close to a solid boundary–details are provided inproblem statement.

Find: Torque to rotate disk.

Assumptions: Linear velocity distribution: A,$A? = ,$? = J3$?!

ANALYSIS

B = 7A,$A?

B = 7J3$?

= 0!01× 5× 3$0!002 = 253 N/m2

A Torque = 3BAI

= 3(103)203A3 = 50033A3

Torque =

0405Z

0

50033A3 = 50034$4¯0450

Torque =2.45×10"4 N·m

47

PROBLEM 2.43

Situation: In order to provide damping for an instrument, a disk is rotated in acontainer of oil.

Find: Derive an equation for damping torque as a function of 1'6' J and 7!

APPROACH

Apply the Newton’s law of viscosity.

ANALYSIS

Shear stress

B = 7A,

A?

=73J

9

Find di!erential torque–on an elemental strip of area of radius 3 the di!erentialshear force will be BAI or B(203A3). The di!erential torque will be the product ofthe di!erential shear force and the radius 3.

A&one side = 3[B(203A3)]

= 3[(73J$9)(203A3)]

= (207J$9)33A3

A&both sides = 4(307J$9)33A3

Integrate

& =

9*2Z

0

(407J$9)33A3

= (1/16)07J14$9

48

PROBLEM 2.44

Situation: One type of viscometer involves the use of a rotating cylinder inside a xedcylinder. The temperature range is 50 to 200!F.

Find: (a) Design a viscometer that can be used to measure the viscosity of motor oil.

Assumptions:

1. Motor oil is SAE 10W-30. Data from Fig A-2: 7 will vary from about 2 ×10"4lbf-s/ft2 to 8× 10"3lbf-s/ft2!

2. Assume the only signicant shear stress develops between the rotating cylinderand the xed cylinder.

3. Assume we want the maximum rate of rotation (J) to be 3 rad/s.

ANALYSIS

One possible design solution is given below.Design decisions:

1. Let M = 4!0 in. = 0.333 ft

2. Let I.D. of xed cylinder = 9.00 in. = 0.7500 ft.

3. Let O.D. of rotating cylinder = 8.900 in. = 0.7417 ft.

Let the applied torque, which drives the rotating cylinder, be produced by a forcefrom a thread or small diameter monolament line acting at a radial distance 3&!Here 3& is the radius of a spool on which the thread of line is wound. The appliedforce is produced by a weight and pulley system shown in the sketch below.

h rc

"rW

Pulley

The relationship between 7' 3&' J' M' and . is now developed.

& = 31H& (1)

where & = applied torque31 = outer radius of rotating cylinder

49

H& = shearing force developed at the outer radius of the rotating cylinder but H& =BI& where I& = area in shear = 2031M

B = 7A,$A? % 7!,$!3 where !, = 31J and !3 = spacing

Then & = 31(7!,$!3)(2031M)

= 317(31J$!3)(2031M) (2)

But the applied torque & =.3& so Eq. (2) become

.3& = 3317J(20)M$!3

Or

7 = (.3&!3)$(20JM331) (3)

The weight. will be arbitrarily chosen (say 2 or 3 oz.) and J will be determined bymeasuring the time it takes the weight to travel a given distance. So 3&J = ,fall orJ = ,fall$3&! Equation (3) then becomes

7 = (.$,()(32&$3

31)(!3$(20M))

In our design let 3& = 2 in. = 0.1667 ft. Then

7 = (.$H()(0!16672$!37083)(0!004167$(20 × !3333)

7 = (.$,()(!02779$!05098)

7 = (.$,()(1!085× 10"3) lbf · s$ft2

Example: If . = 2oz. = 0.125lb. and ,( is measured to be 0.24 ft/s then

7 = (0!125$0!24)(1!085× 10"3)= 0!564× 10"4 lbf · s$ ft2

COMMENTS Other things that could be noted or considered in the design:

1. Specify dimensions of all parts of the instrument.

2. Neglect friction in bearings of pulley and on shaft of cylinder.

3. Neglect weight of thread or monolament line.

4. Consider degree of accuracy.

5. Estimate cost of the instrument.

50

PROBLEM 2.45

Situation: Water in a 1000 cm3 volume is subjected to a pressure of 2× 106N$m2!

Find: Volume after pressure applied.

Properties: From Table A.5, N = 2!2× 109 Pa

ANALYSIS

Modulus of elasticity

N = "!%,—!,—

!,— = "!%

N,—

= "·(2× 106) Pa(2!2× 109) Pa

¸1000 cm3

= "0!9091 cm3

Final volume

,—(#:'2 = ,—+!,—

= (1000" 0!9091) cm3

= 999!1 cm3

,—(#:'2 = 999 cm3

51

PROBLEM 2.46

Situation: Water is subjected to an increase in pressure.

Find: Pressure increase needed to reduce volume by 1%.

Properties: From Table A.5, N = 2!2× 109 Pa!

ANALYSIS

Modulus of elasticity

N = "!%,—!,—

!% = N!,—,—

= "¡2!2× 109 Pa

¢µ"0!01× ,—,—

¶

=¡2!2× 109 Pa

¢(0!01)

= 2! 2× 107 Pa

!% = 22MPa

52

PROBLEM 2.47

Situation: Very small spherical droplet of water.

Find: Pressure inside.

ANALYSIS

Refer to Fig. 2-6(a). The surface tension force, 203O, will be resisted by the pressureforce acting on the cut section of the spherical droplet or

%(032) = 203O

% = 2O$3

= 4O$A

53

PROBLEM 2.48

Situation: A spherical soap bubble has an inside radius ", a wall-thickness P, andsurface tension O.

Find: (a) Derive a formula for the pressure di!erence across the bubble(b) Pressure di!erence for a bubble with a radius of 4 mm.

Assumptions: The e!ect of thickness is negligible, and the surface tension is that ofpure water.

APPROACH

Apply equilibrium, then the surface tension force equation.

ANALYSIS

Force balance

p

2 x 2 R# $

Surface tension force

XH = 0

!%0"2 " 2(20"O) = 0

!% = 4O$"

!%4mm rad. = (4× 7!3× 10"2 N/m)$0!004 m = 73.0 N/m2

54

PROBLEM 2.49

Situation: A water bug with 6 legs, each with a contact length of 5 mm, is balancedon the surface of a water pond.

Find: Maximum mass of bug to avoid sinking.

Properties: Surface tension of water, from Table A.5, O = 0!073 N/m.

APPROACH


ANALYSIS

Force equilibrium

Upward force due to surface tension = Weight of Bug

H- = /)

To nd the force of surface tension (H- ), consider the cross section of one leg of thebug:

!

F F

Surface tensionforce on oneside of leg

Cross sectionof bug leg

Assume is smallThen cos =1; F cos = F

!! !


H- = (2$leg)(6 legs)OL

= 12OL

= 12(0!073 N/m)(0!005 m)

= 0!00438N

Apply equilibrium

H- "/) = 0

/ =H-)=0!00438N

9!81m2$ s

= 0!4465× 10"3 kg

/ = 0!447× 10"3 kg

55

PROBLEM 2.50

Situation: A water column in a glass tube is used to measure pressure.Part of the water column height is due to pressure in a pipe, and part is due tocapillary rise.Additional details are provided in the problem statement.

Find: Height of water column due to surface tension e!ects.

Properties: From Table A.5: surface tension of water is 0.005 lbf/ft.

ANALYSIS


!M = 4O$((A) = 4× 0!005$(62!4× A) = 3!21× 10"4$A ft.A = 1$4 in. = 1$48 ft.; !M = 3!21× 10"4$(1$48) = 0!0154 ft. = 0.185 in.

A = 1$8 in. = 1$96 ft.; !M = 3!21× 10"4$(1$96) = 0!0308 ft. = 0.369 in.

A = 1$32 in. = 1$384 ft.; !M = 3!21× 10"4$(1$384) = 0!123 ft.= 1.48 in.

56

PROBLEM 2.51

Situation: Two vertical glass plates are spaced 1 mm apart.

Find: Capillary rise (M) between the plates.

Properties: From Table A.5, surface tension of water is 7!3× 10"2 N/m.

APPROACH


ANALYSIS

!

$$y

y

Equilibrium

XH3 = 0

Force due to surface tension = Weight of uid that has been pulled upward

(2L)O = (MLP) (

Solve for capillary rise (M)

2OL" MLP( = 0

M =2O

(P

M =2× (7!3× 10"2)9810× 0!0010

= 0!0149 m

= 14.9 mm

57

PROBLEM 2.52

Situation: A spherical water drop has a diameter of 1-mm.

Find: Pressure inside the droplet.

Properties: From Table A.5, surface tension of water is 7!3× 10"2 N/m

APPROACH


ANALYSIS

Equilibrium (half the water droplet)

Force due to pressure = Force due to surface tension

%I = O2

!%0"2 = 20"O

Solve for pressure

!% = 2O$"

!% = 2× 7!3× 10"2$(0!5× 10"3) = 292 N/m2

58

PROBLEM 2.53

Situation: A tube employing capillary rise is used to measure temperature of water.

Find: Size the tube (this means specify diameter and length).

APPROACH

Apply equilibrium and the surface tension force equation.

ANALYSIS

The elevation in a column due to surface tension is

!M =4O

(A

where ( is the specic weight and A is the tube diameter. For the change in surfacetension due to temperature, the change in column elevation would be

!M =4!O

(A=4× 0!01679810× A

=6!8× 10"6

A

The change in column elevation for a 1-mm diameter tube would be 6.8 mm . Spe-cial equipment, such the optical system from a microscope, would have to be used tomeasure such a small change in deection It is unlikely that smaller tubes made oftransparent material can be purchased to provide larger deections.

59

PROBLEM 2.54

Situation: A glass tube is immersed in a pool of mercury–details are provided in theproblem statement.

Find: Depression distance of mercury: A

APPROACH

Apply equilibrium and the surface tension force equation.

ANALYSIS

cos K0AO = !M(0A2

4

Solving for !M results in

!M =4 cos KO

(A

Substitute in values

!M =4× cos 40× 0!514

(13!6× 9810)× 0!001= 0!0118m

!M = 11!8mm

60

PROBLEM 2.55

Situation: A soap bubble and a droplet of water both with a diameter of 2mm, fallingin air. The value of surface tension is equal.

Find: Which has the greater pressure inside.

ANALYSIS

The soap bubble will have the greatest pressure because there are two surfaces (twosurface tension forces) creating the pressure within the bubble. The correct choice isa)

61

PROBLEM 2.56

Situation: A hemispherical drop of water at 20oC is suspended under a surface.

Find: Diameter of droplet just before separation

Properties: Table A.5 (water at 20 !C): ( = 9790N$m3'[for surface tension, seefootnote (2)] O = 0!073N$m! .

ANALYSIS

Equilibrium.

Weight of droplet = Force due to surface tensionµ013

12

¶( = (01)O

Solve for 1

12 =12O

(

=12× (0!073 N/m)9790 N/m3

= 8! 948× 10"5m2

1 = 9! 459× 10"3m

1 = 9!46mm

62

PROBLEM 2.57

Situation: Surface tension is being measured by suspending liquid from a ring witha mass of 10 grams, an outside diameter of 10 cm and an inside diameter of 9.5 cm.Force to pull ring is weight corresponding to 14 gms.

Find: Surface tension

ANALYSIS

Equilibrium.

(Upward force) = (Weight of uid) + (Force due to surface tension)

H = . + O(01# + 01!)

Solve for surface tension

O =H ".

0(1# +1!)

=(0!014" 0!010) kg× 9!81m$ s2

0(0!1 + 0!095)m

= 6! 405× 10"2kg

s2

O = 0!0641 N/m

63

PROBLEM 2.58

Situation: The boiling temperature of water decreases with increasing elevation.Change in vapor pressure with temperature is "341 kPa!)

!Atmospheric pressure (3000 m) is 69 kPa.

Find: Boiling temperature at an altitude of 3000 m.

Properties: Vapor pressure of water at 100!C is 101 kN$m2.

Assumptions: Assume that vapor pressure versus boiling temperature is a linearrelationship.

APPROACH

Develop a linear equation for boiling temperature as a function of elevation.

ANALYSIS

Let D& = "Boiling Temperature." Then, D& as a function of elevation is

D& (3000 m) = BT (0 m) +µ!D&

!%

¶!%

Thus,

D& (3000 m) = 100 !C+

µ"1!0 !C3!1 kPa

¶(101" 69) kPa

= 89! 677 !C

Boiling Temperature (3000 m) = 89!7 !C

64

PROBLEM 3.1

Situation: A Crosby gage tester is applied to calibrate a pressure gage.A weight of 140 N results in a reading of 200 kPa.The piston diameter is 30 mm.

Find: Percent error in gage reading.

APPROACH

Calculate the pressure that the gage should be indicating (true pressure). Comparethis true pressure with the actual pressure.

ANALYSIS

True pressure

%true =H

I

=140N

(0$4× 0!032) m2= 198' 049 kPa

Percent error

% Error =(%recorded " %true) 100

%true

=(200" 198) 100

198= 1!0101%

% Error = 1!01%

65

PROBLEM 3.2

Situation: Two hemispherical shells are sealed together.Exterior pressure is %atm = 14!5 psia! Interior pressure is 0.1patm!Inner radius is 6 in. Outer radius is 6.25 in.Seal is located halfway between the inner and outer radius.

Find: Force required to separate the two shells.

APPROACH

Apply equilibrium to a free body comprised of one shell plus the air inside.

ANALYSIS

Free body diagram

pinsideA

patmAFpull

Equilibrium.

PH3 = 0

Hpull + %#I" %atmI = 0

Solve for force

Hpull = (%atm " %#)I= (1" 0!1)

¡14!5 lbf$ in2

¢ ¡0 × 6!1252 in2

¢

= 1538 lbf

Hpull = 1540 lbf

66

PROBLEM 3.3

Situation: This is an applied problem. To work the problem, we recorded data froma parked vehicle. Relevant information:

• Left front tire of a parked VW Passat 2003 GLX Wagon (with 4-motion).

• Bridgestone snow tires on the vehicle.

• Ination pressure = 36 psig. This value was found by using a conventional"stick-type" tire pressure gage.

• Contact Patch: 5!88 in × 7!5 in. The 7.5 inch dimension is across the tread.These data were found by measuring with a ruler.

• Weight on the front axle = 2514 lbf. This data was recorded from a stickeron the driver side door jamb. The owners manual states that this is maximumweight (car + occupants + cargo).

Assumptions:

1. The weight on the car axle without a load is 2000 lbf. Thus, the load actingon the left front tire is 1000 lbf.

2. The thickness of the tire tread is 1 inch. The thickness of the tire sidewall is1/2 inch.

3. The contact path is at and rectangular.

4. Neglect any tensile force carried by the material of the tire.

Find:(a) Apply engineering principles to estimate the size of the contact patch.(b) Compare the estimated area of contact with the measured area of contact.

APPROACH

To estimate the area of contact, apply equilibrium to the contact patch.

ANALYSIS

Equilibrium in the vertical direction applied to a section of the car tire

%#I# = Hpavement

67

where %# is the ination pressure, I# is the area of the contact patch on the inside ofthe tire and Hpavement is the normal force due to the pavement. Thus,

I# =Hpavement%#

=1000 lbf

36 lbf$ in2

= 27!8 in2

Comparison. The actual contact patch has an area I! = 5!88 in×7!5 in = 44!1 in2!Using the assumed thickness of rubber, this would correspond to an inside contactarea of I! = 4!88 in× 5!5 in = 26!8 in2!Thus, the predicted contact area

¡27!8 in2

¢and the measured contact area

¡26!8 in2

¢

agree to within about 1 part in 25 or about 4%.

COMMENTS

The comparison between predicted and measured contact area is highly dependenton the assumptions made.

68

PROBLEM 3.4

Situation: An air chamber is described in the problem statement.

Find: Number of bolts required at section B-B.

Assumptions: Same force per bolt at B-B.

ANALYSIS

Hydrostatic force

H per bolt at I"I = %(0$4)12$20

%(0$4)12$20 = %(0$4)A2$Q

Q = 20× (A$1)2

= 20× (1$2)2

Q = 5

69

PROBLEM 3.5

Situation: A glass tube is inserted into water.Tube length is 2 = 10 cm! Tube diameter is A = 0!5mm!Depth of insertion is 2 cm. Atmospheric pressure is %atm = 100 kPa.

Find: Location of water line in tube.

Properties: Density of water is # = 1000 kg$m3! Surface tension (from Table A.5;see footnote 2) is O = 0!073N$m!

ANALYSIS

p Ai

p Al

2 cm l

Equilibrium (system is a very thin layer of uid)X

H; = 0

"%#I+ %<I+ O0A = 0 (1)

where %# is the pressure inside the tube and %< is the pressure in water at depth L!

Ideal gas law (constant temperature)

%#,"# = %atm,"tube%# = %atm(,"tube $,"#)

= %atm(0!10Itube$((!08 + L)(Itube))

%# = %atm(0!10$(!08 + L)) (2)

Hydrostatic equation (location 1 is the free surface of the water; location 2 is at adepth L)

%< = %atm + #)L (3)

Solve Eqs. (1) to (3) simultaneously for L' %# and %< (we used TK Solver)

L = 0!019233m

%# = 100772Pa

%< = 100189Pa

L = 1!92 cm

70

PROBLEM 3.6

Situation: A reservoir is described in the problem statement.

Find: Describe the gage pressure along a vertical line.

ANALYSIS

Correct graph is (b).

71

PROBLEM 3.7

Situation: A closed tank with Bourdon-tube gages tapped into it is described in theproblem statement.

Find:(a) Specic gravity of oil.(b) Pressure at C.

APPROACH

Apply the hydrostatic equation.

ANALYSIS

Hydrostatic equation (from oil surface to elevation B)

%= + (R= = %> + (R>

50' 000 N/m2 + (oil (1 m ) = 58,530 N/m2 + (oil (0 m)

(oil = 8530 N/m2

Specic gravity

6 =(oil(water

=8530 N/m2

9810 N/m2

6oil = 0!87

Hydrostatic equation (in water)

%1 = (%btm of oil) + (water (1m)

Hydrostatic equation (in oil)

%btm of oil = (58' 530Pa + (oil × 0!5m)

Combine equations

%1 = (58' 530Pa + (oil × 0!5m) + (water (1m)= (58' 530 + 8530× 0!5) + 9810 (1)= 72' 605 N/m2

%1 = 72!6 kPa

72

PROBLEM 3.8

Situation: A manometer is described in the problem statement.

Find: Water surface level in the left tube as compared to the right tube.

ANALYSIS

(a) The water surface level in the left tube will be higher because of greater surfacetension e!ects for that tube.

73

PROBLEM 3.9

Situation: A force is applied to a piston—additional details are provided in the problemstatement.

Find: Force resisted by piston.

APPROACH

Apply the hydrostatic equation and equilibrium.

ANALYSIS

Equilibrium (piston 1)

H1 = %1I1

%1 =H1I1

=4× 200N0 · 0!042m2

= 1!592× 105 Pa

Hydrostatic equation

%2 + (R2 = %1 + (R1

%2 = %1 + (6(water) (R1 " R2)= 1!592× 105 Pa +

¡0!85× 9810N$m3

¢("2m)

= 1!425× 105 Pa

Equilibrium (piston 2)

H2 = %2I2

=¡1!425× 105N$m2

¢Ã0 (0!1m)2

4

!

= 1119N

H2 = 1120 N

74

PROBLEM 3.10

Situation: A diver goes to a depth of 50 meters.

Find: (a) Gage pressure.(b) Ratio of pressure to normal atmospheric pressure.

APPROACH

Apply the hydrostatic equation.

ANALYSIS


% = (!R = 9790× 50= 489' 500 N/m2

% = 489!5 kPa gage

Calculate pressure ratio

%50%atm

=489!5 + 101!3

101!3

%50$%atm = 5!83

75

PROBLEM 3.11

Situation: Water and kerosene are in a tank. & = 20 !C!The water layer is 1 m deep. The kerosene layer is 0.5 m deep.

Find: Gage pressure at bottom of tank.

Properties: From Table A.5: (water = 9790 N/m3 (kerosene = 8010 N/m

3!

APPROACH

Apply the manometer equation.

ANALYSIS

Manometer equation (add up pressure from the top of the tank to the bottom of thetank).

%atm + (k (0!5m) + (w (1!0m) = %btm

Solve equation

%btm = 0 + (k (0!5m) + (w (1!0m)

=¡8010N$m3

¢(0!5m) +

¡9790N$m3

¢(1!0m)

= 13!8 kPa

%btm = 13!8 kPa-gage

76

PROBLEM 3.12

Situation: A hydraulic lift is being designed.Capacity = 20,000 lbf (10 tons). Weight of lift = 1000 lbf.Lift speed = 6 feet in 20 seconds. D = 2 to 8 inches.Piston pump data. Pressure range 200 to 3000 psig. Capacity = 5, 10 and 15 gpm.

Find: (a) Select a hydraulic pump capacity (gpm).(b) Select a cylinder diameter (1)

APPROACH

Apply equilibrium to nd the smallest bore diameter (D) that works. Then nd thelargest bore diameter that works by considering the lift speed requirement. Selectbore and pump combinations that meet the desired specications.

ANALYSIS

Equilibrium (piston)H = %I

where H = 21' 000 lbf is the load that needs to be lifted and % is the pressure on thebottom of the piston. Maximum pressure is 3000 psig so minimum bore area is

Imin =H

%max

=21' 000 lbf

3000 in2

= 7!0 in2

77

Corresponding minimum bore diameter is

1 =

r4

0I

1min = 2!98 in

The pump needs to provide enough ow to raise the lift in 20 seconds.

I!2 = ,!P

where I is the bore area, !2 is stroke (lift height), , is the volume/time of uidprovided by the pump, and !P is the time. Thus, the maximum bore area is

Imax =,!P

!2

Conversion from gallons to cubic feet¡ft3¢: 7.48 gal=1 ft3! Thus, the maximum

bore diameter for three pumps (to meet the lift speed specication) is given in thetable below.

pump (gpm) pump (cfm) A (ft2) Dmax (in)5 0.668 0.037 2.6110 1.337 0.074 3.6815 2.01 0.116 4.61

Since the minimum bore diameter is 2.98 in., the 5 gpm pump will not work. The 10gpm pump can be used with a 3 in. bore. The 15 gpm pump can be used with a 3or 4 in. bore.

1.) The 10 gpm pump will work with a bore diameter between 3.0 and 3.6 inches.

2.) The15 gpm pump will work with a bore diameter between 3.0 and 4.6 inches.

COMMENTS

1. These are preliminary design values. Other issues such as pressure drop in thehydraulic lines and valves would have to be considered.

2. We recommend selecting the 15 gpm pump and a 4.5 inch bore to providelatitude to handle pressure losses, and to reduce the maximum system pressure.

78

PROBLEM 3.13

Situation: A liquid occupies an open tank.At a depth of 5m' pressure is % = 75 kPa!

Find: Specic weight and specic gravity of the liquid.

APPROACH

Apply the hydrostatic equation between the top surface and a depth of 5 m.

ANALYSIS

Hydrostatic equation. (location 1 is on the top surface; location 2 is at depth of 5m).

%1(+ R1 =

%2(+ R2

%atm(+ 5m =

%2(+ 0m

Since %atm = 0

( =%2(5m)

=75' 000N$m2

(5m)

( = 15 kN$m3

Specic gravity

6 =15 kN$m3

9!8 kN$m3

6 = 1!53

79

PROBLEM 3.14

Situation: A tank with an attached manometer is described in the problem statement.

Find: Increase of water elevation in manometer.

Properties: From Table A.5, (w=9790 N/m3!

Assumptions: Ideal gas.

APPROACH

Apply the hydrostatic equation and the ideal gas law.

ANALYSIS

Ideal gas law (mole form; apply to air in the manometer tube)

%," = Q<&

Because the number of moles (Q) and temperature (& ) are constants, the ideal gasreduces to Boyle’s equation.

%1,"1 = %2,"2 (1)

State 1 (before air is compressed)

%1 = 100' 000 N/m2 abs

,"1 = 1 m×Itube(a)

State 2 (after air is compressed)

%2 = 100' 000 N/m2 + (w(1 m"!L)

,"2 = (1 m"!L)Itube(b)


%1,"1 = %2,"2¡100' 000N$m2

¢(1 m×Itube) =

¡100' 000 N/m2 + (w(1 m"!L)

¢(1 m"!L)Itube

100' 000 = (100' 000 + 9810 (1"!L)) (1"!L)

Solving for !L!L = 0!0826 m

80

PROBLEM 3.15

Situation: A tank tted with a manometer is described in the problem statement.

Find: Deection of the manometer.(!M)

APPROACH

Apply the hydrostatic principle to the water and then to the manometer uid.

ANALYSIS

Hydrostatic equation (location 1 is on the free surface of the water; location 2 is theinterface)

%1(water

+ R1 =%2(water

+ R2

0Pa

9810N$m3+ 0!15m =

%29810N$m3

+ 0m

%2 = (0!15m)¡9810N$m3

¢

= 1471!5Pa

Hydrostatic equation (manometer uid; let location 3 be on the free surface)

%2(man. uid

+ R2 =%3

(man. uid+ R3

1471!5Pa

3 (9810N$m3)+ 0m =

0Pa

(man. uid+!M

Solve for !M

!M =1471!5Pa

3 (9810N$m3)= 0!0500m

!M = 5!00 cm

81

PROBLEM 3.16

Situation: An odd tank is described in the problem statement.

Find:(a) Maximum gage pressure.(b) Where will maximum pressure occur.(c) Hydrostatic force on side C-D.

APPROACH

Apply the hydrostatic equation, and then the hydrostatic force equation.

ANALYSIS


0 + 4× (H2O + 3× 3(H2O = %max

%max = 13× 9' 810= 127' 530 N/m2

%max = 127!5 kPa

Answer ! Maximum pressure will be at the bottom of the liquid that has a specicgravity of 6 = 3.

Hydrostatic force

H)9 = %I

= (127' 530" 1× 3× 9810)× 1 m2

H)9 = 98!1 kN

82

PROBLEM 3.17

Situation: Sea water at a point 6 km deep is described in the problem statement.

Find: % di!erence in sea water density.

APPROACH

Apply the hydrostatic equation to nd the change in pressure. Use bulk modulus torelate change in pressure to change in density.

ANALYSIS


!% = ( (!M)

= 10' 070× 6× 103

Bulk modulus

N? = !%$(A#$#)

(A#$#) = !%$N@

= (10' 070× 6× 103)$(2!2× 109)= 27!46× 10"3

A#$# = 2!75%

83

PROBLEM 3.18

Situation: A steel pipe and chamber weigh 600 lbf!The dimension L = 2!5 ft!

Find: Force exerted on chamber by bolts (H>)

APPROACH

Apply equilibrium and the hydrostatic equation.

ANALYSIS

Equilibrium. (system is the steel structure plus the liquid within)

(Force exerted by bolts) + (Weight of the liquid) +

(Weight of the steel) = (Pressure force acting on the bottom of the free body )

H> +.liquid +.& = %2I2 (1)

Hydrostatic equation. (location 1 is on surface; location 2 at the bottom)

%1(+ R1 =

%2(liquid

+ R2

0 + 5L =%2

1!2(water+ 0

%2 = 1!2(water5L

= 1!2× 62!4× 5× 2!5= 936 psfg

Area

I2 =012

4=0L2

4

=0 × 2!52

4= 4!909 ft2

Weight of liquid

.liquid =

µI2L+

0A2

44L

¶(liquid

=

µI2L+

0L3

16

¶(1!2) (water

=

Ã¡4!909 ft2

¢(2!5 ft) +

0 (2!5 ft)3

16

!(1!2)

µ62!4

lbf

ft3

¶

= 1148! 7 lbf

84

Substitute numbers into Eq. (1)

H> + (1148! 7 lbf) + (600 lbf) =¡936 lbf$ ft2

¢ ¡4!909 ft2

¢

H> = 2846! 1

H> = 2850 lbf

85

PROBLEM 3.19

Situation: A metal dome with water is described in the problem statement.

Find: Force exerted by bolts.

APPROACH


ANALYSIS

Equilibrium (system is comprised of the dome/pipe apparatus plus the water within)

XH; = 0

Hbolt = Hpressure ".H2O ".metal (1)

Weight of water

.H2O = (2$3)063 × 62!4 + 12× (0$4)× (3$4)2 × 62!4= 28' 559 lbf

Hydrostatic equation (location 1 is on free surface; location 2 is at the bottom of thedome).

% (bottom) = (R = (6L

= (62!4) (6) (3)

= 1123!2 lbf$ ft2

Pressure force

HPressure = % (bottom)I

= (1123!2)¡0 · 62

¢

= 127' 030 lbf

Substitute numbers into Eq. (1)

Hbolt = Hpressure ".H2O ".metal

= 127' 030 lbf " 28' 559 lbf " 1300 lbf= 97171

Hbolt = 97' 200 lbf downward

86

PROBLEM 3.20

Situation: A metal dome with water is described in the problem statement.

Find: Force exerted by the bolts.

APPROACH


ANALYSIS

XH; = 0

%bottomIbottom + Hbolts ".H2O ".dome = 0

where %bottomIbottom = 4!8× 9' 810× 0 × 1!62 = 378!7 kN.H2O = 9' 810(3!2× (0$4)× 0!22 + (2$3)0 × 1!63)

= 85!1 kN

Then Hbolts = "378!7 + 85!1 + 6Hbolts = "287!6 kN

87

PROBLEM 3.21

Situation: A tank under pressure with a dome on top is described in the problemstatement.2 = 2 ft! 6 = 1!5! %= = 5 psig. .dome = 1000 lbf!

Find: (a) Vertical component of force in metal at the base of the dome.(b) Is the metal in tension or compression?

APPROACH

Apply equilibrium to a free body comprised of the dome plus the water within. Applythe hydrostatic principle to nd the pressure at the base of the dome.

ANALYSIS

Equilibrium

1000 lbf

WlFdFd

pB

XH; = 0 (1)

HA + %>I".liquid ".dome = 0 (4)

Hydrostatic equation%> + (R> = %= + (R=

%> = %= "¡(62B

¢6!R

= (5 psig)¡144 in2$ ft2

¢"¡62!4 lbf$ ft3

¢(1!5) (3 ft)

= 439!2 psfg

Weight of the liquid

.liquid =¡(62B

¢(6) (Volume)

=¡62!4 lbf$ ft3

¢(1!5)

µ2

3023 ft3

¶

= 1568 lbf

Pressure Force

H> = %>I

= (439!2 psfg)¡0 × 22 ft2

¢

= 5519 lbf

88

Substitute into Eq. (1).

HA = "H> +.liquid +.dome

= " (5519 lbf) + (1568 lbf) + (1000 lbf)= "2951 lbf

HA = 2950 lbf (metal is in tension)

89

PROBLEM 3.22

Situation: A piston system is described in the problem statement.

Find: Volume of oil to be added to raise piston by 1 in.

ANALYSIS

h

Volumeadded

Volume added is shown in the gure. First get pressure at bottom of piston

Hydrostatic force

%,I, = 10 lbf

%, = 10$I,

= 10$((0$4)× 42)= 0!796 psig = 114!6 psfg


(oilM = 114!6 psfg

M = 114!6$(62!4× 0!85) = 2!161 ft = 25.9 in

Finally

,"added = (0$4)(42 × 1 + 12 × 26!9)

,"added = 33!7 in.3

90

PROBLEM 3.23

Situation: An air bubble rises from the bottom of a lake.

Find: Ratio of the density of air within the bubble at 34 ft to the density at 8 ft.

Assumptions: a.) Air is ideal gas. b.) Temperature is constant. c.) Neglect surfacetension e!ects.

APPROACH

Apply the hydrostatic equation and the ideal gas law.

ANALYSIS

Ideal gas law

# =%

"&

#34 =%34"&

; #8 =%8"&

#34#8

=%34%8

where % is absolute pressure (required in ideal gas law).


%8 = %atm + ( (8 ft)

= 2120 lbf$ ft2 +¡62!4 lbf/ft3

¢(8 ft)

= 2619 lbf/ft2

%34 = %atm + ( (34 ft)

= 2120 lbf$ ft2 +¡62!4 lbf/ft3

¢(34 ft)

= 4241!6 lbf/ft2

Density ratio

#34#8

=4241!6 lbf/ft2

2619 lbf/ft2

= 1! 620

#34$#8 = 1!62

91

PROBLEM 3.24

Situation: A liquid’s mass density property is described in the problem statement.

Find: Gage pressure at 10 m depth.

ANALYSIS

# = #water(1 + 0!01A)

or ( = (water(1 + 0!01A)

A%$AR = "(A%$AA = (water(1 + 0!01A)

Integrating% = (water(A+ 0!01A

2$2) + -

For boundary condition %gage = 0 when A = 0 gives - = 0!

% (A = 10m) = (water(10 + 0!01× 102$2)% (A = 10m) = 103 kPa

92

PROBLEM 3.25


Find: Depth where pressure is 60 kPa.

ANALYSIS

# = #water(1 + 0!01A)

or ( = (water(1 + 0!01A)

A%$AR = "(A%$AA = (water(1 + 0!01A)

Integrating% = (water(A+ 0!01A

2$2) + -

For boundary condition %gage = 0 when A = 0 gives - = 0!

% = (water(A+ 0!01 A2$2)

60' 000 N/m2 = (9810 N/m3)(A+ !005 A2)

Solving the above equation for A yields

A = 5!94m

93

PROBLEM 3.26


Find: Pressure at depth of 20 ft.

ANALYSIS

A%$AR = "(= "(50" 0!1 R)

% = ""20Z

0

(50" 0!1 R) AR

= "50 R + 0!1 R2$2 |"200

= 1000 + 0!1× 400$2

% = 1020 psfg

94

PROBLEM 3.27

Situation: A pipe system is described in the problem statement.

Find: Gage pressure at pipe center.

APPROACH


ANALYSIS

Manometer equation. (add up pressures from the pipe center to the open end of themanometer)

%pipe + (0!5 ft)(62!4 lbf/ft3) + (1 ft)(2× 62!4 lbf/ft3)

"(2!5 ft)(62!4 lbf/ft3) = 0

%pipe = (2!5" 2" 0!5) ft (62!4 lbf/ft3) = 0

% (center of pipe) = 0!0 lbf$ ft2

95

PROBLEM 3.28


Find: Gage pressure at pipe center.

APPROACH


ANALYSIS

Manometer equation (from A to the open end of the manometer)

%= + (2!0 ft)(62!3 lbf/ft3)" (2$12 ft)(847 lbf/ft3) = 0

%= = "124!6 lbf/ft2 + 141!2 lbf/ft2 = +16!6 lbf/ft2

%= = +0!12 psi

96

PROBLEM 3.29

Situation: A piezometer (A = 0!5mm) is connected to a pipe. The uid is waterSurface tension is relevant. Liquid level in the piezometer is 15 cm

Find: Estimate gage pressure in pipe A.

Properties: From Table A-5: (62B = 9790N$m3! From the footnote in Table A-5,

O62B = 0!073N$m!

Assumptions: For capillary rise, assume a small contact angle—cos K % 1!

APPROACH

Apply equilibrium to a free body comprised of a 15 cm column of water.

ANALYSIS

Equilibrium (vertical direction)

%=I". + HC = 0 (1)

Weight of the water column. = (

¡0A2$4

¢2 (2)

Force due to surface tensionHC = O0A (3)

Combine Eqs. (1) to (3):

%=¡0A2$4

¢" (

¡0A2$4

¢2+ O0A = 0

Thus%= = (2"

4O

A

Calculations:

%= =¡9790N$m3

¢(0!15m)"

4 (0!073N$m)

0!0005m= 884Pa-gage

%= = 884Pa-gage

97

PROBLEM 3.30


Find: Pressure at the center of pipe B.

APPROACH


ANALYSIS

Manometer equation (add up pressures from the open end of the manometer to thecenter of pipe B).

%> = 0

+¡0!30m× 20' 000N$m3

¢

"¡0!1m× 20' 000N$m3

¢

"¡0!5m× 10' 000N$m3

¢

= "1000Pa

%> = "1!00 kPa-gage

98

PROBLEM 3.31

Situation: A container is described in the problem statement.

Find: Pressure in the air within the container

APPROACH

Apply conservation of mass to nd the decrease in liquid level in the container. Then,apply the hydrostatic equation.

ANALYSIS

Conservation of mass (applied to liquid)

Gain in mass of liq. in tube = Loss of mass of liq. in container

(Volume change in tube) #liquid = (Volume change in container ) #liquid,"tube = ,"container

(0$4)12tube × L = (0$4)12

container × (!M)container

(!M)container =

µ1tube1container

¶2L

(!M)container = (1$8)2 × 40= 0!625 cm


%container = (L sin 10! +!M)#)

= (40 sin 10! + 0!625)× 10"2 × 800× 9!81

%container = 594 Pa

99

PROBLEM 3.32

Situation: A container is described in the problem statement.

Find: Pressure in the air within the container

APPROACH

Apply conservation of mass to nd the decrease in liquid level in the container. Then,apply the hydrostatic equation.

ANALYSIS

Conservation of mass (applied to liquid)

Gain in mass of liq. in tube = Loss of mass of liq. in container

(Volume change in tube) #liquid = (Volume change in container ) #liquid,"tube = ,"container

(0$4)12tube × L = (0$4)12

container × (!M)container

(!M)container =

µ1tube1container

¶2L

(!M)container = (1$10)2 × 3= 0!03 ft


%container = (L sin 10! +!M)(

= (3 sin 10! + !03)× 50= 27! 548 lbf$ ft2

%container = 27!5 psfg

100

PROBLEM 3.33

Situation: A piston scale is described in the problem statement.

Find: Select a piston size and standpipe diameter.

ANALYSIS

First of all neglect the weight of the piston and nd the piston area which will givereasonable manometer deections. Equating the force on the piston, the piston areaand the deection of the manometer gives

. = !M(I

where ( is the specic weight of the water. Thus, solving for the area one has

I =.

(!M

For a four foot person weighing 60 lbf, the area for a 4 foot deection (manometernear eye level of person) would be

I =60

62!4× 4= 0!24 ft2

while for a 250 lbf person 6 feet tall would be

I =250

62!4× 6= 0!66 ft2

It will not be possible to maintain the manometer at the eye level for each person sotake a piston area of 0.5 ft2! This would give a deection of 1.92 ft for the 4-foot, 60lbf person and 8 ft for the 6-foot, 250 lbf person. This is a good compromise.

The size of the standpipe does not a!ect the pressure. The pipe should be big enoughso the person can easily see the water level and be able to read the calibration onthe scale. A 1/2 inch diameter tube would probably su"ce. Thus the ratio of thestandpipe area to the piston area would be

IpipeIpiston

=0!785× 0!52

0!5× 144= 0!0027

This means that when the water level rises to 8 ft, the piston will only have movedby 0!0027× 8 = 0!0216 ft or 0.26 inches.The weight of the piston will cause an initial deection of the manometer. If thepiston weight is 5 lbf or less, the initial deection of the manometer would be

!M! =.piston

(Ipiston= 0!16 ft or 1.92 inches

This will not signicantly a!ect the range of the manometer (between 2 and 8 feet).The system would be calibrated by putting knows weights on the scale and markingthe position on the standpipe. The scale would be linear.

101

PROBLEM 3.34


Find: Gage pressure at center of pipe A.(a) units of pounds per square inch(b) units of kilopascals.

APPROACH


ANALYSIS

Manometer equation

%= = 1!31× 847" 4!59× 62!4= 823!2 psf

%= = 5!72 psig

%= = 0!4× 1!33× 105 " 1!4× 9810%= = 39!5 kPa gage

102

PROBLEM 3.35

Situation: A U-tube manometer is described in the problem statement.

Find: Specic weight of unknown uid.

ANALYSIS

Volume of unknown liquid is V–= (0$4)A2L = 2 cm3

," = (0$4)(0!5)2L = 2

L = 10!186 cm

Manometer equation (from water surface in left leg to liquid surface in right leg)

0 + (10!186 cm - 5 cm)(10"2 m/cm)(9,810 N/m3)"(10!186 cm)(10"2 m/cm)(liq. = 0

508!7 Pa " 0!10186(liq. = 0

(liq. = 4' 995 N/m3

103

PROBLEM 3.36

Situation: A U-tube is described in the problem statement.

Find: (a) Locate the water surface.(b) Locate the mercury surfaces.(c) Find the maximum pressure in tube.

Properties: (a) Mercury from Table A.4: (6D = 847 lbf$ ft3! (b) Water from Table

A.4: (620 = 62!4 lbf$ ft3

APPROACH

Since the mercury column has a length of 1.0 ft, write an equation that involves?E and ?F! Apply the manometer equation to develop a second equation, and thensolve the two equations simultaneously. Apply the hydrostatic equation to nd themaximum pressure.

ANALYSIS

Water

1.5 ft

Hg

y L

yR

Since the column of mercury is 1.0 ft long:

?E + ?F = 1 ft"8 in

12 in$ ft(1)

= 0!333 ft

Manometer equation

0 + (1!0× 62!4) + (?E × 847)" (?F × 847) = 0 (2)

?E " ?F = "0!0737 ft

Combine eqns. (1) and (2):

2?E = 0!333" 0!0737?E = 0!130 ft

104

The water/mercury interface is 0.13 ft above the horizontal leg.

The air/water interface is 1.13 ft above the horizontal leg.

?F = 0!333" ?E= 0!203 ft

The air/mercury interface is 0.203 ft above the horizontal leg.

Hydrostatic Equation.

%max = 0!203× 847%max = 172 psfg

105

PROBLEM 3.37

Situation: A U-tube is described in the problem statement.

Find: (a) Design the manometer.(b) Predict probable degree of accuracy.

ANALYSIS

Consider the manometer shown in the gure.

! h

• Use a manometer uid that is heavier than water. The specic weight of themanometer uid is identied as ("!

• Then !Mmax = !%max$((" " (H2O)!

• If the manometer uid is carbon-tetrachloride ((" = 15' 600)'!Mmax = 60 ×103$(15' 600" 9' 180) = 13!36 m –(too large).

• If the manometer uid is mercury ((" = 133' 000)'!Mmax = 60×103$(1333' 000"9' 810) = 0!487 m–(O.K.). Assume the manometer can be read to ±2 mm.Then % error = ±2$487 = ±0!004 = ±0!4%. The probable accuracy for fulldeection (0.5m) is about 99.6%. For smaller pressure di!erences the possibledegree of error would vary inversely with the manometer deection. For ex-ample, if the deection were 10 cm = 0.1 m, then the possible degree of errorwould be ±2% and the expected degree of accuracy would be about 98%.

COMMENTS

Error analysis is much more sophisticated than presented above; however, this simpletreatment should be enough to let the student have an appreciation for the subject.

106

PROBLEM 3.38


Find: Design a apparatus to measure specic weights from 50 lbf$ ft3 to 100 lbf$ ft3

ANALYSIS

One possible apparatus might be a simple glass U-tube. Have each leg of the U-tubeequipped with a scale so that liquid levels in the tube could be read. The proceduremight be as described in steps below:

1. Pour water into the tube so that each leg is lled up to a given level (for exampleto 15 in. level).

2. Pour liquid with unknown specic weight into the right leg until the water inthe left leg rises to a given level (for example to 27 in. level).

3. Measure the elevation of the liquid surface and interface between the two liquidsin the right tube. Let the distance between the surface and interface be L ft.

4. The hydrostatic relationship will be (H2O(20) = (<L or (< = 2SH2O$L.

5. To accommodate the range of ( specied the tube would have to be about 3 or4 ft. high.

The errors that might result could be due to:

1. error in reading liquid level

2. error due to di!erent surface tension

(a) di!erent surface tension because of di!erent liquids in each leg

(b) one leg may have slightly di!erent diameter than the other one; therefore,creating di!erent surface tension e!ect.

Sophisticated error analysis is not expected from the student. However, thestudent should sense that an error in reading a surface level in the manometerwill produce an error in calculation of specic weight. For example, assumethat in one test the true value of L were 0.28 ft. but it was actually read as 0.29ft. Then just by plugging in the formula one would nd the true value of (would be 7.14 (H2O but the value obtained by using the erroneous reading wouldbe found to be 6.90 (H2O. Thus the manometer reading produced a -3.4% errorin calculated value of (. In this particular example the focus of attention wason the measurement of L. However, the setting of the water surface in the leftleg of the manometer would also involve a possible reading error, etc.

107

COMMENTS

Other things that could be considered in the design are:

1. Diameter of tubing

2. Means of support

3. Cost

4. How to empty and clean tube after test is made.

108

PROBLEM 3.39


Find: Pressure at center of pipe A.

ANALYSIS

Manometer equation

%= = (0!9 + 0!6× 13!6" 1!8× 0!8 + 1!5)9' 810 = 89' 467 Pa%= = 89!47 kPa

109

PROBLEM 3.40


Find: (a) Di!erence in pressure between points A and B.(b) Di!erence in piezometric head between points A and B.

APPROACH


ANALYSIS

Manometer equation

%= " (1m)¡0!85× 9810N$m3

¢+ (0!5m)

¡0!85× 9810N$m3

¢= %>

%= " %> = 4169Pa

%= " %> = 4!169 kPa

Piezometric head

M= " M> = (%=(+ R=)" (

%>(+ R>)

=%= " %>(

+ (R= " R>)

=4169N$m2

0!85× 9810N$m3" 1m

= "0!5 m

M= " M> = "0!50 m

110

PROBLEM 3.41


Find: Manometer deection when pressure in tank is doubled.

ANALYSIS

%" %'5" = (M

For 150 kPa absolute pressure and an atmospheric pressure of 100 kPa,

(M = 150" 100 = 50 kPa

For an absolute pressure of 300 kPa

(M:G% = 300" 100 = 200 kPa

Divide equations to eliminate the specic weight

M:G%M

=200

50= 4!0

soM:G% = 4!0M

111

PROBLEM 3.42

Situation: A manometer tapped into a vertical conduit is described in the problemstatement.

Find: (a) Di!erence in pressure between points A and B(b) Piezometric pressure between points A and B .

Properties: From Table A.4, (HD = 847 lbf/ft3!

(oil = (0!95)(62!4 lbf/ft3)

= 59!28 lbf/ft3

ANALYSIS

Manometer equation

%= + (18$12) ft ((oil) + (2$12) ft. (oil + (3$12) ft (oil"(3$12) ft (Hg " (2$12) ft (oil = %>

thus

%= " %> = ("1!75 ft.)(59.28 lbf/ft3) + (0!25 ft.)(847 lbf/ft3)

%= " %> = 108!01 lbf/ft2

Piezometric head

M= " M> = (%= " %>)$(oil + R= " R>M= " M> = (108!01 lbf/ft)/(59.28 lbf/ft3) + (1!5" 0)

M= " M> = 3!32 ft.

112

PROBLEM 3.43

Situation: Two manometers attached to an air tank are described in the problemstatement.

Find: Di!erence in deection between manometers.

ANALYSIS

The pressure in the tank using manometer > is

%5 = %'5" " (%!M0

and using manometer C is%5 = 0!9%'5" " (%!M'

Combine equations%'5" " (%!M0 = 0!9%'5" " (%!M'

or0!1%'5" = (%(!M0 "!M')

Solve for the di!erence in deection

!M0 "!M' =0!1%'5"(%

=0!1× 105

9!81× 103

!M0 "!M' = 1!02 m

113

PROBLEM 3.44

Situation: A manometer measuring pressure di!erence is described in the problemstatement.

Find: (a) Pressure di!erence.(b) Piezometric pressure di!erence between points A and B.

APPROACH

Apply the manometer equation and the hydrostatic equation.

ANALYSIS

Manometer equation

%> = %= + 0!03(( " 0!03(" " 0!1((

or%> " %= = 0!03((( " (")" 0!1((


%> " %= = 0!03(9810" 3× 9810)" 0!1× 9810

%> " %= = "1!57 kPa

Change in piezometric pressure

%;> " %;= = %> + ((R> " (%= + ((R=)= %> " %= + (((R> " R=)

But R> " R= is equal to 0.1 m so from equation above

%;> " %;= = %> " %= + 0!1((= 0!03(9810" 3× 9810)= "588!6 Pa

%;> " %;= = "0!589 kPa

114

PROBLEM 3.45

Situation: A tank has a small air tube in it to measure the surface level of theliquid—additional details are provided in the problem statement.

Find: Depth of liquid in tank.

Assumptions: Neglect the change of pressure due to the column of air in the tube.

ANALYSIS

%gage " (A" 1)(liquid = 0

20' 000" ((A" 1)× 0!85× 9' 810) = 0

A = (20' 000$(0!85× 9' 810)) + 1

A = 3!40 m

115

PROBLEM 3.46

Situation: The atmosphere is described in the problem statement.

Find: The correct statement.

ANALYSIS

A%$AR = (

Because ( becomes smaller with an increase in elevation the ratio of (A%$AR)’s willhave a value greater than 1.

116

PROBLEM 3.47

Situation: The boiling point of water is described in the problem statement.&sea level = 296 K= 23!C

Find: Boiling point of water at 1500 and 3000 m for standard atmospheric conditions.

APPROACH

Apply the atmosphere pressure variation equation that applies to the troposphere.

ANALYSIS

For standard atmosphereAtmosphere pressure variation (troposphere)

% = %0[(&0 " T(R " R0))$&0]D*HF

= 101!3[296" 5!87(R " R0))$296]D*HF

where)$T" = 9!81$(5!87× 10"3 × 287) = 5!823

So%1I500 = 101!3[(296" 5!87(1!5))$296]54823 = 84!9 kPa%3I000 = 101!3[(296" 5!87(3!0))$296]54823 = 70!9 kPa

From table A-5,

&boiling, 1,500 m ! 95 !C (interpolated)

&boiling, 3,000 m ! 90 !C (interpolated)

117

PROBLEM 3.48

Situation: This problem involves pressure variation from a depth of 10m in a lake to4000m in the atmosphere.

Find: Plot pressure variation.

Assumptions: Atmospheric pressure is 101 kPa. The lake surface is at sea level.

ANALYSIS

Atmosphere pressure variation (troposphere)

%= = 101!3

µ1"

5!87× 10"3 × R296

¶54823

Pressure in water%% = 101!3 + 9!810× R

P re s s u re (k P a )

4 0 6 0 8 0 1 0 0 1 2 0 1 4 0 1 6 0 1 8 0 2 0 0 2 2 0

0

1 0 0 0

2 0 0 0

3 0 0 0

4 0 0 0

118

PROBLEM 3.49

Situation: A woman breathing is described in the problem statement.

Find: Breathing rate at 18,000 ft.

Assumptions: Volume drawn in per breath is the same.Air is an ideal gas.

ANALYSIS

Let bV–# = constant where > = breathing rate = number of breaths for each unit oftime, V–= volume per breath, and # = mass density of air. Assume 1 is sea level andpoint 2 is 18,000 ft. elevation. Then

>1,"1 #1 = >2,"2 #2>2 = >1(,"1 $,"2)(#1$#2)

then >2 = >1(#1$#2) but # = (%$"& )

Thus, >2 = >1(%1$%2)(&2$&1)

%2 = %1(&2$&1)D*HF

%1$%2 = (&2$&1)"D*HF

Then >2 = >1(&2$&1)1"D*HF

Since the volume drawn in per breath is the same

>2 = >1(#1$#2)

Ideal gas law>2 = >1(%1$%2)(&2$&1)%1$%2 = (&2$&1)

"D*HF

>2 = >1(&2$&1)1"D*HF

where >1 = 16 breaths per minute and &1 = 59!H = 519!"

&2 = &1 " T(R2 " R1) = 519" 3!221× 10"3(18' 000" 0) = 461!0 oR>2 = 16(461!0$519)1"3242*(34221×10

"3×1I715)

>2 = 28!4 breaths per minute

119

PROBLEM 3.50

Situation: A pressure gage in an airplane is described in the problem statement.

Find: Elevation and temperature when pressure is 75 kPa.

ANALYSIS


% = %0[(&0 " T(R " R0))$&0]D*HF

75 = 95[(283" 5!87(R " 1))$283]9481*(5487×10"3×287)

R = 2!91 km

& = &0 " T(R " R0)= 10" 5!87(2!91" 1)

& = "1!21oC

120

PROBLEM 3.51

Situation: A pressure gage in an airplane is described in the problem statement.

Find: Elevation when pressure is 10 psia.

ANALYSIS


% = %0[(&0 " T(R " R0))$&0]D*HF

10 = 13!6[((70 + 460)" 3!221× 10"3(R " 2' 000))$(70 + 460)]3242*(34221×10"3×1I715)

R = 10' 452 ft

121

PROBLEM 3.52

Situation: Denver, CO (the mile-high city) is described in the problem statement.

Find: (a) Pressure in both SI and traditional units.(b) Temperature in both SI and traditional units.(c) Density in both SI and traditional units.

ANALYSIS


& = &0 " T(R " R0)= 533" 3!221× 10"3(5' 280" 0) = 516!R= 296" 5!87× 10"3(1' 609" 0)

& = 287 K = 516 !R

% = %0(&$&0)D*HF

= 14!7(516$533)54823

% = 12!2 psia

%' = 101!3(287$296)9481*(5487×10"3×287)

%' = 86!0 kPa = 12!2 psia

Ideal gas law

# = %$"&

= (12!2× 144)$1' 715× 516)= 0.00199 slugs/ft3

# = 86' 000$(287× 287)

# = 1!04 kg/m3 = 0.00199 slugs/ft3

122

PROBLEM 3.53

This problem involves the Martian atmosphere. Some relevant data.

• Temperature at the Martian surface is & = "63 !C = 210K The pressure atthe Martian surface is % = 7 mbar.

• The atmosphere consists primarily of CO2 (95.3%) with small amounts of ni-trogen and argon.

• Acceleration due to gravity on the surface is 3.72 m/s2.

• Temperature distribution. Approximately constant from surface to 14 km.Temperature decreases linearly at a lapse rate of 1.5oC/km from 14 to 34 km.

Find: Pressure at an elevation of 8 km.Pressure at an elevation of 30 km.

Assumptions: Assume the atmosphere is totally carbon dioxide.

Properties: CO2 (from Table A.2): the gas constant is " =189 J/kg·K.

APPROACH

Derive equations for atmospheric pressure variation from rst principles.

ANALYSIS

A.) Elevation of 8 km.

Di!erential equation describing pressure variation in a hydrostatic uid

A%

AR= "#) (1)

Ideal gas law# =

%

"&(2)

Combine Eqs. (1) and (2)A%

AR= "

%

"&) (3)

Integrate Eq. (3) for constant temperature

ln%

%!= "

(R " R!))"&

(4)


ln%

%!= "

(8000m) (3!72m$ s2)

(189 J$ kg · K) (210K)= "0!7498

123

Thus

%

%!= exp("0!7498)

= 0!4725

and

% = (7mbar)× 0!4725= 3!308 mbar

%(R = 8km) = 3!31mbar

B.) Elevation of 30 km.

Apply Eq. (4) to nd the pressure at R = 14 km

%14 km%!

= exp

·"(14000m) (3!72m$ s2)

(189 J$ kg · K) (210K)

¸

= exp("1!3122)= 0!2692

%14 km = (7mbar) (0!2692)

= 1! 884 mbar

In the region of varying temperature Eq. (3) becomes

A%

AR=

%)

"[&! + T(R " R!)]

where the subscript U refers to the conditions at 14 km and T is the lapse rate above14 km. Integrating gives

%

%!=

·&! " T(R " R!)

&!

¸D*HF

Calculations for R = 30km!

%

(1! 884 mbar)=

·210" 0!0015(30000" 14000)

210

¸3472*(040015×189)

= 0!2034

% = (1! 884 mbar) 0!2034

= 0!3832mbar

%(R = 30 km) = 0!383mbar

124

PROBLEM 3.54

Situation: Standard atmospheric conditions are described in the problem statement.

Find: (a) Pressure at 30 km altitude.(b) Density at 30 km altitude.

ANALYSIS

The equation for pressure variation where the temperature increases with altitude is

A%

AR= "( =

%)

"[&! + T(R " R!)]

where the subscript U refers to the conditions at 16.8 km and T is the lapse rate above16.8 km. Integrating this equation gives

%

%!=

·&! + T(R " R!)

&!

¸"D*HF

Substituting in the values gives

%

%!=

·215!5 + 1!38× (30" 16!8)

215!5

¸"9481*(1438×04287)

= 1!084"2448

= 0!134

Thus the pressure is

% = 0!134× 9!85= 1!32 kPa.

Ideal gas law

# =%

"&

=1!32

0!287× 234# = 0!0197 kg/m3

125

PROBLEM 3.55

Situation: The US standard atmosphere from 0 to 30 km is described in the problemstatement.

Find: Design a computer program that calculates the pressure and density.

ANALYSIS

The following are sample values obtained using computer calculations.

altitude (km) temperature (!C) pressure (kPa) density (kg/m3)10 -35.7 27.9 0.40915 -57.5 12.8 0.20825 -46.1 2.75 0.042

126

PROBLEM 3.56

Situation: A submerged gate is described in the problem statement.

Find: (a) Net force on gate.(b) Moment required to keep gate closed.

ANALYSIS

Hydrostatic forceForce of slurry on gate = %&I and it acts to the right. Force of water on gate = %%Iand it acts to the left

Hnet = (%& " %%)I= (8(& " 8(%)I= (8 ft)(16 ft2)(150 lbf/ft3 " 60 lbf/ft3)

Hnet = 11' 520 lbf

Because the pressure is uniform along any horizontal line the moment on the gate iszero; therefore, the moment required to keep the gate closed is zero.

127

PROBLEM 3.57

Situation: Two submerged gates are described in the problem statement.

Find: How the torque changes with increasing water depth 4!

APPROACH

Apply hydrostatic force equation.

ANALYSIS

Let the horizontal gate dimension be given as > and the vertical dimension, M!

Torque (gate A)&= = H (?1, " ?)

where H = the hydrostatic force acting on the gate and (?1, " ?) is the distancebetween the center of pressure and the centroid of the gate. Thus

&= = ((4 " (M$2))(>M)(V$?I)= ((4 " (M$2))(>M)(>M3$12)$(4 " (M$2))(>M))

&= = (>M3$12

Therefore, &= does not change with 4.

Torque (gate B)

&> = H ((M$2) + ?1, " ?)= ((4 " (M$2))(>M)((M$2) + ?1, " ?)= ((4 " (M$2))(>M)((M$2) + V(?I))= ((4 " (M$2))(>M)[(M$2) + (>M3$12)$((4 " (M$2))>M)]= ((4 " (M$2))>M2$2 + (>M3$12

Thus, &= is constant but &> increases with 4.

Case (b) is a correct choice.

Case (c) is a correct choice.

128

PROBLEM 3.58

Situation: Two submerged gates are described in the problem statement.

Find: Choose the statements that are valid.

ANALYSIS

The correct answers obtained by looking at the solution to problem 3.57 are thata, b, and e are valid statements.

129

PROBLEM 3.59


Find: Force of gate on block.

ANALYSIS

Hydrostatic force

Hhs = %I

= ?(I

= (10m)×¡9810N$m3

¢× (4× 4) m2

= 1! 569 6× 106N

Center of pressure

?1, " ? =V

?I

=>M3$12

?I

=(4× 43$12) m4

(10m) (4× 4) m2= 0!133 33m

Equilibrium (sum moments about the pivot)

Hhs (?1, " ?)" Hblock (2m) = 0¡1! 569 6× 106N

¢(0!133 33m)" Hblock (2m) = 0

Hblock = 1!046× 105N (acts to the left)

Hgate = 105 kN (acts to the right)

130

PROBLEM 3.60

Situation: Concrete forms are described in the problem statement.

Find:a.) Hydrostatic force per foot on formb.) Force exerted on bottom tie.

ANALYSIS

Hydrostatic force

Hhs = %I = ?(I

= 4!5 ft× 150 lbf$ ft3 × (9 ft)

Hhs = 6075lbfft

Center of pressure

?1, = ? +V

?I

= 4!5 +(1× 93)$124!5× 9

= 6!00 ft

Equilibrium (sum moments about the top tie)

Hbottom tie =Hhs × ?1,

M

=2 ft× 6075 lbf$ ft× 6!00 ft

9 ft8100 lbf

Hbottom tie = 8100 lbf (tension)

131

PROBLEM 3.61

Situation: A rectangular gate is hinged at the water line. The gate is 4 ft high by12 ft wide.

Find: Force to keep gate closed.

Properties: From Table A.4, (Water = 62!4 lbf/ft3

ANALYSIS

Hydrostatic Force (magnitude):

HJ = %I

=¡(H2O × ?

¢ ¡48 ft2

¢

=¡62!4 lbf$ ft3 × 2 ft

¢ ¡48 ft2

¢

= 5950 lbf

Center of pressure. Since the gate extends from the free surface of the water, HJacts at 2/3 depth or 8/3 ft. below the water surface.

Equilibrium. ( moment center is the hinge)

X+ = 0

(HJ × 8$3 ft)" (4 ft)H = 0

H =5950 lbf × 8$3 ft

4 ft

H = 3970 lbf to the left

132

PROBLEM 3.62

Situation: A submerged gate is described in the problem statement. The gate is 6 ftby 6 ft.

Find: Reaction at point A.

APPROACH

Find the hydrostatic force and the center of pressure. Since the gate is in equilibrium,sum moments about the stop.

ANALYSIS

Hydrostatic force (magnitude)

H = %I

= (3 m + 3 m× cos 30!)(9810 N/m3)× 36 m2

H = 1' 977' 000 N

Center of pressure

? = 3 + 3$ cos 30o

= 6!464 m

?1, " ? =V

?I

=(64$12)m4

6!464 m× 24 m2= 0!696 m

Equilibrium.Take moments about the stop

X+stop = 0

6"= " (3" 0!696)× 1' 977' 000 = 0

"= = 759' 000N

Reaction at point A = 759 kN . This force is normal to gate and acting at an angleof 30! below the horizontal.

133

PROBLEM 3.63


Find: Force * required to begin to open gate.

ANALYSIS

The length of gate is&42 + 32 = 5 m

Hydrostatic force

H = %I = ?(I

= (3)(9810)(2× 5)= 294!3 kN

Center of pressure

?1, " ? =V

?I

=(2× 53) $12(3) (2× 5)

= 0!694 4 m

Equilibrium.

X+hinge = 0

294!3× (2!5 + 0!694 4)" 3* = 0

* = 313 kN

134

PROBLEM 3.64

Situation: A submerged gate opens when the water level reaches a certain value.Other details are given in the problem statement.

Find: M in terms of L to open gate.

APPROACH

As depth of water increase, the center of pressure will move upward. The gate willopen when the center of pressure reaches the pivot..

ANALYSIS

Center of pressure (when the gate opens)

?1, " ? = 0!60L" 0!5L= 0!10L (1)

Center of pressure (formula)

?1, " ? =V

?I

=(L× L3) $12(M+ L$2)L2

(2)


0!10L =(L× L3) $12(M+ L$2)L2

0!10 =L

12(M+ L$2)

M =5

6L"

1

2L

=1

3L

M = L$3

135

PROBLEM 3.65

Situation: A buttery valve is described in the problem statement.

Find: Torque required to hold valve in position.

ANALYSIS

Hydrostatic force

H = %I = ?(I

= (30 ft× 62!4 lb/ft3)(0 ×12$4) ft2)

= (30× 62!4× 0 × 102$4) lb= 147' 027 lb

Center of pressure

?1, " ? = V$?I

= (034$4)$(?032)

= (52$4)$(30$!866)

= 0!1804 ft

Torque

Torque = 0!1804× 147' 027& = 26' 520 ft-lbf

136

PROBLEM 3.66


Find: Will gate fall or stay in position.

ANALYSIS

Hydrostatic force

H = %I

= (1 + 1!5)9' 810× 1× 3×&2

= 104' 050

Center of pressure

?1, " ? =V

?I

=

¡1× (3

&2)3¢$12

(2!5×&2)(1× 3

&2)

= 0!4243 m

Overturning moment

+1 = 90' 000× 1!5= 135' 000 N ·m

Restoring moment

+2 = 104' 050× (3&2$2" 0!424)

= 176' 606N · mF +1

So the gate will stay in position.

137

PROBLEM 3.67


Find: Will gate fall or stay in position.

ANALYSIS

H = (4 + 3!535)62!4× (3× 7!07&2) = 14' 103 lbf

?1, " ? = 3× (7!07&2)3$(12× 7!535

&2× 3× 7!07

&2)

= 0!782 ft

Overturning moment +1 = 18' 000× 7!07$2 = 63' 630 N ·mRestoring moment +2 = 14' 103(7!07

&2$2" 0!782)

= 59' 476 N ·mG+1

So the gate will fall.

138

PROBLEM 3.68


Find: (a) Hydrostatic force (H ) on gate, (b) Ratio ("-$H ) of the reaction force tothe hydrostatic force.

ANALYSIS

H = %I

= (M+ 2M$3)((.M$ sin 60!)$2

H = 5(.M2$3&3

?1, " ? = V$?I =. (M$ sin 60!)3$(36× (5M$(3 sin 60!))× (.M$2 sin 60!))= M$(15

&3)

"+ = 0

"-M$ sin 60! = H [(M$(3 sin 60!))" (M$15

&3)]

"-$H = 3$10

139

PROBLEM 3.69


Find: (a) Magnitude of reaction at A.(b) Comparison to that for a plane gate.

ANALYSIS

a)

H63AK = %I = (0!25L+ 0!5L× 0!707)× W.L = 0!6036(.L2

?1, " ? = V$?I = (.L3$12)$(((0!25L$0!707) + 0!5L)×.L)?1, " ? = 0!0976LX

+hinge = 0

Then " 0!70"=L+ (0!5L+ 0!0976L)× 0!6036(.L2 = 0

"= = 0!510(.L2

b) The reaction here will be less because if one thinks of the applied hydrostaticforce in terms of vertical and horizontal components, the horizontal component willbe the same in both cases, but the vertical component will be less because there isless volume of liquid above the curved gate.

140

PROBLEM 3.70


Find: Force required to hold gate in place.

APPROACH

To develop an equation for the force * , apply equilibrium by summing moments aboutthe hinge. Solving this equation requires the hydrostatic force. The hydrostatic forcecan be found by calculating the pressure at the depth of the centroid and by ndingthe line of action. To nd the line of action, calculate the equivalent depth of liquidthat account for the pressure acting the free surface.

ANALYSIS

Hydrostatic equation (from free surface of the liquid to centroid of the gate)

%1(liquid

+ R1 =%2(liquid

+ R2

%16(water

+³?1 +

?22

´=

%26(water

+ 0

(5× 144) lbf$ ft2

0!8×¡62!4 lbf$ ft3

¢ +µ1 ft +

10 ft

2

¶=

%2

0!8×¡62!4 lbf$ ft3

¢

%2 = 1019!5 lbf$ ft2

Hydrostatic force

H = %I = %2I

=¡1019!5 lbf$ ft2

¢(10 ft× 6 ft)

= 61170 lbf

Line of action of hydrostatic force

?1, " ? =V

?I(1)

V =>M3

12=6 ft (10 ft)3

12= 500 ft4

I = (10 ft× 6 ft) = 60 ft2

To nd ? in Eq. (1), apply the hydrostatic equation to locate an equivalent freesurface where pressure is zero.

0

(liquid+ Mequivalent =

%1(liquid

+ 0

Mequivalent =(5× 144) lbf$ ft2

0!8×¡62!4 lbf$ ft3

¢

= 14! 423 ft

141

? = Mequivalent +10 ft

2= 14! 423 ft +

10 ft

2= 19! 423 ft

Back to Eq. (1)

?1, " ? =V

?I

=500 ft4

(19! 423 ft) 60 ft2

= 0!429 ft

Equilibrium. (sum moments about the hinge)

"*?2 + H³?22+ 0!429 ft

´= 0

* = H

µ1

2+0!429 ft

?2

¶

= 61170 lbf

µ1

2+0!429 ft

10 ft

¶

= 33209 lbf

* = 33' 200 lbf

142

PROBLEM 3.71

Situation: A concrete form is described in the problem statement.

Find: Moment at base of form per meter of length.

ANALYSIS

H = %I = (1!5$2)24' 000× (1!5$ sin 60!) = 31' 177 N?1, " ? = V$?I

= 1× (1!5$ sin 60!)3$(12× (1!5$2 sin 60!))× (1!5$ sin 60!))= 0!2887 m

Sum moment at base

+ = 31' 177× (1!5$2 sin 60! " 0!2887)= 18' 000 N ·m/m

+ = 18 kN·m/m

143

PROBLEM 3.72


Find: Gate is stable or unstable.

ANALYSIS

?1, = (2$3)× (8$ cos 45!) = 7!54 m

Point D is (8/cos 45!) m-3.5 m=7.81 m along the gate from the water surface; there-fore, the gate is unstable.

144

PROBLEM 3.73


Find: Minimum volume of concrete to keep gate in closed position.

ANALYSIS

H = %I = 1× 9' 810× 2× 1 = 19' 620 N?1, " ? = V$?I = (1× 23)$(12× 1× 2× 1) = 0!33 m

. = 19' 620× (1" 0!33)$2!5 = 5' 258 N," = 5' 258$(23' 600" 9' 810)

," = 0!381 m3

145

PROBLEM 3.74


Find: Minimum volume of concrete to keep gate in closed position..

ANALYSIS

H = 2!0× 62!4× 2× 4 = 998!4 lbf?1, " ? = (2× 43)$(12× 2!0× 2× 4) = 0!667 ft

. = 998!4(2!0" 0!667)$5 = 266 lbf," = 266$(150" 62!4)

," = 3!04 ft3

146

PROBLEM 3.75


Find: Length of chain so that gate just on verge of opening.

APPROACH

Apply hydrostatic force equations and then sum moments about the hinge.

ANALYSIS

Hydrostatic force

H6 = %I = 10× 9' 810× 012$4

= 98' 100× 0(12$4)= 77' 048 N

?1, " ? = V$(?I)

= (034$4)$(10× 012$4)

?1, " ? = 32$40 = 0!00625 m

EquilibriumX

+Hinge = 0

H6 × (0!00625 m)" 1× H = 0

But H = Hbuoy ".= I(10 m" L)(H2O " 200= (0$4)(!252)(10" L)(9' 810)" 200= 4815!5 N" 481!5L N" 200 N= (4615!5" 481!5L) N

where L = length of chain

77' 048× 0!00625" 1× (4615!5" 481!5L) = 0

481!55" 4615!5 + 481!5L = 0

L = 8!59 m

cpy -y

F

FH

147

PROBLEM 3.76

Situation: Three submerged gates are described in the problem statement.

Find: Which wall requires the greatest resisting moment.

ANALYSIS

The horizontal component of force acting on the walls is the same for each wall.However, walls I"I0 and - "- 0 have vertical components that will require greaterresisting moments than the wall D"D0. If one thinks of the vertical component as aforce resulting from buoyancy, it can be easily shown that there is a greater ”buoyant”force acting on wall I"I0 than on - 0- 0. Thus,

wall I"I0 will require the greatest resisting moment.

148

PROBLEM 3.77

Situation: A radial gate is described in the problem statement.

Find: Where the resultant of the pressure force acts.

ANALYSIS

dF=pdA

Consider all the di!erential pressure forces acting on the radial gate as shown. Be-cause each di!erential pressure force acts normal to the di!erential area, then eachdi!erential pressure force must act through the center of curvature of the gate. Be-cause all the di!erential pressure forces will be acting through the center of curvature(the pin), the resultant must also pass through this same point (the pin).

149

PROBLEM 3.78

Situation: A curved surface is described in the problem statement.

Find: (a) Vertical hydrostatic force.(b) Horizontal hydrostatic force.(c) Resultant force.

ANALYSIS

H? = 1× 9' 810× 1×+(1$4)0 × (1)2 × 1× 9' 810

H? = 17' 515 N

E = +0$H?

= 1× 1× 1× 9' 810× 0!5 + 1× 9' 810×1Z

0

&1" E2EAE$17' 515

= 0.467 m

H6 = %I

= (1 + 0!5)9' 810× 1× 1

H6 = 14' 715 N

?1, = ? + V$?I

= 1!5 + (1× 13)$(12× 1!5× 1× 1)?1, = 1!555 m

HF =p(14' 715)2 + (17' 515)2

HF = 22' 876 N

tan K = 14' 715$17' 515

K = 40!20

150

PROBLEM 3.79

Situation: A radial gate is described in the problem statement.

Find: Hydrostatic force acting on gate.

ANALYSIS

x

y

60o

From the reasoning given in the solution to problem 3.94, we know the resultantmust pass through the center of curvature of the gate. The horizontal component ofhydrostatic force acting on the gate will be the hydrostatic force acting on the verticalprojection of the gate or:Hydrostatic force

H6 = %I

= 25 ft× 62!4 lb/ft3 × 40 ft× 50 ftH6 = 3' 120' 000 lb

The vertical component of hydrostatic force will be the buoyant force acting on theradial gate. It will be equal in magnitude to the weight of the displaced liquid (theweight of water shown by the cross-hatched volume in the above Fig.).Thus,

H? = (,"where ," = [(60$360)0 × 502 ft2 " (1$2)50× 50 cos 30! ft2]× 40 ft

= 226!5 ft2 × 40 ft= 9600 ft3

Then H? = (62!4 lbf/ft3)(9060 ft3) = 565' 344 lbs

Hresult = 3' 120' 000 i + 565' 344 j lbf

acting through the center of curvature of the gate.

151

PROBLEM 3.80

Situation: A metal surface with liquid inside is described in the problem statement.

Find: Magnitude, direction, and location of horizontal and vertical components.

ANALYSIS

H6 = %I

= "2!5× 50× (3× 1)

H6 = "375 lbf/ft

(force acts to the right)

H? = ," ( = (1× 3 + 0 × 32 ×1

4)50

H? = 503!4 lbf/ft (downward)

?1, = 2!5 + 1× 33$(12× 2!5× 1× 3)?1, = 2!8 ft above the water surface

152

PROBLEM 3.81

Situation: A plug is described in the problem statement.

Find: Horizontal and vertical forces on plug.

ANALYSIS

Hydrostatic force

H7 = %I

= (RI

= 9810× 2× 0 × 0!22

H7 = 2465 N

The vertical force is simply the buoyant force.

H@ = (,"

= 9810×4

6× 0 × 0!253

H@ = 321 N

153

PROBLEM 3.82

Situation: A dome below the water surface is described in the problem statement.

Find: Magnitude and direction of force to hold dome in place.

ANALYSIS

H6 = (1 + 1)9810× 0 × (1)2

= 61' 640 N = 61!64 kN

This 61.64 kN force will act horizontally to the left to hold the dome in place.

(?1, " ?) = V$?I

= (0 × 14$4)$(2× 0 × 12)= 0!125 m

The line of action lies 0.125 m below the center of curvature of the dome.

H? = (1$2)(40 × 13$3)9' 810= 20' 550 N

H? = 20!55 kN

To be applied downward to hold the dome in place.

154

PROBLEM 3.83

Situation: A dome below the water surface is described in the problem statement.

Find: Force on the dome.

ANALYSIS

(4/3)(r/"#

Vertical projection

The horizontal component of the hydrostatic force acting on the dome will be thehydrostatic force acting on the vertical projection of the bottom half of the dome.

Hydrostatic force

H6 = %I

% = (4$3)(5$0) ft (62.4 lbf/ft3)

= 132!4 lbf/ft2

H6 = (132!4 lbf/ft2)(0$8)(102) ft2 = 5' 199 lbf

The vertical component of force will be the buoyant force acting on the dome. Itwill be the weight of water represented by the cross-hatched region shown in the Fig.(below).

Dome

Thus,

H? = (,"= (62!4 lbf/ft3)((1$6)013$4) ft3

H? = 8' 168 lbf

The resultant force is then given below. This force acts through the center of curva-ture of the dome.

Fresult = 5' 199i+ 8' 168j lbf

155

PROBLEM 3.84

Situation: A block of material is described in the problem statement.

Find: Specic weight and volume of material.

ANALYSIS

.in air = 700 N = ," (block (1)

.in water = 400 N = (," (block " ," (water) (2)

(water = 9810 N/m3 (3)

SolveEqs. (1), (2), and (3)

," = 0!0306 m3

(block = 22' 900 N/m3

156

PROBLEM 3.85

Situation: A weather balloon is described in the problem statement.

Find: Maximum altitude of balloon.

Assumptions: &0 = 288 K

ANALYSIS

Initial Volume

,"0 = (0$6)130

= (0$6) (1m)3

= 0!524 m3

Ideal gas law

#0IHe =%0IHe"He&0

=111' 300

(2077)(288)

= 0!186 kg/m3

Conservation of mass/0 = /alt.

,"0 #0IHe = ,"alt. #He,"alt. = ,"0

L0"HeLHe

EquilibriumX

H; = 0

Hbuoy. ". = 0

,"alt. #air) " (/) + .He) = 0

Eliminate V–alt.

(,"0 #0$#He)#air) = (/) + ,"0 #0IHe))

Eliminate #’s with equation of state

(,0#0)(%alt.$"air))

(%alt. + 10' 000)$("He)= (mg + ,0#0))

(0!524)(0!186)(9!81)(2077)%alt.(%alt. + 10' 000)(287)

= (0!1)(9!81) + (0!524)(0!186)(9!81)

Solve%alt. = 3888 Pa

157

Check to see if %alt. is in the troposphere or stratosphere. Using Eq. (3.15) solve forpressure at top of troposphere.

% = %0

·&0 " T(R " R0)

&0

¸D*HF

= 101' 300[(296" 5!87× 10"3)(13' 720)$296]54823

= 15' 940 Pa

Because %alt. G %at top of troposphere we know that %alt. occurs above the stratosphere.The stratosphere extends to 16.8 km where the temperature is constant at -57.5oC.The pressure at the top of the stratosphere is given by Eq. (3.16)

% = %0="(;";0)D*F-

= 15!9 exp("(16' 800" 13' 720)× 9!81$(287× 215!5))= 9!75 kPa

Thus the balloon is above the stratosphere where the temperature increases linearlyat 1.387oC/km. In this region the pressure varies as

% = %0

·&0 + T(R " R0)

&0

¸"D*HF

Using this equation to solve for the altitude, we have

3888

9750=

·215!5 + 1!387× (R " 16!8)

215!5

¸"9481*(04001387×287)

0!399 = [1 + 0!00644× (R " 16!8)]"2446

R = 22!8 km

158

PROBLEM 3.86

Situation: A rock is described in the problem statement.

Find: Volume of rock.

ANALYSIS

," ( = 918 N

," (( " 9' 810) = 609 N

," = (918" 609)$9' 810

," = 0!0315 m3

159

PROBLEM 3.87

Situation: A rod is described in the problem statement.

Find: Describe the liquid.

ANALYSIS

Rod weight = (22I(M + 2I(2#M )))

= 42I(M)

= 42I(M

Buoyancy force

Buoyant force G ," (Liq = 32I(LiqRod weight = Buoyant force

42I(M G 32I(Liq(Liq F (4$3)(M .

The liquid is more dense than water so is answer c).

160

PROBLEM 3.88

Situation: A person oating is a boat with an aluminum anchor.

Find: Change of water level in pond in the pond.

ANALYSIS

Weight anchor = 0.50 ft3 × (2!2× 62!4 lb/ft3) = 68!65 lb.The water displaced by boat due to weight of anchor

= 68!65 lb/(62.4 lb/ft3) = 1!100 ft3

Therefore, when the anchor is removed from the boat, the boat will rise and the waterlevel in the pond will drop:

!M = 1!10 ft3$500 ft2 = 0!0022 ft

However, when the anchor is dropped into the pond, the pond will rise because of thevolume taken up by the anchor. This change in water level in the pond will be:

!M = 0!500 ft3$500 ft2 = !001 ft

Net change =-.0022 ft + .001 ft = -.0012 ft = -.0144 in.The pond level will drop 0.0144 inches.

161

PROBLEM 3.89

Situation: An inverted cone containing water is described in the problem statement.

Find: Change of water level in cone.

ANALYSIS

6 = 0!6 =! (block = 0!6(water

Weight of displaced water = weight of block

,"M (M = ,"0 (0,"M = ((0$(M ),"0,"M = 0!6,"0 = 120 cm3

Then the total volume below water surface when block is oating in water = V–MIorg. + 120 cm3

,"MIorig. = (0$3)(10 cm)3

= 1047!2 cm3

,"nal = 1047!2 cm3 + 120 cm3

(0$3)M3nal = 1167!2 cm3

Mnal = 10!368 cm

!M = 0!368 cm

162

PROBLEM 3.90

Situation: Concrete cylindrical shells are described in the problem statement.

Find: Height above water when erected.

ANALYSIS

The same relative volume will be unsubmerged whatever the orientation; therefore,

$ L

h

sV

usV

,"u.s.,"s

=MI

2I=2Iu.s.2I

or M$2 = Iu.s.$I

Also,

cos K = 50$100 = 0!50

K = 60! and 2K = 120!

SoIu.s. = (1$3)0"

2 "" cos 60!" sin 60!

Therefore

M$2 = "2 [((1$3)0)" sin 60! cos 60!)] $0"2 = 0.195

M = 7!80 m

163

PROBLEM 3.91

Situation: A cylindrical tank is described in the problem statement.

Find: Change of water level in tank.

ANALYSIS

!,"M (M = . block

!,"M = 2 lbf/(62.4 lbf/ft3) = 0!03205 ft3

! !MI- = !,"M!M = !,"M $IN = 0!03205 ft3$((0$4)(12) ft2)!M = 0!0408 ft

Water in tank will rise 0.0408 ft.

164

PROBLEM 3.92

Situation: A oating platform is described in the problem statement.

Find: Length of cylinder so that it oats 1 m above water surface.

ANALYSIS

XH3 = 0

"30' 000" 4× 1' 0002+ 4× (0$4)× 12 × 10' 000(2" 1) = 0

2 = 2!24 m

165

PROBLEM 3.93

Situation: A oating block is described in the problem statement.

Find: Depth block will oat.

Assumptions: The block will sink a distance ? into the uid with 6 = 1!2.

ANALYSIS

XH3 = 0

". + %I = 0

"(62)2 × 32× 0!8(water + (2× (water + ? × 1!2(M )3622 = 0

? = 1!1672

A = 2!1672

166

PROBLEM 3.94

Situation: A cylindrical tank holds water. Water depth is 2 ft (before addition ofice). Cylinder is 4 ft high and 2 ft in diameter. A 5 lbm chunk of ice is added tothe tank.

Find: (a) Change of water level in tank after ice is added, (b) change in water levelafter the ice melts, (c) explain all processes.

ANALYSIS

Change in water level (due to addition of ice)

.ice = Hbuoyancy

= !,—M(M

So

!,—M =.ice

(M=

5 lbf

62!4 lbf$ ft3

= 0!0801 ft3

Rise of water in tank (due to addition of ice)

!M =!,—MIcyl

=0!0801 ft3

(0$4)(2 ft)2= 0!02550 ft = 0!3060 in

!M = 0!306 in G== (due to addition of ice)

Answer ! When the ice melts, the melted water will occupy the same volume ofwater that the ice originally displaced; therefore, there will be no change in watersurface level in the tank after the ice melts.

167

PROBLEM 3.95

Situation: A partially submerged wood pole is described in the problem statement.

Find: Density of wood.

ANALYSIS

+= = 0

".wood × (0!52 cos 30!) + H>. × (5$6)2 cos 30! = 0

"(wood ×I2× (0!52 cos 30!) + ((1$3)I2(H2O)× (5$6)2 cos 30

! = 0

(wood = (10$18)(H2O(wood = 5' 450 N/m

3

#wood = 556 kg/m3

168

PROBLEM 3.96

Situation: A partially submerged wood pole is described in the problem statement.

Find: If pole will rise or fall.

ANALYSIS

Sum moments about I to see if pole will rise or fall. The forces producing momentsabout I will be the weight of the pole and the buoyant force.

W

FB

X+= = "(1$2)(2 cosT)(2(,I) + (3$4)(2 cosT)(2$2)(liqI

= 22I cosT["(1$2)(, + (3$8)(liq]= <("80 + 75)

A negative moment acts on the pole; therefore, it will fall.

169

PROBLEM 3.97

Situation: A oating ship is described in the problem statement.

Find: How much the ship will rise or settle.

ANALYSIS

Draft = (38' 000× 2' 000)$40' 000(

=1900

(ft

Since ( of salt water is greater than ( of fresh water, the ship will take a greater draftin fresh water.

(1900$62!4)" (1900$64!1) = 0.808 ft

170

PROBLEM 3.98

Situation: A submerged spherical buoy is described in the problem statement.

Find: Weight of scrap iron to be sealed in the buoy.

ANALYSIS

XH? = 0;H> " H& " H% " H1 = 0H& = H> " H% " H1

= (4$3)0(0!6)3 × 10' 070" 1600" 4' 500

H& = 3010 N of scrap

171

PROBLEM 3.99

Situation: A balloon used to carry instruments is described in the problem statement.

Find: Diameter of spherical balloon.

Assumptions: Standard atmospheric temperature condition.

APPROACH

Apply buoyancy force and the ideal gas law.

ANALYSIS

W

FB

WI

Ideal gas law

& = 533" 3!221× 10"3 × 15' 000 = 485"#air = (8!3× 144)$(1' 715× 485)

= 0!001437 slugs/ft3

#He = (8!3× 144)$(12' 429× 485)= 0!000198 slugs/ft3

Equilibrium

XH = 0

= HE " H0 " H#= (1$6)013)(#air " #He)" 01

2(0!01)" 10= 13 × 16!88(14!37" 1!98)10"4 "12 × 3!14× 10"2 " 10

1 = 8!35 ft

172

PROBLEM 3.100

Situation: A buoy is described in the problem statement.

Find: Location of water level.

ANALYSIS

The buoyant force is equal to the weight.

H> =.

The weight of the buoy is 9!81× 460 = 4512 N.The volume of the hemisphere at the bottom of the buoy is

, =1

2

0

613 =

0

1213 =

0

12m3

The buoyant force due to the hemisphere is

H> =0

12(9!81)(1010) = 2594 N

Since this is less than the buoy weight, the water line must lie above the hemisphere.Let M is the distance from the top of the buoy. The volume of the cone which liesbetween the top of the hemisphere and the water line is

, =0

332!M! "

0

332M =

0

3(0!52 × 0!866" M3 tan2 30)

= 0!2267" 0!349M3

The additional volume needed to support the weight is

, =4512" 25949!81× 1010

= 0!1936 m3

Equating the two volumes and solving for M gives

M3 =0!0331

0!349= 0!0948 m3

M = 0!456 m

173

PROBLEM 3.101

Situation: A hydrometer is described in the problem statement.

Find: Weight of hydrometer.

ANALYSIS

Hbuoy. = . .

," (M = .

(1 cm3 + (5!3 cm)(0.01 cm2))(0!13) m3$cm3((M ) = . .

(1!53 cm3)(10"6 m3$cm3)(9810 N/m3) = . .

. = 1!50× 10"2 N

174

PROBLEM 3.102


Find: Specic gravity of oil.

ANALYSIS

Hbuoy. = .

(1 cm3 + (6!3 cm)(0.1 cm2))(0!013) m3$cm3(oil = 0!015 N

(1 + 0!63)× 10"6 m3(oil = 0!015 N

(oil = 9202 N/m3

6 = (oil$(M= 9202$9810

6 = 0!938

175

PROBLEM 3.103


Find: Weight of each ball.

ANALYSIS

Equilibrium (for a ball to just oat, the buoyant force equals the weight)

H> =. (1)

Buoyancy force

H> =

µ013

6

¶(uid (2)

Combine Eq. (1) and (2) and let 1 = 0!01m!

. =

µ013

6

¶6(water

=

Ã0 (0!01)3

6

!6 (9810)

= 5!136× 10"36 (3)

The following table (from Eq. 3) shows the weights of the balls needed for the requiredspecic gravity intervals.

ball number 1 2 3 4 5 6sp. gr. 1!01 1!02 1!03 1!04 1!05 1!06

weight (mN) 5!19 5!24 5!29 5!34 5!38 5!44

176

PROBLEM 3.104


Find: Range of specic gravities.

ANALYSIS

When only the bulb is submerged;

H> = . .

(0$4)£0!022 × 0!08

¤× 9810× 6 = 0!035× 9!81

6 = 1!39

When the full stem is submerged;

(0$4)£(0!02)2 × (0!08) + (0!01)2 × (0!08)

¤9' 810× 6

= 0!035× 9!816 = 1!114

Range 1.114 to 1.39

177

PROBLEM 3.105


Find: Design a hydrometer to measure the specic weight of liquids.

Assumptions: The hydrometer will consist of a stem mounted on a spherical ball asshown in the diagram. Assume also for purposes of design than the diameter of thestem is 0.5 in and the maximum change in depth is 2 in.

ANALYSIS

½ in

2 in

Since the weight of the hydrometer is constant, the volumes corresponding to thelimiting uid specic weights can be calculated from

. = (60,"60 = (70,"70or

,"70,"60

=60

70= 0!857

The change in volume can be written as

,"60 ","70 = ,"60 (1","70,"60

) = 0!143,"60

The change in volume is related to the displacement of the uid on the stem by

I!M

,"60= 0!143

For the parameters given above the volume of the hydrometer when immersed inthe 60 lbf/ft3 liquid is 2.74 in3! Assume there is one inch of stem between the lowermarking and the top of the spherical ball so the volume of the spherical ball would be2.55 in3 which corresponds to a ball diameter of 1.7 in. The weight of the hydrometerwould have to be

. = (60,"60 = 0!0347 lbf/in3 × 2!74 in3 = 0!095 lbf

If one could read the displacement on the stem to within 1/10 in, the error would inthe reading would be 5%.Other designs are possible. If one used a longer stem displacement, a larger volumehydrometer would be needed but it would give better accuracy. The design willdepend on other constraints like the volume of uid and space available.

178

PROBLEM 3.106

Situation: A barge is described in the problem statement.

Find: Stability of barge.

ANALYSIS

Draft = 400' 000$(50× 20× 62!4)= 6!41 ft G 8 ft

GM = V00$,""CG=

£(50× 203$12)$(6!41× 50× 20)

¤" (8" 3!205)

= 0!40 ft

Will oat stable

179

PROBLEM 3.107

Situation: A oating body is described in the problem statement.

Find: Location of water line for stability and specic gravity of material.

ANALYSIS

For neutral stability, the distance to the metacenter is zero. In other words

X+ =V!!,""X- = 0

where X- is the distance from the center of gravity to the center of buoyancy.

Moment of inertia at the waterline

V!! =Y32

12

where 2 is the length of the body. The volume of liquid displaced is MY2 so

X- =Y32

12MY2=Y2

12M

The value for X- is the distance from the center of buoyancy to the center of gravity,or

X- =Y

2"M

2

SoY

2"M

2=Y2

12Mor µ

M

Y

¶2"M

Y+1

6= 0

Solving for M$Y gives 0.789 and 0.211. The rst root gives a physically unreasonablesolution. Therefore

M

Y= 0!211

The weight of the body is equal to the weight of water displaced.

(0,0 = ((,

Therefore

6 =(0((=YM2

Y22=M

Y= 0.211

The the specic gravity is smaller than this value, the body will be unstable (oatstoo high).

180

PROBLEM 3.108

Situation: A block of wood is described in the problem statement.

Find: Stability.

ANALYSIS

draft = 1× 7500$9' 810 = 0!7645 mZfrom bottom = 0!7645$2 = 0!3823m

Metacentric height

X = 0!500 m; CG = 0!500" 0!3823 = 0!1177 mX+ = (V$,")"CG

= ((0"4$4)$(0!7645× 0"2))" 0!1177= 0!0818 m" 0!1177 m (negative)

Thus, block is unstable with axis vertical.

181

PROBLEM 3.109

Situation: A block of wood is described in the problem statement.

Find: Stability.

ANALYSIS

draft = 5,000/9,810

= 0.5097 m

Metacentric height

GM = V00$,""CG=

£(0 × 0!54$4)$(0!5097× 0 × 0!52)

¤" (0!5" 0!5097$2)

= "0!122 m, negative

So will not oat stable with its ends horizontal.

182

PROBLEM 3.110

Situation: A oating block is described in the problem statement.

Find: Stability.

ANALYSIS

Analyze longitudinal axis

GM = V00$,""CG= (34(24)3$(12×4 × 24 × 34))"4$2= "4$6

Not stable about longitudinal axis.

Analyze transverse axis.

GM = (24 × (34)3$(12×4 × 24 × 34))" 34$4= 0

Neutrally stable about transverse axis.Not stable

183

PROBLEM 4.1

Situation: The valve in a system is gradually opened to have a constant rate ofincrease in discharge.

Find: Describe the ow at points A and B.

ANALYSIS

B: Non-uniform, unsteady.

A: Unsteady, uniform.

184

PROBLEM 4.2

Situation: Water ows in a passage with ow rate decreasing with time.

Find: Describe the ow.

ANALYSIS

(b) Unsteady and (d) non-uniform.

(a) Local and (b) convective acceleration.

185

PROBLEM 4.3

Situation: A ow pattern has converging streamlines.

Find: Classify the ow.

ANALYSIS

Non-uniform; steady or unsteady.

186

PROBLEM 4.4

Situation: A uid ows in a straight conduit. The conduit has a section with constantdiameter, followed by a section with changing diameter.

Find: Match the given ow labels with the mathematical descriptions.

ANALYSIS

Steady ow corresponds to [,&$[P = 0Unsteady ow corresponds to [,&$[P 6= 0Uniform ow corresponds to ,&[,&$[9 = 0Non-uniform ow corresponds to ,&[,&$[9 6= 0

187

PROBLEM 4.5

Situation: Pathlines are shown in gure. Discharge is constant and ow is nonturbu-lent.


ANALYSIS

True statements: (a), (c).

188

PROBLEM 4.6

Situation: Dye is injected into a ow eld. The streakline is shown.

Find: Draw a pathline of the particle.

ANALYSIS

The streakline shows that the velocity eld was originally in the horizontal directionto the right and then the ow eld changed upward to the left. The pathline startso! to the right and then continues upward to the left.

189

PROBLEM 4.7

Situation: A hypothetical ow has the following characteristics:For 0 ' P ' 5 seconds, @ = 2 m/s, ; = 0For 5 G P ' 10 seconds, @ = 3 m/s, ; = "4 m/s

At time zero a dye streak was started, and a particle was released.

Find: For P = 10 s, draw to scale the streakline, pathline of the particle, and stream-lines.

ANALYSIS

From 0GtG5, the dye in the streakline moved to the right for a distance of 10 m. Atthe same time a particle is released from the origin and travels 10 m to the right.Then from 5GtG10, the original line of dye is transported in whole downward to theright while more dye is released from the origin. The pathline of the particle proceedsfrom its location at t=5 sec downward to the right.At 10 sec, the streamlines are downward to the right.

190

PROBLEM 4.8

Situation: A dye streak is produced in a ow that has a constant speed. The originof the streak is point A, and the streak was produced during a 10 s interval.

Find: (a) Sketch a streamline at P = 8 s.(b) Sketch a particle pathline at P = 10 s (for a particle that was released from pointA at time P = 2 s).

ANALYSIS

At 8 seconds (near 10 sec) the streamlines of the ow are horizontal to the right.

Streamlines at P = 8 s

Initially the ow is downward to the right and then switches to the horizontal directionto the right. Thus one has the following pathline.

Particle pathline for a particle released at P = 2 s

191

PROBLEM 4.9

Situation: A periodic ow eld is described in which streamline pattern changes everysecond and repeats every two seconds.

Find: Sketch a streakline at P = 2!5 s.

ANALYSIS

From time P = 0 to P = 1 s dye is emitted from point A and will produce a streakthat is 10 meters long (up and to the right of A). See Fig. A below. In the nextsecond the rst streak will be transported down and to the right 10 meters and a newstreak, 10 ft. long, will be generated down and to the right of point A (see Fig. Bbelow). In the next 0.5 s streaks in Fig. B will move up and to the right a distanceof 5 meters and a new streak 5 meters in length will be generated as shown in Fig.C.

192

PROBLEM 4.10

Situation: The gure below shows a pathline and a particle line for a ow. Theuid particle was released from point A at P = 0 s. The streakline was produced byreleasing dye from point A from P = 0 to 5 s.

Find: (a)Sketch a streamline for P = 0 s.(b) Describe the ow as steady or unsteady.

ANALYSIS

In the above sketch, the dye released at P = 0 s is now at point B. Therefore, astreamline corresponding to P = 0 s should be tangent to the streakline at point B.We can reach the same conclusion by using the pathline.

In the above sketch, the path of a uid particle at P = 0 s is shown by the dotted lineat point A. There, a streamline corresponding to P = 0 s should be tangent to thepathline at point A. Thus, streamlines at P = 0 appear as shown below:

The ow is unsteady because the streakline, streamlines and pathlines di!er.

193

PROBLEM 4.11

Situation: A velocity eld is dened by @ = 5 m/s and ; = "2P m/s, where t is time.

Find: (a) Sketch a streakline for P = 0 to 5 s.(b) Sketch a pathline for a particle for P = 0 to 5 s. The particle is released from thesame point as the dye source.(c) Sketch streamlines at P = 5 s.

ANALYSIS

Particle pathline.Since @ = AE$AP, we may write AE = @AP! This can be integrated to give the x-positionof a particle at any time P:

E = E! +

Z@AP = E! +

Z5AP

E = E! + 5P

Similarly,

? = ?! +

Z;AP = 0 +

Z"2PAP

? = ?! " P2

Letting E! = ?! = 0'we can construct a table of coordinates

t (s) x (m) y(m)0 0 01 5 -12 10 -43 15 -94 20 -165 25 -25

The (E' ?) data from this table are plotted in the gure below

194

Streakline.

To construct the streakline, solve for the displacement of dye particles. The dyeparticle released at time P = 1 s will reach a position given by

E = E! +

Z 5

1

@AP

= 0 +

Z 5

1

5AP = 21

? = ?! +

Z 5

1

;AP

= 0 +

Z 5

1

"2PAP = 0" P2|51 = "24

The dye particle released at time P = 2 s will reach a position given by

E = 0 +

Z 5

2

5AP = 15

? = 0 +

Z 5

2

"2PAP = "21

Performing similar calculations for each time yields the coordinates of the streakline.These results are plotted in the above gure.Streamlines (at P = 5 s)

Dye released at P = 5 s is at point A in the sketch above. Therefore,a streamlinecorresponding to P = 5 s should be tangent to the streakline at point A. We can reachthe same conclusion by using the pathline. The path of a uid particle at P = 5 s isat point B. There, a streamline corresponding to P = 0 s should be tangent to thepathline at point B. The streamlines are shown below

Animation An animation of the solution can be found athttp://www.justask4u.com/csp/crowe.

195

PROBLEM 4.12

Situation: Fluid ows along a circular path. It rst moves in the clockwise directionat 0 rad/s for 1 second and then reverses direction with the same rate.

Find: (a) Draw a pathline at time P = 2 s!(b) Draw a streakline at time P = 2 s!

ANALYSIS

Pathline

For the rst second the particle will follow the circular streamline (clockwise) throughan angle of 0 radians (1/2 circle). Then for the 2nd second the particle reverses itsoriginal path and nally ends up at the starting point. Thus, the pathline will beshown:

Streakline

For the rst second a stream of dye will be emitted from staring point and the streakfrom this dye will be generated clockwise along the streamline until the entire top halfcircle will have a steak of dye at the end of 1 second. When the ow reverses a newdye streak will be generated on the bottom half of the circle and it will be superposedon top of the streak that was generated in the rst second. The streakline is shownfor t=1/2 sec., 1 sec. & 2 sec.


196

PROBLEM 4.13

Situation: Fluid ows in a three-dimensional ow eld. The uid moves in each ofcoordinate directions at 1 m/s for one second.

Find: (a) Sketch a pathline on a three dimensional coordinate at time P = 3 s!(b) Sketch a streakline at time P = 2 s!

ANALYSIS

The nal pathline and streakline are shown below.

x x

y y

z zPathline Streakline

197

PROBLEM 4.14

Situation: A droplet moves from location A to B in a uniform ow eld leaving atrail of vapor.

Find: Sketch the location of the vapor trail.

ANALYSIS

The vapor will produce a vapor trail as shown.

A

Vapor is transport from the droplet in the ow direction as the droplet proceedsupward to the right.

198

PROBLEM 4.15

Situation: Fluid ows in a two dimensional ow eld with @ = 20P2 and ; = 30P1*2!The period of time is 0 ' P ' 1! The pathline and streakline begin at the origin.

Find: Write a computer program to give the coordinates of (a) streakline and (b)pathline.

ANALYSIS

The computed streaklines and pathline are shown below.

x-coordinate

0 2 4 6 8

y co

ordi

nate

0

5

10

15

20

25

Pathline

Streakline

In FORTRAN:Dimension statementsInitial valuesdo 10 i=1,Nt=t+dtu=20*t**2v=30*sqrt(t)xp(i+1)=xp(i)+u*dtyp(i+1)=yp(i)+v*dtdo 20 j=i,1,-1xs(j+1)=xs(j)+u*dtys(j+1)=ys(j)+v*dt

20 continue10 continue

199

PROBLEM 4.16

Situation: A series of ows are described in the problem statement.

Find: Classify the ows as one dimensional, two dimensional, or three dimensional.

ANALYSIS

a. Two dimensional e. Three dimensionalb. One dimensional f. Three dimensionalc. One dimensional g. Two dimensionald. Two dimensional

200

PROBLEM 4.17

Situation: Flow past a circular cylinder with constant approach velocity.

Find: Describe the ow as:(a) Steady or unsteady.(b) One dimensional, two dimensional, or three dimensional.(c) Locally accelerating or not, and is so, where.(d) Convectively accelerating or not, and if so, where.

ANALYSIS

(a) Steady.(b) Two-dimensional.(c) No.(d) Yes, convective acceleration is present at all locations where the streamlinescurve. Also, convective acceleration is present at each where a uid particles changesspeed as it moves along the streamline.

201

PROBLEM 4.18

Situation: A ow with this velocity eld: @ = EP+ 2?' ; = EP2 " ?P' Y = 0!

Find: Acceleration, a' at location (1,2) and time P = 3 seconds.

ANALYSIS

Acceleration in the x-direction

C$ = @[@$[E+ ;[@$[? + Y[@$[R + [@$[P

= (EP+ 2?)(P) + (EP2 " ?P)(2) + 0 + E

At E = 1 m' ? = 2 m and P = 3 s

C$ = (3 + 4)(3) + (9" 6)(2) + 1 = 21 + 6 + 1 = 28 m/s2

Acceleration in the y-direction

C3 = @[;$[E+ ;[;$[? + Y[;$[R + [;$[P

= (EP+ 2?)(P2) + (EP2 " ?P)("P) + 0 + (2EP" ?)

At E = 1 m' ? = 2 m and P = 3 s

C3 = (3 + 4)(9) + (9" 6)("3) + (6" 2) = 63" 9 + 4 = 58 m/s2

a = 28 i+ 58 j m/s2

202

PROBLEM 4.19

Situation: Air is owing around a sphere. The x-component of velocity along thedividing streamline is given by @ = "\!(1" 33!$E3)!

Find: An expression for the x-component of acceleration (the form of the answershould be C$ = C$ (E' 3!' \!)).

ANALYSIS

C$ = @[@$[E+ [@$[P

= "\0(1" 330$E3)[$[E("\0(1" 330$E

3)) + [$[P("\0(1" 330$E3))

= \20 (1" 330$E

3)("3330$E4) + 0

C$ = "(3\20 330 $ E4)(1" 330 $ E3)

203

PROBLEM 4.20

Situation: A velocity eld at 3 = 10 m where ,O = 10P!

Find: Magnitude of acceleration at 3 = 10 m and P = 1 s.

ANALYSIS

,O = 10P

Ctang. = ,O[,O$[9+ [,O$[P

Ctang. = 0 + 10 m/s2

Cnormal = , 2O $3

= (10P)2$3 = 100P2$10 = 10P2

at P = 19

Cnormal = 10 m/s2

Ctotal =qC2tang. + C

2normal =

&200

Ctotal = 14!14 m/s2

204

PROBLEM 4.21

Situation: Flow occurs in a tapered passage. The velocity is given as

, = ]$I

and] = ]! "]1

P

P!

The point of interest is section AA, where the diameter is 50 cm. The time of interestis 0.5 s.

Find: (a) Velocity at section AA: ,(b) Local acceleration at section AA: C<(c) Convective acceleration at section AA: C1

ANALYSIS

] = ]0 "]1P$P0 = 0!985" 0!5P (given)

, = ]$I (given)[,

[9= +2

msper m (given)

The velocity is

, = ]$I

= (0!985" 0!5× 0!5)$(0$4× 0!52)

, = 3!743 m/s

Local acceleration

C< = [,$[P = [$[P(]$I)

= [$[P((0!985" 0!5P)$(0$4× 0!52))= "0!5$(0$4× 0!52)

C< = "2!55 m/s2

Convective acceleration

C1 = , [,$[9

= 3!743× 2C1 = +7!49 m/s2

205

PROBLEM 4.22

Situation: One-dimensional ow occurs in a nozzle. Velocity varies linearly from 1ft/s at the base to 4 ft/s at the tip. The nozzle is 18 inches long.

Find: (a) Convective acceleration: C1(b) Local acceleration: C<

ANALYSIS

Velocity gradient

A,$A9 = (,tip " ,base)$2= (4" 1)$1!5= 2 s"1

Acceleration at mid-point

, = (1 + 4)$2

= 2!5 ft/s

C1 = ,A,

A9= 2!5× 2

C1=5 ft/s2

Local acceleration

C< = 0

206

PROBLEM 4.23

Situation: One-dimensional ow occurs in a nozzle and the velocity varies linearlywith distance along the nozzle. The velocity at the base of the nozzle is 1P (ft/s) and4P (ft/s) at the tip.

Find: Local acceleration midway in the nozzle: C<

ANALYSIS

C< = [,$[P

, = (P+ 4P)$2

= 2!5P (ft/s)

Then

C< = [$[P(2!5P)

a<=2.5 ft/s2

207

PROBLEM 4.24

Situation: Flow in a two-dimensional slot with

, = 2³^!>

´µ PP!

¶

Find: An expression for local acceleration midway in nozzle: C2

ANALYSIS

, = 2

µ^!>

¶µP

P!

¶but > = D$2

, =

µ4^!D

¶µP

P!

¶

C2 = [,$[P

C2=4^!$(DP!)

208

PROBLEM 4.25

Situation: Flow in a two-dimensional slot and velocity varies as

, = 2³^!>

´µ PP!

¶

Find: An expression for convective acceleration midway in nozzle: C1

ANALYSIS

C1 = , [,$[E

The width varies as> = D " E$8

, = (^0$P0)2P(D " E$8)"1

[,$[E = (^0$P0)2P(1$8)(D " E$8)"2

C1 = , [,$[E = , (^0$P0)24P2(1$8)$(D " (1$8)E)"3

At E = 2D

C1 = (1$2)(^0$P0)2P2$((3$4)D)3

C1 = 32$27(^0$P0)2P2$D3

209

PROBLEM 4.26

Situation: Water ow in a nozzle with

, = 2P$(1" 0!5E$_)2

Find: With 2 = 4 ft, and E = 0!52 and P = 3 s, nd (a) the local acceleration and(b) the convective acceleration

ANALYSIS

C< = [,$[P

= [$[P[2P$(1" 0!5E$2)2]= 2$(1" 0!5E$2)2

= 2$(1" 0!5× 0!52$2)2

C< = 3!56 ft/s2

C1 = , ([,$[E)

= [2P$(1" 0!5E$2)2][$[E[2P$(1" 0!5E$2)2]= 4P2$((1" 0!5E$2)52)= 4(3)2$((1" 0!5× 0!52$2)54)

C1 = 37!9 ft/s2

210

PROBLEM 4.27

Situation: Flow through an inclined pipe at 30o from horizontal and decelerating at0.3g.

Find: Pressure gradient in ow direction.

APPROACH

Apply Euler’s equation.

ANALYSIS

!

30o

Euler’s equation

[$[L(%+ (R) = "#C<[%$[L+ ([R$[L = "#C<

[%$[L = "#C< " ([R$[L= "(($))× ("0!30))" ( sin 30!

= ((0!30" 0!50)

[%$[L = "0!20(

211

PROBLEM 4.28

Situation: Kerosene (S=0.80) is accelerated upward in vertical pipe at 0.2g.

Find: Pressure gradient required to accelerate ow.

APPROACH


ANALYSIS

Applying Euler’s equation in the R-direction

[(%+ (R)$[R = "#C; = "(($))× 0!20)[%$[R + ( = "0!20(

[%$[R = (("1" 0!20)= 0!80× 62!4("1!20)

[%$[R = "59!9 lbf/ft3

212

PROBLEM 4.29

Situation: A hypothetical liquid with zero viscosity and specic weight of 10 kN/m3

ows through a vertical tube. Pressure di!erence is 12 kPa.

Find: Direction of acceleration.

APPROACH


ANALYSIS

Euler’s equation

#C< = "[$[L(%+ (R)C< = (1$#)("[%$[L" ([R$[L)

Let L be positive upward. Then [R$[L = +1 and [%$[L = (%= " %>)$1 = "12' 000Pa/m. Thus

C< = ()$()(12' 000" ()C< = )((12' 000$()" 1)C< = )(1!2" 1!0) m/s2

C< has a positive value; therefore, acceleration is upward. Correct answer is a).

213

PROBLEM 4.30

Situation: A piston and water accelerating upward at 0.5g.

Find: Pressure at depth of 2 ft. in water column.

APPROACH


ANALYSIS

Euler’s equation

#C< = "[$[L(%+ (R)

Let L be positive upward.

#(0!5 g) = "[%$[L" ([R$[L(($g)(0!5g) = "[%$[L" ((1)

[%$[L = "((0!5 + 1) = "1!5(

Thus the pressure decreases upward at a rate of 1.5(! At a depth of 2 ft.:

%2 = (1!5()(2) = 3(

= 3 ft.× 62!4 lbf/ft3

%2 = 187!2 psfg

214

PROBLEM 4.31

Situation: Water stands with depth of 10 ft in a vertical pipe open at top and sup-ported by piston at the bottom.

Find: Acceleration of piston to create a pressure of 9 psig immediately above piston.

APPROACH


ANALYSIS

Euler’s equation

[$[9(%+ (R) = "#C&

Take 9 as vertically upward with point 1 at piston surface and point 2 at water surface.

"!(%+ (R) = #C&!9

"(%2 " %1)" ((R2 " R1) = #C&!9

"(0" 9× 144)" 62!4× 10 = 1!94× 10C&C& = (9× 144" 62!4× 10)$19!4

C& = 34!6 ft/s2

215

PROBLEM 4.32

Situation: Water accelerates at 6m$ s2 in a horizontal pipe.

Find: Pressure gradient.

APPROACH


ANALYSIS

Euler’s equation with no change in elevation

([%$[9) = "#C&= "1' 000× 6

[%$[9 = "6' 000 N/m3

216

PROBLEM 4.33

Situation: Water accelerated from rest in horizontal pipe, 100 m long and 30 cm indiameter, at 6 m/s2. Pressure at downstream end is 90 kPa gage.

Find: Pressure at upstream end.

APPROACH


ANALYSIS

Euler’s equation with no change in elevation

([%$[9) = "#C&= "1' 000× 6= "6' 000 N/m3

%downstream " %upstream = ([%$[9)!9

%upstream = 90' 000 + 6' 000× 100= 690' 000 Pa, gage

%upstream = 690 kPa, gage

217

PROBLEM 4.34

Situation: Water stands at depth of 10 ft in a vertical pipe closed at the bottom bya piston.

Find: Maximum downward acceleration before vaporization assuming vapor pressureis zero (abs).

APPROACH


ANALYSIS

Applying Euler’s equation in the R-direction with % = 0 at the piston surface

[$[R(%+ (R) = "#C;!(%+ (R) = "#C;!R

(%+ (R)at water surface " (%+ (R)at piston = "#C;(Rsurface " Rpiston)%atm " %@ + ((Rsurface " Rpiston) = "12 #C;

14!7× 144" 0 + 62!4(10) = "10× 1!94C;C; = "141!3 ft/s2

218

PROBLEM 4.35

Situation: A liquid with zero viscosity and specic weight of 100 lbf/ft3 ows througha conduit. Pressure are given at two points.

Find: Which statements can be discerned with certainty.

APPROACH


ANALYSIS

Euler’s equation

"[$[L(%+ (R) = #C<"[%$[L" ([R$[L = #C<

where [%$[L = (%> " %=)$L = (100" 170)$2 = "35 lb/ft3 and [R$[L = sin 30! = 0!5!Then

C< = (1$#)(35" (100)(0!5))= (1$#)("15 ) lbf/ft3

• Because C< has a negative value we conclude that Answer ! (d) the accelera-tion is in the negative L direction .

• Answer ! The ow direction cannot be established; so answer (d) is the onlyanswer that can be discerned with certainty.

219

PROBLEM 4.36

Situation: Velocity varies linearly with distance in water nozzle.

Find: Pressure gradient midway in the nozzle.

APPROACH


ANALYSIS

Euler’s equationA$AE(%+ (R) = "#C$

but R =const.; therefore

A%$AE = "#C$C$ = Cconvective = , A,$AE

A,$AE = (80" 30)$1 = 50 s"1,mid = (80 + 30)$2 = 55 ft/s

= (55 ft/s)(50 ft/s/ft) = 2' 750 ft/s2

Finally

A%$AE = ("1!94 slug/ft3)(2' 750 ft/s2)

A%$AE = "5' 335 psf/ft

220

PROBLEM 4.37

Situation: Tank accelerated in x-direction to maintain liquid surface slope at -5/3.

Find: Acceleration of tank.

APPROACH


ANALYSIS

Euler’s equation. The slope of a free surface in an accelerated tank.

tanT = C$$)

C$ = ) tanT

= 9!81× 3$5

C$ = 5!89 m/s2

221

PROBLEM 4.38

Situation: Closed tank full of liquid accelerated downward at 1.5g and to the right at0.9g. Specic gravity of liquid is 1.1. Tank dimensions given in problem statement.

Find: (a) %) " %=!(b) %> " %=!

APPROACH


ANALYSIS

Euler’s equation Take L in the z-direction.

"A%

AL" (

AL

AL= #C2

(A%$AL) = "#() + C<)= "1!1× 1!94(32!2" 1!5× 32!2)= 34!4 psf/ft

%> " %= = "34!4× 4%> " %= = "137!6 psf

Take L in the x-direction. Euler’s equation becomes

"A%

AE= #C$

%) " %> = #C$2

= 1!1× 1!94× 0!9) × 3= 185!5 psf

%) " %= = %) " %> + (%> " %=)%) " %= = 185!5" 137!6

%) " %= = 47!9 lbf/ft2

222

PROBLEM 4.39

Situation: Closed tank full of liquid accelerated downward at 2/3g and to the right atone g. Specic gravity of liquid is 1.3. Tank dimensions given in problem statement

Find: (a) %) " %=!(b) %> " %=!

APPROACH


ANALYSIS

Euler’s equation in R direction

A%$AR + ( = "#C;A%$AR = "#() + C;)A%$AR = "1!3× 1' 000(9!81" 6!54)

= "4' 251 N/m3

%> " %= = 4' 251× 3= 12' 753 Pa

%> " %= = 12!753 kPa

Euler’s equation in E-direction

"A%

AE= #C$

%) " %> = #C$2

= 1!3× 1' 000× 9!81× 2!5= 31' 882 Pa

%) " %= = %) " %> + (%> " %=)%) " %= = 31' 882 + 12' 753

= 44' 635 Pa

%) " %= = 44!63 kPa

223

PROBLEM 4.40

Situation: Truck carrying tank with open top will not accelerate or decelerate morethan 8.02 ft/s2! Tank dimensions given in problem statement.

Find: Maximum depth before spilling.

APPROACH


ANALYSIS

Euler’s equation applied to slope of an accelerated free surface.

tanT = C$$) = 8!02$32!2 = 0!2491

tanT = M$9

M = 9 tanT = 9× 0!2491 = 2!242 ftdmax = 7" 2!242

dmax = 4!758 ft

224

PROBLEM 4.41

Situation: Truck carries cylindrical tank (axis vertical) and will not accelerate ordecelerate more than 1/3g. Truck also goes around unbanked curve with radius of 50m.

Find: Maximum depth that tank can be lled before spilling and maximum speed oncurve.

APPROACH

Apply Euler’s equation on straight section and on the unbanked curve.

ANALYSIS

d

D

D

%

Euler’s equation On straight section, the slope of a free surface is

tanT = C$$)

= (1$3))$)

= 1$3

tanT = 1$3 = (1 " A)$(0!51)thus A = 1 " (1$6)1 = (5$6)1

Tank can be 5/6 full without spilling

On unbanked curve

tanT = 1$3

Then 1$3 = C:$)

C: = (1$3))

, 2$3 = (1$3))

or , =p(1$3))3

, = 12!8 m/s

225

PROBLEM 4.42

Situation: An accelerating tank is described in the problem statement.

Find: Explain the conditions shown.

ANALYSIS

The correct choice is (b). The tank is placed on a vehicle with constant speedmoving about a circular track.

226

PROBLEM 4.43

Situation: Rectangular tank with opening at top corner carries oil (S=0.83) andaccelerates uniformly at 19.62 m/s2!Depth of oil at rest is 2 m. Tank dimensionsgiven in problem.

Find: Maximum pressure in tank during acceleration.

ANALYSIS

Euler’s equation The conguration for the liquid in the tank is shown in the diagram.

3 m

4 m

2 m

x

x+ 1.5x

The liquid surface intersects the bottom at a distance E from the right side. Thedistance in the x direction between the contact surface at the bottom and the top is3$ tanT = 1!5

tan K = C&$) = 2

area of air space = 4× 14 = 3× (E+ 1!5 + E)$2E = 0!583 m

The maximum pressure is at the bottom, left corner and is equal to

%max$( = (4" 0!583)= 0!83× 1000× 19!62× 3!417

%max = 55!6 kPa

227

PROBLEM 4.44

Situation: A water jet is described in the problem statement.

Find: Height M jet will rise.

APPROACH

Apply the Bernoulli equation from the nozzle to the top of the jet. Let point 1 bein the jet at the nozzle and point 2 at the top.

ANALYSIS

Bernoulli equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2

where %1 = %2 = 0 gage

,1 = 20 ft/s

,2 = 0

0 + (20)2$2) + R1 = 0 + 0 + R2

R2 " R1 = M = 400$64!4

M = 6!21 ft

228

PROBLEM 4.45

Situation: A pitot tube measuring airspeed on an airplane at 10,000 ft where thetemperature is 23oF and the pressure is 10 psia. The pressure di!erence is 10 inchesof water.

Find: Airspeed.

APPROACH

Apply the Pitot tube equation.

ANALYSIS

Pitot tube equation

, =p2!%;$#

!%; = (H2OMH2O

= 62!4× (10$12)= 52 psf

Ideal gas law

# = %$("& )

= (10)(144)$((1' 716)(483))

= 0!00174 slugs/ft3

, =

q2× 52 lbf/ft2$(0!00174 slugs/ft3)

, = 244 ft/s

229

PROBLEM 4.46

Situation: A stagnation tube in a tank is rotated in a tank 10 cm below liquid surface.Arm is 20 cm long and specic weight of uid is 10,000 N/m3!

Find: Location of liquid surface in central tube.

APPROACH

Pressure variation equation for rotating ow from pt. 1 to pt. 2 where pt. 1 is atliquid surface in vertical part of tube and pt. 2 is just inside the open end of the pitottube.

ANALYSIS

10 cm

1

Elevation view Plan view

0

2

Pressure variation equation- rotating ow

%1$( " , 21 $2) + R1 = %2$( " , 22 $2) + R20" 0 + (0!10 + L) = %2$( " 32J2$2) " 0 (1)

where R1 = R2! If we reference the velocity of the liquid to the tip of the pitot tubethen we have steady ow and Bernoulli’s equation will apply from pt. 0 (point aheadof the pitot tube) to point 2 (point at tip of pitot tube).

%0$( + ,20 $2) + R0 = %2$( + ,

22 $2) + R2

0!1($( + 32J2$2) = %2$( + 0 (2)

Solve Eqs. (1) & (2) for LL = 0 liquid surface in the tube is the same as the elevation as outside liquid surface.

230

PROBLEM 4.47

Situation: A glass tube with 90o bend inserted into a stream of water. Water in tuberises 10 inches above water surface.

Find: Velocity.

APPROACH

Apply the Bernoulli equation.

ANALYSIS

Hydrostatic equation (between stagnation point and water surface in tube)

%&(= M+ A

where A is depth below surface and M is distance above water surface.

Bernoulli equation (between freestream and stagnation point)

%&(

= A+, 2

2)

M+ A = A+, 2

2)

, 2

2)= M

, = (2× 32!2× 10$12)1*2

, = 7!33 fps

231

PROBLEM 4.48

Situation: A glass tube in a 3 m/s stream of water.

Find: Rise in vertical leg relative to water surface.

APPROACH


ANALYSIS

Apply hydrostatic equation between stagnation point and water surface in tube

%&(= M+ A

From application of the Bernoulli equation

%&(

= A+, 2

2)

M+ A = A+, 2

2)

M =, 2

2)

= 32$(2× 9!81)= 0!459 m

M = 45!9 cm

232

PROBLEM 4.49

Situation: A Bourdon tube gage attached to plate in 40 ft/s air stream.

Find: Pressure read by gage.

ANALYSIS

Because it is a Bourdon tube gage, the di!erence in pressure that is sensed will bebetween the stagnation point and the separation zone downstream of the plate.Therefore

!-, = 1" (-,Iback of plate)!-, = 1" (neg. number)

! !%$(#, 20 $2) = 1 + positive number

!% = (#, 20 $2)(1 + positive number)

Case (c) is the correct choice.

233

PROBLEM 4.50

Situation: An air-water manometer is connected to a Pitot tube to measure air ve-locity. Manometer deects 2 in. The air is at 60oF and 15 psia.

Find: Velocity.

APPROACH

Apply the Pitot tube equation calculate velocity. Apply the ideal gas law to solvefor density.

ANALYSIS

Ideal gas law

# = %$"&

= 15× 144$(1' 715)(60 + 460)= 0!00242 slugs/ft

Pitot tube equation, = (2!%;$#)

1*2

From the manometer equation

!%; = (%!M(1" ('$(%)

but ('$(% ¿ 1 so

, = (2(%!M$#)1*2

= (2× 62!4× (2!0$12)$0!00242)1*2

, = 92!7 fps

234

PROBLEM 4.51

Situation: A ow-metering device is described in the problem. Air has density of 1.2kg/m3 and a 10 cm deection of water measured on manometer.

Find: Velocity at station 2.

APPROACH

Apply the Bernoulli equation and the manometer equation.

ANALYSIS

Bernoulli equation%1$( + ,

21 $2) = %2$( + ,

22 $2) = %5$(

Manometer equation

%1 + 0!1× 9810"neglectz }| {

0!1× 1!2× 9!81 = %5

%5 " %1 = 981 N/m2 = #, 21 $2

, 21 = 2× 981$1!2,1 = 40!4 m/s

,2 = 2,1

V2=80.8 m/s

235

PROBLEM 4.52

Situation: A spherical Pitot tube is used to measure the ow velocity in water. Thevelocity at the static pressure tap is 1.5,!! The piezometric pressure di!erence is 3kPa.

Find: Free stream velocity: ,!

APPROACH

Apply the Bernoulli equation between the two points. Let point 1 be the stagnationpoint and point 2 at 90! around the sphere.

ANALYSIS

Bernoulli equation

%;1 + #,21 $2 = %;2 + #,

22 $2

%;1 + 0 = %;2 + #(1!5,0)2$2

%;1 " %;2 = 1!125#, 20, 20 = 3' 000$(1!125× 1' 000) = 2!67 m2$s2

,0 = 1!63 m/s

236

PROBLEM 4.53

Situation: A device for measuring the water velocity in a pipe consists of a cylinderwith pressure taps at forward stagnation point and at the back on the cylinder in thewake. A pressure di!erence of 500 Pa is measured.

Find: Water velocity: ,!

APPROACH

Apply the Bernoulli equation between the location of the two pressure taps. Let point1 be the forward stagnation point and point 2 in the wake of the cylinder.

ANALYSIS

The piezometric pressure at the forward pressure tap (stagnation point, -, = 1) is

%;1 = %;0 + #, 2

2

At the rearward pressure tap%;2 " %;0#? 202

= "0!3

or

%;2 = %;0 " 0!3#, 202

The pressure di!erence is

%;1 " %;2 = 1!3#, 202

The pressure gage records the di!erence in piezometric pressure so

,0 = (2

1!3#!%)1*2

= (2

1!3× 1000× 500)1*2

= 0!88 m/s

237

PROBLEM 4.54

Situation: The design of a spherical Pitot tube measuring the ow velocity. Velocityvaries as , = ,! sin K!

Find: (a) Angle K for pressure tap.(b) Equation for free-stream velocity.(c) E!ect of o!set angle `!

APPROACH

(a) Apply the Bernoulli equation between the free stream and the location of thepressure tap gives.(b) Apply the Bernoulli equation between the stagnation point, tap A, and pressuretap B.(c) Let the pressure tap on the axis of the probe be tap A and the other one tap B.

ANALYSIS

(a) Bernoulli equation

%! +1

2#, 2! = %+

1

21!52, 2! sin

2 K

But at the pressure tap location % = %! so

2!25 sin2 K = 1

Solving for K gives

K = 41!8!

(b) Bernoulli equation

%= = %> +1

21!52#, 2! sin

2 K = %> +1

21!52#, 2!

1

2!25or

,! =q

2(,#",$)L

(c) The pressure at tap A would be

%= = %! "1

2#, 2! 1!5

2 sin2 ` = %! " 1!125#, 2! sin2 `

The pressure at tap B would be

%> = %! " 1!125#, 2! sin2(` + 41!8!)

The pressure di!erence would be

%= " %> = 1!125#, 2!£sin2(` + 41!8!)" sin2 `

¤

238

Solving for the velocity gives

,! =

s%= " %>

1!125#£sin2(` + 41!8!)" sin2 `

¤

which will designated at the “true” velocity, ,- ! The “indicated” velocity, ,P ' is theone calculated assuming that tap A is at the stagnation point. The ratio of theindicated velocity to the true velocity would be

,P,-=q2!25

£sin2(` + 41!8!)" sin2 `

¤

The error is

error =,- " ,P,-

= 1",P,-

Angle &, degrees

0 2 4 6 8 10 12

0

2

4

6

8

10

12

14

16

Erro

r, %

239

PROBLEM 4.55

Situation: A Pitot tube measuring the ow velocity in water is described in theproblem statement.

Find: Explain how to design the Pitot tube.

ANALYSIS

Three pressure taps could be located on a sphere at an equal distance from thenominal stagnation point. The taps would be at intervals of 120!! Then when theprobe is mounted in the stream, its orientation could be changed in such a way as tomake the pressure the same at the three taps. Then the axis of the probe would bealigned with the freestream velocity.

240

PROBLEM 4.56

Situation: Two Pitot tubes are connected to air-water manometers to measure airand water velocities.

Find: The relationship between ,= and ,M .

, =p2)!M =

p2!%;$#

ANALYSIS

The !%; is the same for both; however,

#% FF #'

Therefore ,= F ,M . The correct choice is b).

241

PROBLEM 4.57

Situation: A Pitot tube measures the velocity of kerosene at center of 12 inch pipe.Deection of mercury–kerosene manometer is 5 inches.

Find: Velocity.Properties From table A.4 #ker = 1!58 slugs/ft

3!(ker = 51 lbf/ft3

APPROACH

Apply the Pitot tube equation and the hydrostatic equation.

ANALYSIS


!%; = !M((HG " (ker)= (5$12)(847" 51)= 332 psf

Pitot tube equation

, = (2!%;$#)1*2

= (2× 332$1!58)1*2

, = 20!5 fps

242

PROBLEM 4.58

Situation: A Pitot tube for measuring velocity of air at 20oC at std. atm. pressure.Di!erential pressure gage reads 3 kPa.

Find: Air velocity.Properties From table A.3 #(20oC)= 1!2 kg/m3

APPROACH


ANALYSIS

Pitot tube equation

, = (2!%;$#)1*2

= (2× 3' 000$1!2)1*2

, = 70!7 m/s

243

PROBLEM 4.59

Situation: A Pitot tube is used to measure the velocity of air at 60oF and std. atm.pressure. A pressure di!erence of 11 psf is measured.

Find: Air velocity.Properties From table A.3 #a(60

oF)= 0!00237 slugs/ft3

APPROACH


ANALYSIS

Pitot tube equation

, =p2!%;$#

, = (2× 11$0!00237)1*2

, = 96!3 fps

244

PROBLEM 4.60

Situation: A Pitot tube measures gas velocity in a duct. The gas density is 0.12lbm/ft3 and the piezometric pressure di!erene is 0.9 psi.

Find: Gas velocity in duct.

APPROACH


ANALYSIS

Pitot tube equation The density is 0.12 lbm/ft3$32!2 = 0!00373 slugs/ft3

, =p2!%;$#

= [2× 0!9× 144$0!00373]1*2

, = 264 ft/s

245

PROBLEM 4.61

Situation: A sphere moving horizontally through still water at 11 ft/s. Velocity atpoint I induced by moving sphere is 1 ft/s with respect to earth.

Find: Pressure ratio: %=$%0

APPROACH


ANALYSIS

A

1 ft/s

By referencing velocities to the spheres a steady ow case will be developed. Thus,for the steady ow case ,0 = 11 ft/s and ,= = 10 ft/s. Then when Bernoulli’sequation is applied between points 0 and I it will be found that %=$%0 F 1 (case c)

246

PROBLEM 4.62

Situation: A body moving horizontally through still water at 13 m/s. Velocity atpoints D and - induced by body are 5 m/s and 3 m/s.

Find: Pressure di!erence: %> " %)

ANALYSIS

Bernoulli equation Refer all velocities with respect to the sphere. Flow is then steadyand the Bernoulli equation is applicable.

%> " %) = (1' 000$2)[(13" 3)2 " (13" 5)2]= 18' 000 Pa

%> " %) = 18 kPa

247

PROBLEM 4.63

Situation: Water in a ume is described in the problem statement.

Find: If gage A will read greater or less than gage B.

ANALYSIS

Both gage A and B will read the same, due to hydrostatic pressure distribution inthe vertical in both cases. There is no acceleration in the vertical direction.

248

PROBLEM 4.64

Situation: An apparatus is used to measure the air velocity in a duct. It is connectedto a slant tube manometer with a 30o leg with the indicated deection. The air inthe duct is 20oC with a pressure of 150 kPa, abs. The manometer uid has a specicgravity of 0.7.

Find: Air velocity

APPROACH


ANALYSIS

The side tube samples the static pressure for the undisturbed ow and the centraltube senses the stagnation pressure.Bernoulli equation

%0 + #,20 $2 = %stagn. + 0

or ,0 =q(2$#)(%stagn. " %0)

But

%stagn. " %0 = (0!067" 0!023) sin 30! × 0!7× 9' 810 = 151!1 Pa# = %$"& = 150' 000$(287× (273 + 20)) = 1!784 kg/m3

Then

,0 =p(2$1!784)(151!1)

V0=13.02 m/s

249

PROBLEM 4.65

Situation: A spherical probe with pressure coe"cients given is used to nd gas ve-locity. The pressure di!erence is 4 kPa and the gas density is 1.5 kg/m3.

Find: Gas velocity.

APPROACH

Apply the denition of pressure coe"cient.

ANALYSIS

Pressure coe"cient

!-, = 1" ("0!4)!-, = 1!4 = (%= " %>)$(#, 20 $2), 20 = 2(4' 000)$(1!5× 1!4)

,0 = 61!7 m/s

250

PROBLEM 4.66

Situation: An instrument used to nd gas velocity in smoke stacks. Pressure coe"-cients are given. Connected to water manometer with 0.8 cm deection. The gas isat 101 kPa, abs and the temperature is 250oC. The gas constant is 200 J/kgK.

Find: Velocity of stack gases.

ANALYSIS

Ideal gas law

# = %$"&

= 101' 000$(200× (250 + 273))= 0!966 kg/m2

Manometer equation!%; = ((% " (')!M

but (% À (' so

!%; = (%!M

= 9790× 0!008= 78!32 Pa

(%= " %>); = (-,= " -,>)#, 20 $2(%= " %>); = 1!3#, 20 $2

, 20 = 2× 78!32$(1!3× 0!966)

,0 = 11!17 m/s

251

PROBLEM 4.67

Situation: A spherical probe is used to measure water velocity. Pressure taps locatedat stagnation point and max width. A deection of 5 cm measured on mercurymanometer. Velocity at maximum width is 1.5 times the free stream velocity.

Find: Free-stream velocity.

APPROACH

Apply the Bernoulli equation between points 1 and 2. Let point 1 be at the stagnationpoint and point 2 be at the 90! position. At the 90! position \ = 1!5\ sin# = 1!5\ .

ANALYSIS

Bernoulli equation

%;1 +

=0z }| {#, 21 $2 = %;2 + #,

22 $2

%;1 " %;2 = #, 22 $2

((6D " (62B)!M = (#$2)(1!5\)2

(((6D$(62B)" 1)!M = (1$2))(1!5\)2

(13!6" 1)× 0!05 = (1$2))(2!25)\2

\ = 2!34 m/s

252

PROBLEM 4.68

Situation: The wake of a sphere which separates at 120o! The free stream velocity ofair (# = 1!2 kg/m3) is 100 m/s.

Find: (a) Gage pressure.(b) Pressure coe"cient.

APPROACH

Apply the Bernoulli equation from the free stream to the point of separation and thepressure coe"cient equation.

ANALYSIS

Pressure coe"cient-, = (%" %0)$(#, 2$2)

Bernoulli equation

%0 + #\2$2 = %+ #@2$2

%" %0 = (#$2)(\2 " @2)

or(%" %0)$(#\2$2) = (1" (@$\)2)

but

@ = 1!5\ sin K

@ = 1!5\ sin 120!

@ = 1!5\ × 0!866

At the separation point

(@$\) = 1!299

(@$\)2 = 1!687

-, = 1" 1!687

-, = "0!687

%gage = -,(#$2)\2

= ("0!687)(1!2$2)(1002)= "4' 122 Pa

%gage = "4!122 kPa gage

253

PROBLEM 4.69

Situation: A pressure transducer is connected between taps of spherical Pitot tubeand reads 120 Pa. Air density is 1.2 kg/m3!

Find: Free-stream velocity.

APPROACH

Apply the Bernoulli equation between the stagnation point (forward tap) and theside tap where @ = 1!5\! Neglect elevation di!erence.

ANALYSIS

@ = 1!5\ sin K

@O=90! = 1!5$1)

= 1!5\

Bernoulli equation

%1 + #,21 $2 = %2 + #,

22 $2

%1 " %2 = (#$2)(, 22 " ,21 )

%1 " %2 = (1!2$2)((1!5$2 " 0)120 = 1!35\2

\ = 9!43 m/s

254

PROBLEM 4.70

Situation: A Pitot tube used to measure the airspeed of an airplane. Calibrated toprovide correct airspeed with & = 17oC and p=101 kPa, abs. Pitot tube indicates 60m/s when pressure is 70 kPa, abs and temperature is -6.3oC.

Find: True airspeed.

APPROACH


ANALYSIS

Pitot tube equation

, = <p2!%;$#

then

,calibr. = (<$p#calibr.)

p2!%;

,true = (<$p#true)

p2!%; (1)

,indic. = (<$p#calib.)

p2!%; (2)

Divide Eq. (1) by Eq. (2):

,true$,indic. =p#calib.$#true

=

s%calib%true

&true&calib

= [(101$70)× (273" 6!3)$(273 + 17)]1*2

= 1!15

,true = 60× 1!15,true = 69 m/s

255

PROBLEM 4.71

Situation: Two pressure taps are located at ±30o from the horizontal plane on acylinder and connected to a water manometer. Air with a density of 1.2 kg/m3

moving at 50 m/s approaches the cylinder at 20o from the horizontal plane.

Find: Deection of water manometer.

APPROACH

Evaluate the pressure coe"cient at the two taps locations to nd pressure di!erence.

ANALYSIS

One pressure tap is located 10o from the stagnation point and the other at 50o.Thepressure coe"cients at the two locations are

-, = 1" 4 sin2 K-,I50 = 1" 4 sin2 50!

= 1" 4(0!766)2 = "1!347-,I10 = 1" 4(0!174)2 = +0!879

Pressure coe"cient di!erence,

-,I10 " -,I50 = 0!879" ("1!347) = 2!226

Equating the pressure di!erence to the manometer deection

!% = !-,#air,20 $2

= 2!226× 1!2× 502$2= 3340 Pa

!M = !%$(H2O= 3340$9810

= 0!340 m

!M = 34!0 cm

256

PROBLEM 4.72

Situation: Check equations for pitot tube velocity measurement provided by instru-ment company.

Find: Validity of pitot tube equations provided.

APPROACH

Apply the Bernoulli equation

ANALYSIS

Applying the Bernoulli equation to the Pitot tube, the velocity is related to the changein piezometric pressure by

!%; = #, 2

2

where !%; is in psf, # is in slugs/ft3 and , is in ft/s. The piezometric pressuredi!erence is related to the "velocity pressure" by

!%;(lbf/ft2) = (%(lbf/ft

3)M@(in)$12(in/ft)

= 62!4× M@$12= 5!2M@

The density in slugs/ft3 is given by

#(slug/ft3) = A (lbm/ft3)$)1(lbm/slug)

= A$32!2

= 0!03106A

The velocity in ft/min is obtained by multiplying the velocity in ft/s by 60. Thus

, = 60

r2× 5!2M@0!03106A

= 1098

rM@A

This di!ers by less than 0.1% from the manufacturer’s recommendations. This couldbe due to the value used for )1 but the di!erence is probably not signicant comparedto accuracy of "velocity pressure" measurement.From the ideal gas law, the density is given by

# =%

"&

where # is in slugs/ft3' % in psfa and & in oR. The gas constant for air is 1716 ft-lbf/slug-oR. The pressure in psfg is given by

% (psfg) = *'(in-Hg)× 13!6× 62!4 (lbf/ft3)$12(in/ft)= 70!72*'

257

where 13.6 is the specic gravity of mercury. The density in lbm/ft3 is

A = )1#

= 32!2×70!72*'1716× &

= 1!327*'&

which is within 0.2% of the manufacturer’s recommendation.

258

PROBLEM 4.73

Situation: The ow of water over di!erent surfaces is described in the problem state-ment.

Find: Relationship of pressures.

ANALYSIS

The ow curvature requires that %> F %9 + (A where A is the liquid depth. Also,because of hydrostatics %) = %9 + (A! Therefore %> F %) . Also %= G %9 + (A so%= G %) ! So %> F %) F %=.The valid statement is (b).

259

PROBLEM 4.74

Situation: The velocity vector V = 10Ei" 10?j describes a ow eld.

Find: Is the ow irrotational?

ANALYSIS

In a two dimensional ow in the E" ? plane, the ow is irrotational if (Eq. 4.34a)

[;

[E=[@

[?

The velocity components and derivatives are

@ = 10E[@

[?= 0

; = "10?[;

[E= 0

Therefore the ow is irrotational.

260

PROBLEM 4.75

Situation: A velocity eld is described by @ = "J? ; = JE

Find: Vorticity and Rate of rotation

ANALYSIS

Rate of rotation

J; = (1$2)([;$[E" [@$[?)= (1$2)(J " ("J))= (1$2)(2J)

J; = J

Vorticity is twice the average rate of rotation; therefore, the vorticity = 2J

261

PROBLEM 4.76

Situation: A two-dimensional velocity eld is given by

@ =-(?2 " E2)(E2 + ?2)2

, ; ="-E?

(E2 + ?2)2

Find: Check if ow is irrotational.

ANALYSIS

Apply equations for ow rotation in E" ? plane.

[;$[E" [@$[? = (2-?$(?2 + E2)2)" (2-(?2 " E2)2?$(?2 + E2)3)+(2-?$(?2 + E2)2)" (4-E?(2E)$(?2 + E2)3)

= 0 The ow is irrotational

262

PROBLEM 4.77

Situation: A velocity eld is dened by @ = EP+ 2?' ; = EP2 " ?P.

Find: (a) Acceleration at E = ? = 1 m and P = 1 s.(b) Is the ow rotational or irrotational?

ANALYSIS

Irrotational ow:[@$[? = 2; [;$[E = P2 [@$[? 6= [;$[E

Therefore, the ow is rotational.Determine acceleration:

C$ = @[@$[E+ ;[@$[? + [@$[P

C$ = (EP+ 2?)P+ 2(EP2 " ?P) + EC3 = @[;$[E+ ;[;$[? + [;$[P

= (EP+ 2?)P2 + (EP2 " ?P)("P) + (2EP" ?)a = ((EP+ 2?)P+ 2P(EP" ?) + E)i+ (P2(EP+ 2?)" P2(EP" ?) + (2EP" ?)) j

Then for E =l m, ? =l m, and P =l s the acceleration is:

a = ((1 + 2) + 0 + 1) i+ ((1 + 2) + 0 + (2" 1)) j m/s

a = 4 i+ 4 j m/s2

263

PROBLEM 4.78

Situation: Fluid ows between two stationary plates.

Find: Find rotation of uid element when it moves 1 cm downstream

APPROACH

Apply equations for rotation rate of uid element..

ANALYSIS

The rate of rotation for this planar (two-dimensional) ow is

J; =1

2([;

[E"[@

[?)

In this problem, ; = 0 so

J; = "1

2

[@

[?= 8?

The time to travel 1 cm is

!P =1

@

=1

2(1" 4?2)

The amount of rotation in 1 cm travel is

!K = J;!P

!K =4?

(1" 4?2)


264

PROBLEM 4.79

Situation: A velocity distribution is provided for a combination of free and forcedvortex.

;O =1

3

£1" exp("32)

¤

Find: Find how much a uid element rotates in one circuit around the vortex as afunction of radius.

ANALYSIS

The rate of rotation is given by

J; =1

3

A

A3(;O3)

J; =1

3

A

A3[1" exp("32)]

= exp("32)

The time to complete one circuit is

!P =203

;O

=2032

[1" exp("32)]

So, the total rotation in one circuit is given by

!K = J;!P

!K

20(rad) = 32

exp("32)1" exp("32)

A plot of the rotation in one circuit is shown. Note that the rotation is 20 for r$ 0(rigid body rotation) and approaches zero (irrotational) as 3 becomes larger.

265


266

PROBLEM 4.80

Situation: Closed tank 4 feet in diameter with piezometer attached is rotated at 15rad/s about a vertical axis.

Find: Pressure at bottom center of tank.

APPROACH

Apply the equation for pressure variation equation- rotating ow.

ANALYSIS


%+ (R " #32J2$2 = %, + (R, " #32,J2$2

where %, = 0' r, = 3 ft and 3 = 0' then

% = "(#$2)(9× 225) + ((R, " R)= (1!94$2)(2025) + 62!4× 2!5= "1808 psfg = "12!56 psig

% = "12!6 psig

267

PROBLEM 4.81

Situation: A tank 1 foot in diameter and 1 foot high with liquid (S=0.8) is rotatedon 2 foot arm. The speed is 20 ft/s and pressure at point A is 25 psf.

Find: Pressure at B.

APPROACH

Apply the pressure variation equation- rotating ow from point I to point D!

ANALYSIS


%= + (R= " #32=J2$2 = %> + (R> " #32>J

2$2

%> = %= + (#$2)(J2)(32> " 3

2=) + ((R= " R>)

where J = ,=$3= = 20$1!5 = 13!333 rad/s and # = 0!8× 1!94 slugs/ft3! Then

%> = 25 + (1!94× 0!80$2)(13!332)(2!52 " 1!52) + 62!4× 0!8("1)= 25 + 551!5" 49!9

%> = 526!6 psf

268

PROBLEM 4.82

Situation: A closed tank with liquid (S=1.2) is rotated about vertical axis at 10 rad/sand upward at 4 m/s2.

Find: Di!erence in pressure between points I and D: %> " %=

APPROACH

Apply the pressure variation equation for rotating ow between points D & -. Letpoint - be at the center bottom of the tank.

ANALYSIS


%> " #32>J2$2 = %) " #32)J

2$2

where 3> = 0!5 m, 3) = 0 and J = 10 rad/s. Then

%> " %) = (#$2)(J2)(0!52)

= (1200$2)(100)(0!25)

= 15' 000 Pa

%) " %= = 2( + #C;L

= 2× 11' 772 + 1' 200× 4× 2= 33' 144 Pa

Then

%> " %= = %> " %) + (%) " %=)= 15' 000 + 33' 144

= 48' 144Pa

%> " %= = 48!14 kPa

269

PROBLEM 4.83

Situation: A U-tube rotating about one leg. Before rotation, the level of liquid ineach leg is 0.25 m. The length of base and length of leg is 0.5 m.

Find: Maximum rotational speed so that no liquid escapes from the leg.

APPROACH

Apply the pressure variation equation for rotating ow. Let point 1 be at top ofoutside leg and point 2 be at surface of liquid of inside leg.

ANALYSIS

At the condition of imminent spilling, the liquid will be to the top of the outside legand the leg on the axis of rotation will have the liquid surface at the bottom of itsleg.Pressure variation equation- rotating ow

%1 + (R1 " #321J2$2 = %2 + (R2 " #322J

2$2

where %1 = %2' R1 = !5 m and R2 = 0

( × 0!5" (($))× !52J2$2 = 0J2 = 4)

= 2&)

J = 6!26 rad/s

270

PROBLEM 4.84

Situation: A U-tube rotating about one leg at 60 rev/min. Liquid at bottom of U-tube has specic gravity of 3.0. There is a 6 inch height of uid in outer leg. Distancebetween legs is 1 ft.

Find: Specic gravity of other uid.

APPROACH

Apply the pressure variation equation for rotating ow between points 1 & 2.

ANALYSIS


3

2 1

S=3.0

%2 + (R2 " #322J2$2 = %1 + (R1 " #321J

2$2

where R2 = R1' 31 = 0' 32 = 1 ft. and J = (60$60)× 20 = 20 rad/s. Then

%2 = (1!94× 3)(12)(20)2$2 = 114!9 psfg (1)

Also, by hydrostatics, because there is no acceleration in the vertical direction

%2 = 0 +1

2× (( (2)

where (( is the specic weight of the other uid. Solve for (( between Eqs. (1) &(2)

(( = 229!8 lbf/ft3

6 = (($(H2O= 229!8$62!4

6 = 3!68

271

PROBLEM 4.85

Situation: A U-tube rotating about one leg at 32.12 rad/s. Geometry given in problemstatement.

Find: New position of water surface in outside leg.

APPROACH

Apply the pressure variation equation for rotating ow between the water surface inthe horizontal part of the tube and the water surface in the vertical part of the tube.

ANALYSIS

A preliminary check shows that the water will evacuate the axis leg. Thus uidconguration is shown by the gure.

d

0.40+d


%1 + (R1 " #321J2$2 = %2 + (R2 " #322J

2$2

where 31 = A' 32 = 0!30 m and (R2 " R1) = 0!50 + A. Then

(#J2$2)(322 " 321) = ((0!50 + A)

(1' 000× 32!122$2)(0!32 " A2) = (0!50 + A)9' 810

Solving for A yields A = 0!274 mThen

R2 = 0!50 + 0!274

z=0.774 m

272

PROBLEM 4.86

Situation: A U-tube is attached to rotating platform and platform rotating at 4 rad/s.

Find: Elevation of liquid in smaller leg of U-tube.

APPROACH

Apply the pressure variation equation for rotating ow between the liquid surface inthe large tube and the liquid surface in the small tube for conditions after rotationoccurs.

ANALYSIS

1 2

Pressure variation equation- rotating owLet 1 designate large tube and 2 the small tube.

(R1 " (#$2)321J2 = (R2 " (#$2)322J

2

R1 " R2 = (#$2()(J2)(321 " 322)

= ((($))$(2())J2(321 " 322)

= (J2$(2)))(0!42 " 0!22)= (42$(2)))(0!12)

= 0!0978 m = 9!79 cm

Because of the di!erent tube sizes a given increase in elevation in tube (1) will beaccompanied by a fourfold decrease in elevation in tube (2). Then R1 " R2 = 5!Rwhere !R = increase in elevation in (1)

!R1 = 9!79 cm/ 5=1.96 cm or R1 = 21!96 cm

Decrease in elevation of liquid in small tube

!R2 = 4!R1 = 7!83

R2 = 20 cm" 7!83 cmz2=12.17 cm

273

PROBLEM 4.87

Situation: A manometer with mercury (S=13.6) at the base is rotated about one leg.Water with height of 10 cm in central leg at is described in the problem statement.The length of the base is one meter. Height of mercury in outer leg is 1 cm.

Find: Rotational speed.

APPROACH

Apply the pressure variation equation for rotating ow between pts. (1) & (2).

ANALYSIS

(2)(1)

10 cm

However %1 = (0!10 m)((H2O) because of hydrostatic pressure distribution in thevertical direction (no acceleration).Pressure variation equation- rotating ow

%1 + (R1 " #321J2$2 = %2 + (R2 " #322J

2$2

where %2 = 0' R2 " R1 = 0!01 m, 31 = 0 and 32 = 1 m. Then

0!1(H2O + 0 + 0 = 0 + (Hg × 0!01" ((Hg$))× 12J2$2

J2 = ((2))(0!01(Hg " 0!1(H2O))$(HgJ = (2× 9!81)(!01" (0!1$13!6))

J = 0!228 rad/s

274

PROBLEM 4.88

Situation: A manometer is rotated about one leg. There is a 25 cm height di!erencein liquid (S=0.8) between the legs. The length of the base is 10 cm.

Find: Acceleration in )’s in leg with greatest amount of oil..

APPROACH

Apply the pressure variation equation for rotating ow between the liquid surfaces of1 & 2Let leg 1 be the leg on the axis of rotation. Let leg 2 be the other leg of themanometer.

ANALYSIS


%1 + (R1 " #321J2$2 = %2 + (R2 " #322J

2$2

0 + (R1 " 0 = (R2 " (($))322J2$2

J2322$(2g) = R2 " R1C: = 3J2

= (R2 " R1)(2g)$3= (0!25)(2g)$32= (0!25)(2g)$0!1

C: = 5)

275

PROBLEM 4.89

Situation: A fuel tank rotated at 3 rev/min in zero-gravity environment. End oftank 1.5 m from axis of rotation and fuel level is 1 m from rotation axis. Pressure innon-liquid region is 0.1 kPa and density of fuel is 800 kg/m3!

Find: Pressure at exit (point I).

APPROACH

Apply the pressure variation equation for rotating ow from liquid surface to pointA. Call the liquid surface point 1.

ANALYSIS


%1 + (R1 " #321J$2 = %= + (R= " #32=J

2$2

%= = %1 + (#J2$2)(32= " 3

21) + ((R1 " R=)

However ((R1 + R=) = 0 in zero-) environment. Thus

%= = %1 + ((800 kg/m3)$2)(60$60 rad/s)2(1!52 " 12)

= 100 Pa+ 49!3 Pa

%= = 149!3 Pa

276

PROBLEM 4.90

Situation: A rotating set of tubes is described in the problem statement.

Find: Derive a formula for the angular speed when the water will begin to spill.

APPROACH

Start with pressure variation equation for rotating ow. Let point 1 be at the liquidsurface in the large tube and point 2 be at the liquid surface in the small tube.

ANALYSIS


%1 + (R1 " %321J2$2 = %2 + (R2 " #322J$2

!

1

2

3.75

The change in volume in leg 1 has to be the same as leg 2. So

!M1A21 = !M2A

22

!M1 = !M2

µA22A21

¶

=!M24

The elevation di!erence between 1 and 2 will be

R2 " R1 = 3L+3L

4= 3!75L

Then %1 = %2 = 0 gage, 32 = L' and R2 " R1 = 3!75L so

#322J2$2 = ((3!75L)

(($(2)))(L2)J2 = 3!75(L

J2 =7!5)

L

J =p7!5)$L

277

PROBLEM 4.91

Situation: Mercury is rotating in U-tube at J and mercury levels shown in diagram.

Find: Level of mercury in larger leg after rotation stops.

APPROACH

Apply the pressure variation equation for rotating ow from the liquid surface in thesmall tube (S) to the liquid surface in the large tube (L).

ANALYSIS


%. + (C. " #32.J2$2 = %E + (RE " #32EJ

2$2

But %. = %E' 3. = 0!5L and 3E = 1!5L. Then

(#$2)J2(32E " 32.) = ((RE " R.)

(($2))J2(1!52L2 " 0!52L2) = ((2L)

J2 = 2)$L

J =p2)$(5$12)

J = 12!43 rad/s

Change in volume of Hg in small tube is same as in large tube. That is

(& = (E!R&0A

2$4 = !RE0(2A)2$4

!R& = 4!RE

Also

!R& +!RE = 2L

4!RE +!RE = 2× (5$12) ft = 0!833 ft!RE = 0!833 ft$5 = 0!167 ft

Mercury level in large tube will drop 0.167 ft from it original level.

278

PROBLEM 4.92

Situation: Water in a 1 cm diameter tube, 40 cm long. Closed at one end and rotatedat 60.8 rad/s.

Find: Force exerted on closed end.

APPROACH

Apply the pressure variation equation for rotating ow from the open end of the tubeto the closed end.

ANALYSIS


%1 = (R1 " #321J2$2 = %2 + (R2 " #322J

2$2

where R1 = R2. Also let point 2 be at the closed end; therefore 31 = 0 and 32 = 0!40m.

%2 = (#$2)(0!402)(60!8)2

= 500× 0!16× 3697= 295!73 kPa

Then

H = %2I = 295' 730× (0$4)(!01)2

H = 23!2 N

279

PROBLEM 4.93

Situation: Mercury in rotating manometer with dimensions shown on gure.

Find: Rate of rotation in terms of ) and L.

APPROACH

Apply the pressure variation equation for rotating ow from the mercury surface inthe left tube to the mercury surface in the right tube. Then %< = %K.

ANALYSIS


(R< " #32<J2$2 = (RK " #32KJ

2$2

J2(($2))(32K " 32< ) = ((RK " R<)J2 = 2)(RK " R<)$(32K " 3

2< )

= 2)(L)$(9L2 " L2)

J =p)$(4L)

280

PROBLEM 4.94

Situation: A U-tube rotated around left leg. Rotated at 5 rad/s and then 15 rad/s.Dimensions given on problem gure.

Find: (a) water level in tube at 5 rad/s.(b) water level for 15 rad/s.

APPROACH

Apply the pressure variation equation for rotating ow between the water surfaceand the left leg and the water surface in the right leg. At these surfaces %< = %K = 0gage.

ANALYSIS

Pressure variation equation- rotating ow(a) Assume that there is uid in each leg of the manometer.

(R2 " #322 J2$2 = (RK " #32KJ

2$2

R2 " RK = "32KJ2$2) = "J2L2$2)

where the subscript _ refers to the left leg and 3 to the right leg. Because the manome-ter rotates about the left leg 32 = 0!Then

R2 " RK = "52 × 0!252

2× 9!81= 0!080 m = 8 cm (1)

Also

R2 + RK = 1!4L

= 35 cm (2)

Solving Eqs. (1) and (2) for R2 and RK yields

z<=13.5 cm and zK=21.5 cm

(b) Assume as before that the liquid exists in both vertical legs. Then

R2 " RK = "J2L2$2)

= "152 × 0!32

2× 9!81= 1!032 m = 103!2 cm (3)

Solving Eqs. 2 and 3 for R2 and RK yields R2 = 69!1 cm and RK = "34!1 cm which is animpossible answer. The uid then must not totally ll the lower leg and must looklike

281

!

1

1.4 +d

Let subscript 1 refer to location of liquid in lower leg as shown. Applying equationfor pressure variation equation- rotating ow gives

(R1 " #321J2$2 = (RK " #32KJ

2$2

where R1 = 0 and 3K = L so

"321J2$2) = RK " L2J2$2)RK = J2$2)(L2 " 321)

The total length of liquid in the legs has to be the same before rotation as after so

L" 31 + RK = 2× 0!7L+ LRK = 1!4L+ 31

One can now write1!4L+ 31 = J

2$2)(L2 " 321)

or

1!4 +31L

=LJ2

2)

·1"

³31L

´2¸

=0!3× 152

2× 9!81

·1"

³31L

´2¸

= 1!032

·1"

³31L

´2¸

Solving the quadratic equation for 31$L gives

31L= 0!638

With L = 30 cm,r1=19.15 cm and zK=61.15 cm

282

PROBLEM 4.95

Situation: U-tube rotated about vertical axis at 8 rad/s and then at 20 rad/s.

Find: Pressures at points I and D!

Assumptions: %? = 0

APPROACH

Apply the pressure variation equation for rotating ow.

ANALYSIS

Pressure variation equation- rotating ow Writing out the equation

%= + (R= " #32=J2$2 = %F + (RF " #32FJ

2$2

where %F = 0 gage, 3= = 0' 3F = 0!64 m and RF " R= = 0!32 m The density is 2000kg/m3 and the specic weight is 2× 9810 = 19620 N/m3! For a rotational speed of 8rad/s

%= = ((RF " R=)" #32FJ2$2

%= = 0!32× 19620" 2000× 0!642 × 82$2= "19' 936 Pa

%= = "19!94 kPa

%> = ((RF " R=)= 0!32× 19620

%> = 6!278 kPa

Now for J = 20 rad/s solve for %= as above.

%= = 19620× 0!32" 2000× 0!642 × 202$2= "157' 560 Pa;

which is not possible because the liquid will vaporize. Therefore the uid musthave the conguration shown in the diagram with a vapor bubble at the center.

0.32 m+r

vapor

r

283

Assume %K = 0!Therefore, %= = %? = "101 kPa abs . Now the equation for rotatingows becomes

%K " #32J2$2 = %> " #× 3F2J2$2

where %K = %? = "101 kPa, The height of the liquid in the right leg is now0!32 + 3! Then

"101' 000" 2000× 20232$2 = 19620× (0!32 + 3)" 2000× 0!642 × 202$2"101' 000" 400' 00032 = 6278 + 196203 " 163' 84032 + 0!049053 " 0!1414 = 0

Solving for 3 yields 3 = 0!352 m. Therefore

%> = (0!32 + 0!352)× 19620= 13' 184Pa

%> = 13!18 kPa

284

PROBLEM 4.96

Situation: Water in U-tube rotated around one leg and end of leg is closed with aircolumn.

Find: Rotational speed when water will begin to spill from open tube.

APPROACH

Apply the pressure variation equation for rotating ow between water surface in legA-A to water surface in open leg after rotation.

ANALYSIS

When the water is on the verge of spilling from the open tube, the air volume in theclosed part of the tube will have doubled. Therefore, we can get the pressure in theair volume with this condition.

%#(# = %(((

and a and b refer to initial and nal conditions

%( = %#(#$(( = 101 kPa×1

2%( = 50!5 kPa, abs = "50!5 kPa, gage


%= + (R= " #32=J2$2 = %open + (Ropen " #3openJ2$2

%= + 0" 0 = 0 + ( × 6L" #(6L)2J2$2"50!5× 103 = 9810× 6× 0!1" 1000× 0!62 × J2$2

"50!5× 103 = 5886" 180J2

Y2 = 313!3

J = 17!7 rad/s

285

PROBLEM 4.97

Situation: A centrifugal pump consisting of a 10 cm disk is rotated at 2500 rev/min.

Find: Maximum operational height: R

APPROACH

Apply the pressure variation equation for rotating ow from point 1 in vertical pipeat level of water to point 2 at the outer edge of the rotating disk.

ANALYSIS

Pressure variation equation- rotating owThe exit pressure of the pump is atmospheric.Let point 1 be the liquid surface where R = 0 and point 2 the pump outlet.

%1 + (R1 " #321J2$2 = %2 + (R2 " #322J

2$2

0 + 0" 0 = 0 + (R2 " #322J2$2

0 = R2 " 0!052J2$2)

The rotational rate is

J = (2' 500 rev/min)(1 min/60 s)(20 rad/rev)=261.8 rad/s

Therefore

R2 = ((0!05)(261!8))2$(2× 9!81)R2 = 8!73 m

286

PROBLEM 4.98

Situation: A tank rotated at 5 rad/s about horizontal axis and water in tank rotatesas a solid body.

Find: Pressure gradient at R = "1' 0'+1!

APPROACH


ANALYSIS


[%$[3 + (([R$[3) = "#3J2

[%$[R = "( " #3J2

when R = "1 m

[%$[R = "( " #J2

= "((1 + J2$))= "9' 810(1 + 25$9!81)

[%$[R = "34!8 kPa/m

when R = +1 m

[%$[R = "( + #J2

= "((1" J2$))= "9810× (1" 25$9!81)

[%$[R = 15!190 kPa/m

At R = 0

[%$[R = "(

[p/[z=-9.810 kPa/m

287

PROBLEM 4.99

Situation: A rotating tank is described in the problem 4.98.

Find: Derive an equation for the maximum pressure di!erence.

APPROACH


ANALYSIS

Below the axis both gravity and acceleration cause pressure to increase with decreasein elevation; therefore, the maximum pressure will occur at the bottom of the cylin-der. Above the axis the pressure initially decreases with elevation (due to gravity);however, this is counteracted by acceleration due to rotation. Where these two ef-fects completely counter-balance each other is where the minimum pressure will occur([%$[R = 0)! Thus, above the axis:

[%$[R = 0 = "( + 3J2# minimum pressure condition

Solving: 3 = ($#J2; %min occurs at Rmin = +)$J2!Using the equation for pressurevariation in rotating ows between the tank bottom where the pressure is a maximum( Rmax = "30) and the point of minimum pressure.

%max + (Rmax " #320J2$2 = %min + (Rmin " #32minJ

2$2

%max " (30 " #320J2$2 = %min + ()$J

2 " #()$J2)2J2$2

%max " %min = !%max = (#J2$2)[320 " ()$J2)2] + ((30 + )$J

2)

Rewriting

!%max =LQ2K202+ (30 +

RD2Q2

288

PROBLEM 4.100

Situation: A tank 4 ft in diameter and 12 feet long rotated about horizontal axis andwater in tank rotates as a solid body. Maximum velocity is 20 ft/s.

Find: Maximum pressure di!erence in tank and point of minimum pressure.

APPROACH

Same solution procedure applies as in Prob. 4.99.

ANALYSIS

From the solution to Prob. 4.99 %minoccurs at R = ($#J2 where J = (20 ft/s)/2.0 ft= 10 rad/s. Then

Rmin = ($#J2

= )$J2

= 32!2$102

Rmin = 0!322 ft above axis

The maximum change in pressure is given from solution of Problem 4.99

!%max =#J23202

+ (30 +()

2J2

=1!94× 102 × 22

2+ 62!4× 2 +

62!4× 32!22× 102

= 388 + 124!8 + 10!0

!pmax=523 lbf/ft2

289

PROBLEM 4.101

Situation: Incompressible and inviscid liquid ows around a bend with inside radiuso1 1 m and outside radius of 3 m. Velocity varies as , = 1$3!

Find: Depth of liquid from inside to outside radius.

APPROACH

Apply the Bernoulli equation between the outside of the bend at the surface (point2) and the inside of the bend at the surface (point 1).

ANALYSIS

Bernoulli equation

(%2$() + ,22 $2) + R2 = (%1$() + ,

21 $2) + R1

0 + , 22 $2) + R2 = 0 + , 21 $2) + R1

R2 " R1 = , 21 $2) " ,22 $2)

where ,2 = (1$3) m/s; ,1 = (1$1) m/s. Then

R2 " R1 = (1$2))(12 " 0!332)R2 " R1 = 0!045 m

290

PROBLEM 4.102

Situation: The velocity at outlet pipe from a reservoir is 16 ft/s and reservoir heightis 15 ft.

Find: Pressure at point I!

APPROACH


ANALYSIS

Bernoulli equation Let point 1 be at surface in reservoir.

(%1$() + (,21 $2)) + R1 = (%=$() + (,

2=$2)) + R=

0 + 0 + 15 = %=$62!4 + 162$(2× 32!2) + 0

%= = (15" 3!98)× 62!4%= = 688 psfg

p==4.78 psig

291

PROBLEM 4.103

Situation: The velocity at outlet pipe from a reservoir is 6 m/s and reservoir heightis 15 m.


APPROACH


ANALYSIS

Bernoulli equation Let point 1 be at reservoir surface.

(%1$() + (,21 $2)) + R1 = (%=$() + (,

2=$2)) + R=

0 + 0 + 15 = %=$9810 + 62$(2× 9!81) + 0

%= = (15" 1!83)× 9810%= = 129' 200 Pa, gage

p==129.2 kPa, gage

292

PROBLEM 4.104

Situation: The ow past a cylinder in a 40 m/s wind. Highest velocity at the maxi-mum width of sphere is twice the free stream velocity.

Find: Pressure di!erence between highest and lowest pressure.

Assumptions: Hydrostatic e!ects are negligible and the wind has density of 1.2 kg/m3.

APPROACH

Apply the Bernoulli equation between points of highest and lowest pressure.

ANALYSIS

The maximum pressure will occur at the stagnation point where , = 0 and the pointof lowest pressure will be where the velocity is highest (,max = 80 m/s).Bernoulli equation

%7 + #,27 $2 = %< + #,

2< $2

%7 + 0 = %< + (#$2)(,2max)

%7 " %< = (1!2$2)(802)

= 3' 840 Pa

%7 " %< = 3!84 kPa

293

PROBLEM 4.105

Situation: Velocity and pressure given at two points in a duct and uid density is1000 kg/m3!


APPROACH

Check to see if it is irrotational by seeing if it satises Bernoulli’s equation.

ANALYSIS

The ow is non-uniform.Bernoulli equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2

(10' 000$9' 810) + (1$(2× 9!81)) + 0 = (7' 000$9' 810) + 22(2× 9!81) + 01!070 6= 0!917

Flow is rotational. The correct choice is c.

294

PROBLEM 4.106

Situation: Water owing from a large orice in bottom of tank. Velocities and eleva-tions given in problem.

Find: %= " %>!

APPROACH


ANALYSIS

Bernoulli equation

%=(+ R= +

, 2=2)

=%>(+ R> +

, 2>2)

%= " %> = ([(, 2> " ,2=)$2) " R=]

= 62!4[(400" 64)$(2× 32!2)" 1]

%= " %> = 263!2 psf

295

PROBLEM 4.107

Situation: Ideal ow past an airfoil in a 80 m/s airstream. Velocities on airfoil are 85and 75 m/s and air density is 1.2 kg/m3.

Find: Pressure di!erence between bottom and top.Assumption: The pressure due to elevation di!erence between points is negligible.

ANALYSIS

The ow is ideal and irrotational so the Bernoulli equation applies between any twopoints in the ow eld

%1 + (R1 + #,21 $2 = %1 + (R1 + #,

21 $2

%2 " %1 = (#$2)(, 21 " ,22 )

%2 " %1 = (1!2$2)(852 " 752)= 960 Pa

%2 " %1 = 0!96 kPa

296

PROBLEM 4.108

Situation: Horizontal ow between two parallel plates and one is xed while othermoves.

Find: Is the Bernoulli equation valid to nd pressure di!erence between plates?

ANALYSIS

This is not correct because the ow between the two plates is rotational and theBernoulli equation cannot be applied across streamlines. There is no acceleration ofthe uid in the direction normal to the plates so the pressure change is given by thehydrostatic equation so

%1 " %2 = (M

297

PROBLEM 4.109

Situation: A cyclonic storm has a wind speed of 15 mph at 3 = 200 mi.

Find: Wind speed at 3 = 50 and 100 miles: ,50 & ,100!

ANALYSIS

, 3 = Const.

(15 mph) (200 mi.) = Const.

,100 = Const./100 mi.

= (15 mph)(200 mi./100 mi.)

,100 = 30 mph

,50 = (15 mph)(200/50)

,50 = 60 mph

298

PROBLEM 4.110

Situation: A tornado is modeled as a combined forced and free vortex and core has adiameter of 10 mi. At 50 mi. from center, velocity is 20 mph. The core diameter is10 miles. The wind velocity is , = 20 mph at a distance of 3 = 50 miles,

Find: (a) Wind velocity at edge of core: ,10(b) Centrifugal acceleration at edge of core: C1

ANALYSIS

The velocity variation in a free vortex is

, 3 = const

Thus,50(50) = ,10(10)

Therefore,10 = ,50

50

10= 5× 20 = 100 mph

Acceleration (Eulerian formulation)

, = 100× 5280$3600 = 147 ft/sC1 = , 2$3

= 1472$(10× 5280)

C1 = 0!409 ft/s2

299

PROBLEM 4.111

Situation: A whirlpool modeled as free and forced vortex. The maximum velocity is10 m/s at 10 m.

Find: Shape of the water surface.

APPROACH

Apply the Bernoulli equation to the free vortex region.

ANALYSIS

Bernoulli equation

R10 +, 2max2)

= R +, 2

2)= 0

The elevation at the juncture of the forced and free vortex and a point far from thevortex center where the velocity is zero is given by

R10 = ", 2max2)

In the forced vortex region, the equation relating elevation and speed is

R10 ", 2max2)

= R ", 2

2)

At the vortex center, , = 0' so

R0 = R10 ", 2max2)

= ", 2max2)

", 2max2)

= ", 2max)

R = "102

9!81= "10!2 m

In the forced vortex region

, =3

1010 m/s = 3

so the elevation is given by

R = "10!2 +32

2)

In the free vortex region

, = 1010

3

so the elevation is given by

300

R = R10 +, 2max2)

"100

2)

µ10

3

¶2="51032

Radius, m

0 10 20 30 40 50-12

-10

-8

-6

-4

-2

0

Ele

vatio

n, m

301

PROBLEM 4.112

Situation: Tornado modeled as combination of forced and free vortex with maximumvelocity of 350 km/hr at 50 m.

Find: Variation in pressure.

APPROACH

Apply the pressure variation equation-rotating ow to the vortex center and theBernoulli equation in the free vortex region.

ANALYSIS

From the pressure variation equation-rotating ow, the pressure reduction from at-mospheric pressure at the vortex center is

!% = "#, 2maxwhich gives

!% = "1!2× (350×1000

3600)2 = "11!3 kPa

or a pressure of %(0) = 100"11!3 = 88!7 kPa. In the forced vortex region the pressurevaries as

%(0) = %" #, 2

2

In this region, the uid rotates as a solid body so the velocity is

, =3

50,max = 1!943

The equation for pressure becomes

% = 88!7 + 2!2632$1000 for 3 ' 50 m

The factor of 1000 is to change the pressure to kPa. A the point of highest velocitythe pressure is 94.3 kPa.Bernoulli equation

%(50) +1

2#, 2max = %+

1

2#, 2

In the free vortex region so the equation for pressure becomes

% = %(50) +1

2#, 2max

·1" (

50

3)2¸

for 3 ) 50 m

% = 94!3 + 5!65×·1" (

50

3)2¸

302

Radius (m)

0 50 100 150 200 25088

90

92

94

96

98

100

102

Pre s

sure

(kPa

)

303

PROBLEM 4.113

Situation: A tornado is modeled as a forced and free vortex.

Find: Pressure coe"cient versus nondimensional radius.

APPROACH

Apply Eq. 4.48 for the vortex center and the Bernoulli equation in the free vortexregion.

ANALYSIS

From Eq. 4.48 in the text, the pressure at the center of a tornado would be "#, 2maxso the pressure coe"cient at the center would be

-, ="#, 2max12#, 2max

= "2

For the inner, forced-vortex region the pressure varies as

%(0) = %"1

2#, 2

so the pressure coe"cient can be written as

-, =%" %!12#, 2max

= "2 + (,

,max)2 for 3 ' 31

-, = "2 + (3

31)2

so the pressure coe"cient at the edge of the forced vortex is -1.Bernoulli equation

%(31) +1

2#, 2max = %+

1

2#, 2

Pressure coe"cient

-, =%" %!12#, 2max

=%(31)" %!12#, 2max

+ [1" (313)2] for 3 ) 31

-, = "1 + [1" (313)2] = "(

313)2

304

0 1 2 3 4 5

-2

-1

0

Pres

sure

coe

ffici

ent

Nondimensional radius (r/r )c

305

PROBLEM 4.114

Situation: A weather balloon in a tornado modeled as a forced-free vortex.

Find: Where the balloon will move.

ANALYSIS

The uid in a tornado moves in a circular path because the pressure gradient pro-vides the force for the centripetal acceleration. For a uid element of volume ( therelationship between the centripetal acceleration and the pressure gradient is

#, 2

3= (

A%

A3

The density of a weather balloon would be less than the local air so the pressuregradient would be higher than the centripetal acceleration so theballoon would move toward the vortex center.

306

PROBLEM 4.115

Situation: The pressure distribution in a tornado.

Find: If the Bernoulli equation overpredicts or underpredicts the pressure drop.

ANALYSIS

As the pressure decreases the density becomes less. This means that a smaller pressuregradient is needed to provide the centripetal force to maintain the circular motion.This means that the Bernoulli equation will overpredict the pressure drop.

307

PROBLEM 4.116

Situation: A two dimensional ow in the E " ? plane is described in the problemstatement.

ANALYSIS

a) Substituting the equation for the streamline into the Euler equation gives

@STS$AE+ @ST

S3A? = ") S7

S$AE

; S@S$AE+ ; S@

S3A? = ") S7

S3A?

or

SS$

³T2

2

ÁE+ S

S3(T

2

2)A? = ") S7

S$AE

SS$

³@2

2

ÁE+ S

S3(@

2

2)A? = ") S7

S3A?

Adding both equations

[

[E

µ@2 + ;2

2

¶AE+

[

[?

µ@2 + ;2

2

¶A? = ")(

[M

[EAE+

[M

[?A?)

or

A(@2 + ;2

2+ )M) = 0

b) Substituting the irrotationality condition into Euler’s equation gives

@STS$+ ; S@

S$= ") S7

S$

; S@S3+ @ST

S3A? = ") S7

S3

or

SS$(T

2+@2

2+ )M) = 0

SS3(T

2+@2

2+ )M) = 0

308

PROBLEM 4.117

Situation: Di!erent ow patterns are created by breathing in or out.

Find: Why it is easier to blow a candle out while exhaling rather than inhaling.

ANALYSIS

The main point to this question is that while inhaling, the air is drawn into yourmouth without any separation occurring in the ow that is approaching your mouth.Thus there is no concentrated ow; all air velocities in the vicinity of your face arerelatively low. However, when exhaling as the air passes by your lips separationoccurs thereby concentrating the ow of air which allows you to easily blow out acandle.

309

PROBLEM 4.118

Situation: High winds can lift roofs from buildings.

Find: Explain why winds lift roofs rather than force them downward.

ANALYSIS

If a building has a at roof as air ows over the top of the building separation willoccur at the sharp edge between the wall and roof. Therefore, most if not all ofthe roof will be in the separation zone. Because the zone of separation will have apressure much lower than the normal atmospheric pressure a net upward force willbe exerted on the roof thus tending to lift the roof.Even if the building has a peaked roof much of the roof will be in zones of separation.These zones of separation will occur downwind of the peak. Therefore, peaked roofbuildings will also tend to have their roofs uplifted in high winds.

310

PROBLEM 5.1

Situation: Water ows in a 25 cm diameter pipe. ] = 0!04 m3/s.

Find: Mean velocity: ,

APPROACH

Apply the ow rate equation.

ANALYSIS

Flow rate equation

, = ]$I

= 0!04$(0$4× 0!252)

, = 0!815 m/s

311

PROBLEM 5.2

Situation: Water ows in a 16 in pipe. , = 3 ft/s.

Find: Discharge in cfs and gpm.

APPROACH


ANALYSIS

Flow rate equation

] = , I

= (3 ft/s)(0$4× 1!3332)

] = 4!19 ft3/s

= (4!17 ft3/s)(449 gpm/ft3/s)

] = 1880 gpm

312

PROBLEM 5.3

Situation: Water ows in a 2 m diameter pipe. , = 4 m/s.

Find: Discharge in m3/s and cfs.

APPROACH


ANALYSIS

Flow rate equation

] = , I

= (4)(0$4× 22)

] = 12!6 m3/s

] = (12!6 m3/s)(1/0.02832)(ft3$s)/(m3$s)

] = 445 cfs

313

PROBLEM 5.4

Situation: An 8 cm. pipe carries air, , = 20 m/s, & = 20!C, % = 200 kPa-abs.

Find: Mass ow rate: /

ANALYSIS

Ideal gas law

# = %$"&

= 200' 000$(287× 293)# = 2!378 c)$/3

Flow rate equation

/ = #, I

= 2!378× 20× (0 × 0!082$4)

/ = 0!239 kg/s

314

PROBLEM 5.5

Situation: A 1 m pipe carries natural gas, , = 20 m/s, & = 15!C, % = 150 kPa-gage.


APPROACH

Apply the ideal gas law and the ow rate equation.

ANALYSIS

Ideal gas law

# = %$"&

= (101 + 150)103$((518)× (273 + 15))= 1!682 kg/m3

Flow rate equation

/ = #, I

= 1!682× 20× 0 × 0!52

/ = 26!4 kg/s

315

PROBLEM 5.6

Situation: A pipe for an aircraft engine test has / = 200 kg/s and , = 240 m/s.% = 50 kPa-abs, & = "18 !C.

Find: Pipe diameter: 1

APPROACH

Apply the ideal gas law and the ow rate equation.

ANALYSIS

Ideal gas law

# = %$"&

= (50× 103)$((287)(273" 18))= 0!683 kg/m3

Flow rate equation/ = #I,

So

I = /$(#, )

= (200)$((0!683)(240))

= 1!22 m2

I = (0$4)12 = 1!22

1 = ((4)(1!22)$0)1*2

1 = 1!25 m

316

PROBLEM 5.7

Situation: Air ows in a rectangular air duct with dimensions 1!0×0!2 m. ] = 1100m3/hr.

Find: Air velocity: ,

APPROACH


ANALYSIS

Flow rate equation

, = ]$I

= 1' 100 (m3$hr)$3600 (sec/hr)$(1× 0!20) m2

, = 1!53 m$ s

317

PROBLEM 5.8

Situation: In a circular duct the velocity prole is ;(3) = ,! (1" 3$") ' where ,! isvelocity at 3 = 0!

Find: Ratio of mean velocity to center line velocity: , $,!

APPROACH


ANALYSIS

Flow rate equation

] =

Z;AI

where AI = 203A3! Then

] =

Z F

0

,0(1" (3$"))203A3

= ,0(20)((32$2)" (33$(3"))) |F0

= 20,0(("2$2)" ("2$3))

= (2$6)0,0"2

Average Velocity

, =]

I,

,0=

]

I

1

,0

=(2$6)0,0"

2

0"21

,0

, $,! = 1$3

318

PROBLEM 5.9

Situation: Water ows in a rectangular channel. The velocity prole is , (E' ?) =,.(1 " 4E2$. 2)(1 " ?2$12)' where . and 1 are the channel width and depth,respectively.

Find: An expression for the discharge: ] = ](,.' 1' . )

APPROACH


ANALYSIS

Flow rate equation

] =

ZV · AA =

Z Z, (E' ?)AEA?

=

Z M*2

"M*2

Z 9

3=0

,.(1" 4E2$. 2)(1" ?2$12)A?AE

] = (4$9),..1

319

PROBLEM 5.10

Situation: Water ows in a 4 ft pipe. The velocity prole is linear. The center linevelocity is ,max = 15 ft/s. The velocity at the wall is ,min = 12 ft/s.

Find: Discharge in cfs and gpm.

APPROACH


ANALYSIS

Flow rate equation

] =

Z

=

, AI

=

Z K0

0

, 203A3

where , = ,max " 33$30!

] =

Z K0

0

(,max " (33$30))203A3

= 20320((,max$2)" (3$3))= 20 × 4!00((15$2)" (3$3))

] = 163!4 cfs

= 163!4× 449] = 73' 370 gpm

320

PROBLEM 5.11

Situation: Water ows in a 2 m pipe. The velocity prole is linear. The center linevelocity is ,max = 8 m/s and the velocity at the wall is ,min = 6 m/s.

Find: (a) Discharge: ](b) Mean velocity: ,

APPROACH


ANALYSIS

Flow rate equation

] = 20320((,max$2)" (2$3)) (see problem 5.10 for derivation)

= 20 × 1((8$2)" (2$3))

] = 20!94 m3/s

, = ]$I = 20!94$(0 × 1)

, = 6!67 m/s

321

PROBLEM 5.12

Situation: Air ows in a square duct with velocity prole shown in the gure.

Find: (a) Volume ow rate: ](b) Mean velocity: ,(c) Mass ow rate: / (if density is 1.2 kg/m3)

ANALYSIS

A] = , AI

A] = (20?)A?

] = 2

Z 045

0

, AI

= 2

Z 045

0

20?A?

= 40?2$2|0450= 20× 0!25

] = 5 m3/s

, = ]$I

= (5 m3$s)/(1 m2)

, = 5 m/s

/ = #]

= (1!2 kg/m3)(5 m3$s)

/ = 6!0 kg/s

322

PROBLEM 5.13

Situation: An open channel ow has a 30! incline. , = 18 ft/s. Vertical depth is 4ft. Width is 25 ft.

Find: Discharge: ]

APPROACH


ANALYSIS

Flow rate equation

] = , ×I= 18× 4 cos 30! × 25

] = 1' 560 cfs

323

PROBLEM 5.14

Situation: An open channel ow has a 30! incline. Velocity prole is @ = ?1*3 m/s.Vertical depth is 1 m. Width is 1.5 m.

Find: Discharge: ]

APPROACH


ANALYSIS

Flow rate equation

] =

Z 04866

0

?1*3(2 dy)

= 1!5

Z 04866

0

?1*3dy

= (1!5$(4$3))?4*3|04866 m0

] = 0!93 m3/s

324

PROBLEM 5.15

Situation: Open channel ow down a 30! incline. Velocity prole is @ = 10 (=3 " 1)m/s. Vertical depth is 1 m and width is 2 m.

Find: (a) Discharge: ](b) Mean velocity: ,

APPROACH


ANALYSIS

Flow rate equation

] =

Z 04866

0

, A?

] =

Z 04866

0

(10)(=3 " 1)2 dy

= [(2)(10)(=3 " ?)]048660

] = 10!23 m3/s

, = ]$I

= (10!23 m3$s)/(2× 0!866 m2)

, = 5!91 m/s

325

PROBLEM 5.16

Situation: Water (20! C, ( = 9790 N/m3) enters a weigh tank for 15 min. Theweight change is 20 kN.

Find: Discharge: ]

ANALYSIS

] = ,$!P

= ./((!P)

= 20' 000$(9790× 15× 60)

] = 2!27× 10"3 m3/s

326

PROBLEM 5.17

Situation: Water enters a lock for a ship canal through 180 ports. Port area is 2× 2ft. Lock dimensions (plan view) are 105 × 900 ft. The water in the lock rises at 6ft/min.

Find: Mean velocity in each port: ,,!K5

APPROACH

Apply the continuity principle.

ANALYSIS

Continuity principle

X,,I, = ,rise ×Irise

180× ,, × (2× 2) = (6$60)× (105× 900)

,port = 13!1 ft/s

327

PROBLEM 5.18

Situation: Water ows through a rectangular and horizontal open channel. Thevelocity prole is @ = @max(?$A):' where y is depth, @max = 3 m/s, A = 1!2 m, andQ = 1$6!

Find: (a) Discharge: ^(m3/s per meter of channel width).(b) Mean velocity: ,

APPROACH


ANALYSIS

Flow rate equation

^ =

Z A

0

@max(?$A):A? = @maxA$(Q+ 1)

= 3× 1!2$((1$6) + 1)

^ = 3!09 m2/s

, = ^$A

= 3!09$1!2

, = 2!57 m/s

328

PROBLEM 5.19

Situation: A ow with a linear velocity prole occurs in a triangular-shaped openchannel. The maximum velocity is 6 ft/s.

Find: Discharge: ]

APPROACH


ANALYSIS

Flow rate equation

] =

Z, AI

where , = 5? ft/s' AI = EA? = 0!5 ?A? ft2

^ =

Z 1

0

(6?)× (0!5?A?)

= (3?3$3)|10^ = 1 cfs

329

PROBLEM 5.20

Situation: Flow in a circular pipe. The velocity prole is , = ,1(1" (3$3!)):.

Find: An expression for mean velocity of the form , = , (,1' Q)!

APPROACH


ANALYSIS

Flow rate equation

] =

Z

=

, AI

=

Z K0

0

,1(1" (3$320)):203A3

= "0320,1Z K!

0

(1" (3$30)2):("23$32!)A3

This integral is in the form of

Z U

0

@:A@ = \:+1$(Q+ 1)

so the result is

] = "0320,1(1" (3$30)2):+1$(Q+ 1)|K00

= (1$(Q+ 1)),10320

, = ]$I

, = (1$(Q+ 1)),1

330

PROBLEM 5.21

Situation: Flow in a pipe has a velocity prole of , = 12(1" 32$32!)

Find: (a) Plot the velocity prole(b) Mean velocity: ,(c) Discharge: ]

APPROACH


ANALYSIS

3$30 1" (3$30)2 , (m/s)0.0 1.00 12.00.2 0.96 11.50.4 0.84 10.10.6 0.64 7.680.8 0.36 4.321.0 0.00 0.0

] =

Z

=

, AI

=

Z K0

0

,1(1" (3$320))203A3

= "0320,1Z K!

0

(1" (3$30)2)("23$32!)A3

= (1$2),10320

, = ]$I

, = (1$2),1

, = 6 m/s

331

Flow rate equation

] = , I

= 6× 0$4× 12

] = 4!71 m3/s

332

PROBLEM 5.22

Situation: Water (60 !F) ows in a 1.5 in. diameter pipe. / = 80 lbm/min.


APPROACH


ANALYSIS

Flow rate equation

, = /$#I

, = (80$60)$£(62!37)× (0$4× (1!5$12)2)

¤

, = 1!74 ft/s

333

PROBLEM 5.23

Situation: Water (20 !C) ows in a 20 cm diameter pipe. / = 1000 kg/min.


APPROACH


ANALYSIS

Flow rate equation

, = /$#I

= (1' 000$60)$£(998)× (0$4× 0!202)

¤

, = 0!532 m/s

334

PROBLEM 5.24

Situation: Water (60 !F) enters a weigh tank for 10 min. The weight change is 4765lbf.

Find: Discharge: ] in units of cfs and gpm

ANALYSIS

] = ,$!P

= !./((!P)

= 4765$(62!37× 10× 60)

] = 0!127 cfs

= 0!127× 449] = 57!0 gpm

335

PROBLEM 5.25

Situation: Water (60 !F) ows in a 4 in. diameter pipe. , = 8 ft/s.

Find: (a) Discharge: ] in units of cfs and gpm(b) Mass ow rate: /

APPROACH


ANALYSIS

Flow rate equation

] = , I

= 8(0$4× (4$12)2)

] = 0!698 cfs

= 0!698× 449] = 313 gpm

Mass ow rate

/ = #]

= 1!94× 0!698/ = 1!35 slugs/s

336

PROBLEM 5.26

Situation: As shown in the sketch below, two round plates, each with speed , , movetogether. At the instant shown, the plate spacing in h. Air ows across sectionA with a speed , 0. Assume , 0 is constant across section A. Assume the air hasconstant density.

r

Find: An expression for the radial component of convective acceleration at section A.

APPROACH

Apply the continuity principle to the control volume dened in the problem sketch.

ANALYSIS


/! " /# = "A$APZ

c.v.#AV–

#, 0I0 = "("2#, I)2, I = , 0I0

The control volume has radius 3 so

, 0 = 2, I$I0 = 2, (032)$(203M) = , 3$M

Convective acceleration

C1 = , 0[$[3(, 0)

= , 3$M [$[3(, 3$M)

= , 23$M2

C1 = ,21$2M2

337

PROBLEM 5.27

As shown in the sketch below, two round plates, each with speed , , move together.At the instant shown, the plate spacing in h. Air ows across section A with a speed, 0. Assume , 0 is constant across section A. Assume the air has constant density.

r

Find: An expression for the radial component of local acceleration at section A.

APPROACH

Apply the continuity principle to the control volume dened in the problem sketch .

ANALYSIS


/! " /# = "A$APZ

c.v.#AV–

#, 0I0 = "("2#, I)2, I = , 0I0

Control volume has radius 3 so

, 0 = 2, I$I0 = 2, (032)$(203M) = , 3$M

Introducing time as a parameter

M = M0 " 2, P

so, 0 = 3,$(M0 " 2, P)

Local acceleration

[, 0$[P = [$[P[3, (M0 " 2, P)"1] = 3, ("1)(M0 " 2, P)"2("2, )[, 0$[P = 23, 2$(M0 " 2, P)2

but M0 " 2, P = M and 3 = " so

[, 0$[P = 2", 2$M2

[, 0$[P = 1, 2$M2

338

PROBLEM 5.28

Situation: Pipe ows A and B merge into a single pipe with area Iexit = 0!1 m2.]= = 0!02P m3$s and ]> = 0!008P2 m3$s.

Find: (a) Velocity at the exit: ,exit(b) Acceleration at the exit: Cexit

Assumptions: Incompressible ow.

APPROACH


ANALYSIS


]exit = ]= +]>

,exit = (1$Iexit)(]= +]>)

= (1$0!01 m2)(!02P m3/s + 0!008P2 m3/s)

= 2P m/s + 0!8P2 m/s

Then at P = 1 sec,,exit = 2!8 m/s

Acceleration

Cexit =[,

[P+ ,

[,

[E

Since V varies with time, but not with position, this becomes

Cexit =[,

[P= 2 + 1!6P m/s

Then at P = 1 secCexit = 3!6 m/s2

339

PROBLEM 5.29

Situation: Air ow downward through a pipe and then outward between to paralleldisks. Details are provided on the gure in the textbook.

Find: (a) Expression for acceleration at point A.(b) Value of acceleration at point A.(c) Velocity in the pipe.

APPROACH


ANALYSIS

Flow rate equation

,K = ]$I = ]$(203M)

C1 = ,K[,K$[3

= (]$(203M))("1)(])$(2032M)

C1 = "]2$(3(203M)2)

When 1 = 0!1 m' 3 = 0!20 m' M = 0!005 m' and ] = 0!380 m3/s

,pipe = ]$Ipipe

= 0!380$((0$4)× 0!12)

,pipe = 48!4 m/s

Then

C1 = "(0!38)2$((0!2)(20 × 0!2× 0!005)2)

C1 = "18' 288 m/s2

340

PROBLEM 5.30

Situation: Air ow downward through a pipe and then outward between to paralleldisks as illustrated on gure in problem.

Find: (a) At P = 2 s, acceleration at point A: C2&(b) At P = 3 s, acceleration at point A: C3&

ANALYSIS

C< = [,$[P = [$[P(]$(203M))

C< = [$[P(]0(P$P0)$(203M))

C< = (]0$P0)$203M

C<;2I3 = (0!1$1)$(20 × 0!20× 0!01) = 7!958 m/s2

From solution to Problem 5.29

C1 = "]2$(3(203M)2)

At P = 29'] = 0!2 m3$s

C1I2& = "1266 m/s2

C2& = C< + C1 = 7!957" 1' 266

C2& = "1' 258 m/s2

At P = 3s, ] = 0!3 m3$s

C1I3& = "2' 850 m/s2

C3& = "2' 850 + 7!957

C3& = "2' 840 m/s2

341

PROBLEM 5.31

Situation: Water ows into a tank through a pipe on the side and then out the bottomof the tank with velocity

&2)M. Water rising in tank at 0.1 cm/s.

Find: Velocity in the inlet: ,in

APPROACH

Apply the continuity principle. Let the control surface surround the liquid in thetank and let it follow the liquid surface at the top.

ANALYSIS


/! " /# = "A

AP

Z

cv

#A(

"#,inIin + #,outIout = "A

AP(#ItankM)

",inIin + ,outIout = "Itank(AM$AP)

",in(!0025) +p2g(1)(!0025) = "0!1(0!1)× 10"2

,in =

&19!62(!0025) + 10"4

0!0025

,in = 4!47 m/s

342

PROBLEM 5.32

Situation: A bicycle tire (( = 0!04 ft3) is inated with air at an inlet ow rate of]in = 1 cfm and a density of 0.075 lbm/ft3. The density of the air in the inatedtire is 0.4 lbm/ft3.

Find: Time needed to inate the tire: P

APPROACH

Apply the continuity principle. Select a control volume surrounding the air withintire.

ANALYSIS


(#])in =A

AP+cv

This equation may be integrated to give

(#])in P =+CV

or

P =+CV

(#])in

=0!04× 0!4

0!075× (1$60)P = 12!8 s

343

PROBLEM 5.33

Situation: Conditions in two ow cases are described in the problem statement.

Find: (a) Value of b.(b) Value of ADsys$AP!(c) Value of

P>#V ·A

(d) Value of A$APRcv >#AV–

ANALYSIS

Case (a) Case (b)1) > = 1 1) D = 12) ADsys$AP = 0 2) ADsys$AP = 03)

P>#V ·A =

P#V ·A 3)

P>#V ·A =

P#V ·A

= "2× 12× 1!5 = 2× 1× 2= "36 slugs/s "1× 2× 2 = 0

4) A$APRcv >#AV– = +36 slugs/s 4) A$AP

Rcv >#AV– = 0

344

PROBLEM 5.34

Situation: Mass is owing into and out of a tank

Find: Select the statement that is true.

ANALYSIS

Mass ow out

/o = (#I, )2

= 2× 0!2× 5= 2 kg/s

Mass ow in

/i = (#I, )1

= 3× 0!1× 10= 3 kg/s

Only selection (b) is valid.

345

PROBLEM 5.35

Situation: The level in the tank (see below) is inuenced by the motion of pistons Aand B. Each piston moves to the left. ,= = 2,>

Find: Determine whether the water level is rising, falling or staying the same.

APPROACH

Apply the continuity principle. Select a control volume as shown above. Assume itis coincident with and moves with the water surface.

ANALYSIS


/! " /# = "A$APZ

cv#AV–

#2,>I= " #,>I> = "#A$APZ

cvAV–

where I= = (0$4)32;I> = (0$4)62 and I= = (1$4)I>! Then

2,>(1$4)I> " ,>I> = "A$APZ

)?

A,

,>I>((1$2)" 1) = "A$APZ

)?

AV–

A$AP

Z

)?

AV– = (1$2),>I>

A$AP(IM) = ($12),>I>

IAM$AP = (1$2),>I>

Because (1$2),>I> is positive AM$AP is positive; therefore, one concludes that thewater surface is rising.

346

PROBLEM 5.36

Situation: A piston in a cylinder is moving up and control consists of volume incylinder.

Find: Indicate which of the following statements are true.

ANALYSIS

a) True b) True c) True d) True e) True

347

PROBLEM 5.37

Situation: A control volume is described in the problem statement.

Find: (a) Value of >.(b) Value of ADsys$AP!(c) Value of

P>#V ·A!

(d) Value of A$APR>#AV–.

ANALYSIS

a) > = 1!0

b) ADsys$AP = 0

c)X

>#V ·A =X

#V ·AX

#V ·A = (1!5 kg/m3)("10 m/s)(0$4)× (0!04)2 m2

+ (1!5 kg/m3)("6 m/s)(0$4)× (0!04)2 m2

+ (1!2 kg/m3)(6 m/s)(0$4)× (0!06)2 m2

= "0!00980 kg/s

d) BecauseX

>#V ·A+ A$APZ>#AV– = 0

Then A$APZ>#AV– = "

X>#V ·A

or A$APR>#A," = +0!00980 kg/s (mass is increasing in tank)

348

PROBLEM 5.38

Situation: A plunger moves downward in a conical vessel lled with oil. At a certaininstant in time, the upward velocity of the oil equals the downward velocity of theplunger.

Find: Distance from the bottom of the vessel: ?

ANALYSIS

Select a control volume surrounding the liquid. The rate at which volume of uid isdisplaced upward is

,up(12 " A2)(0$4)

From the continuity principle

,down × 0A2$4 = ,up(12 " A2)0$4

2A2 = 12

1 =&2A

But ?$1 = 24A$2A so 1 = ?$12 so

? = 12&2A

349

PROBLEM 5.39

Situation: A 6 in. diameter cylinder falls at a speed ,) = 3 ft/s. The containerdiameter is 8 in.

Find: Mean velocity (,- ) of the liquid in the space between the cylinder and the wall.

APPROACH

Apply continuity principle and let the c.s. be xed except at the bottom of thecylinder where the c.s. follows the cylinder as it moves down.

ANALYSIS


0 = A$AP

Z#AV–+ /! " /#

0 = A$AP(V–)+ ,-I=0 = ,)I) + ,- (0$4)(8

2 " 62)0 = "3× (0$4)62 + ,- (0$4)(82 " 62),- = 108$(64" 36)

,- = 3!86 ft/s (upward)

350

PROBLEM 5.40

Situation: A round tank (1 = 4 ft) is being lled with water from a 1 ft diameterpipe. In the pipe, , = 10 ft/s

Find: Rate at which the water surface is rising: ,F

APPROACH

Apply the continuity principle and let the c.s. move up with the water surface in thetank.

ANALYSIS


0 = A$AP

Z

)?

#AV–+ /! " /#

0 = A$AP(MI- )" ((10 + ,F)I,)

where I- = tank area, ,F =rise velocity and I, =pipe area.

0 = I-AM$AP" 10I, " ,FI,

but AM$AP = ,F so

0 = I-,F " 10I, " ,FI,,F = (10I,)$(I- "I,) = 10(0$4)(12)$((0$4)42 " (0$4)12)

,F = (2$3) ft/s

351

PROBLEM 5.41

Situation: An 8 in. sphere is falling at 4 ft/s in a 1 ft diameter cylinder lled withwater

4

Find: Velocity of water at the midsection of the sphere

APPROACH

Apply the continuity equation.

ANALYSIS

As shown in the above sketch, select a control volume that is attached to the fallingsphere. Relative to the sphere, the velocity entering the control volume is ,1 and thevelocity exiting is ,2Continuity equation

"A$APZ

)?

#A, = 0 = /# " /!

I1,1 = I2,2

(0 × 1!02$4)× 4 = ,20(1!02 " !672)$4

,2 = 7!26 fps

The velocity of the water relative to a stationary observer is

, = ,2 " ,&,7GKG, = 7!26" 4!0

= 3.26 ft/s

352

PROBLEM 5.42

Situation: Air ows in a rectangular duct. ] = 1!44 m3$s.

Find: (a) Air speed for a duct of dimensions 20× 50 cm: ,1(b) Air speed for a duct of dimensions 10× 40 cm: ,2

APPROACH


ANALYSIS

Flow rate equation

,1 = ]$I1

= 1!44$(0!2× 0!5)

,1 = 14!4 m/s

,2 = 1!44$(0!1× 0!4)

,2 = 36!0 m/s

353

PROBLEM 5.43

Situation: Flow (] = 0!3 m3/s) enters a pipe that has an inlet diameter of 30 cm.Outlet diameters are 20 and 15 cm. Each outlet branch has the same mean velocity.

Find: Discharge in each outlet branch: ]20 cm, ]15 cm

APPROACH


ANALYSIS

Flow rate equation

, = 0!3$(0$4)(0!22 + 0!152)

= 6!11 m/s

]20 cm = , I20

= 6!11× (0 × 0!1× 0!1)

]20 cm = 0!192 m3/s

]15 cm = , I15

= 6!11× (0 × 0!075× 0!075)

]15 cm = 0!108 m3/s

354

PROBLEM 5.44

Situation: Flow (] = 0!3 m3/s) enters a pipe that has an inlet diameter of 30 cm.Outlet diameters are 20 and 15 cm. In the larger outlet (20 cm) the ow rate is twicethat in the smaller outlet (15 cm).

Find: Mean velocity in each outlet branch: ,15, ,20

ANALYSIS

Continuity principle]tot. = 0!30 m3/s = ]20 +]15

Since ]20 = 2]15

0!30 = 2]15 +]15

]15 = 0!10 m3/s;

]20 = 0!20 m3/s;

Flow rate equation

,15 = ]15$I15

,15 = 5!66 m/s

,20 = 0!20$I20

,20 = 6!37 m/s

355

PROBLEM 5.45

Situation: Water ows through an 8 in. diameter pipe that is in series with a 6 inpipe. ] = 898 gpm.

Find: Mean velocity in each pipe: ,6, ,8

APPROACH


ANALYSIS

Flow rate equation

] = 898 gpm = 2 cfs

,8 = ]$I8

= 2$(0 × 0!667× 0!667$4)

,8 = 5!72 fps

,6 = ]$I6

= 2$(0 × 0!5× 0!5$4)

,6 = 10!19 fps

356

PROBLEM 5.46

Situation: Water ows through a tee as shown in gure in the textbook.

Find: Mean velocity in outlet B: ,>

APPROACH


ANALYSIS


,> = (,=I= " ,1I1)$I>= [(6× 0$4× 42)" (4× 0$4× 22)]$(0$4× 42)

,> = 5!00 m/s

357

PROBLEM 5.47

Situation: Gas ows in a round conduit which tapers from 1.2 m to 60 cm. Detailsare provided on the gure with the problem statement.

Find: Mean velocity at section 2: ,2

APPROACH


ANALYSIS


,2 = (#1I1,1)$(#2I2)

= (#1121,1)$(#21

22)

= (2!0× 1!22 × 15)$(1!5× 0!62)

,2 = 80!0 m/s

358

PROBLEM 5.48

Situation: Pipes A and B are connected to an open tank with surface area 80 ft2.The ow rate in pipe A is ]= = 8 cfm, and the level in the tank is rising at a rate of1.0 in./min.

Find: (a) Discharge in pipe B: ]>(b) If ow in pipe B entering or leaving the tank.

APPROACH

Apply the continuity principle. Dene a control volume as shown in the above sketch.Let the c.s. move upward with the water surface.

ANALYSIS


0 = A$AP

Z

)?

#AV–+X

#V ·A

0 = IAM$AP+]> "]=]> = ]= "I AM$AP

= 8" (80)(1!0$12)

]> = +1!33 cfm

Because ]> is positive ow is leaving the tank through pipe D.

359

PROBLEM 5.49

Situation: A tank with one inow and two outows indicated by diagram with prob-lem statement.

Find: (a) Is the tank lling or emptying.(b) Rate at which the tank level is changing: A7

A5

ANALYSIS

Inow = 10× 0 × 22$144 = 0!8727 cfsOutow = (7× 0 × 32$144) + (4× 0 × 1!52$144) = 1!571 cfsOutow F Inow, Thus, tank is emptying

AM

AP= "]$I

= "(1!571" 0!8727)$(0 × 32)A7A5= "0!0247 ft/s

360

PROBLEM 5.50

Situation: The sketch shows a tank lled with water at time P = 0 s.

Find: (a) At P = 22 s, if the the water surface will be rising or falling.(b) Rate at which the tank level is changing: A7

A5

APPROACH

Apply the continuity principle. Dene a control volume in which the control surface(c.s.) is coincident with the water surface and moving with it.

ANALYSIS


A$AP

Z

1@

#AV– = /# " /!

A$AP(#IM) = (#I, )in " (#I, )outA$AP(#IM) = #(0$4× 12)(1) + #(0$4× 0!52)(2)IAM$AP = (0$4)" (0$8)IAM$AP = (0$8)

Since IAM$AP F 0, the water level must be rising. While the water column occupiesthe 12 in. section, the rate of rise is

AM$AP = (0$8) $I

= 0$(8× 0$4× 12)= 1$2 ft/s

Determine the time it takes the water surface to reach the 2 ft. section:

10 = (AM$AP)P;

P = (10)$(1$2) = 20 secs.

Therefore, at the end of 20 sec. the water surface will be in the 2 ft. section. Thenthe rise velocity will be:

361

AM$AP = 0$(8I)

= 0$(8× 0$4× 22)

AM$AP = 1$8 ft/sec

362

PROBLEM 5.51

Situation: A lake is fed by one inlet, ]#: = 1000 cfs. Evaporation is 13 cfs per squaremile of lake surface. Lake surface area is I(M) = 4!5+5!5M, where M is depth in feet.

Find: (a) Equilibrium depth of lake.(b) The minimum discharge to prevent the lake from drying up.

Assumptions: Equilibrium.

APPROACH


ANALYSIS


]Evap. = ]in.¡13 ft3/s/mi2

¢(4!5 + 5!5M) mi2 = 1' 000 ft3/s

Solve for depth M :M = 13!2 ft. at equilibrium

The lake will dry up when M = 0 and ]Evap. = ]in.. For M = 0,

13(4!5 + 5!5× 0) = ]in.

Lake will dry up when ]in. = 58!5 ft3$s

363

PROBLEM 5.52

Situation: A nozzle discharges water (]! = 5 cfs) onto a plate moving towards thenozzle. Plate speed equals half the jet speed.

Find: Rate at which the plate deects water: ],

APPROACH

Apply the continuity principle. Select a control volume surrounding the plate andmoving with the plate.

ANALYSIS

Continuity principle]#: = ],

Reference velocities to the moving plate. Let ,! be the speed of the water jet relativeto the nozzle. From the moving plate, the water has a speed of ,! + 1$2,! = 3,!$2!Thus

], = ]#:

= ,#:I!

= (3,!$2) (I!) = (3$2) (,!I!)

= (3$2)]!

], = 7!5 cfs

364

PROBLEM 5.53

Situation: A tank with a depth M has one inow (] = 20 ft3/s) and one outowthrough a 1 ft diameter pipe. The outow velocity is

&2)M!

Find: Equilibrium depth of liquid.

APPROACH

Apply the continuity principle and the ow rate equation.

ANALYSIS


]in. = ]out at equilibrium

]out = 20 ft3$ s

Flow rate equation

]out = ,outIout

20 = (p2)M)(0$4× A2out) where A = 1 ft.

Solving for M yieldsM = 10!1 ft.

365

PROBLEM 5.54

Situation: Flows with di!erent specic weights enter a closed tank through ports Aand B and exit the tank through port C. Assume steady ow. Details are providedon gure with problem statement.

Find: At section C:(a) Mass ow rate.(b) Average velocity.(c) Specic gravity of the mixture.

Assumptions: Steady state.

APPROACH

Apply the continuity principle and the ow rate equation.

ANALYSIS


X/# "

X/! = 0

"#=,=I= " #>,>I> + #),)I) = 0

#),)I) = 0!95× 1!94× 3 + 0!85× 1!94× 1

/ = 7!18 slugs/s

Continuity principle, assuming incompressible ow

,)I) = ,=I= + ,>I>

= 3 + 1 = 4 cfs

Flow rate equation

,) = ]$I = 4$[0$4(1$2)2]

=20.4 ft/s

#) = 7!18$4 = 1!795 slugs/ft3

6 = 1!795$1!94

6 = 0!925

366

PROBLEM 5.55

Situation: O2 and CH4 enter a mixer, each with a velocity of 5 m/s. Mixer conditions:200 kPa-abs., 100 !-! Outlet density: # = 2!2 kg/m3! Flow areas: 1 cm2 for theCH4, 3 cm2 for the O2' and 3 cm2 for the exit mixture.

Find: Exit velocity of the gas mixture: ,exit

APPROACH

Apply the ideal gas law to nd inlet density. Then apply the continuity principle.

ANALYSIS

Ideal gas law

#02 = %$"&

= 200' 000$(260× 373)= 2!06 kg/m3

#)64 = 200' 000$(518× 373)= 1!03 kg/m3


X/# =

X/!

#e,eIe = #O2,O2IO2 + #CH4,CH4ICH4,e = (2!06× 5× 3 + 1!03× 5× 1)$(2!2× 3)

,e = 5!46 m/s

367

PROBLEM 5.56

Situation: A 10 m3 tank is lled with air from a compressor with mass ow rate/ = 0!5#!$# and initial density is 2 kg/m

3!

Find: Time to increased the density of the air in the tank by a factor of 2.

APPROACH


ANALYSIS


/! " /# = "A

AP

Z

)?

#A(

"A$AP(#() = " /#

((A#$AP) = 0!5#0$#

Separating variables and integrating

#A# = 0!5#0AP$(#2$2|(0 = 0!5#0!P$(

(#2( " #20)$2 = 0!5#0!P$(!P = (#0

¡(#2($#

20)" 1

¢

= 10(2)(22 " 1)!P = 60s

368

PROBLEM 5.57

Situation: A tire (volume 0.5 ft3) develops a slow leak. In 3 hr, the pressure dropsfrom 30 to 25 psig. The leak rate is / = 0!68%I$

&"& , where I is the area of the

hole. Tire volume and temperature (60 !F) remain constant. %'5" = 14 psia.

Find: Area of the leak.

APPROACH


ANALYSIS

Continuity principle/!T5 = "A$AP(#V–)

Ideal gas law# = %$"&

Combining previous 2 equations

/!T5 = "(V–$"& )(A%$AP)

Let /!T5 = 0!68I$&"& in the above equation

0!68%I$&"& = "(V–$"& )(A%$AP)


(1$%)(A%$AP) = "(0!68I&"& )$V–

LQ(%0$%) = (0!68I&"&P)$V–

Finding area

I = (V–$0!68P&"& )LQ(%0$%)

= (0!5$[(0!68× 3× 3' 600)p1' 716× 520]LQ(44$39)

I = 8!69× 10"9 ft2 = 1!25× 10"6 in2

369

PROBLEM 5.58

Situation: An O2 bottle (18 !C) leaks oxygen through a small orice (A = 0!15mm). As time progresses, the pressure drops from 10 to 5 MPa, abs. The leak rateis / = 0!68%I$

&"& , where I is the area of the orice.

Find: Time required for the specied pressure change.

APPROACH

Apply the continuity principle and the ideal gas law.

ANALYSIS

Continuity principle/!T5 = "A$AP(#V–)

Ideal gas law# = %$"&

Combining previous 2 equations

/!T5 = "(V–$"& )(A%$AP)

Let /!T5 = 0!68I$&"& in the above equation

0!68%I$&"& = "(V–$"& )(A%$AP)


(1$%)(A%$AP) = "(0!68I&"& )$V–

LQ(%0$%) = (0!68I&"&P)$V–

Finding time

P = (V–$0!68I&"& )LQ(%0$%)

= 0!1LQ(10$5)$(0!68(0$4)(1!5× 10"4)2&260× 291) = 21' 000 s

P = 5h 50 min!

370

PROBLEM 5.59

Situation: A 60-cm tank is draining through an orice. The water surface dropsfrom 3 to 0.3 m.

Find: Time required for the water surface to drop the specied distance (3 to 0.5 m).

ANALYSIS

From example 5-7 the time to decrease the elevation from M1 to M is

P = (2I-$p2)I2)(M

1*21 " M1*2)

= 2× (0$4× 0!62)(&3"

&0!53)$(

&2× 9!81×(0$4)× 0!032)

P = 185 s

371

PROBLEM 5.60

Situation: A cylindrical drum of water is emptying through a pipe on the bottom.

1 = 2 ft.' " = 1 ft.,, =p2)M;2 = 4 ft.

A = 2 in. = 0!167 ft.' M0 = 1 ft.

Find: Time to empty the drum.

APPROACH

Apply the continuity principle. Let the control surface surround the water in thetank. Let the c.s. be coincident with the moving water surface. Thus, the controlvolume will decrease in volume as the tank empties. Let y denote elevation, andsituate the origin at the bottom of the tank.

ANALYSIS


/! " /# = "A$APZ

cv#A(

+#, I = "A$APZ

cv#A( (1)

#p2)MI = "#A$AP(() (2)

APp2)MI = "A( (3)

Let A( = "2(2E)A?. Substituted into Eq. (3) we have

APp2)MI = 22EA? (4)

But M can be expressed as a function of ?:

M = "" ?

orAPp2)("" ?)I = 22EA?

Also

"2 = E2 + ?2

E =p?2 ""2 =

p(? "")(? +")

APp2)("" ?)I = 22

p(? "")(? +")A?

AP = (22$(p2)I))

p(? +")A? (5)

Integrate Eq. (5)

P|50 = (22$(p2)I))

Z F

0

p"+ ?A?

= (22$(p2)I))[(2$3)("+ ?)3*2]F0

P = (22$(p2)I))(2$3)((2")3*2 ""3*2)

372

For " = 1P = (22$(

p2)I))(2$3)(23*2 " 1) (6)

In Eq. (5) I = (0$4)A2 = 0!0219 ft2! Therefore

P = (2× 4$&64!4× 0!0219))(2$3)(1!828)

P = 55!5 s

COMMENTS The above solution assumes that the velocity of water is uniformacross the jet just as it leaves the tank. This is not exactly so, but the solutionshould yield a reasonable approximation.

373

PROBLEM 5.61

Situation: A pipe with discharge 0.03 ft3/s lls a funnel. Exit velocity from thefunnel is ,G =

&2)M' and exit diameter is 1 in. Funnel section area is I. = 0!1M2.

Find: Level in funnel at steady state: M

APPROACH


ANALYSIS

Continuity principle (steady state)

/#: = /!T5

or#] = #IG

p2)M

Solving for M gives

M =1

2)

µ]

IG

¶2

=1

2× 32!2

µ!03

0$4× (1$12)2

¶2

M = 0!47 ft

374

PROBLEM 5.62

Situation: Water drains from a pressurized tank. Tank section area: 1 m2.

Exit velocity: ,G =q

2,L+ 2)M. Exit area: 10 cm3!

Supply pressure: % = 10 kPa. Initial tank level: M! = 2 m.

Find: Time for the tank to empty(a) with given supply pressure.(b) if supply pressure is zero.

APPROACH

Apply the continuity principle. Dene a control surface coincident with the tankwalls and the top of the uid in the tank.

ANALYSIS


#A,

AP= "#IG,G

Density is constant. The di!erential volume is IAM so the above equation becomes

"IAM

IG,G= "AP

or

"IAM

IGq

2,L+ 2)M

= AP

Integrating this equation gives

"I

IG

1

)

µ2%

#+ 2)M

¶1*2|07!= !P

or

!P =I

IG

1

)

"µ2%

#+ 2)M!

¶1*2"µ2%

#

¶1*2#

and for I = 1 m2' IG = 10"3 m2' M! = 2 m, % = 10 kPa and # = 1000 kg/m3 resultsin

!P = 329 s or 5!48 min (supply pressure of 10 kPa)

For zero pressure in the tank, the time to empty is

!P =I

IG

s2M!)=639 s or

!P = 10!65 min (supply pressure of zero)

375

PROBLEM 5.63

Situation: A tapered tank drains through an orice at bottom of tank. The watervelocity in the orice is

&2)M. Dimensions of tank provided in the problem statement.

Find: (a) Derive a formula for the time to drain.(b) Calculate the time to drain.

APPROACH


ANALYSIS

From continuity principle

] = "I- (AM$AP)AP = "I-AM$]

where ] =&2)MIV =

&2)M(0$4)A2V

I- = (0$4)(A+ -1M)2 = (0$4)(A2 + 2A-1M+ -

21M

2)

AP = "(A2 + 2A-1M+ -1M2)AM$(p2)M1*2A2V)

P = "Z 7

70

(A2 + 2A-1M+ -21M

2)AM$(p2)M1*2A2V)

P = (1$(A2Vp2)))

Z 70

7

(A2M"1*2 + 2A-1M1*2 + -21M

3*2)AM

P = (2$(A2Vp2)))

£A2M1*2 + (2$3)A-1M

3*2 + (1$5)-21M5*2¤707

Evaluating the limits of integration gives

P = (2$(A2Vp2)))

h(A2(M

1*20 " M1*2) + (2$3)A-1(M

3*20 " M3*2) + (1$5)-21(M

5*20 " M5*2)

i

Then for M0 = 1 m' M = 0!20 m, A = 0!20 m, -1 = 0!3' and AV = 0!05 m

P = 13!6 s

376

PROBLEM 5.64

Situation: Water drains out of a trough and water velocity at bottom of trough is&2)M! Trough dimensions are provided in the problem statement.

Find: (a) Derive a formula for the time to drain to depth h.(b) Calculate the time to drain to 1/2 of the original depth.

APPROACH


ANALYSIS


/! " /# = "A$APZ

)4?4

#AV–

#p2)MIG = "A$AP

Z

)4?4

#AV–

Mass of water in control volume = #D×Face area

+ = #D(.0M+ M2 tanT)

Then

#p2)MIG = "A$AP #D(.0M+ M

2 tanT)p2)MIG = "D.0(AM$AP)" 2DM tanT(AM$AP)

AP = (1$(p2)IG))("D.0M

"1*2AM" 2D tanTM1*2AM)

Integrate

P = (1$p2)IG)

Z 7

70

"D.0M"1*2AM" 2D tanTM1*2AM

P = (1$(p2)IG))("2D.0M

1*2 " (4$3)D tanTM3*2)770

P = (&2DM

3*20 $(

&)IG))((.0$M0)(1" (M$M0)045) + (2$3) tanT(1" (M$M0)145))

For .0$M0 = 0!2' T = 30!' IG)

045$(M1450 D) = 0!01 sec."1 and h/h0 = 0!5 we get

P = 43!5 seconds

377

PROBLEM 5.65

Situation: Water drains out of a spherical tank. Tank diameter: 1 m. Hole diameter:1 cm.Exit velocity: ,G =

&2)M. At time zero, the tank is half full.

Find: Time required to empty the tank.

APPROACH

Apply the continuity principle. Select a control volume that is inside of the tank andlevel with the top of the liquid surface.

ANALYSIS


#A(AP= "#IG,G

LetA(AP=A(IM)

AP= I

AM

AP

Continuity becomesAM

AP= "

IGI

p2)M

The cross-sectional area in terms of " and M is

I = 0["2 " ("" M)2] = 0(2"M" M2)

Substituting into the di!erential equation gives

0("2"M+ M2)IG&2)M

AM = AP

or0

&2)IG

¡"2"M1*2 + M3*2

¢AM = AP

Integrating this equation results in

0&2)IG

µ"4

3"M3*2 +

2

5M5*2

¶|0F= !P

Substituting in the limits yields

0&2)IG

14

15"5*2 = !P

For " = 0!5 m and IG = 7!85× 10"5 m2, the time to empty the tank is

!P = 1491 s or 24!8 min

378

PROBLEM 5.66

Situation: A tank containing oil is described in the problem statement.

Find: Predict the depth of the oil with time for a one hour period.

ANALYSIS

The numerical solution provides the following results:

Time, min

0 10 20 30 40 50 60 70

Dep

th, m

1

2

3

4

5

6

379

PROBLEM 5.67

Situation: An end-burning rocket motor has chamber diameter of 10 cm and nozzleexit diameter of 8 cm. Propellant density is 1800 kg/m3 and regression rate is 1 cm/s.Pressure and temperature at exit plane are 10 kPa abs and 2000oC. Gas constant is415 J/kgK.

Find: Gas velocity at nozzle exit plane: ,G

APPROACH

Apply the continuity principle and the ideal gas law.

ANALYSIS

Ideal gas law

#G = %$"&

= 10' 000$(415× 2273) = 0!0106 kg/m3

The rate of mass decease of the solid propellant is #,I1 3 where #, is the propellantdensity, I1 is the chamber cross-sectional area and 3 is the regression rate. This isequal to the mass ow rate supplied to the chamber or across the control surface.From the continuity principle

,G = #,I1 3$ (#GIG)

= 0!01× 1' 750× (0$4× 0!12)$£0!0106×

¡0$4× 0!082

¢¤

,G = 2' 850 m/s

380

PROBLEM 5.68

Situation: An cylindrical-port rocket motor has internal diameter of 20 cm. Propel-lant with density of 2000 kg/m3 regresses at 1.2 cm/s. Inside propellant diameter is12 cm and length is 40 cm. Diameter of rocket exit is 20 cm and velocity is 2000 m/s.

Find: Gas density at the exit: #G

ANALYSIS

ID = 012+ 2(0$4)(120 "1

2)

= 0 × 0!12× 0!4 + (0$2)(0!22 " 0!122) = 0!191 m2

#G = ,D#DID$(,GIG) = 0!012× 2' 000× 0!191$(2' 000× (0$4)× (0!20)2)

#G = 0!073 kg/m3

381

PROBLEM 5.69

Situation: Mass ow through rocket nozzle is / = 0!65%1I5$&"&1 and regression

rate is 3 = C%:1 ! Operates at 3.5 MPa and Q = 0!3!

Find: (a) Derive a formula for chamber pressure.(b) Calculate the increase in chamber pressure if a crack increases burn area by 20%.

APPROACH


ANALYSIS

Continuity principle. The mass ux o! the propellant surface equals ow rate throughnozzle.

#, 3ID = /

#,C%:1ID = 0!65%1I5$

p"&1

%1":1 = (C#,$0!65)(ID$I5)("&1)1*2

%1 = (C#,$0!65)1*(1":)(ID$I5)

1*(1":)("&1)1*(2(1":))

!%1 = 3!5(1 + 0!20)1*(1"043)

!%1 = 4!54 MPa

382

PROBLEM 5.70

Situation: A piston moves in a cylinder and drives exhaust gas out an exhaust portwith mass ow rate / = 0!65%1I@$

&"&1! Bore is 10 cm and upward piston velocity

is 30 m/s. Distance between piston and head is 10 cm. Valve opening 1 cm2' pressure300 kPa abs, chamber temperature 600oC and gas constant 350 J/kgK.

Find: Rate at which the gas density is changing in the cylinder: A#$AP

Assumptions: The gas in the cylinder has a uniform density and pressure. Ideal gas.

ANALYSIS

Continuity equation. Control volume is dened by piston and cylinder.

A$AP(#, ) + 0!65%1I@$p"&1 = 0

V–A#$AP+ #AV–$AP+ 0!65%1I@$p"&1 = 0

A#$AP = "(#$V–) AV–$AP" 0!65%1I@$V–p"&1

V– = (0$4)(0!1)2(0!1) = 7!854× 10"4 m3

(AV–$AP) = "(0$4)(0!1)2(30) = "0!2356 m3/s# = %$"& = 300' 000$(350× 873)= 0!982 kg/m3

A#$AP = "(0!982$7!854× 10"4)× ("0!2356)

"0!65× 300' 000× 1× 10"4

7!854× 10"4 ×&350× 873

A#$AP = 250 kg/m3·s

383

PROBLEM 5.71

Situation: The ow pattern through a pipe contraction is described in the problemstatement. Discharge of water is 70 cfs and pressure at point A is 3500 psf.

Find: Pressure at point D!

APPROACH

Apply the Bernoulli equation and the continuity principle.

ANALYSIS


,= = ]$I= = 70$(0$4× 62) = 2!476 ft/s,> = ]$I> = 70$(0$4× 22) = 22!28 ft/s

Bernoulli equation

%=$( + ,2=$2) + R= = %>$( + ,

2>$2) + R>

%>$( = 3500$62!4" 2!482$64!4" 22!282$64!4" 4%> = 2775 lbf/ft2

%> = 19!2 lbf/in2

384

PROBLEM 5.72

Situation: The ow of water through a pipe contraction is described in the problemstatement. Velocity at point E is 50 ft/s and pressure and velocity at point C are 15psi and 10 ft/s.

Find: Pressure at point N!

APPROACH


ANALYSIS

Bernoulli equation Bernoulli equation applicable since ow steady, irrotational andnon-viscous.

%)$( + ,2)$(2)) + R) = %W$( + ,

2W$(2)) + RW

(15× 144)$( + 102$(2)) + R1 = %W$( + 502$(2)) + RW

%W$( = ((15× 144)$() + (1$2))(102 " 502) + R1 " RW%W = 15× 144 + (62!4$64!4)("2' 400)) + 62!4(3" 1)

= 2' 160 psf " 2' 325 psf+ 125 psf%W = "40 psf = "0!28 psi

385

PROBLEM 5.73

Situation: An annular venturimeter is mounted in a pipe with air ow at standardconditions. The pipe diameter is 4 in. and the ratio of the diameter of the cylindricalsection to the pipe is 0.8. A pressure di!erence of 2 in. of water is measured betweenthe pipe and cylindrical section. The ow is incompressible, inviscid and steady.

Find: Find the volume ow rate

APPROACH


ANALYSIS

Take point 1 as upstream in pipe and point 2 in annular section. The ow is incom-pressible, steady and inviscid so the Bernoulli equation applies

%1 + (R1 + #, 212= %2 + (R2 + #

, 222

Also R1 = R2! From the continuity equation

I1,1 = I2,2

ButI2 =

0

4(12 " A2)

so

I2I1

= 1"A2

12

= 1" 0!82

= 0!36

Therefore,2 =

,10!36

= 2!78,1

Substituting into the Bernoulli equation

%1 " %2 =#

2(, 22 " ,

21 )

=#

2, 21 (2!78

2 " 1)

= 3!36#, 21

The standard density is 0.00237 slug/ft3 and the pressure di!erence is

!% =2

1262!4

= 10!4 psf

386

Solving for ,1

, 21 =10!4

3!36× 0!00237= 1306

,1 = 36!14 ft/s

The discharge is

] = I1,1

= 36!14×0

4×µ4

12

¶2

= 3!15 cfs

Q=189.2 cfm

387

PROBLEM 5.74

Situation: A venturi-type applicator is used to spray liquid fertilizer. The exit-throatarea ratio is 2 and the exit diameter is 1 cm. Flow in venturi is 10 lpm. The entranceto the feed tube is 10 cm below venturi throat the level in the container is 5 cm abovethe entrance to the feed tube. The ow rate in the feed tube is 0.5

&!M in lpm and

!M is the di!erence in piezometric head in meters. The liquid fertilizer has samedensity as water.

Find: a) The ow rate of liquid fertilizer and b) the mixture ratio of fertilizer to waterat exit.

APPROACH

Use the continuity and Bernoulli equation to nd the pressure at the throat and usethis pressure to nd the di!erence in piezometric head and ow rate.

ANALYSIS

The Bernoulli equation is applicable between stations 1 (the throat) and 2 (the exit).

%1(+ R1 +

, 212)=%2(+ R2 +

, 222)

From the continuity equation

,1 =I2I1,2

= 2,2

Also R1 = R2 so

%1("%2(

=, 222)(1" 22)

= "3, 222)

At the exit %2 = 0 (gage)%1(= "3

, 222)

The ow rate is 10 lpm or

] = 10 lpm×1 min60 s

×10"3 m3

1 l= 0!166× 10"3 m3$s

The exit diameter is 1 cm so

I2 =0

40!012

= 7!85× 10"5 m2

388

The exit velocity is

,2 =]

I2=0!166× 10"3

7!85× 10"5= 2!12 m/s

Therefore

%1(

= "3×2!122

2× 9!81= "0!687 m

Let point 3 be the entrance to the feed tube. Then

!M = M3 " M1=

%3(+ R3 " (

%1(+ R1)

=%3("%1(+ (R3 " R1)

= 0!05" ("0!687)" 0!1= 0!637 m

a) The ow rate in the feed tube is

]( = 0!5&0!637

Q(=0.40 lpm

b) Concentration in the mixture

]2]2 +]%

=0!4

10 + 0!4

X%X%+X&

=0.038 (or 3.8%)

389

PROBLEM 5.75

Situation: Cavitation in a venturi section with inlet diameter of 40 cm and throatdiameter of 10 cm. Upstream pressure is 120 kPa gage and atmospheric pressure is100 kPa. Water temperature is 10oC.

Find: Discharge for incipient cavitation.

APPROACH

Apply the continuity principle and the Bernoulli equation.

ANALYSIS

Cavitation will occur when the pressure reaches the vapor pressure of the liquid(%? = 1' 230 Pa abs).Bernoulli equation

%= + #,2=$2 = %throat + #,

2throat$2

where ,= = ]$I= = ]$((0$4)× 0!402)Continuity principle

,throat = ]$Ithroat = ]$((0$4)× 0!102)#$2(, 2throat " ,

2=) = %= " %throat

(#]2$2)[1$((0$4)× 0!102)2 " 1$[((0$4)× 0!402)2]= 220' 000" 1' 230

500]2(16' 211" 63) = 218' 770

] = 0!165 m3/s

390

PROBLEM 5.76

Situation: Air with density 0.0644 lbm/ft3 ows upward in a vertical venturi witharea ratio of 0.5. Inlet velocity is 100 ft/s. Two pressure taps connected to manometerwith uid specic weight of 120 lbf/ft3.

Find: Deection of manometer.

Assumptions: Uniform air density.

APPROACH

Apply the Bernoulli equation from 1 to 2 and then the continuity principle. Letsection 1 be in the large duct where the manometer pipe is connected and section 2in the smaller duct at the level where the upper manometer pipe is connected.

ANALYSIS


,1I1 = ,2I2

,2 = ,1(I1$I2)

= 100(2)

= 200 ft/s

Bernoulli equation

%;1 + #,21 $2 = %;2 + #,

22 $2

%;1 " %;2 = (1$2)#(, 22 " ,21 )

= (1$2)(0!0644$32!2)(40' 000" 10' 000)= 30 psf

Manometer equation

%;1 " %;2 = !M((liquid " (air)30 = !M(120" !0644)

!M = 0!25 ft.

391

PROBLEM 5.77

Situation: An atomizer utilizing a constriction in an air duct is described in theproblem statement.

Find: Design an operable atomizer.

ANALYSIS

Assume the bottom of the tube through which water will be drawn is 5 in. below theneck of the atomizer. Therefore if the atomizer is to operate at all, the pressure inthe necked down portion must be low enough to draw water 5 in. up the tube. Inother words %neck must be "(5$12)(water = "26 psfg. Let the outlet diameter of theatomizer be 0.5 in. and the neck diameter be 0.25 in. Assume that the change inarea from neck to outlet is gradual enough to prevent separation so that the Bernoulliequation will be valid between these sections. Thus

%: + #,2: $2 = %0 + #,

20 $2

were Q and 0 refer to the neck and outlet sections respectively. But

%: = "26 psfg and %0 = 0

or"26 + #, 20 $2 = #,

20 $2 (1)

,:I: = ,0I0

,: = ,0I0$I: (2)

= ,0(!5$!25)2

,: = 4,0

Eliminate ,: between Eqs. (1) and (2)

"26 + #(4,0)2$2 = #, 20 $2

"26 + 16#, 20 $2 = #, 20 $2

15#, 20 $2 = 26

,0 = ((52$15)$#)1*2

Assume # = 0!0024 slugs/ft2

,0 = ((52$15)$0!0024)1*2

= 38 ft/s

] = , I = 38× (0$4)(!5$12)2

= !052 cfs

= 3!11 cfm

One could use a vacuum cleaner (one that you can hook the hose to the dischargeend) to provide the air source for such an atomizer.

392

PROBLEM 5.78

Situation: A suction device based on a venturi is described in the problem statement.Suction cup is 1 m below surface and venturi 1 m above. Throat area id 1/4 of exitarea and exit area is 0.001 m2! Cup area is 0.1 m2 and water temperature is 15oC.

Find: (a) Velocity of water at exit for maximum lift.(b) Discharge.(c) Maximum load supportable by suction cup.

Properties: From Table A.5 %@(15!) = 1' 700 Pa!From Table A.5 # = 999 kg/m3!

APPROACH


ANALYSIS

Venturi exit area, IG'= 10"3 m2' Venturi throat area, I5 = (1$4)IG' Suction cuparea, I& = 0!1 m2

%atm = 100 kPa

&water = 15! C

Bernoulli equation for the Venturi from the throat to exit with the pressure at thethroat equal to the vapor pressure of the water. This will establish the maximumlift condition. Cavitation would prevent any lower pressure from developing at thethroat.

%@$( + ,25 $2) + R5 = %G$( + ,

2Gmax$2) + RG (1)


,5I5 = ,GIG

,5 = ,G(IG$I5) (2)

,5 = 4,G

Then Eq. (1) can be written as

1' 700$( + (4,Gmax)2$2) = 100' 000$( + , 2Gmax$2)

,Gmax = ((1$15)(2)$()(98' 300))1*2

= ((1$15)(2$#)(98' 300))1*2

,Gmax = 3!62 m/s

]max = ,GIG

= (3!62 m/s)(10"3m2)

]max = 0!00362 m3/s

393

Find pressure in the suction cup at the level of the suction cup.

%5 + (!M = %suction

%suction = 1' 700 Pa+ 9,800× 2= 21' 300 Pa

But the pressure in the water surrounding the suction cup will be %atm + ( × 1 =(100 + 9!80) kPa, or

%water " %suction = (109' 800" 21' 300) Pa= 88' 500 Pa

Thus the maximum lift will be:

Liftmax = !%I& = (%water " %suction)I&= (88' 500 N/m2)(0!1 m2)

Liftmax = 8' 850 N

394

PROBLEM 5.79

Situation: A hovercraft is supported by air pressure.

Find: Air ow rate necessary to support the hovercraft.

APPROACH


ANALYSIS

The pressure di!erential necessary to support the hovercraft is

!%I = .P

!% = 2000 lbf/(15× 7) ft2

= 19!05 psfg

Bernoulli equation applied between the ow under the skirt (1) and chamber underthe hovercraft (2). Assume atmospheric pressure where ow exits under skirt. Alsoassume the air density corresponds to standard conditions.

%1 + #, 212

= %2 + #, 222

#, 212

= %2 " %1

,1 =

s2(%2 " %1)

#

=

s2× 19!05 psf

0!00233 slugs/ft3

= 127!9 ft/s

The discharge is

] = , I

= 127!9 ft/s× 44 ft× 0!25 ft= 1407 cfs

Q=84,400 cfm

395

PROBLEM 5.80

Situation: Water forced out of a cylinder by a piston travelling at 5 ft/s. Cylinderdiameter is 4 in and throat is 2 in.

Find: Force required to drive piston.

APPROACH


ANALYSIS


,1I1 = ,2I2

,2 = ,1(1$A)2 = 5× (4$2)2 = 20 ft/s

Bernoulli equation

%1$( + ,21 $2) = , 22 $2)

%1 =#

2(, 22 " ,

21 )

= 1!94× (202 " 520= 364 psf

Then

Hpiston = %1I1 = 364× (0$4)× (4$12)2

F=31.7 lbf

396

PROBLEM 5.81

Situation: A jet of water owing from a 0.5 ft diameter nozzle with discharge of 20cfs.

Find: Gage pressure in pipe.

APPROACH


ANALYSIS

Bernoulli equation

%1$( + ,21 $2) + R1 = %V$( + ,

2V $2) + RV

where 1 and d refer to conditions in pipe and jet, respectively

,1 = ]$I1

= 20$((0$4)× 1!02) = 25!5 ft/s,VIV = ,1I1;,V = ,1I1$IV

,V = 25!5× 4 = 102 ft/s

Also R1 = RV and %V = 0! Then

%1$( = (, 2V " ,21 )2)

%1 = ((, 2V " ,21 )$2)

= 62!4(1022 " 25!52)$64!4= 9' 451 psfg

%1 = 65!6 psig

397

PROBLEM 5.82

Situation: Airow past a sphere is described in problem 4.19 with \! = 30 m/s and# = 1!2 kg/m3.

Find: Pressure in the air at E = 3!' 1!13! and 23!!

APPROACH


ANALYSIS

Bernoulli equation

%0 + #,20 $2 = %$ + #,

2$ $2

where %0 = 0 gage. Then

%$ = (#$2)(, 20 " ,2$ )

,$ = @ = \0(1" 330$E3)

,$=K0 = \0(1" 1) = 0,$=K0 = \0(1" 1$1!13) = 7!46 m/s,$=2K0 = \0(1" 1$23) = 26!25 m/s

Finally

%$=K0 = (1!2$2)(303 " 0) = 540 Pa, gage

%$=141K0 = (1!2$2)(302 " 7!462) = 507 Pa, gage

%$=2K0 = (1!2$2)(302 " 26!252) = 127 Pa, gage

398

PROBLEM 5.83

Situation: An elbow meter is described in the problem statement where velocity variesas , = <$3.

Find: (a) Develop an equation for the discharge.(b) Evaluate the coe"cient b(31$32)!

ANALYSIS

, = <$3

] =

Z, AI =

Z, 2A3 = 2

Z(<$3)A3 = <2LQ(32$31) (1)

!% = (1$2)#(, 21 " ,22 )

!% = (1$2)#((<2$321)" (<2$322)) = (<

2#$2)((322)" (321))$(3

21322) (2)

Eliminate < between Eqs. (1) and (2) yielding:

(2!%$#) = ((]2)$(22(LQ(3231))2))(322 " 3

21)$(3

21322)

I1 = 2(32 " 31)! 2!%$# = (]2$I21)(32 " 31)

2(322 " 321)$(3

21322(LQ(32$31))

2)

] = I1p2!%$#(3132LQ(32$31))$((32 " 31)(322 " 3

21)045)

] = I1p2!%$#((32$31)LQ(32$31))$((32$31 " 1)((322$321)" 1)045)

For 32$31 = 1!5 the b(32$31) is evaluated

b(32$31) = 1!5LQ1!5$(0!5× 1!25045)

b(32$31) = 1!088

399

PROBLEM 5.84

Situation: A 1 ft diameter sphere moves at 10 ft below surface in water at 50oF.

Find: Speed at which cavitation occurs.

APPROACH

Apply the Bernoulli equation between the freestream and the maximum width.

ANALYSIS

Let %! be the pressure on the streamline upstream of the sphere. The minimumpressure will occur at the maximum width of the sphere where the velocity is 1.5times the free stream velocity.Bernoulli equation

%! +1

2#, 2! + (M! = %+

1

2#(1!5,!)

2 + ((M! + 0!5)

Solving for the pressure % gives

% = %! " 0!625#, 2! " 0!5(

The pressure at a depth of 10 ft is 624 lbf/ft2! The density of water is 1.94 slugs/ft3

and the specic weight is 62.4 lbf/ft3! At a temperature of 50!F, the vapor pressureis 0.178 psia or 25.6 psfa. Substituting into the above equation

25!6 psfa = 624 psfa" 0!625× 1!94× , 2! " 0!5× 62!4567!2 = 1!21, 2!

Solving for ,! gives

,! = 21!65 ft/s

400

PROBLEM 5.85

Situation: A hydrofoil is tested in water at 10oC. Minimum pressure on foil is 70 kPaabs when submerged 1.8 m and moving at 8 m/s.

Find: Speed that cavitation occurs.

Assumptions: %atm = 101 kPa abs; %vapor = 1' 230 Pa abs.

APPROACH

Consider a point ahead of the foil (at same depth as the foil) and the point of minimumpressure on the foil, and apply the pressure coe"cient denition between these twopoints.

ANALYSIS

Pressure coe"cient-, = (%min " %0)$(#, 20 $2)

where

%0 = %atm + 1!8( = 101' 000 + 1!8× 9' 810 = 118' 658 Pa abs.%min = 70' 000 Pa abs; ,0 = 8 m/s

Then-, = (70' 000" 118' 658)$(500× 82) = "1!521

Now use -, = "1!521 (constant) for evaluating , for cavitation where %min is now%vapor:

"1!521 = (1' 230" 118' 658)$((1' 000$2), 20 )

,0 = 12!4 m/s

401

PROBLEM 5.86

Situation: A hydrofoil is tested in water at 10oC. Minimum pressure on foil is 70 kPaabs when submerged 1.8 m and moving at 8 m/s.

Find: Speed that cavitation begins when depth is 3 m.

APPROACH


ANALYSIS

From the solution to Prob. 5.85, we have the same -,' but %0 = 101' 000 + 3( =130' 430. Then:

"1!521 = (1' 230" 130' 430)$((1' 000$2), 20 )

,0 = 14!37 m/s

402

PROBLEM 5.87

Situation: Hydrofoil is tested in water at 50oF. Minimum pressure on foil is 2.5 psivacuum when submerged 4 ft and moving at 20 ft/s.

Find: Speed that cavitation begins.

APPROACH

Consider a point ahead of the foil (at same depth as the foil) and the point of minimumpressure on the foil, and apply the pressure coe"cient denition between these two

ANALYSIS

%min = "2!5× 144 = "360 psf gage%0 = 4( = 4× 62!4 = 249!6 psf

Then

-, = (%min " %0)$(#, 20 $2) = ("360" 249!6)$((1!94$2)× 202)

-, = "1!571

Now let %min = %vapor = 0!178 psia = "14!52 psia = "2' 091 psfgThen

"1!571 = "(249!6 + 2' 091)$((1!94$2), 20 )

,0 = 39!2 ft/s

403

PROBLEM 5.88

Situation: Hydrofoil is tested in water at 50oF. Minimum pressure on foil is 2.5 psivacuum when submerged 4 ft and moving at 20 ft/s..

Find: Speed that cavitation begins when depth is 10 ft.

APPROACH


ANALYSIS

From solution of Prob. 5.87 we have -, = "1!571 but now %0 = 10( = 624 psf. Then:

"1!571 = "(624 + 2' 091)$((1!94$2), 20 )

,0 = 42!2 ft/s

404

PROBLEM 5.89

Situation: A sphere moving in water at depth where pressure is 18 psia. Maximumvelocity on sphere is 1.5 freestream velocity. Water density is 62.4 lbm/ft3 and tem-perature is 50oF.

Find: Speed at which cavitation occurs.

Properties: From Table A.5 %@(50!) = 0!178 psia.

APPROACH

Apply the Bernoulli equation between a point in the free stream to the 90! positionwhere , = 1!5,0. The free stream velocity is the same as the sphere velocity(reference velocities to sphere).

ANALYSIS

Bernoulli equation

#, 20 $2 + %0 = %+ #(1!5,0)2$2

where %0 = 18 psia

#, 20 (2!25" 1)$2 = (18" 0!178)(144), 20 = 2(17!8)(144)$((1!25)(1!94)) ft2/s2

,0 = 46!0 ft/sec

405

PROBLEM 5.90

Situation: Minimum pressure on cylinder moving 5 m/s horizontally in water at 10oCat depth of 1 m is 80 kPa abs. Atmospheric pressure is 100 kPa.

Find: Velocity at which cavitation occurs.

Properties: From Table A.5 %@(10!-) = 1' 230 Pa.

APPROACH

Apply the denition of pressure coe"cient.

ANALYSIS

Pressure coe"cient

-, = (%" %0)$(#, 20 $2)%0 = 100' 000 + 1× 9' 810 Pa = 109' 810 Pa% = 80' 000 Pa

Thus -, = "2!385

For cavitation to occur % = 1' 230 Pa

"2!385 = (1' 230" 109' 810)$(1' 000, 20 $2)

,0 = 9!54 m/s

406

PROBLEM 5.91

Situation: A velocity eld is dened by @ = , (E3 + E?2) and ; = , (?3 + ?E2).

Find: Is continuity satised?

APPROACH


ANALYSIS

Continuity equation

([@$[E) + ([;$[?) + ([Y$[R) = , (3E2 + ?2) + , (3?2 + E2) + 0

6= 0 Continuity is not satised

407

PROBLEM 5.92

Situation: A velocity eld is given as @ = ?$(E2 + ?2)3*2 and ; = "E$(E2 + ?2)3*2.

Find: (a) Check if continuity is satised.(b) Check if ow is rotational or irrotational

ANALYSIS

[@$[E+ [;$[? = "3E?$(E2 + ?2)5*2 + 3E?$(E2 + ?2)5*2

= 0 Continuity is satised

[@$[? " [;$[E = "3?2$(E2 + ?2)5*2 + 1$(E2 + ?2)3*2

= 3E2$(E2 + ?2)5*2 + 1$(E2 + ?2)3*2

6= 0 Flow is not irrotational

408

PROBLEM 5.93

Situation: A @-component of a velocity eld is @ = IE?!

Find: (a) What is a possible ;-component?(b) What must the ;-component be if the ow is irrotational?

ANALYSIS

@ = IE?

[@$[E+ [;$[? = 0

I? + [;$[? = 0

[;$[? = "I?

; = ("1$2)I?2 + -(E)

for irrotationality

[@$[? " [;$[E = 0

IE" [;$[E = 0

[;$[E = IE

; = 1$2IE2 + -(?)

If we let -(?) = "1$2I?2 then the equation will also satisfy continuity.

; = 1$2I(E2 " ?2)

409

PROBLEM 6.1

Situation: A balloon is held stationary by a force H!Data: A = 15 mm, ; = 50 m/s, # = 1!2 kg/m3

Find: Force required to hold balloon stationary: H

Assumptions: Steady ow, constant density.

APPROACH

Apply the momentum principle.

ANALYSIS

Force and momentum diagrams (x-direction terms)

Momentum principle (E-direction)

XH$ =

X

1&

/!;!$ "X

1&

/#;#$

H = /;

= #I;2

= (1!2)

µ0 × 0!0152

4

¶(502)

H = 0!53 8

410

PROBLEM 6.2

Situation: A balloon is held stationary by a force H!Pressure inside the balloon: % = 8 in.-H2O = 1990 PaA = 1 cm, # = 1!2 kg/m3

Find: (a)x-component of force required to hold balloon stationary: H(b)exit velocity: ;

Assumptions: Steady, irrotational, constant density ow.

APPROACH

To nd the exit velocity, apply the Bernoulli equation. To nd the force, apply themomentum principle.

ANALYSIS

Force and momentum diagrams (x-direction terms)

Bernoulli equation applied from inside the balloon to nozzle exit

%$# = ;2$2

; =p2%$# =

p2× 1990$1!2

; = 57!6 m$ s


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

H = /; = #I;2 = (1!2)¡0 × 0!012$4

¢(57!62)

H = 0!31N

411

PROBLEM 6.3

Situation: A water jet is lling a tank. The tank mass is 5 kg.The tank contains 20 liters of water.Data for the jet: A = 30 mm, ; = 15 m/s, & = 15 !C.

Find: (a) Force on the bottom of the tank: 8(b) Force acting on the stop block: H

Properties: Water—Table A.5: # = 999 kg$m3, ( = 9800N$m3.

Assumptions: Steady ow.

APPROACH

Apply the momentum principle in the x-direction and in the y-direction.

ANALYSIS

Force and momentum diagrams


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

H = "(" /; cos 70!)= #I;2 cos 70!

Calculations

#I;2 = (999)

µ0 × 0!032

4

¶(152)

= 158!9 N

H = (158!9 N) (cos 70!)

= 54!3N

H = 54!3N acting to right

412

?-directionX

H3 =X

1&

/!;!3 "X

1&

/#;#3

8 ". = "(" /; sin 70!)8 = . + #I;2 sin 70!

Calculations:

. = .tank +.water

= (5) (9!81) + (0!02)(9800)

= 245!1 N

8 = . + #I;2 sin 70!

= (245!1 N) + (158!9 N) sin 70!

8 = 149 8 acting upward

413

PROBLEM 6.4

Situation: Water jet is lling a tank. Friction acts on the bottom of the tank. Tankmass is 25 lbm; tank contains 5 gallons of water.Jet: A = 2 in., ; = 50 ft/s, & = 70 !F.

Find: Minimum coe"cient of friction (7) so force on stop block is zero.

Assumptions: Steady ow, constant density, steady and irrotational ow.

APPROACH

Apply the momentum principle in the x- and y-directions.

ANALYSIS


Momentum principle (?-direction)

XH3 =

X

1&

/!;!3 "X

1&

/#;#3

8 ". = "(" /; sin 70!)8 = . + #I;2 sin 70!


78 = "(" /; cos 70!) = #I;2 cos 70!

7 =(#I;2 cos 70!)

8

414

Calculations

#I;2 = (1!94)¡0 × (1$12)2

¢(502)

= 105!8 lbf

.620 = (V–

= (62!37)(5)$(7!481)

= 41!75 lbf

. = (41!75 + 25) lbf

= 66!7 lbf

8 = 66!7 + 105!8× sin 70! =166!2 lbf

7 =105!8× cos 70!

166!2

7 = 0!22

415

PROBLEM 6.5

Situation: A design contest features a submarine powered by a water jet.Speed of the sub is ,sub = 1!5m$ s.Inlet diameter is 11 = 25mm! Nozzle diameter is 12 = 5mm!Hydrodynamic drag force (H9) can be calculated using

H9 = -9

µ#, 2sub2

¶I,

Coe"cient of drag is -9 = 0!3! Projected area is I, = 0!28m2!

Find: Speed of the uid jet (,jet)!

Properties: Water—Table A.5: # = 999 kg$m3.

Assumptions: Assume steady ow so that the accumulation of momentum term iszero.

APPROACH

The speed of the uid jet can be found from the momentum principle because thedrag force will balance with the net rate of momentum outow.

ANALYSIS

Momentum equation. Select a control volume that surrounds the sub. Select areference frame located on the submarine. Let section 1 be the outlet (water jet)and section 2 be the inlet. The momentum equation is

XF =

X

1&

/!v! "X

1&

/#v#

HDrag = /2;2 " /1;1$

By continuity, /1 = /2 = #Ijet,jet! The outlet velocity is ;2 = ,jet! The x-component of the inlet velocity is ;1$ = ,sub! The momentum equation simpliesto

416

HDrag = #Ijet,jet(,jet " ,sub)

The drag force is

HDrag = -9

µ#, 2sub2

¶I,

= 0!3

Ã(999 kg$m3) (1!5m$ s)2

2

!¡0!28m2

¢

= 94!4N

The momentum equation becomes

HDrag = #Ijet,jet [,jet " ,sub]94!4N =

¡999 kg$m3

¢ ¡1!96× 10"5m2

¢,jet [,jet " (1!5m$ s)]

Solving for the jet speed gives,jet = 70!2m$ s

COMMENTS

1. The jet speed (70.2 m/s) is above 150 mph. This present a safety issue. Also,this would require a pump that can produce a large pressure rise.

2. It is recommended that the design be modied to produce a lower jet velocity.One way to accomplish this goal is to increase the diameter of the jet.

417

PROBLEM 6.6

Situation: Horizontal round jet strikes a plate.Water at 70!F' # = 1!94 slug/ft3, ] = 2 cfs.Horizontal component of force to hold plate stationary: H$ = 200 lbf

Find: Speed of water jet: ;1

APPROACH

Apply the momentum principle to a control volume surrounding the plate.

ANALYSIS



XH$ = " /;1$H$ = "(" /;1) = #];1

;1 =H$#]

=200

1!94× 2

;1 = 51!5 ft$ s

418

PROBLEM 6.7

Situation: Horizontal round jet strikes a plate.Water at 70 !F' # = 1!94 slug/ft3!Pressure at A is %= = 25 psig.Horizontal component of force to hold plate stationary: H$ = 500 lbf

Find: Diameter of jet: A

APPROACH

Apply the Bernoulli equation, then the momentum principle.

ANALYSIS


Bernoulli equation applied from inside of tank to nozzle exit

%=$# = ;21$2

;1 =

r2%=#

=

r2× 25× 144

1!94= 60!92 ft/s


419

XH$ = " /;1$H$ = "(" /;1) = #I;21

I =H$#;21

=500

1!94× 60!922

I = 0!0694 ft2

A =p4I$0

=p4× 0!0694$0

A = 0!30 ft

420

PROBLEM 6.8

Situation: An engineer is designing a toy to create a jet of water.Force H1 is the force needed to move the piston.Force H2 is the force to hold the handle stationary.Cylinder diameter is 1 = 80mm! Nozzle diameter is A = 15mm!Piston speed is ,piston = 300mm$ s!

Find: (a) Which force (H1 versus H2) is larger? Explain your answer using conceptsof the momentum principle.(b) Calculate H1!(c) Calculate H2!

Assumptions: 1.) Neglect friction between the piston and the wall. (2.) Assume theBernoulli equation applies (neglect viscous e!ects; neglect unsteady ow e!ects).

Properties: Table A.5 (water at 20 !C): # = 998 kg$m3!

APPROACH

To nd the larger force, recognize that the net force must be in the direction of accel-eration. To solve the problem, apply the momentum equation, continuity equation,equilibrium equation, and the Bernoulli equation.

ANALYSIS

Finding the larger force (H1 versus H2)! Since the uid is accelerating to the rightthe net force must act to the right. Thus, H1 is larger than H2. This can also beseen by application of the momentum equation.

Momentum equation (E-direction) applied to a control volume surrounding the toy.

XH$ = /;out

H1 " H2 = /;out

H1 " H2 = #

µ0A2

4

¶, 2out (1)

Notice that Eq. (1) shows that H1 F H2.

421

Continuity equation applied to a control volume situated inside the toy.

]in = ]outµ012

4

¶,piston =

µ0A2

4

¶,out

,out = ,piston12

A2

= (0!3m$ s)

µ80mm

15mm

¶2

,out = 8!533m$ s

Bernoulli equation applied from inside the toy to the nozzle exit plane.

%inside +#, 2piston2

=#, 2out2

%inside =#¡, 2out " , 2piston

¢

2

=(998 kg$m3)

¡(8!533m$ s)2 " (0!3m$ s)2

¢

2= 36!29 kPa

Equilibrium applied to the piston (the applied force H1 balances the pressure force).

H1 = %inside

µ012

4

¶

= (36290Pa)

Ã0 (0!08m)2

4

!

H1 = 182N

Momentum principle (Eq. 1)

H2 = H1 " #µ0A2

4

¶, 2out

= 182N"¡998 kg$m3

¢Ã0 (0!015m)2

4

!(8!533m$ s)2

H2 = 169N

COMMENTS

1. The force H1 is only slightly larger than H2.

2. The forces (H1 and H2) are each about 40 lbf. This magnitude of force may betoo large for users of a toy. Or, this magnitude of force may lead to materialfailure (it breaks!). It is recommended that the specications for this productbe modied.

422

PROBLEM 6.9

Situation: Water jet from a re hose on a boat.Diameter of jet is A = 3 in., speed of jet is , = 70 mph = 102.7 ft/s.

Find: Tension in cable: &

Properties: Table A.5 (water at 50 !F): # = 1!94 slug$ ft3!

APPROACH


ANALYSIS


Flow rate

/ = #I,

=¡1!94 slug$ ft3

¢ ¡0 × (1!5$12 ft)2

¢(102!7 ft$ s)

= 9!78 slug$ s


XH = / (;!)$

& = /, cos 60!

& = (9!78 slug$ s)(102!7 ft$ s) cos 60!

= 502! 2 lbf

& = 502 lbf

423

PROBLEM 6.10

Situation: Water jet (5 !C) from a re hose on a boat with velocity, ; = 50 m/s, anddensity, # = 1000 kg/m3!Allowable load on cable: & = 5!0 kN.

Find: (a) Mass ow rate of jet: /(b)Diameter of jet: A

APPROACH

Apply the momentum principle to nd the mass ow rate. Then, calculate diameterusing the ow rate equation.

ANALYSIS



XH = / (;!)$

& = /; cos 60!

/ = &$ (; cos 60!) = 5000$(50× cos 60!)

/ = 200 kg$ s

Flow rate

/ = #I; = #0A2;$4

A =

s4 /

#0;

=

r4× 200

1000× 0 × 50= 7! 136× 10"2m

A = 7!14 cm

424

PROBLEM 6.11

Situation: Water (60 !F) ows through a nozzle.A1 = 3 in, A2 = 1 in., %1 = 2000 psfg, %2 = 0 psfg

Find: (a) Speed at nozzle exit: ;2(b) Force to hold nozzle stationary: H

Assumptions: Neglect weight, steady ow.

APPROACH

Apply the continuity principle, then the Bernoulli equation, and nally the momen-tum principle.

ANALYSIS



I1;1 = I2;2

;1 = ;2

µA2A1

¶2(1)

Bernoulli equation applied from 1 to 2

%1#+;212=;222

(2)

Combining Eqs. (1) and (2)

%1 = #

µ;222

¶Ã1"

µA2A1

¶4!

2000 = 1!94×µ;222

¶×

Ã1"

µ1

3

¶4!

;2 = 45!69 ft/s

From Eq. (1)

;1 = ;2

µA2A1

¶2

= 45!69×µ1

3

¶2

= 5!077 ft/s

425

Flow rate

/1 = /2 = /

= (#I;)2

= 1!94×

Ã0

4×µ1!0

12

¶2!× 45!69

= 0!4835 slug/s


XH$ = / [(;!)$ " (;#)$]

H + %1I1 = / (;2 " ;1)H = "%1I1 + / (;2 " ;1)

H = "(2000 lbf$ ft2)×

Ã0

4×µ3

12

¶2!ft2

+(0!4835 slug$ s)× (45!69" 5!077) ft$ s= "78!5 lbf

Force on nozzle = 78.5 lbf to the left

426

PROBLEM 6.12

Situation: Water (15 !C) ows through a nozzle, # = 999 kg/m3!A1 = 10 cm., A2 = 2 cm., ;2 = 25 m/s.

Find: (a)Pressure at inlet: %1(b)Force to hold nozzle stationary: H

Assumptions: Neglect weight, steady ow, %2 = 0 kPa-gage.

APPROACH


ANALYSIS



I1;1 = I2;2

;1 = ;2 (A2$A1)2

= 25× (2$10)2

= 1!0 m/s

/1 = /2

= (#I;)2

= 999×µ0 × 0!022

4

¶× 25

= 7!85 kg/s

Bernoulli equation applied from 1 to 2

%1$#+ ;21$2 = ;22$2

%1 =³#2

´ ¡;22 " ;

21

¢

=

µ999

2

¶(252 " 12)

= 3! 117× 105 Pa

%1 = 312 kPa

427


XH$ = / [(;!)$ " (;#)$]

H + %1I1 = / (;2 " ;1)H = "%1I1 + / (;2 " ;1)

H = "(311!7× 103)µ0 × 0!12

4

¶+ (7!85) (25" 1)

= "2259! 7N

Force on nozzle = 2.26 kN to the left

428

PROBLEM 6.13 The problem involves writing a program for the ow in a nozzleand applying it to problems 6.12 and 6.14. No solution is provided.

429

PROBLEM 6.14

Situation:Pressurized air drives a water jet out of a tank. The thrust of the water jet reducesthe tension in a supporting cable.. = 200N (water plus the container). Tension in cable: & = 10N!Nozzle diameter (A = 12mm) ! 4 = 425mm.

Find: The pressure in the air that is situated above the water.

Assumptions: Assume that the Bernoulli equation can be applied (i.e. assume irro-tational and steady ow).

APPROACH

Apply the momentum equation to nd the exit velocity. Then, apply the Bernoulliequation to nd the pressure in the air.

ANALYSIS

Section area of jet

I2 =0A2

4

=0 (0!012m)2

4= 1! 131 × 10"4m2

Momentum equation (cv surrounding the tank; section 2 at the nozzle)

XF = /!v!

"& +. = /;2

("10 + 200) N = #I2;22

430

Solve for exit speed (;2)

190N =¡999 kg$m3

¢ ¡1!131× 10"4m2

¢;22

;2 = 41!01m$ s

Bernoulli equation (location 1 is on the water surface, location 2 is at the water jet).

%air +#;212+ #)R1 = %2 +

#;222+ #)R2

Let ;1 % 0' %2 = 0 gage and !R = 0!425m.

%air =#;222" #)!R

=(999 kg$m3) (41!01m$ s)2

2"¡999 kg$m3

¢(9!81m$ s2) (0!425m)

= (835' 900Pa)

µ1!0 atm

101!3 kPa

¶

%air = 8!25 atm

431

PROBLEM 6.15

Situation: Free water jet from upper tank to lower tank, lower tank supported byscales A and B.] = 2 cfs, A1 = 4 in., M = 1 ft, 4 = 9 ftWeight of tank: .- = 300 lbf, surface area of lower tank: 4 ft2

Find: (a) Force on scale A: H=(b) Force on scale B: H>

Properties: Water at 60 !F: # = 1!94 slug/ft3' ( = 62!37 lbf/ft3!

APPROACH


ANALYSIS


Flow rate

/ = #]

= 1!94× 2!0= 3!88 slug/s

;1 =]

I1=4]

012

=4× 2!0

0 × (4$12)2

= 22!9 ft/s

Projectile motion equations

432

;2$ = ;1 = 22!9 ft/s

;23 =p2)4

=&2× 32!2× 9

= 24!1 ft/s


XH$ = / [(;!)$ " (;#)$]

"H> = " / (;2$)"H> = "3!88× 22!9

H> = 88!9 lbf


XH3 = /

h(;!)3 " (;#)3

i

H= ".62B ".- = " / (;23)H= = .62B +.- " / (;23)

H= = (62!37× 4× 1) + 300" (3!88× ("24!1))

H= = 643!0 lbf

433

PROBLEM 6.16

Situation: Gravel (( = 120 lbf/ft3) ows into a barge that is secured with a hawser.] = 50 yd3$min = 22.5 ft3/s, ; = 10 ft/s

Find: Tension in hawser: &


APPROACH


ANALYSIS



XH$ = / (;!)$ " / (;#)$

"& = " /(; cos 20) = "(($))](; cos 20)& = (120$32!2)× 22!5× 10× cos(20) = 788 lbf

& = 788 lbf

434

PROBLEM 6.17

Situation: A xed vane in the horizontal plane; oil (6 = 0!9).;1 = 18 m/s, ;2 = 17 m/s, ] = 0!15 m3/s

Find: Components of force to hold vane stationary: H$' H3

APPROACH


ANALYSIS


Mass ow rate

/ = #]

= 0!9× 1000× 0!15= 135 kg/s


XH$ = / (;!)$ " / (;#)$

H$ = /(";2 cos 30)" /;1

H$ = "135(17 cos 30 + 18)

H$ = "4!42 kN (acts to the left)


435

XH3 = / (;!)3 " / (;#)3

H3 = / (";2 sin 30)= 135 ("17 sin 30)= "1!15 kN

H3 = "1!15 kN (acts downward)

436

PROBLEM 6.18

Situation: A xed vane in the horizontal plane; oil (6 = 0!9).;1 = 90 ft/s, ;2 = 85 ft/s, ] = 2!0 cfs

Find: Components of force to hold vane stationary: H$' H3

APPROACH


ANALYSIS


Mass ow rate

/ = #] = 0!9× 1!94× 2!0 = 3!49 slug/s


XH$ = / (;!)$ " / (;#)$

H$ = /(";2 cos 30)" /;1

H$ = "3!49(85 cos 30 + 90)

H$=-571 lbf (acts to the left)

?-direction

XH3 = / (;!)3 " / (;#)3

H3 = / (";2 sin 30) = 3!49 ("85 sin 30) = "148 lbf

H3 = "148 lbf (acts downward)

437

PROBLEM 6.19

Situation: A horizontal, two-dimensional water jet deected by a xed vane, # = 1!94slug/ft3!;1 = 40 ft/s, width of jets: Y2 = 0!2 ft, Y3 = 0!1 ft.

Find: Components of force, per foot of width, to hold the vane stationary: H$' H3

Assumptions: As the jet ows over the vane, (a) neglect elevation changes and (b)neglect viscous e!ects.

APPROACH

Apply the Bernoulli equation, the continuity principle, and nally the momentumprinciple.

ANALYSIS


Bernoulli equation

;1 = ;2 = ;3 = ; = 40 ft/s


Y1;1 = Y2;2 + Y3;3

Y1 = Y2 + Y3 = (0!2 + 0!1) = 0!3 ft


XH$ =

X/! (;!)$ " /# (;#)$

"H$ = /2; cos 60 + /3("; cos 30)" /1;

H$ = #;2("I2 cos 60 +I3 cos 30 +I1)H$ = 1!94× 402 × ("0!2 cos 60 + 0!1 cos 30 + 0!3)

H$ = 890 lbf/ft (acts to the left)


438

XH3 =

X/! (;!)3

H3 = /2; sin 60 + /3("; sin 30)= #;2(I2 sin 60"I3 sin 30)= 1!94× 402 × (0!2 sin 60" 0!1 sin 30)

H3 = 382 lbf/ft (acts upward)

439

PROBLEM 6.20

Situation: A water jet is deected by a xed vane, / = 25 lbm/s = 0.776 slug/s.;1 = 20 ft/s

Find: Force of the water on the vane: F

APPROACH

Apply the Bernoulli equation, and then the momentum principle.

ANALYSIS


Bernoulli equation;1 = ;2 = ; = 20 ft/s


XH$ = /! (;!)$ " /# (;#)$

"H$ = /; cos 30" /;

H$ = /;(1" cos 30) = 0!776× 20× (1" cos 30)H$ = 2!08 lbf to the left

?-directionX

H3 = /! (;!)3

"H3 = /("; cos 60) = "0!776× 20× sin 30H3 = 7!76 lbf downward

Since the forces acting on the vane represent a state of equilibrium, the force of wateron the vane is equal in magnitude & opposite in direction.

F = "H$i"H3j

= (2!08 lbf)i+(7!76 lbf)j

440

PROBLEM 6.21

Situation: A water jet strikes a block and the block is held in place by friction—however, we do not know if the frictional force is large enough to prevent the blockfrom sliding.;1 = 10 m/s, / = 1 kg/s, 7 = 0!1' mass of block: / = 1 kg

Find:(a) Will the block slip?(b) Force of the water jet on the block: F

Assumptions:1.) Neglect weight of water.2.) As the jet passes over the block (a) neglect elevation changes and (b) neglectviscous forces.

APPROACH


ANALYSIS


Bernoulli equation;1 = ;2 = ; = 10 m/s


XH$ = /! (;!)$ " /# (;#)$

"H( = /; cos 30" /;

H( = /;(1" cos 30)= 1!0× 10× (1" cos 30)

H( = 1!34 N

441

?-directionX

H3 = /! (;!)3

8 ". = /(; sin 30)

8 = /) + /(; sin 30)

= 1!0× 9!81 + 1!0× 10× sin 30= 14!81 N

Analyze friction:

• H( (required to prevent block from slipping) = 1!34 N

• H( (maximum possible value) = 78 = 0!1× 14!81 = 1!48 N

block will not slip

Equilibrium of forces acting on block gives

F = (Force of the water jet on the block)

= "(Force needed to hold the block stationary)= "H( i+ (. "8)j

So

F =(1!34N) i+("5!00N) j

442

PROBLEM 6.22

Situation: A water jet strikes a block and the block is held in place by friction,7 = 0!1!/ = 1 kg/s, mass of block: / = 1 kg

Find: Maximum velocity (;) such that the block will not slip.

Assumptions: Neglect weight of water.

APPROACH


ANALYSIS


Bernoulli equation;1 = ;2 = ;

Momentum principle (E-direction)X

H$ =X

1&

/!;!$ "X

1&

/#;#$

"78 = /; cos 30" /;

8 = /; (1" cos 30) $7

?-directionX

H3 =X

1&

/!;!3 "X

1&

/#;#3

8 ". = /(; sin 30)

8 = /) + /(; sin 30)

Combine previous two equations

/; (1" cos 30) $7 = /) + /(; sin 30)

; = /)$ [ / (1$7" cos 30$7" sin 30)]; = 1× 9!81$ [1× (1$0!1" cos 30$0!1" sin 30)]

; = 11!7m$ s

443

PROBLEM 6.23

Situation: A water jet strikes plate I and a portion of this jet passes through thesharp-edged orice at the center of the plate.; = 30 m/s, 1 = 5 cm, A = 2 cm

Find: Force required to hold plate stationary: H

Properties: # = 999 kg/m3

Assumptions: Neglect gravity.

APPROACH


ANALYSIS

Force and momentum diagrams (only x-direction vectors shown)


XF =

X

1&

/!v! "X

1&

/#v#

"H = /2; " /1;

H = #I1;2 " #I2;2

= #;2³04

´(12 " A2)

= 999× 302 ×0

4× (0!052 " 0!022)

H=1.48 kN (to the left)

444

PROBLEM 6.24

Situation: 2D liquid jet strikes a horizontal surface.;1 = ;2 = ;3 = ;

Find: Derive formulas for A2 and A3 as a function of >1 and K!

Assumptions: Force associated with shear stress is negligible; let the width of the jetin the z-direction = w.

APPROACH

Apply the continuity principle, then the momentum principle.Continuity principle

/1 = /2 + /3

#Y>1; = #YA2; + #YA3;

>1 = A2 + A3



XH$ =

X

1&

/!v! "X

1&

/#v#

0 = ( /3; + /2(";))" /1; cos K

0 =¡#YA3;

2 " #YA2;2¢" #Y>1;2 cos K

0 = A3 " A2 " >1 cos K

Combining x-momentum and continuity principle equations

A3 = A2 + >1 cos K

A3 = >1 " A2A2 = >1(1" cos K)$2

A3 = >1(1 + cos K)$2

445

PROBLEM 6.25

Situation: A 2D liquid jet impinges on a vertical wall.;1 = ;2 = ;

Find: (a) Calculate the force acting on the wall (per unit width of the jet): H$Y(b) Sketch and explain the shape of the liquid surface.

Assumptions: 1.) Steady ow. 2.) Force associated with shear stress is negligible.

APPROACH


ANALYSIS

Let w = the width of the jet in the z-direction. Force and momentum diagrams


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

"H = " /;1 sin 45!

H = #YP;2 sin 45!

The force on that acts on the wall is in the opposite direction to force pictured onthe force diagram, thus

H$Y = #P;2 sin 45! (acting to the right)

446

?-directionX

H3 =X

1&

/!;!3 "X

1&

/#;#3

". = / (";)" / (";) cos 45!

. = /;(1" cos 45!)

COMMENTS

Thus, weight provides the force needed to increase y-momentum ow. This weightis produced by the uid swirling up to form the shape show in the above sketches.

447

PROBLEM 6.26

Situation: A jet engine (ramjet) takes in air, adds fuel, and then exhausts the hotgases produced by combustion.At the inlet: ;1 = 225 m/sAt the exit: #2 = 0!25 kg/m

3 I2 = 0!5 m2

Find: Thrust force produced by the ramjet: &

Assumptions: 1.) Neglect the mass addition due to the fuel (that is, /in = /out =/ = 50 kg/s). 2.) Assume steady ow.

APPROACH


ANALYSIS


where H is the force required to hold the ramjet stationary.

Calculate exit velocity

/2 = #2I2;2

;2 = /2$(#2I2) = 50$(0!25× 0!5) = 400 m/s


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

H = /(;2 " ;1) = 50(400" 225)

& = 8!75 kN (to the left)

448

PROBLEM 6.27

Situation: A horizontal channel is described in the problem statement.

Find: Develop an expression for ?1!

APPROACH


ANALYSIS

Momentum principle (E-direction) (cs passes through sections 1 and 2)

XH$ = /;2

(%I)1 " (%I)2 = #];2

(D?21($2)" (D?22($2) = #](]$?2D)

?1=p?22 + (2$()?2))× (]$D)2

449

PROBLEM 6.28

Situation: An end section of a pipe has a slot cut in it—additional information isprovided in the problem statement.

Find: (a)How the pressure will change in the pipe from E = 0 to E = 2.(b) Devise a way to solve for the pressure distribution.

Assumptions: Neglect viscous resistance.

APPROACH

Apply the momentum principle and the continuity principle.

ANALYSIS

Obtain the pressure variation along the pipe by applying the momentum equation insteps along the pipe (numerical scheme). The rst step would be for the end segmentof the pipe. Then move up the pipe solving for the pressure change (!%) for eachsegment. Then %end +

P!% would give the pressure at a particular section. The

momentum equation for a general section is developed below.


"H$ =P1&

/!,!$ "P1&

/#,#$

%1I1 " %2I2 = #]2(]2$I2)" #]1(]1$I1)but I1 = I2 = I so we get

%1 " %2 = (#$I2)(]22 "]21) (1)

As section 1 approaches section 2 in the limit we have the di!erential form

"A% = (#$I2)A]2 = 2(#$I2)]A]


]1 "]2 = !?p2%$#!E

]1 = ]2 +!?p2%$#AE

450

In the limit at !E$ 0 we have

A] = "!?p2%$#AE

The di!erential equation for pressure becomes

A% = 2(#$I2)A]2 = 2(#$I2)]!?p2%$#AE

Integrating the momentum equation to evaluate ] at location E we have

] = "!?Z $p

2%$#AW

so the equation for pressure distribution is

% |!E0 = (4$I2)!?2Z !E

0

%1*2·Z $

0

%1*2AW

¸AE

where 2 is some distance along the pipe.

COMMENTS

This equation has to be integrated numerically. One can start at the end of the pipewhere the pressure is known (atmospheric pressure). The one can assume a linearpressure prole over the interval !2! An iterative solution would be needed for eachstep to select the slope of the pressure curve (pressure gradient). The pressure willdecrease in the direction of ow.

451

PROBLEM 6.29

Situation: A cone is supported by a vertical jet of water.Weight of the cone is. = 30N! Speed of the water jet as it emerges from the oriceis ,1 = 15m$ s!Jet diameter at the exit of the orice is A1 = 2cm!

Find: Height to which cone will rise: M!

Assumptions: Based on application of the Bernoulli equation, assume that the speedof the uid as it passes by the cone is constant (,2 = ,3) !

APPROACH

Apply the Bernoulli equation and the momentum principle.

ANALYSIS

c.s.

V1

1

2

3

60 o

Bernoulli equation

, 212)+ 0 =

, 222)+ M

, 22 = (15)2 " 2)M, 22 = (15)2 " 2)M = 225" 2× 9!81M, 22 = 225" 19!62M

Momentum principle (?-direction). Select a control volume surrounding the cone.

XH3 = /!;!3 " /#;#3

". = /(;33 " ;2)"30 = 1000× 15× 0 × (0!01)2(,2 sin 30! " ,2)

Solve for the ,2,2 = 12!73 m/s

452

Complete the Bernoulli equation calculation

, 22 = 225" 19!62M(12!73)2 = 225" 19!62M

M = 3!21m

453

PROBLEM 6.30

Situation: A 180! pipe bend (6 in. diameter) carries water.] = 6 cfs % = 20 psi gage

Find: Force needed to hold the bend in place: H$ (the component of force in thedirection parallel to the inlet ow)

APPROACH


Assumptions: The weight acts perpendicular to the ow direction; the pressure isconstant throughout the bend.

ANALYSIS


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

2%I" H$ = "2 /;

Calculations

%I = (20× 144)¡0$4× 0!52

¢= 565!5 lbf

/; = #]2$I = 1!94× 62$(0$4× 0!52) = 355!7 lbfH$ = 2(%I+ /;) = 2× (565!5 + 355!7) lbf

H$=1840 lbf (acting to the left, opposite of inlet ow)

454

PROBLEM 6.31

Situation: Hot gas ows through a return bend–additional details are provided inthe problem statement.

Find: Force required to hold the bend in place: H$

APPROACH

Apply the continuity principle, then the momentum principle.

ANALYSIS

100 ft/s

x

2

1

/ = 1 lbm/s = 0!0311 slugs/s

At section (1):

;1 = 100 ft/s

#1 = 0!02 lbm/ft3 = 0!000621 slugs/ft3

At section (2):#2 = 0!06 lbm/ft

3 = 0!000186 slugs/ft3


#1;1I1 = #2;2I2

;2 = (#1$#2)(I1$I2);1

;2 = (0!02$0!06)(1$1);1

= 33!33 ft/s


XH$ =

X

$&

/!;!' "X

1&

/#;#'

= /(;2 " ;1)H$ = 0!0311("33!33" 100)

H$ = "4!147 lbf

455

PROBLEM 6.32

Situation: Fluid (density #, discharge ], and velocity , ) ows through a 180! pipebend–additional details are provided in the problem statement.. Cross sectionalarea of pipe is I.

Find: Magnitude of force required at anges to hold the bend in place.

Assumptions: Gage pressure is same at sections 1 and 2. Neglect gravity.

APPROACH


ANALYSIS


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

%1I1 + %2I2 + H$ = /(;2 " ;1)

thus

H$ = "2%I" 2 /,H$ = "2%I" 2#],

Correct choice is (d)

456

PROBLEM 6.33

Situation: Water ows through a 180! pipe bend–additional details are provided inthe problem statement.

Find: External force required to hold bend in place.

APPROACH


ANALYSIS

Flow rate equation

; = ]$I = 20$(0 × 0!5× 0!5) = 25!5 fps


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

%1I1 + %2I2 + H$ = /(;2 " ;1)

thus

H$ = "2%I" 2 /;= "2(15× 144(0$4× 12) + 1!94× 20× 25!5)= "5' 370 lbf


XH3 = 0

".bend ".620 + H3 = 0

H3 = 200 + 3× 62!4 = 387!2 lbf

Force requiredF = ("5370i+ 387j) lbf

457

PROBLEM 6.34


Find: Force that acts on the anges to hold the bend in place.

APPROACH

Apply the continuity and momentum equations.

ANALYSIS

Flow rate

;1 =]

I

=4× 0!3m3$ s0 × (0!2m)2

= 9!549 m/s

Continuity. Place a control volume around the pipe bend. Let section 2 be the exitand section 1 be the inlet

] = I1;1 = I2;2

thus ;1 = ;2

Momentum principle (E-direction). Place a control volume around the pipe bend.Let section 2 be the exit and section 1 be the inlet.

XH$ =

X

1&

/!;!$ "X

1&

/#;#$

2%I+ H$ = #] (";2)" %];1H$ = "2%I" 2#];

Calculations

2%I = (2)(100' 000)(0

4)(0!22)

= 6283N

2#], = (2)(1000)(0!3)(9!55)

= 5730N

H$ = " (2%I+ 2#];)= " (6283N + 5730N)= "12!01 kN

Momentum principle (R-direction). There are no momentum ow terms so the mo-mentum equation simplies to

458

H; = .bend +.water

= 500 + (0!1)(9810)

= 1!481 kN

The force that acts on the anges is

F = ("12!0i+ 0j+1!48k) kN

459

PROBLEM 6.35

Situation: A 90! pipe bend is described in the problem statement.

Find: Force on the upstream ange to hold the bend in place.

APPROACH


ANALYSIS

Velocity calculation

; = ]$I = 10$((0$4× 1!02) = 12!73 ft/s


XH$ =

X

1&

/!;!$ "X

1&

/#;#$

%I+ H$ = #](0" ;)H$ = 1!94× 10(0" 12!73)" 4× 144× 0$4× 12 = "699 lbf

?-direction

H3 = #]("; " 0)H3 = "1!94× 10× 12!73 = "247 lbf

R-directionX

H; = 0

"100" 4× 62!4 + H; = 0H; = +350 lbf

The force isF = ("699i " 247j + 350k) lbf

460

PROBLEM 6.36

Situation: A 900 pipe bend is described in the problem statement.

Find: E"component of force applied to bend to hold it in place: H$

APPROACH


ANALYSIS


; = ]$I = 10$(0 × 12$4) = 12!73 m/s


XH$ =

X

1&

/;!$ "X

1&

/;#$

%I+ H$ = #](0" ;)

300' 000× 0 × 0!52 + H$ = 1000× 10× (0" 12!73)H$ = "362' 919 N = -363 kN

461

PROBLEM 6.37


Find: Vertical component of force exerted by the anchor on the bend: H'

APPROACH


ANALYSIS


; = ]$I

= 31!4$(0 × 1× 1)= 9!995 ft/sec


XH3 = #](;23 " ;13)

H' ".water ".bend " %2I2 sin 30! = #](; sin 30! " ; sin 0!)H' = 0 × 1× 1× 4× 62!4 + 300

+8!5× 144× 0 × 1× 1× 0!5+1!94× 31!4× (9!995× 0!5" 0)

H' = 3310 lbf

462

PROBLEM 6.38

Situation: Water ows through a 60! pipe bend and jets out to atmosphere–additionaldetails are provided in the problem statement.

Find: Magnitude and direction of external force components to hold bend in place.

APPROACH


ANALYSIS

Flow rate equation

;1 = 10$4 = 2!5 m/s

] = I1;1 = 0 × 0!3× 0!3× 2!5 = 0!707 m3$s

Bernoulli equation

%1 = %2 + (#$2)(;22 " ;

21)

= 0 + (1000$2)(10× 10" 2!5× 2!5)= 46' 875 Pa


H$ + %1I1 = #](";2 cos 60! " ;1)H$ = "46' 875× 0 × 0!3× 0!3 + 1000× 0!707× ("10 cos 60! " 2!5)

= "18' 560 N

?-direction

H3 = #](";2 sin 60! " ;1)H3 = 1000× 0!707× ("10 sin 60! " 0)

= "6123 N

R-direction

H; ".H20 ".bend = 0

H; = (0!25× 9' 810) + (250× 9!81) = 4' 905 N

Net forceF = ("18!6i" 6!12j+ 4!91k) kN

463

PROBLEM 6.39

Situation: Water ows through a nozzle–additional details are provided in the prob-lem statement.

Find: Vertical force applied to the nozzle at the ange: H3

APPROACH

Apply the continuity principle, then the Bernoulli equation, and then the momentumprinciple.

ANALYSIS


;1I1 = ;2I2

;1 = ;2I2$I1 = 65 ft/s

] = ;2I2 = (130 ft/s)(0!5 ft2)

= 65 cfs

Bernoulli equation

%1$( + ;21$2) + R1 = %2$( + ;

22$2) + R2

%1$( = 0 + (1302$2)) + 2" (652$2))%1 = 62!4(262!4 + 2" 65!6)%1 = 12' 400 lbf/ft2


%1I1 ".620 ".nozzle + H3 = #](;2 sin 30! " ;1) (1)

Momentum ow terms

#](;2 sin 30! " ;1) = (1!94)(62!5) [(130 sin 30!)" 65]

= 0 lbf

Thus, Eq. (1) becomes

H3 = .620 +.nozzle " %1I1= (1!8× 62!4) + (100)" (12400× 1)= "12' 190 lbf

H3 = 12' 200 lbf (acting downward)

464

PROBLEM 6.40

Situation: Gasoline ows through a 135! pipe bend–additional details are providedin the problem statement.

Find: External force required to hold the bend: H

APPROACH


ANALYSIS

1

2

45o

y

x

Flow rate

] = ;I = 15× 0$4× 12

= 11!78 cfs


H$ = #](;2$ " ;1$)

%1I1 + %2I2 cos 45! + H$ = #](";2 cos 45! " ;1)

H$ = "%I(1 + cos 45!)" #];(1 + cos 45!)= "(1440)× (0$4× 12)(1 + cos 45!)

"(0!8× 1!94)(11!78)(15)(1 + cos 45!)= "2400 lbf

Momentum principle (?-direction)X

H3 = #](;23 " ;13)

%2I2 sin 45! + H3 = #](";2 sin 45! " 0)

H3 = "%I sin 45! " #];2 sin 45!

H3 = "(1440)(0$4× 12) sin 45! " (0!8× 1!94)(11!78)(15) sin 45!

H3 = "994 lbf

Net forceF = ("2400i" 994j) lbf

465

PROBLEM 6.41

Situation: Gasoline ows through a 135! pipe bend–additional details are providedin the problem statement.

Find: External force required to hold the bend in place: H

APPROACH


ANALYSIS

Discharge

] = 8× 0$4× 0!15× 0!15= 0!141 m3/s


XH$ = /(;2$ " ;1$)

%1I1 + %2I2 cos 45! + H$ = #](";2 cos 45! " ;1)

H$ = "%I(1 + cos 450)" #];(1 + cos 45!)= "(100' 000)(0$4× 0!152)(1 + cos 45!)

"(1000× 0!8)(0!141)(8)(1 + cos 45!)= "4557 N

Momentum principle ?-direction

XH3 = #](;23 " ;13)

%2I2 sin 45! + H3 = "#];2 sin 45!

= "(100' 000)(0$4× 0!152) sin 45!

"(1' 000× 0!8)(0!141)(8) sin 45!

= "1' 888 N

Net forceF = ("4!56i" 1!89j) kN

466

PROBLEM 6.42

Situation: Water ows through a 60! reducing bend–additional details are providedin the problem statement.

Find: Horizontal force required to hold bend in place: H$

APPROACH


ANALYSIS

Bernoulli equation

;1 = ;2I2$I1

= 50(1$10)

= 5 m/s

%1 + #;21$2 = %2 + #;

22$2

Let %2 = 0, then

%1 = "(1' 000)$2)(52) + (1' 000$2)(502)%1 = 1237!5 kPa


XH$ = /(;2$ " ;1$)

%1I1 + H$ = #I2;2(;2 cos 60! " ;1)

H$ = "1' 237' 000× 0!001 + 1' 000× 0!0001× 50(50 cos 60! " 5)H$ = 1140 N

467

PROBLEM 6.43

Situation: Water ows through a three dimensional pipe bend–additional details areprovided in the problem statement.

Find: Force that the thrust block exerts on the bend: F

APPROACH

Apply the momentum principle in each coordinate direction (x, y and z). To keeptrack of directions (this is a problem in three dimensions), use unit vectors to representthe velocity and pressure terms.

ANALYSIS

2

V1

V2

p A2 2

p A1 1

1

Flow speed

, =]

I

=4× 16m3$ s0 (1!3m)2

= 12!05m$ s

Inlet velocity vectors (written using direction cosines)

v1 = ,1[(13$L1)j" (10$L1)k]

where L1 =&132 + 102! Thus

v1 = (12!05m$ s) [0!793j" 0!6097k]

Exit velocity vector (written using direction cosines)

v2 = ,2[(13$L2)i+ (19$L2)j" (20$L2)k]

where L2 =&132 + 192 + 202! Then

v2 = (12!05m$ s) [0!426i+ 0!623j" 0!656k]

468

Pressure forces (written using direction cosines)

F,1 = %1I1(0!793j" 0!6097k)F,2 = %2I2("0!426i" 0!623j+ 0!656k)

Weight

W = (.water +.metal)k

= (("3× 9810)" 10000)k= ("39 430N)k

Momentum equation (E-direction)X

H$ = #](;2$ " ;1$)

H$ " 0!426× %2I2 = #][(12!05m$ s) (0!426)" 0]

where

%2I2 = 25' 000× (0$4)× (1!3)2

= 33' 183N

#] = 1000× 16= 16' 000 kg$ s

Thus

H$ = (%2I2) (0!426) + (#]) (12!05m$ s) (0!426)

= (33' 183N) (0!426) + (16' 000 kg$ s) (12!05m$ s) (0!426)

= 96' 270N

Momentum equation (?-direction)X

H3 = #](;23 " ;13)

H3 + %1I1 (0!793)" %2I2 (0!623) = #][0!623,2 " 0!793,1]

where

%1I1 = 20' 000× (0$4)(1!3)2

= 26' 546 N

#][0!623,2 " 0!793,1] = 16' 000 [(0!623) (12!05)" (0!793) (12!05)]= "32' 780N

Thus

H3 = "%1I1 (0!793) + %2I2 (0!623) + #][,2 (0!623)" ,1 (0!793)]= " (26' 546N) (0!793) + (33' 183N) (0!623)" (32' 780N)= "33' 160N

469

Momentum equation (R-direction)

XH; = #](;2; " ;1;)

H; " %1I1 (0!6097) + %2I2(0!656)". = #] [,2 ("0!656)" ,1 ("0!6097)]

Evaluate the momentum ow terms

#] [,2 (0!656)" ,1 ("0!6097)] = 16' 000 [12!05 ("0!656)" 12!05 ("0!6097)]= "8927N

The momentum equation becomes

H; = %1I1 (0!6097)" %2I2(0!656) +. + #] [(,2 (0!656)" ,1 ("0!6097)]H; = (26' 546N) (0!6097)" (33' 183N) (0!656) + (39 430N)" (8927N)

H; = 24' 920N

Net forceF = (96!3i" 33!2j+ 24!9k) kN

470

PROBLEM 6.44

Situation: Water ows through a tee–—additional details are provided in the problemstatement.

Find: Pressure di!erence between sections 1 and 2.

APPROACH


1 2

3 500 kg/s

ANALYSIS


/1 + 500 kg/s = /2

/1 = (10 m/s)(0!10 m2)(1000 kg/m3) = 1000 kg/s

/2 = 1000 + 500 = 1500 kg/s

;2 = ( /2)$(#I2) = (1500)$((1000)(0!1)) = 15 m/s


XH$ = /2;2$ " /1;1$ " /3;3$

%1I1 + %2I2 = /2;2 " /1;1 " 0I(%1 " %2) = (1500)(15)" (1000)(10)%1 " %2 = (22' 500" 10' 000)$0!10

= 125' 000 Pa

= 125 kPa

471

PROBLEM 6.45

Situation: Water ows through a wye–additional details are provided in the problemstatement.

Find: E"component of force to hold wye in place.

APPROACH


1 2

330o

x

Flow rate

;1 = ]1$I1 = 20 ft/s

;2 = ]2$I2 = 12 ft/s

]3 = 20" 12 = 8 ft3/s;3 = ]3$I3 = 32 ft/s


XH$ = /2;2 + /3;3 cos 30

! " /1;1

H$ + %1I1 " %2I2 = (20#)("20) + (12#)(+12) + (32 cos 30!)(#)(8)H$ + (1000)(1)" (900)(1) = "400#+ 144#+ #(8)(32)(0!866)

H$ = "100 + 1!94("34!3)

H$ = "166!5 lbf (acting to the left)

472

PROBLEM 6.46

Situation: Water ow through a horizontal bend and T section–additional detailsare provided in the problem statement.

y

x

1

2

3

/1 = 10 lbm/s

/2 = /3 = 5 lbm/s

I1 = I2 = I3 = 5 in2

%1 = 5 psig

%2 = %3 = 0

Find: Horizontal component of force to hold tting stationary: H$

APPROACH


ANALYSIS

Velocity calculations

;1 = /1$#I1

= (10$32!2)$ [(1!94)(5$144)]

= 4!61 ft/s

;2 = /2$#I2

= (5$32!2)$ [(1!94)(5$144)]

= 2!31 ft/s


XH$ = " /2;2 " /1;1

%1I1 + H$ = " /2;2 " /1;1

H$ = "%1I1 " /2;2 " /1;1

= "(5× 5)" (5$32!2)(2!31)" (10$32!2)(4!61)

H$ = "26!8 lbf

473

PROBLEM 6.47

Situation: Water ows through a horizontal bend and T section–additional detailsare provided in the problem statement.

y

x

1

2

3

;1 = 6 m/s %1 = 4!8 kPa

;2 = ;3 = 3 m/s %2 = %3 = 0

I1 = I2 = I3 = 0!20 m2

Find: Components of force (H$' H3) needed to hold bend stationary.

APPROACH


ANALYSIS

Discharge

]1 = I1;1 = 0!2× 6 = 1!2 m3/s]2 = ]3 = I2;2 = 0!2× 3 = 0!6 m3/s


XH$ = " /2;2 " /1;1

%1I1 + H$ = "#(]2;2 +]1;1)H$ = "%1I1 " #(]2;2 +]1;1)

= "4800× 0!2" 1000(0!6× 3 + 1!2× 6)


?-directionX

H3 = /3(";3)H3 = "#]3;3 = "1000× 0!6× 3

H3=-1.8 kN (acts downward)

474

PROBLEM 6.48

Situation: Water ows through a horizontal tee–additional details are provided inthe problem statement.

Find: Components of force (H$' H3) needed to hold the tee in place.

APPROACH


ANALYSIS

Velocity calculations

,1 =0!25

(0 × 0!075× 0!075)= 14!15 m/s

,2 =0!10

(0 × 0!035× 0!035)= 25!98 m/s

,3 =(0!25" 0!10)

(0 × 0!075× 0!075)= 8!49 m/s

Momentum equation (E-direction)

H$ + %1I1 " %3I3 = /3,3 " /1,1

H$ = "%1I1 + %3I3 + #,3]" #,1]H$ = " (100' 000× 0 × 0!075× 0!075) + (80' 000× 0 × 0!075× 0!075)

+ (1000× 8!49× 0!15)" (1000× 14!15× 0!25)H$ = "2617N

Momentum equation ?-direction

H3 + %3I3 = "#,3]H3 = "#,3]" %3I3H3 = "1000× 25!98× 0!10" 70' 000× 0 × 0!035× 0!035

= "2867 N

Net forceF = ("2!62i" 2!87j) kN

475

PROBLEM 6.49

Situation: Water ows through an unusual nozzle–additional details are provided inthe problem statement.

y

x

Find: Force at the ange to hold the nozzle in place: F

APPROACH


APPROACH


ANALYSIS


;,I, =X

;VIV

;, = 2× 30× 0!01$0!10= 6!00 m/s

Bernoulli equation%pipe$( + ;

2,$2) = %jet$( + ;

2V$2)

Then

%, = (($2))(;2V " ;2,)

= 500(900" 36)= 432' 000 Pa


%,I, + H$ = ";,#;,I, + ;V#;VIVH$ = "1000× 62 × 0!10 + 1' 000× 302 × 0!01" 432' 000× 0!1H$ = "37' 800 N

?-direction

H3 = /(";V) = ";V#;VI= "30× 1000× 30× 0!01= "9000 N

476

R-directionX

H; = 0

"200" (V–+ H; = 0

H; = 200 + 9810× 0!1× 0!4= 592 N

Net forceF = ("37!8i" 9!0j+ 0!59k) kN

477

PROBLEM 6.50

Situation: Water ows through a converging nozzle–additional details are providedin the problem statement.

1 2

xv1

v2

Find: Force at the ange to hold the nozzle in place: H

APPROACH

Apply the Bernoulli equation to establish the pressure at section 1, and then applythe momentum principle to nd the force at the ange.

ANALYSIS

Continuity equation (select a control volume that surrounds the nozzle).

]1 = ]2 = ] = 15 ft3$ s

Flow rate equations

;1 =]

I1=4×]012

1

=4×

¡15 ft3$ s

¢

0 (1 ft)2

= 19!099 ft$ s

;2 =]

I2=4×]012

2

=4×

¡15 ft3$ s

¢

0 (9$12 ft)2

= 33! 953 ft$ s

Bernoulli equation

%1 +#;212

= %2 +#;222

%1 = 0 +#(;22 " ;21)

2

=1!94 slug$ ft3(33! 9532 " 19!0992) ft2$ s2

2= 764!4 lbf$ ft2


%1I1 + H = /;2 " /;1

478

Calculations

%1I1 = (764!4 lbf$ ft2)(0$4)(1 ft)2

= 600!4 lbf

/;2 " /;1 = #] (;2 " ;1)= (1!94 slug$ ft3)(15 ft3$ s) (33! 953" 19!098) ft$ s= 432! 3 lbf

Substituting numerical values into the momentum equation

H = "%1I1 + ( /;2 " /;1)

= "600!4 lbf + 432! 3 lbf= "168! 1 lbf

H = "168 lbf (acts to left)

479

PROBLEM 6.51

Situation: Water ows through a converging nozzle–additional details are providedin the problem statement.

Find: Force at the ange to hold the nozzle in place: H$

APPROACH


ANALYSIS


;1 = 0!3$(0 × 0!15× 0!15) = 4!244 m/s;2 = 4!244× 9 = 38!196 m/s

Bernoulli equation

%1 = 0 + (1' 000$2)(38!1962 " 4!2442) = 720 kPa


H$ = "720' 000× 0 × 0!152 + 1' 000× 0!3(38!196" 4!244)


480

PROBLEM 6.52

Water ows through a nozzle with two openings–additional details are provided inthe problem statement

Find: E-component of force through ange bolts to hold nozzle in place.

APPROACH


ANALYSIS


;= = ;> = 16× 144$[(0$4)(4× 4 + 4!5× 4!5)]= 80!93 fps

;1 = 16$(0 × 0!5× 0!5)= 20!37 fps

Bernoulli equation

%1 = 0 + (1!94$2)(80!93× 80!93" 20!37× 20!37) = 5951 psf


H$ = "5' 951× 0 × 0!5× 0!5× sin 30! " 80!93× 1!94× 80!93× 0 × 2×2$144" 20!37× 1!94× 16!0 sin 30!

H$ = "3762 lbf

481

PROBLEM 6.53

Situation: Water ows through a nozzle with two openings–additional details areprovided in the problem statement.

Find: E-component of force through ange bolts to hold nozzle in place: H$

APPROACH


ANALYSIS


;= = ;> = 0!5$(0 × 0!05× 0!05 + 0 × 0!06× 0!06) = 26!1 m/s;1 = 0!5$(0 × 0!15× 0!15) = 7!07 m/s

Bernoulli equation

%1 = (1' 000$2)(26!12 " 7!072) = 315' 612 Pa


XH$ = /!;!$ "/#;#$

H$ + %1I1 sin 30 = " /;= " /;# sin 30

H$ = "315' 612× 0 × 0!152 × sin 30! " 26!1× 1' 000× 26!1

×0 × 0!052 " 7!07× 1000× 0!5 sin 30! = "18' 270 N = -18.27 kN

482

PROBLEM 6.54

Situation: Water ows through a nozzle that is bolted onto a pipe–additional detailsare provided in the problem statement.

1 2

Find: Tension load in each bolt: &

APPROACH


ANALYSIS

Continuity principle;2 = (I1$I2);1 = 4;1

Bernoulli equation

(;21$2)) + (%1$() = (;22$2)) + (%2$()

15(;21$2)) = (200' 000$9810)

;1 = 5!16 m/s

;2 = 20!66 m/s

] = 0!365 m3/s


XH$ = /!;!$ " /#;#$

Hbolts + %1I1 = #](;2 " ;1)

Thus

Hbolts = "%1I1 + #](;2 " ;1)Hbolts = "200' 000× 0 × 0!152 + 1000× 0!365(20!66" 5!16)

= "8440 N

Force per bolt = 1413 N

483

PROBLEM 6.55

Situation: Water jets out of a two dimensional slot.Flow rate is ] = 5 cfs per ft of slot width. Slot spacing is 4 = 8 in! Jet height is> = 4 in!

Find: (a)Pressure at the gage.(b)Force (per foot of length of slot) of the water acting on the end plates of the slot.

APPROACH

To nd pressure at the centerline of the ow, apply the Bernoulli equation. To ndthe pressure at the gage (higher elevation), apply the hydrostatic equation. To ndthe force required to hold the slot stationary, apply the momentum principle.

ANALYSIS

Continuity. Select a control volume surrounding the nozzle. Locate section 1 acrossthe slot. Locate section 2 across the water jet.

]1 = ]2 = ] =5 ft3$ s

ft

Flow rate equations

,1 =]

I1=

5 ft2$ s

(8$12) ft

= 7!5 ft$ s

,2 =]

I2=

5 ft2$ s

(4$12) ft

= 15! ft$ s

Bernoulli equation

%1 =#

2(, 22 " ,

21 )

=1!94 slug$ ft3

2(152 " 7!52)

ft2

s2

%1 = 163! 69 lbf$ ft2

Hydrostatic equation. Location position 1 at the centerline of the slot. Locateposition 3 at the gage.

%1(+ R1 =

%3(+ R3

163! 69 lbf$ ft2

62!4 lbf$ ft3+ 0 =

%3

62!4 lbf$ ft3+(8$12) ft

2

%3 = 142! 89 psf

484

%3 = 143 lbf$ ft2 = 0!993 lbf$ in2


XH$ = /,2 " /,1

H$ + %1I1 = #](,2 " ,1)H$ = "%1I1 + #](,2 " ,1) (1)

Calculations

%1I1 =¡163! 69 lbf$ ft2

¢(8$12 ft)

= 109! 13 lbf$ ft (a)

#](,2 " ,1) =¡1!94 slug$ ft3

¢ ¡5 ft2$ s

¢(15! ft$ s" 7!5! ft$ s)

= 72! 75 lbf$ ft (b)


H$ = " (109! 13 lbf$ ft) + 72! 75 lbf$ ft

= "36! 38lbf

ft

The force acting on the end plates is equal in magnitude and opposite in direction(Newton’s third law).

Hwater on the end plates = 36! 38lbfftacting to the right

485

PROBLEM 6.56

Situation: Water is discharged from a two-dimensional slot–additional details areprovided in the problem statement

Find: (a)Pressure at the gage.(b)Force (per foot of length of slot) on the end plates of the slot.

APPROACH

Apply the Bernoulli equation, then the hydrostatic equation, and nally the momen-tum principle.

ANALYSIS


;0 = 0!4$0!07 = 5!71 m/s

;> = 0!40$0!20 = 2!00 m/s

Bernoulli equation

%> = (1000$2)(5!712 " 2!002) = 14!326 kPa


%gage = 14' 326" 9810× 0!1 = 13.3 kPa


XH$ = /!;!$ " /#;#$

H$ + %>I> = #](;0 " ;>)

thus

H$ = "14' 326× 0!2 + 1000× 0!4(5!71" 2!00)= "1' 381 N

= -1.38 kN/m

486

PROBLEM 6.57

Situation: Water ows through a spray head–additional details are provided in theproblem statement.

v1

v2

30o

Find: Force acting through the bolts needed to hold the spray head on: H3

APPROACH


ANALYSIS


;1 = ]$I1 = 3$(0$4× 0!52) = 15!28 ft/s

Bernoulli equation

%1 =#

2

¡;22 " ;

21

¢

=1!94

2

¡652 " 15!282

¢

= 3872!


XH3 = /!;!3 " /#;#3

H3 + %1I1 = #](";2 sin 30! " ;1)H3 = ("3872)(0$4× 0!52) + 1!94× 3("65 sin 30! " 15!28)

= -1040 lbf

487

PROBLEM 6.58

Situation: An unusual nozzle creates two jets of water–additional details are providedin the problem statement.

Find: Force required at the ange to hold the nozzle in place: F

APPROACH


ANALYSIS


;1 =]

I

=2× 80!2× 0$4× 12

0$4× 42= 10!025 fps


PH$ =

P/!$ " /#;#$

%1I1 + H$ = /2;2$ + /3;3$ " /1;1$

H$ = "43× 0 × 22 + 1!94× 80!22 × 0 × !52$144"(1!94× 80!2× 0 × 0!52$144)× 80!2 sin 30"(1!94× 10!025× 0 × 0!16672)× 10!025

= "524!1 lbf


XH3 = /!3 " /#;#3

H3 = /3;33 = #I;3(";3 cos 30!)= "1!94(0$4× (1$12)2)80!22 cos 30!

= "58!94 lbf

Net forceF=(-524.1i-58.9j) lbf

488

PROBLEM 6.59

Situation: Liquid ows through a ”black sphere”–additional details are provided inthe problem statement.

v2

y

x

v1v3

30o

Find: Force in the inlet pipe wall required to hold sphere stationary: F

APPROACH


ANALYSIS


I1;1 = I2;2 +I3;3

;3 = ;1I1I3" ;2

I2I3

= 50 ft$ s

µ22

12

¶" 100 ft$ s

µ12

12

¶

= 100 ft$ s


H$ = /3;3$

= "#I3;23 sin 30!

= "(1!94× 1!2)

Ã0 (1$12)2

4

!(1002) sin 30!

= "63!49 lbf

?-directionH3 ". + %1I1 = /2;23 + /3;33 " /1;13

thusH3 =. " %1I1 + /2;2 "/3;3 cos 30

! " /1;1

489

Calculations

.1 " %1I1 = 200" 60× 0 × 12

= 11!50 lbf

/2;2 = #I2;22

= (1!2× 1!94)

Ã0 (1$12)2

4

!(1002)

= 126!97 lbf

/3;3 cos 30! = #I3;

23 cos 30

!

= (1!2× 1!94)

Ã0 (1$12)2

4

!(100)2 cos 30!

= 109!96 lbf

/1;1 = #I1;21

= (1!2× 1!94)

Ã0 (2$12)2

4

!(502)

= 126! 97 lbf

thus,

H3 = (. " %1I1) + /2;2 " (/3;3 cos 30!)" /1;1

= (11!50) + 126!97" (109!96)" 126! 97= "98! 46 lbf

Net ForceF = ("63!5i" 98!5j) lbf

490

PROBLEM 6.60

Situation: Liquid ows through a ”black sphere”–additional details are provided inthe problem statement.

Find: Force required in the pipe wall to hold the sphere in place: F

APPROACH


ANALYSIS


;3 = (10× 52 " 30× 2!52)$(2!52)= 10 m/s


H$ = "10 sin 30! × 1500× 10× 0 × 0!01252

= "36!8 N


H3 = "400' 000× 0 × 0!0252 + 600 + (15000)×("102 × 0!0252 + 302 × 0!01252

"102 × 0!01252 cos 30!)= 119 N

Net ForceF = ("36!8i+ 119j) N

491

PROBLEM 6.61

Situation: Liquid ows through a “black box”–additional details are provided in theproblem statement.

1

2

3

x

4

Find: Force required to hold the “black box” in place: F

APPROACH


ANALYSIS


]4 = 0!6" 0!10= 0!50 m3/s


H$ = " /1;1' " /3;3'= " /1;1 + /3;3

= 0

?-direction

H3 = /2;2( + /4;4(H3 = #]2;2 " #]4;4

= (2!0× 1000)(0!1)(20)" (2!0× 1000)(0!5)(15)= "11!0 kN

Net ForceF = (0i" 11!0j) kN

492

PROBLEM 6.62

To verify Eq. (6.11) the quantities ]' ;1' ;2I>' ?1' ?2 and HJ will have to be measured.Since a laboratory is available for your experiment it is assumed that the laboratoryhas equipment to obtain ]. The width > can be measured by a suitable scale. Thedepths ?1 and ?2 can be measured by means of piezometer tubes attached to openingsin the bottom of the channel or by means of point gages by which the actual level ofthe surface of the water can be determined. Then ;1 and ;2 can be calculated from; = ]$I = ]$(>?).

The force on the gate can be indirectly evaluated by measuring the pressure distribu-tion on the face of the gate. This pressure may be sensed by piezometers or pressuretransducer attached to small openings (holes) in the gate. The pressure taps on theface of the gate could all be connected to a manifold, and by appropriate valvingthe pressure at any particular tap could be sensed by a piezometer or pressure trans-ducer. The pressures at all the taps should be measured for a given run. Then byintegrating the pressure distribution over the surface of the gate one can obtain HJ.Then compare the measured HJ with the value obtained from the right hand side ofEq. (6.11). The design should be such that air bubbles can be purged from tubesleading to piezometer or transducer so that valid pressure readings are obtained.

493

PROBLEM 6.63

Situation: Water ows through a sluice gate––additional details are provided inthe problem statement.

2F

0.6 ft

3 ft

1

Find: Force of water (per unit width) acting on the sluice gate.

APPROACH


ANALYSIS

Bernoulli equation

;21$2) + R1 = ;22$2) + R2

(0!6$3)2;22$2) + 3 = ;22$2) + 0!6

;2 = 12!69 fps

;1 = 2!54

] = 7!614 cfs/ft


XH$ = #](;2$ " ;1$)

H$ + %1I1 " %2I2 = #](;2 " ;1)H$ = "62!4× 3!0× 3!0$2 + 62!4× 0!6× 0!6$2 + 1!94× 7!614

×(12!69" 2!54) = -120 lbf/ft

494

PROBLEM 6.64

Situation: A ow in a pipe is laminar and fully developed–additional details areprovided in the problem statement.

Find: Derive a formula for the resisting shear force (HN) as a function of the parame-ters 1' %1' %2' #' and \!

APPROACH

Apply the momentum principle, then the continuity principle.

ANALYSIS


XH$ =

Z

1&

#;(; · AI)

%1I1 " %2I2 " HN =

Z

=2

#@22AI" (#I@1)@1

%1I" %2I" HN = "#@21I+Z

=2

#@22AI (1)

Integration of momentum outow term

@2 = @max(1" (3$30)2)2

@22 = @2max(1" (3$30)2)2

Z

=2

#@22AI =

K0Z

0

#@2max(1" (3$30)2)2203A3

= "#@2max0320

K0Z

0

(1" (3$30)2)2("23$320)A3

To solve the integral, let

@ = 1"µ3

3!

¶2

Thus

A@ =

µ"23

32!

¶A3

495

The integral becomes

Z

=2

#@22AI = "#@2max0320

0Z

1

@2A@

= "#@2max0320

µ@3

3|01

¶

= "#@2max0320

µ0"

1

3

¶

=+#@2max03

20

3(2)


\I =

Z@AI

=

K0Z

0

@max(1" (3$30)2)20 rdr

= "@max0320

K0Z

0

(1" (3$30)2)("23$320) dr

= "@max0320(1" (3$30)2)2$2|K00

= @max0320$2

Therefore@max = 2\

Substituting back into Eq. 2 givesZ

=2

#@22AI = 4#\20320$3

Finally substituting back into Eq. 1, and letting @1 = \ , the shearing force is givenby

HN=892

4[%1 " %2 " (1$3)#\2]

496

PROBLEM 6.65

Situation: A swamp boat is powered by a propeller–additional details are providedin the problem statement.

1

2

Find: (a) Propulsive force when the boat is not moving.(b) Propulsive force when the boat is moving at 30 ft/s.

Assumptions: When the boat is stationary, neglect the inlet ow of momentum—thatis, assume ;1 ! 0!

APPROACH


ANALYSIS

a.) Boat is stationary

Momentum principle (E-direction) Select a control volume that surrounds the boat.

XH$ = /;2 " /;1

Hstop % /;2

Mass ow rate

/ = #I2;2

=¡0!00228 slug$ ft3

¢Ã0 (3 ft)2

4

!(90 ft$ s)

= 1!451 slug$ s

Thus

Hstop = /;2

= (1!451 slug$ s) (90 ft$ s)

= 130!59 lbf

Force (stationary boat) = 131 lbf

b.) Boat is moving

497

Momentum principle (E-direction). Select a control volume that surrounds the boatand moves with the speed of the boat. The inlet velocity is ;1 = 30 ft/s

XH$ = / (;2 " ;1)

= (1!451 slug$ s) (90" 30) ft$ sH$ = #](;2 " ;1)H$ = 0!00228× 636!17(90" 30)

= 87!1 lbf

Force (moving boat) = 87.1 lbf

498

PROBLEM 6.66

Situation: Air ows through a windmill–additional details are provided in the prob-lem statement.

Find: Thrust on windmill.

APPROACH


ANALYSIS

Continuity principle;2 = 10× (3$4!5)2 = 4!44 m/s


XH$ = /(;2 " ;1)

H$ = /(;2 " ;1)= (1!2)(0$4× 33)(10)(4!44" 10)

H$ = "472!0 N (acting to the left)

& = 472 N (acting to the right)

499

PROBLEM 6.67

Situation: A jet pump is described in the problem statement.

Find: (a) Derive a formula for pressure increase across a jet pump.(b) Evaluate the pressure change for water if IV$I! = 1$3' ;V = 15 m/s and ;! = 2m/s.

APPROACH


ANALYSIS


;1 = ;0120$(1

20 "1

2V ) (1)

;2 = (;0120 + ;V1

2V )$1

20 (2)


XH$ = /(;2 " ;1)

(%1 " %2)0120$4 = "#;

210(1

20 "1

2V )$4" #;

2V01

2V$4 + #;

2201

20$4

thus,

(%2 " %1) = #;21(120 "12

V )$120 + #;

2V ×12

V$120 " #;22 (3)

Calculations

;1 = ;0$(1" (1V$10)2)= 2$(1" (1$3))= 3 m/s

;2 = ;0 + ;V(12V$1

20)

= 2 + 15(1$3)

= 7 m/s

from Eq. (3)

%2 " %1 = #£;21¡1" (1V$10)2

¢+ ;2V (1V$10)

2 " ;22¤

= 1000£32(1" (1$3)) + 152(1$3)" 72

¤

= 32 kPa

500

PROBLEM 6.68

Situation: The problem statement describes a jet pump.2 6v /2g2

xvj

4ft y"

1

v= 1 ft/s

Find: Develop a preliminary design by calculating basic dimensions for a jet pump.

APPROACH

Apply the momentum principle, then the continuity principle.

ANALYSIS

Momentum principle (E-direction)Carry out the analysis for a section 1 ft wide (unit width) and neglect bottom friction.

XH$ = /2;2 " /1;1 " /V;V

(?21$2" (?22$2 = "1#(1× (4"!?))" ;V#(;V!?) + ;2#(;2?2) (1)

but ?2 = 4 ft + 6 ;2$2)

= 4 + 6$2) = 4!0932 ft


;2?2 = ;1(4"!?) + ;V!?;2 = ;1(4"!?)$?2 + ;V!?$?2

Assume!? = 0!10 ft

Then;2 = 1(3!9)$(4!093) + ;V × 0!1$4!0392 = 0!9528 + 0!02476;V (2)


;2V " (0!9528 + 0!02476;2V × 40!932 = 5)(?22 " ?

21)" 39!0

= 82!44 ft2$s2

Solving:;V = 12!1 ft/s IV = 0!10 ft2

If circular nozzles were used, then IV = (0$4)A2V ; AV = 4!28 in. Therefore, one could

use 8 nozzles of about 4.3 in. in diameter discharging water at 12.1 ft/s

COMMENTS

Like most design problems, this problem has more than one solution. That is, othercombinations of AV' ;V and the number of jets are possible to achieve the desiredresult.

501

PROBLEM 6.69

Situation: Lift and drag forces are being measured on an airfoil that is situated in awind tunnel–additional details are provided in the problem statement.

y pu

8 m/s

2

12 m/s0.25 m

0.25 mL

D

0.5 m

x pl

1m

c.v. c.s.

1

10 m/s

Find: (a) Lift force: 2(b) Drag force: 1

APPROACH


ANALYSIS


H$ =X

1&

/;0 " /1;1

"1 + %1I1 " %2I2 = ;1("#;1I) + ;'(#;'I$2) + ;0(#;0I$2)"1$I = %2 " %1 " #;21 + #;

2'$2 + #;

20$2

where

%1 = %T(E = 0) = %<(E = 0) = 100 Pa, gage

%2 = %T(E = 1) = %<(E = 1) = 90 Pa, gage

then

"1$I = 90" 100 + 1!2("100 + 32 + 72)"1$I = "5!2

1 = 5!2× 0!52 = 1.3 N


H3 = 0

"2+Z 2

1

%<DAE"Z 1

0

%TDAE = 0 where D is depth of tunnel

"2+Z 1

0

(100" 10E+ 20E(1" E))0!5AE"Z 1

0

(100" 10E" 20E(1" E))0!5AE = 0

"2+ 0!5(100E" 5E2 + 10E2 " (20$3)E3)|10 " 0!5(100E" 5E2 " 10E2 + (20$3)E3|10 = 0

502

thus,

"2+ 49!167" 45!833 = 0

2 = 3!334 N

503

PROBLEM 6.70

Situation: A torpedo-like device is being tested in a wind tunnel–additional detailsare provided in the problem statement.

c.s.Drag

Force of device on air= -Drag

Find: (a) Mass rate of ow.(b)Maximum velocity at the outlet section.(c)Drag on the device and support vanes.

APPROACH


ANALYSIS

Mass ow rate

/ = #;I

=¡0!0026 slug$ ft3

¢× (120 ft$ s)×

µ0(3!0 ft)2

4

¶

= 2!205 slug$ s

/ = 2!205 slug/ s

At the outlet section Z 0

0

;AI = ]

But ; is linearly distributed, so ; = ;max(3$30)! ThusZ K0

0

µ;max

3

3!

¶203A3 = ;I

2;max320

3= ;320

;max =3;

2

=3 (120 ft$ s)

2;max = 180 ft$ s

;max = 180 ft$ s

504


H$ =

Z K0

0

#;22AI" /;1 (1)

a.) Forces analysis XH$ = %1I1 " %2I2 "1 (a)

b.) Outlet velocity prole

;2 = ;max3

3!

=

µ3;

2

¶µ3

3!

¶(b)

c.) Outlet momentum owZ K0

0

#;22AI =

Z K0

0

#

·µ3;

2

¶µ3

3!

¶¸2203A3

= 20#

µ3;

2

¶2 Z K0

0

µ3

3!

¶23A3

= 20#

µ3;

2

¶2µ32!4

¶(c)

Substituting Eqns. (a) and (c) into the momentum equation (1) givesX

H$ =

Z K0

0

#;22AI" /;1

%1I1 " %2I2 "1 = 20#

µ3;

2

¶2µ32!4

¶" /;1

1 = %1I1 " %2I2 " 20#µ3;

2

¶2µ32!4

¶+ /;1 (2)

Calculations (term by term)

%1I1 = (144× 0!24)×µ0 × 32

4

¶

= 244!3 lbf

%2I2 = (144× 0!1)×µ0 × 32

4

¶

= 101! 9 lbfZ K0

0

#;22AI = 20#

µ3;

2

¶2µ32!4

¶

= 20 (0!0026)

µ3 (120)

2

¶2µ1!52

4

¶

= 297! 7 lbf

/;1 = (2!205) (120)

= 264! 6 lbf

505

Substituting numerical values into Eq. (2)

1 = %1I1 " %2I2 " 20#µ3;

2

¶2µ32!4

¶+ /;1

= 244!3 lbf " 101! 9 lbf " 297! 7 lbf + 264! 6 lbf= 109!3 lbf

1 = 109!3 lbf

506

PROBLEM 6.71

Situation: A tank of water rests on a sled–additional details are provided in theproblem statement.

Find: Acceleration of sled at time P

APPROACH


ANALYSIS

This type of problem is directly analogous to the rocket problem except that theweight does not directly enter as a force term and %G = %atm. Therefore, the appro-priate equation is

+ A;&$AP = #;2GIG " H(C = (1$+)(#;2G(0$4)A

2G " 7. )

where 7 =coe"cient of sliding friction and . is the weight

. = 350 + 0!1× 1000× 9!81 = 1331 NC = ()$. )(1' 000× 252(0$4)× 0!0152 " (1331× 0!05))= (9!81$1' 331)(43!90) m/s2

= 0.324 m/s2

507

PROBLEM 6.72

Situation: A uid jet strikes a wave that is moving at a speed ;@ = 7 m/s. 11 = 6cm. Speed of the uid jet is 20 m/s, relative to a xed frame.

45o

x

y

v1

vv

v2

Find: Force of the water on the vane.

ANALYSIS

Force and momentum diagramsSelect a control volume surrounding and moving with the vane. Select a referenceframe attached to the moving vane.


H$ = /;2$ " /;1$

"H$ = " /;2 cos 45! " /;1


H3 = /;23 " /;13

H3 = /;2 sin 45!

Velocity analysis

• ;1 is relative to the reference frame = (20" 7) = 13.

• in the term / = #I; use ; which is relative to the control surface. In this case; = (20" 7) = 13 m/s

• ;2 is relative to the reference frame ;2 = ;1 = 13 m/s

Mass ow rate

/ = #I;

= (1' 000 kg)(0$4× 0!062)(13)= 36!76 kg/s

508

Evaluate forces

H$ = /;1(1 + cos 45)

= 36!76× 13(1 + cos 45) = 815!8 N

which is in the negative E"direction.

H3 = /;2 sin 45

= 36!76× 13 sin 45 = 338!0 N

The force of the water on the vane is the negative of the force of the vane on thewater. Thus the force of the water on the vane is

F = (815!8i"338j) N

509

PROBLEM 6.73

Situation: A cart is moving with steady speed–additional details are provided in theproblem statement.

17 m/s

17 m/s

17 m/s

y

x

2

3

1

45o

Find: Force exerted by the vane on the jet: F

APPROACH


ANALYSIS

Make the ow steady by referencing all velocities to the moving vane and let the c.v.move with the vane as shown.


H$ = /2;2$ " /1;1

H$ = (172 cos 45!)(1000)(0$4)(0!12)$2" (17)(1000)(17)(0$4)(0!12)= +802" 2270 = "1470 N


H3 = /2;23 " /;33

= (17)(1' 000)(sin 45!)(17)(0$4)(0!12)$2" (17)2(1000)(0$4)(0!12)$2= "333 N

F(water on vane) = (1470i+ 333j) N

510

PROBLEM 6.74

Situation: A cart is moving with steady speed–additional details are provided in theproblem statement.

Find: Rolling resistance of the cart: HK!22#:D

ANALYSIS

Let the control surface surround the cart and let it move with the cart at 5 ft/s.Then we have a steady ow situation and the relative jet velocities are shown below.

55 ft/s

55 ft/s

55 ft/s

y

x

2

3

1

45o


XH$ = /2;2' " /1;1

Calculations

/1 = #I1,1

= (1!94)(0$4× 0!12)55= 0!838 kg/s

/2 = 0!838$2

= 0!419 kg/s

Hrolling = /1;1 " /2;2 cos 45!

= 0!838× 55" 0!419× 55 cos 45!

Hrolling = 29!8 lbf (acting to the left)

511

PROBLEM 6.75

Situation: A water is deected by a moving cone.Speed of the water jet is 25m$ s (to the right). Speed of the cone is 13m$ s (to theleft). Diameter of the jet is 1 = 10 cm!Angle of the cone is K = 50!.

Find: Calculate the external horizontal force needed to move the cone: H$

Assumptions: As the jet passes over the cone (a) assume the Bernoulli equationapplies, and (b) neglect changes in elevation.

APPROACH


ANALYSIS

Select a control volume surrounding the moving cone. Select a reference frame xedto the cone. Section 1 is the inlet. Section 2 is the outlet.Inlet velocity (relative to the reference frame and surface of the control volume).

;1 = ,1 = (25 + 13) m$ s

38m$ s

Bernoulli equation. Pressure and elevation terms are zero, so

,1 = ,2 = ;2 = 38m$ s


H$ = /(;2$ " ;1)= #I1,1 (;2 cos K " ;1)= #I1,

21 (cos K " 1)

=

µ1000

kg

m3

¶×µ0 × (0!1m)2

4

¶× (38m$ s)2 (cos 50! " 1)

= "4!051 kN

H$ = 4!05 kN (acting to the left)

512

PROBLEM 6.76

Situation: A jet of water is deected by a moving van–additional details are providedin the problem statement.

Find: Power (per foot of width of the jet) transmitted to the vane: *

APPROACH


ANALYSIS

Select a control volume surrounding the moving cone. Select a reference frame xedto the cone.

Velocity analysis

;1 = ,1 = 40 ft/s

;2 = 40 ft/s


XH$ = /(;2$ " ;1)

H$ = 1!94× 40× 0!3× (40 cos 50" 40)= "332!6 lbf

Calculate power

* = H;

= 332!6× 60= 19,956 ft-lbf/s = 36.3 hp

513

PROBLEM 6.77

Situation: A sled of mass /& = 1000 kg is decelerated by placing a scoop of widthY = 20 cm into water at a depth A = 8 cm.

Find: Deceleration of the sled: C&

ANALYSIS

Select a moving control volume surrounding the scoop and sled. Select a stationaryreference frame.


0 =A

AP(/&;&) + /;2' " /;1'

Velocity analysis

;1' = 0

,1 = 100 m/s

,2 = 100 m/s

v2 = 100 m/s[" cos 60i+ sin 60j] + 100i m/s;2$ = 50 m/s

The momentum principle equation simplies to

0 = /&C& + /;2' (1)

Flow rate

/ = #I1,1

= 1000× 0!2× 0!08× 100= 1600 kg/s

From Eq. (1).

C& = "/;2$/&

=("1600)(50)

1000

= -80 m/s2

514

PROBLEM 6.78

Situation: A snowplow is described in the problem statement.

Find: Power required for snow removal: *

APPROACH


ANALYSIS

Momentum principle (E-direction)Select a control volume surrounding the snow-plow blade. Attach a reference frameto the moving blade. X

H$ = #](;2$ " ;1)

Velocity analysis

,1 = ;1 = 40 ft/s

;2' = "40 cos 60! cos 30!

= "17!32 ft/s

CalculationsX

H$ = 1!94× 0!2× 40× 2× (1$4)("17!32" 40)= "444!8 lbf

Power

* = H,

= 444!8× 40= 17' 792 ft-lbf/s

* = 32!3 hp

515

PROBLEM 6.79

Maximum force occurs at the beginning; hence, the tank will accelerate immediatelyafter opening the cap. However, as water leaves the tank the force will decrease,but acceleration may decrease or increase because mass will also be decreasing. Inany event, the tank will go faster and faster until the last drop leaves, assuming noaerodynamic drag.

516

PROBLEM 6.80

Situation: A cart is moving with a steady speed along a track.Speed of cart is 5m$ s (to the right). Speed of water jet is 10m$ s.Nozzle area is I = 0!0012m2!

Find: Resistive force on cart: HK

APPROACH


ANALYSIS

Assume the resistive force (HK) is caused primarily by rolling resistance (bearingfriction, etc.); therefore, the resistive force will act on the wheels at the groundsurface. Select a reference frame xed to the moving cart. The velocities andresistive force are shown below.

5 m/s

5 m/s

2

1

c.s.

x

Fr

Velocity analysis

,1 = ;1 = ;2 = 5 m/s

/ = %I1,1

= (1000)(0!0012)(5)

= 6 kg/s


XH$ = /(;2 " ;1)

"HK = 6("5" 5) = "60 N

HK = 60 N (acting to the left)

517

PROBLEM 6.81

Situation: A jet with speed ;V strikes a cart (+ = 10 kg), causing the cart toaccelerate.The deection of the jet is normal to the cart [when cart is not moving].Jet speed is ;V = 10 m/s. Jet discharge is ] = 0!1 m3/s.

Find: (a)Develop an expression for the acceleration of the cart.(b)Calculate the acceleration when ;1 = 5 m/s.

Assumptions: Neglect rolling resistance.Neglect mass of water within the cart.

APPROACH


ANALYSIS

Select a control surface surrounding the moving cart. Select a reference frame xedto the nozzle. Note that a reference frame xed to the cart would be non-inertial.


y

x

m v2 2

.

m v1 1

.

W

N

(Mv )=Mac c

ddt


XH$ =

A

AP(/;1) + /2;2' = " /1;1

Momentum accumulation

Note that the cart is accelerating. Thus,

A

AP

Z

1@

;$#A, =A

AP;1

Z

1@

#A, =A

AP(+;1)

= /C1

518

Velocity analysis

,1 = ;V " ;1 [relative to control surface];1 = ;V [relative to reference frame (nozzle)]

from conservation of mass

;23 = (;V " ;1);2' = ;1

/2 = /1

Combining terms

XH$ =

A

AP(+;1) + /(;2$ " ;1)

0 = +C1 + #I1(;V " ;1)(;1 " ;V)

C1 =(LX*@))(@)"@*)2

Y

Calculations

C1 =1' 000× 0!1$10(10" 5)2

10

C1 = 25 m$ s2 (when ;1 = 5m$ s)

519

PROBLEM 6.82

Situation: A hemispherical nozzle sprays a sheet of liquid at a speed ; through a 180!

arc. Sheet thickness is P.

v

yd!

!

Find: An expression for the force in ?-direction to hold the nozzle stationary.The math form of the expression should be H3 = H3(#' ;' 3' P)!

APPROACH


ANALYSIS


H3 =

Z

1&

;3#V · dA

=

Z 8

0

(; sin K)#;(P3AK)

= #;2P3

Z 8

0

sin KAK

H3=2#, 2P3

520

PROBLEM 6.83

Situation: The problem statement describes a planar nozzle.

Find:a.) Derive an expression for b (K)b.) Derive an expression for e (K)

ANALYSIS

Dene IG as the projection of the exit area on the ? plane. Use the momentumequation to solve this problem and let the control surface surround the nozzle andfuel chamber as shown above. The forces acting on the system are the pressure forcesand thrust, & . The pressure forces in the E-direction are from %0 and %G. Writingthe momentum equation in the E-direction we have:

& + %0IG " %GIG =

Z

=

;$#V · dA

& + %0IG " %GIG =

Z2(; cos K)#(";2"AK)

& + %0IG " %GIG = "2;2#2"Z O

0

cos KAK

& + %0IG " %GIG = "2;2#2" sin K

But

/ = 2

Z O

0

#;AI = 2

Z O

0

#;2"AK = 2#;2"K

& + %0IG " %GIG = "2#;22"K(sin K$K)& + %0IG " %GIG = "; / sin K$K

& = /;(" sin K$K) + %GIG " %0IG& = /;b(K) +IG(%G " %0)e(K)

where

b(K) = " sin K/K

e(K) = 1

521

PROBLEM 6.84

Situation: Air ows through a turbofan engine. Inlet mass ow is 300 kg$ s!Bypass ratio is 2.5. Speed of bypass air is 600m$ s!Speed of air that passes through the combustor is 1000m$ s!

m2

m1

.

.

300 m/s

600 m/s

1,000 m/s

c.s.

A B

Additional details are given in the problem statement.

Find: Thrust (& ) of the turbofan engine.

Assumptions: Neglect the mass ow rate of the incoming fuel.

APPROACH

Apply the continuity and momentum equations.

ANALYSIS

Continuity equation/= = /> = 300 kg$ s

also

/> = /combustor + /bypass

= /combustor + 2!5 /combustor

/> = 3!5 /combustor

Thus

/combustor =/>

3!5=300 kg$ s

3!5= 85! 71 kg$ s

/bypass = /> " /combustor

= 300 kg$ s" 85! 71 kg$ s= 214!3 kg$ s

522


XH$ =

X/;out " /;in

H$ = [ /bypass,bypass + /combustor,combustor]" /=,=

= [(214!3 kg$ s) (600m$ s) + (85! 71 kg$ s) (1000m$ s)]" (300 kg$ s) (300m$ s)= 124' 290N

& = 124' 300 N

523

PROBLEM 6.85

Situation: A problem in rocket-trajectory analysis is described in the problem state-ment.

Find: Initial mass of a rocket needed to place the rocket in orbit.

ANALYSIS

+0 = +( exp(,00e$& )

= 50 exp(7200$3000)

= 551.2 kg

524

PROBLEM 6.86

Situation: A toy rocket is powered by a jet of water–additional details are providedin the problem statement.

Find: Maximum velocity of the rocket.

Assumptions: Neglect hydrostatic pressure; Inlet kinetic pressure is negligible.

ANALYSIS

Newtons 2:A law.X

H = /C

& ". = /C

where & =thrust and . =weight

& = /;G

/;G "/) = /A;F$AP

A;F$AP = (&$/)" )= (&$(/# " /P))" )

A;F = ((&AP)$(/# " /P))" )AP;F = ("&$ /)LQ(/# " /P)" )P+ const.

where ;F = 0 when P = 0! Then

const. = (&$ /) ln(/#)

;F = (&$ /) ln((/#)$(/# " /P))" )P;Fmax = (&$ /) ln(/#$/()" )P(&$ / = /;G$ / = ;G

Bernoulli equation(neglecting hydrostatic pressure)

%# + #(;2# $2 = %G + #(;

2G$2

The exit pressure is zero (gage) and the inlet kinetic pressure is negligible. So

;2G = 2%#$#(= 2× 100× 103$998= 200 m2$s2

;G = 14!14 m/s

/ = #G;GIG

= 1000× 14!14× 0!1× 0!052 × 0$4= 2!77 kg/s

525

Time for the water to exhaust:

P = /%$ /

= 0!10$2!77

= 0!0369

Thus

;max = 14!14 ln((100 + 50)$50)" (9!81)(0!036)

= 15.2 m/s

526

PROBLEM 6.87

Situation: A rocket with four nozzles is described in the problem statement.

Find: Thrust of the rocket (all four nozzles).

APPROACH


ANALYSIS

Momentum principle (R-direction)

XH; = /;;[per engine]

& " %'IG cos 30! + %GIG cos 30! = ";G cos 30!#;GIG& = "1× 0!866

×(50' 000" 10' 000 + 0!3× 2000 × 2000)= "1!074× 106 N

Thrust of four engines

&total = 4× 1!074× 106

= 4!3× 1068= 4.3 MN

527

PROBLEM 6.88

Situation: A rocket nozzle is connected to a combustion chamber.Mass ow: / = 220 kg$ s. Ambient pressure: %! = 100 kPa!Nozzle inlet conditions: I1 = 1m2' @1 = 100m$ s, %1 = 1!5MPa-abs.Nozzle exit condition? I2 = 2m2' @2 = 2000m$ s, %2 = 80 kPa-abs.

Assumptions: The rocket is moving at a steady speed (equilibrium).

Find: Force on the connection between the nozzle and the chamber.

APPROACH

Apply the momentum principle to a control volume situated around the nozzle.

ANALYSIS


XH$ = /!;!$ " /#;#$

H + %1I1 " %2I2 = /(;2 " ;1)

where H is the force carried by the material that connects the rocket nozzle to therocket chamber.

Calculations (note the use of gage pressures).

H = /(;2 " ;1) + %2I2 " %1I1= (220 kg$ s) (2000" 100)m$ s +

¡"20' 000N$m2

¢ ¡2m2

¢

"¡1' 400' 000N$m2

¢ ¡1m2

¢

= "1!022× 106 N= "1!022 MN

The force on the connection will be

H = 1!022 MN

The material in the connection is in tension.

528

PROBLEM 6.89

Situation: A problem related to the design of a conical rocket nozzle is described inthe problem statement.

Find: Derive an expression for the thrust of the nozzle.

APPROACH


ANALYSIS


XF =

Zv#v · dA

& =

Z H

0

;G cos K#;G

Z 2+

0

sin K3Af3AK

& = 2032#;2G

Z H

0

cos K sin KAK

= 2032#;2G sin2 T$2

= #;2G2032(1" cosT)(1 + cosT)$2

Exit Area

IG =

Z H

0

Z 28

0

sin K3Af3AK = 2032(1" cosT)

& = #;2GIG(1 + cosT)$2 = /;G(1 + cosT)$2

529

PROBLEM 6.90

Situation: A valve at the end of a gasoline pipeline is rapidly closed–additionaldetails are provided in the problem statement.

Find: Water hammer pressure rise: !%

ANALYSIS

Speed of sound

Z =pN@$#

= ((715)(106)$(680))045

= 1025 m/s

Pressure rise

!% = #;Z

= (680)(10)(1025)

= 6.97 MPa

530

PROBLEM 6.91

Situation: A valve at the end of a long water pipeline is rapidly closed–additionaldetails are provided in the problem statement.

Find: Water hammer pressure rise: !%

ANALYSIS

Z =

sN@#

=

r2!2× 109

1000= 1483 m/s

Pcrit = 22$Z

= 2× 10' 000$1483= 13!5 s F 10 s

Then

!% = #;Z

= 1000× 4× 1483= 5' 932' 000 Pa

= 5.93 MPa

531

PROBLEM 6.92

Situation: A valve at the end of a water pipeline is instantaneously closed–additionaldetails are provided in the problem statement.

Find: Pipe length: 2

ANALYSIS

Determine the speed of sound in water

Z =

sN@#

=

r2!2× 109

1000= 1483 m/s

Calculate the pipe length

P = 42$Z

3 = 42$1483

2=1112 m

532

PROBLEM 6.93

Situation: A valve at the end of a water pipeline is closed during a time period of 10seconds.Additional details are provided in the problem statement.

Find: Maximum water hammer pressure: !%max

ANALYSIS

Determine the speed of sound in water

Z =

sN@#

Z =

r320' 000

1!94= 4874 ft/s

Determine the critical time of closure

Pcrit = 22$Z

= 2× 5× 5280$4874= 10!83 s F 10 s

Pressure rise

!%max = #;Z

= 1!94× 8× 4874= 75,644 psf = 525 psi

533

PROBLEM 6.94

Situation: A valve at the end of a long water pipe is shut in 3 seconds–additionaldetails are provided in the problem statement

Find: Maximum force exerted on valve due to the waterhammer pressure rise: H@'2@G

ANALYSIS

Pcrit =22

Z

=2× 40001485!4

= 5!385 s F 3 s

Hvalve = I!%

= I#(]$I)Z

= #]Z

= 998× 0!03× 1483= 44!4 kN

534

PROBLEM 6.95

Situation: The easy way to derive the equation for waterhammer pressure rise is touse a moving control volume.

Find: Derive the equation for waterhammer pressure rise (Eq. 6.12).

ANALYSIS

V+c cP'

p+v=c= +

"'

' ' "'

Continuity equation

(; + Z)# = Z(#+!#)

! !# = ;#$Z


XH$ =

X;$#v ·A

%I" (%+!%)I = "(, + Z)#(, + Z)I+ Z2(#+!#)I!% = 2#;Z" Z2!#+ ;2#

= 2#;Z" Z2;#$Z+ ;2#= #;Z+ #;2

Here #;2 is very small compared to #;Z

! !% = #;Z

535

PROBLEM 6.96

Situation: The problem statement describes a waterhammer phenomena in a pipe.

Find: Plot a pressure versus time trace at point B for a time period of 5 seconds.Plot a pressure versus distance trace at P = 1!5 s!

ANALYSIS

; = 0!1m$ s

Z = 1483m$ s

%pipe = 10( " #;2pipe$2% 98' 000 Pa

!% = #;Z

= 1000× 0!10× 1483!% = 148' 000 Pa

Thus

%max = %+!%

= 98' 000 + 148' 000

= 246 kPa- gage

%min = %"!% = "50 kPa gageThe sequence of events are as follows:

!P "!PPressure wave reaches pt B. 1000$1483 = 0!674 s 0.67 s

Time period of high pressure at B 600$1483 = 0!405 s 1.08 sTime period of static pressure at B 2000$1483 = 1!349 s 2.43 s

Time period of negative pressure at B 600$1483 = 0!405 s 2.83 sTime period of static pressure at B 2000$1483 = 1!349 s 4.18 sTime period of high pressure at B 600$1' 483 = 0!405 s 4.59 sTime period of static pressure atB 2000$1483 = 1!349 s 5.94 s

Results are plotted below:

200 kPa

100 kPa

0

0 1 2 3 4 5

536

200 kPa

100 kPa

500 m 1,000 m0

Figure 1:

At P = 1!5 s high pressure wave will have travelled to reservoir and static wave willbe travelling toward valve.

Time period for wave to reach reservoir = 1300$1483 = 0!877 s. Then static wavewill have travelled for 1!5 " 0!877 s = 0!623 s. Distance static wave has travelled= 0!623 s ×1' 483 m/s = 924 m. The pressure vs. position plot is shown below:

537

PROBLEM 6.97

Situation: A water hammer phenomenon occurs in a steel pipe–additional detailsare provided in the problem statement.

Find: (a) The initial discharge.(b) Length from I to D!

ANALYSIS

Z = 1483 m/s

!% = #!;Z

P = 2$Z

2 = PZ = 1!46 s × 1' 483 = 2165 m

!; = !%$#Z

= (2!5" 0!2)× 106 Pa$1!483× 106 kg/m2s = 1!551 m/s

] = ;I = 1!551× 0$4 = 1.22 m3/s

538

PROBLEM 6.98

Situation: Water is discharged from a slot in a pipe–additional details are providedin the problem statement.

Find: Reaction (Force and Moment) at station I"I

APPROACH

Apply the momentum principle and the moment of momentum principle.

ANALYSIS

y

x

Plan View

;3 = "(3!1 + 3E) m/s


XH3 =

Z;3#v · dA

H3 = "Z 143

043

(3!1 + 3E)× 1' 000× (3!1 + 3E)× 0!015AE = "465 N

"3 = 465 N

Flow rate

] =

Z;AI = 0!015

Z 143

043

(3!1 + 3E)AE = 0!0825 m3/s

;1 = ]$I = 0!0825$(0 × 0!042) = 16!4 m/s

Momentum principle (R-direction)

XH; = " /1;1

H; " %=I= ".( = " /;1H; = 30' 000× 0 × 0!042 + 0!08× 0 × 0!042 × 9' 810+1!3× 0 × 0!0252 × 9' 810 + 1000× 0!0825× 16!4

= 1530 N

"; = "1530 N

539

Moment-of-momentum (R"direction)

&; =

Z

1&

3;#v · dA

= 15

Z 143

043

(3!1 + 33)23A3 = 413!2 N ·m

Moment-of-momentum (?-direction)

&3 +.31" = 0

where .=weight, 31"=distance to center of mass

&3 = "1!30 × 0!0252 × 9810× 0!65 = "16!28 N ·m

Net reaction at A-A

F = (465j" 1530k)N

T = (16!3j" 413k)N · m

540

PROBLEM 6.99

Situation: Water ows out a pipe with two exit nozzles–additional details are pro-vided in the problem statement.

30o

1

2

3

Find: Reaction (Force and Moment) at section 1.

APPROACH

Apply the continuity equation, then the momentum principle and the moment ofmomentum principle.

ANALYSIS

Continuity principle equation

;1 = (0!1× 50 + 0!2× 50)$0!6 = 25 ft/s


XH$ = /3;3$ + /2;2$

H$ = "20× 144× 0!6" 1!94× 252 × 0!6 + 1!94× 502 × 0!2+1!94× 502 × 0!1× cos 60! = "1,243 lbf

Momentum equation (?-direction)

XH3 = /2;23

H3 = 1!94× 50× 50× 0!1× cos 30! = 420 lbf


32 /2;23 = (36$12)(1!94× 0!1× 50)50 sin 60! = 1260 ft-lbf

Reaction at section 1

F = (1243i" 420j)lbf

M = ("1260k) ft-lbf

541

PROBLEM 6.100

Situation: Water ows out a pipe with two exit nozzles–additional details are pro-vided in the problem statement.

2 y

x3

30o

Find: Reaction (Force and Moment) at section 1.

APPROACH

Apply the continuity principle, then the momentum principle and the moment ofmomentum principle.

ANALYSIS

Continuity principle equation

,1 = (0!01× 20 + 0!02× 20)$0!1 = 6 m/s


XH$ =

X/!;!$ "

X/#;#$

H$ + %1I1 = /3;3 + /2;2 cos 30" /1;1

H$ = "200' 000× 0!1" 1000× 62

×0!1 + 1000× 202 × 0!02+1000× 202 × 0!01× cos 30!

= -12,135 N

Momentum equation (?-direction)

H3 ". = /2;2 sin 30!

Weight

. = .H2O +.pipe

= (0!1)(1)(9810) + 90

= 1071 N

542

thus

H3 = 1000× 202 × 0!01× sin 30! + 1' 071

= 3071 N


+; ".31" = 32 /2;23

+; = (1071× 0!5) + (1!0)(1000× 0!01× 20)(20 sin 30!)= 2535 N ·m

Reaction at section 1

F = (12!1i" 3!1j) kN

M = ("2!54k) kN · m

543

PROBLEM 6.101

Situation: A reducing pipe bend held in place by a pedestal. Water ow. No forcetransmission through the pipe at sections 1 and 2.Assume irrotational ow. Neglect weight

1

22 ft

y

x

thus

H$ = "2' 880× 0!196" 2' 471× 0!0873" 3!875(10!19 + 22!92) = "909!6 lbf


/; " 3%1I1 + 3%2I2 = "3 /;2 + 3 /;1/; = 3(%1I1 " %2I2)" 3 /(;2 " ;1)

where 3 = 1!0 ft.

+; = 1!0(2' 880× 0!196" 2' 471× 0!08753)" 1!0× 3!875(22!92" 10!19)= 300!4 ft-lbf

Moment-of-momentum (?-direction)

+3 + %1I133 + %2I233 = "33 /;2 " 33 /1;1

where 33 = 2!0 ft.

+3 = "33[%1I1 + %2I2 + /(;1 + ;2)]

= "2!0× 909!6+3 = "1819 ft-lbf

Net force and moment at 3

F = "910i lbf

M = ("1820j+ 300k) ft-lbf

545

PROBLEM 6.102

Situation: Arbitrary contol volume with length !9!

Find: Derive Euler’s equation using the momentum equation.

ANALYSIS

v

pA

"s

p

W= A s% !

$1

2(p + p )/21 2

Continuity equationA

AP

Z#A, + /! " /# = 0

For a control volume that is xed in spaceZ[#

[PA, + /! " /# = 0

For the control volume shown above the continuity equation is expressed as

[#

[PI!9+ (#;I)2 " (#;I)1 = 0

where I is the average cross-sectional area between 1 and 2 and the volume of thecontrol volume is I!9! Dividing by !9 and taking the limit as !9$ 0 we have

I[#

[P+[

[9(#;I) = 0

In the limit the average area becomes the local area of the stream tube.

The momentum equation for the control volume is

XH& =

A

AP

Z#;A, + /!;! " /#;#

For a control volume xed in space, the accumulation term can be written as

A

AP

Z#;A, =

Z[

[P(#;)A,

546

The forces are due to pressure and weight

XH& = %1I1 " %2I2 + (

%1 + %22

)(I2 "I1)" (I!9 sin K

where the third term on the right is the pressure force on the sloping surface and K isthe orientation of control volume from the horizontal. The momentum equation forthe control volume around the stream tube becomes

[

[P(#;)I!9+ #I;2;2 " #I;1;1 = (%1 " %2)I" (I!9 sin K

Dividing by !9 and taking limit as !9$ 0' we have

I[

[P(#;) +

[

[9(#I;2) = "

[%

[9I" (I sin K

By di!erentiating product terms the left side can be written as

I[

[P(#;) +

[

[9(#I;2) = ;[I

[#

[P+[

[9(#;I)] +I#

[;

[P+I#;

[;

[9

The rst term on the right is zero because of the continuity equation. Thus themomentum equation becomes

#[;

[P+ #;

[;

[9= "

[%

[9" ( sin K

But sin K = [R$[9 and [;$[P + ;[;$[9 = C&' the acceleration along the path line.Thus the equation becomes

#C& = "[

[9(%+ (R)

which is Euler’s equation.

547

PROBLEM 6.103

Situation: A helicopter rotor uses two small rockets motors–details are provided inthe problem statement.

Find: Power provided by rocket motors.

APPROACH

Apply the momentum principle. Select a control volume that encloses one motor,and select a stationary reference frame.

ANALYSIS

Velocity analysis

;# = 0

,# = 3Y

= 3!5× 20= 21!991 m$ s

,0 = 500 m$ s

;0 = (500" 21!99) m$ s= 478!01m$ s

Flow rate

/ = #I#,#

= 1!2× !002× 21!991= 0!05278 kg/s


H$ = /;0 " /;#

= /;0

= 0!05278× 478= 25!23 N

Power

* = 2H3Y

= 2× 25!23× 3!5× 20= 1110W

* = 1!11 kW

548

PROBLEM 6.104

Situation: A rotating lawn sprinkler is to be designed.The design target is 0.25 in. of water per hour over a circle of 50-ft radius.

Find: Determine the basic dimensions of the lawn sprinkler.

Assumptions:1.) The Bernoulli equation applies.2.) Assume mechanical friction is present.

APPROACH


ANALYSIS

Flow rate. To supply water to a circle 50 ft. in diameter at a 1/4 inch per hourrequires a discharge of

] = MI

= (1$48)0(502$4)$3600

= 0!011 cfs

Bernoulli equation. Assuming no losses between the supply pressure and the sprinklerhead would give and exit velocity at the head of

, =

r2%

#

=

r2× 50× 144

1!94= 86 ft/s

If the water were to exit the sprinkler head at the angle for the optimum trajectory(45!), the distance traveled by the water would be

9 =, 2G2)

The velocity necessary for a 25 ft distance (radius of the spray circle) would be

, 2G = 2)9 = 2× 32!2× 25 = 1610,G = 40 ft/s

This means that there is ample pressure available to do the design. There will belosses which will a!ect the design. As the water spray emerges from the spray head,

549

atomization will occur which produces droplets. These droplets will experience aero-dynamic drag which will reduce the distance of the trajectory. The size distributionof droplets will lead to small droplets moving shorted distances and larger dropletsfarther which will contribute to a uniform spray pattern.

The sprinkler head can be set in motion by having the water exit at an angle withrespect to the radius. For example if the arm of the sprinkler is 4 inches and theangle of deection at the end of the arm is 10 degrees, the torque produced is

+ = #]3,G sin K

= 1!94× 0!011× 40× sin 10= 0!148 ft-lbf

The downward load on the head due to the discharge of the water is

H3 = #],G sin 45

= 1!94× 0!011× 40× sin 45= 0!6 lbf

The moment necessary to overcome friction on a at plate rotating on another atplate is

+ = (2$3)7H:3!

where 7 is the coe"cient of friction and 3! is the radius of the plate. Assuming a 1/2inch radius, the limiting coe"cient of friction would be

7 =3

2

+

H:3!

=3

2

0!148

0!6× (1$24)= 8!9

This is very high, which means there is adequate torque to overcome friction.

These are initial calculations showing the feasibility of the design. A more detaileddesign would now follow.

550

PROBLEM 6.105

Following the same development in the text done for the planar case, there will beanother term added for the two additional faces in the R-direction. The rate of changeof momentum in the control volume plus the net e#ux through the surfaces becomes

1

!E!?!R

Z

1@

[

[P(#@)A, +

#uu$+!$*2 " #uu$"!$*2!E

+#@;3+!3*2 " #@;3"!3*2

!?+#@Y;+!;*2 " #@Y;"!;*2

!R

where Y is the velocity in the R"direction and !R is the size of the control volumein the R"direction. Taking the limit as !E' !?' and !R $ 0 results in

[

[P(#@) +

[

[E(#uu) +

[

[?(#@;) +

[

[R(#@Y)

In the same way, accounting for the pressure and shear stress forces on the three-dimensional control volume leads to an additional shear stress term on the R-face.There is no additional pressure force because there can only be a force due to pressureon the faces normal to the E-direction. The force terms on the control volume become

%$"!$*2 " %$+!$*2!E

+B$$ |$+!$*2 "B$$ |$"!$*2

!E

+B 3$ |3+!3*2 "B 3$ |3"!3*2

!?+B ;$ |;+!;*2 "B ;$ |;"!;*2

!R

Taking the limit as !E' !?' and !R $ 0 results in

"[%

[E+[B$$[E

+[B 3$[?

+[B ;$[R

The body force in the E-direction is

#)$!V–!E!?!R

= #)$

551

PROBLEM 6.106

Substituting in the constitutive relations gives

[B$$[E

+[B 3$[?

+[B ;$[R

= 27[2@

[E2+ 7

[

[?([@

[?+[;

[E) + 7

[

[R([@

[R+[Y

[E)

This can be written as

[B$$[E

+[B 3$[?

+[B ;$[R

= 7([2@

[E2+[2@

[?2+[2@

[R2) + 7

[

[E([@

[E+[;

[?+[Y

[R)

The last term is equal to zero from the Continuity principle equation for an incom-pressible ow, so

[B$$[E

+[B 3$[?

+[B ;$[R

= 7([2@

[E2+[2@

[?2+[2@

[R2)

552

PROBLEM 7.1

Situation: An engineer is estimating power for a water pik.Water jet diameter is A = 1$8 in! = 3!175mm. Exit speed is ,2 = 40m$ s.

Find: Estimate the minimum electrical power in watts.

Properties: At 10 !C' density of water is # = 1000 kg$m3!

Assumptions: 1.) Neglect all energy losses in the mechanical system—e.g. motor,gears, and pump.2.) Neglect all energy losses in the uid system—that is, neglect losses associated withviscosity.3.) Neglect potential energy changes because these are very small.4.) Assume the velocity distribution in the water jet is uniform (T = 1) !

APPROACH

In the water pik, electrical energy is converted to kinetic energy of the water. Balanceelectrical power with the rate at which water carries kinetic energy out of the nozzle.

ANALYSIS

Power =Amount of kinetic energy that leaves the nozzle

Each interval of time

=!/

? 222

!P

where!/ is the mass that has owed out of the nozzle for each interval of time (!P) !Since the mass per time is mass ow rate: (!/$!P = / = #I2,2)

Power =/, 222

=#I2,

32

2

553

Exit area

I2 = 0$4×¡3!175× 10"3m

¢2

= 7!917× 10"6m2

Thus!

Power =(1000 kg$m3) (7!917× 10"6m2) (40m$ s)3

2

Power = 253W

COMMENTS

Based on Ohm’s law, this device would draw about 2 amps on a standard householdcircuit.

554

PROBLEM 7.2

Situation: A turbine is described in the problem statement.

Find: Power output.

APPROACH

Apply the energy principle.

ANALYSIS

Energy principle

]" .& = /[(M2 " M1) + (, 22 " ,21 )$2]

"10" .& = 4' 000[(2' 621" 3' 062) + (502 " 102)$(2× 1' 000)] kJ/hr.& = 489 kW

555

PROBLEM 7.3

Situation: A small wind turbine is being developed.Blade diameter is 1 = 1!0m! Design wind speed is , = 15 mph = 6!71m$ s!Air temperature is & = 50 !F = 10 !C. Atmospheric pressure is % = 0!9 bar = 90 kPa.Turbine e"ciency is g = 20%!

Find: Power (* ) in watts that can be produced by the turbine.

APPROACH

Find the density of air with the idea gas law. Then, nd the kinetic energy of thewind and use 20% of this value to nd the power that is produced.

ANALYSIS

Ideal gas law

# =%

"&

=90' 000Pa

(287 J$ kg · K) (10 + 273) K= 1!108 kg$m3

ANALYSIS

Rate of KE =Amount of kinetic energy

Interval of time

=!/, 2$2

!P

556

where !/ is the mass of air that ows through a section of area I = 012$4 foreach unit of time (!P) ! Since the mass for each interval of time is mass ow rate:(!/$!P = / = #I, )

Rate of KE =/, 2

2

=#I, 3

2

The area is I2 = 0$4× (1!0m)2 = 0!785m2

Rate of KE =(1!103 kg$m3) (0!785m2) (6!71m$ s)3

2

Rate of KE = 130!9W

Since the output power is 20% of the input kinetic energy

* = (0!2) (130!9W)

* = 26!2W

COMMENTS

The amount of energy in the wind is di!use (i.e. spread out). For this situation,a wind turbine that is 1 m in diameter in a moderately strong wind (15 mph) onlyprovides enough power for approximately one 25 watt light bulb.

557

PROBLEM 7.4

Situation: A compressor is described in the problem statement.

Find: Power required to operate compressor.

APPROACH

Apply the energy principle.

ANALYSIS

Energy principle

. = ]+ /(, 21 $2" ,22 $2 + M1 " M2)

The inlet kinetic energy is negligible so

. = /(", 22 $2 + M1 " M2)= 1!5("2002$2 + 300× 103 " 500× 103)

. = -330 kW

558

PROBLEM 7.5

Situation: Flow through a pipe is described in the problem statement.

Find: (a)Velocity and (b)temperature at outlet.

ANALYSIS

M1 + ,21 $2 = M2 + ,

22 $2 (1)

M1 " M2 = , 22 $2" ,21 $2

/ = #1,1I = (%1$"&1),1I

or&1 = %1,1I$(" /)

where

I = (0$4)× (0!08)2 = 0!00503 m2

M1 " M2 = Z,(&1 " &2) = [Z,%1,1I$(" /)]" [Z,%2,2I$(" /)] (2)

Z,%1I(" /) = 1' 004× 150× 103 × 0!00503$(287× 0!5)= 5' 279 m/s

andZ2%2I$(" /) = (100$150)× (5' 279) = 3' 519 m/s

,1 = /$#1I

where#1 = 150× 10

3$(287× 298) = 1!754 kg/m3

Then,1 = 0!50$(1!754× 0!00503) = 56!7 m/s (3)

Utilizing Eqs. (1), (2) and (3), we have

56!7× 5' 279" 3' 519,2 = (, 22 $2)" (56!72$2) (4)

Solving Eq. (4) yields ,2 = 84!35 m/s

Z,(&1 " &2) = (84!352 " 56!72)$2 = 1' 950 m2/s2

&2 = &1 " (1' 950$Z,)= 20!C" 1' 905$1' 004

&2 = 18.1!C

559

PROBLEM 7.6

Situation: A hypothetical velocity distribution in a pipe is described in the problemstatement.

Find:(a) Kinetic energy correction factor: T(b)Mean velocity in terms of ,max!

ANALYSIS

Denition of average velocity

, =1

I

Z

=

, (3)AI

Velocity prole

, (3) = ,max " 0!5,max3

30

, (3) = ,max(1" 0!53

30)

Then

, =

µ,max0320

¶Z K0

0

µ1"

3

230

¶203A3

=

µ20,max0320

¶Z K0

0

µ3 "

32

230

¶A3

=

µ20,max0320

¶µ32!2"33!630

¶

= , = 23,max

Kinetic energy correction factor

T =

µ1

0320

¶Z K0

0

!

"

³1" K

2K0

´,max

23,max

#

$3

203A3

=

µ2

320

¶µ3

2

¶3 Z K0

0

µ1"

3

230

¶33A3

560

Performing the integration (we used the computer program Maple)

T =351

320

orT = 1! 097

561

PROBLEM 7.7

Situation: A hypothetical velocity distribution in a rectangular channel is describedin the problem statement.

Find: Kinetic energy correction factor: T

ANALYSIS

, = ,max$2 and , = ,max?$A


T = (1$A)

Z A

0

(,max?$((,max$2)A))3A?

= (1$A)

Z A

0

(2?$A)3A?

T = 2

562

PROBLEM 7.8

Situation: Velocity distributions (a) through (d) are described in the problem state-ment.

Find: Indicate whether T is less than, equal to, or less than unity.

ANALYSIS

a) T = 1!0 ; b) T F 1!0 ; c) T F 1!0 ; d) T F 1!0

563

PROBLEM 7.9

Situation: A velocity distribution is shown in case (c) in problem 7.8.


ANALYSIS


T = (1$I)

Z

=

(,$, )3AI

Flow rate equation

, = ," " (3$30),", = ,"(1" (3$30))

] =

Z, AI

=

Z K0

0

, (203A3)

=

Z K0

0

,"(1" 3$30)203A3

= 20,"

Z K0

0

[3 " (32$30)]A3

Integrating yields

] = 20,"[(32$2)" (33$(330))]K00

] = 20,"[(1$6)320]

] = (1$3),"I

Thus, = ]$I = ,"$3


T = (1$I)

Z K0

0

[,"(1" 3$30)$((1$3),")]3203A3

= (540$0320)

Z K0

0

(1" (3$30))33A3

T =2.7

564

PROBLEM 7.10

Situation: A velocity distribution is shown in case (d) in problem 7.8.


ANALYSIS

Flow rate equation

, = c3

] =

Z K0

0

, (203A3)

=

Z K0

0

20c32A3

= 20c330$3

, = ]$I

= ((2$3)c0330)$0320

= 2$3 c 30


T = (1$I)

Z

=

(,$, )3AI

T = (1$I)

Z K0

0

(c3$(2$3 c30))3203A3

T = ((3$2)320$(0320))

Z K0

0

(3$30)33A3

T = ((27$4)$320)(350$(53

30))

T = 27/20

565

PROBLEM 7.11

Situation: The kinetic energy correction factor for ow in a pipe is 1.08.

Find: Describe the ow (laminar or turbulent).

ANALYSIS

b) turbulent

566

PROBLEM 7.12

Situation: The velocity distribution in a pipe is described in the problem statement.

Find: Derive formula for kinetic energy correction factor as a function of Q.

ANALYSIS

Flow rate equation

@$@max = (?$30): = ((30 " 3)$30): = (1" 3$30):

] =

Z

=

@AI

=

Z K0

0

@max(1" 3$30):203A3

= 20@max

Z K0

0

(1" 3$30):3A3

Upon integration] = 20@max3

20[(1$(Q+ 1))" (1$(Q+ 2))]

Then

, = ]$I = 2@max[(1$(Q+ 1))" (1$(Q+ 2))]= 2@max$[(Q+ 1)(Q+ 2)]


T =1

I

Z K0

0

[@max(1" 3$30):$(2@max$((Q+ 1)(Q+ 2)))]3203A3

Upon integration one gets

C = (1$4)[((Q+ 2)(Q+ 1))3$((3Q+ 2)(3Q+ 1))]

If Q = 1$6, then

T = (1$4)[((1$6 + 2)(1$6 + 1))3$((3× 1$6 + 2)(3× 1$6 + 1))]T = 1!077

567

PROBLEM 7.13

Situation: The velocity distribution in a pipe is described in the problem statement.

@$@max = (?$A):

Find: Derive formula for kinetic energy correction factor.

ANALYSIS

Solve for ^ rst in terms of @max and A

^ =

Z A

0

@A? =

Z A

0

@max(?$A):A? = @max$A

:

Z A

0

?:A?

Integrating:

^ = (@max$A:)[?:+1$(Q+ 1)]A0

= @maxA:+1A":$(Q+ 1)

= @maxA$(Q+ 1)

Then@ = ^$A = @max$(Q+ 1)


T = (1$I)

Z

=

(@$@)3AI

= 1$A

Z A

0

[@max(?$A):$(@max$(Q+ 1))]

3A?

= ((Q+ 1)3$A3:+1)

Z A

0

?3:A?

Integrating

T = ((Q+ 1)3$A3:+1)[A3:+1$(3Q+ 1)]

= (Q+ 1)3$(3Q+ 1)

When Q = 1$7

T = (1 + 1$7)3$(1 + 3$7)

T = 1.045

568

PROBLEM 7.14

Situation: Flow though a pipe is described in the problem statement.

Find: Kinetic energy correction factor: T.

ANALYSIS


T =1

I

Z

=

µ,

,

¶3AI

The integral is evaluated using

Z

=

µ,

,

¶3AI '

1

, 3

X

#

0(32# " 32#"1)(

;# + ;#"12

)3

The mean velocity is 24.32 m/s and the kinetic energy correction factor is 1.187.

569

PROBLEM 7.15

Situation: Water ows from a pressurized tank, through a valve and out a pipe.

Section 1 (air/water interface in tank): %1 = 100 kPa, R1 = 12m.Section 2 (pipe outlet): %2 = 0kPa' R2 = 0m, ,2 = 10m$ s!Head loss for the system depends on a minor loss coe"cient (<E) ! The equation forhead loss is:

ME = <E, 2

2)

Find: Find the value of the minor loss coe"cient (<E) !

Properties: Water @ 15 !C from Table A.5: ( = 9800N$m3!

Assumptions: 1.) Assume steady ow.2.) Assume the outlet ow is turbulent so that T2 = 1!0!3.) Assume water temperature is 15 !C.4.) Assume the velocity at section 1 is negligible—that is ,1 % 0!

APPROACH

Apply the energy equation to a control volume surrounding the water. Analyze eachterm and then solve the resulting equation to nd the minor loss coe"cient.

ANALYSIS

Energy equation

%1(+ T1

, 212)+ R1 + M, =

%2(+ T2

, 222)+ R2 + M5 + ME (1)

Analyze each term:

• At the inlet. %1 = 100 kPa, ,1 % 0, R1 = 12m

• At the exit , %2 = 0kPa' ,2 = 10m$ s' T2 = 1!0.

• Pumps and turbines. M, = M5 = 0

570

• Head loss. ME = <E? 2

2D

Eq. (1) simplies to

%1(+ R1 = T2

, 222)+<E

, 222)

(100' 000Pa)

(9800N$m3)+ 12m =

(10m$ s)2

2 (9!81m$ s2)+<E

(10m$ s)2

2 (9!81m$ s2)

22!2m = (5!097m) +<E (5!097m)

Thus<E = 3!35

COMMENTS

1. The minor loss coe"cient (<E = 3!35) is typical of a valve (this information ispresented in Chapter 10).

2. The head at the inlet³,1R+ R1 = 22!2m

´represents available energy. Most of

this energy goes to head loss³<E

? 222D= 17!1m

´! The remainder is carried as

kinetic energy out of the pipe³T2

? 222D= 5!1m

´!

571

PROBLEM 7.16

Situation: Water owing from a tank is described in the problem statement.

Find: Pressure in tank.

APPROACH

Apply the energy equation from the water surface in the tank to the outlet.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

%1$( = , 22 $2) + ME " R1 = 6,22 $2) " 10

,2 = ]$I2 = 0!1$((0$4)(1$12)2) = 18!33 ft/s

%1$( = (6(18!332)$64!4)" 10 = 2!13 ft%1 = 62!4× 21!3 = 1329 psfg

%1 = 9.23 psig

572

PROBLEM 7.17

Situation: A pipe draining a tank is described in the problem statement.

Find: Pressure at point I and velocity at exit.

Assumptions: T2 = 1

APPROACH

To nd pressure at point A, apply the energy equation between point A and the pipeexit. Then, then apply energy equation between top of tank and the exit.

ANALYSIS

Energy equation (point A to pipe exit).

%=(+ R= + T=

, 2=2)+ M, =

%2(+ R2 + T2

, 222)+ M5 + ME

Term by term analysis: ,= = ,2 (continuity); %2 = 0-gage; (R= " R>) = ?; M, = 0'M5 = 0' ME = 0! Thus

%= = "(?= "62!4× 4

%= = -250 lb/ft2

Energy equation (top of tank and pipe exit)

%1$( + T1,21 $2) + R1 + M, = %2$( + T2,

22 $2) + R2 + M5 + ME

R1 = , 22 $2) + R2

,2 =p2)(R1 " R2)

=&2× 32!2× 14,2 = 30.0 ft/s

573

PROBLEM 7.18

Situation: A pipe draining a tank is described in the problem statement.

Find: Pressure at point I and velocity at the exit.

Assumptions: T1 = 1!

APPROACH

To nd pressure at point A, apply the energy equation between point A and the pipeexit. Then, then apply energy equation between top of tank and the exit.

ANALYSIS

Energy equation

%=(+ R= + T=

, 2=2)+ M, =

%2(+ R2 + T2

, 222)+ M5 + ME

Term by term analysis: ,= = ,2 (continuity); %2 = 0-gage; (R= " R>) = ?; M, = 0'M5 = 0' ME = 0! Thus

%= = "(?%= = "

¡9810N$m3

¢(2m)

%= = "19!6 kPa

Energy equation

%1$( + T1,21 $2) + R1 + M, = %2$( + T2,

22 $2) + R2 + M5 + ME

R1 = , 22 $2) + R2

,2 =p2)(R1 " R2)

=&2× 9!81× 10,2 = 14.0 m/s

574

PROBLEM 7.19


Find: Pressure di!erence between I and D!

ANALYSIS

Flow rate equation

,= = ]$I1

= 1!910m/s

,> =

µ20

12

¶2× ,=

= 5!31 m/s

Energy equation

%= " %> = 1( + (#$2)(, 2> " ,2=) ;

%= " %> = 1× 9810× 0!9 + (900$2)(5!312 " 1!912)

= 19.88 kPa

575

PROBLEM 7.20


Find: Discharge in pipe.

Assumptions: T = 1.

APPROACH

Apply the energy equation from the water surface in the reservoir (pt. 1) to the outletend of the pipe (pt. 2).

ANALYSIS

Energy equation%1(+, 212)+ R1 =

%2(+, 222)+ R2 + ME

Term by term analysis:

%1 = 0; %2 = 0

R2 = 0; ,1 ' 0

The energy equation becomes.

R1 =, 222)+ ME

11m =, 222)+ 5

, 222)= 6

, 222)

, 22 =

µ2)

6

¶(11)

,2 =

sµ2× 9!81m$ s2

6

¶(11m)

,2 = 5!998 m/s

Flow rate equation

] = ,2I2

= (5!998m$ s)¡9 cm2

¢µ10"4m2cm2

¶

= 5! 398 2× 10"3m3

s

] = 5!40× 10"3 m3$ s

576

PROBLEM 7.21

Situation: An engineer is estimating the power that can be produced by a smallstream.Stream discharge: ] = 1!4 cfs. Stream temperature: & = 40 !F.Stream elevation: 4 = 34 ft above the owner’s residence!

Penstock

Turbine & Generato

H

Find: Estimate the maximum power in kilowatts that can be generated.(a) The head loss is 0.0 ft, the turbine is 100% e"cient and the generator is 100%e"cient.(b) The head loss is 5.5 ft, the turbine is 70% e"cient and the generator is 90%e"cient.

APPROACH

To nd the head of the turbine (Mt), apply the energy equation from the upper watersurface (section 1) to the lower water surface (section 2). To calculation power,use * = g( /)Mt)' where g accounts for the combined e"ciency of the turbine andgenerator.

ANALYSIS

Energy equation

%1(+ T1

, 212)+ R1 =

%2(+ T2

, 222)+ R2 + M5 + ME (1)

Term by term analysis

%1 = 0; ,1 % 0%2 = 0; ,2 % 0

R1 " R2 = 4

Eq. (1) becomes

4 = Mt + ME

Mt = 4 " ME

Flow rate

/) = (]

=¡62!4 lbf$ ft3

¢ ¡1!4 ft3$ s

¢

= 87!4 lbf$ s

577

Power (case a)

* = /)M5

= /)4

= (87!4 lbf$ s) (34 ft) (1!356 J/ ft · lbf)= 4!02 kW

Power (case b).

* = g /) (4 " ME)= (0!7)(0!9) (87!4 lbf$ s) (34 ft" 5!5 ft) (1!356 J/ ft · lbf)= 2!128 kW

Power (case a) = 4.02 kW

Power (case b) = 2.13 kW

COMMENTS

1. In the ideal case (case a), all of the elevation head is used to make power.When typical head losses and machine e"ciencies are accounted for, the powerproduction is cut by nearly 50%.

2. From Ohm’s law, a power of 2.13 kW will produce a current of about 17.5 ampsat a voltage of 120V. Thus, the turbine will provide enough power for about1 typical household circuit. It is unlikely the turbine system will be practical(too expensive and not enough power for a homeowner).

578

PROBLEM 7.22

Situation: Flow in a pipe is described in the problem statement.

Find: Pressure at station 2.

APPROACH

Apply ow rate equation and then the energy equation.

ANALYSIS

Flow rate equation

,1 =]

I1=

6

0!8= 7!5 ft/s

, 212)

= 0!873 ft

,2 =]

I2=

6

0!2= 30 ft/s

, 222)

= 13!98 ft

Energy equation

%1(+, 212)+ R1 =

%2(+, 222)+ R2 + 6

15× 1440!8× 62!4

+ 0!873 + 12 =%2(+ 13!98 + 0 + 6

%2(

= 36!16 ft

%2 = 36!16× 0!8× 62!4= 1185 psfg

%2 = 8!23 psig

579

PROBLEM 7.23


Find: (a) Discharge in pipe(b) Pressure at point B.

Assumptions: ( = 9810 N/m

APPROACH

Apply the energy equation.

ANALYSIS

Energy equation

%reser.$( + ,2K $2) + RK = %outlet$( + ,

20 $2) + R0

0 + 0 + 5 = 0 + , 20 $2)

,0 = 9!90 m/s

Flow rate equation

] = ,0I0

= 9!90× (0$4)× 0!202

] = 0!311 m3/s

Energy equation from reservoir surface to point B:

0 + 0 + 5 = %>$( + ,2>$2) + 3!5

where

,> = ]$,> = 0!311$[(0$4)× 0!42] = 2!48 m/s, 2>$2) = 0!312 m

%>$( " 5" 3!5 = 0!312

%> = 11!7 kPa

580

PROBLEM 7.24

Situation: A microchannel is described in the problem statement.

Find: Pressure in syringe pump.

APPROACH

Apply the energy equation and the ow rate equation.

ANALYSIS

Energy equation

%1(

= ME + T2, 2

2)

=3272,

(12+ 2

, 2

2)(1)

Flow rateThe cross-sectional area of the channel is 3.14×10"8 m2! A ow rate of 0.1 7l/s is10"7 l/s or 10"10 m3/s. The ow velocity is

, =]

I

=10"10

3!14× 10"8= 0!318× 10"2 m/s= 3!18 mm/s

Substituting the velocity and other parameters in Eq. (1) gives

%1(

=32× 1!2× 10"3 × 0!05× 0!318× 10"2

7' 850× 4× (10"4)2+ 2×

(0!318× 10"2)2

2× 9!81= 0!0194 m

The pressure is

%1 = 799 kg/m3 × 9!81 m/s2 × 0!0194 m

%1 =152.1 Pa

581

PROBLEM 7.25

Situation: A re hose is described in the problem statement.

Find: Pressure at hydrant.

APPROACH

Apply the energy equation.

ANALYSIS

Energy equation

%1(+ R1 =

, 222)+ R2 + ME

where the kinetic energy of the uid feeding the hydrant is neglected. Because of thecontraction at the exit, the outlet velocity is 4 times the velocity in the pipe, so theenergy equation becomes

%1(

=, 222)+ R2 " R1 + 10

, 2

16× 2)

%1 =

µ1!625

2), 2 + 50

¶(

=

µ1!625

2× 9!81× 402 + 50

¶9810

= 1! 791× 106 Pa

%1 = 1790 kPa

582

PROBLEM 7.26

Situation: A siphon is described in the problem statement.

Find: Pressure at point D!

ANALYSIS

Flow rate equation

,1 = ]$I2

,1 = 2!8$((0$4)× (8$12)2)= 8!02 ft/s

Energy equation (from reservoir surface to -)

%1$( + ,21 $) + R1 = %1$( + ,

21 $2) + R1 + ME

0 + 0 + 3 = 0 + 8!022$64!4 + 0 + ME

ME = 2!00 ft

Energy equation (from reservoir surface to D).

0 + 0 + 3 = %>$( + ,2>$2) + 6 + (3$4)× 2 ; ,> = ,) = 8!02 ft/s

%>$( = 3" 1" 6" 1!5 = "5!5 ft%> = "5!5× 62!4

= "343 psfg%> = -2.38 psig

583

PROBLEM 7.27


Find: Force on pipe joint.

APPROACH

Apply the momentum principle, then the energy equation.

ANALYSIS

xVx

p =02

Fj

p A1 1

Momentum Equation

XH$ = /,!I$ " /,#I$

HV + %1I1 = "#, 2$I+ #,2$I

HV = "%1I1

Energy equation

%1(+, 212)+ R1 =

%2(+, 222)+ R2 + ME

%1 " %2 = (ME%1 = ((3) = 187!2 psfg

HV = "187!2× (9

144)

HV = "11!7 lbf

584

PROBLEM 7.28


Find:a.) Discharge.b.) Pressure at point D!

APPROACH

Apply the energy equation from A to C, then from A to B.

ANALYSIS

Head loss

M<pip e =, 2,2)

Mtotal = M<pip e + M<outlet = 2, 2,2)

Energy equation (from A to C)

0 + 0 + 30 = 0 + 0 + 27 + 2, 2,2)

,, = 5!42 m/s

Flow rate equation

] = ,,I,

= 5!42× (0$4)× 0!252

] = 0!266 m3/s

Energy equation (from A to B)

30 =%>(+, 2,2)+ 32 + 0!75

, 2,2)

%>(

= "2" 1!75× 1!497 m

%> = "45!3 kPa, gage

585

PROBLEM 7.29


Find: Depth of water in upper reservoir for incipient cavitation.

APPROACH

Apply the energy equation from point A to point B.

ANALYSIS

Flow rate equation

, = ]$I

=8× 10"4 m3/s1× 10"4 m2

= 8m$ s

Calculations

, 2$2) = 82$(2× 9!81) = 3!262 mMEI=#> = 1!8, 2$2) = 5!872 m

Energy equation (from A to B; let R = 0 at bottom of reservoir)

%=$( + ,2=$2) + R= = %>$( + ,

2>$2) + R> + ME

100' 000$9' 810 + 0 + R= = 1' 230$9' 810 + 3!262 + 10 + 5!872

R= = depth = 9.07m

586

PROBLEM 7.30


Find: Direction of ow.

Assumptions: Assume the ow is from A to B.

APPROACH

Apply the energy equation from A to B.

ANALYSIS

Energy equation

%=$( + ,2=$2) + R= = %>$( + ,

2>$) + R> + ME

(10' 000$9' 810) + 10 = (98' 100$9' 810) + 0 + ME

ME = 1!02 + 10 = 10!0

= +1!02

Because the value for head loss is positive it veries our assumption of downwardow. Correction selection is b)

587

PROBLEM 7.31

Situation: A system with a machine is described in the problem statement.

Find: Pressures at points I and D!

Assumptions: Machine is a pump

APPROACH

Apply the energy equation between the top of the tank and the exit, then betweenpoint B and the exit, nally between point A and the exit.

ANALYSIS

Energy equation

R1 + M, =, 222)+ R2

Assuming the machine is a pump. If the machine is a turbine, then M, will be negative.The velocity at the exit is

,2 =]

I2=

10840!52

= 50!93 ft/s

Solving for M, and taking the pipe exit as zero elevation we have

M, =50!932

2× 32!2" (6 + 12) = 22!3 ft

Therefore the machine is a pump.Applying the energy equation between point B and the exit gives

%>(+ R> = R2

Solving for %> we have

%> = ((R2 " R>)%> = "6× 62!4 = "374 psfg

%>=-2.6 psig

Velocity at A

,= =

µ6

12

¶2× 50!93 = 12!73 ft/s

Applying the energy equation between point A and the exit gives

588

%=(+ R= +

, 2=2)=, 222)

so

%= = ((, 222)" R= "

, 2=2))

= 62!4× (50!932 " 12!732

2× 32!2" 18)

= 1233 psfg

%= = 8.56 psig

589

PROBLEM 7.32

Situation: A system is described in the problem statement.

Find: Pressure head at point 2.

ANALYSIS

Let ,: = velocity of jet from nozzle:Flow rate equation

,: =]

I:=

0!10

((0$4)× 0!102)= 12!73 m/s

, 2:2)

= 8!26 m

,2 =]

I2=

0!10

((0$4)× 0!32)= 1!41 m/s

, 222)

= !102 m

Energy equation

%2(+ 0!102 + 2 = 0 + 8!26 + 7

,2R= 13!16 m

590

PROBLEM 7.33

Situation: A pump draws water out of a tank and moves this water to elevation C.Diameter of inlet pipe is 8 in! Diameter of outlet pipe is 1) = 4 in!Speed of water in the 4 in pipe is ,) = 12 ft$ s! Power delivered to the pump is 25 hp!Pump e"ciency is g = 60%! Head loss in pipe (between A & C) is ME = 2, 2)$2)!

Find: Height (M)above water surface.

APPROACH

Apply the energy equation from the reservoir water surface to the outlet.

ANALYSIS

Energy equation

%1(+ T1

, 212)+ R1 + M, =

%2(+ T2

, 222)+ R2 + ME

0 + 0 + 0 + M, = 0 +, 212)+ M+ 2

, 212)

M, = M+ 3, 212)

(1)

Velocity head, 212)=122

64!4= 2!236 ft (2)

Flow rate equation

] = ,)I)

=

µ12 ft

s

¶Ã0 (4$12 ft)2

4

!

= 1!047 ft3$ s

Power equation

* ( hp) =](M,550g

M, =* (550) g

](

=25 (550) 0!6

1!047 (62!4)

= 126!3 ft (3)

591

Substitute Eqs. (2) and (3) into Eq. (1)

M, = M+ 3, 212)

126!3 ft = M+ (3× 2! 236) ftM = 119!6 ft

M = 120 ft

592

PROBLEM 7.34

Situation: A system with pump is described in the problem statement.

Find: Height above water surface.

ANALYSIS

Energy equation

0 + 0 + 0 + M, = 0 + M+ 3!0, 212)

, 2)2)

=32

(2× 9!81)= 0!459 m

* =](M,0!6

M, =25' 000× 0!6

(3× 0$4× 0!102 × 9' 810)= 64!9 m

M = 64!9" 3!0× !459M = 63!5 m

593

PROBLEM 7.35


Find: Horsepower delivered by pump.

APPROACH

Apply the ow rate equation, then the energy equation from A to B. Then apply thepower equation.

ANALYSIS

Flow rate equation

,= =]

I==

3!0

((0$4)× 12)= 3!82 ft/sec

, 2=2)

= 0!227 ft

,> =]

I>=

3!0

((0$4)× 0!52)= 15!27 ft/s

, 2>2)

= 3!62 ft

Energy equation

%=(+, 2=2)+ R= + M, =

%>(+, 2>2)+ R>

5×144

62!4+ 0!227 + 0 + M, = 60×

144

62!4+ 3!62 + 0

M, = 130!3 ft

Power equation

* (hp) =](M,550

= 3!0× 62!4×130!3

550

* = 44!4 hp

594

PROBLEM 7.36


Find: Power supplied to ow.

APPROACH

Apply the ow rate equation. Then apply the energy equation from reservoir surfaceto end of pipe. Then apply the power equation.

ANALYSIS

Flow rate equation

, = ]$I

= 8$((0$4)× 12)= 10!2 m/s

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 + ME

0 + 0 + 40 + M, = 0 + , 2$2) + 20 + 7, 2$2)

, 2$2) = 10!22$(2× 9!81) = 5!30 m

Then

40 + M, = , 2$2) + 20 + 7, 2$2)

M, = 8× 5!30 + 20" 40= 22!4 m

Power equation

* = ](M,

= 8× 9810× 22!4* = 1!76 MW

595

PROBLEM 7.37


Find: Power pump must supply.

APPROACH

Apply the ow rate equation, then the energy equation from reservoir surface to the10 m elevation. Then apply the power equation.

ANALYSIS

Flow rate equation

, = ]$I

= 0!25$((0$4)× 0!32)= 3!54 m/s

, 2$2) = 0!639 m

Energy equation

0 + 0 + 6 + M, = 100' 000$9810 + , 2$2) + 10 + 2!0, 2$2)

M, = 10!19 + 10" 6 + 3!0× 0!639M, = 16!1 m

Power equation

* = ](M,

= 0!25× 9!180× 16!1* = 39!5 kW

596

PROBLEM 7.38


Find: Horsepower pump supplies.

APPROACH

Apply the ow rate equation, then the energy equation. Then apply the powerequation.

ANALYSIS

Flow rate equation

,12 = ]$I12 = 6$((0$4)× 12) = 7!64 ft/sec, 212$2) = 0!906 ft

,6 = 4,12 = 30!56 ft/sec

, 26 $2) = 14!50 ft

Energy equation

(%6$( + R6)" (%12$( + R12) = (13!55" 0!88)(46$12)$0!88(%12$( + R12) + ,

212$2) + M, = (%6$( + R6) + ,

26 $2)

M, = (13!55$0!88" 1)× 3!833 + 14!50" 0!906M, = 68!8 ft

Power equation

* (hp) = ](M,$550

* = 6× 0!88× 62!4× 68!8$550

* = 41!2 hp

597

PROBLEM 7.39

Situation: A system with a turbine is described in the problem statement.

Find: Power output from turbine.

APPROACH

Apply the energy equation from the upstream water surface to the downstream watersurface. Then apply the power equation.

ANALYSIS

Energy equation

%1(+, 212)+ R1 =

%2(+,22)+ R2 + ME + M-

0 + 0 + 35 = 0 + 0 + 0 + 1!5, 2

2)+ M-

, =]

I=

400

((0$4)× 72)= 10!39 ft/s

, 2

2)= 1!68 ft

M5 = 35" 2!52 = 32!48 ft

Power equation

* (hp) = ](M5 ×0!9

550

=(400)(62!4)(32!48× 0!9)

550

* = 1326 hp

598

PROBLEM 7.40


Find: Power produced by turbine.

Assumptions:(a) All head loss is expansion loss.(b) 100% e"ciency.

APPROACH

Apply the energy equation from the upstream water surface to the downstream watersurface. Then apply the power equation.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + M5 + ME

0 + 0 + 15 m = 0 + 0 + 0 + M5 + ,2$2)

M5 = 15 m" (52$2))= 13!73 m

Power equation

* = ](M5

= (1 m3/s)(9810 N/m3)(13!73 m)

* = 134!6 kW

599

PROBLEM 7.41


Find:(a) Power generated by turbine.(b) Sketch the EGL and HGL.

APPROACH

Apply the energy equation from the upper water surface to the lower water surface.Then apply the power equation.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME + M5

0 + 0 + 100 ft = 0 + 0 + 4 ft + M5M5 = 96 ft

Power equation

* = (](M5)(e!.)

* (hp) = ](M5(e!.)$550 = 1' 000× 62!4× 96× 0!85$550

* = 9258 hp

Turbine

EGL

HGL

600

PROBLEM 7.42

Situation: A system with a pump is described in the problem statement.

Find: Power delivered by pump.

APPROACH

Apply the energy equation from the reservoir water surface to point B. Then applythe power equation.

ANALYSIS

Energy equation

%$( + , 2$2) + R + M, = %>$( + ,2>$2) + R>

0 + 0 + 40 + M, = 0 + 0 + 64; M, = 25 m

Flow rate equation

] = ,VIV = 25× 10"4 m2 × ,Vwhere ,V =

p2) × (65" 35) = 24!3 m/s

] = 25× 10"4 × 24!3 = 0!0607 m3/s

Power equation

* = ](M,

* = 0!0607× 9' 810× 25* = 14!89 kW

601

PROBLEM 7.43


Find: Power delivered by pump.

ANALYSIS

0 + 0 + 110 + M, = 0 + 0 + 200; M, = 90 ft

* (hp) = ](M,$550

] = ,VIV = 0!10 ,V

,V =p2) × (200" 110) = 76!13 ft/s

] = 7!613 ft3/s

Power equation

* = ](M,

* = 7!613× 62!4× 90$550

* = 77!7 hp

602

PROBLEM 7.44


Find: Power required for pump.

ANALYSIS

Energy equation

M, = R2 " R1 + ME

Expressing this equation in terms of pressure

(M, = (R2 " (R1 +!%2!&&

Thus pressure rise across the pump is

(M, = 53 lbf/ft3 × 200 ft +60× 144 lbf/ft2 = 19' 240 psf

Flow rate equation

] = , ×I

] = 3500 gpm× 0!002228ft3$sgpm

= 7!80 cfs

Power equation

. = ](M,

= 7!80× 19' 240$550. = 273 hp

603

PROBLEM 7.45


Find: Time required to transfer oil.

APPROACH

Apply the energy equation between the top of the uid in tank A to that in tank B.

ANALYSIS

Energy equation

M, + R= = R> + ME

or

M, + R= = R> + 20, 2

2)+, 2

2)

Solve for velocity

, 2 =2)

21(M, + R= " R>)

, 2 =2× 9!8121

(60 + R= " R>)

, = 0!966 6 (60 + R= " R>)1*2

The sum of the elevations of the liquid surfaces in the two tanks is

R= + R> = 21

So the energy equation becomes

, = 0!9666(81" 2R>)1*2

Continuity equation

AR>AP

= ,IpipeItank

= ,(0!2m)2

(12m)2

=¡2!778× 10"4

¢,

=¡2!778× 10"4

¢0!9666(81" 2R>)1*2

= 2!685× 10"4(81" 2R>)1*2

Separate variables

AR>(81" 2R>)1*2

= 2!685× 10"4AP

604

Integrate

20 ftZ

1

AR>(81" 2R>)1*2

=

!5Z

0

2!685× 10"4AP

¡"&81" 2R>

¢20 ft1 ft

=¡2!685× 10"4

¢!P

³"p81" 2 (20) +

p81" 2 (1)

´=

¡2!685× 10"4

¢!P

2! 485 1 =¡2!685× 10"4

¢!P

!P = 9256 s

!P = 9260 s = 2!57 hr

605

PROBLEM 7.46


Find:(a) Write a computer program to show how the pressure varies with time.(b) Time to pressurize tank to 300 kPa!

APPROACH

Apply the energy equation between the water surface at the intake and the watersurface inside the tank.

ANALYSIS

Energy equation

M, + R1 =%2(+ R2 + ME

Expressing the head loss in terms of the velocity allows one to solve for the velocityin the form

, 2 =2)

10(M, + R1 " R5 "

%5()

Substituting in values

, = 1!401(46" R5 " 10!193

4" R5)1*2

The equation for the water surface elevation in the tank is

!R5 = ,I,I5!P =

,

2500!P

A computer program can be written taking time intervals and nding the uid leveland pressure in the tank at each time step. The time to reach a pressure of 300 kPaabs in the tank is 698 seconds or 11.6 minutes. A plot of how the pressure varieswith time is provided.

606

Time (sec)

0 200 400 600 800

Pres

sure

(kPa

)

50

100

150

200

250

300

350

607

PROBLEM 7.47

Situation: A system with two tanks connected by a pipe is described in the problemstatement.

Find: Discharge between two tanks: ]

APPROACH

Apply the energy equation from water surface in A to water surface in B.

ANALYSIS

Energy equation

%=$( + ,2=$2) + R= = %>$( + ,

2>$2) + R> +

XME

%= = %> = %atm and ,= = ,> = 0

Let the pipe from A be called pipe 1. Let the pipe from B be called pipe 2Then X

ME = (,1 " ,2)2$2) + , 22 $2)


,1I1 = ,2I2

,1 = ,2(I2$I1)

However I2 = 2I1 so ,1 = 2,2! Then the energy equation gives

R= " R> = (2,2 " ,2)2$2) + , 22 $2)= 2, 22 $2)

,2 =p)(R= " R>)

=p10) m/s

Flow rate equation

] = ,2I2

=³p

10) m/s´(20 cm2)(10"4 m2/cm2)

] = 0!0198 m3/s

608

PROBLEM 7.48


Find:a) Horizontal force required to hold transition in place.b) Head loss.

APPROACH

Apply the ow rate equation, the sudden expansion head loss equation, the energyequation, and the momentum principle.

ANALYSIS

Flow rate equation

,40 = ]$I40 = 1!0$((0$4)× 0!402) = 7!96 m/s, 240$2) = 3!23 m

,60 = ,40 × (4$6)2 = 3!54 m/s, 260$2) = 0!639 m

Sudden expansion head loss equation

ME = (,40 " ,60)2$2)= 0.996 m

Energy equation

%40$( + ,240$2) = %60$( + ,

260$2) + ME

%60 = 70' 000 + 9810(3!23" 0!639" 0!996) = 85' 647 Pa

Momentum principle

p A1 1p A22

Fx

XH$ = /!,$I! " /#,$I#

70' 000× 0$4× 0!42 " 85' 647× 0$4× (0!62) + H$ = 1000× 1!0× (3!54" 7!96)H$ = "8796 + 24' 216" 4' 420

= 11' 000 N

H$ = 11.0 kN

609

PROBLEM 7.49


Find: Head loss.

APPROACH

Apply the continuity principle, then the sudden expansion head loss equation.

ANALYSIS


,8I8 = ,15I15

,15 =,8I8I15

= 4× (8$15)2 = 1!14 m/s


ME =(,8 " ,15)2

(2))

ME =(4" 1!14)2

(2× 9!81)

ME = 0.417 m

610

PROBLEM 7.50


Find: Head loss

APPROACH

Apply the ow rate equation, then the sudden expansion head loss equation.

ANALYSIS

Flow rate equation

,6 = ]$I6 = 5$((0$4)× (1$2)2) = 25!46 ft/s;,12 = (1$4),6 = 6!37 ft/s


ME = (,6 " ,12)2$(2))= (25!46" 6!37)2$(2× 32!2)

ME = 5.66 ft

611

PROBLEM 7.51


Find:(a) Horsepower lost.(b) Pressure at section 2.(c) Force needed to hold expansion.

APPROACH

Find the head loss by applying the sudden expansion head loss equation, rst solvingfor ,2 by applying the continuity principle. Then apply the power equation, theenergy equation, and nally the momentum principle.

ANALYSIS

Continuity equation

,2 = ,1(I1$I2)

= 25(1$4)

= 6!25 ft/s


ME = (,1 " ,2)2$(2))ME = (25" 6!25)2$64!4

= 5!46 ft

a)Power equation

* (hp) = ](M$550

] = , I = 25(0$4)(52) = 490!9 ft3$s

* = (490!9)(62!4)(5!46)$550

* = 304 hp

b)Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

(5× 144)$62!4 + 252$64!4 = %2$( + 6!252$64!4 + 5!46

%2$( = 15!18 ft

%2 = 15!18× 62!4= 947 psfg

%2 = 6.58 psig

612

c)Momentum equation

XH$ = /!,$I! " /#,$I#

/ = 1!94× (0$4)× 52 × 25= 952!3 kg/s

%1I1 " %2I2 + H$ = /(,2 " ,1)(5)(14)0$4)(52)" (6!57)(144)(0$4)(102) + H$ = 952!3× (6!25" 25)

H$ = 42,426 lbf

613

PROBLEM 7.52


Find: Longitudinal force transmitted through pipe wall.

APPROACH

Apply the energy equation, then the momentum principle.

ANALYSIS

1

2

c.s.

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

but ,1 = ,2 and %2 = 0! Therefore

%1$( = "50 + 10%1 = "2496 lbf/ft2

Momentum principle

XH3 = /,3I! " /,3I# = #](,23 " ,13)

"%1I1 " (I2" 22+ Hwall = 0

Hwall = 1!52+ (I12" %1I1= 75 + (0$4)× 0!52(62!4× 50" 2' 496)= 75 + 122!5

Hwall = 197.5 lbf

614

PROBLEM 7.53


Find: (a) Pressure at outlet of bend.(b) Force on anchor block in the E-direction.

APPROACH

Apply the energy equation, then the momentum principle.

ANALYSIS

Energy equation

%50$( + ,250$2) + R50 = %80$( + ,

280$2) + R80 + ME

where %50 = 650' 000 Pa and R50 = R80Flow rate equation

,80 = ]$I80 = 5$((0$4)× 0!82) = 9!947 m/s, 280$2) = 5!04 m

Continuity equation

,50 = ,80 × (8$5)2 = 25!46 m/s, 250$2) = 33!04 m

ME = 10 m

Then

%80$( = 650' 000$( + 33!04" 5!04" 10%80 = 650' 000 + 9' 810(33!04" 5!04" 10) = 826' 600 Pa

%80 = 826.6 kPa

Momentum principle

XH$ = /,! " /,# = #](,80I$ " ,50I$)

%80I80 + %50I50 × cos 60! + H$ = 1' 000× 5("9!947" 0!5× 25!46)H$ = "415' 494" 63' 814" 113' 385

= "592' 693NH$ = -592.7 kN

615

PROBLEM 7.54


Find: Head loss at pipe outlet.

APPROACH


ANALYSIS

Flow rate equation

, = ]$I

= 10((0$4)× 12)= 12!73 ft/sec


ME = , 2$2)

ME = 2.52 ft

616

PROBLEM 7.55


Find: Head loss at pipe outlet.

APPROACH


ANALYSIS

Flow rate equation

, = ]$I

= 0!50$((0$4)× 0!52)= 2!546 m/s


ME = , 2$2)

= (2!546)2$(2× 9!81)

ME =0.330 m

617

PROBLEM 7.56


Find: Maximum allowable discharge before cavitation.

Properties: From Table A.5 %@ = 2340 Pa, abs.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2

0 + 0 + 5 = %2Ivapor$( + ,22 $2) + 0

%2Ivapor = 2340" 100' 000 = "97' 660 Pa gage

Then

, 22 $2) = 5 + 97' 660$9' 790 = 14!97 m

,2 = 17!1 m/s

Flow rate equation

] = ,2I2

= 17!1× 0$4× 0!152

] = 0.302 m3/s

618

PROBLEM 7.57


Find:a.) Head (4) at incipient cavitation.b) Discharge at incipient cavitation.

Properties: From Table A.5 %@ = 2340 Pa, abs.

APPROACH

First apply the energy equation from the Venturi section to the end of the pipe. Thenapply the energy equation from reservoir water surface to outlet:

ANALYSIS

(b) Energy equation from Venturi section to end of pipe:

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

%vapor$( + ,21 $2) = 0 + , 22 $2) + 0!9,

22 $2)

%vapor = 2' 340 Pa abs. = "97' 660 Pa gage


,1I1 = ,2I2

,1 = ,2I2$I1

= 2!56,2

Then, 21 $2) = 6!55,

22 $2)

Substituting into energy equation

"97' 660$9' 790 + 6!55, 22 $2) = 1!9, 22 $2)

,2 = 6!49 m/s

Flow rate equation

] = ,2I2

= 6!49× 0$4× 0!42

] = 0.815 m3/s

Energy equation from reservoir water surface to outlet:

R1 = , 22 $2) + ME

4 = 1!9, 22 $2)

4 = 4.08 m

619

PROBLEM 7.58

Situation: A system with a machine is described in the problem statement.

Find: (a) Direction of ow.(b) What kind of machine is at point A.(c) Compare the diameter of pipe sections.(d) Sketch the EGL.(e) If there is a vacuum at anywhere, if so where it is.

ANALYSIS

(a) Flow is from right to left.(b) Machine is a pump.(c) Pipe CA is smaller because of steeper H.G.L.(d)

EGL

(e) No vacuum in the system.

620

PROBLEM 7.59

Situation: A system with a reservoir, pipe, and nozzle is described in the problemstatement.

Find:(a) Discharge (]) !(b) Draw the HGL and EGL.

APPROACH

Apply the energy equation from the reservoir surface to the exit plane of the jet.

Assumptions:

ANALYSIS

Energy equation. Let the velocity in the 6 inch pipe be ,6! Let the velocity in the12 inch pipe be ,12!

%1$( + ,21 $2) + R1 = %2$( + ,

26 $2) + R2 + ME

0 + 0 + 100 = 0 + , 26 $2) + 60 + 0!025(1000$1),212$2)


,6I6 = ,12I12

,6 = ,12(I12$I6)

,6 = ,12122

62= 4,12

, 26 $2) = 16, 212$2)

Substituting into energy equation

40 = (, 212$2))(16 + 25)

, 212 = (40$41)2× 32!2,12 = 7!927 ft/s

Flow rate equation

] = ,12I12

= (7!927)(0$4)(12)

] = 6!23 ft3/s

621

EGL

HGL

622

PROBLEM 7.60

Situation: A hydroelectric power plant is described in example 7.5.

Find: Draw the HGL and EGL.

ANALYSIS

EGL

HGL

1.52 m

EGL

HGL

V /2g2

E1.=610m

623

PROBLEM 7.61

Situation: A ow system is described in problem 7.57.


ANALYSIS

EGL

HGL0.71m

10 m (approx.)

624

PROBLEM 7.62

Situation: A reservoir and pipe system is described in example 7.3.


ANALYSIS

37.2 m

42.6 m

2000 m

80 m

625

PROBLEM 7.63

Situation: A system with a black box is described in the problem statement.

Find: What the black box could be.

ANALYSIS

Because the E.G.L. slopes downward to the left, the ow is from right to left. Inthe ”black box” there could either be a turbine, an abrupt expansion or a partiallyclosed valve. Circle b, c, d.

626

PROBLEM 7.64

Situation: A system with an HGL is described in the problem statement.

Find: Whether this system is possible, and if so under what conditions.

ANALYSIS

This is possible if the uid is being accelerated to the left.

627

PROBLEM 7.65

Situation: A system with two tanks connected by a tapered pipe is described in theproblem statement.


ANALYSIS

EGL

HGL

628

PROBLEM 7.66

Situation: A system with an HGL and EGL is described in the problem statement.

Find: See problem statement.

ANALYSIS

(a) Solid line is EGL, dashed line is HGL.(b) No; AB is smallest.(c) From B to C.(d) pmax is at the bottom of the tank.(e) pmin is at the bend C.(f) A nozzle.(g) above atmospheric pressure.(h) abrupt expansion.

629

PROBLEM 7.67

Situation: A system with two tanks connected by a pipe is described in the problemstatement and gure 7.8.

Find: Discharge of water in system

APPROACH

Apply energy equation from upper to lower reservoir.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 100 = 0 + 0 + 070 +X

MEX

ME = 30 m

ME = !02× (2$1)(, 2$2))30 = 0!02× (200$0!3)(, 2T $2)) + (0!02(100$0!15) + 1!0),

2A $2) (1)

Flow rate equation

,T = ]$IT = ]$((0$4)× 0!32) (2)

,A = ]$IA = ]$((0$4)× 0!152) (3)

Substituting Eq. (2) and Eq. (3) into (1) and solving for ] yields:

] = 0.110 m3/s

630

PROBLEM 7.68


Find:(a) Power supplied to the pump.(b) Sketch the HGL and EGL.

APPROACH

Apply the ow rate equation to nd the velocity. Then calculate head loss. Nextapply the energy equation from water surface to water surface to nd the head thepump provides. Finally, apply the power equation.

ANALYSIS

Flow rate equation

, = ]$I

= 3!0$((0$4)× (8$12)2)= 8!594 ft/sec

Head loss

ME =

µ0!018

2

1

, 2

2)

¶+

µ, 2

2)

¶

= 0!018

µ3000

8$12

¶8!5942

2 (32!2)+8!5942

2 (32!2)

= 94! 04 ft

Energy equation

%1(+ T1

, 212)+ R1 + M, =

%2(+ T2

, 222)+ R2 + ME

0 + 0 + 90 + M, = 0 + 0 + 140 + 94!04

M, = 144!0 ft

Power equation

* = ](M,

= 3!0× 62!4× 144

= 26957ft lbf

s

= 26957ft lbf

s

µft · lbf550 hp · s

¶

* = 49!0 hp

631

EGL

HGL

EGL

HGL

632

PROBLEM 7.69

Situation: A system with two tanks connected by a pipe is described in the problemstatement.

Find: (a) Discharge in pipe.(b) Pressure halfway between two reservoirs.

APPROACH

To nd the discharge, apply the energy equation from water surfaceI to water surfacein D! To nd the pressure at location P, apply the energy equation from water surfaceI to location * .

ANALYSIS

Energy equation

%=$( + ,2=$2) + R= = %>$( + ,

2>$2) + R> + ME

0 + 0 +4 = 0 + 0 + 0 + 0!01× (300$1), 2, $2) + ,2, $2)

16 = 4, 2, $2)

,, =&4× 2× 9!81 = 8!86 m/s

Flow rate equation

] = , I

= 8!86× (0$4)× 12

] = 6.96 m3/s

Energy equation between the water surface in I and point * :

0 + 0 +4 = %,$( + ,2, $2) " M+ 0!01× (150$1),

2, $2)

16 = %,$( " 2 + 2!5, 2, $2)

where , 2, $2) = 4 m. Then

%, = 9' 810(16 + 2" 10)

%, = 78.5 kPa

EGLHGL

p/(

V /2g=4 m2

633

PROBLEM 7.70

Situation: A system with two reservoirs connected by a pipe is described in theproblem statement.

Find: Elevation in left reservoir.

APPROACH

Apply the energy equation from the left reservoir to the right reservoir.

ANALYSIS

Energy equation

%E$( + ,2E$2) + RE = %F$( + ,

2F$2) + RF + ME

0 + 0 + RE = 0 + 0 + 110 + 0!02(200$1!128)(, 21 $2))

+0!02(300$1!596)(, 22 $2)) + (,1 " ,2)2$2) + , 22 $2)

Flow rate equation

,1 = ]$I1

= 16$1 = 16 ft/s

,2 = 8 ft/s

Substituting into the energy equation

RE = 110 + (0!02$2))((200$1!238)(162) + (300$1!596)(82)) + ((16" 8)2$64!4) + 82$64!4= 110 + 16!58 + 0!99 + 0!99

RE = 128.6 ft

EGLHGL

V /2g12

V /2g22

V1 V2

634

PROBLEM 7.71


Find: (a) Pump power.(b) Sketch the HGL and EGL.

APPROACH

Apply the energy equation from the upper reservoir surface to the lower reservoirsurface.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 + ME

0 + 0 + 150 + M, = 0 + 0 + 250 +X

0!018(2$1)(, 2$2)) + , 2$2)

Flow rate equation

,1 = ]$I1 = 3$((0$4)× 12) = 3!82 m/s, 21 $2) = 0!744 m

,2 = ]$I2 = 4,1 = 15!28 m/s

, 22 $2) = 11!9 m


M, = 250" 150 + 0!018[(100$1)× 0!744 + (1' 000$0!5)× 11!9] + 11!9= 541!6 m

Power equation

* = ](M,$e!.

= 3× 9' 810× 541!6$0!74

* = 21.54 MW

EGL

HGLh =535 mp

11.9 m

0.794 m

635

PROBLEM 7.72

Situation: A system showing the HGL and EGL is described in the problem statementand in Figure 7.9.

Find: (a) Water discharge in pipe(b) Pressure at highest point in pipe.

APPROACH

First apply energy equation from reservoir water surface to end of pipe to nd the ,to calculate the ow rate. Then to solve for the pressure midway along pipe, applythe energy equation to the midpoint:

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

0 + 0 + 200 = 0 + , 2$2) + 185 + 0!02(200$0!30), 2$2)

14!33, 2$2) = 15

, 2$2) = 1!047

, = 4!53 m/s

Flow rate equation

] = , I

= 4!53× (0$4)× 0!302

] = 0.320 m3/s

Energy equation to the midpoint:

%1$( + ,21 $2) + R1 = %"$( + ,

2"$2) + R" + ME

0 + 0 + 200 = %"$( + ,2"$2) + 200 + 0!02(100$0!30),

2$2)

%"$( = "(, 2$2)(1 + 6!667)= ("1!047)(7!667) = "8!027 m

%" = "8!027(= "78' 745 Pa

%" = -78.7 kPa

636

PROBLEM 7.73


Find: Time required to ll tank to depth of 10 m.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 + ME

but %1 = %2 = 0' R1 = 0' ,1 = 0' ,2 ' 0! The energy equation reduces to

0 + 0 + 0 + M, = 0 + 0 + (2 m+ M) + ME

where M =depth of water in the tank

20" (4)(104)]2 = M+ 2 + , 2$2) + 10, 2$2)

where , 2$2) is the head loss due to the abrupt expansion. Then

18 = (4)(104)]2 + 11(, 2$2)) + M

, = ]$I

(11, 2)$2) = (11$2))(]2$I2) = (1!45)(105)]2

18 = 1!85× 105]2 + M]2 = (18" M)$((1!85)(105))] = (18" M)045$430

But ] = I-AM$AP where I- = tank area, so

! AM$AP = (18" M)045$((430)(0$4)(5)2) = (18" M)045$8' 443AM$(18" M)045 = AP$8' 443

Integrate:"2(18" M)045 = (P$8' 443) + const.

But P = 0 when M = 0 so const. = "2(18)045! Then

P = (18045 " (18" M)045)(16' 886)

For M = 10 m

P = (18045 " 8045)(16' 886)= 23' 880 s

P = 6.63 hrs

637

PROBLEM 7.74

Situation: A system showing the HGL and EGL is described in the problem state-ment.

Find:(a) Direction of ow.(b) Whether there is a reservoir.(c) Whether the diameter at E is uniform or variable.(d) Whether there is a pump.(e) Sketch a physical set up that could exist between - and 1.(f) Whether there is anything else revealed by the sketch.

ANALYSIS

(a) Flow is from I to N because EGL slopes downward in that direction.(b) Yes, at 1, because EGL and HGL are coincident there.(c) Uniform diameter because , 2$2) is constant (EGL and HGL uniformly spaced).(d) No, because EGL is always dropping (no energy added).(e)

(f) Nothing else.

638

PROBLEM 7.75

Situation: A system with a reservoir and a pipe is described in the problem statement.

Find:(a) Discharge(b) Draw HGL and EGL(c) location of maximum pressure(d) location of minimum pressure(e) values for maximum and minimum pressure

APPROACH

Apply the energy equation from reservoir water surface to jet.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

0 + 0 + 100 = 0 + , 22 $2) + 30 + 0!014(2$1)(,2, $2))

100 = 0 + , 22 $2) + 30 + 0!014(500$0!60),2, $2)

Continuity equation

,2I2 = ,,I,

,2 = ,,I,$IE

,2 = 4,,

Then

, 2, $2)(16 + 11!67) = 70

,, = 7!045 m/s

, 2, $2) = 2!53 m

Flow rate equation

] = ,,I,

= 7!045× (0$4)× 0!602

] = 1.992 m3/s

639

pmin

EGLHGL

pmax

40.5 m

%min : 100 = %min$( + ,2, $2) + 100 + 0!014(100$0!60),

2, $2)

100 = %min$( + 100 + 3!33× 2!53

%min = -82.6 kPa, gage

%max$( = 40!5" 2!53 m

%max = 372.5 kPa

640

PROBLEM 7.76

Situation: A wind mill is described in problem 6.66.

Find: Power developed by windmill.

Assumptions: Negligible head loss.

APPROACH

Apply energy equation from upstream end to downstream end and the continuityprinciple to nd the head delivered to the turbine. Then apply the power equation.

ANALYSIS

Energy equation

%1$( + ,21 $2) = %2$( + ,

22 $2) + M5

M5 = , 21 $2) " ,22 $2)


,2 = ,1I1$I2 = ,1(3$4!5)2 = 0!444,1

, 22 $2) = 0!197, 21 $2)

Then substituting into the energy equation

M5 = 102$(2× 9!81)[1" 0!197]= 4!09 m

Power equation

* = ](M5

= 10(0$4)× 32 × 1!2× 9!81× 4!09

* = 3.40 kW

641

PROBLEM 7.77

Situation: A design of a subsonic wind tunnel is described in the problem statement.

Find: Power required.

APPROACH

To nd the head provided by the pump, apply the energy equation from upstreamend to downstream end . Then apply the power equation.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 + ME

0 + 0 + 0 + M, = 0 + , 22 $2) + 0 + 0!025,2- $2)


,-I- = ,2I2

,2 = ,-I-$I2

= ,- × 0!4, 22 $2) = 0!16, 2- $2)


M, =, 2-2)(0!185)

=602

2× 9!81(0!185)

M, = 33!95 m

Power equation

* = ](M,

= (, I) (#))M,

= (60× 4) (1!2× 9!81) (33!95)

* = 95.9 kW

642

PROBLEM 7.78

Situation: Flow through a pipe accelerated around a disk—additional details are pro-vided in the problem statement.

Find:(a) Develop an expression for the force required to hold the disk in place in terms ofU, D, d, and #!(b) Force required under given conditions.

APPROACH

Apply the energy equation from section (1) to section (2), and apply the momentumprinciple.

ANALYSIS

Control volume

U2

F disk on fluid

U1

Energy equation

%1 + #\21$2 = %2 + #\

22$2

%1 " %2 = #\22$2" #\21$2

but

\1I1 = \2(0$4)(12 " A2)

\2 = \112$(12 " A2) (1)

Then%1 " %2 = (#$2)\21 [(1

4$(12 " A2)2 " 1] (2)

Momentum principle for the C.V.

XH$ = /!\! " /#\# = #](\2$ " \1$)

%1I" %2I+ Hdisk on uid = #](\2 " \1)Huid on disk = HA = #](\1 " \2) + (%1 " %2)I

643

Eliminate %1 " %2 by Eq. (2), and \2 by Eq. (1):

HA = #\I(\1 " \112$(12 " A2)) + (#\2$2)[(14$(12 " A2)2 " 1]I

HA = #\2012$8[1$(12$A2 " 1)2]

When \ = 10 m/s, 1 = 5 cm, A = 4 cm and # = 1!2 kg/m3

HA = (1!2× 1020 × (0!05)2$8)[1$((0!05$0!04)2 " 1)2]HA = 0.372 N

644

PROBLEM 8.1

Situation: Consider equations: (a) ] = (2$3)-2&2)43*2, (b) , = (1!49$Q)"2*361*2,

(c) M( = b(2$1), 2$2), (d) 1 = 0!074""042G DE#, 2$2!

Find: Determine which equations are homogeneous.

a

] = (2$3)-2p2)43*2

[]] = 23$& = 2(2$& 2)1*223*2

23$& = 23$& homogeneous

b

, = (1!49$Q)"2*361*2

[, ] = 2$& = 2"1*622*3 not homogeneous

c

M( = b(2$1), 2$2)

[M( ] = 2 = (2$2)(2$& )2$(2$& 2) homogeneous

d

1 = 0!074""042G DE#, 2$2

[1] = +2$& 2 = 2× 2× (+$23)(2$& )2 homogeneous

645

PROBLEM 8.2

Situation: Consider variables: (a) & (torque), (b) #, 2$2 (c)pB$# (d) ]$813

Find: Determine the dimensions of the variables.

a [& ] =+2$& 2 × 2 = +22$& 2

b [#, 2$2] = (+$23)(2$& )2 = +$2& 2

c [pB$#] =

p(+2$& 2)$22)$(+$23) = 2$&

d []$813] = (23$& )$(&"123) = 1$ Dimensionless

646

PROBLEM 8.3

Situation: Liquid is draining out of a tank–details are provided in the problemstatement.

Find: The 0"groups.

APPROACH

Use the step-by-step method.

ANALYSIS

!M 2 !7A

0 !7A

0 !7A

0P & P & P &# Y

E3#A3 +

1 2 9A

0 9A

0 9A

0A 2

( YE2- 2

(A2 Y- 2

RLA

1- 2

R52

LA0

M1 2 71A

0 71A

0 71A

0

In the rst step, length is taken out with A! In the second step, mass is taken outwith #A3! In the third step, time is taken out with P! The functional relationship is

!7A= b(9

A,R5

2

LA,71A)

This can also be written as

!M

A= b(

A

1')P2

A'M1A)

647

PROBLEM 8.4

Situation: Small amplitude waves move on a liquid surface–details are provided inthe problem statement.

Find: Dimensionless functional form for wave celerity.

APPROACH

Use the exponent method.

ANALYSIS

, = b(M' O' (' ))

where [, ] = 2$&' [M] = 2' [O] =+$& 2' [(] =+$(22& 2)' [)] = 2$& 2

[, ] = [M'O0(1)A]

2$& = (2')(+ 0$& 20)(+ 1$(221& 21)(2A$& 2A)

2 : 1 = C" 2Z+ A+ : 0 = >+ Z

& : 1 = 2>+ 2Z+ 2A

Determine the exponents b, c & d in terms of a

0" 2Z+ A = 1" C>+ Z+ 0 = 0

2>+ 2Z+ 2A = 1

Solution yields: > = "Z' A = 1$2

"2Z+ 1$2 = 1" C =! "2Z = 1$2" C =! Z = "1$4 + C$2> = 1$4" C$2

Thus

, = M'O(1*4"'*2)(("1*4+'*2))1*2

=¡)1*2O1*4$(1*4

¢(M(1*2$O1*2)'

Which can also be written as

, 4($()2O) = b(M2($O)

Alternate forms:

(, 4($()2O))(O$M2() = b(M2($O)

, 2$()M)2 = b(M2($O)

or,$p)M = b(M2($O)

648

PROBLEM 8.5

Situation: Capillary rise in a tube is described in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

M 2 7A

0 7A

0A 2O Y

- 2O Y

- 2CRA2

0

( YE2-2

(A2 Y- 2

In the rst step, A was used to remove length and in the second (A2 was used toremove both length and time. The nal functional form is

7A= b( C

RA2)

649

PROBLEM 8.6

Situation: Drag force on a small sphere is described in the problem statement.

Find: The relevant 0"groups.

APPROACH


ANALYSIS

H9YE-2

Z,A

Y- 2

Z,/A2

1-

Z,/? A

0

, E-

?A

1-

?A

1-

7 YE-

7A Y-

A 2

In the rst step, length is removed with A! In the second, mass is removed with 7Aand in the third time is removed with ,$A! Finally

Z,/? A

= -

650

PROBLEM 8.7

Situation: Drag on a rough sphere is described in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

H9YE- 2

Z,9

Y- 2

Z,L94

1- 2

Z,L? 292 0

1 2# Y

E3#13 +

7 YE-

71 Y-

/L92

1-

/L? 9

0

, E9

?9

1-

?9

1-

c 2 [9

0 [9

0 [9

0

In the rst step, length is removed with 1! In the second step, mass is removed with#13 and in the nal step time removed with ,$1! The nal functional form is

Z,L? 292 = b(

L? 9/' [9)

Other forms are possible.

651

PROBLEM 8.8

Situation: A spinning ball is described in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

H YE- 2

Z9

Y-2

ZL94

1-2

ZL? 292 0

1 2, E

-?9

1-

?9

1-

# YE3

#13 +7 Y

E-71 Y

-/L92

1-

/L? 9

0

c& 2 [-9

0 [-9

0 [-9

0J 1

-J 1

-J 1

-Q9?

0

Length is removed in the rst step with 1' mass in the second step with #13 andtime in the third step with ,$1! The functional form is

ZL? 292 = b(

L? 9/' [-9' Q9?)

There are other possible forms.

652

PROBLEM 8.9

Situation: Drag on a square plate is described in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

H9YE- 2

Z,>

Y- 2

Z,L>4

1-2

Z,L? 2>2

0

, E-

?>

1-

?>

1-

# YE3

#D3 +D 27 Y

E-7D Y

-/L>2

1-

/L? >

0

@0 E-

T0

>1-

T0

>1-

T0

?0

2$ 2 E'>

0 E'>

0 E'>

0

Length is removed in rst step with D' mass is removed in second with #D3 and timeis removed in the third with ,$D! The function form is

Z,L? 2>2

= b( /L? >

' T0

?' E'>)

Other forms are possible.

653

PROBLEM 8.10

Situation: Flow through a small horizontal tube is described in the problem state-ment.

Find: The 0"groups.

APPROACH


ANALYSIS

!,!<

YE2- 2

!,!<12 Y

- 2!,!<

9/

1-

!,!<

92

/?0

7 YE-

71 Y-

, E-

?9

1-

?9

1-

1 2

Length is removed in the rst step with 1' mass is removed in the second with 71and time is removed in the third with ,$1! Finally we have

!,!<

92

/?= -

or

!%

!L= -

7,

12

654

PROBLEM 8.11

Situation: A centrifugal pump is described in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

!% YE- 2

!%1 Y- 2

!,L92

1- 2

!,:L92 0

1 2Q 1

-Q 1

-Q 1

-

] E3

-X93

1-

X93

1-

X:93

0# Y

E3#13 +

In the rst step, length is removed with 1! In the second step, mass is removed with#13 and time is removed in the third step with Q! The functional form is

!,:L92 = b(

X:93)

655

PROBLEM 8.12

Situation: A bubble is oscillating in an inviscid uid–additional details are providedin the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

b 1-

b 1-

b 1-

bq

LF2

,0

% YE- 2

,L

E2

- 2,LF2

1- 2

" 2 " 2# Y

E3

c 0 c 0 c 0 c 0

In the rst step, mass is removed with #! In the second step, length is removed with" and, nally, in third step time is removed with %$#"2! The nal functional form is

b"q

L,= b(c)

656

PROBLEM 8.13

Situation: The problem statement describes force on a satellite in the earth’s upperatmosphere.

Find: The nondimensional form of equation.

APPROACH


ANALYSIS

H = e'#011ZA

+2$& 2 = 2'(+$23)021(2$& )A

= 2'"30+1+A+ 0&"A

Equating powers of +' 2 and & , we have

& : A = 2

+ : > = 1

2 : 1 = C" 3 + Z+ Z1 = C" 3 + Z+ 2

C+ Z = 2

C = 2" Z

Therefore,

H = e(2"1)#11Z2

H$(#Z2e2) = b(1$e)

Another valid answer would be

H$(#Z212) = b(1$e)

657

PROBLEM 8.14

Situation: The problem statement describes the velocity of ripples moving on thesurface of a small pond.

Find: An expression for ,!

APPROACH


ANALYSIS

, E-

, E-

?<

1-

? <1.2L1.2

C1.20

L 2 L 2

# YE3

LC

-2

E3LCL3 & 2

O Y-2

In the rst step, mass is removed with O. In the second step, length is removed withL and in the third step, time is removed with #L3$O! The functional form is

,q

<LC= -

or

, = -

rO

#L

658

PROBLEM 8.15

Situation: A circular plate rotates with a speed J–additional details are provided inthe problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

& YE2

- 2-92

Y- 2

-/93

1-

-/93Q

0

7 YE-

71 Y-

J 1-

J 1-

J 1-

6 2 .9

0 .9

0 .9

01 2

In the rst step, length is removed with 1. In the second step, mass is removed with71 and in the last step, time is removed with J! The nal functional form is

-/93Q

= b( .9)

659

PROBLEM 8.16

Situation: A study involves capillary rise of a liquid in a tube–additional details areprovided in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

M 2 7A

0 7A

0 7A

0P & P & P &

O Y-2

O Y-2

CLA3

1- 2

C52

LA30

# YE3

#A3 +

( YE2- 2

(A2 Y-2

RLA

1- 2

R52

LA0

7 YE-

7A Y-

/LA2

1-

/5LA2

0

A 2

In the rst step, length is removed with A! In the second step, mass is removed with#A3 and in the nal step, time is removed with P! The nal functional form is

7A= b(C5

2

LA3,R5

2

LA, /5LA2)

660

PROBLEM 8.17

Situation: An engineer characterizing power * consumed by a fan.Power depends on four variables: * = b (#'1']' Q)

# is the density of air1 is the diameter of the fan impeller] is the ow rate produced by the fanQ is the rotation rate of the fan.

Find:(a) Find the relevant 0-groups.(b) Suggest a way to plot the data.

APPROACH

Apply the 0-Buckingham theorem to establish the number of 0-groups that need tobe found. Apply the step-by-step method to nd these groups and then use the0-groups to decide how a plot should be made.

ANALYSIS

0-Buckingham theorem. The number of variables is Q = 5. The number of primarydimensions is / = 3!

Number of 0-group = Q"/= 5" 3= 2

Step by step method. The variable of interest are * = b (#'1']' Q) ! The step-by-step process is given in the table below. In the rst step, the length dimension iseliminated with 1. In the second step, the mass dimension is eliminated with #13.In the third step, the time dimension is eliminated with 1$Q!

661

* YE2

-3\92

Y- 3

\L95

1- 3

\L95:3

0

# YE3

#13 +1 2

] E3

-X93

1-

X93

1-

X:93

0Q 1

-Q 1

-Q 1

-

The functional form of the equation using 0-groups to characterize the variables is:

\L95:3

= b¡X:93

¢

Answer part b ==F Plot dimensionless pressure (*$#15Q3)on the vertical axis,dimensionless ow rate (]$Q13)on the horizontal axis.

662

PROBLEM 8.18

Situation: A gas-particle mixture that is owing in a tube is causing erosion of thewall–additional details are provided in the problem statement.

Find: Determine a set of 0-groups. Express the answer as

=,

N= b (01' 02' 03' 04)

APPROACH


ANALYSIS

= = b(D3' O'N' ,' A' +,'1)

where

[=] = +$(22& ) ; [D3] = dimensionless

[N] = +$(2& 2) ; [O] =+$(2& 2)

[, ] = 2$& ; [A] = 2 ; [ +,] =+$& ; [1] = 2

! [=] = [NHO], RA^ +_,1

`]

+(22& ) = (+$(2& 2))H(+$(2& 2))](2$& )R2^(+$& )_2`

+ : 1 = T+ ` + h

2 : 2 = T+ ` " ( " i " e& : 1 = 2T+ 2` + ( + h

Use T' ( and h as unknowns

T+ 0 + h = 1" ` (5)

T" ( + 0 = 2" ` + i + e (6)

2T+ ( + h = 1" 2` (7)

(1) : T+ h = 1" `(2) + (3) : 3T+ h = 3" 3` + i + e

(2) + (3)" (1) : 2T = 2" 2` + i + e

663

T = 1" ` + (i + e)$2h = "T+ 1" ` = "1 + ` " ((i + e)$2) + 1" ` = "(i + e)$2= T" 2 + ` " i " e= 1" ` + ((i + e)$2)" 2 + ` " (i + e) = "1" ((i + e)$2)

= = b(N(1"]+((^+`)*2)T], "1"((^+`)*2A^ +"((^+`)*2), 1`' Br

or

=,$N = b(O$N'NA2$(, +,)' N12$( +,, )' D3)

Alternate form:=,$N = b(O$N'NA2$, +,' A$1'D3)

664

PROBLEM 8.19

Situation: The problem statement describes the ow of water or oil through an abruptcontraction.

Find: The 0-groups that characterize pressure drop. Express the answer as

!%A4

#]2= b(01' 02)

APPROACH


ANALYSIS

!% YE- 2

!%A Y- 2

!,LA2

1-2

!,A4

LX20

] E3

-XA3

1-

XA3

1-

# YE3

#A3 +

7 YE-

7A Y-

/LA2

1-

/ALX

0

1 2 9A

0 9A

0 9A

0A 2

Length is removed with A in the rst step, mass with #A3 in the second step and timewith ]$A3 in the third step. The nal form is

!,A4

LX2= b( /A

LX,9A)

665

PROBLEM 8.20

Situation: Flow through a transition section (large diameter to small diameter) in apipe where the Reynolds number is very large.

Find: Compare viscous forces to inertial forces.

ANALYSIS

Reynolds number %inertial forcesviscous forces

Thus, if Reynolds number is large, the viscous forces are small compared to the inertialforces.

Answer ==FViscous forces are relatively small as compared to the inertial forces.

666

1.

PROBLEM 8.21

Situation: A solid particle falls through a viscous uid–additional details are pro-vided in the problem statement.

Find: Find the 0"groups–express the answer in the form:

,&)1

= b (01' 02)

APPROACH


ANALYSIS

, ' = #0(#1,7A1G)(

Writing out the dimensions

µ2

&

¶'=

µ+

23

¶0µ+

23

¶1µ+

2&

¶A(2)G

µ2

& 2

¶(

Setting up the equations for dimensional homogeneity

2 : C = "3>" 3Z" A+ =+ b+ : 0 = >+ Z+ A& : C = A+ 2b

Substituting the equation for & into the one for 2 gives

0 = "3>" 3Z" 2A+ =" b0 = >+ Z+ A

Solving for = from the rst equation and Z from the second equation

= = 3>+ 3Z+ 2A+ bZ = "A" >

and the equation for = becomes

= = "A+ b

Substituting into the original equation

, A+2( = #0(#"A"0, 7A1"A+()(

Collecting terms

667

µ, #,1

7

¶A=

µ1)

, 2

¶( µ#(#,

¶0

The functional equation can be written as

?%D9= b

³? L/9

/'L0L/

´

668

PROBLEM 8.22

Situation: A bubble is rising in a liquid–additional details are provided in the prob-lem statement.

Find: The 0"groups.

APPROACH


ANALYSIS

The functional relationship is

, = b(#2' 72'1' O' ))

Using the step-by-step method

, E-

?9

1-

?9

1-

?%D9

0

#2YE3

#213 +

72YE-

721Y-

/%L%9

21-

/%L%9

3.2D1.20

1 2O Y

-2O Y

- 2C

L%93

1- 2

CL%9

2D0

) E-2

D9

1- 2

D9

1- 2

In the rst step, 1 was used to remove the length dimension. In the second step,#21

3 was used to remove the mass dimension and nally, in the third step,p)$1

was used to remove the time dimension. The nal functional form can be expressedas

?%D9= b

³/2%

L2%93D' CL%9

2D

´

669

PROBLEM 8.23

Situation: The problem statement describes a ow meter.

Find: The 0"groups.

APPROACH


ANALYSIS


/ = b(1'7'!%' #)

Using the exponent method, we have

/' = 1071!%A#G

Writing out the dimensional equation

+

&

'

= 20µ+

2&

¶1µ+

2& 2

¶Aµ+

23

¶G

and the equations for the dimensions are

2 : 0 = >" Z" A" 3=+ : C = Z+ A+ =& : C = Z+ 2A

Substituting the equation for time into the equation for mass yields two equations

0 = >" Z" A" 3=0 = "A+ = or A = =

and the rst equation becomes

0 = >" Z" 4A or > = Z+ 4A

Substituting back into the original equation

/1+2A = 11+4A71!%A#A

Collecting like powers gives

µ/2

14#!%

¶A=

µ71

/

¶1

A functional relationship is

/&#!%12

= b(71

/)

670

The functions can be combined to form

"%L!,92 = b(

/%L!,9

)

671

PROBLEM 8.24

Situation: The problem statement describes a torpedo-like device.

Find:Identify which 0"groups are signicant.Justify the answer.

ANALYSIS

• Viscous stresses inuence drag. Thus, Reynolds number is signicant.

• Because the body in near the surface, the motion will produce waves. Thesewaves will inuence drag. Thus, the Froude number is important.

• A major design consideration is the drag force on the object. The appropriate0"group is the coe"cient of drag (-9)which is dened by

-9 =Hdrag

#, 2$2IK

Answer ==F Signicant 0"groups are Reynolds number, Froude number and thecoe"cient of drag.

672

PROBLEM 8.25

Situation: Liquid is moving through a bed of sand–additional details are providedin the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS


!% = b(1'2' T' 7' #)

Using the step-by-method

!% YE- 2

!%!9 Y- 2

!,L!&2

1- 2

!,L!&2

/20

1 2 9!&

0 9!&

0 9!&

0!9 2T 0 T 0 T 0 T 07 Y

E-7!9 Y

-/

L!&21-

# YE3

#!93 +

In the rst step, the length was removed with !9! In the second step, the mass wasremoved with #!93! In the third step, time was removed with 7$#!92! Finally thefunctional form is

%L!,!&/

= b( 9!&' T)

673

PROBLEM 8.26

Situation: An oscillating n is being tested in a wind tunnel–additional details areprovided in the problem statement.

Find: The 0"groups.

APPROACH


ANALYSIS


H9 = b(#' ,' 6' J)

Writing out the dimensional parameters using the exponent method

H '9 = #0, 16AJG

Including the dimensions

µ+2

& 2

¶'=

µ+

23

¶0µ2

&

¶122A

µ1

&

¶G

Writing the equations for dimensional homogeneity,

+ : C = >2 : C = "3>+ Z+ 2A& : 2C = Z+ =

Solving for C' > and Z in terms of A' and = gives

C = A" =$2> = A" =$2Z = 2A" 2=

Substituting into the original equation

HA"G*29 = #A"G*2, 2A"2G6AJG

µH9#, 26

¶A=

ÃH1*29 J

#1*2, 2

!G

so

H9#, 26

= b(H1*29 J

#1*2, 2)

674

It is standard practice to eliminate H9 from the right side of the equation. To dothis, we may use the concept that 0-groups may be combined by multiplication ordivision. The result is

Z,L? 2.

= b³Q2.? 2

´

675

PROBLEM 8.27

Situation: The problem statement describes a centrifugal pump.

Find: The 0-groups.

APPROACH


ANALYSIS

] = b(8'1' M,' 7' #' ))

] E3

-X93

1-

X93

1-

Xa93 0

8 1-

8 1-

1-

1-

1 2

M, 2 7/9

0 7/9

0 7/9

07 Y

E-71 Y

-/L92

1-

/La92 0

# YE3

#13 +) E

- 2D9

1- 2

D9

1- 2

Da29

0


Xa93 = b(

7/9, /La92 ,

Da29

)

Some dimensionless variables can be combined to yield a di!erent form

]

813= b(

M,)

8212'

7

#812')

821)

676

PROBLEM 8.28

Situation: Drag force on a submarine is studied using a 1$15 scale model–additionaldetails are provided in the problem statement.

Find: (a) Speed of water in the tunnel for dynamic similitude.(b) The ratio of drag forces (ratio of drag force on the model to that on the prototype).

APPROACH

Dynamic similarity is achieved when the Reynolds numbers are the same.

ANALYSIS

Match Reynolds number

Re" = Re,

," =2,2"

:":,,,

," = 15×1× 10"6

1!4× 10"6× 2 = 21.4 m/s

The ratio of the drag force on the model to that on the prototype is

H9I"H9I,

=#"#,

µ,",,

¶2µ_"_,

¶2

=998

1015

µ21!4

2

¶2µ1

15

¶2

= 0.500

677

PROBLEM 8.29

Situation: The problem statement describes ow (oil and water) in a pipe.

Find: Velocity of water for dynamic similarity.

APPROACH


ANALYSIS


Re% = Re0,%A

:%=

,0A

:0

,% =,0:%:0

= 0!5 m/s (10"6

10"5)

= 0.05 m/s

678

PROBLEM 8.30

Situation: The problem statement describes ow (oil and water) in a pipe.

Find: Velocity of water for dynamic similarity.

APPROACH


ANALYSIS


Re5 = Re15,515:5

=,15115:15

,5 = ,15(11515)(:5:15)

= (2 m/s)(15

5)

µ10"6

4× 10"6

¶

,5 = 1!5 m$ s

679

PROBLEM 8.31

Situation: The problem statement describes a venturi meter.

Find:a.) The discharge ratio (]"$],)b.) Pressure di!erence (!%,) expected for the prototype.

ANALYSIS


Re" = Re,

,"2"$:" = ,,2,$:,

,"$,, = (2,$2")(:"$:,) (1)

Multiply both sides of Eq. (1) by I"$I, = 22"$22,:

(,"I")$(,,I,) = (2,$2")× (1)× 22"$22,

]"$], = 2"$2,

]"$], = 1/10

-,1 = -,/(!%$#, 2)" = (!%$#, 2),

!%, = !%"(#,$#")(,,$,")2

= !%"(1)(2"$2,)2

= 300× (1$10)2 = 3.0 kPa

680

PROBLEM 8.32

Situation: The problem statement describes vortex shedding from a cylinder.

Find: The 0"groups.

APPROACH


ANALYSIS

Q 1-

Q 1-

Q 1-

:A?

0, E

-, E

-?A

1-

A 2 A 2

# YE3

L/

-E2

LA2

/& ? AL

/0

7 YE-

Mass is removed with 7 in the rst step, length with A in the second step and timewith ,$A in the last step. The nal functional form is

:A?= b(? AL

/)

681

PROBLEM 8.33

Situation: Drag is to be measured with a scale model (1$5) of a bathysphere.

Find: The ratio of towing speeds (ratio of speed of the model to the speed of theprototype).

APPROACH

Dynamic similarity based on matching Reynolds number of the model and prototype.

ANALYSIS

Reynolds number

Re" = Re,,"2":"

=,,2,:,

Assume :" = :,

,"2" = ,,2,,",,=2,2"

= 5,,

,"$,, = 5

682

PROBLEM 8.34

Situation: A spherical balloon is tested by towing a 1/3 scale model in a lake–additional details are provided in the problem statement.

1" = 1 ft; 1, = 3 ft; :, = 1!58× 10"4 ft2$sec;:" = 1!22× 10"5 ft2$sec; ," = 5 ft/sec; H" = 15 lbf

Find: Drag force on the prototype (operates in air).

APPROACH

Dynamic similarity based on Reynolds number and on pressure coe"cient.

ANALYSIS

Match Reynolds numbers

Re" = Re,

,"1"$:" = ,,1,$:,

or,,," = (1"$1,)(:,$:") = (1$3)(1!58× 10"4$1!22× 10"5) (1)

Match pressure coe"cients

-,1 = -,/!%"$(#",

2"$2) = !%,$(#,,

2, $2)

!%,$!%" = (#,$#")(,2, $,

2")

H,$H" = (!%,I,)$(!%"I") = (I,$I")(#,$#")(,2, $,

2") (2)

Combine Eq. (1) and (2)

H,$H" = (#,$#")(:,$;")2 = (0!00237$1!94)(1!58× 10"4$1!22× 10"5)2

= 0!2049

H, = 15× 0!2049

= 3!07 lbf = 13!7N

683

PROBLEM 8.35

Situation: An engineer needs a value of lift force(HE) for an airplane.Coe"cient of lift: -E = 0!4.Denition of coe"cient of lift.

-E = 2HE#, 26

Density of ambient air: # = 1!1 kg$m3!Speed of the air relative to the airplane: , = 80m$ s!Planform area (i.e. area from a top view): I = 15m2.

Find: The lift force in units of Newtons.

APPROACH

Use the specied value of -E = 0!4 along with the denition of this 0-group.

ANALYSIS

From the denition of -E:

HE = -E

µ#, 2

2

¶6

= (0!4)(1!1 kg$m3) (80m$ s)2

2

¡15m2

¢

= 21' 100N

HE = 21!1 kN

COMMENTS

This lift force is about 4750 lbf.

684

PROBLEM 8.36

Situation: A 1/5 scale model of a plane is tested in a wind tunnel–additional detailsare provided in the problem statement.

Find: Density of the air in tunnel.

APPROACH

Dynamic similarity based on matching Reynolds number and Mach number.

ANALYSIS


Re" = Re,

(, 1$:)" = (, 1$:),

(,"$,,) = (1,$1")(:"$:,)

:"$:, = (,"1"$,,1,)

(7"#,$7,#") = (,"1"$,,1,)

#" = #,(7"$7,)(,,$,")(1,$1") (1)

Match Mach number

+" = +,

(,$Z)" = (,$Z),

(,"$,,) = Z"$Z,

= ((&c"& )"$(

&c"& ),)

=q&"$&, = (298$283)

1*2 (2)

Combining Eqs. (1) and (2):

#" = 1!26(1!83× 10"5$1!76× 10"5)(283$298)1*2(5)

= 6.38 kg/m3

685

PROBLEM 8.37

Situation: Flow in a pipe is being tested with air and water.

Find: Velocity ratio: ,air$,water

ANALYSIS


Re= = ReM

,=2=$:= = ,M2M$:M ; but 2=$2M = 1

! ,=,M = :=$:M % (1!6)(10"4)$(1!2)(10"5)(at 60!H ),=$,M F 1

The correct choice is c)

686

PROBLEM 8.38

Situation: Pipe ow is being studied–additional details are provided in the problemstatement.

Find: Mean velocity of water in model to insure dynamic similarity.

ANALYSIS


Re" = Re,

,"A"#"$7" = ,,A,#,$7,," = ,,(A,$A")(#,$#")(7"$7,)

," = (3 ft/s)(48$4)(1!75$1!94)((2!36× 10"5)$(4× 10"4))

," = 1!92 ft/s

687

PROBLEM 8.39

Situation: A student team is designing a radio-controlled blimp.

Drag force is characterized with a coe"cient of drag:!

-9 = 2H9

#, 2I,= 0!3

Blimp speed is , = 750mm$ s. Maximum diameter of the blimp is 1 = 0!475m.Projected area is I, = 012$4!

Find:a.) Reynolds number.b.) Force of drag in newtons.c.) Power in watts.

Properties: Air at & = 20 !C: # = 1!2 kg$m3' 7 = 18!1× 10"6N · s$m2.

Assumptions: Assume the blimp cross section is round.

APPROACH

Find the Reynolds number by direct calculation. Find the drag force using thedenition of-9! Find power (* ) by using the product of force and speed: * = HDrag,!

ANALYSIS

Reynolds number

Re =, 1#

7

=(0!75m$ s) (0!475m) (1!2 kg$m3)

(18!1× 10"6N · s$m2)Re = 23' 600

Projected area

I, =012

4=0 (0!475m)2

4= 0!177m2

688

Drag force

H9 = -9

µ#, 2

2

¶I,

= (0!3)(1!2 kg$m3) (0!75m$ s)2

2

¡0!177m2

¢

H9 = 17!9× 10"3N

Power

* = H9,

=¡17!9× 10"3N

¢(0!75m$ s)

* = 13!4× 10"3W

COMMENTS

1. The drag force is about 1/50th of a Newton, which is about 1/200th of a lbf.

2. The power is about 10 milliwatts. The supplied power would need to be higherto account for factors such as propeller e"ciency and motor e"ciency.

689

PROBLEM 8.40

Situation: A 1/1 scale model of a torpedo is being tested in a wind tunnel–additionaldetails are provided in the problem statement.

Find: Air velocity in wind tunnel.

APPROACH

Dynamic similarity based on Reynolds number.

ANALYSIS

Match the Reynolds number of the model and prototype. This leads to.

,air = (10)(1$1)(1!41× 10"5$1!31× 10"6)

= 107.6 m/s

690

PROBLEM 8.41

Situation: The problem statement describes ow in a conduit (on earth) to be usedto characterize a prototype that will be build on the moon.

Find: Kinematic viscosity of uid for model on earth.

APPROACH

Dynamic similarity based on Reynolds number and Froude number.

ANALYSIS

Match Froude number

H3moon = H3earth

(,$p)2)/ = (,$

p)2)=

,G$," = ()G$)")045(2G$2")

045

= (5)045(1)


Re" = ReG

(, 2$:)" = (, 2$:)G

:G = (,G$,"):" = (5)0450!5× (10"5) m2/s

:G = 1!119× 10"5 m2/s

691

PROBLEM 8.42

Situation: The problem statement describes a 1/15 scale model of a drying tower.

Find: Entry velocity of the model uid (water).

APPROACH

Dynamic similarity based on Reynolds number.

ANALYSIS


Re" = Re,,"2":"

=,,2,:,

," = (2,2")(:":,),,

= (15)

µ1× 10"6

4× 10"5

¶(12 m/s)

," = 4!50 m/s

692

PROBLEM 8.43

Situation: A 1/5 scale model is being used to characterize a discharge meter–additional details are provided in the problem statement.

Find:a.) Velocity for the prototype.b.) Pressure di!erence for the prototype.

APPROACH

Dynamic similarity based on Reynolds number and pressure coe"cients.

ANALYSIS


Reprot. = Remodel

,prot. = ,model(2model$2prot.)(:prot.$:model)

,prot. = 1(1$5)(10"5$10"6) = 2.0 m/s


-,I" = -,I,(!%$#, 2)" = (!%$#,

2),!%, = !%"(#,$#")(,,$,")

2

= 3!0× (860$998)× (2!0$1!0)2

= 10.3 kPa

693

PROBLEM 8.44

Situation: Water owing through a rough pipe is to be characterized by using air owthrough the same pipe–additional details are provided in the problem statement.

Find:(a) Air velocity to achieve dynamic similarity.(b) Pressure di!erence for the water ow.

APPROACH

Dynamic similitude based on Reynolds number and pressure coe"cients.

ANALYSIS


Reair = Rewater

(, 1#$7)air = (, 1#$7)water

,' = ,%(1%$1')(#%$#')(7'$7%)

#% = 1' 000 kg/m3

# = #'I std. atm. × (150 kPa/101 kPa)

= 1!20× (150$101) = 1!78 kg/m3

7' = 1!81× 10"5 N · s/m2

7% = 1!31× 10"3 N · s/m2

Then

,' = 1!5 m/s (1,000/1.78)(1.81× 10"5$1!31× 10"3)

,'=11.6 m/s


-,& = -,2(!%$#, 2)% = (!%$#, 2)'

!%% = !%'(#%$#')(,%$,')2

= 780× (1' 000$1!78)(1!5$11!6)2

= 7' 330 Pa = 7.33 kPa

694

PROBLEM 8.45

Situation: A device for a minesweeper (a noisemaker) will be studied by using a1/5 scale model in a water tunnel–additional details are provided in the problemstatement.

Find:(a) Velocity to use in the water tunnel.(b) Force that will act on the prototype.

APPROACH

Dynamic similarity based on matching Reynolds number and pressure coe"cient.

ANALYSIS


Retunnel = Reprototype

,tunnel = ,prot.(5

1)(:tunnel:prot.

)

,tunnel = 5(5

1)(1)

,tunnel = 25 m/s


-,tunnel = -,prototyp eµ!%

#, 2

¶

tunnel

=

µ!%

#, 2

¶

prototypeµ!%tunnel!%prot.

¶=

µ#tunnel#prot.

¶µ, 2tunnel, 2prot.

¶

Multiply both sides of the equation by Itunnel$Iprot. = 225$22,!

(!%×I)tunnel(!%×I)prot.

=

µ#tunnel#prot.

¶×µ, 2tunnel, 2prot.

¶×µ252,

¶2

HtunnelHprot.

= (1

1)(5)2(

1

5)2

Htunnel = Hprot. = 2400 N

695

PROBLEM 8.46

Situation: Air forces on a building are to be characterized by using a 1/100 scalemodel–additional details are provided in the problem statement.

Find: (a) Density needed for the air in the wind tunnel.(b) Force on the full-scale building (prototype).

ANALYSIS

Reynolds number

Re" = Re,

(#, 2$7)" = (#, 2$7),

#"$#, = (,,$,")(2,$2")(7"$7,)

= (25$300)(100)(1)

= 8!33

#" = 8!33#, = 0.020 slugs/ft3

H"$H, = (!%"$!%,)(I"$I,) (1)

-,I"-,I,

=

µ!%"#",

2"

¶µ#,,

2,

!%,

¶

1 =

µ!%"!%,

¶µ#,#"

¶µ, 2,, 2"

¶

=

µ!%"!%,

¶µ1

8!33

¶µ25

300

¶2

Then!%"$!%, = 1' 200 (2)

solve Eqs. (1) and (2) for H"$H,

H"$H, = 1' 000I"$I,

= 1200(1$104) = 0!12

H, =H"0!12

= 417 lbf

696

PROBLEM 8.47

Situation: Performance of a large valve will be characterized by recording data on a1/3 scale model–additional details are provided in the problem statement.

Find:a) Flow rate to be used in the model (laboratory) test.b) The pressure coe"cient for the prototype.

ANALYSIS

Re" = Re, or (, 1#$7)" = (, 1#$7),

Then

,"$,, = (1,$1")(#,$#")(7"$7,)

Multiply both sides of above equation by I"$I, = (1"$1,)2

(I"$I,)(,"$,,) = (1,$1")(1"$1,)2(#,$#")(7"$7,)

]"$], = (1"$1,)(#,$#")(7"$7,)

= (1$3)(0!82)(10"3$(3× 10"3))]"$], = 0!0911

or ]" = ], × 0!0911

]" = 0!50× 0!0911 m3$s = 0.0455 m3/s

-,=1.07

697

PROBLEM 8.48

Situation: The moment acting on the rudder of submarine will be studied using a1/60 scale model–additional details are provided in the problem statement.

Find:(a) Speed of the prototype that corresponds to the speed in the water tunnel.(b) Moment that corresponds to the data from the model.

ANALYSIS


-,1 = -,/(!%$#, 2)" = (!%$#, 2),

or!%"$!%, = (#",

2")$(#,,

2, ) (1)

Multiply both sides of Eq. (1) by (I"$I,)× (2"$2,) = (2"$2,)3 and obtain

Mom."$Mom., = (#"$#,)(,"$,,)2(2"$2,)

3 (2)


,"2"$:" = ,,2,$:,,"$,, = (2,$2")(:"$:,)

(3)

Substitute Eq. (3) into Eq. (2) to obtain

+"$+, = (#"$#,)(:"$:,)2(2"$2,)

+, = +"(#,$#")(:,$:")2(2,$2")

= 2(1' 026$1' 000)(1!4$1!31)2(60)

= 141 N·m

Also

,, = 10(1$60)(1!41$1!310)

= 0.179 m/s

698

PROBLEM 8.49

Situation: A model hydrofoil is tested in a water tunnel–additional details are pro-vided in the problem statement.

Find: Lift force on the prototype.

ANALYSIS


-,1 = -,/(!%$#, 2)" = (!%$#, 2),

!%"$!%, = (#"$#,)(,2"$,

2, )

Multiply both sides of the above equation by I"$I, = (2"$2,)2

(!%"$!%,)(I"$I,) = (#"$#,)(,2"$,

2, )(2

2"$2

2,) = H"$H, (1)


(, 2#$7)" = (, 2#$7),

(,,$,")2 = (2"$2,)

2(#"$#,)2(7"$7")

2 (2)

Eliminating (,,$,")2 between Eq. (1) and Eq. (2) yields

H,$H" = (#"$#,)(7,$7")2

Then if the same uid is used for models and prototype, we have

H,$H" = 1

or

H, = 25 kN

699

PROBLEM 8.50

Situation: A 1/8 scale model of an automobile will be tested in a pressurized windtunnel–additional details are provided in the problem statement.

Find: Pressure in tunnel test section.

ANALYSIS

Match Mach number

+" = +,

,"$Z" = ,,$Z,;,"$,, = Z"$Z, (1)


Re" = Re,

,"2"#"$7" = ,,2,#,$7,

or,"$,, = (2,$2")(#,$#")(7"$7,) (2)

Eliminate ,"$,, between Eqs. (1) and (2) to obtain

Z"$Z, = (2,$2")(#,$#")(7"$7,) (3)

ButZ =

pN? $# =

pc%$# =

pc%$(%$"& ) =

&c"&

Therefore Z"$Z, = 1' then from Eq. (3)

1 = (8)(#,$#")(1)

or#" = 8#,

But

# = %$"&

so

(%$"& )" = 8(%$"& ),

%" = 8%,

= 8 atm

= 0.808 MPa abs.

700

PROBLEM 8.51

Situation: A 1/8 scale model of an automobile will be tested in a pressurized windtunnel–additional details are provided in the problem statement.

Find:a) Speed of air in the wind tunnel to match the Reynolds number of the prototype.b) Determine if Mach number e!ects would be important in the wind tunnel.

ANALYSIS


Re" = Re,

,"2"#"$7" = ,,2,#,$7,; But #"$7" = #,$7,

so," = ,,(2,$2") = 80× 10 = 800 km/hr = 222 m/s

Mach number

+ = ,$Z = 222$345 = 0!644

Because + ) 0!3' Mach number e!ects would be important .

701

PROBLEM 8.52

Situation: A satellite is entering the earth’s atmosphere–additional details are pro-vided in the problem statement.

Find: Determine if the ow is rareed.

APPROACH

Use the ratio of Mach number and Reynolds number.

ANALYSIS

Mach number and Reynolds number

+$Re = (,$Z)(7$#, 1) = (7)$(#Z1)

where

# = %$"& = 22$(1716× 393) = 3!26× 10"5 slugs/ft3

and Z = 975 ft/s and 7 = 3!0× 10"7 lbf-s/ft2 so

+$Re = 3!0× 10"7$(3!26× 10"5 × 975× 2) = 4!72× 10"6 G 1Not rareed

702

PROBLEM 8.53

Situation: Water droplets are in an air stream.Breakup occurs when .$

&Re = 0!5.

,air = 25m$ s' %air = 1!01 kPa' O = 0!073N$m!

Find: Droplet diameter for break up.

APPROACH

Apply the .$&Re = 0!5 criteria, combined with the equations for Weber number

and Reynolds number.

ANALYSIS

Weber number and Reynolds number

.$&Re =

#A, 2&:

O&, A

=, 3*2

&#A7

O

So breakup occurs when, 3*2

&#A7

O= 0!5

Solve for diameter

A =

·0!5O

, 3*2&#7

¸2

=0!25O2

, 3#7

Calculations

A =0!25O2

, 3#7

=0!25× 0!0732

253 × 1!2× (18!1× 10"6)= 3.93 mm

703

PROBLEM 8.54

Situation: The problem statement describes breakup of a liquid jet of heptane..

Find: Diameter of droplets.

Properties: From Table A.3, # = 0!95 kg/m3!

ANALYSIS

Weber number. = 6!0 = #1, 2$O

1 = 6O$#, 2 = 6× 0!02$(0!95× (30)2) = 1!40× 10"4 m = 140 7m

704

PROBLEM 8.55

Situation: The problem statement describes breakup of a jet of water into droplets.

Find: Estimated diameter of droplets.

Properties: From Table A.3 # = 1!20 kg/m3 and from Table A.5 O = 0!073 N/m.

ANALYSIS

Weber number

. = 6!0 =#1, 2

O

1 =6O

#, 2=

6× 0!073(1!2× (20)2)

= 9!125× 10"4 m = 0.91 mm

705

PROBLEM 8.56

Situation: A model test is described in the problem statement.

Find: Relationship between kinematic viscosity ratio and scale ratio.

ANALYSIS

Match Froude numbers

H" = H,; (,&)2)" = (

,&)2),

or,",,

=

s)"2"),2,

(1)


Re" = Re,; (, 2

:)" = (

, 2

:), or

,",,= (

2,2")(:":,) (2)

Eliminate ,"$,, between Eqs. (1) and (2) to obtain:s)"2"),2,

= (2,2")(:":,)' but )" = ),

Therefore: :"$:, = (2"$2,)3*2

706

PROBLEM 8.57

Situation: The spillway of a dam is simulated using a 1/20 scale model–additionaldetails are provided in the problem statement.

Find:a) Wave height (prototype).b) Wave period (prototype).

APPROACH

Dynamic similarity based on Froude number.

ANALYSIS

Match Froude number

P,P"= (

2,2")1*2

Thenwave periodprot = 2× (20)

1*2 = 8!94 s

andwave heightprot = 8 cm × 20 = 1!6 m

707

PROBLEM 8.58

Situation: A prototype of a dam is represented with a 125scale model. Other details

are provided in the problem statement.

Find:a) Velocity for prototype.b) Discharge for prototype.

APPROACH


ANALYSIS

Match Froude number

H3" = H39

,"$(()")(2"))045 = ,,$((),)(2,))

045

,,$," = (2,$2")045 = 5 (1)

,, = (2!5)(5) m/s

= 12.5 m/s

Discharge for the prototype is], = ,,I, (2)

From Eq. (1)

,, = ,"

µ2,2"

¶045(3)

From geometric similarity

I, = I"

µ2,2"

¶2(4)

Combining Eqs. 2, 3 and 4 gives

], = ,"

µ2,2"

¶045I"

µ2,2"

¶2

= ,"I"

µ2,2"

¶245

=¡0!1m3$ s

¢(25)245

= 312. 5 m3

s

708

PROBLEM 8.59

Situation: A seaplane model has a 1$12 scale.

Find: Model speed to simulate a takeo! condition at 125 km/hr.

Assumptions: Froude number scaling governs the conditions.

ANALYSIS

Match Froude number

," = ,,

s2"2,

= 125

r1

12= 36.1 m/s

709

PROBLEM 8.60

Situation: A model spillway has a 136scale.

Discharge for the prototype is 3000 m3$ s!

Find: (a) Velocity ratio.(b) Discharge ratio.(c) Model discharge

APPROACH


ANALYSIS

Match Froude number

,"$,, =q2"$2, (1)

or for this case

,"$,, =p1$36 = 1/6

Multiply both sides of Eq. (1) by I"$I, = (2"$2,)2

,"I"$,,I, = (2"$2,)1*2(2"$2,)

2

]"$], = (2"$2,)5*2

or for this case

]"$], = (1$36)5*2 = 1/7,776

]" = 3000$7776 = 0.386 m3/s

710

PROBLEM 8.61

Situation: Flow in a river is to be studied using a 1/64 scale model–additional detailsare provided in the problem statement.

Find: Velocity and depth in model at a corresponding point to that specied for theprototype.

ANALYSIS

Match Froude number

H3" = H3,

,"$(()")(2"))045 = ,,$((),)(2,))

045

," = ,,(2"$2,)045 = ,,(1$8) = 1.875 ft/s

Geometric similitude

A"$A, = 1$64

A" = (1$64)A,

= (1$64)(20) = 0.312 ft

711

PROBLEM 8.62

Situation: Details are provided in the problem statement..

Find: Velocity and discharge for prototype.

ANALYSIS

Match Froude number

,, = ,"

q2,$2" (1)

= 7!87&30

= 43.1 ft/s

Multiply both sides of Eq. (1) by I,$I" = (2,$2")2

,,I,,"I"

=

µ2,2"

¶5*2

So

],$]" = (2,$2")5*2

], = 3!53× (30)5*2

= 17,400 ft3/s

712

PROBLEM 8.63

Situation: Flow around a bridge pier is studied using a 1/10 scale model.

Find: (a) Velocity and (b)wave height in prototype.

APPROACH

Use Froude model law.

ANALYSIS


,, = ,"

q2,$2" = 0!90

&10 = 2.85 m/s

2,$2" = 10; therefore, wave heightprot. = 10× 2!5 cm = 25 cm

713

PROBLEM 8.64

Situation: A 1/25 scale model of a spillway is tested–additional details are providedin the problem statement.

Find: Time for a particle to move along a corresponding path in the prototype.

ANALYSIS


,,$," =q2,$2"

or

(2,$P,)$(2"$P") = (2,$2")1*2

Then

P,$P" = (2,$2")(2"$2,)1*2

P,$P" = (2,$2")1*2

P, = 1×&25 = 5 min

Also

],$]" = (2,$2")5*2

], = 0!10× (25)5*2 = 312.5 m3/s

714

PROBLEM 8.65

Situation: A tidal estuary is modeled using a 1/250 scale–additional details areprovided in the problem statement.

Find: Velocity and period in the model.

ANALYSIS

Match Froude number

H3" = H3,

or

µ,&)2

¶

"

=

µ,&)2

¶

,

(1)

,",,

= (2"2,)1*2

because )" = ),! Then

(E151)

(E/5/)=

µ2"2,

¶1*2

or

P"P,=

µ2"2,

¶1*2(2)

Then from Eq. (1)

," = ,,

µ2"2,

¶1*2= 4!0× (1$250)1*2 = 0.253 m/s

From Eq. (2)

P" = (12!5 hr) (1$250)1*2 = 0!791hr = 47.4 min

715

PROBLEM 8.66

Situation: The maximum wave force on a 1/36 scale sea wall was 80 N.& = 10! (model and prototype).

Find: Force on the wall (for the full scale prototype).

Assumptions: Fresh water (model) and seawater (prototype).

APPROACH

Dynamic similarity based on pressure coe"cient and Froude number.

ANALYSIS


-,1 = -,/ ; (!%$#,2)" = (!%$#,

2),

!%"$!%, = (#"$#,)(,"$,,)2 (1)

Multiply both sides of Eq. (1) by I"$I, = 22"$22,

(!%"I")$(!%,I,) = (#"$#,)(2"$2,)2(,"$,,)

2


,"$,, =q2"$2, (2)

Eliminating ,"$,, from Eqs. (1) and (2) yields

H"$H, = (#"$#,)(2"$2,)2(2"$2,)

H"$H, = (#"$#,)(2"$2,)3

H, = H"(#,$#")(2,$2")3 = 80(1' 026$1' 000)(36)3 = 3.83 MN

716

PROBLEM 8.67

Situation: A model of a spillway is built at a 1/25 scale–additional details areprovided in the problem statement.

Find:(a) Water discharge in model for dynamic similarity.(b) Force on the prototype.

APPROACH

Dynamic similitude based on matching pressure coe"cients and Froude numbers.

ANALYSIS


-,1 = -,/ ; (!%$#,2)" = (!%$#,

2),

!%"$!%, = (#"$#,)(,"$,,)2


(!%"I")$(!%,I,) = (#"$#,)(2"$2,)2(,"$,,)

2

H"H,

= (#"$#,)(2"$2,)2(,"$,,)

2 (1)

Match Froude number,",,=

s2"2,

(2)

Eliminate ,"$,, from Eqs. (1) and (2)

H,H"

=

µ#,#"

¶µ2,2"

¶3

H, = (22N)

µ1

1

¶µ25

1

¶3

= 344!8N

H, = 345N


,"22"

,,22,=

µ2"2,

¶5*2

]"],

=

µ2"2,

¶5*2

]" =¡150m3$ s

¢µ 125

¶5*2

= 0!048m3$ s

717

Match Froude number

,,$," =q2,$2"

]"$], = (2"$2,)5*2

]" = 150× (1$25)5*2 = 0.048 m3/s

From solution to Prob. 8.66 we have:

H, = H"(#,$#")(2,$2")3

= 22(1$1)(25)3 = 344 kN

718

PROBLEM 8.68

Situation: A scale model of a dam will be constructed in a laboratory.

Find: The largest feasible scale ratio.

ANALYSIS

Check the scale ratio as dictated by ]"$], (see Problem 8.64)

]"$], = 0!90$5' 000 = (2"$2,)5*2

or

2"$2, = 0!0318

Then with this scale ratio

2" = 0!0318× 1' 200 m = 38.1 m

." = 0!0318× 300 m = 9.53 m

Therefore, model will t into the available space, so use

2"$2, = 1$31!4 = 0!0318

719

PROBLEM 8.69

Situation: A scale model of a ship is tested in a towing tank–additional details areprovided in the problem statement.

Find: Speed for the prototype that corresponds to the model test.

APPROACH


ANALYSIS

Match Froude number

,"$p)"2" = ,,$

p),2,

,, = ,"p2,$

p2"

= (4 ft/s) (150/4)1*2

,,=24.5 ft/s

720

PROBLEM 8.70

Situation: A scale model (1/25) of a ship is described in the problem statement.

Find: (a) Velocity of the prototype.(b) Wave resistance of the prototype.

ANALYSIS

Follow the solution procedure of Prob. 8.66:

,"$,, =q2"$2,; ,, = 5×

&25 = 25 ft/s

H"$H, = (2"$2,)3; H, = 2(25)

3 = 31,250 lbf

721

PROBLEM 8.71

Situation: A scale model (1/20) of a ship is described in the problem statement.

Find: (a) Velocity of the prototype.(b) Wave resistance of the prototype.

ANALYSIS

Match Froude number

H3" = H3,,"

()"2")045=

,,(),2,)045

,, = ,"

µ2,2"

¶045= 17.9 m/s

Calculate force

H, = (25 N)µ2,2"

¶3

= (25)(20)3

H, = 200' 000N

H, = 200 kN

722

PROBLEM 8.72

Situation: A scale model¡120

¢of a building is being tested–details are provided in

the problem statement.

Find: Drag on the prototype building.

Assumptions: -,1 = -,/ , #" = #,

ANALYSIS


(!%$(#, 2$2)" = (!%$(#, 2$2),

!%"$!%, = (#"$#,)(,2"$,

2, )

Assuming #" = #,

H"$H, = (!%"$!%,)(I"$I,) = (,"$,,)2(2"$2,)

2

(H,$H") = (40$20)2(20)2

H, = (200 N)(4)(400) = 320' 000 N = 320 kN

Choice (d) is the correct.

723

PROBLEM 8.73

Situation: A scale model¡1250

¢of a building is being tested–details are provided in

the problem statement.

Find:(a) Pressure values on the prototype.

• windward wall

• side wall

• leeward wall

(b)Lateral force on the prototype in a 150 km/hr wind.

Assumptions: -,Imodel = -,Iprot.

ANALYSIS

Match pressure coe"cients-,Imodel = -,Iprot.

then

!%,$((1$2)#,,2, ) = -,/ = -,1

or

!%, = -,1((1$2)#,,2, )

= -,1 × (1$2)× 1!25× (150' 000$3' 600)2

%" %0 = 1085!6-,1

but %0 = 0 gage so

% = 1085!6-,1 Pa

Extremes of pressure are therefore:

%windward wall = 1!085 kPa

%side wall = 1085!6× ("2!7) = -2.93 kPa

%leeward wall = 1085× ("0!8) = -868 Pa

Lateral Force:!%"$!%, = ((1$2)#",

2")$((1$2)#,,

2, )

724

Multiply both sides of equation by I"$I, = 22"$22,

(!%"I")$(!%,I,) = (#"$#,)(,2"$,

2, )(2

2"$2

2,) = H"$H,

H,$H" = (#,$#")(,2, $,

2")(2

2,$2

2")

H, = 20(1!25$1!20)((150' 000$3' 600)2$(20)2)(250)2

H, = 5!65 MN

725

PROBLEM 8.74

Situation: Drag force is measured in a water tunnel and a wind tunnel–details areprovided in the problem statement.

Find:(a) Find the relevant 0-groups.(b) Write a computer program and reduce the given data.(c) Plot the data using the relevant 0-groups.

ANALYSIS

Performing a dimensional analysis shows that

Z,L? 292 = b(

L? 9/)

The independent variable is the Reynolds number. Plotting the data using the di-mensionless numbers gives the following graph.

Reynolds number

0 2e+5 4e+5 6e+5 8e+5 1e+6

Forc

e co

effic

ient

0.005

0.010

0.015

0.020

0.025

0.030

0.035

Water Air

726

PROBLEM 8.75

Situation: Pressure drop is measured in a pipe with (a) water and (b) oil. Detailsare provided in the problem statement.

Find:(a) Find the relevant 0-groups(b) Write a computer program and reduce the given data(c) Plot the data using the relevant 0-groups

ANALYSIS

Performing a dimensional analysis on the equation for pressure drop shows

!,9EL? 2

= b(L? 9/)

where the independent parameter is Reynolds number. Plotting the data using thedimensionless parameters gives the following graph.

Reynolds number

0 1e+5 2e+5 3e+5 4e+5 5e+5 6e+5

Dim

ensi

onle

ss p

ress

ure

drop

0.006

0.008

0.010

0.012

0.014

0.016

0.018

0.020

Water Crude oil

727

PROBLEM 9.1

Situation: A block sliding on an oil lm is described in the problem statement.

Find: Terminal velocity of block.

APPROACH

Apply equilibrium. Then relate shear force (viscous drag force) to viscosity and solvethe resulting equation.

ANALYSIS

Equilibrium

Hshear = . sin K

B = Hshear$I& =. sin K$22

Shear stressB = 7A,$A? = 7× ,$!?

W

!

Fshear

L

L

"y

or

, = B!?$7

Then

, = (. sin K$22)!?$7

, = (150 sin 10!$0!352)× 1× 10"4$10"2

, = 2.13 m/s

728

PROBLEM 9.2

Situation: A board sliding on an oily, inclined surface is described in the problemstatement.

Find: Dynamic viscosity of oil

ANALYSIS

From the solution to Prob. 9.1, we have

7 = (. sin K$22)!?$,

7 = (40× (5$13)$32)× (0!02$12)$0!5

7 = 5!70× 10"3 lbf-s/ft2

729

PROBLEM 9.3

Situation: A board sliding on an oily, inclined surface is described in the problemstatement.

Find: Dynamic viscosity of oil.

ANALYSIS


7 = (20× (5$13)$12)× 5× 10"4$0!12

7 = 3!20× 10"2 N·s/m2

730

PROBLEM 9.4

Situation: Uniform, steady ow occurs between two plates—additional details areprovided in the problem statement.

Find: (a) Other conditions present to cause odd velocity distribution.(b) Location of minimum shear stress.

ANALYSIS

Upper plate is moving to the left relative to the lower plate.

Minimum shear stress occurs where the maximum velocity occurs (where A@$A? =0).

731

PROBLEM 9.5

Situation: A laminar velocity distribution is described in the problem statement.

Find: Whether statements (a) through (e) are true or false.

ANALYSIS

a). True b). False c). False d). False e). True

732

PROBLEM 9.6

Situation: A plate being pulled over oil is described in the problem statement.

Find: (a) Express the velocity mathematically in terms of the coordinate systemshown.(b) Whether ow is rotation or irrotational.(c) Whether continuity is satised.(d) Force required to produce plate motion.

ANALYSIS

By similar triangles @$? = @max$!P

y "y

Umax

U

U

or

@ = (@max$!?)?

@ = (0!3$0!002)? m/s

@ = 150 y m/s

; = 0

For ow to be irrotational [@$[? = [,$[E here [@$[? = 150 and [,$[E = 0 . Theequation is not satised; ow is rotational .[@$[E+ [;$[? = 0 (continuity equation) [@$[E = 0 and [;$[? = 0so continuity is satised.Use the same formula as developed for solution to Prob. 9-1, but . sin K = Hshear!Then

H& = I7,$P

H& = 0!3× (1× 0!3)× 4$0!002

H&=180 N

733

PROBLEM 9.7

Situation: The gure in problem 2.30 is for the velocity distribution in a liquid suchas oil.

Find: Whether each of the statements (a) though (e) is true or false.

ANALYSIS

Valid statements are (c), (e).

734

PROBLEM 9.8

Situation: A wire being pulled though a tube is described in the problem statement.

Find: Viscous shear stress on the wire compared to that on the tube wall.

ANALYSIS

The shear force is the same on the wire and tube wall; however, there is less area inshear on the wire so there will be a greater shear stress on the wire.

735

PROBLEM 9.9

Situation: Two plates in oil are described in the problem statement.

Find: Derive an equation for the velocity of the lower plate.

Assumptions: A linear velocity distribution within the oil.

ANALYSIS

The velocity distribution will appear as below:

Equilibrium

(Force on top of middle plate) = (Force on bottom of middle plate)

B 1I = B 2I

B 1 = B 2

71!,1$P1 = 72!,2$P2

71 × (, " ,lower)$P1 = 72,lower$P2

, 71$P1 " 71,lower$P1 = 72,lower$P2

,lower(72$P2 + 71$P1) = , 71$P1

,lower = (, 71$P1)$(72$P2 + 71$P1)

736

PROBLEM 9.10

Situation: A disk in oil is described in the problem statement.

Find: Torque required to rotate disc.

ANALYSIS

B = 7A;$A?

B = 73J$!?

A& = 3AH

A& = 3BAI

A& = 3(73J$!?)203A3

! y

dr

Plan View

&

Then

& =

Z K

0

A& =

Z K0

0

(7J$!?)2033A3

& = (207J$!?)34$4|K00 = 207J340$(4!?)

For

!? = 0!001 ft; 30 = 6” = 0!50 ft; J = 180× 20$60 = 60 rad/s7 = 0!12 lbf-s/ft2

& = (2× 0!12× 60$0!001)(0!54$4)& = 222 ft-lbf

737

PROBLEM 9.11

Situation: A disk in oil is described in the problem statement.

Find: Torque required to rotate disk.

ANALYSIS


& = 207J340$(4!?)

where

3 = 0!10 m

!? = 2× 10"3 mJ = 10 rad/s

7 = 6 N · s/m2

& = 20 × 8× 10× 10"4$(4× (2× 10"3))

& = 6.28 N·m

738

PROBLEM 9.12

Situation: A cone in oil is described in the problem statement.

r& !

T)

r0

Find: Derive an equation for the torque in terms of the other variable.

Assumptions: K is very small.

ANALYSIS

A& = (7@$9)AI× 3= 73J sin`2032A3$(3K sin`)

= 207J32A3$K

& = (7J$K)(2033$3)|K00& = (2/3)03307J$K

739

PROBLEM 9.13

Situation: A plate in glycerin is described in the problem statement.

Find:a) Sketch the velocity distribution at section I"I.b) Force required to pull plate.

Properties: Glycerin (Table A.4): 7 = 1!41N · s$m2!

ANALYSIS

Velocity distribution:

V=0.4 m/s

H = BI

= 7A,

A?I

=¡1!41N · s$m2

¢µ0!4m$ s0!002m

¶× 1m× 2m× 2 sides

= 1128N

H = 1130 N

740

PROBLEM 9.14

Situation: A bearing is described in the problem statement.

Find: Torque required to turn bearing.

ANALYSIS

B = 7,$i

& = BI3

where & = torque, I = bearing area = 203>

& = B203>3 = B2032>

= (7,$i)(2032>)

where V=rJ! Then

= (7$i)(3J)(2032>)

= (7$i)(20J)33>

= (0!1$0!001)(20)(200)(0!009)3(0!1)

& = 9!16× 10"4N · m

741

PROBLEM 9.15

Situation: A shaft turning inside a stationary cylinder is described in the problemstatement.

Find: Show that the torque per unit length acting on the inner cylinder is given by& = 407J32&$(1" (32&$32!)!

ANALYSIS

Subscript 9 refers to inner cylinder. Subscript U refers to outer cylinder. The cylinderis unit length into page.

*

"rr

*+

s

&& = B(203)(3)

&! = B(203)(3) + A$A3(B203 · 3)!3&& " &! = 0

A$A3(B2032L)!3 = 0; A$A3(B32) = 0

Since there is no angular acceleration, the sum of the torques must be zero. Therefore

&& " &! = 0

A$A3(B2032)!3 = 0

A$A3(B32) = 0

Then

B32 = -1

B = 73(A$A3)(,$3)

So

733(A$A3(,$3)) = -1

7(A$A3(,$3)) = -13"3

Integrating,7;$3 = ("1$2)-13"2 + -2

742

At 3 = 3!, ; = 0 and at 3 = 3&, ; = 3&J so

-1 = 2-2320

7J = -2(1" 320$32&)

-2 = 7J$(1" 320$32&)

Then

B & = -13"2& = 2-2(30$3&)

2 = 27J320$(32& " 3

20) = 27J$((3

2&$3

20)" 1)

So&& = B203

2& = 407J3

2&$((3

2&$3

20)" 1)

which is the torque on the uid. Torque on shaft per unit length

& = 407J32&$(1" (32&$3

20)

743

PROBLEM 9.16

Situation: A shaft turning inside a stationary cylinder is described in the problemstatement.

Find: Power necessary to rotate shaft.

APPROACH

Apply the equation developed in Problem 9.15.

ANALYSIS

& = 407JL32&$(1" (32&$3

20))

= 40 × 0!1× (50)(0!01)20!03$(1" (1$1!1)2)= 0!00109 N ·m

* = &J

= (0!00109 N ·m) (50 s"1)

* = 0.0543 W

744

PROBLEM 9.17

Situation: A viscosity measuring device is described in the problem statement.

Find: Viscosity of uid.

APPROACH

Apply the equation developed in Problem 9.15.

ANALYSIS

& = 0!6(0!02) = 0!012 N ·m7 = & (1" 32&$3

20)$(40JL3

2&)

= 0!012(1" 22$2!252)$(40(20)(20$60)(0!1)(0!02)2)

7 = 2.39 N·s/m2

745

PROBLEM 9.18

Situation: Oil ows down an inclined surface —additional details are provided in theproblem statement.

Find: (a) Maximum and (b) mean velocity of ow.

ANALYSIS

@ = () sin K$2:)?(2A" ?)

@max occurs at the liquid surface where ? = A' so

@max = () sin K(2:))A2

where K = 30!' : = 10"3 m2/s and A = 2!0× 10"3 m@max = (9!81× sin 30!$(2× 10"3))× (2!0× 10"3)2

= 9!81× 10"3 m/s

@max = 9!81mm$ s

, = ()A2 sin K)$(3:)

= 9!81× (2!0× 10"3)2 sin 30!$(3× 10"3)

, = 6!54mm$ s

746

PROBLEM 9.19

Situation: SAE 30W oil (100 !F) ows down an inclined surface (K = 20!) !The Reynolds number is 200.

Find: (a) Depth of oil(b) discharge per unit width.

Properties: SAE 30W oil (100 !F) properties (from Table A.4) are ( = 55!1 lbf$ ft3,7 = 0!002 lbf · s$ ft2' : = 0!0012 ft2$ s!

ANALYSIS

Flow rate equation.^ = , A (1)

Reynolds number

Re =, A

:(2)

Combine Eqs. (1) and (2)Re =

^

:

^ = Re×:= 200×

¡0!0012 ft2$ s

¢

= 0!240 ft2$ s

^ = 0!240 ft2/ s

Since the ow is laminar, apply the solution for ow down an inclined plane.

^ =

µ1

3

¶µ(

7

¶A3 sin (K)

0!24 ft2$ s =

µ1

3

¶µ55!1 lbf$ ft3

0!002 lbf · s$ ft2

¶A3 sin (20!)

Solving for depth (A)A = 0!0424 ft = 0!509 in

747

PROBLEM 9.20

Situation: Water ows down a roof —additional details are provided in the problemstatement.2 = 15 ft; "K = 0!4 in./hr. = 9!26 × 10"6 ft/s, 7 = 2!73 × 10"5 lb-s/ft2; ( = 62!4lbf/ft3; K = 10!!

Find: (a) Depth.(b) Average velocity.

ANALYSIS

Flow rate equationTotal discharge per unit width of roof is:

^ = 2× 1×"K (1)

where "K = rainfall rate. But Eq. 9.8

^ = (1$3)(($7)A3 sin K

orA = (3^7$(( sin K))1*3 (2)

Combining equations 1 and 2, gives

A = (32"K7$(( sin K))1*3

A = (3× 15× 9!26× 10"6 × 2!73× 10"5$(62!4× sin 10!))1*3

= 1!02× 10"3 ftA = 0.012 in.

Using Eq. 9.9a, = 0!137 ft/s

748

PROBLEM 9.21

Situation: Flow occurs between two plates—additional details are provided in theproblem statement.

Find: Shear (drag) force on lower plate.

ANALYSIS

@ = "(($27)(D? " ?2)AM$A9

@max occurs at ? = D$2 so

@max = "(($27)(D2$2"D2$4)AM$A9 = "(($27)(D2$4)AM$A9

From problem statement A%$A9 = "1200 Pa/m and AM$A9 = (1$()A%$A9! Also D = 2mm= 0!002 m and 7 = 10"1N·s/m2! Then

@max = "(($27)(D2$4)((1$()("1' 200))= (D2$87)(1' 200)

= (0!0022$(8× 0!1))(1' 200)= 0!006 m/s

@max = 6.0 mm/s

H& = BI = 7(A@$A?)× 2× 1!5B = 7× ["(($27)(D " 2?)AM$A9]

but Bplate occurs at ? = 0! So

H& = "7× (($27)×D × ("1' 200$()× 3 = (D$2)× 1' 200× 3= (0!002$2)× 1' 200× 3

H& = 3.6 N

749

PROBLEM 9.22


Find: Maximum uid velocity in E"direction.

APPROACH


ANALYSIS


@max = "((D2$87)((1$()(A%$A9))= "(0!012$(8× 10"3))("12)

@max = 0.150 ft/s

750

PROBLEM 9.23


Find: Direction of ow.

APPROACH

Flow will move from a location high energy to a location of low energy. For steadyow in a constant area pipe, energy is proportional to piezometric head (M) !

ANALYSIS

M= = (%=$() + R= = (150$100) + 0 = 1!5

M> = (%>$() + R> = (100$100) + 1 = 2

M> F M=

Therefore ow is from D to I: downward

751

PROBLEM 9.24

Situation: Glycerin ows downward between two plates—additional details are pro-vided in the problem statement.

Find: Discharge per unit width.

Properties: Table A.4 (Glycerin) 7 = 1!41 N·s/m2 and : = 1!12× 10"3m2$ s!

Assumptions: Flow will be laminar.

ANALYSIS

^ = "D3(

127

AM

A9

AM$A9 = A$A9(%$( + R)

= (1$()A%$A9+ AR$A9

= "1

Then

^ = "µD3(

127

¶("1)

= "µ0!0043 × 12' 30012× 1!41

¶("1)

^ = 4! 65× 10"5 m2/s

Now check to see if the ow is laminar (Reynolds number G 1' 000)

Re = , D$: = ^$:

=4! 65× 10"5 m2$ s1!12× 10"3m2$ s

Re = 0!0415 * Laminar

Therefore, the original assumption of laminar ow was correct.

752

PROBLEM 9.25


Find: Maximum uid velocity in R"direction.

ANALYSIS

The expression for @max is

@max = "D2(

87

AM

A9

where

AM$A9 = AM$AR = A$AR(%$( + R)

= (1$()A%$AR + 1

= (1$(0!8× 62!4))("8) + 1 = "0!16 + 1 = 0!840

Then

@max = "((0!8× 62!4× 0!012)$(8× 10"3)(0!840)

@max = "0!524 ft/s

Flow is downward.

753

PROBLEM 9.26


Find: Maximum uid velocity in R-direction.

ANALYSIS

@max = "D2(

87

AM

A9

where

AM$A9 = AM$AR = A$AR(%$( + R)

= (1$()A%$AR + 1

= (1$(0!85× 9' 810)("10' 000) + 1= "0!199

Then

@max = "(0!85× 9' 810× 0!0022)$(8× 0!1)("0!199)= 0!0083 m/s

@max = 8.31 mm/s

Flow is upward.

754

PROBLEM 9.27


Find: Maximum uid velocity in R-direction.

ANALYSIS

From solution to Prob. 9.21 we have

@max = "D2(

87

AM

A9

where

AM$A9 = AM$AR = A$AR(%$( + R)

= (1$()A%$AR + 1

= (1$(0!8× 62!4))("60)) + 1 = "0!202

Then

@max = "(0!8× 62!4× 0!012)$(8× 0!001)("0!202)

@max = +0.126 ft/s

The ow is upward.

755

PROBLEM 9.28


Find: Pressure gradient in the direction of ow.

Properties: From Table A.4 7 = 2× 10"3 lbf·s/ft2; ( = 55!1 lbf/ft3!

ANALYSIS

Flow rate and maximum velocity

, = ^$D

= 0!009$(0!09$12)

= 1!20 ft/s

@max = (3$2), = 1!8 ft/s

60o

@max = "D2(

87

AM

A9

AM

A9= "

µ87@max(D2

¶

= "µ8× (2× 10"3)× 1!855!1× (0!09$12)2

¶

= "9! 29

ButAM$A9 = (1$()A%$A9+ AR$A9

where AR$A9 = "0!866! Then

"9! 29 = (1$()A%$A9" 0!866A%$A9 = (("9! 29 + 0!866)

= 55!1("9! 29 + 0!866)= "464! 1

756

A%$A9 = -464 psf/ft

757

PROBLEM 9.29


Find: Pressure gradient in direction of ow.

ANALYSIS


, = ^$D

= 24× 10"4$(0!002)= 1!2 m/s

@max = (3$2), = 1!8 m/s

AM

A9= "

87@max(D2

AM$A9 = "8× 0!1× 1!8$(0!8× 9' 810× 0!0022) = "45!87

ButAM$A9 = (1$()A%$A9+ AR$A9

where AR$A9 = "0!866!Then

"45!87 = (1$()A%$A9" 0!866A%$A9 = (("45!87 + 0!866)

A%$A9 = -353 kPa/m

758

PROBLEM 9.30


Find: (a) Derive an expression for the velocity distribution between the plates as afunction of (' ?' 2' 7' and \!(b) Determine the plate velocity as a function of (' 2' and 7 for which the dischargeis zero.

ANALYSIS

Consider the uid element between the plates

y

Fluid element

Us

Consider the forces on the uid element

* "y s * "y+ y" s

W

"B 3!9+ (3+!&?!9" (!9!? = 0

Divide by !9!?"B 3!?

+B 3+!3!?

" ( = 0

Take the limit as !? approaches zero

AB$A? = (

ButB = 7A@$A?

759

SoA

A?(7A@$A?) = (

Integrate

7A@$A? = (? + -1

A@$A? =(

7? + -1

Integrate again

@ =(

7

?2

2+ -1? + -2

Boundary Conditions: At ? = 0' @ = 0 and at ; = 2' @ = \! Therefore,

-2 = 0 and -1 =\

2"(

7

2

2

@ = R/32

2+³UE+ R

/E2

´?

The discharge per unit dimension (normal to page) is given by

^ =

Z E

0

@A?

=

Z E

0

·(

7

?2

2+

µ\

2"(

7

2

2

¶?

¸A?

=(

7

?3

6+\?2

22"(

7

2?2

4|E0

=(

7

23

6+\2

2"(

7

23

4

For zero discharge\2

2=(23

47"(

7

23

6or

\ = 16R/22

760

PROBLEM 9.31

Situation: The ow of mud is described in the problem statement.

Find: (a) Relationships between variables and determine velocity eld.(b) Determine the velocity eld when there is ow.

Assumptions: Unit dimension normal to page.

ANALYSIS

(a) First consider the forces on an element of mud !E long and ?0 deep as shownbelow.

"x

yo

W sin!* ", x

W

There will be no motion if (?0 sin K G B 0(b) Consider forces on the element of mud shown below.

"x

W sin!-*.x

"y ( +(d /dy) y)* * "

XH$ = 0

"B!E+ (B + (AB$A?)!?)!E = 0(AB$A?)!? " ( sin K!? = 0

AB$A? = "( sin K

B = "Z( sin KA? + -

= "( sin K? + -

when ? = 0' B = 0 so

- = ( sin K?0

B = "( sin K? + ( sin K?0 (1)

andB = ( sin K(?0 " ?)

761

But for the mudB = B 0 + gA@$A? (2)

Eliminate B between equations (1) and (2)

B 0 + gA@$A? = ( sin K(?0 " ?)A@$A? = [( sin K(?0 " ?)" B 0] $g (3)

Upon integration

@ = (1$g)£( sin K(?0? " ?2$2)" B 0?

¤+ -

when? = 0' @ = 0 =! - = 0

If B G B 0' A@$A? = 0! Transition point is obtained from Eq. (3)

0 = (( sin K(?0 " ?)" B 0)B 0 = ( sin K(?0 " ?)B 0 = ( sin K?0 " ( sin K?

? =( sin K?0 " B 0

( sin K(4)

?T = ?0 " (B 0$( sin K) (5)

When 0 G ? G ?5K' B F B 0 and

@ = [( sin K(??0 " ?2$2)" B 0?] $g (6)

When ?5K G ? G ?0' B G B 0 so @ = @max = @5K and the velocity distribution is shownon the gure.

yutr

uytr

y0

762

PROBLEM 9.32

Situation: Glycerin ows between two cylinders —additional details are provided inthe problem statement.

Find: Discharge.

Properties: Table A.4 (Glycerin) 7 = 1!41 N·s/m2 and : = 1!12× 10"3m2$ s!

ANALYSIS

Discharge per unit width between two stationary plates is given by Eq. 9.12. Multiplethis by the average width of the channel

¡01¢to give

] = "µD3(

127

¶µAM

A9

¶01

The change in piezometric head (M) with position (9) is given by

AM

A9=

A(\R+ R)

A9

=AR

A9= "1

Combining equations gives

] =

µD3(

127

¶01

=

µ(0!0013m3) (12' 300N$m3)

12× (1!41N · s$m2)

¶× 0 × (0!029m)

= 6!62× 10"8m3$ s

] = 6!62× 10"8m3$ s

763

PROBLEM 9.33

Situation: A bearing is described in the problem statement.

Find: Amount of oil pumped per hour.

ANALYSIS

H = %avg. ×I= 1$2 %max ×I= 1$2 %max × 0!3 m× 1 m

or

%max = 2H$0!3 m2 = 2× 50' 000$0!30= 333' 333 N/m2

Then A%$A9 = "333' 333 N/m2$0!15 m = "2' 222' 222 N/m3! For ow between wallswhere sin K = 0, we have

@max = "(($27)(D ×D$2"D2$4)(A$A9(%$())@max = "(D2$87)A%$A9,avg. = 2$3@max

= "(1$12)(D2$7)A%$A9

Then^per side = , D = "(1$12)(D3$7)A%$A9

and

^total = 2, D = "(1$6)(D3$7)A%$A9= "(1$6)× ((6× 10"4 m)3$(0!2 N · s/m2))×"2' 222' 222 N/m3)= 4!00× 10"4 m3/s

^ = 1!44 m3/hr

764

PROBLEM 9.34

Situation: Couette ow —described in section 9.2.

Find: Velocity distribution.

APPROACH

Apply the continuity principle and Navier-Stokes equation.

ANALYSIS

The ow is steady and incompressible. There is no pressure gradient in the owdirection. Let E be in the ow direction and ? is the cross-stream direction. In theCouette ow problem

[@

[E= 0

so from the continuity principle

[;

[?= 0

or ; =constant. The constant must be zero to satisfy the boundary conditions.The E-component of the Naiver Stokes equation reduces to

A2@

A?2= 0

Integrating twice gives

@ = -1? + -2

Applying the boundary conditions that @(0) = 0 and @(2) = \ gives

@ = \ 3E

765

PROBLEM 9.35

Situation: This problem involves an Ei!el-type wind tunnel.

Test section width (square) is . = 457mm! Test section length is 2 = 914mm.

Find: Find the ratio of maximum boundary layer thickness to test section width(i (E = 2) $. ) for two cases:(a) Minimum operating velocity (\! = 1m$ s).(b) Maximum operating velocity (\! = 70m$ s).

Properties: Air properties from Table A.3. At & = 20 !C and % = 1atm' : =15!1× 10"6m2$ s!

APPROACH

Calculate the Reynolds number to establish if the boundary layer ow is laminar orturbulent. Then, apply the appropriate correlation for boundary layer thickness (i.e.for i)!

ANALYSIS

Reynolds number for minimum operating velocity

ReE =\!2

:

=(1m$ s) (0!914m)

(15!1× 10"6m2$ s)= 60' 530 (minimum operating velocity)

Since ReE ' 500' 000, the boundary layer is laminar.

Correlation for boundary layer thickness (laminar ow)

i =5E

Re1*2$

=5× (0!914m)&60' 530

= 18!57mm

766

Ratio of boundary layer thickness to width of the test section

i

.=18!57mm

457mm

i$. = 0!0406 (minimum operating velocity)

Reynolds number (maximum operating velocity)

ReE =\!2

:

=(70m$ s) (0!914m)

(15!1× 10"6m2$ s)= 4' 237' 000 (maximum operating velocity)

Since ReE ) 500' 000, the boundary layer is turbulent.

Correlation for boundary layer thickness (turbulent ow):

i =0!16E

Re1*7$

=0!16× (0!914m)(4' 237' 000)1*7

= 16!53mm

Ratio of boundary layer thickness to width of the test section

i

.=16!53mm

457mm

i$. = 0!036 (maximum operating velocity)

COMMENTS

1. Notice that the boundary layer is slightly thinner for the maximum velocity.

2. In both cases (maximum and minimum velocity), the boundary layer thicknessis only a small fraction of the width.

767

PROBLEM 9.36

Situation: A uid ows over a horizontal plate, giving the shear stress distributionshown in the sketch.

The speed of the uid free stream is \! = 2!4m$ s.The plate is an isosceles triangle with 2 = 1!5m!

Find: Find the viscous drag force in newtons on the top of the plate.

APPROACH

Since shear stress (B !) is the tangential force per unit area, integrate over area to ndthe drag force.

ANALYSIS

Viscous drag force (H&)

H& =

Z

Area

B !(E)AI

AI = .AE

H& =

EZ

0

B !(E). (E)AE

Plate width. (E) = 2" E

Shear stress distribution (C = 10Pa and > = 8Pa)

B !(E) = C" >E

2

768

Combine equations & integrate

H& =

EZ

0

B !(E). (E)AE

=

EZ

0

³C" >

E

2

´(2" E) AE

=

EZ

0

µC2" CE" >E+

>E2

2

¶AE

=

µC

2">

6

¶22

=

µ10

2"8

6

¶Pa× (1!5m)2

H& = 8!25N

769

PROBLEM 9.37

Situation: A thin plate is held stationary in a stream of water—additional details areprovided in the problem statement.

Find: (a) Thickness of boundary layer.(b) Distance from leading edge.(c) Shear stress.

APPROACH

Find Reynolds number. Then, calcuate the boundary layer thickness and shear stresswith the appropriate correlations

ANALYSIS

Reynolds number

Re = \0E$:

E = Re :$\0

= 500' 000× 1!22× 10"5$5

E = 1!22 ft

Boundary layer thickness correlation

i = 5E$Re1*2$ (laminar ow)

= 5× 1!22$(500' 000)1*2

= 0!0086 ft

i = 0.103 in.

Local shear stress correlation

B 0 = 0!3327(\0$E) Re1*2$

= 0!332× 2!36× 10"5(5$1!22)× (500' 000)1*2

B 0 = 0.0227 lbf/ft2

770

PROBLEM 9.38

Situation: Flow over a smooth, at plate —additional details are provided in theproblem statement.

Find: Ratio of the boundary layer thickness to the distance from leading edge justbefore transition.

ANALYSIS

Boundary layer thickness

i$E = 5$Re1*2$ (laminar ow)

= 5$(500' 000)1*2

i$E = 0.0071

771

PROBLEM 9.39

Situation: A horizontal plate (part of an engineered system for sh bypass) dividesa ow of water into two streams.

Water temperature is & = 40 !FFree stream velocity is \! = 12 ft$ s!Plate dimensions are 2 = 8 ft and . = 4 ft!

Find: Calculate the viscous drag force on the plate (both sides).

Properties: From Table A.5. Kinematic viscosity is : = 1!66× 10"5 ft2$ s! Densityis # = 1!94 slug$ ft3!

APPROACH

Find the Reynolds number to establish whether the boundary layer is laminar ormixed. Select the appropriate correlation for average resistance coe"cient (-() !Then, calculate the shear (i.e. drag) force (H&).

ANALYSIS

Reynolds Number.

ReE =\!2

:

=(12 ft$ s) (8 ft)¡16!6× 10"6 ft2$ s

¢ = 5' 783' 000

Thus, the boundary layer is mixed.

Average shear stress coe"cient

-( =0!523

ln2 (0!06ReE)"1520

ReE

=0!523

ln2 (0!06× 5' 783' 000)"

1520

5' 783' 000= 0!00295

Surface resistance (drag force)

H& = -(#, 2

2I

= 0!00295

¡1!94 slug$ ft3

¢(12 ft$ s)2

2(2× 8 ft× 4 ft)

= 26!38 lbf

H& = 26!4 lbf

772

PROBLEM 9.40


Find: Ratio of shear stress at edge of boundary layer to shear stress at the platesurface: B ^$B 0

ANALYSIS

At the edge of the boundary layer the shear stress, B ^, is approximately zero. There-fore, B ^$B 0 ! 0! Choice (a) is the correct one.

773

PROBLEM 9.41

Situation: Air ows over a device that is used to measure local shear stress—additionaldetails are provided in the problem statement.

Find: Force due to shear stress on the device

Assumptions: Over the length of the device (1 cm), assume that the local shear stresscoe"cient (Z() equals the average shear stress coe"cient (-() !

ANALYSIS

Reynolds number

Re$ =\E

:

=(25m$ s)× (1m)(1!5× 10"5m2$ s)

= 1!667× 106

Local shear stress coe"cient (turbulent ow)

Z( =0!455

ln2 (0!06Re$)

=0!455

ln2 (0!06× 1!667× 106)= 0!003433


H& = -(#\2!2I

= Z(#\2!2I

= 0!003433(1!2 kg$m3) (25m$ s)2

2(0!01m)2

= 1! 287× 10"4N

H& = 1!29× 10"4N

774

PROBLEM 9.42

Situation: The velocity prole and shear stress for ow over a at plate are describedin the problem statement.

Find: Equation for boundary layer thickness.

ANALYSIS

@$\0 = (?$i)1*2

B 0 = 1!66\07$i

B 0 = #\20A$AE

Z ^

0

(@$\0(1" @$\0))A?

= #\20A$AE

Z ^

0

((?$i)1*2 " (?$i))A?

= #\20A$AE[(2$3)(?$i)3*2 " 1$2(?$i)2]^0

1!66\07$i = (1$6)#\20Ai$AE

iAi$AE = 9!967$(#\0)

i2$2 = 9!967E$(#\0) = 9!96E2$Re$

i = 4!46E$Re1*2$

For the Blasius solution i = 5E$Re1*2

775

PROBLEM 9.43

Situation: Flow over a at plate —additional details are provided in the problemstatement.

Find: Liquid velocity 1 m from leading edge and 1 mm from surface.

APPROACH

Calculate Reynolds number and then use gure 9-6.

ANALYSIS

Reynolds numberRe$ = , E$: = 1× 1$2× 10"5 = 50' 000

The boundary layer is laminar. Use Fig. 9-6 to obtain @$\0

?Re045$ $E = 0!001(5× 104)045$1 = 0!224

Then from Fig. 9.6 @$\0 ! 0!075 ; @ = 0!075 m/s

776

PROBLEM 9.44

Situation: Flow over a thin, at plate —additional details are provided in the problemstatement.

Find: Skin friction drag on one side of plate.

ANALYSIS

Reynolds number

ReE = 1!5× 105

-( = 1!33$Re045E= 0!00343


H$ = -(D2#\2$2

= !00343× 1× 3× 1' 000× 12$2

H$ = 5!15 N

777

PROBLEM 9.45


Find: Velocity 1 m downstream and 3 mm from plate.

ANALYSIS

Reynolds number

Re$ = \E$:

= 5× 1$10"4

= 5× 104

Since Re$ ' 500' 000, the boundary layer is laminar.

Laminar velocity prole (use Fig. 9-6 to obtain @$\0)

?Re045$E

=(0!003)(5× 104)045

1= 0!671

Then from Fig. 9-6 @$\0 = 0!23! Therefore

@ = 5× 0!23@ = 1!15 m/ s

778

PROBLEM 9.46


Find: Oil velocity 1 m from leading edge and 10 cm from surface.

APPROACH

Calculate Reynolds number and apply gure 9-6.

ANALYSIS

Reynolds numberRe$ = 1× 1$10"4 = 104

The boundary layer is laminar. Use Fig. 9-6 to obtain @$\0

?Re045$ $E = 0!10× 102$1 = 10

Therefore the point is outside the boundary layer so @ = \0 = 1 m/s.

779

PROBLEM 9.47

Situation: Water ows over a submerged at plate.Plate length is 2 = 0!7m and the width is . = 1!5m!Free stream velocity is \! = 1!5m$ s!

Find:(a) Thickness of boundary layer at the location where "=$ = 500' 000!(b) Distance from leading edge.where the Reynolds number reaches 500,000.(c) Local shear stress.at the location where "=$ = 500' 000!

Properties: Table A.5 (water at 10 !C): # = 1000 kg$m3' 7 = 1!31 × 10"3N · s$m2,: = 1!31× 10"6m2$ s!

APPROACH

Calculate Reynolds number. Next calculate boundary layer thickness and local shearstress.

ANALYSIS

Reynolds number

"=$ = 500' 000

500' 000 =\0E

:

E =500' 000:

\0

=500000× (1!31× 10"6m2$ s)

1!5m$ s

= 0!436 7m

b.) E = 0!437m

Boundary layer thickness correlation

i =5E

Re1*2$....Laminar ow

=5× 0!436 7m&500000

= 3!09× 10"3 m

a.) i = 3!09 mm

Local shear stress correlation

B 0 = 0!3327(\0$E)Re1*2$

= 0!332× 1!31× 10"3(1!5$0!4367)× (500' 000)1*2

c.) B 0 = 1!06 N/m2

780

PROBLEM 9.48

Situation: Water ows over a submerged at plate.Plate length is 2 = 0!7m and the width is . = 1!5m!Free stream velocity is \! = 1!5m$ s!

Find: (a) Shear resistance (drag force) for the portion of the plate that is exposed tolaminar boundary layer ow.(b) Ratio of laminar shearing force to total shearing force.

Properties: Table A.5 (water at 10 !C): # = 1000 kg$m3' 7 = 1!31 × 10"3N · s$m2,: = 1!31× 10"6m2$ s!

ANALYSIS

For the part of the plate exposed to laminar boundary layer ow, the average shearstress coe"cient (-() is

-( =1!33&ReE

(laminar BL ow)

=1!33

&500000

= 0!00188

Transition occurs when Reynolds number is 500,000.

500000 =\!Etransition

:

500000 =(1!5m$ s)× (Etransition)1!31× 10"6m2$ s

Solving for the transition location gives

Etransition = 0!436 7m

Surface resistance (drag force) for the part of the plate exposed to laminar boundarylayer is

H& = -(#\2!2I

= 0!00188

Ã1000 kg$m3 × (1!5m$ s)2

2

!(0!436 7m× 1!5m)

= 1! 385N

Reynolds number for the plate

ReE = \0 × 2$:= 1× 0!7$(1!31× 10"6)= 8!015× 105

781

Thus, the boundary layer is mixed. The average shear stress coe"cient (-() is

-( =0!523

ln2 (0!06ReE)"1520

ReE(mixed BL ow)

=0!523

ln2 (0!06× 8!015× 105)"

1520

8!015× 105= 0!00260

Surface resistance (drag force) for the whole plate is

H&total = -(

µ#\202

¶I

= 0!00260

Ã1000 kg$m3 × (1!5m$ s)2

2

!(0!7m× 1!5m)

= 3! 071N

The ratio of drag forces is

H& (laminar ow)H& (total)

=1! 385N

3! 071N= 0!451 0

H&lam .$H&total = 0!451

782

PROBLEM 9.49

Situation: Flow over an airplane wing is described in the problem statement.

Properties: From Table A.3 : = 1!6× 10"5 m3$s and # = 1!17 kg/m3.

Find: (a) Friction drag on wing.(b) Power to overcome friction drag.(c) Fraction of chord which is laminar ow.(d) Change in drag if boundary tripped at leading edge.

APPROACH

(a) Calculate friction drag.(b) Find power as the product of drag force and speed: * = H&,(c) Calculate the critical length at a Reynolds number of Re = 5× 105.(d) Compare the average shear stress coe"cients for a mixed boundary layer andall-turbulent boundary layer.

ANALYSIS

\0 = (200 km/hr)(1,000 m/km)/(3,600 s/hr)

\0 = 55!56 m/s

Reynolds number

Re = \02$:

= (55!56)(2)$(1!6× 10"5)= 6!9× 106

From Fig. 9.14, the ow is mixed laminar and turbulent


H& = -(D2#\20$2

-( =0!523

ln2(0!06Re)"1520

Re

= 0!00290

Wing has two surfaces so

H&Iwing = 2× -(D2#\20$2= (2)(0!00290)(11)(1!17)(55!56)2

H&Iwing = 230 N (a)

783

Power

* = Hs,wing\0

= 230× 55!56* = 12!78 kW (b)

Critical laminar Re = 5× 105 = \0E$:

E1K = 5× 105:$\0= (5× 105)(1!6× 10"5)$55!56

E1K = 14 cm

b3CZ! = E1K$2

= !14$2

b3CZ = !07 (c)

If all of boundary layer is turbulent then

-( = 0!074$Re042

-( = 0!00317

Then

Htripped B.L.$Hnormal = 0!00317$0!00290

= 1!093

Change in drag with tripped B.L. is 9!3N increase.

784

PROBLEM 9.50

Situation: Turbulent ow over a at plate —additional details are provided in theproblem statement.

Properties: From Table A.5 # = 998 kg/m3; : = 10"6 m2$s.

Find: Velocity 1 cm above plate surface.

ANALYSIS

Local shear stress

@& = (B 0$#)045 = (0!1$998)045 = 0!01 m/s

@&?$: = (0!01)(0!01)$(10"6) = 102

From Fig. 9-10 for @&?$: = 100 it is seen that Eq. 9-34 applies

@$@& = 5!57 log(?@&$:) + 5!56

= 5!75 log(100) + 5!56 = 17!06

@ = @&(17!06) = 0!01(17!06)

@ = 0!171 m/s

785

PROBLEM 9.51


Find: (a) Resistance of plate.(b) Boundary layer thickness at trailing edge.

ANALYSIS

Reynolds number

ReE = \02$:

= 0!15× 1!5$(10"6)= 2! 25× 105

ReE ' 500,000; therefore, laminar boundary layer


i = 5E$Re1*2$= 5× 1!5$(2! 25× 105)1*2 = 1! 581 1× 10"2 m

i = 15!8 mm


-( = 1!33$Re1*2E

= 1!33$(2! 25× 105)1*2

= 0!00280


H& = -(I#\20$2

= 0!00280× 1!0× 1!5× 2× 1000× 0!152$2

H& = 0!094 5 N

786

PROBLEM 9.52


Find: (a) Skin friction drag per unit width of plate.(b) Velocity gradient at surface 1 m downstream from leading edge.

ANALYSIS

Reynolds number

ReE = \02#$7

= 20× 2× 1!5$10"5

= 6× 106


-( =0!523

ln2(0!06Re)"1520

Re

= 0!00294


H& = -((2D2)#\20$2

= 0!00294× (2× 1× 2)(1!5× 202$2)H& = 3!53 N

Reynolds number

Re1" = 6× 106 × (1$2)= 3× 106

Local shear stress coe"cient

Z( = 0!455$ ln2(0!06Re1")

= 0!455$ ln2(0!06× 3× 106)= 0!0031

Local shear stress

B 0 = Z(#\20$2

= 0!0031× 1!5× 202$2= 0!93 N/m2

B 0 = 7A@$A?

or

A@$A? = B 0$7

= 0!93$10"5

A@$A? = 9!3× 104 s"1

787

PROBLEM 9.53

Situation: Start with equation 9.44

Find: Carry out the steps leading to equation 4.47

ANALYSIS

Equation 9.44 is

B 0#=7

72\20Ai

AE

Substituting in Eq. 9.46 gives

0!010\20

µ:

\0i

¶1*6=7

72\20Ai

AE

Cancelling the \0’s and rearranging gives

72

7× 0!010

µ:

\0

¶1*6= i1*6

Ai

AE

Separate variables

0!1028

µ:

\0

¶1*6AE = i1*6Ai

Integrate

6

7i7*6 = 0!1028

µ:

\0

¶1*6E+ -

But i(0) = 0 so the constant is zero. Solving for i gives

i = (7

6× 0!1028)6*7

µ:

\0

¶1*7E6*7

Dividing through by E results in

^$= 0416

Re1.7'

788

PROBLEM 9.54

Situation: Flow over an airplane wing is described in the problem statement.

Find: (a) Speed at which turbulent boundary layer appears.(b) Total drag at this speed.

ANALYSIS

Reynolds number

Returb = 5× 105

=\Z

:

\ =(5× 105);

Z

=(5× 105)(1!58× 10"4)

5$12

= 189! 6 ft$ s

\ = 190 ft/s


-( = 1!33$(5× 105)045

= 0!00188


H& = -((#\2$2)I

= (0!00188)((0!00237)(189! 6)2$(2))(2)(3)(5$12)

H& = 0!200 lbf

789

PROBLEM 9.55


Find: (a) Skin friction drag on top per unit width.(b) Shear stress on plate at downstream end.

APPROACH

Apply the momentum principle to the c.v. shown. Then calculate the local shearstress.

ANALYSIS

c.s.

.

shear stress

y

Momentum principle

XH$ =

Z

c.v.,$#V · AA

H&Iplate on c.v. = "#, 21 i +Z#, 22 AI+ #,1^top

where

,2 = (,max$i)? = ,1?$i

^top = ,1i "Z ^

0

,2A? = ,1i "Z ^

0

,1?$iA?

^top = ,1i " ,1?2$2i|^0 = ,1i " 0!5,1i = 0!5,1i

Then

H& = "#, 21 i +Z ^

0

#(,1?$i)2A? + 0!5#, 21 i

= "#, 21 i + #,21 i$3 + 0!5#,

21 i

= #, 21 i("1 + (1$3) + (1$2)) = "0!1667#,21 i

For ,1 = 40 m/s, # = 1!2 kg/m3' and i = 3× 10"3 m we have

H& = "0!1667× 1!2× 402 × 3× 10"3

= "0!9608

790

or the skin friction drag on top side of plate is H& = +0!960 N.Local shear stress

B 0 = 7A,$A?

= 1!8× 10"5 × 40$(3× 10"3)

B 0 = 0!24 N/m2

791

PROBLEM 9.56

Situation: Start with Eq. 9.43

Find: Perform the integration and simplify to obtain Eq. 9.44.

ANALYSIS

Equation 9.43 is

B 0#= \20

A

AE

Z ^

0

³?i

´1*7 ·1"

³?i

´1*7¸A?

Changing the variable of integration to

g =³?i

´

the integral becomes

Z ^

0

³?i

´1*7 ·1"

³?i

´1*7¸A? = i

Z 1

0

g1*7£1" g1*7

¤Ag

= i

Z 1

0

[g1*7 " g2*7]Ag

Integrating we have

i

Z 1

0

[g1*7 " g2*7]Ag = i[7

8g8*7 "

7

9g9*7]10 =

7

72i

The equation then becomes

B 0#=7

72\20Ai

AE

792

PROBLEM 9.57

Situation: The velocity prole in a boundary layer is replaced by a step prole—additional details are provided in the problem statement.

Find: Derive an equation for displacement thickness.

ANALYSIS

/ =

Z ^

0

#@A? =

Z ^

^##'\'A? = #'\'(i " i

&)

#'\'i& = #'\'i "

Z ^

0

#@A?

= #'\'

Z ^

0

(1" (#@)$#'\')A?

! i& =Z ^

0

(1" (#@)$(#'\'))A?

793

PROBLEM 9.58

Situation: Displacement thickness is described in the problem statement.

Find: Magnitude of displacement thickness.

ANALYSIS

The streamlines will be displaced a distance i& = ^defect$,1 where

^defect =

Z ^

0

(,1 " ,2)A? =Z ^

0

(,1 " ,1?$i)A?

Then

i& = [

Z ^

0

(,1 " ,1?$i)A?]$,1

=

Z ^

0

(1" ?$i)A?

= i " i$2= i$2

i& = 1!5 mm

794

PROBLEM 9.59

Situation: Relationship between shear stress and boundary layer thickness:

B 0#= !0225\20 (

:

\0i)1*4

Find: (a) The variation of boundary layer thickness with E and Re$.(b) The variation of Local shear stress coe"cient with Re$.(c) The variation of average shear stress coe"cient with ReE.

APPROACH

Apply the integral method represented by Eq. 9.44 and the relationship betweenshear stress and boundary layer thickness (above).

ANALYSIS

Evaluating the integral for the 1/7th power prole gives

B 0#=7

72\20Ai

AE

Substituting in the expression for shear stress gives

0!0225:1*4

\1*40

=7

72i1*4

Ai

AE

Integrating and using the initial condition at i(0) = 0 gives

^$= 0437

Re1.5'

Substituting the equation for i into the equation for shear stress gives

Z( =04058

Re1.5'

Integrating this over a plate for the average shear stress coe"cient gives

-( =1

2

Z E

0

Z(AE

-( =04072

Re1.53

795

PROBLEM 9.60

Situation: Flow over two at plates —additional details are provided in the problemstatement.

Find: Ratio of skin friction drag on two plates.

ANALYSIS


H& = -(D2#\20$2

where -( = 04523ln2(0406×Re3)

" 1520Re3

Reynolds number

ReEI30 = 30× 10$10"6 = 3× 108

ReEI10 = 108

Then

-(I30 = 0!00187

-(I10 = 0!00213

Then

H&I30$H&I10 = (0!00187$0!00213)× 3

H&I30$H&I10 = 2!59

796

PROBLEM 9.61

Situation: A sign being pulled through air is described in the problem statement.

Properties: From Table A.3 : = 1!41× 10"5 m2$s and # = 1!25 kg/m3!

Find: Power required to pull sign.

APPROACH

Find the average shear stress coe"cient (-() and then calculate the surface resistance(drag force). Find power using the product of speed and drag force (* = H&, ).

ANALYSIS

Reynolds number

ReE =,02

:

=35× 30

1!41× 10"5ReE = 7! 447× 107

Average shear stress coe"cient (Eq. 9.54 or Fig. 9.14)

-( =0!523

ln2 (0!06ReE)"1520

ReE(turbulent ow)

=0!523

ln2 (0!06× 7! 447× 107)"

1520

7! 447× 107= 0!00221


H& = -(I#\20$2

H& = 0!00221× 2× 30× 2× 1!25× 352$2= 203!0 N

* = H&, = 203!0× 35* = 7!11 kW

797

PROBLEM 9.62

Situation: A plastic panel being lowered in the ocean.Panel dimensions are 2 = 1m, . = 3m, and P = 0!003m!Other data is provided in the problem statement.

Find: Tension in cable.

APPROACH

Apply equilibrium to the panel. Apply the surface resistance equation and thebuoyancy force equation to calculate the unknown forces.

ANALYSIS

EquilibriumX

H; = 0

& + H& = HBuoy. ". = 0

& =. " H& " HBuoy. (1)

T

Fs

W

F .Buoy

Buoyancy force

HBuoy. = (water,"= 0!003× 3× 10' 070= 90!6 N


H& = -(I#\20$2

Reynolds number

"G3 = , 2$:

= 2× 1$(1!4× 10"6)= 1! 429× 106

From Fig. 9-14 or Eq. 9.54,-( = 0!00299

798

So

H& = 0!00299× 2× 3× 1026× 9$2= 82!83 N

Eq. (1) gives

& = 250" 82!83" 90!6& = 76!6 N

799

PROBLEM 9.63

Situation: A plate falling though water is described in the problem statement.

Find: Falling speed in fresh water.

APPROACH

Apply equilibrium with the weight, buoyancy and drag force.

ANALYSIS

Equilibrium. "D = H&

. " (water," =1

2-(I#\

20

23!5" 998× 9!81× 0!002 =1

2× 1000× 2× 2× -( × \20

or

\20 =0!001962

-(

Using the equation for the average resistance coe"cient (-( ) and solving gives

\0 = 0!805 m/s

800

PROBLEM 9.64


Properties: From Table A.5 : = 10"6 m2/s.

Find: (a) Thickness of viscous sublayer 1 m downstream from leading edge.(b) Would a roughness element 100 7m high a!ect the local skin friction coe"cient,if so why?

ANALYSIS

i0 = 5:$@&

where @& = (B 0$#)045 andLocal shear stress

B 0 = Z(#\20$2

B 0$# = [0!455$ ln2(0!06Re$)]\20$2

Reynolds number

Re$ = \0E$:

= (5)(1)$10"6

= 5× 106

Then

B 0$# = [0!455$ ln2(0!06× 5× 106)](25$2)B 0$# = 0!0357 m2$s2

@& = (B 0$#)045 = 0!189 m/s

Finallyi0 = 5:$@& = (5)(10

"6)$(0!189)

i0 = 26!4× 10"6 m

Roughness element size of 100 microns is about 4 times greater than the thickness ofthe viscous sublayer; therefore, it would denitely a!ect the skin friction coe"cient.

801

PROBLEM 9.65

Situation: A model plane falling though air is described in the problem statement.

Properties: From Table A.3 # = 1!2 kg/m3; : = 1!51× 10"5 m2$s.

Find: Falling speed.

APPROACH

Determine the drag force (surface resistance) and apply equilibrium.

ANALYSIS


H& = -(#(\20$2)I

-( = 0!074$Re042

Equilibrium

. . = H&

3 = 2(0!074$(\0 × 0!1$(1!51× 10"5))042)(1!2)(\20$2)(1× 0!1)

Solving for \0 yields \0 = 67!6 m/s.

802

PROBLEM 9.66


Find: Total drag force on plate.

ANALYSIS

The drag force (due to shear stress) is

H& = -(1

2#\2!D2

The density and kinematic viscosity of air at 20oC and atmospheric pressure is 1.2kg/m3 and 1.5×10"5 N·s/m2' respectively. The Reynolds number based on the platelength is

ReE =15× 1

1!5× 10"5= 106

The average shear stress coe"cient on the “tripped” side of the plate is

-( =0!074

(106)1*5= 0!0047

The average shear stress coe"cient on the “untripped” side is

-( =0!523

ln2(0!06× 106)"1520

106= 0!0028

The total force is

H& =1

2× 1!2× 152 × 1× 0!5× (0!0047 + 0!0028)

H& = 0!506 N

803

PROBLEM 9.67

Situation: Flow Through two at plates is described in the problem statement.

Find: (a) Length where boundary layers merge.(b) Shearing force per unit depth.

Properties: The density and kinematic viscosity of water at these conditions are 1000kg/m3 and 10"6 m2/s.

APPROACH

Apply the correlation for boundary layer thickness for a tripped leading edge.

ANALYSIS


i =0!37E

Re1*5$(boundary layer tripped at leading edge)

=0!37E4*5¡U!+

¢1*5

Setting i = 0!002 m and E = 2 gives

24*5 =0!002

0!37

µ10

10"6

¶1*5= 0!136

or

2 = 0!0826 m

Check the Reynolds number

Re$ =0!0826× 1010"6

= 8!26× 105

so the equations for the tripped boundary layer (Re$ G 107) are valid.


-( =0!074

¡040826×1010"6

¢1*5

= 0!00485

Surface resistance (drag force).

804

H&D

= 2×1

2#\2!-(2

= 998× 102 × 0!00485× 0!0826Z->= 40!0 N/m

805

PROBLEM 9.68

Situation: Develop a computer program with input of Reynolds number and natureof boundary layer.

Find: Boundary layer thickness, Local shear stress coe"cient, and average shearstress coe"cient.

ANALYSIS

Typical results from program.Normal boundary layer

Reynolds number i$E Z( -(5× 105 0!00707 0!000939 0!0018811!0× 106 0!0222 0!00376 0!0028011!0× 107 0!01599 0!00257 0!002803

Tripped boundary layer

Reynolds number i$E Z( -(1!0× 106 0!0233 0!!336 !0046691!0× 108 0!0115 0!00186 0!00213

806

PROBLEM 9.69

Situation: A boat planes in water at a temperature of 60 !F . Boat speed is \0 = 70mph = 102!7 ft$ s!Model the boat hull as a at plate with length 2 = 8 ft and width . = 3 ft!

Find: Power required to overcome skin friction drag.

Properties: From Table A.5 : = 1!22× 10"5 ft2$s and # = 1!94 slug$ ft3!

APPROACH

Power is the product of drag force and speed (* = H&\0) ! Find the drag force usingthe appropriate correlation.

ANALYSIS

Reynolds number

ReE =\02

:

=(102!7 ft$ s) (8 ft)

1!22× 10"5 ft2$ s)

= 6!73× 107

Thus, the boundary layer is mixed. From Fig. 9-14 or Eq. 9.54 -( = 0!00224!


H& = -(

µ#\202

¶I

= 0!00224

Ã¡1!94 slug$ ft3

¢(102!7 ft$ s)2

2

!(8 ft× 3 ft)

= 549!4 lbf.

Power

* = H&\0

= (549!4 lbf) (102!7 ft$ s)

= 56' 420ft-lbfs

=

µ56' 420

ft-lbfs

¶µs · hp

550 ft · lbf

¶

* = 103 hp

807

PROBLEM 9.70

Situation: A javelin moving through air is described in the problem statement.

Find: (a) Deceleration.(b) Drag.(c) Acceleration in head and tail wind .(d) Maximum distance.

Properties: From Table A.3 : = 1!51× 10"5 m2$s and # = 1!20 kg/m3.

Assumptions: Turbulent boundary layer where I& = 012 = 0×0!025×2!65 = 0!208m2;

ANALYSIS

Surface resistance

H& = -(I&#\20$2

Reynolds number

ReE = \02$: = 30× 2!65$(1!51× 10"5)= 5!3× 106

Then from Fig. 9-14, -( = 0!00297! Then

H& = 0!00297× 0!208× 1!2× 302$2= 0!334 N

H = /C

or

C = H$/ = 0!334$(8!0$9!81)

C = 0!410 m/s2

With tailwind or headwind -( will still be about the same value: -( ! 0!00297! Then

H&Iheadwind = 0!334× (35$30)2

H&Iheadwind = 0!455 N

H&Itailwind = 0!334× (25$30)2

H&Itailwind = 0!232 N

As a rst approximation for maximum distance, assume no drag or lift. So formaximum distance, the original line of ight (from release point) will be at 45! withthe horizontal—this is obtained from basic mechanics. Also, from basic mechanics:

? = ")P2$2 + ,0P sin K

808

andE = ,0P cos K

or upon eliminating P from the above with ? = 0, we get

E = 2, 20 sin K cos K$)

= 2× 322 × 0!7072$9!81E = 104!4 m

ThenP = E$,0 cos K = 104!4$(32× 0!707) = 4!61 s

Then the total change in velocity over 4!6 s! 4!6× C& = 4!6× ("0!41) = "1!89 m/sand the average velocity is , = (32 + 30!1)$2 = 31 m/s. Then, a better estimate ofdistance of throw is: E = 312$9!81 = 98!0 m

809

PROBLEM 9.71

Situation: A log is being pulled through water—additional details are provided in theproblem statement.

Find: Force required to overcome surface resistance.

Properties: From table A.5 : = 1!31× 10"6 m2/s.

ANALYSIS

Reynolds number

ReE = 1!7× 50$(1!31× 10"6)= 6!49× 107

From Fig. 9-14 -( = 0!00225Surface resistance

H& = -(I&#,20 $2

= 0!00225× 0 × 0!5× 50× 1' 000× 1!72$2H& = 255 N

810

PROBLEM 9.72

Situation: A passenger train moving through air is described in the problem state-ment.

Find: power required.

Properties: From Table A.3 : = 1!41× 10"5 m2$s.

ANALYSIS

Reynolds number

ReE = \02$: = (100' 000$3' 600)× 150$(1!41× 10"5)Re100 = 2!95× 108

Re200 = 5!9× 108

-(100 = 0!00187

-(200 = 0!00173

Surface resistance equation

H& = -(I#\20$2

H&100 = 0!00187× 10× 150× 1!25× (100' 000$3' 600)2$2

H&100 = 1' 353 N

H&200 = 5' 006 N

Power

*100 = 1' 353× (100' 000$3' 600)

*100 = 37!6 kW

*200 = 5' 006× (200' 000$3' 600)*200 = 278 kW

811

PROBLEM 9.73

Situation: A boundary layer next to the smooth hull of a ship is described in theproblem statement.

Find: (a) Thickness of boundary layer at E = 100 ft!(b) Velocity of water at ?$i = 0!5.(c) Shear stress on hull at E = 100 ft!

Properties: Table A.5 (water at 60 !F): # = 1!94 slug$ ft3 ' ( = 62!37 lbf$ ft3'

7 = 2!36× 10"5 lbf · s$ ft2, : = 1!22× 10"5 ft2$ s!

ANALYSIS

Reynolds number

Re$ =\E

:

=(45)(100)

1!22× 10"5= 3!689× 108

Local shear stress coe"cient

Z( =0!455

ln2(0!06Re$)=

0!455

ln2(0!06 # 3!689× 108)= 0!001591

Local shear stress

B 0 = Z(

µ#\202

¶

= (0!001591)

µ1!94× 452

2

¶

B 0 = 3!13 lbf/ft2 (c)

Shear velocity

@& = (B 0$#)045

= (3!13$1!94)045

= 1!270 ft/s

Boundary layer thickness (turbulent ow)

i$E = 0!16Re"1*7$ = 0!16¡3!689× 108

¢"1*7

= 0!009556

i = (0!009556)(100)

i = 0!956 ft (a)

i$2 = 0!48 ft

812

From Fig. 9-12 at ?$i = 0!50, (\0 " @)$@& ! 3 Then

(45" @)$1!27 = 3

@ (? = i$2) = 41!2 ft/s (b)

813

PROBLEM 9.74

Situation: A ship moving through water is described in the problem statement.

Find: Skin friction drag on ship.

Properties: From Table A.5 : = 1!41× 10"5 ft2$s and # = 1!94 slugs/ft3!

ANALYSIS

Reynolds number

ReE = \02$:

= (30)(600)$(1!41× 10"5)= 1!28× 109

From Fig. 9-14 -( = 0!00158!

Surface resistance equation.

H& = -(I&#\20$2

= (0!00158)(50' 000)(1!94)(30)2$2

H& = 68' 967 lbf

814

PROBLEM 9.75

Situation: A barge in a river is described in the problem statement.

Find: Shear (drag) force.

Properties: : = 1!2× 10"5 ft2$s and # = 1!94 slugs/ft3!

ANALYSIS

Reynolds number

ReE = , 2$:

= 10× 208$(1!2× 10"5)= 1!73× 108

From Fig. 9-14 -( = 0!00199.


H& = -(D2#,20 $2

= (0!00199)(44)(208)(1!94$2)(102)

H& = 1' 767 lbf

815

PROBLEM 9.76

Situation: A supertanker in open seas is described in the problem statement.

Find: (a) Skin friction drag.(b) Power required.(c) Boundary layer thickness 300 m from bow.

Properties: From Table A.4 : = 1!4× 10"6 m2$s and # = 1026 kg/m3!

APPROACH

Find Reynolds number, and then calculate the average shear stress coe"cient (-().Next, nd the drag force and calculate power as the product of drag force and speed(* = H& × , ) ! To nd boundary layer thickness, apply the correlation for a turbulentboundary layer.

ANALYSIS

Reynolds number

ReE =\02

:

=(18× 0!515)× 325

1!4× 10"6= 2! 152× 109

Average shear stress coe"cient (-() (from Eq.9.54 or Fig. 9.14)

-( =0!523

ln2 (0!06ReE)"1520

ReE(turbulent ow)

=0!523

ln2 (0!06× 2! 152× 109)"

1520

2! 152× 109= 0!001499


H& = -(I#\20$2

= 0!001499× 325(48 + 38)× 1026× (18× 0!515)2$2= 1! 847× 106N

H& = 1!85 MN

Power

* = 1! 847× 106 × (18× 0!515)

* = 17!1 MW

816

Reynolds number

Re300 =\0E

:

=18× 0!515× 3001!4× 10"6

= 1! 986× 109

Thus, turbulent boundary layer

Correlation for boundary layer thickness (turbulent ow)

i

E=

0!16

Re1*7$

=0!16

(1! 986× 109)1*7

= 7! 513× 10"3

i = 300m× !007513i = 2!25 m

817

PROBLEM 9.77

Situation: A model test is to be done to predict the drag on a ship—additional detailsare provided in the problem statement.

Find: Wave drag on actual ship.

Properties: From Table A.5 : = 1!22× 10"5 ft2$s and # = 1!94 slugs/ft3!

ANALYSIS

Equilibrium

H3" = H3,

2"$2, = 1$100

,"$()2")045 = ,,$()2,)

045

,"$,, = (2"$2,)045 = 1$10

, 2"$,2, = 1$100

," = (1$10)(30 ft/s) = 3 ft/s

Viscous drag on model:

ReE = , 2$:

= (3)(5)$(1!22× 10"5)= 1!23× 106

-( = 0!00293 from Fig. 9-14


H&I" = -((1$2)#,2I

= (0!00293)(1$2)(1!94)(32)(2!5)

= 0!0639 lbf

! Hwave,m = 0!1" 0!0639 = 0!0361 lbf

Assume, for scaling up wave drag, that

(-,)" = (-,),

(!%$(#, 2$2))" = (!%$(#, 2$2)),

!%"$!%, = (#"$#,)(,2"$,

2, )

But

H"$H, = (!%"$!%,)(I"$I,) = (#"$#,)(,2"$,

2, )(I"$I,)

= (#"$#,)(2"$2,)3 = (1!94$1!99)(1$100)3

H, = H"(1!99$1!94)(100)3 = 0!0361(1!99$1!94)(106)

H, = 3!70× 104 lbf

818

PROBLEM 9.78

Situation: A model test is done to predict the drag on a ship—additional details areprovided in the problem statement.

Find: (a) Speed of prototype.(b) Model skin friction and wave drag.(c) Ship drag in salt water.

Properties: From Table A.5 :" = 1!00× 10"6 m2/s and #" = 998 kg/m3!From Table A.4 :, = 1!4× 10"6 m2$s and #" = 1026 kg/m3!

ANALYSIS

," = 1!45 m/s

,, = (2,$2")1*2 × ,"

=&30× 1!45," = 7!94 m/s

Re" = 1!45(250$30)$(1!00× 10"6) = 1!2× 107

Re, = 7!94× 250$1!4× 10"6 = 1!42× 109

-( =0!523

ln2(0!06Re)"1520

Re

-(" = 0!00275

-(, = 0!00157


H&" = -("I#,2$2

= 0!00275(8' 800$302)998× 1!452$2

H&" = 28!21 N

Hwave1 = 38!00" 28!21Hwave1 = 9!79 N

Hwave/ = (#,$#")(2,$2")3Hwave1 = (1' 026$998)30

3(9!79) = 272 kN

H&, = -(/I#,2$2 = 0!00157(8' 800)1' 026× 7!942$2 = 447 kN

H, = Hwave/ + H&, = 272 + 447

H, = 719 kN

819

PROBLEM 9.79

Situation: A hydroplane skims across a lake—additional details are provided in theproblem statement.

Find: Minimum shear stress on smooth bottom.

APPROACH

Minimum B 0 occurs where Z( is minimum. Two points to check: (1) where Re$ ishighest; i.e., Re$ = ReE and (2) Transition point at Re$ = 5× 105 (this is the end ofthe laminar boundary layer).

ANALYSIS

(1) Check end of plate

ReE = \02$:

= 15× 3$10"6

= 4!5× 107

Z( !0!455

ln2(0!06Re$)= 0!00207

(2) Check transitionRe$ = 5× 105

Z( = 0!664$Re1*2$= 0!00094

Local shear stress

B 0m in = Z(min#\20$2

= 0!00094× 998× 152$2

B 0min = 106 N/m2

820

PROBLEM 9.80

Situation: A water skier is described in the problem statement.

Find: Power to overcome surface resistance.

Properties: From Table A.5 : = 1!2× 10"5 ft2$s and # = 1!94 slugs/ft3.

ANALYSIS

Reynolds number

ReE = , 2$:

= 44× 4$1!2× 10"5

= 147(105) = 1!47(107)

From Fig. 9.14 -( = 0!0027!


H9 (per ski) = 0!0027(4)(1$2)(1!94)(442$2) = 10!14 lbf

H9 (2 skis) = 20!28 lbf

Power

* (hp) = 20!28× 44$550

* = 1!62 hp

821

PROBLEM 9.81

Situation: A ship is described in the problem statement.

Find: (a) Surface drag.(b) Thickness of boundary layer at stern.


APPROACH

Apply the surface resistance equation by rst nding Reynolds number and-( . Thenapply the correlation for boundary layer thickness.

ANALYSIS

Reynolds number

ReE = \02$: = 10× 80$(1!4× 10"6)ReE = 5!7× 108

From Fig. 9-14 -( = 0!00173!

Surface resistance

H9 = -(I#\20$2

= 0!00173× 1' 500× 1' 026× 102$2

H9 = 133 kN


i$E =0!16

Re1*7$i$E = 0!0090

i = 80× 0!0090i = 0!72 m

822

PROBLEM 10.1

Situation: Mean-velocity proles are described in the problem statement.

Find: Match the proles with the descriptions.

ANALYSIS

a. (3) b. (1) c. (2) d.(1) e. (3) f. (2)

823

PROBLEM 10.2

Situation: Liquid ows in a vertical pipe–details are provided in the problem state-ment

Find: (a) Determine the direction of ow.(b) Calculate the mean uid velocity in pipe.

ANALYSIS

Energy equation

%0$( + T!,20 $2) + R0 = %10$( + T10,

210$2) + R10 + ME

To evaluate, note that T!, 20 $2) = T10,210$2)! Substituting values gives

200' 000$8000 + 0 = 110' 000$8000 + 10 + M(

M( = 1!25m

Because ME is positive, the ow must be upward.

Head loss (laminar ow)

M( =3272,

(12

, =M((1

2

3272

=1!25× 8000× 0!012

32× (3!0× 10"3)× 10= 1!042m$ s

, = 1!04m$ s

824

PROBLEM 10.3

Situation: A viscous oil draining is described in the problem statement.

Find: Valid characterization at the time when the oil surface reaches level of section2.

ANALYSIS

Valid statements are (a), (d) and (e).

825

PROBLEM 10.4

Situation: Oil is pumped through a 2 in. pipe. ] = 0!25 cfs.

Find: Pressure drop per 100 feet of level pipe.

Properties: Oil Properties: 6 = 0!97' 7 = 10"2 lbf · s$ ft2

ANALYSIS

Flow rate equation

, = ]$I

= 0!05$((0$4)× (1$12)2)= 9!17 ft/sec

Reynolds number

Re = , 1#$7

= 9!17× (1$12)× 0!97× 1!94$10"2

= 144 (thus, ow is laminar)

Pressure Drop

!% =3272,

12

=32× 10"2 × 100× 9!17

(1$12)2

= 42' 255psf100 ft

= 293 psi/100 ft

826

PROBLEM 10.5

Situation: Liquid ows downward in a smooth vertical pipe. 1 = 1 cm , = 2!0m$ s%1 = 600 kPa

Find: Pressure at a section that is 10 feet below section 1.

Properties: # = 1000 kg$m3 7 = 0!06N · s$m2

ANALYSIS

Reynolds number

Re =, 1#

7

=2× 0!01× 1000

0!06= 333

Since Re G 2000, the ow is laminar.

Energy principle

%1$( + T1,21 $2) + R1 = %2$( + T1,

22 $2) + R2 + ME

Since ,1 = ,2, the velocity head terms (i.e. kinetic energy terms) cancel. The energyequation becomes

600' 000$(9!81× 1000) + 10 = %2$( + 0 + 3272,$(12

%2$( = 600' 000$( + 10" 32× 0!06× 10× 2$(((0!01)2)%2 = 600' 000 + 10× 9810" 384' 000

= 314 kPa

827

PROBLEM 10.6

Situation: A liquid ows in a pipe. 1 = 8mm' , = 1m$ s!

Find: (a) Determine if the velocity distribution will be logarithmic or parabolic. (b)Calculate the ratio of shear stress 1 mm from the wall to the shear stress at the wall..

Properties: # = 1000 kg$m3' 7 = 10"1 N · s$m2' : = 10"5m2$ s!

ANALYSIS

Reynolds number

Re =, 1#

7

=(1)(0!008)(1000)

10"1

= 80 (laminar)

Because the ow is laminar, the velocity distribution will be parabolic. For a par-abolic velocity distribution

, = ,1(1" 32$"2)

Velocity gradient

A,$A3 = "23,1$"2

Shear stress

B = 7A,

A3

Ratio of shear stress

B 3 mmB 4 mm

=

¡7A?AK

¢3 mm¡

7A?AK

¢4 mm

=" (723,1$"2)3 mm" (723,1$"2)4 mm

=(3)3 mm(3)4 mm

Therefore

B 3 mmB 4 mm

=3

4

= 0.75

828

PROBLEM 10.7

Situation: Glycerin ows in a tube–other details are provided in the problem state-ment.

Find: Pressure drop in units of pascals per 10 m.

Properties: Glycerin at 20 !C from Table A.4: 7 = 1!41N · s$m2, : = 1!12 ×10"3m2$ s!

ANALYSIS

, =]

I

=8× 10"6

(0$4)× 0!0302

= 0!01132 m/s

Reynolds number

Re =, 1

:

=0!01132× 0!0301!12× 10"3

= 0!3032 (laminar)

Then

!%( =3272,

12

=32× 1!41× 10× 0!01132

0!0302

= 5675Pa per 10m of pipe length

5.68 kPa per 10 m of pipe length

829

PROBLEM 10.8

Situation: Kerosene ows out a tank and through a tube–other details are providedin the problem statement.

Find: (a) Mean velocity in the tube.(b) Discharge.

Assumptions: Laminar ow so T = 2!

APPROACH

Apply the energy equation from the surface of the reservoir to the pipe outlet.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + 2,

2$2) + R2 + 3272,$((12)

0 + 0 + 0!50 = 0 + , 2$) + 3272,$((12)

Thus

, 2$) + 3272,$((12)" 0!50 = 0

, 2$32!2 + 32(4× 10"5)(10),$(0!80× 62!4× (1$48)2)" 0!50 = 0

, 2 + 19!0, " 16!1 = 0

Solving the above quadratic equation for , yields:

, = 0!81 ft/s

Check Reynolds number to see if ow is laminar

Re = , 1#$7

= 0!81× (1$48)(1!94× 0!8)$(4× 10"5)Re = 654!8 (laminar)

] = , I

= 0!81× (0$4)(1$48)2

= 2!76× 10"4 cfs

830

PROBLEM 10.9

Situation: Oil is pumped through a horizontal pipe–other details are provided in theproblem statement.

Find: Pressure drop per 10 m of pipe.

ANALYSIS

Re = , 1#$7

= 0!7× 0!05× 940$0!048= 685

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + 3272,$(1

2

Simplify

%1 " %2 = 3272,$12

= 32× 0!048× 10× 0!7$(0!05)2

%1 " %2 = 4301Pa

%1 " %2 = 4!30 kPa

831

PROBLEM 10.10

Situation: SAE 10-W oil is pumped through a tube–other details are provided inthe problem statement

Find: Power to operate the pump.

ANALYSIS

Energy equation

%1$( + R1 + T1,21 $2) + M, = %2$( + T2,

22 $2) + R2 + ME

SimplifyM, = ME = b(2$1)(,

2$2))

Flow rate equation

, = ]$I = 7!85× 10"4$((0$4)(0!01)2) = 10 m/s

Reynolds number

Re = , 1$: = (10)(0!01)$(7!6× 10"5) = 1316 (laminar)

Friction factor (b)

b =64

Re

=64

1316= 0!0486

Head of the pump

M, = b(2$1)(, 2$2))

= 0!0486(8$0!01)(102$((2)(9!81))

= 198 m

Power equation

* = M,(]

= 198× 8630×¡7!85 · 10"4

¢

= 1341W

832

PROBLEM 10.11

Situation: Oil ows downward in a pipe–other details are provided in the problemstatement

Find: Pressure gradient along the pipe.

ANALYSIS

Re = , 1$:

= (2)(0!10)$(0!0057)

= 35!1 (laminar)

"A$A9(%+ (R) = 327,$12

"A%$A9" (AR$A9 = (32)(10"2)(2)$0!12

"A%$A9" (("0!5) = 64

A%$A9 = (0!5)(0!9)(62!4)" 64A%$A9 = 28!08" 64

= -35.9 psf/ft

833

PROBLEM 10.12

Situation: Fluid ows in a smooth pipe–other details are provided in the problemstatement

Find: (a) Magnitude of maximum velocity, (b) Resistance coe"cient, (c) Shear ve-locity, and (d) Shear stress 25 mm from pipe center.

ANALYSIS

Reynolds number

Re =, 1#

7

=0!05× 0!1× 800

0!01= 400

Therefore, the ow is laminar

,max = 2, = 10 cm/s

b = 64$Re

= 64$400

= 0.16

@&$, =pb$8

@& =p0!16$8× 0!05

= 0.00707 m/s

B 0 = #@2&= 800× 0!007072

= 0!040 N/m2

Get B K=04025 by proportions:

0!025$0!05 = B$B 0; B = 0!50B 0

B = 0!50× 0!040= 0.020 N/m2

834

PROBLEM 10.13

Situation: Kerosene ows in a pipe.& = 20!C, ] = 0!02m3$9' 1 = 20 cm

Find: Determine if the ow is laminar or turbulent.

ANALYSIS

Re = , 1#$7

= (]$I)1$:

= 4]$(01:)

= 4× 0!04$(0 × 0!25× 2!37× 10"6)= 85' 957

Flow is turbulent

835

PROBLEM 10.14

Situation: Fluid ows out of a tank through a pipe that has a contraction in diameterfrom 2 to 1 m.Each pipe is 100 m long. Friction factor in each pipe is b = 0!01

Find: Ratio of head lossME (1-m pipe)ME (2-m pipe)

ANALYSIS

ME = b12111

, 212)

ME (1-m pipe)ME (2-m pipe)

=

µb121,

21 $ (11)

b222, 22 $ (12)

¶

= (12$11)(,21 $,

22 )

,1I1 = ,2I2

,1$,2 = I2$I1 = (12$11)2

(,1$,2)2 = (12$11)

4

Thus

ME (1-m pipe)ME (2-m pipe)

= (12$11)(12$11)4

= (12$11)5

= 25

= 32

Correct choice is (d).

836

PROBLEM 10.15

Situation: Glycerin ows in a pipe1 = 0!5 ft' & = 68!F, , = 2 ft$ s

Find: (a) Determine if the ow is laminar or turbulent. (b) Plot the velocity distrib-ution.

Properties: Glycerin at 68 !F from Table A.4: 7 = 0!03 lbf · s$ ft2, : = 1!22 ×10"2 ft2$ s!

ANALYSIS

Re =, 1

:

=2× 0!5

1!22× 10"2

= 81!97 (laminar)

To nd the velocity distribution, begin with Eq. (10.7)

, (3) =32! " 32

47

·"A

A9(%+ (R)

¸

From Eq. (10.10) ·"A

A9(%+ (R)

¸=87,

32!

Combine equations

, (3) =32! " 32

47

·87,

32!

¸

= 2,

µ1"

32

32!

¶

= (4 ft$ s)

Ã1"

µ3

3!

¶2!

Create a table of values and plot

3 (in) 3$30 , (3) (ft/s)0 0 40.5 1/6 3.891.0 1/3 3.561.5 1/2 3.002 2/3 2.222.5 5/6 1.223 1 0

837

Velocity (ft/s)

0 1 2 3 4 50

1

2

3

Rad

ius

(in)

838

PROBLEM 10.16

Situation: Glycerin (20!C) ows through a funnel–details are provided in the prob-lem statement.

Find: Mean velocity of glycerine.

Properties: Glycerin at 20 !C from Table A.4: # = 1260 kg$m3, ( = 12' 300N$m3,7 = 1!41N · s$m2, : = 1!12× 10"3m2$ s!

Assumptions: Assume laminar ow (T2 = 2!0) !

ANALYSIS

Energy equation (Let section 1 be the surface of the liquid and section 2 be the exitplane of the funnel).

%1(+ T1

, 212)+ R1 + M, =

%2(+ T2

, 222)+ R2 + M5 + ME

!

0 + 0 + 0!30 = 0 + 2!0

µ, 222)

¶+ 0 +

3272,2(12

0!30 = 2!0

µ, 22

2× 9!81

¶+

µ32× 1!41× 0!2× ,212300× 0!012

¶

Solve quadratic equation.

,2 = "72! 01,2 = ,2 = 4! 087× 10"2

Select the positive root

,2 = 0!0409 m/s

Check the laminar ow assumption

Re =, 1#

7

=0!0409× 0!01× 1260

1!41= 0!365

Since Re ' 2000, the laminar ow assumption is valid.

839

PROBLEM 10.17

Situation: Castor oil ows in a steel pipe.Flow rate is ] = 0!2 ft3$ s! Pipe length is 2 = 0!5mi = 2640 ft!Allowable pressure drop is 10 psi.

Find: Diameter of steel pipe.

Properties: Viscosity of castor oil is 7 = 8!5× 10"3 lbf-s/ft2!Specic gravity of castor oil is 6 = 0!85!

Assumptions: Assume laminar ow.

ANALYSIS

!%( =3272,

12

or

!%( =3272]

(0$4)×14

Then

14 =12872]

0!%(

=128× 8!5× 10"3 × 2640× 0!2

0 × 10× 14414 = 0!126 98

1 ) 0!5969 ft

Find velocity.

, =]

I

=0!2

0$4× 0!59692

= 0!7147 ft/sec.

Check Reynolds number

Re =, 1#

7

=0!7147× 0!5969× (0!85× 1!94)

8!5× 10"3= 82!76

Thus, the initial assumption of laminar ow is valid. Use a pipe with an insidediameter of

1 ) 0!597 ft

840

PROBLEM 10.18

Situation: Mercury ows downward through a long round tube. & = 20!CThe tube is oriented vertically and open at both ends.

Find: Largest tube diameter so that the ow is still laminar.

Properties: From Table A.4: 7 = 1!5 × 10"3 N·s/m2' : = 1!2 × 10"7 m2/s, ( =133' 000 N/m3

Assumptions: The tube is smooth.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + ME


%1 = %2; ,1 = ,2; T1 = T2; R1 " R2 = 2

The energy equation2 = ME (1)


ME = M( =3272,

(12(2)

Combining Eqs. (1) and (2)

ME(12

327,= ME

(12

327,= 1 (3)

Reynolds number

Re =, 1

:= 2000

, =2000:

1(4)

Combining Eqs. (3) and (4)(13

64' 0007:= 1

or

1 = 3

r64' 0007:

(

= 3

s(64' 000) (1!5× 10"3) (1!2× 10"7)

133' 000

= 4!43× 10"4m

841

PROBLEM 10.19

Situation: Glycerin ows in a steel tube—additional details are provided in the problemstatement

Find: (a) Determine if the ow is laminar or turbulent, (b) Will pressure increaseor decrease in direction of ow? (c) Calculate the rate of change of pressure in thedirection of ow, (d) Calculate shear stress at the center of the tube and (e) Calculateshear stress at the wall.

Properties: Glycerin at 20 !C from Table A.4: # = 1260 kg$m3, ( = 12' 300N$m3,7 = 1!41N · s$m2, : = 1!12× 10"3m2$ s!

ANALYSIS

Re =, 1

:

=0!40× 0!041!12× 10"3

= 14!29

Answer ==F Since " ' 2000, the ow is laminar.

From solution to Problem 10-11

AM

A9="327,(12

AM

A9=

"327,(12

="32× 1!41× 0!412300× 0!042

= "0!917 1

or

(1$()A%$A9+ AR$A9 = "0!917 1

Because ow is downward, AR$AR = "1! Then

A%$A9 = 12300[1" 0!917 1]= 1019! 7

= 1.02 kPa/m

Answer ! Pressure increases in the direction of ow (downward).

842

From Eq. 10-3B = ((3$2)["AM$A9]

or

B = 12' 300(3$2)× 0!917 1

At the center of the pipe (3 = 0)

B K=0=0

At the wall (3 = 2cm)

Bwall = B 0 = 12' 300(0!02$2)× 0!917 1

Bwall = 113N$m2

843

PROBLEM 10.20

Situation: The design might have a physical conguration as shown below. Thedesign should be based upon solving Eq. 10.17 (M( = 3272,$((12)) for the viscosity7. Since this is for laminar ow, the size of pipe and depth of liquid in the tank shouldbe such that laminar ow will be assured ("G G 1000). For the design suggested here,the following measurements, conditions, and calculations would have to be made:

Pump

A. Measure tube diameter by some means.B. Measure ( or measure temperature and get ( from a handbook.C. Establish steady ow by having a steady supply source (pump liquid from areservoir).D. Measure ]. This could be done by weighing an amount of ow for a givenperiod of time or by some other means.E. Measure M($2 by the slope of the piezometric head line as obtained frompiezometers. This could also be obtained by measuring!% along the tube by meansof pressure gages or pressure transducers from which M($2 could be calculated.F. Solve for 7 with Eq. 10.17.

844

PROBLEM 10.21

Situation: Velocity measurements are made in a 1-ft diameter pipe. Other detailsare provided in the problem statement.

Find: Kinematic viscosity of uid.

ANALYSIS

Since the velocity distribution is parabolic, the ow is laminar. Then

!%( = 3272,$12

: = 7$# = !%(12$(322, #)

: = 15× 12$(32× 100× 2$2× 0!9× 1!94)

= 0.00268 ft2/s

845

PROBLEM 10.22

Situation: Velocity measurements are made in a 30-cm diameter pipe. Other detailsare provided in the problem statement.

Find: Kinematic viscosity of uid.

ANALYSIS

Following the solution for Problem 10.21,

: = !%(12$(322, #)

= 1' 900× (0!3)2$(32× 100× 0!75× 800)

= 8!91× 10"5 m2/s

846

PROBLEM 10.23

Situation: Water is pumped through tubes in a heat exchanger–other details areprovided in the problem statement

Find: Pressure di!erence across heat exchanger

ANALYSIS

Reynolds number (based on temperature at the inlet)

Re20! =, 1

:=0!12× 0!005

10"6= 600

Since Re ' 2000' the ow is laminar. Thus,

!% = 3272,$12

Assume linear variation in 7 and use the temperature at 25!C. From Table A.5

7avg. = 725!

= 8!91× 10"4 N · s/m2

and

!% = 3272,$12

= 32× 8!91× 10"4 × 5× 0!12$(0!005)2

= 684 Pa

847

PROBLEM 10.24

Situation: Oil ows through a 2-in. diameter smooth pipe—details are provided in theproblem statement.

Find: (a) The direction of the ow.(b) Resistance coe"cient.(c) Nature of the ow (laminar or turbulent).(d) Viscosity of oil.

ANALYSIS

Based on the deection on the manometer, the static pressure within the rightside of the pipe is larger than the pressure on the left end. Thus, the ow isdownward (from right to left).

Energy principle%2(+ T2

, 222)+ R2 =

%1(+ T1

, 212)+ R1 + ME

Term by term analysisT1,1 = T2,2; R2 " R1 = 2 ft

Darcy Weisbach equationME = b(2$1),

2$(2))

Combine equations%2 " %1(oil

= ("2 ft) + b2

1

, 2

2)(1)

Manometer equation

%2 + (4 ft) (oil + (0!33 ft) (oil " (0!33 ft) (Hg " (2 ft) (oil = %1

Calculate values

%2 " %1(oil

= " (4 ft)" (0!33 ft) + (0!33 ft)(Hg(oil

+ (2 ft)

= " (2 ft) + (0!33 ft)µ6Hg6oil

" 1¶

= " (2 ft) + (0!33 ft)µ13!6

0!8" 1¶

%2 " %1(oil

= 3!28 ft (2)

848

Substitute Eq. (2) into (1)

(3!28 ft) = ("2 ft) + b2

1

, 2

2)or

b = 5!28

µ1

2

¶µ2)

, 2

¶

= 5!28

µ1$6

30

¶µ2× 32!252

¶

b = 0!076

Since the resistance coe"cient (b) is now known, use this value to nd viscosity.

Resistance coe"cient (b) (assume laminar ow)

b =64

Re

0!076 =647

#, 1or

7 =0!076#, 1

64

=0!076× (0!8× 1!94)× 5× (1$6)

64

= 0.00154 lbf · s/ ft2

Check laminar ow assumption

Re =, 1#

7

=5× (1$6)× (0!8× 1!94)

0!00154= 840

Answer ! Flow is laminar.

849

PROBLEM 10.25

Situation: Oil ows through a 5-cm. diameter smooth pipe—details are provided inthe problem statement.

Find: (a) Flow direction.(b) Resistance coe"cient.(c) Nature of ow (laminar or turbulent).(d) Viscosity of oil.

ANALYSIS

Based on the deection on the manometer, the piezometric head on the right sideof the pipe is larger than that on the left side. Since the velocity at 1 and 2 is thesame, the energy at location 2 is higher than the energy at location 1. Since the auid will move from a location of high energy to a location of low energy, the ow isdownward (from right to left).Energy principle

%2(+ T2

, 222)+ R2 =

%1(+ T1

, 212)+ R1 + ME

Assume T1,1 = T2,2! Let R2 " R1 = 1m! Also the head loss is given by the DarcyWeisbach equation: M( = b(2$1), 2$(2))! The energy principle becomes

%2 " %1(oil

= ("1m) + b2

1

, 2

2)(1)

Manometer equation

%2 + (2m) (oil + (0!1m) (oil " (0!1m) (Hg " (1m) (oil = %1

Algebra gives

%2 " %1(oil

= " (2m)" (0!1m) + (0!1m)(Hg(oil

+ (1m)

= " (1m) + (0!1m)µ6Hg6oil

" 1¶

= " (1m) + (0!1m)µ13!6

0!8" 1¶

%2 " %1(oil

= 0!6m (2)

850

Substituting Eq. (2) into (1) gives

(0!6m) = ("1m) + b2

1

, 2

2)or

b = 1!6

µ1

2

¶µ2)

, 2

¶

= 1!6

µ0!05

12

¶µ2× 9!811!22

¶

b = 0!0908

Since the resistance coe"cient is now known, this value can be used to nd viscosity.To perform this calculation, assume the ow is laminar, and apply Eq. (10.23).

b =64

Re

0!0908 =647

#, 1or

7 =0!0908#, 1

64

=0!0908× (0!8× 1000)× 1!2× 0!05

64

= 0.068 N · s/m2

Now, check Reynolds number to see if laminar ow assumption is valid

Re =, 1#

7

=1!2× 0!05× (0!8× 1000)

0!068= 706

Thus, ow is laminar.

851

PROBLEM 10.26

Situation: A liquid ows through a 3-cm diameter smooth pipe.The ow rate is doubled.Other details are provided in the problem statement.

Find: Determine if the head loss would double.

ANALYSIS

M(2

= 2

=b

1

µ, 2

2)

¶

=b

0!03

µ12

2× 9!81

¶

= 1!699b

Rearrange

1!699b = 2

b = 1!177

Assume laminar ow:b = 64$Re

or

Re = 64$1!177 = 54!4 (laminar)

Indeed, the ow is laminar and it will be laminar if the ow rate is doubled.

Answer ! The head loss varies linearly with , (and ]); therefore, the head losswill double when the ow rate is doubled.

852

PROBLEM 10.27

Situation: Oil ows in a 12-in. smooth tube–other details are provided in the prob-lem statement.

Find: Viscous shear stress on wall.

ANALYSIS

As shown in Eq. (10.21), the resistance coe"cient is dened by

B ! =b

4

µ#, 2

2

¶

So

B ! =0!017

4

µ(0!82× 1!94) 62

2

¶

= 0!122 lbf$ ft2

853

PROBLEM 10.28

Situation: Fluids (oil and a gas) ow through a 10-cm. smooth tube–other detailsare provided in the problem statement.

Find: Velocity ratio: (,max,oil$,max,gas)!

ANALYSIS

Reoil =, 1#

7

=(1)(0!1)(900)

10"1

= 900

Since ow at this Reynolds number is laminar, the centerline velocity is twice themean velocity, or

,max , oil = 2,

For the gas

Regas =, 1#

7

=(1!0)(0!1)(1)

10"5

= 104

This corresponds to turbulent ow–Thus,

,max , gas % 1!08,

Therefore

,max,oil,max,gas

%2

1!08F 1

So, case (a) is the correct answer.

854

PROBLEM 10.29

Situation: Water ows with a through a horizontal run of PVC pipeSpeed of water: , = 5 ft$ s! Length of the pipe: 2 = 100 ft!Pipe is a 2.5" schedule 40: V1 = 2!45 in! = 0!204 ft

Find: (a) Pressure drop in psi.(b) Head loss in feet.(c) Power in horsepower needed to overcome the head loss.

Properties: Water @ 50 !F from Table A.5: # = 1!94 slug$ ft3, ( = 62!4 lbf$ ft3,: = 14!1× 10"6 ft2$ s!

Assumptions: 1.) Assume c& = 0!2.) Assume T1 = T2, where subscripts 1 and 2 denote the inlet and exit of the pipe.

APPROACH

To establish laminar or turbulent ow, calculate the Reynolds number. Then ndthe appropriate friction factor (b) and apply the Darcy-Weisbach equation to ndthe head loss. Next, nd the pressure drop using the energy equation. Lastly, ndpower using * = /)M( !

ANALYSIS

Reynolds number

Re =, 1

:

=(5 ft$ s) (0!204 ft)¡14!1× 10"6 ft2$ s

¢

= 72' 400

Thus, ow is turbulent.

Friction factor (b) (Swamee-Jain correlation)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³5474

72I400049

í2

= 0!0191

Darcy-Weisbach equation

M( = b2

1

, 2

2)

= 0!0191

µ100 ft

0!204 ft

¶(5 ft$ s)2

2× 32!2 ft$ s2= 3!635 ft

855

M( = 3!64 ft (part b)

Energy equation

%1(+ T1

, 212)+ R1 + M, =

%2(+ T2

, 222)+ R2 + M5 + ME

Select a control volume surrounding the pipe. After analysis of each term, the energyequation simplies to

%1(

=%2(+ M(

or !% = (M(

=¡62!4 lbf$ ft3

¢(3!635 ft)

= 227 psf

= 227

µlbf

ft2

¶µft2

144 in2

¶

!% = 1!58 psi (part a)

Flow rate equation

/ = #I,

=¡1!94 slug$ ft3

¢Ã0 (0!204 ft)2

4

!(5 ft$ s)

= 0!317 slug$ s

Power equation

. = /)M(

= (0!317 slug$ s)¡32!2 ft$ s2

¢(3!635 ft)

µ1!0 hp

550 ft · lbf$ s

¶

= 0!06746 hp

Power to overcome head loss = 0!0675 hp (part c)

COMMENTS

1. The pressure drop for a 100 ft run of pipe (!% = 227 psf % 1!6 psi )could bedecreased by selecting a larger pipe diameter.

2. The power to overcome the frictional head loss is about 1/15 of a horsepower.

856

PROBLEM 10.30

Situation: Water ows with a through a horizontal run of PVC pipeSpeed of water: , = 2m$ s! Length of the pipe: 2 = 50m!Pipe is a 2.5" schedule 40: V1 = 2!45 in! = 0!0622m!

Find: (a) Pressure drop in kPa.(b) Head loss in meters.(c) Power in watts needed to overcome the head loss.

Properties: Water @ 10 !C from Table A.5: # = 1000 kg$m3, ( = 9810N$m3, : =1!31× 10"6m2$ s!

Assumptions: 1.) Assume c& = 0!2.) Assume T1 = T2, where subscripts 1 and 2 denote the inlet and exit of the pipe.

APPROACH

To establish laminar or turbulent ow, calculate the Reynolds number. Then ndthe appropriate friction factor (b) and apply the Darcy-Weisbach equation to ndthe head loss. Next, nd the pressure drop using the energy equation. Lastly, ndpower using * = /)M( !

ANALYSIS

Reynolds number

Re =, 1

:

=(2m$ s) (0!0622m)

(1!31× 10"6m2$ s)= 94' 960

Thus, ow is turbulent.

Friction factor (b) (Swamee-Jain equation)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³5474

94I960049

í2

= 0!0181


M( = b2

1

, 2

2)

= 0!0181

µ50m

0!0622m

¶(2m$ s)2

2× 9!81m$ s2= 2!966m

857

M( = 2!97m (part b)

Energy equation

%1(+ T1

, 212)+ R1 + M, =

%2(+ T2

, 222)+ R2 + M5 + ME

Select a control volume surrounding the pipe. After analysis of each term, the energyequation simplies to

%1(

=%2(+ M(

or !% = (M(

=¡9810N$m3

¢(2!966m)

= 29' 096 kPa

!% = 29!1 kPa (part a)

Flow rate equation

/ = #I,

=¡1000 kg$m3

¢Ã0 (0!0622m)2

4

!(2m$ s)

= 6!077 kg$ s

Power equation

. = /)M(

= (6!077 kg$ s)¡9!81m$ s2

¢(2!966m)

= 176!8W

Power to overcome head loss = 177W (part c)

COMMENTS

1. The pressure drop (29 kPa) is about 1/3 of an atmosphere This value could bedecreased by increasing the pipe diameter to lower the speed of the water.

2. The power to overcome the frictional head loss is small, about 1/4 of a horse-power.

858

PROBLEM 10.31

Situation: Water @ 70!F ows through a pipe. 1 = 6 in ] = 2 cfs

Find: Resistance coe"cient.

Properties: From Table A.5 :(70oF)= 1!06× 10"5 ft2$s

ANALYSIS

Reynolds number

Re =4]

01:

=4× 2

0 × (6$12)× (1!06× 10"5)= 4!8× 105

From Fig. 10.8 or the Swamee and Jain correction (Eq. 10.26)

b = 0!013

859

PROBLEM 10.32

Situation: Water @ 10!C ows through a pipe. 1 = 25 cm ] = 0!06m3$ s!


Properties: From Table A.5 :(10oC) = 1!31× 10"6 m2$s.

ANALYSIS

Re =4]

01:

=4× 0!06

0 × 0!25× (1!31× 10"6)= 2!33× 105

From Fig. 10.8 or the Swamee and Jain correction (Eq. 10.26)

b = 0!015

860

PROBLEM 10.33

Situation: Air (20!C) ows through a smooth tube.] = 0!015m3$ s 1 = 3cm % = 110 kPa-absolute

Find: Pressure drop per meter of tube length

Properties: From Table A.3 7(20o) = 1!81× 10"5 N·s/m2!

ANALYSIS

, =]

I

=0!015

0$4× 0!032

= 21!2 m/s

# =%

"&

=110' 000

287× 293= 1!31 kg/m3

Re =, 1#

7

=21!2× 0!03× 1!311!81× 10"5

= 46031

Friction factor (b) (Moody diagram-Fig. 10-8)

b = 0!0212

Darcy Weisbach equation

M( = b2

1

, 2

2)

= 0!0212

µ1m

0!03m

¶Ã(21!2m$ s)2

2× 9!81m$ s2

!

= 16!19m for a 1.0 m length of pipe

Pressure drop is given by applying the energy equation to a 1.0 m length of pipe

!% = M(#)

= (16!19m)¡1!31 kg$m3

¢ ¡9!81m$ s2

¢

= 207!6Pafor a 1.0 m length of pipe

!,E= 208 Pa

m

861

PROBLEM 10.34

Situation: Glycerin ows through a commercial steel pipe–other details are providedin the problem statement.

Find: Height di!erential between the two standpipes.

Properties: Glycerin at 20 !C from Table A.4: # = 1260 kg$m3, 6 = 1!26' ( =12' 300N$m3, 7 = 1!41N · s$m2, : = 1!12× 10"3m2$ s!

ANALYSIS

Energy equation (apply from one standpipe to the other)

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + ME

%1$( + R1 = %2$( + R2 + ME

((%1$() + R1))" ((%2$() + R2) = ME

!M = ME

Reynolds number

Re =, 1

:

=(0!6)(0!02)

1!12× 10"3= 10!71

Since Re G 2000' the ow is laminar. The head loss for laminar ow is

ME =3272,

(12

=(32)(1!41)(1)(0!6)

12300× 0!022= 5! 502m

Energy equation

!M = ME

= 5.50 m

862

PROBLEM 10.35

Situation: Air ows through a smooth tube–other details are provided in the problemstatement.

Find: Pressure drop per foot of tube.

Properties: From Table A.3 7(80oF) = 3!85× 10"7 lbf-s/ft2!

ANALYSIS

, = ]$I = 25× 4$(60× 0 × (1$12)2) = 91!67 ft/s# = %$("& ) = 15× 144$(1716× 540) = 0!00233 slugs/ft3

Re = , 1#$7 = 91! 67× (1$12)× 0!00233$(3!85× 10"7)= 4!623× 104

Resistance coe"cient (b) (Swamee-Jain correlation; turbulent ow)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³5474

(44623×104)049

í2

= 0!0211

Pressure drop

!% = b2

1

µ#, 2

2

¶

= 0!0211

µ1 ft

1$12 ft

¶µ0!00233× 91!672

2

¶

= 2! 479 psf/ft

!% = 2!48 psf/ft

863

PROBLEM 10.36

Situation: A pipe is being using to measure viscosity of a uid–details are providedin the problem statement

Find: Kinematic viscosity.

ANALYSIS

M( = b(2$1)(, 2$2))

0!50 = b(1$0!01)(32$(2× 9!81))b = 0!0109

At this value of friction factor, Reynolds number can be found from the Moodydiagram (Fig. 10.8)—the result is

Re = 1!5× 106

Thus

: =, 1

Re

=(3)(0!01)

1!5× 106

= 2!0× 10"8 m2/s

864

PROBLEM 10.37

Situation: Water ows through a pipe–details are provided in the problem statement.


ANALYSIS

!M = M( = 0!90(2!5" 1) = 1!35 ft of waterM( = b(2$1), 2$2)

b = 1!35× (0!05$4)× 2× 9!81$32

= 0.037

865

PROBLEM 10.38

Situation: Water ows through a cast-iron pipe. 1 = 10 cm , = 4m$ s

Find: (a) Calculate the resistance coe"cient.(b) Plot the velocity distribution.

Properties: From Table A.5 :(10oC) = 1!31× 10"6 m2$s.

ANALYSIS

Re =, 1

:

=4(0!1)

1!31× 10"6= 3! 053× 105

Sand roughness height

c&1

=0!00026

0!1= 0!002 6

Resistance coe"cient (Swamee-Jain correlation; turbulent ow)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³04002 6347

+ 5474(34 053×105)049

í2

= 0!0258

b = 0!0258

Velocity prole (turbulent ow)

@

@&= 5!75 log (

?

c&) + 8!5

Friction velocity(@&)

@& =pB 0$# (1)

Resistance coe"cient

B ! =b

4

µ#, 2

2

¶(2)


866

@& = ,

rb

8

= 4

r0!0258

8= 0!227 2m$ s

Velocity prole

@ = (0!227 2m$ s)h5!75 log

³ ?

0!00026

´+ 8!5

i

The distance from the wall (?) is related to pipe radius (") and distance from thecenterline (3) by

? = "" 3

Velocity Prole

@(3) = (0!227 2m$ s)

·5!75 log

µ0!025" 30!00026

¶+ 8!5

¸

Plot

1.5

2

2.5

3

3.5

4

4.5

y

0 0.005 0.01 0.015 0.02x

867

PROBLEM 10.39

Situation: Flow passes through a pipe–details are provided in the problem statement.


ANALYSIS

Re = , A$:

= (1)(0!10)$(10"4)

= 103 (laminar)

b = 64$Re

= 64$1000

= 0.064

Case (a) is correct

868

PROBLEM 10.40

Situation: Water (20!-) ows through a brass tube. Smooth walls (c& = 0) !Tube diameter is 1 = 3 cm! Flow rate is ] = 0!002m3$ s!

Find: Resistance coe"cient

ANALYSIS

Flow rate equation

, =]

I

=0!002

0$4× 0!032

= 2!83 m/s

Reynolds number

Re = , 1$:

= 2!83× 0!03$10"6

= 8!49× 104

Friction factor (b) (Swamee-Jain correlation—Eq. 10.26)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³0 + 5474

(8449×104)049

í2

b = 0!0185

869

PROBLEM 10.41

Situation: A train travels through a tunnel.Air in the tunnel (assume & = 60!H ) will modeled using pipe ow concepts.Additional details are provided in the problem statement

Find: (a) Change in pressure between the front and rear of the train.(b) Power required to produce the air ow in the tunnel.(c) Sketch an EGL and a HGL.

Properties: From Table A.3 ( = 0!0764 lbf/ft3 and : = 1!58× 10"4 ft2$s

APPROACH

Apply the energy equation from front of train to outlet of tunnel.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + ME

%1$( + ,21 $2) = 0 + 0 + 0 + , 22 $2) + b(2$1),

22 $2)

%1$( = b(2$1), 2$2)

c&$1 = 0!05$10 = 0!005

Re = , 1$: = (50)(10)$(1!58× 10"4) = 3!2× 106

Resistance coe"cient (from Moody diagram, Fig. 10.8)

b = 0!030


%1 = (b(2$1)(, 2$2))

= (0!0764)(0!03)(2' 500$10)(502$(64!4))

%1 = 22!24 psfg

Energy equation (from outside entrance to rear of train)

870

%3$( + T3,23 $2) + R3 = %4$( + T4,

24 $2) + R4 +

XME

0 + 0 + 0 = %4$( + ,24 $2) + 0 + (<G + b(2$1)),

2$2)

%4$( = "(, 2$2))(1!5 + b(2$1))= "(502$2))(1!5 + 0!03(2' 500$10))

%4 = "((349!4) = "26!69 psf!% = %1 " %4

= 22!24" ("26!69)

= 48.93 psf

Power equation

* = H,

= (!%I)(50)

= (48!93× 0$4× 102)(50)= 192' 158 ft-lbf/s

= 349 hp

EGLHGL

EGLHGL

Train

871

PROBLEM 10.42

Situation: A siphon tube is used to drain water from a jug into a graduated cylinder.Atube = 3$16 in. = 0!01562 ft 2tube = 50 in.Additional details are provided in the problem statement.

Find: Time to ll cylinder.

Assumptions: & ' 60oF with : = 1!2× 10"5 ft2$s.Neglect head loss associated with any bend in the Tygon tube.

ANALYSIS

Energy equation (from the surface of the water in the jug to the surface in the grad-uated cylinder)

%V$( + TV,2V $2) + RV = %1$( + T1,

21 $2) + R1 +

XME (1)

Assume that the entrance loss coe"cient is equal to 0.5. It could be larger than 0.5,but this should yield a reasonable approximation. Therefore

XME = (0!5 + b2$1 +<W),

2$2)

The exit loss coe"cient, <W, is equal to 1.0. Therefore, Eq. 1 becomes

!R = RV " R1 = (, 2$2))(1!5 + b2$1)

or , =p2)!R$(1!5 + b2$1) (1)

=p2)!R$(1!5 + b × 267)

Assume b = 0!03 and let !R = (21" 2!5)$12 = 1!54 ft. Then

, =p(2))(1!54)$(1!5 + 10!7)

= 2! 85 ft/s

Re =, 1

:

=2!85× !015621!2× 10"5

= 3710

Resistance coe"cient (recalculate)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

£log10

¡0 + 5474

3710049

¢¤2

= 0!040

872

Repeat calculations with a new value of friction factor.

, =p2) × 1!54$(1!5 + 10!68)

= 2!85 ft/s

Re =, 1

:= 3710

h

dh

Fig. A

Use b = 0!040 for nal solution. As a simplifying assumption assume that as thecylinder lls the level of water in the jug has negligible change. As the cylinder isbeing lled one can visualize (see gure) that in time AP a volume of water equal to]AP will enter the cylinder and that volume in the cylinder can be expressed asI1AM'that is

]AP = I1AM

AP = (I1$])AM

But] = ,5I5 (3)

so

AP = ((I1$I5)$, )AM

Substitute , of Eq. (1) into Eq. (2):

AP = (I1$I5)$(2)!R$(1!5 + 267b))1*2AM

(1 = !500 liter = 0!01766 ft3

or

0!01766 = I1 × (11!5 in.$12)I1 = 0!01842 ft2

Itube = (0$4)((3$16)$12)2 = 0!0001917 ft2

I1$I5 = 96!1

The di!erential equation becomes

873

AP = 96!1$(2)!R$(1!5 + 10!9))1*2AM

Let M be measured from the level where the cylinder is 2 in full. Then

!R = ((21 in" 2!5 in)$12)" M!R = 1!542" M

Now we have

AP = 96!1$(2)(1!54" M)$12!2)1*2AMAP = 42!2$(1!54" M)1*2AMAP = "42!2$(1!54" M)1*2("AM)

Integrate:

P = "42!2(1!54" M)1*2$(1$2)|70= "84!4(1!54" M)1*2|04750

= "84!4[(0!79)1*2 " (1!54)1*2]= "84!4(0!889" 1!241)= 29.7 s

COMMENTS

Possible problems with this solution: The Reynolds number is very close to the pointwhere turbulent ow will occur and this would be an unstable condition. The owmight alternate between turbulent and laminar ow.

874

PROBLEM 10.43

Situation: Water ows from an upper reservoir to a lower reservoir—additional detailsare provided in the problem statement.

Find: (a) Elevation of upper reservoir. (b) Sketch the HGL and EGL. (c) Locationof minimum pressure; value of minimum pressure and (d) What is the type of pipe?

APPROACH

Apply the energy equation between water surfaces of the reservoirs. Then to deter-mine the magnitude of the minimum pressure, write the energy equation from theupstream reservoir to just downstream of bend.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + R1 = 0 + 0 + 100 +X

ME

where XME = (<G + 2<0 +<W + b2$1)(,

2$2))

and <G = 0!50; <0 = 0!40 (assumed); <W = 1!0; b2$1 = 0!025× 430$1 = 10!75

, = ]$I = 10!0$((0$4)× 12) = 12!73 ft/sthen

R1 = 100 + (0!5 + 2× 0!40 + 1!0 + 10!75)(12!732)$64!4= 133 ft

Answer ! The point of minimum pressure will occur just downstream of the rstbend as shown by the hydraulic grade line (below).

H.G.L.E.G.L.

pmin.

Energy equation

R1 = R0 + %0$( + ,2$2) + (b2$1), 2$2) +<G,

2$2) +<0,2$2)

%0$( = 133" 110!70" (12!732$64!4)(1!9 + 0!025× 300$1) = "1!35 ft%> = "1!35× 62!4 = -84 psfg = -0.59 psigRe = , 1$: = 12!73× 1$(1!41× 10"5) = 9!0× 105

875

With an b of 0.025 at a Reynolds number of 9 × 105 a value for c&$1 of 0.0025(approx) is read from Fig. 10-8. Answer ! From Table 10.2 the pipe appears to befairly rough concrete pipe.

876

PROBLEM 10.44

Situation: Water ows out of reservoir, through a steel pipe and a turbine.Additional details are provided in the problem statement.

Find: Power delivered by turbine.

Properties: From Table A.5 :(70oF)= 1!06× 10"5 ft2$s

Assumptions: turbulent ow, so T2 % 1.

APPROACH

Apply the energy equation from the reservoir water surface to the jet at the end ofthe pipe.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + M- +

XME

0 + 0 + R1 = 0 + T2,22 $2) + R2 + M- + (<G + b2$1),

2$2)

R1 " R2 = M- + (1 + 0!5 + b2$1),2$2)

100 ft = M- + (1!5 + b2$1),2$2)

But

, = ]$I = 5$((0$4)12) = 6!37 ft/s

, 2$2) = 0!629 ft

Re = , 1$: = 6!0× 105

From Fig. 10.8 b = 0!0140 for c&$1 = 0!00015! Then

100 ft = M- + (1!5 + 0!0140× 1' 000$1)(0!629)M- = (100" 9!74) ft

Power equation

* = ](M- × e!= 5× 62!4× 90!26× 0!80= 22' 529 ft · lbf/s

= 40.96 horsepower

877

PROBLEM 10.45

Situation: A uid ows in a smooth pipe. 7 = 10"2N · s$m2 # = 800 kg$m3

1 = 100mm , = 500mm$ s

Find: (a) Maximum velocity.(b) Resistance coe"cient.(c) Shear velocity.(d) Shear stress 25 mm from pipe center.(e) Determine if the head loss will double if discharge is doubled.

ANALYSIS

Reynolds number

Re =, 1#

7

=(0!5)(0!1)(800)

10"2

= 4000

Because Re F 2000' assume the ow is turbulent.

a) Table 10.1 relates mean and centerline velocity. From this table,

,max = , $0!791

= 0!50$0!791

= 0.632 m/s

b) Resistance coe"cient (from Moody diagram, Fig. 10.8)

b = 0!041

c) Shear velocity is dened as

@& =

rB !#

(1)

Wall shear stress

B ! =b

4

#, 2

2Combine equations

@& = ,

µb

8

¶045

= (0!5)

µ0!041

8

¶045

r0!041× 0!52

8

= 0!0358m/ s

878

d) In a pipe ow, shear stress is linear with distance from the wall. The distanceof 25 mm from the center of the pipe is half way between the wall and thecenterline. Thus, the shear stress is 1/2 of the wall value:

B 25 mm =B !2

The shear stress at the wall is given by Eq. (1)

B ! = #@2&= 800× 0!03582

= 1!025N$m2

Thus

B 25 mm =B !2

=1!025N$m2

2

= 0.513 N/m2

e) If ow rate (]) is doubled, the velocity will also double. Thus, head loss will begiven by

M( = bnew

µ2

1

¶(2, )2

2)

The increase in velocity will increase Reynolds number, thereby decreasing thefriction factor so that bnew G !boriginal Overall the head loss will increase byslightly less than a factor of 4.0.

No, the increase in head loss will be closer to a factor of 4.0

879

PROBLEM 10.46

Situation: This problem involves an energy grade line for steady ow in a pipe inwhich no pumps or turbines are present.

Find: Which statements are true about this EGL.

ANALYSIS

The valid statements are: a, b, d. For cases c & e:

Re = , 1$:

= (1)(1)$(10"6)

= 106

Since Re F 3000' the ow at 1 m/s is in the turbulent range; therefore, the head losswill be more than doubled with a doubling of the velocity.

880

PROBLEM 10.47

Situation: A gure with an EGL and an HGL is missing physical details in somesections.

Find: (a) What is at points A and C.(b) What is at point B.(c) Complete the physical setup after point D.(d) The other information indirectly revealed by the EGL and HGL.

ANALYSIS

a) Pumps are at A and Cb) A contraction, such as a Venturi meter or orice, must be at B.c)

converging pipe

ReservoirC D

d) Other information:(1) Flow is from left to right(2) The pipe between AC is smaller than before or directly after it.(3) The pipe between BC is probably rougher than AB.

881

PROBLEM 10.48

Situation: Water (20oC) ows in cast iron pipe. 1 = 15 cm ] = 0!05m3$ s c& =0!26mmfrom Table A.5 :(20oC)= 10"6 m2$s

Find: (a) Shear stress at the wall.(b) Shear stress 1 cm from wall.(c) Velocity 1 cm from wall.

Properties: Table A.5 (water at 20 !C): # = 998 kg$m3 ' : = 1!00× 10"6m2$ s!

ANALYSIS

Flow rate equation

, =]

I=

0!05

(0$4)× 0!152

= 2!83 m/s

Reynolds number

Re =, 1

:=2!83× 0!1510"6

= 4!2× 105

Relative roughness

c&1

=0!26mm

150mm= 1! 733× 10"3

Resistance coe"cient (Swamee Jain correlation)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³14 733×10"3

347+ 5474

(442×105)049

í2

= 0!0232

Eq. (10-21)

B 0 = b#, 2$8

B 0 = 0!0232× 998× 2!832$8

= 23.2 N/m2

882

In a pipe ow, the shear stress variation is linear; thus,

B 1 = (6!5$7!5)× B 0= 20.0 N/m2

Velocity distribution (turbulent ow)

@& =

rB 0#=

r23!2

998

= 0!1524 m/s

@

@&= 5!75 log

µ?

c&

¶+ 8!5

@ = @&

µ5!75 log

µ?

c&

¶+ 8!5

¶

= 0!1524

µ5!75 log

µ0!01

0!00026

¶+ 8!5

¶

= 2!684m$ s

@ = 2!68m$ s

883

PROBLEM 10.49

Situation: Water ows from one reservoir to another–additional details are given inthe problem statement.

Find: Design a conduit system.

ANALYSIS

One possibility is shown below:

E1 = 100 m

E1 = 85 m-

H.G.L.

E.G.L.

E1 = 70 m

E1 = 55 m-

20 m

Assume that the pipe diameter is 0!50 m. Also assume <0 = 0!20' and b = 0!015!Then

100" 70 = (0!5 + 2× 0!20 + 1 + 0!015× 130$0!5), 2$2), 2$2) = 5!17

The minimum pressure will occur just downstream of the rst bend and its magnitudewill be as follows:

%min$( = 100" 85" (0!5 + 0!20 + 1 + ((0!015× 80$0!5) + 1), 2$2)= "6!20 m

%min = "6!20× 9' 810= -60.8 kPa gage

884

PROBLEM 10.50

Situation: Water is pumped through a vertical steel pipe to an elevated tank on theroof of a building–additional details are provided in the problem statement.

Find: Pressure at point 80 m above pump.

ANALYSIS

Re = 4]$(01:)

= 4× 0!02$(0 × 0!10× 10"6) = 2!55× 105

c&$1 = 4!6× 10"2$100 = 4!6× 10"4

Resistance coe"cientb = 0!0185

Then

M( = (b(2$1),2$2)

where

, = 0!02$((0$4)× 0!12) = 2!546 m/sM( = 0!0185× (80$0!10)× 2!5462$(2× 9!81) = 4!89 m

Energy equation (from pump to location 80 m higher)

%1$( + T1,21 $2) + R1 = %2$( + T2,$2) + R2 + M(

1!6× 106$9' 790 + , 21 $2) = %2$( + ,22 $2) + 80 + 4!89

,1 = ,2

%2 = 769 kPa

885

PROBLEM 10.51

Situation: Water drains from a tank through a galvanized iron pipe. 1 = 1 in!Total elevation change is 14 ft. Pipe length = 10 ft.

Find: Velocity in pipe.

Properties: Kinematic viscosity of water is 1!22× 10"5 ft2$ s! From Table 10.3 <G =0!5! From Table 10.3, c& = 0!006 inches.

Assumptions: Assume turbulent ow (check after calculations are done). AssumeT1 % 1!00!

APPROACH

Apply the energy equation from the water surface in the tank to the outlet of thepipe. Use the Darcy-Weisbach equation for head loss. Assume turbulent ow andthen solve the resulting equations using an iterative approach.

ANALYSIS

Energy equation

%1(+ T1

, 212)+ R1 =

%2(+ T2

, 222)+ R2 +

XME

0 + 0 + 14 = 0 +, 222)+ 0 + (<G + b

2

1), 222)

14 ft =

µ1 +<G + b

2

1

¶, 222)

14 ft =

µ1 + 0!5 + b

(120 in)

(1 in)

¶, 222)

(1)

Eq. (1) becomes

, 2 =2× (32!2 ft$ s2)× (14 ft)

1!5 + 120× bGuess b = 0!02 and solve for ,

, 2 =2× (32!2 ft$ s2)× (14 ft)

1!5 + 120× 0!02, = 15!2 ft$ s

Reynolds number (based on the guessed value of friction factor)

Re =, 1

:

=(15!2 ft$ s) (1$12 ft)

1!22× 10"5 ft2$ s= 103' 856

886

Resistance coe"cient (new value)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

£log10

¡04006347

+ 5474103856049

¢¤2

= 0!0331

Recalculate , based on b = 0!0331

, 2 =2× (32!2 ft$ s2)× (14 ft)1!5 + 120× 0!0331

, = 12!82 ft$ s

Reynolds number (recalculate based on , = 12!82 ft$ s)

Re =(12!8 ft$ s) (1$12 ft)

1!22× 10"5 ft2$ s= 874' 316

Recalculate b based on Re = 874' 316

b =0!25

£log10

¡04006347

+ 5474874316049

¢¤2

= 0!0333

Recalculate , based on b = 0!0333

, 2 =2× (32!2 ft$ s2)× (14 ft)1!5 + 120× 0!0333

, = 12!80 ft$ s

Since velocity is nearly unchanged, stop!

, = 12!80 ft$ s

1. The Reynolds number 874,000 is much greater than 3000, so the assumption ofturbulent ow is justied.

2. The solution approach, iteration with hand calculations, is straightforward.However, this problem can be solved faster by using a computer program thatsolves simultaneous, nonlinear equations.

887

PROBLEM 10.52

Situation: Water drains from a tank, passes through a pipe and then jets upward.Additional details are provided in the problem statement.

Find: (a) Exit velocity of water.(b) Height of water jet.

Properties: From Table 10.2 c& = 0!15 mm = 0!015 cm.From Table 10.3 <0 = 0!9 and <G = 0!5!

Assumptions: The pipe is galvanized iron.The water temperature is 20oC so : = 10"6 m2$s.Relative roughness c&$1 = !015$1!5 = 0!01. Start iteration at b = 0!035!

APPROACH

Apply the energy equation from the water surface in the tank to the pipe outlet.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + 5 = 0 + T2,22 $2) + 0 + (<G + 2<0 + b2$1),

22 $2)

5 = (, 22 $2))(1 + 0!5 + 2× 0!9 + !035× 10$0!015)5 = (, 22 $ (2× 9!81))(26!6),2 = 1!920 m/s

Reynolds number

Re = , 1$:

= 1!92× 0!015$10"6

= 2!88× 104!

Resistance coe"cient (new value)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

£log10

¡0401347+ 5474

28800049

¢¤2

= 0!040

Recalculate ,2 with this new value of b

,2 = 1!81 m/s

888

Energy equation (from the pipe outlet to the top of the water jet)

M = , 2$2)

= (1!81)2$(2× 9!81)= 0.1670 m

= 16.7 cm

889

PROBLEM 10.53

Situation: Water (60!F) is pumped from a reservoir to a large pressurized tank.Additional details are given in the problem statement.

Find: Power to operate the pump.

Properties: From Table A.5 : = 1!22× 10"5 ft2$sFrom Table 10.2 c& = 0!002 in = 1! 67× 10"5 ftFrom Table 10.3 <G = 0!03

Assumptions: Assume the entrance is smooth.

ANALYSIS

Flow rate equation

, = ]$I = 1!0$((0$4)12)

= 1!0$((0$4)(1$3)2)

= 11!46 ft/s

Then

Re = 11!46× (1$3)$(1!22× 10"5) = 3!13× 105

c&$1 = 4!5× 10"4


b = 0!0165

Then

b2$1 = 0!0165× 300$(1$3) = 14!86

Energy equation (from water surface I to water surface D)

%=$( + T=,2=$2) + R= + M, = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + 0 + M, = (10× 144$62!4) + 0 + (<G +<W + b2$1),2$2)

Thus

M, = 23!08 + (0!03 + 1 + 14!86)(11!462$64!4)

= 55!48 ft

Power equation

890

* =](M,g

=1!0× 62!4× 55!48

0!9= 3847 ft · lbf/s= 6.99 horsepower

891

PROBLEM 10.54

Situation: A pump operates between a reservoir and a tank.Additional details are provided in the problem statement

Find: Time to ll tank.

Properties: From Table 10.3 <G = 0!5 and <W = 1!0!

APPROACH

Apply the energy equation from the reservoir water surface to the tank water surface.The head losses will be due to entrance, pipe resistance, and exit.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + R1 + M, = 0 + 0 + R2 + (<G + b2$1 +<W),2$2)

M, = (R2 " R1) + (0!5 + (0!018× 30$0!9) + 1!0), 2$2)M, = M+ (2!1), 2$2)

But the head supplied by the pump is M!(1" (]2$]2max)) so

M!(1"]2$]2max)) = M+ 1!05, 2$)

50(1"]2$4) = M+ 1!05]2$()I2)

50" 12!5]2 = M+ 1!05]2$()I2)

AreaI = (0$4)12 = (0$4)(0!92) = 0!63m2

So

50" 12!5]2 = M+ 0!270]2

50" M = 127!77]2&50" M = 3!57]

The discharge into the tank and the rate of water level increase is related by

] = ItankAM

APso

&50" M = 3!57Itank

AM

APor

892

AP = 3!57Itank(50" M)"1*2AM

Integrating

P = 2× 3!57Itank(50" M)1*2 + -

when P = 0, M = 0 and Itank = 100 m2 so

P = 714(7!071" (50" M)1*2)

When M = 40 m

P = 2791 s

= 46.5 min

893

PROBLEM 10.55

Situation: Kerosene is pumped through a smooth pipe. 1 = 3 cm , = 4m$ s!Additional details are provided in the problem statement

Find: Ratio of head loss for laminar ow to head loss for turbulent ow.

(ME)Laminar ow(ME)Turbulent ow

ANALYSIS

Reynolds number

Re =, 1

:

=4× 0!032× 10"6

= 6× 104

If the ow is laminar at this Reynolds number

blam =64

Re

=64

6× 104= 1! 07× 10"3

Resistance coe"cient (from Moody diagram, Fig.10-8)

bturb = 0!020

Then

(ME)Laminar ow(ME)Turbulent ow

=M(lamM(turb

=blambturb

=0!00107

0!02

= 0.0535

894

PROBLEM 10.56

Situation: Water ows in a uncoated cast iron pipe. 1 = 4 in ] = 0!02 ft3$ s!

Find: Resistance coe"cient b!

Properties: From Table A.5 : = 1!22× 10"5 ft2$sFrom Table 10.2 c& = 0!01 in

ANALYSIS

Reynolds number

Re =4]

01:

=4× 0!02

0 × (4$12)× (1!22× 10"5)= 6!3× 103


c&1

=0!01

4= 0!0025


b = 0!038

895

PROBLEM 10.57

Situation: Fluid ows in a concrete pipe. 1 = 6 in 2 = 900 ft ] = 3 cfs !7 =#: = 0!005 lbf-s/ft2

Additional details are provided in the problem statement

Find: Head loss.

ANALYSIS

Reynolds number

Re = 4]$(01:)

= 4(3!0)$(0(1$2)3!33× 10"3)= 2294 (laminar)

Flow rate equation

, = ]$(012$4)

= 3!0$(0$4× 0!52)= 15!28 ft/s


M( = 3272,$((12)

= 32(5× 10"3)900(15!28)$(1!5× 32!2× (1$2)2)

= 182.2 ft

896

PROBLEM 10.58

Situation: Crude oil ows through a steel pipe. 1 = 15 cm ] = 0!03m3$ s!Points A and B are 1 km apart. %> = 300 kPaAdditional details are provided in the problem statement.


Properties: From Table 10.2 c& = 4!6× 10"5 m.

ANALYSIS

Reynolds number

Re = , 1$:

= 4]$(01:)

= 4× 0!03$(0 × 0!15× (10"2$820))2!09× 104 (turbulent)


c&$1 = 4!6× 10"5$0!15= 3!1× 10"4

Flow rate equation

, = ]$I

= 0!03$(0 × 0!152$4)= 1!698 m/s


b = 0!027


M( = b2

1

, 2

2)

= 0!027

µ1000

0!15

¶µ1!6982

2× 9!81

¶

= 26!4 m

Energy equation

%=$( + T=,2=$2) + R= = %>$( + T>,

2>$2) + R> + M(

%= = 0!82× 9810[(300000$(0!82× 9810)) + 20 + 26!41]= 673 kPa

897

PROBLEM 10.59

Situation: Water exits a tank through a short galvanize iron pipe. 1tank = 2m 1pipe =26mm 2pipe = 2!6mFully open angle valve: <v = 5!0

Find: (a) Time required for the water level in tank to drop from 10 m to 2 m.

Assumptions: The pipe entrance is smooth: <e % 0The kinetic energy correction factor in the pipe is T2 = 1!0

APPROACH

Apply the energy equation from the top of the tank (location 1) to the exit of theangle valve (location 2).

ANALYSIS

Energy equation

M = T2, 2

2)+, 2

2)(<e +<v + b

2

1)


T2 = 1!0

<e % 0' <v = 5!0

2$1 = 2!6$0!026 = 100!0

Combine equation and express , in terms of M

, =

s2)M

6 + 100× b


c&1=0!15

26= 5!8× 10"3

Reynolds number

Re =, × 0!02610"6

= 2!6× 104,

Rate of decrease of height

AM

AP= "

]

I= "

0!000531

3!14, = "0!000169,

A program was written to rst nd , iteratively for a given M using Eq. 10.26 forthe friction factor. Then a new M was found by

M: = M:"1 " 0!000169,!P

898

where !P is the time step. The result was 1424 sec or 23.7 minutes.

COMMENTS

1. When valves are tested to evaluate<valve the pressure taps are usually connectedto pipes both upstream and downstream of the valve. Therefore, the head lossin this problem may not actually be 5, 2$2)!

2. The velocity exiting the valve will probably be highly non-uniform; therefore,this solution should be considered as an approximation only.

899

PROBLEM 10.60

Situation: Water ows from point A to B in a cast iron pipe.Additional information is provided in the problem statement.

Find: Direction and rate of ow.

Properties: From Table A.5 : = 1!41× 10"5 ft2$s.From Table 10.2 c& = 0!01 in = 0!000833 ft.

Assumptions: Flow is from A to B.

ANALYSIS

M( = !(%$( + R)

= ("20× 144$62!4) + 30= "16!2 ft

Therefore, ow is from B to A.

Parameters for the Moody diagram

Re b1*2 = (13*2$:)(2)M($2)1*2

= (23*2$(1!41× 10"5)× 64!4× 16!2$(3× 5' 280))1*2

= 5!14× 104

c&$1 = 4! 2× 10"4

Resistance coe"cient (from the Moody diagram,Fig. 10.8)

b = 0!0175


, =qM(2)1$b2

=p(16!2× 64!4× 2)$(0!0175× 3× 5' 280)

= 2!74 ft/s

Flow rate equation

^ = , I

= 2!74× (0$4)× 22

= 8.60 cfs

900

PROBLEM 10.61

Situation: Water ows between two reservoirs. ] = 0!1m3$ s!The pipe is steel. 1 = 15 cm!Additional details are provided in the problem statement

Find: Power that is supplied to the system by the pump.

Properties: From Table 10.2 c& = 0!046 mm.

ANALYSIS

Flow rate equation

, = ]$I

= 0!10$((0$4)× 0!152)= 5!66 m/s

, 2$2) = 1!63 m

c&$1 = 0!0046$15 = 0!0003

Reynolds number

Re = , 1$: = 5!66× 0!15$(1!3× 10"6)= 6!4× 105

Resistance coe"cient (from the Moody diagram, Fig. 10.8)

b = 0!016

Energy equation (between the reservoir surfaces)

%1$( + T1,21 $2) + R1 + M, = %2$( + T2,

22 $2) + R2 +

XME

M, = R2 " R1 +, 2

2)(<G + b(2$1) +<E)

= 13" 10 + 1!63(0!1 + 0!016× 80$(0!15) + 1)= 3 + 15!7 = 18!7 m

Power equation

* = ](M,

= 0!10× 9810× 18!7= 18' 345W

= 18.3 kW

901

PROBLEM 10.62

Situation: Water ows between two reservoirs in a concrete pipe.Other details are provided in the problem statement.

Find: (a) Discharge (concrete pipe).(b) Discharge (riveted steel).(c) Pump power for uphill ow (concrete pipe).


Assumptions: Based on data in Table 10.2, for concrete pipe c& = 0!3 mm, and forriveted steel c& = 0!9 mm

APPROACH

Apply the energy equation from upstream reservoir water surface to downstreamwater surface.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + ME

R1 = R2 + M(

100 m = (b2$1), 2$2)

c&$1 = 0!3$103 = 0!0003


b = 0!016

Then100m = (0!016× 10' 000$1), 2$2)

, = (100(2))$(160))1*2 = 3!50 m/s

Reynolds number

Re = , 1$: = (3!50)(10)$(1!31× 10"6)= 2!67× 106

Check b from Fig. 10.8 (b = 0!0155) and solve again:

, = 3!55 m/s

]concrete = , I

= (3!55)(0$4)12

]concrete = 2!79 m3/s

902

For riveted steel: c&$1 = 0!9$1000 ' 001 and from Fig. 10.8 b = 0!0198!

]F4.$]1 =p0!0155$0!0198 = 0!885

]Riveted.Steel = 2!47 m3/s

Head of the pump

M, = (R1 " R2) + ME= 100 m+ 100(2!8$2!79)2

= 201 m

Power equation

* = ](M,

= (2!8)(9' 810)(201)

= 5.52 MW

903

PROBLEM 10.63

Situation: A uid ows through a pipe made of galvanized iron. 1 = 8cm : =10"6m2$ s # = 800 kg$m3!Additional details are provided in the problem statement

Find: Flow rate.


ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 + M(

150' 000$(800× 9!81) + , 21 $2) + 0 = 120' 000$(800× 9!81) + , 22 $2) + 3 + M(M( = 0!823

((13*2)$(:))× (2)M($2)1*2 = ((0!08)3*2$10"6)× (2× 9!81× 0!823$30!14)1*2

= 1!66× 104

Relative roughness

c&$1 = 1!5× 10"4$0!08 = 1!9× 10"3

Resistance coe"cient. From Fig. 10-8 b = 0!025! Then

M( = b(2$1)(,2$2))

Solving for ,

, =q(M($b)(1$2)2)

=p(0!823$0!025)(0!08$30!14)× 2× 9!81 = 1!312 m/s

] = , I

= 1!312× (0$4)× (0!08)2

= 6!59× 10"3 m3/s

904

PROBLEM 10.64

Situation: Oil is pumped from a lower reservoir to an upper reservoir through a steelpipe. 1 = 30 cm ] = 0!20m3$ s!From Table 10.2 c& = 0!046 mmAdditional details are provided in the problem statement

Find: (a) Pump power.(b) Sketch an EGL and HGL.

APPROACH

Apply the energy equation between reservoir surfaces .

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 + M, = %2$( + T2,

22 $2) + R2 +

XME

100 + M, = 112 + , 2$2)(<G + b2$1 +<W)

M, = 12 + (, 2$2)) (0!03 + b2$1 + 1)

Flow rate equation

, = ]$I

= 0!20$((0$4)× 0!302)= 2!83 m/s

, 2$2) = 0!408m

Reynolds number

Re = , 1$:

= 2!83× 0!30$(10"5)= 8!5× 104

c&$1 = 4!6× 10"5$0!3= 1!5× 10"4


b = 0!019

Then

M, = 12 + 0!408(0!03 + (0!019× 150$0!3) + 1!0)= 16!3 m

905

Power equation

* = ](M,

= 0!20× (940× 9!81)× 16!3 = 2!67× 104W= 30.1 kW

EGLHGL

906

PROBLEM 10.65

Situation: In a pipe, the resistance coe"cient is b =0.06, 1 = 40 cm' , =3m$ s' : = 10"5m2$ s .

Find: Change in head loss per unit meter if the velocity were doubled.

ANALYSIS

Reynolds number

Re = , 1$:

= 3× 0!40$10"5

= 1!2× 105

Since Re F 3000, the ow is turbulent and obviously the conduit is very rough(b = 0!06); therefore, one would expect b to be virtually constant with increasedvelocity. Since M( = b(2$1) (, 2$2)), we expect, M( + , 2, so if the velocity isdoubled, the head loss will be quadrupled.

907

PROBLEM 10.66

Situation: A cast iron pipe joins two reservoirs. 1 = 1!0 ft 2 = 200 ft!Additional information is provided in the problem statement.

Find: (a) Calculate the discharge in the pipe.(b) Sketch the EGL and HGL.

Properties: From Table 10.2 c& = 0!01 in

Assumptions: Water temperature is 60oF: : = 1!22× 10"5 ft2$s 7 = 2!36× 10"5N ·s$m2 # = 1!94 slug$m3

APPROACH

Apply the energy equation from the water surface in the upper reservoir to the watersurface in the lower reservoir.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + 100 = 0 + 0 + 40 + (<G + 2<v ++<W + b2$1),22$)

100 = 40 + (0!5 + 2× 0!2 + 1!0 + b × 200$1), 2$2)

The equation for , becomes

, 2

2)=

60

1!9 + 200b(1)

Relative roughness

c&1

=0!01

12= 8! 3× 10"4

Reynolds number

Re =, 1

:

=, × 1!0

1!22× 10"5=

¡8!20× 104 × ,

¢(2)

Friction factor (Swamee-Jain correlation—Eq. 10.26)

b =0!25

hlog10

³84 3×10"4

347+ 5474

(8420×104×? )049

í2 (3)

908

Solve Eqs. (1) to (3) simultaneously (we applied a computer program, TK Solver)

, = 26!0m$ s

Re = 2' 130' 000

b = 0!019

Flow rate equation

] = , I

= 26!0(0$4× 12)

= 20.4 cfs

909

PROBLEM 10.67

Situation: A small stream lls a reservoir—water from this reservoir is used to createelectrical power.Discharge is ] = 2 cfs. Elevation di!erence is 4 = 34 ft.Maximum acceptable head loss in the penstock is M( =3 ft.Penstock length is 2 = 87 ft.Penstock is commercial-grade, plastic pipe.

Find: Find the minimum diameter for the penstock pipe.

Properties: Water @ 40 !F from Table A.5: : = 1!66× 10"5 ft2$ s!

Assumptions: 1.) Neglect minor losses associated with ow through the penstock.2.) Assume that pipes are available in even sizes—that is, 2 in., 4 in., 6 in., etc.3.) Assume a smooth pipe— c& = 0!4.) Assume turbulent ow (check this after the calculation is done).

APPROACH

Apply the Darcy-Weisbach equation to relate head loss (M() to pipe diameter. Applythe Swamee-Jain correlation to relate friction factor (b) to ow velocity. Also, writeequations for the Reynolds number and the ow rate. Solve these four equationssimultaneously to give values of 1' ,' b' and Re.

ANALYSIS


M( = b2

1

, 2

2)(1)

Resistance coe"cient (Swamee-Jain correlation; turbulent ow)

b =0!25

£log10

¡5474Re049

¢¤2 (2)

910

Reynolds number

Re =, 1

:(3)

Flow rate equation

] = ,012

4(4)

Solve Eqs. (1) to (4) simultaneously. The computer program TKSolver was used forour solution.

b = 0!01448

, = 9!026 ft$ s

1 = 6!374 in

Re = 289' 000

Recommendation

Select a pipe with 1 = 8 in!

COMMENTS

With an 8-inch-diameter pipe, the head loss associated with ow in the pipe will beless than 10% of the total available head (34 ft). If an engineer selects a pipe that islarger that 8 inches, then cost goes up.

911

PROBLEM 10.68

Situation: Commercial steel pipe will convey water.Design head loss: ME = 1 ft per 1000 ft of pipe length.

Find: Pipe diameter to produce specied head loss.

Properties: From Table A.5 : = 1!22× 10"5 ft2/s.From Table 10.2 c& = 0!002 in = 1! 7× 10"4 ft!Assumptions: The pipes are available in even inch sizes (e.g. 10 in., 12 in., 14 in.,etc.)

ANALYSIS


M( = b2

1

, 2

2)

= b2

1

]2

2)I2

= b82]2

)0215

Solve for diameter

1 =

µb82]2

)02M(

¶1*5

Assume b = 0!015

1 =

Ã0!015

8 (1000) (300)2

32!2× 02 × 1

!1*5

= 8!06 ft

Now get a better estimate of b :

Re = 4]$(01:) = 3!9× 106

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³04002*12347×8406 +

5474(349×106)049

í2

= 0!0104

Compute 1 again:

1 =

Ã0!0104

8 (1000) (300)2

32!2× 02 × 1

!1*5

= 7!49 ft

Thus, specify a pipe with 1 = 90 in

912

PROBLEM 10.69

Situation: A steel pipe will carry crude oil. 6 = 0!93 : = 10"5m2$ s ] =0!1m3$ s!Available pipe diameters are 1 = 20' 22' and 24 cm!Specied head loss: ME = 50m per km of pipe length.

Find: (a) Diameter of pipe for a head loss of 50 m.(b) Pump power.


ANALYSIS


M( = b2

1

, 2

2)

= b2

1

]2

2)I2

= b82]2

)0215

Solve for diameter

1 =

µb82]2

)02M(

¶1*5

Assume b = 0!015

1 =

Ã0!015

8 (1000) (0!1)2

9!81× 02 × 50

!1*5

= 0!19m

Calculate a more accurate value of b

Re = 4]$(01:)

= 4× 0!1$(0 × 0!19× 10"5)= 6!7× 104

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

£log10

¡04046347×190 +

547467000049

¢¤2

= 0!021

Recalculate diameter using new value of b

913

1 = (0!021$0!015)1*5 × 0!19= 0!203 m = 20!3 cm

Use the next larger size of pipe; 1 = 22 cm.

Power equation (assume the head loss is remains at ME % 50 m/1,000 m)

* = ](M(

= 0!1× (0!93× 9810)× 50

= 45.6 kW/km

914

PROBLEM 10.70

Situation: Design a pipe to carry water (] = 15 cfs) between two reservoirs.Distance between reservoirs = 3 mi.Elevation di!erence between reservoirs = 30 ft.

Find: Pipe diameter.

Assumptions: & = 60oF, : = 1!22× 10"5 ft2$s.Commercial steel pipe c& = 0!002 in = 0!00017 ft.

ANALYSIS

Energy equation30 = (<G +<W + b2$1)(]

2$I2)$2)

Assume b = 0!015! Then

30 = (1!5 + 0!015× 3× 5' 280$1)(]2$((0$4)214)$2)

30 = (1!5 + 237!6$1)(152$(0!61714)$64!4

30 = (1!5 + 237!6$1)(5!66$14)

Neglect the entrance and exit losses and solve

1 = 2!15 ft

Re = 4]$(01:)

= 7!3× 105

c&$1 = 0!002$(2!15× 12)= 0!000078


b = 0!0135

Solve again

30 = (1!5 + 214$1)(5!66$14)

1 = 2!10 ft = 25!2 in.

Use 26 in. steel pipe. (one possibility)

915

PROBLEM 10.71

Situation: Problem 7.78 shows a device that can be used to demonstrate cavitation.Let 1 equal diameter of pipe

Find: Design a device that will visually demonstrate cavitation.

Assumptions: water main has a pressure of 50 psig.

ANALYSIS

First you might consider how to physically hold the disk in the pipe. One way todo this might be to secure the disk to a rod and then secure the rod to streamlinedvanes in the pipe such as shown below. The vanes would be attached to the pipe.

Vane

Rod

Side viewEnd view

Disk

To establish cavitation around the disk, the pressure in the water at this sectionwill have to be equal to the vapor pressure of the water. The designer will have todecide upon the pipe layout in which the disk is located. It might be somethinglike shown below. By writing the energy equation from the disk section to the pipeoutlet one can determine the velocity required at the disk to create vapor pressureat that sectional. This calculation will also establish the disk size relative to thepipe diameter. Once these calculations are made, one can calculate the requireddischarge, etc. Once that calculation is made, one can see if there is enough pressurein the water main to yield that discharge with the control valve wide open. If not,re-design the system. If it is OK, then di!erent settings of the control valve will yielddi!erent degrees of cavitation.

Outlet

Water mainValve

Reservoir

Elevation View

916

PROBLEM 10.72


Find: Discharge.

Properties: From Table 10.2 c& = 4× 10"4 ft.From Table A.5 : = 1!41× 10"5 ft2$s.From Table 10.3 <G = 0!5!

APPROACH

Apply the energy equation from water surface in reservoir to the outlet.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + ME

0 + 0 + 120 = 0 + , 2$2) + 70 + (<G ++<W + b(2$1)),2$2)

(, 2$2))(1!5 + b(2$1)) = 50 ft, 2

2)=

50

1!5 + 200b(1)

Sand roughness heightc&$1 = 4× 10"4$0!5 = 0!0008

Reynolds number

Re = 3!54× 104 × , (2)

Solve Eq. 10.26 (for b)'Eq. (1) and (2) simultaneously (we used a hand calculator).The result is

, = 24!6 ft/s

Flow rate equation

] = , I

= 24!6(0$4)(0!52)

= 4.83 cfs

EGL

HGL

917

PROBLEM 10.73


Find: Minimum pressure in pipe.

Properties: From Table A.5 : = 1!41× 10"5 ft2$s.

Assumptions: <G = 0!10

APPROACH

Apply the energy equation from water surface in reservoir to the outlet.

ANALYSIS

Flow rate equation

, = ]$I

= 50 ft/s

Reynolds number

Re = , 1$:

= (50)(2)$(1!41× 10"5)= 7!1× 106

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 600 = 0 + , 22 $2) + 200 + (<G + b(2$1)),2$2)

400 = (, 2$2))(1!10 + b(1' 200$2))

400 = (502$64!4)(1!10 + 600b)

b = 0!0153

From Fig. 10.8 c&$1 = 0!00035 so

c& = 0!00070 ft

The minimum pressure in the pipe is at the pipe outlet.

918

PROBLEM 10.74

Situation: A heat exchanger is described in the problem statement.

Find: Power required to operate heat exchanger with:(a) clean tubes.(b) scaled tubes.


ANALYSIS

/$tube = 0!50 kg/s

]$tube = 0!50$860 = 5!8139× 10"4 m3/s, = ]$I = 5!8139×10"4$((0$4)× (2× 10"2)2) = 1!851 m/sRe = , 1#$7 = 1!851× 0!02× 860$(1!35× 10"4) = 2!35× 105

c&$1 = 0!15$20 % 0!007

From Fig. 10.8 b = 0!034. Then

M( = b(2$1),2$2) = 0!034(5$0!02)× (1!8512$2× 9!81) = 1!48 m

a) * = ](M( = 5!8139× 10"4 × 860× 9!81× 1!48× 100

= 726 W

b) c&$1 = 0!5$16

= 0!031

so from Fig. 10.8 b = 0!058

* = 728× (0!058$0!034)× (20$16)4 = 3.03 kW

919

PROBLEM 10.75


Find: Pump power required.

Assumptions: Smooth bends of 180!'<0 % 0!7

ANALYSIS

Examination of the data given indicates that the tubing in the exchanger has an3$A % 1.Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 + ME

But ,1 = ,2 and %1 = %2 so

M, = ME + (R1 " R2)

The average temperature = 50!- so : = 0!58× 10"6 m2/s

, = ]$I = 3× 10"4$(0$4(0!02)2) = 0!955 m/sRe = , 1$: = 0!955(0!02)$(0!58× (10"6) = 3!3× 104

b = 0!023

ME = (b2$1 + 19<0),2$2) = (0!023(20)$0!02 + 19× 0!7)0!9552$(2× 9!81)

= 1!69 m

M, = R2 " R1 + ME = 0!8 + 1!69 = 2!49 m* = (M,] = 9' 685(2!49)3× 10"4

= 7.23 W

920

PROBLEM 10.76


Find: Power required to operate pump.

Properties: From Table A.5 : = 6!58× 10"7 m2$s.From Table 10.2 c& = 0!0015 mm.

ANALYSIS

Reynolds number

Re =0!02× 106!58× 10"7

= 3!04× 105

Flow rate equation

] =0

4× 0!022 × 10 = 0!00314 m3/s

Relative roughness (copper tubing)

c&1=1!5× 10"3 mm

20 mm= 7!5× 10"5

Resistance coe"cient (from Moody diagram)

b = 0!0155

Energy equation

M, =, 2

2)(b2

1+X

<E)

=102

2× 9!81(0!0155×

10 m0!02 m

+ 14× 2!2) = 196 m

Power equation

* =(]M,g

=9732× 0!00314× 196

0!8= 7487W

* = 7!49 kW

921

PROBLEM 10.77


Find: System operating points.

Properties: From Table 10.2 c& = 1!5× 10"3 mm.

ANALYSIS

Energy equation

M, =, 2

2)(X

<E + b2

1)

Substitute in the values for loss coe"cients, 2$1 and the equation for M,

M,0

"1"

µ]

]max

¶3#=, 2

2)(14× 2!2 + b × 1000)

Flow rate equation

] = , I

= 1!767× 10"4,

Combine equations

M,0

"1"

µ]

]max

¶3#= 1!632× 106]2(30!8 + b × 1000) (1)

Relative roughness

c&1=1!5× 10"3

15= 10"4

Reynolds number

Re =, 1

:

=, × 0!0156!58× 10"7

= 2!28× 104, = 1!29× 108]

Eq. (1) becomes

H (]) = M,0

"1"

µ]

]max

¶3#" 1!632× 106]2(30!8 + b × 1000)

A program was written to evaluate H (]) by inputting a value for ] and tryingdi!erent ]’s until H (]) = 0! The results are

922

M,0 (m) ] (m3/s)2 0.00035610 0.00062920 0.000755

923

PROBLEM 10.78

Situation: A system with a reservoir and free jet is described in the problem state-ment.

Find: The discharge.(b) Points of maximum pressure.(c) Point of minimum pressure.

Assumptions: & = 60oF and : = 1!22× 10"5 ft2/s.3$A = 2 and <0 = 0!2!b = 0!028

ANALYSIS

c&$1 = 0!004

Energy equation

%1$( + R1 + ,21 $2) = %2$( + R2 + ,

22 $2) +

XME

100 = 64 + (, 2$2))(1 + 0!5 +<0 + b × 2$1)= 64 + (, 2$2))(1 + 0!5 + 0!2 + 0!028× 100$1)

36 = (, 2$2))(4!5)

, 2 = 72)$4!5 = 515 ft2$s2

, = 22!7 ft/s

Reynolds number

Re = 22!7(1)$(1!22× 10"5) = 1!9× 106

b = 0!028

Flow rate equation

] = 22!7(0$4)12

= 17.8 cfs

, 2$2) = 36$4!5 = 8!0 ft

EGL

HGL

Elev. = 64 ft

c) maximum pressure

minimum pressure

924

%min$( = 100" 95" (, 2$2))(1 + 0!5) = 5" 8(1!5) = "7 ft%min = "7(62!4) = "437 psfg = -3.03 psig

%max$( + ,2"$2) + R" = %2$( + R2 + ,

22 $2) +

XME

%max$( = 64" 44 + 8!0(0!2 + 0!028(28$1)) = 27!9 ft%max = 27!9(62!4) = 1' 739 psfg = 12.1 psig

925

PROBLEM 10.79

Situation: Gasoline being pumped from a gas tank is described in the problem state-ment.

Find: Pump power.

Properties: From Fig. A.2 6 = 0!68' : = 5!5× 10"6 ft2/sec.

ANALYSIS

] = 0!12 gpm = 2!68× 10"4 cfsA1 = (1$4)(1$12) = 0!0208 ft

A2 = (1$32)(1$12) = 0!0026 ft

A2$A1 = (1$32)$(1$4) = 0!125

( = 62!4(0!68) = 42!4 lbf/ft3

,1 = ]$I = 2!68× 10"4$(0$4(1$48)2) = 0!786 ft/s, 21 $2) = 0!00959 ft

,2 = (32$4)2 × 0!786 = 50!3 ft/s, 22 $2) = 39!3 ft

Re1 = ,111$:

= 0!786(0!0208)$(5!5× 10"6)= 2' 972

From Fig. 10.8 b % 0!040

%1 = 14!7 psia

R2 " R1 = 2 ft

%2 = 14!0 psia

ME = (b2$1 + 5<0),21 $2)

= (0!040× 10$0!0208 + 5× 0!21)0!00959 = 0!194 ftM, = (%2 " %1)$( + R2 " R1 + , 22 $2) + ME

= (14!0" 14!7)144$42!4 + 2 + 39!3 + 0!194 = 39!1 ft

Power equation

* = (M,]$(550=) = 42!4(39!1)0!000268$(550× 0!8)

= 10.1×10"4 hp

926

PROBLEM 10.80

Situation: A partially-closed valve is described in the problem statement.from Table 10.2 c& = 0!046 mm

Find: Loss coe"cient for valve.

APPROACH

First nd ] for valve wide open. Assume valve is a gate valve.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

2 = 0 + 0 + 0 + (, 2$2))(0!5 + 0!9 + 0!2 + 0!9 + 1 + b2$1)

, 2 = 4)$(3!5 + b2$1)


, = [4× 9!81$(3!5 + 0!015× 14$0!1)]1*2 = 2!65 m/sc&$1 ' 0!0005

Re = 2!65× 0!10$(1!3× 10"6) = 2!0× 105

From Fig. 10.8 b = 0!019! Then

, = [4× 9!81$(3!5 + 0!019× 14$0!10)]1*2 = 2!52 m/sRe = 2!0× 105 × 2!52$2!65 = 1!9× 105; O.K.

This is close to 2.0×105 so no further iterations are necessary. For one-half thedischarge

, = 1!26 m/s

Re = 9!5× 104

and from Fig. 10.8 b = 0!021! So

, 2 = 1!588 = 4× 9!81$(3!3 +<@ + 0!021× 14$0!1)3!3 +<@ + 2!94 = 24!7

<@ = 18!5

927

PROBLEM 10.81

Situation: A water main is described in the problem statement.

Find: The pipe size.

Properties: From Table 10.2 c& = 0!15 mm. Table A.5 (water at 10 !C): ( =9810N$m3' : = 1!31× 10"6m2$ s!

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 + M(

(300' 000$9' 810) + 0 = (60' 000$9' 810) + 10 + M(

M( = 14!46 m

b(2$1)(]2$I2)$2) = 14!46

b(2$1)[]2$((0$4)12)2$2)] = 14!46

(42b2]2$02)$2)15 = 14!46

1 = [(8$14!46)b2]2$(02))]1*5


1 = [(8$14!46)× 0!02× 140× (0!025)2$(02 × 9!81)]1*5

= 0!1027 m

Relative roughness

c&1

=0!15

103= 0!00146

Reynolds number

Re =4]

01:

=4× (0!025m3$ s)

0 × (0!1027m)× (1!31× 10"6m2$ s)= 2!266× 105


b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³400146347

+ 5474(24266×105)049

í2

= 2! 271 7× 10"2

928

Recalculate pipe diameter

1 = 0!1027× (0!0227$0!020)1*5

= 0!105 m

Specify a 12-cm pipe

929

PROBLEM 10.82

Situation: A two reservoir system is described in the problem statement.

Find: The discharge.

Properties: From Table 10.3 <02 = 0!35; <02 = 0!16; <1 = 0!39' <G = 0!5 and<W = 1!0!From Table A.5 : = 1!22× 10"5 ft2$s.From Table 10.2 c& = 1!5× 10"4 ft.

ANALYSIS

Energy equation

%1$( + R1 + ,21 $2) = %2$( + R2 + ,

22 $2) +

XME

11 =X

ME = (,21 $2))(<G + 3<01 + b1 × 50$1)

+(, 22 $2))(<1 + 2<02 +<W + b2 × 30$(1$2))

Assume b1 = 0!015; b2 = 0!016

11× 2) = , 21 (0!5 + 3× 0!35 + 0!015(50)) + ,22 (0!39 + 2× 0!16 + 1!0 + 0!016(60))

708 = , 21 (2!3) + ,22 (2!67) = ]

2(2!3$((0$4)2(1)4) + 2!67$((0$4)2(1$2)4)) = 73!0]2

]2 = 708$73!0 = 9!70

] = 3!11 cfs

Re = 4]$(01:)

Re1 = 4(3!11)$(0(1!22× 10"5)) = 3!2× 105

c&$11 = 1!5× 10"4$1 = 0!00015Re2 = 6!5 105; c&$12 = 0!0003

From Fig. 10.8 b1 = 0!016 and b2 = 0!016! No further iterations are necessary so

] = 3!11 cfs

930

PROBLEM 10.83

Situation: A steel pipe is described in the problem statement.from Table A.5 : = 1!31× 10"6 m2$sprovided in problem statement

Find: (a) Discharge and(b) Pressure at point A.

ANALYSIS

Energy equation

%1$( + R1 + ,21 $2) = %2$( + R2 + ,

22 $2) +

XME

0 + 12 + 0 = 0 + 0 + (, 2$2))(1 +<G +<@ + 4<0 + b × 2$1)

Using a pipe diameter of 10 cm and assuming b = 0!025

24) = , 2(1 + 0!5 + 10 + 4(0!9) + 0!025× 1' 000$(0!10)), 2 = 24)$265!1 = 0!888 m2/s2

, = 0!942 m/s

] = , I

= 0!942(0$4)(0!10)2

= 0.0074 m3/s

Re = 0!942× 0!1$1!31× 10"6 = 7× 104

From Fig. 10.8 b % 0!025

%=$( + R= + ,2$2) = %2$( + R2 + ,

2$2) +X

ME

%=$( + 15 = , 2$2)(2<0 + b × 2$1)%=$( = (0!888$2))(2× 0!9 + 0!025× 500$0!1)" 15 = "9!26 m%= = 9810× ("9!26)

= -90.8 kPa

Note that this is not a good installation because the pressure at I is near cavitationlevel.

931

PROBLEM 10.84

Situation: Air ows through a horizontal, rectangular, air-conditioning ductDuct length is 2 = 20m. Section area is 4 by 10 inches (102 by 254 mm).Air speed is , =10 m/s! Sand roughness height for the duct material is c& =0!004mm!

Find: (a) The pressure drop in inches of water.(b) The power in watts needed to overcome head loss.

Properties: Air at 20 !C from Table A.3: # = 1!2 kg$m3, ( = 11!8N$m3. : =15!1× 10"6m2$ s!

Assumptions: 1.) Neglect all head loss associated with minor losses.2.) Assume T1 = T2, where T is the kinetic energy correction factor and sections 1and 2 correspond to the duct inlet and outlet, respectively.

APPROACH

To account for the rectangular section, use hydraulic diameter. Calculate Reynoldsnumber and then choose a suitable correlation for the friction factor (b) ! Apply theDarcy-Weisbach equation to nd the head loss (M(). Apply the energy equation tond the pressure drop, and calculate power using * = /)M( !

ANALYSIS

Hydraulic diameter (16) (four times the hydraulic radius)

16 =4I

*

=4 (0!102m) (0!254m)

(0!102m + 0!102m + 0!254m + 0!254m)= 0!1456m

Reynolds number

Re =, 16:

=(10m$ s) (0!1456m)

(15!1× 10"6m2$ s)= 96' 390


b =0!25

hlog10

³[-

34795+ 5474

Re049

í2

=0!25

hlog10

³4×10"6m

347×(041456m) +5474

96I390049

í2

= 0!0182

932


M( = b2

1

, 2

2)

= 0!0182

µ20m

0!1456m

¶µ102m2$ s2

2× 9!81m$ s2

¶

= 12!72m

Energy equation (section 1 and 2 are the inlet and exit of the duct)

µ%

(

¶

1

=

µ%

(

¶

2

+ ME

Thus

!% = (airM(

=¡11!8N$m3

¢(12!72m)

= 150Pa

= 150Pa

µ1!0 in!-H2O248!8Pa

¶

!% = 0!6 in!-H2O

Power equation

* = (]M(

= !%I,

= (150Pa) (0!102m× 0!254m) (10m$ s)* = 38!9W

COMMENTS

The power to overcome head loss is small (39W)–this is equivalent to the powerrequired to light a small light bulb.

933

PROBLEM 10.85

Situation: An electrostatic air lter is being tested.

Pressure drop is !% =3 in.-H20. Air speed is , = 10m$ s!

Find: The minor loss coe"cient (<) for the lter.

Properties: Air @ 20 !C from Table A.3: # = 1!2 kg$m3, ( = 11!8N$m3. : =15!1× 10"6m2$ s!

APPROACH

Apply the energy equation to relate the pressure drop to head loss. Then, nd theminor loss coe"cient using ME = <, 2$2)!

ANALYSIS

Energy equation (select a control volume surrounding the lter)

µ%

(

¶

1

=

µ%

(

¶

2

+ ME

Thus

ME =!%

(air

=(3 in.-H2O)

³249!2 Pa

in.-H2O

´

11!8N$m3

= 63!36m

Head loss

ME =<, 2

2)

< =2)ME, 2

(1)

=2 (9!81m$ s2) (63!36m)

(10m$ s)2

= 12!43

< = 12!4 (2)

934

COMMENTS

1.) This minor loss coe"cient is larger than the coe"cient for any components listedin Table 10.3.2.) Combining Eqs. (1) and (2) gives < = !%$(#, 2$2)! Thus, the pressure drop forthe lter is about 12 times larger that the pressure change that results when the owis brought to rest.

935

PROBLEM 10.86

Situation: A system with two tanks is described in the problem statement.

Find: The pump power.

Properties: From Table 10.3 <G = 0!03;<0 = 0!35;<W = 1!0!From Table A.5 : = 10"6 m2$s.From Table 10.2 c& = 0!046 mm.

APPROACH

Apply the energy equation from the water surface in the lower reservoir to the watersurface in the upper reservoir.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 200 m+ M, = 0 + 0 + 235 m + (, 2$2))(<G +<0 +<W + b2$1)

Flow rate equation

, = ]$I

= 0!314$((0$4)× 0!32)= 4!44 m/s

, 2$2) = 1!01 m

Reynolds number

Re = , 1$:

= 4!44× 0!3$10"6

= 1!33× 106

c&$1 % 0!00015


b = 0!00014

So

b2$1 = 0!014× 140$0!3 = 6!53M, = 235" 200 + 1!01(0!03 + 0!35 + 1 + 6!53)

= 43!0 m

Power equation

* = ](M,

= 0!314× 9' 790× 43!0= 132 kW

936

PROBLEM 10.87

Situation: A two-tank system with the pump from Fig. 10.16 is described in theproblem statement.

Find: Discharge.

APPROACH


ANALYSIS


M, = 35 + 8!38,2$2)

M, = 35 + 8!38[(]$((0$4)× 0!32)2$2)] = 35 + 85!6]2

System data computed and shown below:

](m3s) $ 0.05 0.10 0.15 0.20 .30M,(m) $ 35.2 35.8 36.9 38.4 42.7

Then, plotting the system curve on the pump performance curve of Fig.10-16 yieldsthe operating point

] = 0!25 m3$s

937

PROBLEM 10.88

Situation: A system with an injector pipe is described in the problem statement.

Find: If the system will operate as a pump.

ANALYSIS

For the system to operate as a pump, the increase in head produced by the jet must begreater than 9 ft (the di!erence in elevation between the lower and upper reservoir).Consider the head change between a section just to the right of the jet and far to theright of it with zero ow in the lower pipe. Determine this head change by applyingthe momentum equation.

1 2

x

,1 = 60 ft/s

] = ,1I1 = 2!94 cfs

,2 = ]$I2 = (60)(0$4)(32)$((0$4)(122))

,2 = 60(32$122) = 3!75 ft/sXH$ = /!,! " /#,#

%1I1 " %2I2 = (3!75)(1!94)(3!75× (0$4)(12))" (60)(1!94)(60× (0$4)(1$4)2)I(%1 " %2) = 1!94("176!7 + 11!04)%2 " %1 = 321 psf

M2 " M1 = (321 lbf/ft2)$(62!4 lbf/ft3) = 5!15 ft

The change in head of 5.15 ft is not enough to overcome the static head of 9.0 ft.;therefore, the system will not act as a pump.

938

PROBLEM 10.89

Situation: A pump is described in the problem statement.

Find: Discharge.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 10 + M, = 0 + 0 + 20 + , 22 $2)(<G + b2$1 + c0)

M, = 10 + (]2$(2)I2))(0!1 + 0!02× 1' 000$(10$12) + 1)I = (0$4)× (10$12)2 = 0!545 ft2

M, = 10 + 1!31]2cfs1 cfs = 449 gpm

M, = 10 + 1!31]2gpm$(449)2

M, = 10 + 6!51× 10"6]2gpm

]$ 1,000 2,000 3,000M$ 16.5 36.0 68.6

Plotting this on pump curve gure yields ] % 2' 950 gpm

939

PROBLEM 10.90


Find: Pumping rate.

ANALYSIS

M, = 20 ft - 10 ft = 10 ftThen from the pump curve for 10.89 one nds ] = 4' 700 gpm.

940

PROBLEM 10.91

Situation: Water pumping from a reservoir is described in the problem statement.

Find: Pump power.


Assumptions: From Table A.5 : = 1!31× 10"6 mm.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 100 + M, = 0 + , 22 $2) + 140 + ,22 $2)(0!03 + b2$1)

Flow rate

,2 = ]$I,

= 25$((0$4)× 1!52)= 14!15 m/s

Reynolds number

Re =, 1

:

=14!15× 1!51!31× 10"6

= 1!620× 107c&1

=0!046

1500= 0!00003

Friction factor (Moody Diagram) or the Swamee-Jain correlation:

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³0400003347

+ 5474(14620×107)049

í2

= 0!009995

% 0!01

Then

M, = 140" 100 +, 222)

µ1!03 + 0!010×

300

1!5

¶

= 140" 100 +14!152

2× 9!81

µ1!03 + 0!010×

300

1!5

¶

M, = 70!92m

941

Power equation

* = ](M,

=¡25m3$ s

¢×¡9810N$m3

¢× (70!92m)

= 17.4 MW

942

PROBLEM 10.92

Situation: Two pipes and their reservoirs are described in the problem statement.

Find: Di!erence in water surface between two reservoirs.

Assumptions: & = 20oC so : = 10"6 ft2/s .

ANALYSIS

c&$115 = 0!1$150 = 0!00067

c&$130 = 0!1$300 = 0!00033

,15 = ]$I15 = 0!1$((0$4)× 0!152) = 5!659 m/s,30 = 1!415 m/s

Re15 = , 1$: = 5!659× 0!15$10"6 = 8!49× 105

Re30 = 1!415× 0!3$10"6 = 4!24× 105

Resistance Coe"cient (from the Moody diagram, Fig. 10-8)

b15 = 0!0185

b30 = 0!0165

Energy equation

R1 " R2 =X

ME

R1 " R2 = (, 215$2))(0!5 + 0!0185× 50$0!15)+(, 230$2))(1 + 0!0165× 160$0!30) + (,15 " ,30)

2$2)

R1 " R2 = (5!6592$(2× 9!81))(6!67)+((1!4152$(2× 9!81))(9!80) + (5!659" 1!415)2$(2× 9!81)

R1 " R2 = 1!632(6!67) + 1!00 + 0!918 = 12.80 m

943

PROBLEM 10.93

Situation: Two pipes and their reservoirs are described in the problem statement.

Find: Di!erence in water surface elevation between two reservoirs.

Properties: From Table 10.3 <G = 0!5 and <W = 1.0.

Assumptions: & = 68!H so : = 1!1× 10"5 ft2/s.

APPROACH

Apply the energy equation from the water surface in the tank at the left to the watersurface in the tank on the right.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

R1 = R2 + (<G + b121$11),21 $2) + (,1 " ,2)

2$2) + ((b222$12) +<W),22 $2)

Calculate velocities and Reynolds number

,1 = ]$I1 = ]$((0$4)(1$2)2) = 25!48 ft/s

Re1 = 25!48× (1$2)$1!1× 10"5) = 1!16× 106

, 21 $2) = 10!1 ft

,2 = ,1$4 = 6!37 ft/s

Re2 = 6!37× 1$1!1× 10"5 = 5!8× 106

, 22 $2) = 0!630

c&$11 = 4× 10"4$0!5 = 8× 10"4

c&$12 = 4× 10"4

From Fig. 10.8 b1 = 0!019 and b2 = 0!016

R1 " R2 = M = (0!5 + !019× 150$(1$2))10!1 + (25!48" 6!37)2$64!4+((0!016× 500$1) + 1)0!630

= 62!6 + 5!7 + 5!7

= 74.0 ft

944

PROBLEM 10.94

Situation: Oil owing through a pipe is described in the problem statement.

Find: Discharge of oil.

Properties: From Table 10.3 <G = 0!50;<@ = 5!6!From Table 10.2 c& = 1!5× 10"4 ft.

APPROACH

Apply the energy equation from reservoir water surface to pipe outlet.

ANALYSIS

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 100 ft = 0 + , 22 $2) + 64 + (,2$2))(<G +<@ + b2$1)

Assume b = 0!015 for rst trial. Then

(, 2$2))(0!5 + 5!6 + 1 + 0!015× 300$1) = 36, = 14!1 ft/s

Re = , 1$: = 14!1× 1$10"4 = 1!4× 105

c&$1 = 0!00015

From Fig. 10.8 b % 0!0175!Second Trial:

, = 13!7 ft/s

Re = 1!37× 105

From Fig. 10.8 b = 0!0175!so

] = , I

= 13!7× (0$4)× 12

= 10.8 ft3/s

EGLHGL

945

PROBLEM 10.95

Situation: A system with a reservoir and a smooth pipe is described in the problemstatement.

Find: (a) Pump horsepower.(b) Pressure at midpoint of long pipe.

Properties: From Table 10.3 <0 = 0!19!From Table A.5 : = 1!22× 10"5 ft2$s.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 30 + 0 + M, = 0 + 60 + (,2$2))(1 + 0!5 + 4<0 + b2$1)

, = ]$I = 2!0$((0$4)× (1$2)2) = 10!18 ft/sec, 2$2) = 1!611 ft

Re = 4]$(01:) = 4× 2$(0 × (1$2)× 1!22× 10"5)= 4!17× 105

From Table 10.8 b = 0!0135 so

M, = 30 + 1!611(1 + 0!5 + 4× 0!19 + 0!0135× 1' 700$(1$2)) = 107!6 ft* = ](M,$550

= 24.4 horsepower

Pressure at midpoint of long pipe

%"$( + R" = R2 + ME

%" = ([(R2 " R") + ME]%" = 62!4[(60" 35) + 0!0135× (600$0!5)× 1!611]

%" = 3' 189 psf = 22.1 psig

EGL

HGL

HGL

EGL

946

PROBLEM 10.96

Situation: A pump system is described in the problem statement.

Find: Pump power.


ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 20 + M, = 0 + 0 + 40 + , 2$2)(<G + 2<0 +<0 + b2$1)

M, = 20 + , 2$2)(0!5 + 2× 0!19 + 1 + b2$1), = ]$I = 1!2$((0$4× 0!62) = 4!25 m/s

, 2$2) = 0!921 m

Re = , 1$: = 4!25× 0!6$(5× 10"5) = 5!1× 104

c&$1 = 0!00008


b = 0!021

SoM, = 20 + 0!921(0!5 + 0!38 + 1 + 6!65) = 27!9 m

Power equation

* =](M,g

=1!2× 0!94× 9810× 27!9

0!80

= 386 kW

947

PROBLEM 10.97

Situation: A system with an upstream reservoir is described in the problem statement.

Find: Elevation of water surface in upstream reservoir.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + R1 = 0 + 0 + 12 + (, 230$2))(0!5 + b2$1) + (,215$2))(<1 + b(2$1) + 1!0)(8)

,30 = ]$I30 = 0!15$((0$4)× 0!302) = 2!12 m/s, 230$2) = 0!229 m (9)

,15 = 4,30 = 8!488 m/s

, 215$2) = 3!67 m

12$11 = 15$30 = 0!5$ <) = 0!37

Then

R1 = 12 + 0!229[0!5 + 0!02× (20$0!3)] + 3!67[0!37 + 0!02(10$0!15) + 1!0]R1 = 22!3 m

H.G.L.

E.G.L.

E.G.L.H.G.L.

V /2g2

p/g

z0

A

A

948

PROBLEM 10.98

Situation: A tank with piping system is described in the problem statement.

Find: (a) Sketch the EGL and HGL.(b) Where cavitation might occur.

ANALYSIS

H.G.L.E.G.L.

Cavitation could occur in the venturi throat section or just downstream of the abruptcontraction (where there will be a contraction of the ow area).

949

PROBLEM 10.99

Situation: A system with a steel pipe is described in the problem statement.

Find: Pressure at point A.

Properties: From Table 10.3 <0 = 0!9' <@ = 10!From Table 10.2 c& = 5× 10"4 ft.From Table A.5 : = 1!41× 10"5 ft2$s.

ANALYSIS

Energy equation

%=$( + ,2=$2) + R= = %2$( + R2 + ,

22 $2) +

XME

%=$( + 20 + 0 = 0 + 90 + 0 + , 2$2)(0!5 + 2<0 +<@ + b(2$1) + 1)

, = ]$I = (50$449)$((0$4)(2$12)2) = 5!1

, 2$2) = 5!12$64!4 = 0!404

Re = 5!1(2$12)$(1!41× 10"5) = 6× 104

c&$1 = 5× 10"4 × 12$2 = 0!003


b = 0!028

Energy equation becomes

%= = ([70 + 0!404(0!5 + 2× 0!9 + 10 + (0!028× 240$(2$12)) + 1!0)]= 62!4× 91!7 = 5722 psfg = 39.7 psig

950

PROBLEM 10.100

Situation: A system with two reservoirs is described in the problem statement.

Find: Water surface elevation in reservoir I!

Properties: (a) From Table 10.2 c& = 0!26 mm. (b) From Table A.5 : = 1!3 × 10"6m2$s.

ANALYSIS

c&$120 = 0!26$200 = 0!0013

c&$115 = 0!0017

,20 = ]$I20 = 0!03$((0$4)× 0!202) = 0!955 m/s]$I15 = 1!697 m/s

Re20 = , 1$: = 0!955× 0!2$(1!3× 10"6) = 1!5× 105

Re15 = 1!697× 0!15$1!3× 10"6 = 1!9× 105

From Fig. 10-8: b20 = 0!022; b15 = 0!024

R1 = R2 +X

ME

R1 = 110 + , 220$2)(0!5 + 0!022× 100$0!2 + 0!19)+, 215$2)[(0!024× 150$0!15)+1!0 + 0!19)]

= 110 + 0!0465(11!7) + 0!1468(25!19)

= 110 + 0!535 + 3!70 = 114.2 m

951

PROBLEM 10.101

Situation: A pipe system must supply water ow from an elevated tank to thereservoir—additional details are provided in the problem statement.

Find: Design the pipe system.

ANALYSIS

One possible design given below:

2 % 300 + 50 + 50 = 400 m; <0 = 0!19

50 =X

ME = ,2$2)(<G + 2<0 + b(2$1) + 1!0) = ,

2$2)(1!88 + b(2$1))

50 = []2$(2)I2)](b(2$1) + 1!88) = [2!52$(2× 9!81×I2)]((400 b$1) + 1!88)


50 = [0!318$((0$4)2 ×14)](0!015× (400$1)) + 1!88)

Solving, one gets1 % 0!59m = 59 cm

Try commercial size 1 = 60 cm. Then

,60 = 2!5$((0$4)× 0!62) = 8!84 m/sRe = 8!8× 0!6$10"6 = 5!3× 106; c&$1 = 0!0001 and b % 0!013

Since b = 0!13 is less than originally assumed b , the design is conservative. So use1 = 60 cm and 2 % 400 m.

60 cm steel pipe

952

PROBLEM 10.102

Situation: A pipe system must supply water ow from an elevated tank to thereservoir—additional details are provided in the problem statement.

Find: Design the system.

Assumptions: Steel pipe will be used.

APPROACH

First write the energy equation from the reservoir to the tank and assume that thesame pipe conguration as used in the solution to P10-99 is used. Also a pump, twoopen gate valves, and two bends will be in the pipe system.

ANALYSIS

Assume 2 % 400 ft.

%1$( + ,21 $2) + R1 + M, = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 450 + M, = 0 + 0 + 500

+(, 2$2))(<G + 2<0 + 2<@ +<W + b2$1)

Assume , % 2 m/s; I = ]$, = 1!0$2 = 0!50 m2

I = (0$4)12 = 0!50 or 1 = !799 m Choose a pipe size of 0.80 m

Then

, = ]$I = 1!0$((0$4)× 0!82) = 1!99 m/s and , 2$2) = 0!202 mc&$1 = 0!00006; Re = , 1$: = 1!6$10"6 = 1!6× 106

Then b = 0!012 (from Fig. 10-8)

M, = 50 + (, 2$2))(0!5 + 2× 0!2 + 2× 0!19 + 1!0 + 0!012× 400$1)= 50 + 1!43 = 51!43 m

* = ](M, = 2!0× 9' 810× 51!43= 1!01 MW

Design will include 0.80 m steel pipe and a pump with output of 1.01 MW

COMMENTS

An innite number of other designs are possible. Also, a design solution wouldinclude the economics of the problem to achieve the desired result at minimum cost.

953

PROBLEM 10.103

Situation: Design lab equipment to illustrate cavitation.Use a venturi nozzle to create the low pressure.Assume a water source with a pressure of % = 50 psig.

Find: Specify the components, the primary dimensions and parameters (ow rates)

ANALYSIS

There are many possible design solutions. The venturi nozzle should be fabricatedfrom clear material so that cavitation can be observed.

954

PROBLEM 10.104

Situation: The guidelines for an experiment to verify the momentum principle aredescribed in the problem statement.

Find: Design the equipment and the experimental procedure.

APPROACH

Because you want to design equipment to illustrate cavitation, it would be desirableto make the ow restriction device from clear plastic so that one may observe theformation of cavitation bubbles. The design calculation for pressure and dischargewould be the same as given for 10.71.

ANALYSIS

Equipment for the momentum experiment is shown below:

Counter Wgt

Pivot point VaneScops

Weight

Necessary measurements and calculations:

a) Discharge. This could be done by using a scale and tank to weigh the ow ofwater that has occurred over a given period of time.

b) The velocity in the jet could be measured by means of a stagnation tube or solvingfor the velocity by using Bernoulli’s equation given the pressure in the nozzlefrom which the jet issues.

c) Initially set the counter balance so that the beam is in its horizontal equilibriumposition. By opening a valve establish the jet of water. Apply necessaryweight at the end of the beam balance to bring the beam back to horizontalequilibrium. By calculation (using moment summation) determine the forcethat the jet is exerting on the vane. Compare this force with the calculatedforce from the momentum equation (using measured ]' , , and vane angle).

955

PROBLEM 10.105


Find: Ratio of discharge in line D to that in line I.

ANALYSIS

ME= = ME>

0!2, 2=$2) = 10, 2>$2) (1)

,= =&50,>

]>$]= = ,>I>$,=I=

= ,>I>$,=((1$2)I>) (2)

]>$]= = 2,>$,=

Solve Eqs. (1) and (2) for ]>$]=:

]>$]= = 2× ,>$&50,>

= 0.283

956

PROBLEM 10.106

Situation: Divided ow is described in the problem statement.

Find: Ratio of velocity in line I to D!

ANALYSIS

Globe valve wide openB

AGate valve half open

XME> =

XME=

MEIglobe + 2MEIelbow = MEIgate + 2MEIelbow

10, 2>$2) + 2(0!9,2>$2)) = 5!6, 2=$2) + 2(0!9,

2=$2))

11!8, 2>$2) = 7!4, 2=$2)

,=$,> = 1!26

957

PROBLEM 10.107

Situation: A parallel piping system is described in the problem statement.

Find: Division of ow of water.

ANALYSIS

,1$,2 = [(b2$b1)(22$21)(11$12)]1*2

Initially assume b1 = b2Then

,1$,2 = [(1' 500$1' 000)(0!50$0!40)]1*2

= 1!369

,1 = 1!37,2

1!2 = ,1I1 + ,2I2

1!2 = 1!37,2 × (0$4)× 0!52 + ,2 × (0$4)× 0!42

,2 = 3!04 m/s

Then ,1 = 1!37× 3!04 = 4.16 m/s

]1 = ,1I1

= 4!16(0$4)× 0!52

= 0.816 m3/s

]2 = 0.382 m3/s

958

PROBLEM 10.108


Find: Discharge in pipe 1.

ANALYSIS

M(I1 = M(I2

b(2$1)(, 21 $2)) = b(42$1)(, 22 $2))

, 21 = 4, 22,1 = 2,2

Thus

]1 = 2]2

= 2 cfs

959

PROBLEM 10.109


Find: The pipe having the greatest velocity.

ANALYSIS

M,I= = M(I> = M(I)

b(2$1)(, 2$2))= = b(2$1)(, 2$2))> = b(2$1)(,2$2)))

0!012(6' 000$1!5), 2= = 0!02(2' 000$!5), 2> = !015(5' 000),2)

48, 2= = 80, 2> = 75,2)

Therefore, ,= will have the greatest velocity. Correct choice is a).

960

PROBLEM 10.110


Find: Ratio of discharges in two pipes.

ANALYSIS

(,1$,2) = [(b2$b1)(22$21)(11$12)]1*2

Let pipe 1 be large pipe and pipe 2 be smaller pipe. Then

(,1$,2) = [(0!014$0!01)(2$32)(21$1)]1*2 = 0!966

(]1$]2) = (,1$,2)(I1$I2) = 0!966× (21$1)2 = 3!86(]large$]small) = 3.86

961

PROBLEM 10.111


Find: (a) Division of ow.(b) Head loss.

ANALYSIS

]18 +]12 = 14 cfs

ME18 = ME12b18(218$118)(,

218$2)) = b12(212$112)(,

212$2))

b18 = 0!018 = b12

so

218]218$1

518 = 212]

212$1

512

]218 = (118$112)5(212$218)]

212

= (18$12)5(2' 000$6' 000)]212= 2!53]212

]18 = 1!59]12

1!59]12 +]12 = 14

2!59]12 = 14

Q12 = 5.4 cfs

]18 = 1!59]12

= 1!59(5!4)

= 8.6 cfs

,12 = 5!4$((0$4(1)2) = 6!88

,18 = 8!6$((0$4)(18$12)2) = 4!87

ME12 = 0!018((2' 000)$1)(6!88)2$64!4 = 26!5

ME18 = 0!018(6' 000$1!5)(4!872$64!4) = 26!5

Thus' ME#"$ = 26.5 ft

962

PROBLEM 10.112


Find: (a) Division of ow.(b) Head loss.

ANALYSIS

] = ]14 +]12 +]16

25 = ,14 × (0$4)× (14$12)2 + ,12 × (0$4)× 12 + ,16 × (0$4)× (16$12)2; (1)

Also, M(14 = M(12 = M(16 and assuming b = 0!03 for all pipes

(3000$14), 214 = (2000$12),212 = (3000$16),

216 (2)

, 214 = 0!778,212 = 0!875,

216

From Eq(1)

25 = 1!069,14 + 0!890,14 + 1!49,14

,14 = 7!25 ft/s

and ,12 = 8!22, ,14 = 7!25 ft/s; ,16 = 7!25 ft/s

]12 = 6.45 ft3/s

]14 = 7.75 ft3/s

]16 = 10.8 ft3/s

,24 = ]$I24 = 25$(0$4× 22) = 7!96 ft/s;,30 = 5!09 ft/s

MEAB = (0!03$64!4)[(2' 000$2!00)(7!96)2 + (2' 000$1)× (8!21)2

+(3' 000$(30$12)× (5!09)2] = 106.8 ft

963

PROBLEM 10.113


Find: (a) Division of ow between pipes.(b) Head loss.


ANALYSIS

Call pipe A-B pipe and pipe ACB pipe 2. Then

M(I1 + M, = M(I2

c&$1 = 0!046$500 ' 0!0001

Assume b1 = b2 = 0!013 (guess from Fig. 10-8)

b(21$11)(,21 $2)) + M, = b(22$12)(,

22 $2))

0!013(2' 000$0!5)(, 21 $2)) + M, = 0!013(6' 000$0!5)(, 22 $2))

2!65, 21 + M, = 7!951, 22 (1)


(,1 + ,2)I = 0!60 m3/s

,1 + ,2 = 0!60$I = 0!6$((0$4)(0!52)) = 3!0558

,1 = 3!0558" ,2 (2)

By iteration (Eqs. (1), (2) and pump curve) one can solve for the division of ow:

]1 = 0.27 m3/s

]2 = 0.33 m3/s

Head loss determined along pipe 1

ME = b(2$1)(, 21 $2))

,1 = ]1$I = 0!27$((0$4)(0!52)) = 1!38 m/s

M2 = 0!013(2000$0!5)(1!382$(2× 9!81))= 5.05 m

964

PROBLEM 10.114


Find: Discharge through pump and bypass line.

ANALYSIS

21

Valve

Pump

], = ]@ + 0!2

(%2 " %1)$( = M,

I = (0$4)(0!12)

= 0!00785 m2

<@,2@ $2) = <@]

2@$(2)I

2) = M,

M, = 100" 100(]@ + 0!2)(0!2)(]2@)$(2× 9!81× (0!00785)

2) = 100" 100]@ " 20165]2@ + 100]@ " 80 = 0

Solve by quadratic formula

]@ = 0.456 m3/s

], = 0!456 + 0!2

= 0.656 m3/s

965

PROBLEM 10.115

Situation: Air and water ow are described in the problem statement.

Find: The relation of the two hydraulic radii.

ANALYSIS

"7 = I$*

"7I= = (I$* )= = 16$16 = 1

"7IM = (I$* )M = 8$8 = 1

! "7I= = "7IM

The correct choice is (a).

966

PROBLEM 10.116

Situation: Air owing through a horizontal duct is described in the problem state-ment.

Find: Pressure drop over 100 ft length.

Properties: From Table A.3 : = 1!58× 10"4 ft2$s and # = 0!00237 slug/ft3!From Table 10.2 c& = 0!0005 ft.

ANALYSIS

M = (6 in)(cos 30!) = 5!20

I = (6)(5!20)$2 = 15!6 in2 = 0!108 ft2

"7 = I$* = 15!6 in2$(3× 6) = 0!867 in.4"7 = 3!47 in. = 0!289 ft.

c&$4"7 = 0!0005/0!289. = 0.00173

Re = (, )(4"7)$: = (12)(0!289)$(1!58× 10"4) = 2!2× 104

From Fig. 10.8 b = 0!030 so the pressure drop is

!%( = (b(2$4"7)(#,2$2)

!%( = 0!030(100$0!289)(0!00237× 122$2)

!%( = 1!77 lbf/ft2

967

PROBLEM 10.117

Situation: Uniform ow of water in two channels is described in the problem state-ment.

Find: Relate ow rates of two channels.

ANALYSIS

] = (1!49$Q)I"2*37 61*2

]=$]> = "2*37I=$"

2*37I> = ("7I=$"7I>)

2*3

where "7I= = 50$20 = 2!5; "7I> = 50$(3× 7!07) = 2!36"7I= F "7I>

! ]= F ]>

The correct choice is (c).

968

PROBLEM 10.118

Situation: A cold-air duct is described in the problem statement.

Find: Power loss in duct.

Properties: From Table A.3 : = 1!46× 10"5!From Table A.2 # = 1!22 kg/m3!

Assumptions: c& = !15 mm= 1!5× 10"4 m

ANALYSIS

Hydraulic radius

I = 0!15 m2

* = 2!30 m

" = I$* = 0!0652 m

4" = 0!261 m

Flow rate equation

, = ]$I

= 6$0!15

= 40 m/s

Reynolds number

Re = , × 4"$:= 40× 0!261$(1!46× 10"5)= 7!15× 105

Friction factor (b) (turbulent ow: Swamee-Jain equation)

b =0!25

£log10

¡[-3479

+ 5474Re049

¢¤2

=0!25

hlog10

³145×10"4347×04261 +

5474(7415×105)049

í2

= 0!01797 % 0!018


M( = b(2$1)(, 2$2))

= 0!018× (100$0!261)(402$(2× 9!81))= 562!4 m

969

Power equation

*loss = ](M(

= 6× 1!22× 9!81× 562!4= 40.4 kW

970

PROBLEM 10.119

Situation: An air conditioning system is described in the problem statement.

Find: Ratio of velocity in trapezoidal to rectangular duct.

ANALYSIS

!Mrect = !Mtrap

! M(Irect = M(Itrap(b02$4"0),

20 $2) = (b'2$4"'),

2' $2)

"0 = I0$*0 = 2$6 = 0!333 ft

"' = I'$*' = 1!4$6 = 0!233 ft

, 2' $,20 = "'$"0 = 0!70

,trap$,rect = 0!84

971

PROBLEM 10.120

Situation: Water owing though a concrete duct is described in the problem state-ment.

Find: Estimate resistance coe"cient.

ANALYSIS

b = b(Re' c&$4")

" = I$* = 0!7 m2$3!4 m = 0.206 m

Re = , (4")$:

= 10× 4× !206× 106

= 8!2× 106

c&$4" = 10"3 m$0!824 m

= 1!2× 10"3

= !0012

From Fig. 10.8: b % 0!020 Choice (b) is the correct one.

972

PROBLEM 10.121

Situation: A wood ume is described in the problem statement.

Find: Discharge of water.

Assumptions: Q = 0!012

APPROACH

Apply Manning’s formula.

ANALYSIS

Manning’s formula

] = (1$Q)I"2*37 6

1*20

I = (1)(2)$2 = 1 m2

"7 = I$*

= 1$2(12 + 12)045 = 0!35 m

] = (1$0!012)(1)(0!35)2*3(0!0015)045

] = 1.60 m3/s

973

PROBLEM 10.122

Situation: A rock-bedded stream is described in the problem statement.

Find: Discharge.

Assumptions: c& = 30 cm.

ANALYSIS

From Fig. 10.8 b % 0!060

" = I$* % 2!21 mc&$4" = 0!034

from Fig. 10.8 b % 0!060

- =p8)$b

= 36!2 m1*2s"1

] = -I&"6

= 347 m3/s

974

PROBLEM 10.123

Situation: A concrete channel is described in the problem statement.

Find: Discharge.

Assumptions: c& = 10"3 m

ANALYSIS

I = 4!5 m2

* = 6 m

" = I$* = 0!75 m

c&$4" = 0!333× 10"3

From Fig. 10.8 b = 0!016

M($2 = b, 2$(2)4")

, =p(8)$b)"6 = 1!92 m/s

Re = 1!92× 3$(1!31× 10"6) = 4!4× 106 b = 0!015

From Fig. 10.8 b = 0!015Then

, = 1!92×p0!016$0!015 = 1!98 m/s

Finally,

] = 1!98× 4!5

= 8.91 m3/s

975

PROBLEM 10.124


Find: Discharge.

Assumptions: c& = 0!003

ANALYSIS

" = I$* = 4× 12$(12 + 2× 4) = 2!4c&$(4") = 0!003$(4× 2!4) = 0!00031

Reb1*2 = ((4")3*2$:)(2)6)1*2 × (2) × 5$8' 000)1*2

= 4!9× 105;

From Fig. 10.8 b = 0!015

, =p8)"6$b

=p8) × 2!4× 5$(0!015× 8' 000)

= 5!07

] = 5!07(4)12

= 243 cfs

Alternate solution:Assume Q = 0!015

] = (1!49$Q)I"2*361*2

= (1!49$0!015)4× 12(2!4)2*3 (5$8' 000)1*2

= 214 cfs

976

PROBLEM 10.125

Situation: Channels of rectangular cross section are described in the problem state-ment.

Find: Cross-sectional areas for various widths.


ANALYSIS

] = 100 cfs

6 = 0!001

] = (1!49$Q)I"046676045

or ]Q$(1!496045) = I"04667

31!84 = I"04667

31!84 = (>?)(>?$(>+ 2?))04667

For di!erent values of > one can compute ? and the area >?. The following tableresults

> (ft) ? (ft) I (ft2) ?$>2 16.5 33.0 8.24 6.0 24.0 1.56 3.8 22.5 0.638 2.8 22.4 0.3510 2.3 23.3 0.2315 1.7 25.5 0.11

Area (m2 )

22 23 24 25 260

1

2

Minimum area at y/b=0.5 (verified)

y/b

977

PROBLEM 10.126

Situation: Sewer partially lls a concrete pipe.The slope is 1 foot of drop per 1000 feet of length.Pipe diameter is 1 = 3 ft! Depth of sewer is ? = 1!5 ft!


Assumptions: Assume that the properties of the sewer are those of clean water.Assume an Manning’s n-value of Q = 0!013!

APPROACH

Using Manning’s equation (traditional units).

ANALYSIS

Hydraulic radius

"7 =Ic*wet

=012$8

01$2=1

4

=3 ft

4= 0!75 ft

Flow area

I =012

8=0 (3 ft)2

8= 3! 534 ft2

Manning’s equation (traditional units)

] =1!49

QI"

2*37

p6!

=1!49

0!013× 3! 534× 0!752*3

r1 ft

1000 ft

= 10! 57 ft3$ s

] = 10! 6 ft3$ s

978

PROBLEM 10.127

Situation: A sewer pipe is described in the problem statement.



ANALYSIS

] = (1!49$Q)I"046677 60450

2’ 2’

1.5’

2.5’

2.5’

!

cos K = 1!5 ft/2.5 ft

K = 53!13!

I = 032((360! " 2× 53!13!)$360) + 0!5× 4 ft× 1!5 ftI = 16!84 ft2

* = 01((360! " 2× 53!13!)$360) = 11!071 ft"7 = I$* = 1!521 ft

"046677 = 1!323

Then ] = (1!49$0!012)(16!84)(1!323)(0!001)045

] = 87!5 cfs

979

PROBLEM 10.128


Find: Average velocity and discharge.

Assumptions: c& = 0!003 ft: = 1!41× 10"5 ft2$s\

ANALYSIS

" = I$* = (10 + 12)6$(10 + 6&5× 2) = 132$36!8 = 3!58

(c&$4") = 0!003$(4× 3!58) = 0!00021Reb1*2 = ((4")3*2$:)(2)6)1*2 = [(4× 3!58)3*2$1!41× 10"5](2)$2000)1*2

= 6!9× 105

From Fig. 10.8 b = 0!014! Then

, =p8)"6$b

=p8) × 3!58$(2000× 0!014)

= 5.74 ft/s

] = , I

= 5!74× 132= 758 cfs

Alternate method, assuming Q = 0!015

, = (1!49$Q)"2*361*2

= (1!49$0!015)(3$3!58)2*3(1$2' 000)1*2

= 5.18 fps

] = 5!18(132)

= 684 cfs

980

PROBLEM 10.129


Find: Depth of ow in trapezoidal channel.


APPROACH

Using Manning’s equation (traditional units).

ANALYSIS

Flow area

I1 =

µ10 ft + (10 ft + 2A)

2

¶A

= 10A+ A2

Wetted perimeter

*wet = 10 ft + 2×&2A2

= 10 + 2&2A

Hydraulic radius

"7 =Ic*wet

=10A+ A2

10 + 2&2A

Manning’s equation (traditional units)

] =1!49

QI1"

2*37

p6!

1000 =1!49

0!012×¡10A+ A2

¢×µ10A+ A2

10 + 2&2A

¶2*3r1 ft

500 ft

Solve this equation (we used a computer program—Maple) to give A = 5! 338 ft!

A = 5! 34 ft

981

PROBLEM 10.130

Situation: A channel is described in the problem statement.

Find: Discharge in trapezoidal channel.


ANALYSIS

] = (1!49$Q)I"2*361*2

I = 10× 5 + 52' * = 10 + 2&52 + 52 = 24!14 ft

" = I$* = 75$24!14 = 3!107 ft

Then

] = (1!49$0!012)(75)(3!107)2*3(4$5' 280)1*2

= 546 cfs

982

PROBLEM 10.131


Find: The uniform ow depth.


ANALYSIS

] = (1$Q)I"2*361*2

25 = (1!0$0!015)4A(4A$(4 + 2A))2*3 × 0!0041*2

Solving for A yields: A = 1!6 m

983

PROBLEM 10.132


Find: The depth of ow.


ANALYSIS

] = (1!49$Q)I"2*361*2

500 = (1!49$0!012)12A(12A$(12 + 2A))2*3 × (10$8' 000)1*2

Solving for A yields: A = 4!92 ft

984

PROBLEM 10.133


Find: Depth of ow in channel.


ANALYSIS

] = (1!49$Q)I "2*37 6

1*20

3' 000 = ((1!49)$(0!015))(10A+ 2A2)((10A+ 2A2)$(10 + 2&5A))2*3(0!001)1*2

955 = (10A+ 2A2)((10A+ 2A2)$(10 + 2&5A))2*3

Solving for A gives A = 10!1 ft

985

PROBLEM 10.134

Situation: A canal is described in the problem statement.

Find: Design a canal having the best hydraulic section for the design ow.

ANALYSIS

For best hydraulic section, the shape will be a half hexagon as depicted belowassume Q = 0!015 (concrete, wood forms unnished - Table 10.3)

b

b

y45 deg

Manning’s equation] = (1!49$Q)I"046677 60450

Then

900 = (1!49$0!015)I"046677 (0!002)045

I"046677 = 202!6

But I = by +?2 where ? = > cos 45! = 0!707b

I = 0!707b2 + 0!50>2 = 1!207b2

"7 = I$* = 1!207b2$3> = 0!4024>

Thus

I"046677 = 202!6 = 1!207>2(0!4024>)04667

>24667 = 308; > = 8.57 ft

986

PROBLEM 10.135

Situation: Sources and loads are described in the problem statement.

Find: Load distribution and pressure at load points.

ANALYSIS

An assumption is made for the discharge in all pipes making certain that the conti-nuity equation is satised at each junction. The following gure shows the networkwith assumed ows.

A

B C

D E

10 cfs 10 cfs

5 cfs

5 cfs 0 cf

s

0 cf

s

k=0.3021

k=0.3021

k=0.00944

k=1.059 k=2.

294

k=0.

7516

15 cfs 10 cfs

5 cfs


M( = b

µ2

1

¶µ, 2

2)

¶

= 8

µb2

)1502

¶]2

= c]2!

where c = 8³

(ED9582

´! The loss coe"cient, c, for each pipe is computed and shown

in Fig. A. Next, the ow corrections for each loop are calculated as shown in theaccompanying table. Since Q = 2 (exponent on ]), Qc]:"1 = 2c]. When thecorrection obtained in the table are applied to the two loops, we get the pipe dischargesshown in Fig. B. Then with additional iterations, we get the nal distribution ofow as shown in Fig. C. Finally, the pressures at the load points are calculated.

Loop ID-Pipe M( = c]

2 2c]ID +0.944 0.189I1 -26.475 10.590D1 0 0Pc]21 "

Pc]211 -25.53

P2<] = 10!78

!Q=-22.66/9.062=2.50 cfs

987

Loop D-1NPipe M( 2c]D- +30.21 6.042D1 0 0-N 0 01N -7.55 3!02

+22.66 9.062!Q=-25.53/10.78=-2.40 cfs

AB C

D E

12.4 cfs 7.5 cfs

7.5 cfs

2.6 cfs

4.9

cfs

2.5

cfs

10 cfs

5 cfs

15 cfs

AB C

D E

11.4 cfs 9.0 cfs

6.0 cfs

3.60 cfs

2.40

cfs

1.0

cfs

10 cfs

5 cfs

15 cfs

%) = %= " ((c=>]2=> + c>)]2>))

= 60 psi × 144 psf/psi" 62!4(0!00944× 11!42 + 0!3021× 9!02)= 8640 psf " 1603 psf= 7037 psf

= 48.9 psi

%W = 8640" ((c=9]2=9 + c9W]29W)

= 8640" 62!4(1!059× 3!52 + 0!3021× 62)= 7105 psf

= 49.3 psi

988

PROBLEM 10.136

Situation: A platform is described in the problem statement.

Find: Scope the system and make enough calculations to justify the feasibility.

Assumptions: Assume that the equipment will have a maximum weight of 1,000 lbfand assume that the platform itself weighs 200 lbf. Assume that the platform willbe square and be 5 ft on a side.

ANALYSIS

The plan and elevation view are shown below:

A=9 sq. ft.

View A-A

A A

Assume that a plenum 1 ft inside the perimeter of the platform will be the source ofair for the underside of the platform.Now develop the relationship for pressure distribution from plenum to edge of plat-form. The ow situation is shown below.

Plenum

1 ft

! y

Determine the M( from the plenum to the edge of the platform:

M( = b(2$1),2$2)

Assume b = 0!02' " = I$* = !?D$2D = !?$2 and 2 = 1 ft.

M( = (0!02× 1$(!?$2)), 2$2)= (0!02$!?), 2$)

= 0!02, 2$(!?))

Multiply both sides by (

!%( = (M( = (0!02$!?)#,2

Assume # = 0!0023 slugs/ft3!Then

989

!%( = (0!02$!?)(!0023), 2

= (46, 2$!?)× 10"6

%avg.(over 4 ft2 area) = (23 , 2$!?)× 10"6

Also determine the !% due to the change in momentum as the ow discharges fromthe plenum.

p 1 p 2

Momentum equation (x-direction)

XH$ = /!,! " /#,#

D!?(%1 " %2) = , (#, D!?)

!%mom = #, 2

The pressure force on the platform is given byThe pressure within the 9 ft2 interior area of the platform will be

!%mom +!%( = ,2(!0023 + (46$!?)× 10"6)

The pressure force on platform is given by

H = 9 ft2 × (!%mom +!%() +!%(Iavg. × 12 ft2

H = 9× , 2[!0023 + (46$!?)× 10"6)] + 12, 2[(23, 2$!?)× 10"6]H = , 2[9× !0023 + (9× 46$!?)× 10"6 + 12× 23× 10"6$!?]H = , 2[9× !0023 + 690× 10"6$!?]

Let !? = 1$8 in.= 0!01042 ft

990

b = , 2[9× !0023 + 690× 10"6$!01042]= , 2[0!0207 + 0!662]

H = !0869, 2

1200 = !0869, 2

, 2 = 13' 809 ft2$s2

, = 117!5 ft/s

] = 117!5×!? × 12 = 14!69 ft3$s!% = , 2(!0023 + 46× 10"6$!?)

= , 2(!0023 + 46× 10"6$0!01042)= , 2(!0023 + !00441)

= 92!7 psf

Power equation

* = ]!%$550

= 14!69× 92!7$550= 2!48 hp

Assume 50% e"ciency for blower, so required power % 5 horsepower. Blower couldbe driven by gasoline engine and also be located on the platform.

991

PROBLEM 10.137

Situation: A system for measuring the viscosity of a gas is described in the problemstatement.

Find: Design the system.

ANALYSIS

There are two design constraints; 1) the Reynolds number in the tube should be lessthan 1000 to insure that the ow in laminar and a closed form expression is availableto the viscosity and 2) the pressure di!erential along the tube should su"ciently lowthat compressibility e!ects on the gas will not be important yet large enough thata measurement can be made with acceptable accuracy. Although not stated in theproblem assume that the density of the gases ranges from 0.8 kg/m3 to 1.5 kg/m3!As a start assume the tube has a 1 mm internal diameter. The Reynolds numbercorresponding to the highest density and lowest viscosity would be

Re =, × 10"3 × 1!5

10"5= 150,

The maximum velocity should not exceed 6 m/s. The pressure drop for laminar owin a pipe is

!% = 3272,

12

Assume the length of the tube is 500 mm (0.5 m), the pressure drop for the largestviscosity would be given by

!% = 321!5× 10"5 × 0!5,

10"6= 240,

For a velocity of 6 m/s, the pressure drop would be 1,440 Pa or 0.2 psig. or about5 in of water. If the initial pressure were atmospheric, this would represent abouta 1% change in pressure which would be acceptable to avoid compressibility e!ects.Compressibility e!ect could also be reduced by operating at a higher pressure wherethe percentage change in pressure would be even smaller.This design could now be rened to conform with the equipment available for mea-suring pressure. Another issue to consider is the design of the entrance to the tubeto minimize entrance losses and exit losses such as a sudden expansion. There is alsothe problem of measuring a small discharge. An idea to consider would be attachingthe end of the tube to an inatable bag immersed in water and measuring the dis-placement of the water with time. Another idea is measuring the pressure drop in atank supplying the tube and calculating the mass change with time.

992

PROBLEM 10.138

Situation: A problem associated with a pressure tap is described in the problemstatement.

Find: Develop ideas to avert the problem.

ANALYSIS

One idea is to use a purge line as shown in the gure. There is a continuous owof gas out the pressure tap which keeps the tap clean. The ow rate should be highenough to keep the tap clean and low enough not of a!ect the readings. The purgegases would be introduced close to the tap so the head loss associated with frictionwould be minimized. The largest pressure drop would be the sudden expansion lossat the tap exit. If %! is the nominal pressure being measured at the tap, then theratio of the sudden expansion losses to the nominal pressure is

#, 2

2%!

and this ratio should be kept as small as possible. If the ratio is 0.01 then an errorof 1% would be produced in the pressure measurement. The ow rate should be justsu"cient to keep the taps clean. This value will depend on the experimental condi-tions.

pressuregage

purgegas

Flow

993

PROBLEM 11.1

Situation: A hypothetical pressure-coe"cient distribution acts on an inclined plate.Other details are provided in the problem statement

Find: Coe"cient of drag.

Assumptions: Viscous e!ects are negligible.

ANALYSIS

Force normal to plate

H: = !%average ×I= -,I'@G#,

20 $2× Z× 1

= 1!5× #, 20 $2× Z× 1

For unit depth of plate and a length Z. Force parallel to free stream direction is thedrag force and is equal to

H9 = Hnormal cos 60!

= (1!5#, 20 $2)× Z× 1$2

The drag coe"cient is dened from the drag force as

-9 =H9

12#, 20 I

=(1!5#, 20 $2)× Z× 1$2

12#, 20 × Z× 1

= 1.5

994

PROBLEM 11.2

Situation: Fluid ow past a square rod. The pressure coe"cient values are shown inthe problem statement.

Find: Direction from which the ow is coming.

ANALYSIS

Flow is from the N.E. direction.Correct choice is d)

995

PROBLEM 11.3

Situation: A pressure distribution is described in the problem statement.

Find: Drag coe"cient for rod.

APPROACH

Apply drag force.

ANALYSIS

The drag coe"cient is based on the projected area of the block from the direction ofthe ow which is the area of each face of the block. The force contributing to dragon the downstream face is

H9 = 0!5I,#,2! $2

The force on each side face is

H& = 0!5I,#,2! $2

Then the drag force on one side is

H& sinT = 0!5I,#,2! $2× 0!5

The total drag force is

H9 = 2((0!5I,#,2! $2)× 0!5) + 0!5I,#,

2! $2 = -9I,#,

2! $2

Solving for -9 one gets -9 = 1!0

996

PROBLEM 11.4

Situation: A pressure distribution is described in the problem statement.

Find: Drag coe"cient for the block.

ANALYSIS

The drag coe"cient is based on the projected area of the block from the ow direction,I,. The drag force on the windward side is

H% = 0!8×1

2#, 20 I,

The force on each of the two sloping sides is

H& = "1!2×1

2#, 20 I,

The total drag force on the rod is

H9 = 0!8×1

2#, 20 I, " 2("1!2×

1

2#, 20 I,) sin 30

!

=1

2#, 20 I,(0!8 + 1!2)

The drag coe"cient is

-9 =H9

12#, 20 I,

= 2.0

997

PROBLEM 11.5

Situation: A wind tunnel can produce air velocity of 100ft/s, 3ft×3ft test section.

Find: The design objective is to design an experiment to measure the drag coe"cientof spheres of varying surface roughness.

ANALYSIS

The drag force equation is

H9 = -9I,#,2$2

or -9 = H9$(I,#,2$2)

Thus H9' I,' and , will have to be measured. The air density # can be obtainedby measuring the air temperature with a thermometer and the air pressure with abarometer and solving for # by the equation of state.

You will need to decide how to position the sphere in the wind tunnel so that its sup-port does not have an inuence on ow past the sphere. One possible setup mightbe as shown below.

Rod

Force dynamometer

Pitot tube

The sphere is attached to a rod and the rod in turn is attached to a force dynamometeras shown. Of course the rod itself will produce drag, however; its drag can beminimized by enclosing the vertical part of the rod in a streamlined housing. Thehorizontal part of the rod would have negligible drag because much of it would bewithin the low velocity wake of the sphere and the drag would be skin friction dragwhich is very small. The air velocity approaching the sphere could be measuredby a Pitot tube inserted into the wind tunnel. It would be removed when thedrag of the sphere is being measured. The projected area of the sphere would beobtained by measuring the sphere diameter and then calculating the area. Thepressure transducer is placed outside the wind tunnel. Blockage e!ects could also beaddressed in the design of this experiment.

Another design consideration that could be addressed is size of sphere. It should belarge enough to get measurable drag readings but not so large as to produce signicantblockage.

998

PROBLEM 11.6

Situation: A runner is competing in a 10 km race.Running speed is a 6:30 pace (i.e. each mile takes six minutes and 30 seconds). Thus,, = 4!127m$ s!The product of frontal area and coe"cient of drag is -9I = 8!0 ft

2 = 0!743m2!One “food calorie” is equivalent to 4186 J.

Find: Estimate the energy in joules and kcal (food calories) that the runner needs tosupply to overcome aerodynamic drag.

Properties: Density of air is 1!22 kg$m3!

Assumptions: Assume that the air is still—that is, there is no wind.

APPROACH

Energy is related to power (* ) and time (P) by N = *P. Find power using theproduct of speed and drag force (* = , HDrag) ! Find time by using distance (A) andspeed (A = , P) !

ANALYSIS

Find the time to run 10 km!

P =A

,

=10' 000m

4!127m$ s

= 2423 s (40min and 23 s)

Drag force

HDrag = -9IRef

µ#, 2

2

¶

=¡0!743m2

¢Ã(1!22 kg$m3) (4!127m$ s)2

2

!

= 7!72N

Power

* = HDrag,

= (7!72N) (4!127m$ s)

= 31!9W

Energy

N = *P

= (31!9 J$ s) (2423 s)

= 77!2 kJ

999

Energy = 77!2 kJ = 18!4 Food Calories

COMMENTS

1. The drag force (7!72N) is small, about 1!7 lbf!

2. The power to overcome drag is small (31!9W) ! Based on one of the author’s(DFE) experience in sports, a t runner might supply 180W to run at a 6:30pace. Thus, the power to overcome drag is about 1/6 of the total power thatthe runner supplies.

3. The energy that the runner expends (18!4 Food Calories) can be acquired byeating a small amount of food. For example, a small piece of candy. .

1000

PROBLEM 11.7

Situation: Wind (,! = 35m$ s) acts on a tall smokestack.Height is M = 75m! Diameter is 1 = 2!5m!

Find: Overturning moment at the base.

Assumptions: Neglect end e!ects—that is the coe"cient of drag from a cylinder ofinnite length is applicable.

Properties: Air at 20 !C from Table A.3: # = 1!2 × 99$101!3 = 1! 17 kg$m3, : =1!51× 10"5m2$ s!

ANALYSIS

Reynolds number

Re =,!1

:

=(35m$ s)× (2!5m)1!51× 10"5m2$ s

= 5!79× 106

Drag forceFrom Fig. 11.5 -9 ! 0!62 so

H9 = -9I,#, 202

= 0!62× (2!5× 75m2)×(1! 17 kg$m3)× (35m$ s)2

2= 83!31 kN

Equilibrium. Sketch a free-body diagram of the stack—the overturning moment +!

is

+! = M$2× H9+! = (75$2) m× (83!31 kN)

= 3.12 MN·m

1001

PROBLEM 11.8

Situation: Wind acts on a ag pole. Additional details are provided in the problemstatement.

Find: Moment at bottom of ag pole.

Properties: From Table A.3 : = 1!51× 10"5 m2$s and # = 1!20 kg/m3!

ANALYSIS

Reynolds number

Re = , 1$: = 25× 0!10$(1!51× 10"5) = 1!66× 105

Drag forceFrom Fig. 11-5: -9 = 0!95 so the moment is

+ = H94$2 = -9I,#(,20 $2)×4$2

= 0!95× 0!10× (352$2)× 1!2× 252$2

= 21.8 kN·m

1002

PROBLEM 11.9

Situation: Flow from 2 to 6 m3$ s though a 50cm diameter pipe.

Find: Design a ow measuring device that consists of a small cup attached to acantilevered support.

ANALYSIS

The cup, sphere or disk should probably be located at the center of the pipe (asshown below) because the greatest velocity of the air stream in the pipe will be atthe center.

cup

Force dynamometer

Streamlined support strut

You want to correlate , and ] with the force acting on your device. First, neglectingthe drag of the support device, the drag force is given as

H9 = -9I,#,20 $2

or ,0 = (2H9$(-9I,))1*2

You can measure temperature, barometric pressure, and gage pressure in the pipe.Therefore, with these quantities the air density can be calculated by the equationof state. Knowing the diameter of the cup, sphere or disk you can calculate I,!Assume that -9 will be obtained from Table 11.1 or Fig. 11.11. Then the otherquantity that is needed to estimate ,0 is the drag H9. This can be measured by aforce dynamometer as indicated on the sketch of the device. However, the supportstrut will have some drag so that should be considered in the calculations. Anotherpossibility is to minimize the drag of the support strut by designing a housing to taround, but be separate from the vertical part of the strut thus eliminating most ofthe drag on the strut. This was also suggested for Problem 11.5.

Once the centerline velocity is determined it can be related to the mean velocity inthe pipe by Table 10.1 from which the ow rate can be calculated. For example, ifthe Reynolds number is about 105 then , $,max ! 0!82 (from Table 10.1) and

] = , I

] = 0!82,maxI

There may be some uncertainty about -9 as well as the drag of the support rod;therefore, the device will be more reliable if it is calibrated. This can be done as

1003

follows. For a given ow make a pitot-tube-velocity-traverse across the pipe fromwhich ] can be calculated. Also for the given run measure the force on the forcedynamometer. Then plot H vs. ]. Do this for several runs so that a curve of H vs.] is developed (calibration completed).

1004

PROBLEM 11.10

Situation: Wind acts on a cooling tower. Height is 4 = 350 ft!Average diameter is 1 = 250 ft! Wind speed is ,! = 200 mph = 293!3 ft$ s!

Find: Drag (H9) acting on the cooling tower.

Properties: Air at 60 !F (Table A.3) has properties of # = 0!00237 slugs/ft3; : =1!58× 10"4 ft2/s.

Assumptions: 1.) Assume the coe"cient of drag of the tower is similar to the coe"-cient of drag for a circular cylinder of innite length (see Fig. 11.5).2.) Assume the coe"cient of drag for a cylinder is constant at high Reynolds numbers.

ANALYSIS

Reynolds number

Re =,!1

:

=293!3× 2501!58× 10"4

= 4! 641× 108

From Fig. 11-5 (extrapolated) -9 ! 0!70! The drag force is given by

H9 = -9IRef#, 2

2

= 0!70× (250 ft× 350 ft)¡0!00237 slugs/ft3

¢(293!3 ft$ s)2

2

= 6! 244× 106slug · fts2

H9 = 6! 24× 106 lbf

1005

PROBLEM 11.11

Situation: A cylindrical rod is rotated about its midpoint–additional details areprovided in the problem statement.

Find: a) Derive an equation for the power to rotate rod.b) Calculate the power.

ANALYSIS

For an innitesimal element, A3' of the rod

AH9 = -9(A3)A#,2rel.$2

where ,rel. = 3J! Then

A& = 3AH9 = -9#A(,2rel.$2)3A3

&total = 2

Z K0

0

A& = 2

Z K0

0

-9A#((3J)2$2)3A3

&total = -9A#J2

Z K0

0

33A3 = -9A#J2340$4

but 30 = 2$2 so

&total = -9A#J224$64

or

* = &J = -9A#J324$64

d

r ro

&

& r

Then for the given conditions:

* = 1!2× 0!02× 1!2× (50)3 × 1!54$64

= 285 W

1006

PROBLEM 11.12

Situation: A ping-pong ball is supported by an air jet.Mass of the ball is / = 2!6× 103 kg!Diameter of the ball is 1 = !038m! Air temperature is & = 18 !C = 291!2K. Airpressure is % = 27 inches-Hg. = 91!4 kPa!

Find: The speed of the air jet.

Properties: Gas constant for air from Table A.2 is 287 J$ kg · K! Air from Table A.3:7 = 1!80× 10"5N · s$m2!

Assumptions: Assume the ping-pong ball is stationary (stable equilibrium).

APPROACH

For the ball to be in equilibrium, the drag force will balance the weight. Relatethe drag force to the speed of the air and apply the Cli! and Gauvin correlation toestimate the coe"cient of drag. Solve the resulting system of equation to nd thespeed of the air jet.

ANALYSIS

Ideal gas law

# =%

"&

=91' 400Pa

(287 J$ kg · K) (291!2K)= 1!094 kg$m3

Equilibrium/) = HDrag (1)

Drag force

HDrag = -9IRef

µ#, 2

2

¶

= -9

µ012

4

¶µ#, 2

2

¶(2)

Cli! and Gauvin correlation (drag on a sphere)

-9 =24

Re9

¡1 + 0!15Re046879

¢+

0!42

1 + 4!25× 104Re"1416(3)

Reynolds Number

Re =, 1#

7(4)

1007

Solve Eqs. (1) to (4) simultaneously. The computer program TKSolver was used forour solution.

Re = 21' 717

HDrag = 0!026N

-9 = 0!46

, = 9!45m$ s

,jet = 9!45m$ s

1008

PROBLEM 11.13

Situation: Vortices are shed from a agpole–additional details are provided in theproblem statement.

Find: Frequency of vortex shedding

ANALYSIS

From Problem 11.8 Re = 1!66× 105! From Fig. 11-10 6P = 0!21

6P = QA$,0

or

Q = 6P,0$A

= 0!21× 25$0!1 = 52.5 Hz

1009

PROBLEM 11.14

Situation: Wind acts on a billboard–additional details are provided in the problemstatement.

Find: Force of the wind.

Properties: From Table A.3 : = 1!58× 10"4 ft2/s; # = 0!00237 slugs/ft3!

ANALYSIS

Reynolds number

,0 = 50 mph = 73 ft/s

Re = ,0>$:

= 73× 10$(1!58× 10"4)= 4!6× 106

Drag forceFrom Table 11-1 -9 = 1!19! Then

H9 = -9I,#,20 $2

= 1!19× 300× 0!00237× 732$2

= 2250 lbf

1010

PROBLEM 11.15

Situation: A 8 ft by 8 ft plate is immersed in a ow of air (60 !F).Wind speed is ,! = 100 ft$ s! Flow direction is normal to the plate.

Find: Drag force on the plate.

Properties: From Table A.3 for air at 60 !F: # = 0!00237 slugs/ft3!

APPROACH

Apply drag force equation.

ANALYSIS

From Table 11-1,-9 = 1!18

Drag force

H9 = -9I,

µ#, 202

¶

H9 = (1!18)(8× 8)(0!00237)(1002)

2

H9 = 895 lbf

1011

PROBLEM 11.16

Situation: A 2m by 2m square plate is towed through water , = 1m$ s! Theorientation is (a) normal and then (b) edgewise.

Find: Ratio of drag forces (normal to edgewise orientation).


ANALYSIS

Drag force

Hedge = 2-(I#,2$2

Hnormal = -9I#,2$2

Then

Hnormal$Hedge = -9$2-(

Re = ReE = , D$: = 1× 2$(1!31× 10"6)= 1!53× 106

From Fig. 9-14 -( = 0!0030 and from Table 11-1 -9 = 1!18! So

Hnormal$Hedge = 1!18$(2× 0!0030) = 197

1012

PROBLEM 11.17

Situation: A round disk (1 = 0!5m) is towed in water (, = 3m$ s)!The disk is oriented normal to the direction of motion.

Find: Drag force.

APPROACH

Apply the drag force equation.

ANALYSIS

From Table 11.1 (circular cylinder with _$A = 0)

-9 = 1!17

Drag force

H9 = -9I,

µ#, 202

¶

= 1!17

µ0 × 0!52

4

¶µ1000× 32

2

¶

= 1033! 8N

H9 = 1030N

1013

PROBLEM 11.18

Situation: A circular billboard is described in the problem statement.

Find: Force on the billboard.

Properties: From Table A.3, # = 1!25 kg/m3!

APPROACH


ANALYSIS

Drag forceFrom Table 11.1 -9 = 1!17

H9 = -9I,#,2$2

= 1!17× (0$4)× 62 × 1!25× 302$2 = 18,608 N

= 18.6 kN

1014

PROBLEM 11.19

Situation: Wind acts on a sign post (see the problem statement for all the details).

Find: Moment at ground level.

Properties: From Table A.3 # = 1!25 kg/m3!

ANALYSIS

Drag forceFrom Table 1.1 -9 = 1!18 Then

+ = 3× H9 = 3× -9I,#, 2$2= 3× 1!18× 22 × 1!25× 402$2

= 14.16 kN·m

1015

PROBLEM 11.20

Situation: A truck carries a rectangular sign.Dimensions of the sign are 1!83m by 0!46m!Truck speed is , = 25m$ s!

Find: Additional power required to carry the sign.

Assumptions: Density of air # = 1!2 kg/m3!

APPROACH

Apply the drag force equation. Then, calculate power as the product of force andspeed.

ANALYSIS

Drag forceFrom Table 11-1 for a rectangular plate with an aspect ratio of _$A = 3!98:

-9 % 1!20

Drag Force

H9 = -9I,#,2$2

= 1!2× 1!83× 0!46× 1!2× 252$2= 379 N

Power

* = H9 × ,= 379× 25

* = 9!47 kW

1016

PROBLEM 11.21

Situation: A cartop carrier is used on an automobile (see the problem statement forall the details).

Find: Additional power required due to the carrier.

Assumptions: Density, # = 1!20 kg/m3. -9 will be like that for a rectangular plate:L$> = 1!5$0!2 = 7!5

ANALYSIS

From Table 11-1-9 ! 1!25

The air speed (relative to the car) is

, = 100 km/hr

= 27! 78 m/s

The additional power is!* = H9,

Substituting drag force

!* = -9I,(#,2$2),

= 1!25× 1!5× 0!2× 1!20× 27!782$2× 80000$3600= 3.86 kW

1017

PROBLEM 11.22

Situation: The problem statement describes motion of an automobile.

Find: Percentage savings in gas mileage when travelling a 55 mph instead of 65 mph.

ANALYSIS

The energy required per distance of travel = H × 9 (distance). Thus, the energy, N,per unit distance is simply the force or

N$9 = H

Substituting drag force

N$9 = 7×. + -9I,#,2$2

N$9 = 0!02× 3' 000 + 0!3× 20× (0!00237$2), 2

For

, = 55 mph = 80.67 ft/sec

N$9 = 106!3 ft-lbf

For

, = 65 mph = 95.33 ft/sec

N$9 = 124!6 ft-lbf

Then energy savings are

(124!6" 106!3)$124!6 = 0!147 or 14.7%

1018

PROBLEM 11.23

Situation: A car (. = 2500 lbf) coasting down a hill (Slope = 6%)has reached steadyspeed.7rolling = 7 = 0!01

-9 = 0!32 I\ = 20 ft2

#air = # = 0!002 slug$ ft3

Find: Maximum coasting speed.

ANALYSIS

Slope of a hill is rise over run, so the angle of the hill is

tan K = 0!06

or

K = arctan (0!06)

= 0!0599 rad = 3!43!

Equate forcesH9 + HK =. × sin 3!43o

where H9 =drag force, HK =rolling friction and . =weight of car!

Insert expressions for drag force and rolling friction.

-9I,#,2$2 +. × 0!01× cos 3!43! =. × sin 3!43!

, 2 =2. (sin 3!43! " !01× cos 3!43!)

-9I,#

, 2 =2× 2500(0!0599" 0!00998)

0!32× 20× 0!002= 1! 95× 104 ft2/s2

, = 139!6 ft/s = 95.2 mph

1019

PROBLEM 11.24

Situation: The problem statement describes a car being driven up a hill


ANALYSIS

The power required is the product of the forces acting on the automobile in thedirection of travel and the speed. The drag force is

H9 =1

2#, 2-9I =

1

2× 1!2× 302 × 0!4× 4 = 864 N

The force due to gravity is

HD =+) sin 3o = 1000× 9!81× sin 3o = 513 N

The force due to rolling friction is

HK = 7+) cos 3o = 0!02× 1000× 9!81× cos 3o = 196 N

The power required is

* = (H9 + HK + H(), = 1573× 30 = 47.2 kW

1020

PROBLEM 11.25

Situation: The problem statement describes a car traveling on a level road.


ANALYSIS

Power* = H,

where H = H9 + HK!Drag force

H9 = -9I,#,20 $2

= 0!4× 2× 1!2× 402$2= 972 N

Friction force

HK = 0!02 .

= 0!02× 10' 000 N= 200 N

Power

* = (972 + 200)× 30= 35.2 kW

1021

PROBLEM 11.26

Situation: The problem statement describes the wind force on a person.

Find: Wind force (the person is you).

Assumptions: -9 is like a rectangular plate: -9 ! 1!20! Height is 1.83 meters;width is .3 meters.

APPROACH

Apply the ideal gas law, then the drag force equation.

ANALYSIS

Ideal gas law

# = %$"&

= 96' 000$(287× (273 + 20))= 1!14 kg/m3

Drag force

H9 = -9I,#,2$2

= 1!2× 1!83× 0!30× 1!14× 302$2

= 338 N

COMMENTS

1. H9 will depend upon -9 and dimensions assumed.

1022

PROBLEM 11.27

Situation: A boxcar is described in the problem statement.

Find: Speed of wind required to blow boxcar over.

Assumptions: & = 10!-; # = 1!25 kg/m3!

ANALYSIS

Take moments about one wheel for impending tipping.X

+ = 0

Fwind

W

W

0.72 m2.51 m

. × 0!72" H9 × 2!51 = 0

H9 = (190' 000× 1!44$2)$2!51 = 54' 500 N = -9I,#,2$2

From Table 11-1, assume -9 = 1!20! Then

, 2 = 54' 500× 2$(1!2× 12!5× 3!2× 1!25)

,=42.6 m/s

1023

PROBLEM 11.28

Situation: A bicyclist is coasting down a hill–additional details are provided in theproblem statement

Find: Speed of the bicycle.

ANALYSIS

Consider a force balance parallel to direction of motion of the bicyclist:

XH = 0

+Hwgt. comp. " H9 " Hrolling resist. = 0

. sin 8! " -9I,#, 2F.$2" 0!02 . cos 8! = 0

. sin 8! " 0!5× 0!5× 1!2, 2F$2" 0!02. cos 8! = 0

. = 80) = 784!8 N

. sin 8! = 109!2 N

. cos 8! = 777!2 N

Then109!2" 0!15, 2F " !02× 777!2 = 0

,F = 25!0 m/s = ,bicycle + 5 m/s

Note that 5 m/s is the head wind so the relative speed is ,bicycle + 5!

,bicycle=20.0 m/s

1024

PROBLEM 11.29

Situation: A bicyclist is traveling into a 3 m/s head wind. Power of the cyclist is* = 175W!Frontal area is I, = 0!5m3! Coe"cient of drag is -9 = 0!3!

Find: Speed of the bicyclist.

Properties: Air density is 1!2 kg$m3!

APPROACH

The drag force depends on the wind speed relative to the cyclist. Use this fact, andapply the power and drag force equation to give a cubic equation.

ANALYSIS

Drag force

H9 = -9I,

µ#, 2F2

¶

,F = (,1 + 3)

H9 = -9I,

Ã# (,1 + 3)

2

2

!

Power

* = H9 × ,1

= -9I,

Ã# (,1 + 3)

2

2

!,1

175 = 0!3× 0!5

Ã1!2 (,1 + 3)

2

2

!,1

Solving the cubic equation (we used a computer program) for speed gives two complexroots and one real root: ,1 = 10! 566!

,1 = 10!6 m$ s

1025

PROBLEM 11.30

Situation: The problem statement describes a sports car with (a) the roof closed and(b) the roof open

Find: (a) Maximum speed with roof closed.(b) Maximum speed with roof open.

Properties: From Table A.3 # = 1!2 kg/m3!

ANALYSIS

* = H, = (7roll+) + -9I,#,20 $2),

* = 7roll+) ,0 + -9I,#,30 $2

Then

80' 000 = 0!05× 800× 9!81, + -9 × 4× (1!2$2), 3

80' 000 = 392!4, + 2!40-9,3

Solving with -9 = 0!30 (roof closed) one nds

,=44.3 m/s (roof closed)

Solving with -9 = 0!42 (roof open) one nds

,=40.0 m/s (roof opened)

1026

PROBLEM 11.31

Situation: An automobile is traveling into a head wind–additional details are pro-vided in the problem statement.

Find: Velocity of the head wind.

Assumptions: Gas consumption is proportional to power.

ANALYSIS

Gas consumption is proportional to H9, where , is the speed of the automobile andH9 is the total drag of the auto (including rolling friction).Drag force

H9 = -9I,#,20 $2 + 0!1+)

= 0!3× 2× 1!2, 20 $2 + 0!1× 500× 9!81= 0!360, 20 + 490!5 N

,0Istill air = (90' 000$3' 600) = 25!0 m/s

Then

H9Istill air = 0!36× 252 = 490!5 = 715!5 N*still air = 715!5× 25 = 17!89 kW

*head wind = 17' 890× 1!20 = (0!36, 20 + 490!5)(25)

where

,0 = ,headwind + 25 = 32 m/s

,headwind=7 m/s

1027

PROBLEM 11.32

Situation: The problem statement describes a 1932 Fiat Balillo that is “souped up”by the addition of a 220-bhp engine.

Find: Maximum speed of “souped up” Balillo.

ANALYSIS

From Table 11.2, -9 = 0!60!

* = (H9 + HK),

, = 60 mph = 88 ft/s

HK = (*$, )" H9 = (*$, )" -9I,#, 2$2= ((40)(550)$88)" (0!60)(30)(0!00237)(882)$2= 250" 165 = 85 lbf

“Souped up” version:

(H9 + 85), = (220)(550)

((-9I,#,2$2) + 85), = (220)(550)

(-9I,#,3$2) + 85, = (220)(550)

0!0213, 3 + 85, " 121' 000 = 0

Solve for , :

, = 171!0 ft/s

= 117 mph

1028

PROBLEM 11.33

Situation: To reduce drag, vanes are added to truck–additional details are providedin the problem statement.

Find: Reduction in drag force due to the vanes.

Assumptions: Density, # = 1!2 kg/m3!

APPROACH


ANALYSIS

H9 = -9I,#,2$2

H9reduction = 0!25× 0!78× 8!36× 1!2(100' 000$3' 600)2$2

H9reduction = 755 N

1029

PROBLEM 11.34

Situation: The problem statement describes a dirigible.

Find: Power required for dirigible.

ANALYSIS

Reynolds number

Re = ,0A$: = (25)(100)$(1!3× 10"4) = 1!92× 107

Drag force

From Fig. 11.11 (extrapolated) -9 = 0!05

H9 = -9I,#,20 $2

= (0!05)(0$4)(1002)(0!07$32!2)(252)$2

= 267 lbf

Power

* = H9,0

= (267)(25)

= 6,670 ft-lbf/s = 12.1 hp

1030

PROBLEM 11.35

Situation: To reduce drag, vanes are added to truck–additional details are providedin the problem statement.

Find: Percentage savings in fuel.

Assumptions: Density, # = 1!2 kg/m3!

ANALYSIS

Assume that the fuel savings are directly proportional to power savings.

* = H,

* = -9 × 8!36× 1!2, 3$2 + 450,

At 80 km/hr:

*w/o vanes = 0!78× 8!36× 1!2, 3$2 + 450, = 52!9 kW*with vanes = 42!2 kW

which corresponds to a 20.2% savings.At 100 km/hr:

*w/o vanes = 96!4 kW

*with vanes = 75!4 kW

which corresponds to a 21.8% savings.

1031

PROBLEM 11.36

Situation: A train is described in the problem statement.

Find: Percentage of resistance due to bearing resistance, form drag and skin frictiondrag.

Assumptions: Density, # = 1!25 kg/m3 and velocity, : = 1!41× 10"5 m2$s.

ANALYSIS

Drag force

H9form = -9I,#,20 $2

H9form = 0!80× 9× 1!25× , 20 $2 = 4!5,20

H9sk in = -(I#,20 $2

Reynolds number

ReE = , 2$: = , × 150$(1!41× 10"5)ReEI100 = (100' 000$3' 600)× 150$(1!41× 10"5) = 2!9× 108

ReEI200 = 5!8× 108

From Eq. (9.54), -(I100 = 0!00188; -(I200 = 0!00173!

, = 100 km/hr , = 200 km/hrH9Iform,100 = 3' 472 N H9Iform,200 = 13' 889 NH9,skin,100 = 1' 360 N H9Iskin,200 = 5006 NHbearing = 3' 000 N Hbearing = 3' 000 NHtotal = 7' 832 N Htotal = 21' 895 N

44% form, 17% skin, 39% bearing 63% form, 23% skin, 14% bearing

1032

PROBLEM 11.37

Situation: Viscosity of liquids—water, kerosene, glycerin.

Find: (b) Design equipment to measure the viscosity of liquids using Stoke’s law.(b) Write instructions for use the equipment.

ANALYSIS

Stoke’s law is the equation of drag for a sphere with a Reynolds number less than 0.5:

H9 = 307,0A

or 7 = H9$(30,0A)

One can use this equation to determine the viscosity of a liquid by measuring the fallvelocity of a sphere in a liquid. Thus one needs a container to hold the liquid (forinstance a long tube vertically oriented). The spheres could be ball bearings, glassor plastic spheres. Then one needs to measure the time of fall between two points.This could be done by measuring the time it takes for the sphere to drop from onelevel to a lower level. The diameter could be easily measured by a micrometer andthe drag, H9' would be given by

H9 =. " Hbuoyant

If the specic weight of the material of the sphere is known then the weight of thesphere can be calculated. Or one could actually weigh the sphere on an analyticbalance scale. The buoyant force can be calculated if one knows the specic weightof the liquid. If necessary the specic weight of the liquid could be measured with ahydrometer.

To obtain a reasonable degree of accuracy the experiment should be designed so thata reasonable length of time (not too short) elapses for the sphere to drop from onelevel to the other. This could be assured by choosing a sphere that will yield a fairlylow velocity of fall which could be achieved by choosing to use a small sphere overa large one or by using a sphere that is near the specic weight of the liquid (forinstance, plastic vs. steel).

COMMENTS

1. Other items that should be or could be addressed in the design are:

A. Blockage e!ects if tube diameter is too small.B. Ways of releasing sphere and retrieving it.C. Possibly automating the measurement of time of fall of sphere.D. Making sure the test is always within Stoke’s law range (Re G 0!5)E. Making sure the elapsed time of fall does not include the time when the

sphere is accelerating.

1033

PROBLEM 11.38

Situation: A 1-ft diameter sphere moves through oil–additional details are providedin the problem statement.

Find: Terminal velocity.

APPROACH

Apply equilibrium involving the weight, drag force and buoyant force.

ANALYSIS

Buoyancy force

Hbuoy. = , (oil= (4$3)0 × (1$2)3 × 0!85× 62!4= 27!77 lbf

Under non-accelerating conditions, the buoyancy is equal to the drag force plus theweight.

Fbuoy.

W+FD

H9 = ". + Hbuoy.

= "27!0 + 27!77 lbf= 0!77 lbf upward

Assume laminar ow. Then

H9 = 0!77 = 3071,0

,0 = 0!77$(3017)

,0 = 0!77$(30 × 1× 1)

,0 = 0.082 ft/s upward

Check laminar ow assumption with Reynolds number

Re = ,0A#$7 = 0!082× 1× 1!94× 0!85$1= 0!14 G0.5

Therefore the assumption is valid.

1034

PROBLEM 11.39

Situation: A sphere 2 cm in diameter rises in oil at a velocity of 1.5 cm/s.

Find: Specic weight of the sphere material.

APPROACH

Apply equilibrium to balance the buoyant force with the drag force and weight.

ANALYSIS

Equilibrium

XH = 0 = "H9 ". + Hbuoyancy

H9 = Hbuoyancy ". (1)

Reynolds number

Re =, 1#

7

=0!015× 0!02× 900

0!096= 2!812

Then from Fig. 11.11-9 ! 10!0

Substitute drag force, weight and Buoyancy force equations into Eq. (1)

-9I,#,20 $2 = , ((oil " (sphere) (2)

Sphere volume is

, = (4/3)033

= 4!19× 10"6 m3

Eq. (2) becomes

10× 0 × 0!012 × 900× 0!0152$2 = 4!19× 10"6(900× 9!81" (sphere)(sphere = 8753N$m3

(sphere = 8753N/m3

1035

PROBLEM 11.40

Situation: The problem statement describes a 1.5-mm sphere moving in oil.

Find: Terminal velocity of the sphere.

APPROACH

Apply the equilibrium principle. To nd the drag force, assume Stokes drag.

ANALYSIS

Equilibrium. Since the ball moves at a steady speed, the sum of forces is zero.

. = H> + H9 (1)

where . is weight, H> is the buoyant force and H9 is drag.

Because the viscosity is large, it is expected that the sphere will fall slowly, so assumethat Stoke’s law applies. Thus, the drag force is

H9 = 307,01

= 30:#,01

Buoyant force

H> = (oil

µ013

6

¶

Equilibrium (Eq. 1) becomes

. = H> + H9

(sphere

µ013

6

¶= (oil

µ013

6

¶+ 30:#,01

µ013(water

6

¶(6sphere " 6oil) = 30:#,01

Ã0 (0!0015m)3 × 9810N$m3

6

!(1!07" 0!95) = 30:#,01

2! 080× 10"6N = 30¡10"4m2$ s

¢ ¡950 kg$m3

¢,! (0!0015m)

2! 080× 10"6N =¡1! 343× 10"3 kg$ s

¢,!

The solution is,! = 1!55mm$ s


Re =,01

:

=(0!00155m$ s)× (0!0015m)

10"4m2$ s= 0!023 25

1036

COMMENTS

The value of Re is within Stokes’ range (Re ' 0!5), so the use of Stokes’ law is valid.

1037

PROBLEM 11.41

Situation: A 2cm plastic ball with specic gravity of 1.2 is released from rest in water(T=20 !C)–additional details are provided in the problem statement.

Find: Time and distance to achieve 99% of terminal velocity.

ANALYSIS

The equation of motion for the plastic sphere is

/A;

AP= "H9 +. " H>

The drag force can is expressed as

H9 =1

2#;2-9

0

4A2 =

-9Re

24307A;

The equation of motion becomes

/A;

AP= "

-9Re

24307A; + #0() " #%)(

Dividing through by the mass of the ball gives

A;

AP= "

-9Re

24

187

#0A2; + )(1"

#%#0)

Substituting in the values

A;

AP= "0!0375

-9Re

24; + 1!635

Eq. 11.10 can be rewritten as

-9Re

24= 1 + 0!15Re04687 +

0!0175Re

1 + 4!25× 104Re"1416

This equation can be integrated using the Euler method

;:+1 = ;: +

µA;

AP

¶

:

!P

9:+1 = 9: + 0!5(;: + ;:+1)!P

The terminal velocity is 0.362 m/s. The time to reach 99% of the terminal velocityis 0.54 seconds and travels 14.2 cm .

1038

PROBLEM 11.42

Situation: A small air bubble is rising in a very tall column of liquid—additional detailsare provided in the problem statement.

Find: (a)Acceleration of the bubble.(b)Form of the drag (mostly skin-friction or form).

ANALYSIS

Equating the drag force and the buoyancy force.

H9 = -1(liq.13 = -21

3

AlsoH9 = -9I,#,

2$2 = -312, 2

Eliminating H9 between these two equations yields

, 2 = -41 or , =p-41

As the bubble rises it will expand because the pressure decreases with an increasein elevation; thus, the bubble will accelerate as it moves upward. The drag willbe form drag because there is no solid surface to the bubble for viscous shear stressto act on.

COMMENTS

As a matter of interest, the surface tension associated with contaminateduids createsa condition which acts like a solid surface.

1039

PROBLEM 11.43

Situation: A 120 lbf (534N) skydiver is free-falling at an altitude of 6500 ft (1981m).Maximum drag conditions: -9I = 8 ft2 (0.743 m2)!Minimum drag conditions, -9I = 1 ft

2 (0.0929 m2)!Pressure and temperature at sea level are 14.7 psia (101 kPa) and 60 !F (15 !C).Lapse rate for the U.S. Standard atmosphere is T = 0!00587K$m!

Find: Estimate the terminal velocity in mph.a.) Case A (maximum drag) -9I = 8 ft

2 (0!743m2)!b.) Case B (minimum drag) -9I = 1 ft2 (0!0929m2)!

APPROACH

At terminal velocity, the force of drag will balance weight. The only unknown is uiddensity—this can be found by using the ideal gas law along with the equations fromchapter 3 that describe the US Standard atmosphere. Use SI units throughout.

ANALYSIS

Atmospheric pressure variation (troposphere)

& = &! " T(R " R!)= (288!1K)" (0!00587K$m)× (1981" 0) m= 276!5K

%

%!=

·&! " T(R " R!)

&!

¸ 678

%

101 kPa=

·276!5K

288!1K

¸ 9481(0400587)(287)

so% = 79!45 kPa

Ideal gas law

# =%

"&

=79' 450

287× 276!5= 1!001 kg$m3

Equilibrium

Weight = Drag

Case A

. = -9I#, 202

534N =¡0!743m2

¢ (1!001 kg$m3), 202

1040

Calculations give,B = 37!9m$ s

,B = 84!7 mph for maximum drag conditions

Case B.Since -9I decreases by a factor of 8, the speed will increase by a factor of

&8!

,B = (84!7mph)&8

,B = 240 mph for minimum drag conditions

1041

PROBLEM 11.44

Situation: Assume Stoke’s law is valid for a Reynolds number below 0.5.

Find: Largest raindrop that will fall in the Stokes’ ow regime.

Assumptions: &air = 60!H ; #air = 0!00237 slugs/ft3; 7air = 3!74× 10"7 lbf-sec/ft2!

APPROACH

Apply Stoke’s law and the equilibrium principle.

ANALYSIS

Drag force is

H9 = 307,01

The equilibrium principle is

013(water6

= 307air,01

12(water = 187air,0

Reynolds number limit for Stokes ow

,01$: = 0!5

,0 =0!5 :air1

Combining equations

12(water = 187air

µ0!5 :air1

¶

13 = 97air:air

1(water

Solving for 1

13 =972air

#air(water

=9× (3!74× 10"7)2

0!00237× 62!4= 8!51× 10"12 ft3

1 = 2!042× 10"4 ft= 0.0024 in.

1042

PROBLEM 11.45

Situation: A falling hail stone is described in the problem statement.

Find: Terminal velocity of hail stone.

APPROACH

Apply the ideal gas law, then calculate the drag force and apply the equilibriumprinciple.

ANALYSIS

Ideal gas law# = %$"& = 96' 000$(287× 273) = 1!23 kg/m3

Equilibrium

XH = 0 = H9 ".H9 = .

Substitute for drag force and weight

-9I,#,2$2 = , U_ × 6' 000

Assume -9 = 0!5

0!5× (0A2$4)× 1!23, 2$2 = (1$6)0A3 × 6' 000, =

pA× 1' 000× 16$1!23

, =p5× 16$1!23 = 8!06 m/s


Re = 8!06× 0!005$(1!3× 10"5)= 3100

From Fig. 11-11 -9 = 0!39 so

, = 8!06× (0!5$0!39)1*2

= 9.13 m/s

COMMENTS

The drag coe"cient will not change with further iterations.

1043

PROBLEM 11.46

Situation: The problem statement describes a rock falling in water.

Find: Terminal velocity of the rock.

APPROACH

Apply equilibrium with the drag force and buoyancy force. Use an iterative solutionto nd terminal velocity.

ANALYSIS

Buoyancy force

.air = , (rock35 = , (rock

Hbuoy = (35" 7)= , (water= , × 9790

Solving for (rock and A: (rock = 12' 223 N$m3 and A = 0!1762 m.

Under terminal velocity conditions

H9 + Hbuoy = .

H9 = 35" 28 = 7 N

Drag forceH9 = -9I,#,

20 $2

or

, 20 = 2H9$(-9I,#)

, 20 = 2× 7$(-9 × 0!176220$4× 998)

,0 = 0!575$p-9

Assume -9 = 0!4 so,0 = 0!91 m/s

Calculate the Reynolds number

Re = (, 1$:)

= 0!91(0!176)$10"6

= 1!60× 105

From Fig. 11.11, try -9 = 0!45' ,0 = 0!86 m/s, Re = 1!51 × 105! There will be nochange with further iterations so

, = 0!86 m/s

1044

PROBLEM 11.47

Situation: A drag chute is used to decelerate an airplane–additional details areprovided in the problem statement.

Find: Initial deceleration of aircraft.

Assumptions: Density, # = 0!075 lbm/ft3 = 0!0023 slug$ ft3!

ANALYSIS

Drag forceH9 = -9I,#,

20 $2 =+C

thenC = -9I,#,

20 $(2+)

where + = 20' 000$32!2 = 621!1 slugs. From Table 11.1 -9 = 1!20 .

I, = (0$4)12 = 113!1 ft2

Then

C = 1!20× 113!1× 0!0023× 2002$(2× 621!1)

= 10.5 ft/s2

1045

PROBLEM 11.48

Situation: A paratrooper falls using a parachute—additional details are provided inthe problem statement

Find: Descent rate of paratrooper.

Assumptions: Density, # = 1!2 kg/m3

APPROACH

In equilibrium, drag force balances weight of the paratrooper.

ANALYSIS

Equilibrium

. = H9

Drag Force

H9 = -9I,#, 202

From Table 11.1 -9 = 1!20. Thus

. = H9 = -9I,#,20 $2

,0 =q2.$(-9I,#)

=p2× 900$(1!2× (0$4)× 49× 1!2)

= 5.70 m/s

1046

PROBLEM 11.49

Situation: A weighted wood cylinder falls through a lake (see the problem statementfor all the details).

Find: Terminal velocity of the cylinder.

Assumptions: For the water density, # = 1000 kg/m3!

APPROACH

Apply equilibrium with the drag force and buoyancy force.

ANALYSIS

Buoyancy force

Hbuoy = , (water= 0!80× (0$4)× 0!202 × 9810= 246!5 N

Then the drag force is

H9 = Hbuoy ".= 246!5" 200= 46!5 N

From Table 11-1 -9 = 0!87! Then

46!5 =-9I,#,

20

2or

,0 =

s2× 46!5-9I,#

,0 =

s2× 46!5

0!87× (0$4)× 0!22 × 1000

= 1.84 m/s

1047

PROBLEM 11.50

Situation: A weighted cube falls through water (see the problem statement for all thedetails).

Find: Terminal velocity in water.

Assumptions: Density of water: # = 1000 kg/m3!

ANALYSIS

Drag forceFrom Table 11-1, -9 = 0!81

H9 = -9I,#,20 $2

I, = (2)(2 cos 45!)(2) = 1!41422

Equilibrium

H9 = . " Hbuoy= 19!8" 9' 81023 = 19!8" 9' 810× (10"1)3 = 10 N

10 = (0!81)(1!414× 10"2)(1' 000)(, 20 )$2

,0=1.32 m/s

1048

PROBLEM 11.51

Situation: A helium-lled balloon moves through air (see the problem statement forall the details).

Find: Terminal velocity of the balloon.

Properties: at & = 15!-: #air % 1!22 kg/m3; #He = 0!169 kg/m3

APPROACH

Apply equilibrium with the weight, drag force and buoyancy force.

ANALYSIS

Velocity from drag force,0 = (2H9$(-9I#))

1*2

Equilibrium

Hnet = H9 ".balloon ".helium + Hbuoy = 0

H9 = +0!15" (1$6)013((air " (He)= +0!15" (1$6)0 × (0!50)39!81(#air " #He)

H9 = +0!15" (1$6)0(0!50)3 × 9!81(1!22" 0!169) = "0!52 N

Assume -9 ! 0!40 Then

,0 = ((2× 0!52$(0!40× (0$4)× 0!52 × 1!22))1*2

= 3!29 m/s

Check Re and -9:

Re = , 1$: = 3!29× 0!5$(1!46× 10"5) = 1!13× 105

From Fig. 11-11, -9 ! 0!45 so one additional iteration is necessary.

,0 = 3!11 m/s upward

1049

PROBLEM 11.52

Situation: The balloon from problem 11.51

Find: Time for the balloon to reach 5000m in altitude.

Assumptions: Balloon does not change in size. Negligible e!ects of change in viscositywith temperature.

ANALYSIS

The equation of motion is obtained by equating the mass times acceleration to theforces acting on the balloon.

/A;

AP= "H9 ". + H>

The mass of the balloon is the sum of the mass associated with the “empty” weight,.0' and the helium.

/ =.0

)+ #6(

= #6((1 +.

#6())

The drag force can be expressed as

H9 =1

2#;2-9

0

4A2 =

-9Re

24307A;

The buoyancy force is

H> = #')(

Substituting the values into the equation of motion, we have

/A;

AP= "

-9Re

24307A; "/) + #')(

Dividing through by the mass, we get

A;

AP= "

-9Re

24

187

#6A2

1

H; " ) +

#'#6)1

H

where

H = 1 +.

#6()

The density of helium at 23oC and atmospheric pressure is 0.1643 kg/m3! Substitute

1050

A;

AP= "

-9Re

24

0!0219

3!19; " 9!81(1"

#'0!1643× 3!19

)

A;

AP= "

-9Re

240!00686; " 9!81(1"

#'0!524

) (1)

The value for -9Re/24 is obtained from Eq. 11.10.

-9Re

24= 1 + 0!15Re04687 +

0!0175Re

1 + 4!25× 104Re"1416

The value for the air density is obtained from the relations for a standard atmosphere.

& = 296" 5!87× 10"3M

and

% = 101!3(1"&

294)54823

and the density is obtained from the ideal gas law.

Eq (1) can be integrated using the Euler method:

;:+1 = ;: +

µA;

AP

¶

:

!P

M:+1 = M: + 0!5(;: + ;:+1)!P

The time to climb to 5000 m is 3081 seconds or 51.3 minutes . Other methods maylead to slightly di!erent answers.

1051

PROBLEM 11.53

Situation: A helium-lled balloon moves through air (see the problem statement forall the details).

Find: Terminal velocity of balloon.

APPROACH


ANALYSIS

Equilibrium

H9 = Hbuoy ".He ".balloon

Substitute buoyancy force and weight

H9 = "0!01 + (1$6)0 × 13((air " (air × 1716$12' 419)H9 = "0!01 + (1$6)0 × 13 × 0!0764(1" 0!138)H9 = "0!010 + 0!0345

= 0!0245 lbf

Also

,0 =q2H9$(-9I,#) =

p2× 0!0245$((0$4)× 0!00237-9)

=p26!3$-9

Assume -9 = 0!40 Then

,0 =p26!3$0!4 = 8!1 ft/s upward

Check Reynolds number and -9!

Re = , 1$: = 8!1× 1$(1!58× 10"4) = 5!1× 104; -9 = 0!50

From Fig. 11-11, -9 = 0!50! Recalculate velocity

,0 =p26!3$0!5 = 7.25 ft/s

No further iterations are necessary.

1052

PROBLEM 11.54

Situation: A man in a boat is pulling up an anchor–additional details are providedin the problem statement.

Find: Tension in rope to pull up anchor.

Assumptions: Density of water: # = 1000 kg/m3!

APPROACH

Apply equilibrium with the tension, weight, drag force and buoyancy force.

ANALYSIS

EquilibriumX

H3 = 0

& ". " H9 " Hbuoy. = 0

Solve for T

& =. + H9 + Hbuoy.

Substitute drag force, buoyancy force, and weight

& = (0$4)× 0!32 × 0!3(15' 000" 9' 810) +-9(0$4)× 0!32 × 1' 000× 1!02$2

From Table 11-1 -9 = 0!90! Then

& = 110 + 31!8 = 141.8 N

1053

PROBLEM 11.55

Situation: The problem statement describes a small spherical pebble falling throughwater.

Find: Terminal velocity of spherical pebble.

Assumptions: Water: : = 10"5 ft2/s. Pebble: (& = 3!0!

APPROACH

Apply equilibrium to balance buoyancy force, weight and drag force. Guess a coe"-cient of drag and iterate to nd the solution.

ANALYSIS

Assume -9 = 0!5

,0 = [((& " (%)(4$3)1$(-9#%)]1*2

,0 = [62!4(3!0" 1)(4$3)× (1$(4× 12))$(0!5× 1!94)]1*2

,0 = 1!891 ft/s

Reynolds number

Re = 1!891× (1$48)$10"5 = 3940

Recalculate the coe"cient of drag. From Fig. 11-11 -9 = 0!39! Then

,0 = 1!891× (0!5$0!38)1*2 = 2.14 ft/s


1054

PROBLEM 11.56

Situation: A 10-cm diameter ball (weight is 15 N in air) falls through 10!C water.

Find: Terminal velocity of the ball.

APPROACH


ANALYSIS

Equilibrium

H9 = . " HbuoyH9 = 15" 9' 810× (1$6)013 = 15" 9' 810× (1$6)0 × 0!13

= 9!86 N

Buoyant force is less than weight, so ball will drop.

9!86 = -9(012$4)× 1' 000, 2$2

, =p9!86× 8$(0-9 × 1' 000× 0!12) = 1!58$

p-9

Assume -9 = 0!4! Then, = 2!50 m/s

Check Reynolds number and -9!

Re = , 1$: = 2!50× 0!1$(1!3× 10"6) = 1!9× 105

From Fig. 11-11 -9 = 0!48! So

, = 1!58$&0!48

= 2.28 m/s downward

1055

PROBLEM 11.57

Situation: A helium-lled balloon is ascending in air (see the problem statement forall the details).

Find: Ascent velocity of balloon..

APPROACH


ANALYSIS

Equilibrium

0 = ".balloon ".He + Hbuoy " H9H9 = "3 + (1$6)013((air " (He)

= "3 + (1$6)0 × 23 × (air(1" 287$2077)= "3 + (1$6)0 × 8× 1!225(1" 0!138)= "3 + 4!422= 1!422 N

Then drag force

H9 = -9I,#,20 $2

,0 =p1!422× 2$((0$4)× 22 × 1!22-9)

=p0!739$-9

Assume -9 = 0!4 then,0 =

p0!739$0!4 = 1!36 m/s

Check Reynolds number and -9

Re = , 1$: = 1!36× 2$(1!46× 10"5) = 1!86× 105

From Fig. 11-11 -9 = 0!42 so

,0 =p0!739$0!42 = 1.33 m/s upward


1056

PROBLEM 11.58

Situation: A spherical meteor (#meteor = 3000 kg$m3) enters the earth’s atmosphere.

Find: Diameter of the meteor.

Properties: Air: % = 20 kPa & = "55!C.

APPROACH

Apply the drag force equation and equilibrium.

ANALYSIS

H9 = .

-9I,#,20

2= .

-9I,c%+2

2= .

! I, =. × 2-9c%+2

From Fig. 11-12 -9 = 0!80

012

4=

. × 2-9c%+2

=(3000013$6)(9!81)× (2)(0!8)(1!4)(20× 103)(12)

0!785412 = 1!37613

so

1 =0!785 4

1!376

= 0.571 m

1057

PROBLEM 11.59

Situation: A sphere is being sized to have a terminal velocity of 0.5 m/s when fallingin water (20!C).The diameter should be between 10 and 20 cm.

Find: Characteristics of sphere falling in water.

APPROACH

Apply equilibrium with the drag force and buoyancy force.

ANALYSIS

Drag force

H9 = -9I,#,20 $2

((& " (%)0A3$6 = -9(0$4)A

2 × 998, 20 $2

Assume -9 = 0!50! Then(& = (93!56$A) + (%

Now determine values of (& for di!erent A values. Results are shown below for a -9of 0.50

A(cm) 10 15 20 Re= , 1$: = 0!5× 0!1$10"6 = 5× 104(&(N/m

3) 10,725 10,413 10,238 -9 = 0!5 O.K.

1058

PROBLEM 11.60

Situation: A rotating sphere is described in the problem statement.

Find: Lift force on the sphere.

APPROACH

Use data shown in Fig. 11.17. Calculate lift force using coe"cient of lift equation.

ANALYSIS

Rotational 0-group.

3J

,0=

(0!15 ft) (50 rad$ s)

3 ft$ s= 2!50

From Fig. 11-17 -E = 0!43

Lift force

HE = -EI,#,20 $2

HE = (0!43)(0$4)(0!32)(1!94)(32)$2

HE = 0.265 lbf

1059

PROBLEM 11.61

Situation: A spinning baseball is thrown from west to east–additional details areprovided in the problem statement.

Find: Direction the baseball will ”break.”

ANALYSIS

It will ”break” toward the north. The correct answer is a) .

1060

PROBLEM 11.62

Situation: A rotating baseball is described in the problem statement.

Find: (a) Lift force on the baseball.(b) Deection of the ball from its original path.

Properties: From table A.3, # = 0!0023 slugs/ft3!

Assumptions: Axis of rotation is vertical, standard atmospheric conditions (& =70!H )!.

ANALYSIS

Rotational parameter

,0 = 85 mph = 125 ft/s

3J$,0 = (9$(12× 20))× 35× 20$125 = 0!21

From Fig. 11-17 -E = 3× 0!05 = 0!15Lift force

HE = -EI#,20 $2

= 0!15× (9$120)2 × (0$4)× 0!0023× 1252$2= 0.121 lbf

Deection will be i = 1$2 CP2 where C is the acceleration

C = HE$+

P = 2$,0 = 60$125 = 0!48 s

C = HE$+ = 0!121$((5$16)$(32!2)) = 12!4 ft/s2

Theni = (1$2)× 12!4× 0!482 = 1.43 ft

1061

PROBLEM 11.63

Situation: A circular cylinder in a wind tunnel is described in the problem statement.

Find: Force vector required to hold the cylinder in position.

APPROACH

Apply lift force and drag force.

ANALYSIS

Correct choice is force vector a) .

1062

PROBLEM 11.64

Situation: Air speed is being determined in a popcorn popper.Additional information is provided in problem statement.

Find: Range of airspeeds for popcorn popper operation.

Properties: Air properties from Table A.3 at 150!C # = 0!83 kg/m3 and : =2!8× 10"5 m2$s.

ANALYSIS

Before corn is popped, it should not be thrown out by the air. Thus, let

,max =

s2H9

-9I,#air

where H9 is the weight of unpopped corn

H9 = /)

= 0!15× 10"3 × 9!81= 1! 472× 10"3N

The cross-section area of the kernels is

I, = (0$4)× (0!006)2 m2

= 2!83× 10"5m2

Assume -9 w 0!4! Then

,max =

s2H9

-9I,#air

=

r2× 1! 472× 10"3

0!4× 2!83× 10"5 × 0!83= 17.7 m/s

Check Reynolds number and -9:

Re =, 1

:

=17!7× 0!0062!8× 10"5

= 3800

From Fig. 11-11 -9 % 0!4 so solution is valid.

1063

For minimum velocity let popped corn be suspended by stream of air. Assume onlythat diameter changes. So

,min = ,max × (IT$I,)1*2

= ,max1T1,

where1, = diameter of popped corn and1T = diameter of unpopped corn17!7¡6mm18mm

¢=

5! 9

,min w ,max1T1,

= 17!7

µ6mm

8mm

¶

,min=5.9m/ s

1064

PROBLEM 11.65

Situation: Wind loads act on a ag pole that is carrying an 6 ft high American Flag.

Find: Determine a diameter for the pole.

Assumptions: The failure mechanism is yielding due to static loading.

ANALYSIS

An American ag is 1.9 times as long as it is high. Thus

I = 62 × 1!9 = 68!4 ft2

Assume

& = 60!H' # = 0!00237 slugs/ft3

,0 = 100 mph = 147 ft/s

Compute drag force on ag

H9 = -9I#,20 $2

= 0!14× 68!4× 0!00237× 1472$2= 244 lbf

Make the ag pole of steel using one size for the top half and a larger size for thebottom half. To start the determination of A for the top half, assume that the pipediameter is 6 in. Then

Hon pipe = -9I,#,20 $2

Re = , 1$: = 147× 0!5$(1!58× 10"4)= 4!7× 105

With an Re of 4!7×105' -9 may be as low as 0.3 (Fig. 11-5); however, for conservativedesign purposes, assume -9 = 1!0! Then

Hpipe = 1× 50× 0!5× 0!00237× 1472$2 = 640 lbf+ = 244× 50× 12 + 640× 25× 12 = 338' 450 in.-lbf

Assume that the allowable stress is 30' 000 psi.

V

Z=

+

Omax

=338' 450

30' 000

= 11!28 in3

From a handbook it is found that a 6 in. double extra-strength pipe will be ade-quate.

1065

Bottom half, Assume bottom pipe will be 12 in. in diameter.

Hag = 224 lbf

H6 in.pipe = 640 lbf

H12 in.pipe = 1× 50× 1× 0!00237× 1472$2= 1' 280 lbf

+ = 12(244× 100 + 640× 75 + 1' 280× 25)= 1' 253' 000 in.-lbf

+& = 41!8 in.3 = V$Z

Handbook shows that 12 in. extra-strength pipe should be adequate.

COMMENTS

Many other designs are possible.

1066

PROBLEM 11.66

Situation: A plate is angled 30! relative to the direction of an approaching ow. Apressure distribution is specied in the problem statement.

Find: Lift coe"cient on plate.

ANALYSIS

Force normal to plate will be based upon the -,Inet' where -,Inet is the average net -,producing a normal pressure on the plate. For example, at the leading edge of theplate the -,Inet = 2!0+1!0 = 3!0! Thus, for the entire plate the average net -, = 1!5!

Then

Hnormal to plate = -,InetIplate#,20 $2

= 1!5Iplate#,20 $2

The force normal to ,0 is the lift force.

HE = (Hnormal to plate)(cos 30!)

-E6#,20 $2 = (1!5)(Iplate)(#,

20 $2) cos 30

!

-E = 1!5 cos 30! = 1.30

based on plan form area. However if -E is to be based upon projected area where

Iproj = Iplate sin 30! then

-E=2.60

1067

PROBLEM 11.67

Situation: An airplane wing has a chord of 4 ft. Air speed is ,! = 200 ft$ s!The lift is 2000 lbf. The angle of attack is 3!!The coe"cient of lift is specied by the data on Fig. 11.23.

Find: The span of the wing.

Properties: Density of air is 0.0024 slug/ft3!

APPROACH

Guess an aspect ratio, look up a coe"cient of lift and then calculate the span. Then,iterate to nd the span.

ANALYSIS

Lift forceFrom Fig. 11-23 assume -E ! 0!60

HE = -EI#, 202

2000 = (0!60)(4>)(0!0024)(2002)

2> = 17!4 ft

>$Z = 17!4$4 = 4!34

From Fig. 11-23, -E = 0!50! Recalculate the span

> = (17!4 ft)

µ0!60

0!50

¶

= 20!9 ft

> = 20!9 ft

1068

PROBLEM 11.68

Situation: A lifting vane for a boat of the hydrofoil type is described in the problemstatement.

Find: Dimensions of the foil needed to support the boat.

ANALYSIS

Use Fig. 11-23 for characteristics; >$Z = 4 so -E = 0!55

HE = -EI#,20 $2

10' 000 = 0!55× 4Z2 × (1!94$2)× 3' 600Z2 = 1!30 ft

Z = 1.14 ft

> = 4Z = 4!56 ft

Use a foil 1.14 ft wide × 4!56 ft long

1069

PROBLEM 11.69

Situation: Two wings, A and B, are described in the problem statement.

Find: Total lift of wing B compared to wing A.

ANALYSIS

-E increases with increase in aspect ratio. The correct choice is (d) .

1070

PROBLEM 11.70

Situation: An aircraft increases speed in level ight.

Find: What happens to the induced drag coe"cient.

ANALYSIS

-9# =-2E

0(>2$6)

In the equation for the induced drag coe"cient (above) the only variable for a givenairplane is -E; therefore, one must determine if -E varies for the given conditions. Ifthe airplane is in level ight the lift force must be constant. Because HE = -EI#, 2$2it is obvious that -E must decrease with increasing , . This would be accomplishedby decreasing the angle of attack. If -E decreases, then Eq. (11.19) shows that -9#also must decrease. The correct answer is (b) .

1071

PROBLEM 11.71

Situation: An airplane wing is described in the problem statement.

Find: (a) An expression for , for which the power is a minimum.(b) , for minimum power

ANALYSIS

.$6 =1

2#-E,

2

or

-E = (2$#)(1$, 2)(.$6)

* = H9,

= (-9! + -2E$0$)(1$2)#,

36

* =1

2#, 36-9! + (4$#

2)(1$, 4). 2$62)(1$(0$))(1

2#, 36)

* =

·1

2, 3-9! + (2$#)(1$(0$, )(.

2$62)

¸6

A*$A, = ((3$2)#, 2-9! " (2$#)(1$(0$, 2))(.$6)2)6

For minimum power A*$A, = 0 so

(3$2)#, 2-9! = (2$#)(1$(0$, 2)(.$6)2

,=£43(.$6)2(1$(0$#2-90))

¤1*4

For # = 1 kg/m3'$ = 10' .$6 = 600 and -9! = 0!2

, =

·4

3(6002)(1$(0 × 10× 12 × 0!02))

¸1*4

= 29.6 m/s

1072

PROBLEM 11.72

Situation: The airstream a!ected by the sing of an airplane is described in the problemstatement.

Find: Show that -9# = -2E$(0$)!

ANALYSIS

Take the stream tube between sections 1 and 2 as a control volume and apply themomentum principle

V

V

Patm

Patm

Patm Patm

y

x

!

For steady ow the momentum equation is

XH3 = /2,23 " /1,13

Also ,1 = ,2 = ,! The only H3, is the force of the wing on the uid in the controlvolume:

H3 = (", sin K) / = (", sin K)#, I= "#, 2I sin K

But the uid acting on the wing in the ? direction is the lift HE and it is the negativeof H3! So

HE = #, 2I sin K

-E = 2HE$(#,26)

Eliminate HE between the two equations yields

-E = 2#, 2I sin K$(#, 26)

-E = 2I sin K$6

= 2(0$4)>2 sin K$6

-E = (0$2) sin K(>2$6)

But sin K ! K for small angles. Therefore

-E = (0$2)K(>2$6)

or

K = 2-E$(0>2$6)

-9##,26$2 = (-E#,

26$2)(K$2)

1073

Eliminating K between the two equations gives

-9##,26$2 = (-E#,

26$2)(-E$(0>2$6))

-9# = -2E$(0$)

1074

PROBLEM 11.73

Situation: The problem statement provides data describing aircraft takeo! and land-ing.

Find: (a)Landing speed.(b) Stall speed.

ANALYSIS

-Emax = 1!40 which is the -E at stall. Thus, for stall

. = -Emax6#,2& $2

= 1!46#, 2& $2

For landing. = 1!26#, 2E$2

But,E = ,& + 8

so. = 1!2I#(,& + 8)

2$2

Therefore

1!2(,& + 8)2 = 1!4, 2&

,& = 99!8 m/s

,E = ,& + 8

,E = 107!8 m/s

1075

PROBLEM 11.74

Situation: An aircraft wing is described in the problem statement.

Find: Total drag on wing and power to overcome drag.

ANALYSIS

Calculate % and then #:

% = %0[&0 " T(R " R0))$&0]D*HF

% = 101!3[(296" (5!87× 10"3)(3' 000))$296](9481*(5487×10"3×287)) = 70!1 kPa

& = 296" 5!87× 10"3 × 3' 000 = 278!4 K

Then

# = %$"&

= 70' 100$(287× 278!4)= 0!877 kg/m3

-E = (HE$6)$(#,20 $2)

= (1' 200× 9!81$20)$(0!877× 602$2)= 0!373

Then

-99 = -2E$(0(>2$6))

= 0!3732$(0$(142$20))

= 0!0045

Then the total drag coe"cient

-9 = -99 + 0!01

= 0!0145

Total wing drag

H9 = -9I,#,20 $2

H9 = 0!0145× 20× 0!877× 602$2

= 458 N

Power

* = 60× 458= 27.5 kW

1076

PROBLEM 11.75

Situation: The problem statement provides data for a Gottingen 387-FB lifting vane.

Find: (a) Speed at which cavitation begins.(b) Lift per unit length on foil.

ANALYSIS

Cavitation will start at point where -, is minimum, or in this case, where

-, = "1!95-, = (%" %0)$(#, 20 $2)

Also%0 = 0!70× 9' 810 Pa gage

and for cavitation

% = %vapor = 1' 230 Pa abs

%0 = 0!7× 9' 810 + 101' 300 Pa abs

So

"1!95 = [1' 230" (0!7× 9' 810 + 101' 300)]$(1' 000, 20 $2)

,0=10.5 m/s

By approximating the -, diagrams by triangles, it is found that -,avg. on the top ofthe lifting vane is approx. -1.0 and -,avg.,b ottom ! +0!45

Thus, !-,avg. ! 1!45! Then

HE = -EI,#,20 $2

HE*length = 1!45× 0!20× 1' 000× (10!5)2$2

HE*length=16,000 N/m

1077

PROBLEM 11.76

Situation: The distribution of -, on the wing section in 11.75 is described in theproblem statement.

Find: Range that -E will fall within.

ANALYSIS

The correct choice is (b) .

1078

PROBLEM 11.77

Situation: The drag coe"cient for a wing is described in the problem statement.

Find: Derive an expression for the -E that corresponds to minimum -9$-E and thecorresponding -E$-9!

ANALYSIS

-9$-E = (-90$-E) + (-E$(0$))

A$A-E(-9$-E) = ("-90$-2E) + (1$(0$)) = 0

-E =p0$-90

-9 = -90 + 0$-90$(0$) = 2-90

Then-E$-9 = (1$2)

p0$$-90

1079

PROBLEM 11.78

Situation: A glider at elevation of 1000 m descends to sea level—see the problemstatement for all the details.

Find: Time in minutes for the descent.

ANALYSIS

L = 1' 000$(sin 1!7!) = 33' 708 m

HE = . = (1$2)#, 2-E6

200× 9!81 = 0!5× 1!2× , 2 × 0!8× 20so

, = 14!3 m/s

Then

P = 33' 708 m/(14.3 m/s)

= 2357 s

= 39.3 min

1080

PROBLEM 11.79

Situation: An aircraft wing is described in the problem statement.

Find: Drag force on the wing.

APPROACH

Apply coe"cients for lift and drag forces.

ANALYSIS

Lift force

HE = -E6#, 202

HE$6 = -E#, 202

Thus

#, 202

=HE$6

-E#, 202

=2000N$m2

0!3= 6667N$m2

From Fig. 11-24 at -E = 0!30' -9 % 0!06

Drag force

H9 = -96#, 202

= (0!06)¡10m2

¢ ¡6667N$m2

¢

= 4000 N

1081

PROBLEM 11.80

Situation: The problem statement describes an ultralight airplane.

Find: (a) Angle of attack.(b) Drag force on wing.

ANALYSIS

Lift force

. = -E6#,20 $2

-E = .$(6#, 20 $2) = (400)$((200)(0!002)(502)$2) = 0!80

From Fig. 11-23 -9 = 0!06 andT = 7!

The drag force is

H9 = -96#,20 $2

= (0!06)(200)(0!002)(502)$2

= 30 lbf

1082

PROBLEM 11.81

Situation: The parameters for a human-powered aircraft are given in the problemstatement.

Find: Design the human-powered aircraft using the characteristics of the wing in Fig.11.23.

ANALYSIS

There are several ways to address this design problem. One approach would be toconsider the wing area and velocities necessary to meet the power constraint. Thatis,

225 = (0!05 + -9)1

2(0!00238 slugs/ft3), 30 6

Make plots of ,0 versus 6 with -9 as a parameter. Then use the constraint of thelift equaling the weight.

40 + 0!12× 6 = -E1

2(0!00238 slugs/ft3), 20 6

Make plots of ,0 versus 6 with -E as a parameter. Where these curves intersectwould give values where both constraints are satised. Next you can plot the curvefor the pairs of -9 and -E where the curves cross. You can also plot -9 versus -E(drag polar) for the airfoil and see if there is a match. If there is no match, the airfoilwill not work. If there is a match, you should try to nd the conguration that willgive the minimum weight.

1083

PROBLEM 12.1

Situation: A sound wave travels in methane at 0 !C!

Find: Speed of wave.

ANALYSIS

Speed of sound

Z =&c"&

=&1!31× 518× 273Z = 430 m/s

1084

PROBLEM 12.2

Situation: A sound wave travels in helium at 50 !C.


ANALYSIS

Speed of sound

Z =&c"&

=p1!66× 2077× (50 + 273)

Z = 1055 m/s

1085

PROBLEM 12.3

Situation: A sound wave travels in hydrogen at 68 !F.


ANALYSIS

Speed of sound

Z =&c"&

=p1!41× 24' 677× (460 + 68)

Z = 4286 ft/s

1086

PROBLEM 12.4

Situation: A sound wave travels in helium and another in nitrogen both at 20 !C.

Find: Di!erence in speed of sound.

ANALYSIS

Speed of sound

ZHe =p(c")He&

=&1!66× 2077× 293

= 1005 m/s

ZN2 =p(c")N2&

=&1!40× 297× 293

= 349 m/s

ZHe " ZN2 = 656 m/s

1087

PROBLEM 12.5

Situation: A sound wave travels in an ideal gas.

Find: Speed of sound for an isothermal process.

ANALYSIS

Z2 = [%$[#; % = #"&

If isothermal, &=const.

! [%$[# = "&

! Z2 = "&

Z =&"&

1088

PROBLEM 12.6

Situation: The relationship between pressure and density for sound travelling througha uid is described in the problem statement.

Find: Speed of sound in water.

ANALYSIS

%" %! = N? ln(#$#!)

Z2 =[%

[#=N@#

Z =pN@$#

Z =p2!20× 109$103

Z = 1483 m/s

1089

PROBLEM 12.7

Situation: An aircraft ying in air at Mach 1!5 is described in the problem statement.

Find: (a) Surface temperature.(b) Airspeed behind shock.

Properties: (a) FromTable A.1 at+1 = 1!5 ' &$&5 = 0!6897; . +2 = 0!7011' &2$&1 =1!320. (b) Air (Table A.2) c = 1!4 and " = 287 J/kg/K.

ANALYSIS

Total temperature will develop at exposed surface

&

&5= 0!6897

&5 =(273" 30)0!6897

= 352!3K = 79!2 !C

Temperature (behind shock)

&2&1

= 1!320

&2 = 1!320× (273!15" 30)= 320!96K

Speed of sound (behind shock)

Z2 =pc"&2

=p(1!4) (287) 320!96

= 359!1m$ s

Mach number (behind shock)

+2 =,2Z2

,2 = Z2+2

= (359!1) (0!7011)

= 251! 77m$ s

,2 = 252m$ s = 906 km$h

1090

PROBLEM 12.8

Situation: A ghter is ying at Mach 2 though air at 273 !F.

Find: Temperature on nose.

Properties: From Table A.1 &$&5 = 0!5556 at + = 2!0

ANALYSIS

&5 = (1$0!5556)(273)

&5 = 491K = 218!C

1091

PROBLEM 12.9

Situation: An aircraft is ying at Mach 1.8 through air at 10000m, 30!5 kPa, and"44 !C.

Find: (a) Speed of aircraft.(b) Total temperature.(c) Total pressure.(d) Speed for + = 1!

ANALYSIS

Speed of sound (at 10,000 m)

Z =&c"&

Z =p(1!40)(287)(229)

Z = 303!3m$ s

Mach number

, = (1!8)(303!3)(3' 600$1' 000)

= 1' 965 km/hr

Total temperature

&5 = 229(1 + ((1!4" 1)$2)× 1!82)= 377< = 104 oC

Total pressure

%5 = (30!5)(1 + 0!2× 1!82)(144*(144"1))

= 175 kPa

Mach number

+ = 1; , = 1× Z = Z, = (303!3)(3' 600$1' 000)

= 1092 km/hr

1092

PROBLEM 12.10

Situation: An airplane is travelling at sea level—additional details are provided in theproblem statement.

Find: Speed of aircraft at altitude where & = "40oC.

APPROACH

Apply the Mach number equation and the speed of sound equation.

ANALYSIS

At sea levelSpeed of sound (sea level)

Z =&c"&

=p(1!4)(287)(288)

= 340!2 m/s

Mach number(sea level)

, = 800 km/hr = 222!2 m/s

+ = 222!2$340!2 = 0!653

Speed of sound (at altitude)

Z =p(1!4)(287)(233)

= 306!0 m/s

Mach number (at altitude)

, = +Z

= 0!653× 306, = 200m$ s = 719 km$h

1093

PROBLEM 12.11

Situation: An aircraft ying through air is described in the problem statement.

Find: Wing loading.

ANALYSIS

Kinetic pressure

^ = (c$2)%+2

= (1!4$2)(30)(0!95)2

= 18!95 kPa

Lift force

HE = -E^6

. = HE$6 = -E^

= (0!05)(18!95)

= 0!947 kPa

. = 947 Pa

1094

PROBLEM 12.12

Situation: An object immersed in airow is described in the problem statement.

Find: (a) Pressure.(b) Temperature at stagnation point.

ANALYSIS

Speed of sound

Z =&c"&

=p(1!4)(287)(293)

= 343 m/s

Mach number

+ = 250$343

= 0!729

Total propertiesTemperature

&5 = (293)(1 + 0!2× (0!729)2)= 293× 1!106= 324 K

&5 = 51!C

Pressure

%5 = (200)(1!106)345

%5 = 284!6 kPa

1095

PROBLEM 12.13

Situation: An airow through a conduit is described in the problem statement.

Find: Mass ow rate through conduit.

APPROACH

Apply the ow rate equation, the ideal gas law, Mach number, speed of sound, andthe total properties equations.

ANALYSIS

Total temperature equation

& =&5¡

1 + ["12+2¢

=283!15K¡

1 +¡144"12

¢0!752

¢

= 254!5K

Total pressure equation

% =%5

¡1 + ["1

2+2¢ ::"1

=360 kPa

¡1 +

¡144"12

¢0!752

¢ 144144"1

= 247!9 kPa

Speed of sound

Z =&c"&

Z =p(1!4)(287)(254!5)

= 319!8m$ s

Mach number

, = +Z

, = (0!75)(319!8)

= 239!9 m/s

Ideal gas law

# =%

"&

=247!9× 103

287× 254!5= 3!394 kg/m3

1096

Flow rate equation

/ = , #I

= (239!9)(3!394)(0!0050)

/ = 4!07 kg/s

1097

PROBLEM 12.14

Situation: Oxygen ows through a reservoir—additional details are provided in theproblem statement.

Find: (a) Velocity.(b) Pressure.(c) Temperature.

ANALYSIS

Total properties

&5 = 200!C = 473 K

& = 473$(1 + 0!2× 0!92)= 473$1!162

& = 407 K

%5 = 300 kPa

% = 300$(1!162)345

% = 177!4 kPa

Speed of sound

Z =&c"&

Z = [(1!4)(260)(407)]1*2

= 384!9 m/s

Mach number

, = +Z

= (0!9)(384!9)

, = 346!4 m/s

1098

PROBLEM 12.15

Situation: High Mach number ow from a reservoir—additional details are providedin the problem statement.

Find: Mach number condensation will occur.

APPROACH

Apply total temperature equation setting the & to 50K and &5 to 300K.

ANALYSIS

Total temperature equation

&0$& = 1 + ((c " 1)$2)+2

300$50 = 6 = 1 + 0!2+2

+ = 5

1099

PROBLEM 12.16

Situation: Hydrogen ow from a reservoir—additional details are provided in the prob-lem statement.

Find: (a) Temperature.(b) Pressure.(c) Mach number.(d) Mass ow rate.

ANALYSIS

&5 = 20oC = 293 K

*5 = 500 kPa

Z,& + ,2$2 = Z,&0

& = &5 " , 2$(2Z,)= 293" (250)2$((2)(14' 223))

& = 290!8 K

Speed of sound

Z =&c"&

=p(1!41)(4' 127)(290!8)

= 1' 301 m/s

Mach number

+ = 250$1301

= 0!192

Total properties (pressure)

% = 500$[1 + (0!41$2)× 0!1922](1441*0441)

% = 487!2 kPa

Ideal gas law

# = %$"&

= (487!2)(103)$(4' 127× 290!8)= 0!406 kg/m3

Flow rate equation

/ = #I,

= (0!406)(0!02)2(0$4)(250)

/ = 0!032 kg/s

1100

PROBLEM 12.17

Situation: A sphere in a Mach-2.5 wind tunnel is described in the problem statement.

Find: Drag on the sphere.

ANALYSIS

% = %5$[1 + ((c " 1)$2)+2][*(["1)

= 600$[1 + 0!2(2!5)2]345

= 35!1 kPa

(1$2)#\2 = c%+2$2

= 1!4× 35!1× 2!52$2= 153!6 kPa

Drag force

H9 = -9(1$2)#\2I

= (0!95)(153!6× 103)(0!02)2(0$4)

H9 = 45!8 N

1101

PROBLEM 12.18

Situation: Eq. 12.27

Find: (a) Expression for pressure coe"cient.(b) Values for pressure coe"cient

ANALYSIS

%5 = (%)[1 + (c " 1)$2×+2]([*(["1))

-, = (%5 " %)#\2$2= (%5 " %)$c%+2$2

= (2$c+2)[(%5$%)" 1]

-, = 2$(c+2)[(1 + (c " 1)+2$2)([*(["1)) " 1]

-,(2) = 2!43

-,(4) = 13!47

-,inc. = 1!0

1102

PROBLEM 12.19

Situation: With low velocities, one can write %5$% = 1 + hAdditional details are provided in the problem statement.

Find: Show that Mach number goes to zero as j goes to zero, and that Eq. 12.32reduces to + = [(2$c)(%5$%" 1)]1*2

ANALYSIS

%5$% = [1 + (c " 1)+2$2][*(["1)

+ =q(2$(c " 1))[(%5$%)(["1)*[ " 1]

%5$% = 1 + h; (%5$%)(["1)*[ = (1 + h)(["1)*[ = 1 + ((c " 1)$c)h+ 0(h2)

(%5$%)(["1)*[ " 1 w ((c " 1)$c)h+ 0(h2)

Neglecting higher order terms

+ = [(2$(c " 1))((c " 1)$c)h]1*2

+ = [(2$c)(%5$%" 1)]1*2 as h$ 0

1103

PROBLEM 12.20

Situation: A normal shock wave is described in the problem statement.

Find: (a) Mach number.(b) Pressure downstream of wave.(c) Temperature downstream of wave.(d) Entropy increase.

ANALYSIS

Speed of sound

Z1 =&c"&

=p(1!4)(297)(223)

= 304!5 m/s

Mach number

+1 = ,$Z

= 500$304!8

= 1!64

Normal shock wave (Mach number)

+22 = [(c " 1)+2

1 + 2]$[2c+21 " (c " 1)]

= [(0!4)(1!64)2 + 2]$[(2)(1!4)(1!64)2 " 0!4]+2 = 0!657

Normal shock wavePressure ratio

%2 = %1(1 + c1+21 )$[(1 + c1+

22 )]

= (70)(1 + 1!4× 1!642)$(1 + 1!4× 0!6572)

%2 = 208 kPa

Temperature ratio

&2 = &1(1 + ((c " 1)$2)+21 )$(1 + ((c " 1)$2)+

22 )

= 223[1 + 0!2× 1!642]$[1 + 0!2× 0!6572]&2 = 316K = 43

!C

Entropy

!9 = "LQ[(%1$%2)(&2$&1)[*(["1)]

= "[LQ(%1$%2) + (c$(c " 1))LQ(&2$&1)]= 297[LQ(70$208) + 3!5LQ(315$223)]

!9 = 35!6 J/kg K

1104

PROBLEM 12.21


Find: (a) Mach number downstream of shock wave.(b) Pressure downstream of shock wave.(c) Temperature downstream of shock wave.

ANALYSIS

Mach number (downstream)

+22 = [(c " 1)+2

1 + 2]$[2c+21 " (c " 1)]

+2 = 0!577

Temperature ratio

(&2$&1) = [1 + ((c " 1)$2)+21 ]$[1 + ((c " 1)$2)+

22 ]

= (1 + (0!2)(4))$(1 + (0!2)(0!577)2) = 1!688

&2 = 505× 1!69&2 = 851!7

!R = 392 !F

Pressure ratio

%2$%1 = (1 + c+21 )$(1 + c+

22 )

= (1 + 1!4× 4)$(1 + 1!4× (0!577)2)= 4!50

%2 = (4!50)(30)

%2 = 135 psia

1105

PROBLEM 12.22


Find: Mach number

APPROACH

Find pressure ratios and apply the compressible ow tables.

ANALYSIS

%52$%1 = 150$40 = 3!75 = (%52$%51)(%51$%1)

Using compressible ow tables:

+ %52$%51 %1$%51 %52$%11.60 0.8952 0.2353 3.801.50 0.9278 0.2724 3.401.40 0.9582 0.3142 3.041.35 0.9697 0.3370 2.87

pt2

p1

Therefore, interpolating, + = 1!59

1106

PROBLEM 12.23

Situation: A shock wave is described in the problem statement.

Find: (a)The downstream Mach number.(b) Static pressure.(c) Static temperature.(d) Density.

Properties: From Table A.2 c = 1!31

APPROACH

Apply the Normal shock wave equations to nd Mach number, pressure, and temper-ature. Apply the ideal gas law to nd density.

ANALYSIS

Normal shock waveMach number

+22 = [(c " 1)+2

1 + 2]$[2c+21 " (c " 1)]

= ((0!31)(9) + 2)$((2)(1!31)(9)" 0!31) =0!2058

+2 = 0!454

Pressure ratio

%2$%1 = (1 + c+21 )$(1 + c+

22 )

= (1 + 1!31× 9)$(1 + 1!31× 0!2058) = 10!07

%2 = 1' 007 kPa, abs

Temperature ratio

&2$&1 = [1 + ((c " 1)$2)+21 ]$[1 + ((c " 1)$2)+

22 ]

= 2!32

&2 = (293)(2!32)

&2 = 680K = 407!C

Ideal gas law

#2 = %2$("&2)

= (1' 007)(103)$((518)(680))

#2 = 2!86 kg/m3

1107

PROBLEM 12.24

Situation: A shock wave is described in the problem statement.

Find: Velocity upstream of wave

Properties: From Table A.2 c = 1!66; " = 2' 077 J/kg/K.

ANALYSIS

Normal shock waveMach number

+21 = [(c " 1)+2

2 + 2]$[2c+22 " (c " 1)]

= 1!249

+1 = 1!12

Temperature ratio

&1$&2 = [1 + ((c " 1)$2)+22 ]$[1 + ((c " 1)$2)+

21 ]

= 0!897

&1 = (0!897)(373) = 335 K

Speed of sound

Z1 =&c"&

= (1!66× 2' 077× 335)1*2

Z1 = 1' 075 m/s

Mach number

,1 = Z1+1

= (1' 075)(1!12)

,1 = 1' 204 m/s

1108

PROBLEM 12.25


Find: (a) Lowest Mach number possible downstream of shock wave(b) Largest density ratio possible(c) Limiting values of +2 and #2$#1 for air.

ANALYSIS

+22 = ((c " 1)+

21 + 2)$(2c+

21 " (c " 1))

Because

+1 FF 1' (c " 1)+21 À 2

2c+21 À (c " 1)

So in limit

+22 $ ((c " 1)+2

1 )$2c+21 = (c " 1)$2c

! +2 $p(c " 1)$2c

#2$#1 = (%2$%1)(&1$&2)

= ((1 + c+21 )$(1 + c+

22 ))(1 + ((c " 1)$2)+

22 )$(1 + ((c " 1)$2)+

21 )

in limit +22 $ (c " 1)$2c and +1 $,

! #2$#1 $ [(c+21 )$((c " 1)$2)+

21 ][(1 + (c " 1)

2$4c)$(1 + c(c " 1)$2c)]#2$#1 $ (c + 1)$(c " 1)+2(air) = 0!378

#2$#1(air) = 6!0

1109

PROBLEM 12.26

Situation: A weak shock wave is described in the problem statement.

Find: (a) Approximation for Mach number downstream of wave.(b) Compare+2 computed with equation from (a) with values in table A.1 for+1 = 1'1!05' 1!1' and 1!2.

ANALYSIS

+22 = [(c " 1)+2

1 + 2]$[2c+21 " (c " 1)]

= [(c " 1)(1 + h) + 2]$[2c(1 + h)" (c " 1)] = [c + 1 + (c " 1)h]$[c + 1 + 2ch]= [1 + (c " 1)h$(c + 1)]$[1 + (2ch)$(c + 1)]! [1 + (c " 1)h$(c + 1)][1" (2ch)$(c + 1)]! 1 + (c " 1" 2c)h$(c + 1)! 1" h! 1" (+2

1 " 1)! 2"+2

1

+1 +2 +2 (Table A-1)1.0 1.0 1.01.05 0.947 0.9531.1 0.889 0.9121.2 0.748 0.842

1110

PROBLEM 12.27

Situation: A truncated nozzle is described in the problem statement.Inputs: total pressure, total temperature, back pressure, ratio of specic heats, gasconstant, and nozzle diameter.

Find: (a) Develop a computer program for calculating the mass ow.(b) Compare program with Example 12.12 with back pressures of 80' 90' 100' 110'120' and 130 kPa and make a table.

ANALYSIS

The computer program shows the ow is subsonic at the exit and the mass ow rateis 0.239 kg/s. The ow rate as a function of back pressure is given in the followingtable.

Back pressure, kPa Flow rate, kg/s80 0!24390 0!242100 0!239110 0!229120 0!215130 0!194

COMMENTS

One notes that the mass ow rate begins to decrease more quickly as the back pressureapproaches the total pressure.

1111

PROBLEM 12.28

Situation: A truncated nozzle is described in the problem statement.

Find: Mass ow rate

ANALYSIS

I- = 3 cm2 = 3× 10"4 m2

%5 = 300 kPa; &5 = 20! = 293 K

%0 = 90 kPa

%0$%5 = 90$300 = 0!3

Because %0$%5 G 0!528' sonic ow at exit.

Laval nozzle ow rate equation

/ = 0!685%5I&$p"&5

= (0!685)(3× 105)(3× 10"4)$p(287)(293)

/ = 0!212 kg/s

1112

PROBLEM 12.29


Find: (a)Mass ow rate of methane.(b) Mass ow rate if Bernoulli equation is valid.

Properties: From Table A.2 c = 1!31; " = 518 J/kgK.

ANALYSIS

I- = 3 cm2 = 3× 10"4m2

I, = 12 cm2 = 12× 10"4m2

%5 = 150 kPa; &5 = 303 K

%0 = 100 kPa;

%0$%5 = 100$150 = 0!667

%&$%5|methane = (2$(c + 1))[*(["1) = 0!544

%0 F %&' subsonic ow at exit

Mach number

+G =q(2$(c " 1))[(%5$%0)(["1)*[ " 1]

=p6!45[(1!5)042366 " 1]

= 0!806

Temperature

&G = 303 K$(1 + (0!31$2)× (0!806)2)= 275 K

Speed of sound

ZG =pc"&G

=p(1!31)(518)(275)

= 432 m/s

Ideal gas law

#G = %0$("&G)

= 100× 103$(518× 275)= 0!702 kg/m3

1113

Flow rate equation

/ = #G,GI-

= (0!702)(0!806)(432)(3× 10"4)

/ = 0!0733 kg/s

Assume the Bernoulli equation is valid,

%5 " %0 = (1$2)#, 2G

,G =p2(150" 100)103$0!702

= 377 m/s

/ = (0!702)(377)(3× 10"4)

/ = 0!0794 kg/s

Error = 8!3% (too high)

1114

PROBLEM 12.30


Find: The total pressure.

ANALYSIS

Speed of sound

ZG =pc"&G

=p(1!4)(287)(283)

= 337 m/s

Ideal gas law (assume sonic ow at the exit so %G = 100 kPa)

#G = %G$"&G

= 100× 103$(287× 283)= 1!23 kg/m3

Flow rate equation

/ = #GIGZG

= (1!23)(4× 10"4)(337)= 0!166 kg/s

Because the mass ow is too low, ow must exit sonically at pressure higher than theback pressure.

Flow rate equation

#G =/

ZGIG

=0!30

337× (4× 10"4)= 2!226 kg/m3

Ideal gas law

%G = #G"&G

= 2! 226× 287× 283 = 1! 808× 105 Pa

Then%5%G

= ((c + 1)$2)[*(["1)

= (1!2)345 = 1!893

%5 = 1!893× 1! 808× 105 Pa%5 = 3! 423× 105 Pa

%5 = 342 kPa

1115

PROBLEM 12.31


Find: Mass ow rate of helium.

Properties: From Table A.2 c = 1!66!

ANALYSIS

(a) %5 = 130 kPaIf sonic at exit,

%& = [2$(c + 1)][*(["1)%5

= 0!487× 130 kPa= 63!3 kPa

Flow must exit subsonicallyTotal propertiesFind Mach number

+2G = (2$(c " 1))[(%5$%0)(["1)*[ " 1]= 3!03[(130$100)044 " 1] = 0!335

+G = 0!579

Temperature

! &G = &5$(1 + ((c " 1)$2)+2)

= 301$(1 + (1$3)(0!335))

= 271K

Ideal gas law

#G = %$"&G

= 100× 103$[(2' 077)(271)]= 0!178 kg/m3

Flow rate equation/ = #GIG,G

Substituting Mach number and Speed of sound equations for ,G

/ = #GIG+G

pc"&G

= (0!178)(12× 10"4)(0!579)p(1!66)(2' 077)(271)

/ = 0!120 kg/s

1116

b)

%5 = 350 kPa

! %& = (0!487)(350) = 170 kPa

! Flow exits sonically

Flow rate equation from (a)

/ = 0!727%5I&$p"&5

= (0!727)(350)103(12× 10"4)$p2' 077× 301

/ = 0!386 kg/s

1117

PROBLEM 12.32


Find: Pressure required for isokinetic sampling.

Properties: From Table A.2 " = 287 J/kgK; c = 1!4!

ANALYSIS

Ideal gas law

# = %$"&

= 100× 103$(287)(873)= 0!399 kg/m3

Flow rate equation

/ = #, I

= (0!399)(60)(0$4)(4× 10"3)2

/ = 0!000301 kg$ s

Mach number

+ = ,$&c"&

= 60$p(1!4)(287)(873)

= 0!101

Total properties

%5 = (100)[1 + (0!2)(0!101)2]345

= 100!7 kPa

&5 = 875 K

Laval nozzle ow rate equation (assume sonic ow)

/ = 0!685%5I&$p"&5

= 0!685(100!7× 103)(0$4)(2× 10"3)2$p(287)(875)

/ = 0!000432 kg$ s

Thus, ow must be subsonic at constriction and solution must be found iteratively.Assume + at constriction and solve for / in terms of +!

Total properties

#G = #5(1 + ((c " 1)$2)+2)("1*(["1) = #5(1 + 0!2+

2)"245

ZG = Z5(1 + ((c " 1)$2)+2)"1*2 = Z5(1 + 0!2+2)"045

1118

Flow rate/ = #GIGZG+G

Combine equations

/ = IG+G#5Z5(1 + 0!2+2)"3

#5 = (0!399)[1 + (0!2)(0!101)2]245 = 0!401 kg/m3

Speed of sound

Z5 =pc"&5

=p(1!4)(287)(875) = 593 m/s

! / = 7!47× 10"4+(1 + 0!2+2)"3

+ /× 104

0.5 3.220.4 2.710.45 2.980.454 3.0040.455 3.01 (correct ow rate)

! %0 = (100!7)(1 + 0!2× 0!4552)"345

%0 = 87!2 kPa

1119

PROBLEM 12.33

Situation: Inputs of Mach number ratio (run with Mach number of 2) and specicheats (run with 1.4, 1.3 and 1.67).

Find: Develop a computer program that outputs: area ratio, static to total pressureratio, static to total temperature ratio, density to total density ratio, and before andafter shock wave pressure ratio.

ANALYSIS

The following results are obtained from the computer program for a Mach number of2:

I$I& 1!69 1!53 1!88&$&5 0!555 0!427 0!714%$%5 0!128 0!120 0!132#$#5 0!230 0!281 0!186+2 0!577 0!607 0!546%2$%1 4!5 4!75 4!27

1120

PROBLEM 12.34

Situation: Inputs: area ratio (run with 5), specic heats (run with 1.4, 1.67, and1.31), and ow condition.

Find: Develop a computer program that outputs Mach number.

ANALYSIS

The following results are obtained for an area ratio of 5:

c +subsonic +supersonic

1!4 0!117 3!171!67 0!113 3!811!31 0!118 2!99

1121

PROBLEM 12.35

Situation: A supersonic wind tunnel is described in the problem statement.

Find: The area ratio and reservoir conditions.


ANALYSIS

Mach number-area ratio relationship

I$I& = (1$+)[(1 + ((c " 1)$2)+2)$((c + 1)$2)]([+1)*(2(["1))

= (1$3)[(1 + 0!2× 32)$1!2]3

I$I& = 4!23

From Table A.1, %$%5 = 0!02722; &$&5 = 0!3571

%5 = 1!5 psia $0!0585

=1!5 psia0!02722

%5 = 55!1 psia

&5 =450 !R0!3571

&5 = 1260!R = 800 !F

1122

PROBLEM 12.36

Situation: The design of a Laval nozzle is described in the problem statement.

Find: The nozzle throat area.

Properties: From Table .2 c = 1!4; " = 297 J/kgK.

ANALYSIS

Find Mach number

+G =q(2$(c " 1))[(%5$%G)(["1)*[1]

=p5[(1' 000$30)04286 " 1]

= 2!94


IG$I& = (1$+)[(1 + ((c " 1)$2)+2)$((c + 1)$2)]([+1)*(2(["1))

= (1$2!94)[(1 + (0!2)(2!94)2)$1!2]3

IG$I& = 4!00

Flow rate equation for Laval nozzle

/ = 0!685%5I-$p"&5

I- = /p"&5$(0!685× %5)

= 5×p(297)(550)$((0!685)(106))

= 0.00295 m2

I- = 29!5 cm2

1123

PROBLEM 12.37

Situation: A rocket nozzle with the following properties is described in the problemstatement.I$I& = 4; %5 = 1!3 MPa = 1!3× 106 Pa; %0 = 35 kPa;c = 1!4.

Find: The state of exit conditions.

ANALYSIS

From Table A1:

+G ! 2!94 =F %G$%5 ! 0!030! %G = 39 kPa

! %G F %0 under expanded

1124

PROBLEM 12.38

Situation: Same as problem 12.37, but the a ratio of specic heats of 1.2.

Find: State of exit conditions.

ANALYSIS

Running the program from Problem 12.33 with c = 1!2 and I$I& = 4 gives %5$% =23!0! Thus the exit pressure is

%G =1!3 MPa23

= 56 kPa

Therefore the nozzle is underexpanded.

1125

PROBLEM 12.39

Situation: A Laval nozzle is described in the problem statement.

Find: (a) Reservoir pressure.(b) Static pressure and temperature at throat.(c) Exit conditions.(d) Pressure for normal shock at exit.

ANALYSIS

a) % = %5 in reservoir because , = 0 in reservoir%$%5 = 0!1278 for I$I& = 1!688 and + = 2 (Table A.1)

%5 = %$0!1278

= 100$0!1278

%5 = 782!5 kPa

b) Throat conditions for + = 1:

%$%5 = 0!5283

&$&5 = 0!8333

% = 0!5283(782!5)

% = 413!4 kPa

& = 0!8333(17 + 273)

= 242K

& = "31 !C

c) Conditions for %5 = 700 kPa:

%$%5 = 0!1278

% = 0!1278(700) = 89!5 kPa =! 89!5 kPa G 100 G kPa

overexpanded exit conditiond) %5 for normal shock at exit:Assume shock exists at + = 2; we know %2 = 100 kPa.From table A.1: %2$%1 = 4!5

%1 = %2$4!5 = 22!2 kPa

%$%5 = 0!1278

%5 = %$0!1278

= 22!2$0!1278

%5 = 173!7 kPa

1126

PROBLEM 12.40


Find: (a) Mach number.(b) Area ratio.

ANALYSIS

Find Mach number

^ = (c$2)%+2

= (c$2)%5[1 + ((c " 1)$2)+2]"[*(["1)+2

LQ^ = LQ(c%5$2)" (c$(c " 1))LQ(1 + ((c " 1)$2)+2) + 2LQ+

([$[+)LQ^ = (1$^)([^$[+)

= ("c$(c " 1))[1$(1 + ((c " 1)$2)+2)][(c " 1)+ ] + 2$+0 = ["c+ ]$[1 + ((c " 1)$2)+2] + (2$+)

= [("c+2 + 2 + (c " 1)+2)$[(1 + ((c " 1)$2)+2)+ ]

0 = 2"+2

+ =&2


I$I& = (1$+)[1 + ((c " 1)$2)+2]$[(c + 1)$2]([+1)*2(["1)

= (1$&2)[(1 + 0!2(2))$1!2]3

I$I& = 1!123

1127

PROBLEM 12.41

Situation: A rocket motor is described in the problem statement.

Find: (a) Mach number, pressure and density at exit.(b) Mass ow rate.(c) Thrust.(d) Chamber pressure for ideal expansion.

ANALYSIS


I$I& = (1$+G)((1 + 0!1×+2G )$1!1)

545 = 4

a) Solve for + by iteration:

+G I$I&3.0 6.732.5 3.422.7 4.452.6 3.902.62 4.0

! +G = 2!62

Total propertiesPressure

%G$%5 = (1 + 0!1× 2!622)"6 = 0!0434! %G = (0!0434)(1!2× 106)

%G = 52!1× 103 Pa

Temperature

&G$&5 = (1 + 0!1× 2!622)"1 = 0!593&G = (3' 273× 0!593)

= 1' 941 K

Ideal gas law

#G = %G$("&G)

= (52!1× 103)$(400× 1' 941)

%G = 0!0671kg/m3

Speed of sound

ZG =&c"&

=p(1!2× 400× 1' 941)

= 965 m/s

1128

Mach number

,G = (965)(2!62)

,G = 2' 528 m/s

b) Flow rate equation

/ = #GIG,G

= (0!0671)(4)(10"2)(2' 528)

/ = 6!78 kg/s

c) Momentum principle

H- = (6!78)(2' 528) + (52!1" 25)× 103 × 4× 10"2

H- = 18!22 kN

d)

%5 = 25$0!0434

%5 = 576 kPa

/ = (25$52!1)(6!78) = 3!25 kg/s

H- = (3!25)(2' 528)

H- = 8!22 kN

1129

PROBLEM 12.42

Situation: A rocket motor design is described in the problem statement.

Find: (a) Nozzle expansion ratio for ideal expansion.(b) Thrust if expansion ratio reduced by 10%.

ANALYSIS

%5$%G = (1 + ((c " 1)$2)+2)[*(["1)

= (1 + 0!1+2)6

+G =q10[(%5$%G)1*6 " 1]

=q10[(2' 000$100)1*6 " 1]

= 2!54


IG$I& = (1$+G)[(1 + 0!1+2G )$1!1]

545

IG$I& = 3!60

Total properties (temperature)

&G = 3' 300$(1 + (0!1)(2!54)2)

= 2006 K

Ideal gas law

#G = 100× 103$(400× 2' 006)= 0!125 kg/m3

Speed of sound

ZG =p(1!2)(400)(2006)

= 981 m/s

Flow rate equation

/ = #GIG,G

= (0!125)(3!38)(10"3)(981)(2!54)

= 1!053 kg/s

1130

Momentum principle

H- = (1!053)(981)(2!54)

H- = 2624 N

(b)

IG$I& = (0!9)(3!60) = 3!24

3!42 = (1$+G)((1 + 0!1+2G )$1!1)

545

Solve by iteration:

+G I$I&2.4 3.0112.5 3.4202.45 3.2042.455 3.2282.458 3.241

! +G = 2!46

%G$%5 = (1 + 0!1+2G )"6 = 0!0585

%G = (0!0585)(2!0× 106) = 117 kPa&G = 3' 300$(1 + 0!1× 2!462) = 2' 056 K

Speed of sound

ZG =pc"&G

=p(1!2)(400)(2056)

= 993 m/s

Momentum principle

H- = (1!053)(993)(2!46) + (117" 100)× 103 × 3!24× 10"3

H- = 2627 N

1131

PROBLEM 12.43


Find: Area ratio where shock occurs in nozzle.

ANALYSIS

%0$%5 = 0!5

Solution by iteration:Choose +Determine I$I&

Find %52$%51 = I&1$I&2(IG$I&)2 = 4(I&1$I&2)Find +G

%G$%51 = (%G$%52)(%52$%51) and converge on %G$%51 = 0!5

A T

Ac

1

2

+ I$I& *52$%51 (IG$I&) +G %G$%512 1.69 0.721 2.88 0.206 0.72.5 2.63 0.499 2.00 0.305 0.4682.4 2.40 0.540 2.16 0.28 0.511 ! I$I& = 2!462.43 2.47 0.527 2.11 0.287 0.4972.425 2.46 0.530 2.12 0.285 0.50

1132

PROBLEM 12.44

Situation: A rocket nozzle is described in the problem statement.

Find: Area ratio and location of shock wave.

ANALYSIS

Use same iteration scheme as problem 12-43 but with c = 1!2 to nd I$I& of shock:

%0$%5 = 100$250 = 0!4 IG$I- (8$4)2 = 4

+ I$I& *52$%51 (IG$I&)2 +G %G$%512.0 1.88 0.671 2.68 0.227 0.5682.4 3.01 0.463 1.85 0.341 0.4322.5 3.42 0.416 1.65 0.385 0.380 ! I$I& = 3!252.46 3.25 0.434 1.74 0.366 0.400

From geometry: A = A5 + 2× tan 15!

A$A5 = 1 + (2E$A5) tan 15!

I$I& = (A$A5)2 = 3!25

= [1 + (2E$A5)(0!268)]2

= [1 + (0!536E$A5)]2

! E$A5 = 1!498

E = (1!498)(4)

E = 5!99 cm

15

d t

o

d

x

1133

PROBLEM 12.45

Situation: A normal shock wave occurs in a nozzle—additional details are provided inthe problem statement.

Find: Entropy increase.


ANALYSIS

I

I&= (1$+)((1 + 0!205×+2)$1!205)24939

Solve iteratively for + (to give I$I& = 4)

+ I$I&2.5 2.612.8 3.453.0 4.162.957 4.0

+1 = 2!957

+22 = ((c " 1)+2

1 + 2)$(2c+21 " (c " 1))

+2 = 0!4799

%2$%1 = (1 + c+21 )$(1 + c+

22 ) = 10!06

%5$%|1 = (1 + ((c " 1)$2)+21 )[*(["1) = 34!20

%5$%|2 = 1!172

%52$%51 = (%52$%2)(%2$%1)(%1$%51) = 0!3449

!9 = " ln(%51$%52) = 4127 ln(1$0!3449)

!9 = 4390 J/kgK

1134

PROBLEM 12.46

Situation: Airow in a channel is described in the problem statement.

Find: (a) Mach number.(b) Static pressure.(c) Stagnation pressure at station 3.

Properties: From Table A.1 + = 2!1' I$I& = 1!837' %$%5 = 0!1094!

ANALYSIS

I& = 100$1!837 = 54!4

%5 = 65$0!1094 = 594 kPa

I2$I& = 75$54!4 = 1!379

+ = 1!74$ %2$%5 = 0!1904$ %2 = 0!1904(594) = 113 kPa

after shock, +2 = 0!630; %2 = 3!377(113) = 382 kPa

I2$I& = (1$+)((1 + 0!2+2)$1!2)3

= 1!155

%5$%2 = (1 + 0!2+2)345 = 1!307

I& = 75$1!155 = 64!9; %5 = 382(1!307) = 499 kPa

I3$I& = 120$64!9 = 1!849; from Table A.1, +3 = 0!336

%3$%5 = 0!9245; %3 = 0!9245(499) = 461 kPa

1135

PROBLEM 12.47

Situation: A shock wave in air is described in the problem statement.+1 = 0!3; I$I& = 2!0351; I& = 200$2!0351 = 98!3 cm2!

Find: Atmospheric pressure for shock position.

ANALYSIS

%$%5 = 0!9395

%5 = 400$0!9395

= 426 kPa

I&$I& = 120$98!3

= 1!2208

By interpolation from Table A.1:

+&1 = 1!562; %1$%5 = 0!2490$ %1 = 0!249(426) = 106 kPa

+&2 = 0!680; %&2$%1 = 2!679$ %&2 = 2!679(106) = 284 kPa

I&$I&2 = 1!1097$ I&2 = 120$1!1097 = 108 cm2

%&2$%52 = 0!7338; %52 = 284$0!7338 = 387 kPa

I2$I&2 = 140$108 = 1!296$+2 = 0!525

%2$%52 = 0!8288

%2 = 0!8288(387)

%2 = 321 kPa

1136

PROBLEM 12.48

Situation: Inputs: b(E"E&)$1 (run for 1, 10, and 100 and c = 1!4) for a compressible,adiabatic ow in a pipe.

Find: Develop a computer program that outputs: Mach number and the ratio ofpressure to the pressure at sonic conditions (%Y$%&).

ANALYSIS

Running the program for initial Mach number given a value of b (E& " E) $1 resultsin

b (E& " E) $1 c = 1!4 c = 1!31+ %Y$%& + %Y$%&

1 0!508 2!10 0!520 2!0210 0!234 4!66 0!241 4!44100 0!0825 13!27 0!0854 12!57

1137

PROBLEM 12.49

Situation: The design of a piping system is described in the problem statement.

Find: Pipe diameter.

Assumptions: +G = 1; %G = 100 kPa; &G = 373(0!8333) = 311 K

ANALYSIS

Speed of sound

ZG =pc"&G

=p1!4(287)311

= 353 m/s

Ideal gas law

#G = 100× 103$(287× 311)= 1!12 kg/m3

Flow rate

I = /$(#, )

= 0!2$(1!12× 353) = 5!06× 10"4 m2 = 5!06 cm2

Solve for 1

1 = ((4$0)I)1*2

= 2!54 cm

Reynolds number

Re = (353× 0!0254)$(1!7× 10"5)= 5!3× 105 $ b = 0!0132

b!E$1 = (0!0132× 10)$0!0254 = 5!20

from Fig. 12.19+1 = 0!302from Fig. 12.20 %$%& = 3!6

%1 = 100(3!6) = 360 kPa F 240 kPa

! Case B

Solve by iteration.

+G &G ZG ,G #G I× 104 Re×10"5 +1 %1$%G0.8 331 365 292 1.054 6.51 4.54 0.314 2.550.7 340 369 259 1.026 7.54 4.11 0.322 2.18

1138

By interpolation, for %1$%G = 2!4' +G = 0!76

&G = 334 K; ZG = 367 m/s; ,G = 279 m/s; #G = 1!042 kg/m3

I = 6!89× 10"4 m2; 1 = 0!0296 mI = 2!96cm

1139

PROBLEM 12.50

Situation: Air entering a steel pipe is described in the problem statement.

Find: (a) Length of pipe for sonic ow.(b) Pressure at pipe exit.

Properties: From Table A.2 " = 1' 716 ft-lbf/slug.

ANALYSIS

& = 67!F = 527!R

Speed of sound

Z =&c"&

=p(1!4)(1' 716)(527)

= 1' 125 ft/sec

+1 = 120$1' 125

= 0!107

Ideal gas law

# = %$"&

= (30× 144)$(1' 716× 527)= 0!00478 slug/ft3

Reynolds number

7 = 3!8× 10"7 lbf-sec/ft2

Re = (120× 1$12× 0!00478)$(3!8× 10"7) = 1!25× 105

From Figs. 10-8 and Table 10.2, b = 0!025

b(E& " EY)$1 = (1"+2)$c+2 + ((c + 1)$2c)LQ[(c + 1)+2$(2 + (c " 1)+2)] = 62!0

! E& " EY = 2 = (62!0)(1$b) = (62!0× 1$12)$0!025 = 207 ft

from Eq. 12.79

%Y$%& = 10!2

%& = 30$10!2

%& = 2!94 psia

1140

PROBLEM 12.51

Situation: Air ows out of a brass tube—additional details are provided in the problemstatement.

Find: Distance upstream where + = 0!2!

Properties: From Table A.2 " = 287 J/kgK.

ANALYSIS

Total properties (temperature

&G = 373$(1 + 0!2× 0!92)= 321 K

Speed of sound

ZG =pc"&G

=p(1!4)(287)(321)

= 359 m/s

Mach number

, = +G$ZG

= (0!9)(359)

= 323 m/s

Ideal gas law

7G = 2!03× 10"5N · s/cm2

# = %$"&G

= (100× 103)$(287× 321) = 1!085 kg/m3

Reynolds number

Re = (323)(1!085)(3× 10"2)$(2!03× 10"5) = 5!18× 105

from Figs. 10-8 and Table 10.2 b = 0!0145

b(E& " E049)$1 = 0!014b(E& " E042)$1 = 14!5

! b(E048 " E042)$1 = 14!49 = b2$1! 2 = (14!49)(3× 10"2)$0!0145

2 = 30!0m

1141

PROBLEM 12.52

Situation: The inlet and exit of a pipe are described in the problem statement.

Find: Friction factor: b !

ANALYSIS

Eq. (12-75)

+ = 0!2

b(EP " E042)$1 = 14!53

+ = 0!6

b(E& " E047)$1 = 0!2

b(E046 " E042)$1 = 14!33

b = 14!33(0!5)$(20× 12)

b = 0!0298

1142

PROBLEM 12.53

Situation: Oxygen owing through a pipe is described in the problem statement.

Find: Mass ow rate in pipe.

Properties: From Table A.2 c = 1!4; " = 260 J/kgK.

Assumptions: Sonic ow at exit.

ANALYSIS

Temperature

&G = 293$1!2

= 244K = "29 !C

Speed of sound

ZG = ,G =p(1!4)(260)(244)

= 298 m/s

Reynolds number

:G w 1× 10"5 m2$s (Fig. A3)Re = (298× 2!5× 10"2)$(1× 10"5) = 7!45× 105

From Figs. 10-8 an Table 10.2, b = 0!024

b(E& " EY)$1 = (10× 0!024)$0!025 = 9!6

From Fig. 12-19 + at entrance = 0.235

%Y$%& = 4!6

%1 = 460 kPa F 300 kPa

Therefore ow must be subsonic at exit so %G$%1 = 100$300 = 0!333! Use iterativeprocedure:

+1(($#"$; )

9Re×105 b b2$1 (($#"$<)

9+G %G$%1

0.20 14.5 6.34 0.024 9.6 4.9 0.31 0.6410.22 11.6 6.97 0.024 9.6 2.0 0.42 0.5160.23 10.4 7.30 0.024 9.6 0.8 0.54 0.4160.232 10.2 7.34 0.024 9.6 0.6 0.57 0.3960.234 10.0 7.38 0.024 9.6 0.4 0.62 0.3660.2345 9.9 7.40 0.024 9.6 0.3 0.65 0.348

1143

For +1 near 0.234, %Y$%& = 4!65

%G$%& = (%Y$%&)(%G$%Y)

%G$%& = (4!65)(0!333) = 1!55

which corresponds to +G = 0!68

Total temperature

&G = 293$(1 + (0!2)(0!68)2)

= 268 K

Speed of sound

ZG =pc"&G

=p(1!4)(260)(268)

= 312 m/s

,G = 212 m/s

Ideal gas law

# = %$"&G

= 105$(260× 268)= 1!435 kg/m3

Flow rate equation

! / = (1!435)(212)(0$4)(0!025)2

/ = 0!149kg/s

1144

PROBLEM 12.54

Situation: Same as 12.53.


ANALYSIS

From the solution to prob. 12.53, we know ow at exit must be sonic since %1 F 460kPa. Use an iterative solution. Guess b = 0!025

b(E& " EY)$1 = 10

+ = 0!23

&5 = 293$(1 + 0!2(0!23)2) = 290 K

Z1 =p(1!4)(290)(260) = 325 m/s

#1 = (500× 103)$(260× 290) = 6!63 kg/m3

Assuming 7 not a function of pressure

71 = 1!79× 10"5N · s/m2

Re = (0!23)(325)(6!63)(2!5× 10"2)$(1!79× 10"5) = 6!9× 105

From Fig. 10.8 and Table 10.2b = 0!024

Try

b = 0!024

b(E& " EY)$1 = 9!6; + = 0!235; &5 w 290 KZ1 = 325 m/s; #1 = 6!63 kg/m

3; 71 w 1!79× 10"5 N · s/m2;Re = 7× 105

gives same b of 0.024. For + = 0!235' %Y$%& = 4!64

%& = 107!8 kPa

&G = 293$1!2 = 244 K

ZG = 298 m/s

#G = (107!8× 103)$(260× 244) = 1!70 kg/m3

! / = (1!70)(298)(0$4)(0!025)2

/ = 0!248 kg/s

1145

PROBLEM 12.55

Situation: A pressure hose connected to a regulator valve is described in the problemstatement.

Find: Hose diameter.

Assumptions: +G = 1; %G = 7 psia.

ANALYSIS

Speed of sound

&G = 560(0!8333) = 467!"

ZG =pc"&G

=p1!4(1' 776)467 = 1' 077 ft/s

Ideal gas law

#G = %$"&

= 7(144)32!2$(1' 776× 467)= 0!039 lbm/ft3

Flow rate equation

I = /$(#, )

= 0!06$(0!039× 1' 077)= 1!43× 10"3 ft2

1 = 0!0427 ft = 0!51 in.

Reynolds number

Re = (1' 077)(0!0427)(0!039)$(1!36× 10"7 × 32!2) = 4!1× 105

c&$1 = 0!0117; b = 0!040

b!E$1 = (0!04× 10)$0!0427 = 9!37

From Fig. 12-19 +1 = 0!24! From Fig. 12-20, %1$%& = 4!54

%1 = 31!8 psia G 45 psia

Therefore Case D applies so + = 1 at exit and %G F 7 psia.Solve by iteration:

+1 &1 ,1 #1 1 Re×10"5 b +1 %G0.24 553 281 0.212 0.0358 1.62 0.040 0.223 9.160.223 554 262 0.212 0.0371 1.56 0.040 0.223 9.16

1 = 0!0371 ft1 = 0!445 in

1146

PROBLEM 12.56

Situation: The design of an air blower and pipe system is described in the problemstatement.

Find: (a)Pressure.(b) Velocity.(c) Density at pipe inlet.

Assumptions: Viscosity of particle-laden ow is same as air.

ANALYSIS

Speed of sound

Z =pc"&G

=p1!4(287)288

= 340 m/s

Mach number

+G = ,$Z

= 50$340

= 0!147

Find +1

$ b(E& " E04147)$1 = 29!2%G$%& = 7!44

Re = 50(0!2)$(1!44× 10"5) = 6!94× 105

c&$1 = 0!00025; b = 0!0158

b!E$1 = [b(E& " EY)$1]" [b(E& " E04147)$1]= 0!0158× 120$0!2 = 9!48

b(E& " EY)$1 = 29!2 + 9!48 = 38!7$+1 = 0!14

Pressure ratio

%1$%& = 7!81

%1$%G = (%1$%&)(%&$%G) = 7!81$7!44 = 1!050

%1 = 1!05(100)

%1 = 105 kPa

Mach number

,1 = 0!14(340)

,1 = 47!6 m/s

1147

Total properties

&1 = &5$(1 + 0!2+21 )

= 288$(1 + 0!2(0!14)2)

= 287

Ideal gas law

#1 = %$"&

= (105× 103)$(287× 287)

#1 = 1!27 kg/m3

1148

PROBLEM 12.57

Situation: Methane is pumped into a steel pipe—additional details are provided in theproblem statement.

Find: Pressure 3 km downstream.

ANALYSIS

Speed of sound

Z1 =pc"&G

=p1!31(518)320

= 466 m/s

Ideal gas law

#1 =%

"&

=1!2× 106

518× 320= 7!24 kg/m3

Mach number

+ =,

Z1=20

466= 0!043

By Eq. 12-75b(E& " E04043)$1 = 407

and by Eq. 12-79

%1$%& = 25!0

Re = 20(0!15)7!24$(1!5× 10"5)= 1!448× 106; c&$1 = 0!00035

b = 0!0160

b!E$1 = 0!0160(3000)$0!15 = 320

[b(E& " E04043)$1]" [b(E& " EY)$1] = b!E$1

b(E& " EY)$1 = 407" 320 = 83$+G = 0!093

By Eq. 12-79

%G$%& = 11!5

%G = (%G$%&)(%&$%1)%1

= (11!5$25!0)¡1!2× 106

¢

%G = 552 kPa

1149

PROBLEM 12.58

Situation: Hydrogen is transported in a n underground pipeline—additional detailsare provided in the problem statement.

Find: Pressure drop in pipe.

Properties: From Table A.2 " = 4' 127 J/kgK; c = 1!41; : = 0!81× 10"4m2!

APPROACH

Find the speed of sound at entrance

ANALYSIS

Speed of sound

Z =pc"&G

=p(1!41)(4' 127)(288)

= 1' 294m$ s

Mach number

! + = 200$1' 294 = 0!154

! c+2 = !0334;&c+ = 0!183

Reynolds number(200)(0!1)$(0!81× 10"4) = 2!5× 105

From Fig. 10-8 and Table 10.2b = 0!018

At entrance

b(E" " E1)$1 = LQ(0!0334) + (1" 0!0334)$0!0334 = 25!5

At exit

b(E" " E2)$1 = b(EY " E1)$1 + b(E1 " E2)$1 = 25!5" (0!018)(50)$0!1= 25!5" 9!0 = 16!5

From Fig. 12-22c+2 = 0!05U3

&c+ = 0!2236

Then

%2$%1 = (%"$%1)(%2$%")

= 0!183$0!2236 = 0!818

! %2 = 204!5 kPa

!% = 45!5 kPa

1150

PROBLEM 12.59

Situation: Helium ows in a tube—additional details are provided in the problemstatement.


Properties: From Table A.2 " = 2077 J/kgK; c = 1!66; : = 1!14× 10"4 m2/s.

ANALYSIS

Speed of sound

Z =pc"&G

=p(1!66)(2077)(288) = 996 m/s

%2$%1 = 100$120 = 0!833

Iterative solution:

,1 +1 Re×10"4 b c+21

(($="$; )9

(($="$<)9

c+22 %2$%1

100 0.100 4.4 0.022 0.0166 55.1 11.1 0.0676 0.49550 0.050 2.2 0.026 0.00415 234.5 182.5 0.0053 0.88555 0.055 2.4 0.025 0.00502 192.9 149.2 0.006715 0.86460 0.060 2.6 0.25 0.00598 161.1 111.1 0.008555 0.83661 0.061 2.6 0.25 0.00618 155.8 105.8 0.00897 0.83060.5 0.0605 2.6 0.25 0.006076 158.5 108.5 0.00875 0.833

Ideal gas law

# = %$"&

= 120× 103$(2' 077)(288)

= 0!201 kg/m3

Flow rate equation

/ = #, I

= (0!201)(60!6)(0$4)(0!05)2

/ = 0!0239 kg/s

1151

PROBLEM 12.60

Situation: The design of a supersonic wind tunnel is described in the problem state-ment.

Find: Do a preliminary design of a the system.

ANALYSIS

The area of the test section is

I- = 0!05× 0!05 = 0!0025 m2

From Table A.1, the conditions for a Mach number of 1.5 are

%$%5 = 0!2724' &$&5 = 0!6897 I$I& = 1!176

The area of the throat is

I& = 0!0025$1!176 = 0!002125 m2

Since the air is being drawn in from the atmosphere, the total pressure and totalpressure are 293 K and 100 kPa. The static temperature and pressure at the testsection will be

& = 0!6897× 293 = 202 K, % = 0!2724× 100 = 27!24 kPa

The speed of sound and velocity in the test section is

Z =&c"& =

&1!4× 287× 202 = 285 m/s

; = 1!5× 285 = 427 m/s

The mass ow rate is obtained using

/ = 0!685%5I&&"&5

= 0!685105 × 0!002125&287× 293

= 0!502 kg/s

The pressure and temperature in the vacuum tank can be analyzed using the re-lationships for an open, unsteady system. The system consists of a volume (thevacuum tank) and an inlet coming from the test section. In this case, the rst law ofthermodynamics gives

/2@2 "/1@1 = /#:(M#: + ;2#:$2) +1 ]2

Assume that the heat transfer is negligible and that the tank is initially evacuated.Then

/2@2 = /2(M#: + ;2#:$2)

1152

since /#: = /2! Thus the temperature in the tank will be constant and given by

Z@& = Z,&#: + ;2#:$2

717× & = 1004× 202 + 4272$2& = 410 K

The continuity equation applied to the vacuum tank is

,A#

AP= /

The density from the ideal gas law is

# =%

"&

which gives

,A%

AP= /"&

or

, =/"&

A%$AP

Assume the nal pressure in the tank is the pressure in the test section. Thus therate of change of pressure will be

A%

AP=27!24

30= 0!908 kPa/s

The volume of the tank would then be

, =0!502× 0!287× 410

0!908= 65 m3

This would be a spherical tank with a diameter of

1 =3

r6,

0= 5!0 m

COMMENTS

1. The tank volume could be reduced if the channel was narrowed after the testsection to reduce the Mach number and increase the pressure. This wouldreduce the temperature in the tank and increase the required rate of pressureincrease.

2. The tunnel would be designed to have a contour between the throat and testsection to generate a uniform velocity prole. Also a buttery valve would haveto be used to open the channel in minium time.

1153

PROBLEM 12.61

Situation: The design of a test system involving truncated nozzles is described in theproblem statement.

Find: Explain how to carry out the test program.

ANALYSIS

A truncated nozzle is attached to a storage tank supplied by the compressor. Thetemperature and pressure will be measured in the tank. These represent the totalconditions. The nozzles will be sonic provided that the tank pressure is greater than14.7/0.528=33 psia (or 18 psig).

Ideal Gas Law

# =%

"&=14!7× 1441716× 520

= 0!00237 slugs/ft3

A mass ow rate of 200 scfm corresponds to

/ = 200× 0!00237$60 = 0!00395 slugs/s

The ow rate is given by

/ = 0!685%5I&&"&5

Using 120 psig and a ow rate of 200 scfm gives a throat area of

I& =/&"&5

0!685%5

=0!00395×

&1716× 520

0!685× 134× 144= 2!82× 10"4 ft2

This area corresponds to an opening of

1 =

r4

0× 2!82× 10"4

= 0!0189 ft = 0.23 in

COMMENTS

1. This would represent the maximum nozzle size. A series of truncated nozzleswould be used which would yield mass ows of 1/4,1/2 and 3/4 of the maximumow rate. The suggested nozzle diameters would be 0.11 in, 0.15 in and 0.19 in.Another point would be with no ow which represents another data point.

2. Each nozzle would be attached to the tank and the pressure and temperaturemeasured. For each nozzle the pressure in the tank must exceed 18 psig toinsure sonic ow in the nozzle. The mass ow rate would be calculated foreach nozzle size and these data would provide the pump curve, the variation ofpressure with ow rate. More data can be obtained by using more nozzles.

1154

PROBLEM 13.1

Situation: A stagnation tube (A = 1 mm) is used to measure air speed.

Find: Velocity such that the measurement error is ' 2!5%.

Properties: : = 1!46× 10"5 m2/s.

ANALYSIS

Algebra using the coe"cient of pressure (from the vertical axis of Fig.13.1) gives

,! =q2!%$(#-,)

The allowable error is 2.5%, thus

,! =

s2!%

#-,= (1" 0!025)

s2!%

#

Thuss1

-,= 0!975

1

-,= 0!9752

-, =1

0!9752= 1!052

Thus when -, % 1!05, there will be a 2.5% error in ,!!

From Fig. 13-1, when -, = 1!05, then Re % 35

,!A

:= Re

,!A

:= 35

,! =35:

A

=35× (1!46× 10"5m2$ s)

0!001m= 0!511m$ s

,! = 0!511m$ s

1155

PROBLEM 13.2

Situation: A stagnation tube (A = 1 mm) is used to measure the speed of water.

Find: Velocity such that the measurement error is ' 1%.

ANALYSIS

Algebra using the coe"cient of pressure (from the vertical axis of Fig.13.1) gives,! =

p2!%$(#-,)! The allowable error is 1%, thus

,! =

s2!%

#-,= 0!99

s2!%

#

This simplies tos1

-,= 0!99

1

-,= 0!99

-, =1

0!992= 1!020

Thus when -, % 1!02, there will be a 1% error in ,!!

From Fig. 13-1, when -, = 1!02, then Re % 60. Thus

Re =, A

:= 60

, =60:

A

=60× (10"6m2$ s)

0!001m= 0!06m$ s

, ) 0!06 m$ s

1156

PROBLEM 13.3

Situation: A stagnation tube (A = 2 mm) is used to measure air speed.Manometer deection is 1 mm-H2O.

Find: Air Velocity: ,

ANALYSIS

#air = 1!25 kg/m3

!Mair = 0!001× 1000$1!25= 0!80 m

From Bernoulli equation applied to a stagnation tube

, =p2)!M = 3.96 m/s

Reynolds number

Re = , A$:

= 3!96× 0!002$(1!41× 10"5)= 563

Pressure coe"cient

-, % 1!001

, = 3!96$p-,

= 3!96$&1!001

= 3.96 m/s

1157

PROBLEM 13.4

Situation: A stagnation tube (A = 2 mm) is used to measure air speed (, = 12m/s).

Find: Deection on a water manometer: !M

Properties: For air, : = 1!4× 10"5 m2/s.

ANALYSIS

Determine -,

Re = , A$:

= 12× 0!002$(1!4× 10"5)= 1714

From Fig. 13.1 -, % 1!00Pressure drop calculationBernoulli equation applied to a stagnation tube

!% = #, 2$2

Ideal gas law

# =%

"&

=98' 000

287× (273 + 10)= 1!21 kg$m3

Then

!% = 9810!M

= 1!21× 122$2= 8!88× 10"3 m= 8.88 mm

1158

PROBLEM 13.5

Situation: A stagnation tube (A = 2 mm) is used to measure air speed.Air kinematic viscosity is 1!55× 10"5

Find: Error in velocity if -, = 1 is used for the calculation.

Properties: Stagnation pressure is !% = 5 Pa.

APPROACH

Calculate density of air by applying the ideal gas law. Calculate speed of air byapplying the Bernoulli equation to a stagnation tube. Then calculate Reynoldsnumber in order to check -,!

ANALYSIS

Ideal gas law

# =%

"&

=100' 000

287× 298= 1!17 kg/m3

Bernoulli equation applied to a stagnation tube

, =

s2!%

#

=

r2× 51!17

= 2!92 m/s

Reynolds number

Re =, A

:

=2!92× 0!0021!55× 10"5

= 377

Thus, -, = 1!002

% error = (1" 1$&1!002)× 100

= 0.1%

1159

PROBLEM 13.6

Situation: A probe for measuring velocity of a stack gas is described in the problemstatement.

Find: Stack gas velocity: ,!

ANALYSIS

Pressure coe"cient

-, = 1!4 = !%$(#, 20 $2)

Thus ,0 =

s2!%

1!4#

# =%

"&

=100' 000

410× 573= 0!426 kg$m3

Calculate pressure di!erence

!% = 0!01 m× 9810= 98!1 Pa

Substituting values

,0 =

s2!%

1!4#

=

r2× 98!11!4× 0!426

= 18.1 m/s

1160

PROBLEM 13.7

Situation: In 3.5 minutes, 14 kN of water ows into a weigh tank.

Find: Discharge: ]

Properties: (water 20!) = 9790 N/m3

ANALYSIS

. =.

!P

=14' 000

3!5× 60= 66!67N$ s

But ( = 9790 N/m3 so

] =.

(

=66!67N$ s

9790N$m3

] = 6!81× 10"3 m3/s

1161

PROBLEM 13.8

Situation: In 5 minutes, 80 m3 of water ows into a weigh tank.

Find: Discharge: ] in units of (a) m3$s, (b) gpm and (c) cfs.

ANALYSIS

] =V–P

=80

300

= 0.267 m3/s

] = 0!267 (m3$s)$(0!02832 m3$s/cfs)

= 9.42 cfs

] = 9!42 cfs × 449 gpm/cfs= 4230 gpm

1162

PROBLEM 13.9

Situation: Velocity data in a 24 inch oil pipe are given in the problem statement.

Find: (a) Discharge.(b) Mean velocity.(c)Ratio of maximum to minimum velocity.

ANALYSIS

Numerical integration

3(m) , (m/s) 20, 3 area (by trapezoidal rule)0 8.7 00.01 8.6 0.54 0.00270.02 8.4 1.06 0.00800.03 8.2 1.55 0.01300.04 7.7 1.94 0.01750.05 7.2 2.26 0.02100.06 6.5 2.45 0.02360.07 5.8 2.55 0.02500.08 4.9 2.46 0.02500.09 3.8 2.15 0.02311.10 2.5 1.57 0.01860.105 1.9 1.25 0.00700.11 1.4 0.97 0.00560.115 0.7 0.51 0.00370.12 0 0 0.0013

Summing the values in the last column in the above table gives ] = 0!196 m3/s.Then,

,mean = ]$I

= 0!196$(0!785(0!24)2)

= 4.33 m/s

,max$,mean = 8!7$4!33

= 2.0

This ratio indicates the ow is laminar. The discharge is

]=0.196 m3/s

1163

PROBLEM 13.10

Situation: Velocity data in a 16 inch circular air duct are given in the problemstatement.% = 14!3 psia, & = 70 !F

Find: (a) Flow rate: ] in cfs and cfm.(b) Ratio of maximum to mean velocity.(c) Whether the ow is laminar or turbulent.(d) Mass ow rate: /!

APPROACH

Perform numerical integration to nd ow rate (]). Apply the ideal gas law tocalculate density. Find mass ow rate using / = #].

ANALYSIS

Numerical integration

?(in.) 3(in.) , (ft/s) 203, (ft2$s) area (ft3$s)0.0 8.0 0 00.1 7.9 72 297.8 1.240.2 7.8 79 322.6 2.580.4 7.6 88 350.2 5.610.6 7.4 93 360.3 5.921.0 7.0 100 366.5 12.111.5 6.5 106 360.8 15.152.0 6.0 110 345.6 14.723.0 5.0 117 306.3 27.164.0 4.0 122 255.5 23.415.0 3.0 126 197.9 18.896.0 2.0 129 135.1 13.887.0 1.0 132 69.4 8.518.0 0.0 135 0 2.88

Summing values in the last column of the above table gives ] = 152!1 ft3$ s = 9124 cfmFlow rate equation

,mean = ]$I

= 152!1$(0!785(1!33)2)

= 109 ft/s

,max$,mean = 135$109

= 1.24

1164

which suggests turbulent ow .Ideal gas law

# =%

"&

=(14!3) (144)

(53!3) (530)

= 0!0728 lbm/ft3

Flow rate

/ = #]

= 0!0728(152!1)

= 11.1 lbm/s

1165

PROBLEM 13.11

Situation: A heated gas ows through a cylindrical stack–additional information isprovided in the problem statement.

Find: (a) The ratio 3"$1 such that the areas of the ve measuring segments areequal.(b) The location of the probe expressed as a ratio of 31$1 that corresponds to thecentroid of the segment(c) Mass ow rate

ANALYSIS

(a)

032" = (0$4)£(1$2)2 " 32"

¤

(3"$1)2 = 1$16" (3"$1)2(1$4)

5$4(3"$1)2 = 1$16

5(3"$1)2 = 1$4

3"$1 =

r1

20

= 0.224

b)

31I =

Z 9*2

0422369

[3 sin(T$2)$(T$2)](0$4)23A3 = 0!9(0$2)(33$3)|04590422369

(31)(0$4)[(1$2)2 " (0!22361)2] = 0!90(0$6)[(0!51)3 " (0!22361)3]

31$1 = 0!341

c)Ideal gas law

# = %$("& )

= 110× 103$(400× 573)= 0!480 kg/m3

Bernoulli equation applied to a stagnation tube

, =q2!%$#D

=q(2)#%)!M$#D

=p(2)(1' 000)(9!81)$0!48

&!M

= 202!2&!M

Values for each section are

1166

Station !M ,1 0.012 7.002 0.011 6.713 0.011 6.714 0.009 6.075 0.0105 6.55

Mass ow rate is given by

/ =X

Isector#,sector = I-#(X

,$5)

= (022$4)(0!480)(6!61) = 9.96 kg/s

1167

PROBLEM 13.12

Situation: A heated gas ows through a cylindrical stack–additional information isprovided in the problem statement.

Find: (a) The ratio 3"$1 such that the areas of the measuring segments are equal(b) The location of the probe expressed as a ratio of 31$1 that corresponds to thecentroid of the segment(c) Mass ow rate

ANALYSIS

Schematic of measurement locations

a)

032" = (0$6)[(1$2)2 " 32"]7$6(3"$1)

2 = (1$6)(1$4)

(3"$1)2 = 1$28

3"$1 = 0!189

b)

31I = 1$6

Z 0459

041899

[3 sin(T$2)$(T$2)]203 A3

(031$6)[(1$2)2 " (3")2] = 0!955(0$3)(33$3)|04509041899

31(0!52 " 0!1892) = 0!955(6$9)[0!53 " 0!1893]1

31$1 = (0!955)6(0!118)$(9(0!2143)) = 0.351

c)

# = %$"& = 115× 103$((420)(250 + 273)) = 0!523 kg/m3

, =q2)#%!M$#D =

p(2)(9!81)(1' 000)$0!523

&!M = 193!7

&!M

Calculating velocity from !M data gives

1168

Station !M(mm) ,1 8.2 17.542 8.6 17.963 8.2 17.544 8.9 18.275 8.0 17.326 8.5 17.867 8.4 17.75

From the above table, ,'@D = 17!75 m/s, ThenFlow rate equation

/ = (012$4)#,avg.

= ((0)(1!5)2$4)(0!523)(17!75)

= 16.4 kg/ s

1169

PROBLEM 13.13

Situation: Velocity data for a river is described in the problem statement.

Find: Discharge: ]

ANALYSIS

Flow rate equation

] =X

,#I#

, I , I1.32 m/s 7.6 m2 10.01.54 21.7 33.41.68 18.0 30.21.69 33.0 55.81.71 24.0 41.01.75 39.0 68.21.80 42.0 75.61.91 39.0 74.51.87 37.2 69.61.75 30.8 53.91.56 18.4 28.71.02 8.0 8.2

Summing the last column gives

] =549.1 m3/s

1170

PROBLEM 13.14

Situation: Velocity is measured with LDV. e = 4880 Å, 2K = 15!! On the Dopplerburst, 5 peaks occur in 12 7s.

Find: Air velocity: ,

ANALYSIS

Fringe spacing

!E =e

2 sin K

=4880× 10"10

2× sin 7!5!= 1!869× 10"6m

Velocity

!P = 12 7s$4 = 3 7s

, =!E

!P

=1!869× 10"6m3× 10"6 s

= 0.623 m/s

1171

PROBLEM 13.15

Situation: A jet and orice are described in the problem statement.

Find: Coe"cients for an orice: -@' -1' -A!

Assumptions: ,V =&2) × 1!90

ANALYSIS

-@ = ,V$,theory =p2) × 1!90$

p2) × 2

-@ =p1!90$2!0 = 0.975

-1 = IV$I0 = (8$10)2 = 0.640

-A = -@-1 = 0!975× 0!64 = 0.624

1172

PROBLEM 13.16

Situation: A uid jet discharges from a 3 inch orice. At the vena contracta, A = 2!6cm.

Find: Coe"cient of contraction: -1

ANALYSIS

-1 = IV$I0

= (2!6$3)2

= 0.751

1173

PROBLEM 13.17

Situation: A sharp edged orice is described in the problem statement.

Find: Flow coe"cient: <

ANALYSIS

If the angle is 90!' the orice and expected ow pattern is shown below in Fig. A.

We presume that the ow would separate at the sharp edge just as it does for theorice with a knife edge. Therefore, the ow pattern and ow coe"cient < shouldbe the same as with the knife edge.

However, if the orice were very thick relative to the orice diameter (Fig. B), thenthe ow may reattach to the metal of the orice thus creating a di!erent ow patternand di!erent ow coe"cient < than the knife edge orice.

1174

PROBLEM 13.18

Situation: Aging changes in an orice are described in the problem statement.

Find: Explain the changes and how they e!ect the ow coe"cients.

ANALYSIS

Some of the possible changes that might occur are listed below:

1. Blunting (rounding) of the sharp edge might occur because of erosion or corro-sion. This would probably increase the value of the ow coe"cient because -1would probably be increased.

2. Because of corrosion or erosion the face of the orice might become rough.This would cause the ow next to the face to have less velocity than when itwas smooth. With this smaller velocity in a direction toward the axis of theorice it would seem that there would be less momentum of the uid to producecontraction of the jet which is formed downstream of the orice. Therefore, asin case A, it appears that < would increase but the increase would probably bevery small.

3. Some sediment might lodge in the low velocity zones next to and upstream of theface of the orice. The ow approaching the orice (lower part at least) wouldnot have to change direction as abruptly as without the sediment. Therefore,the -1 would probably be increased for this condition and < would also beincreased.

1175

PROBLEM 13.19

Situation: Water (60 !F, ] = 3 cfs) ows through an orice (A = 5 in.) in a pipe(1 = 10 in.). A mercury manometer is connected across the orice.

Find: Manometer deection

Properties: Table A.5 (water at 60 !F): # = 1!94 slug$ ft3 ' ( = 62!37 lbf$ ft3'

7 = 2!36 × 10"5 lbf · s$ ft2, : = 1!22 × 10"5 ft2$ s! Table A.4 (mercury at 68 !F):6 = 13!55!

APPROACH

Find K, and then apply the orice equation to nd the pressure drop across the oricemeter. Apply the manometer equation to relate the pressure drop to the deectionof the mercury manometer.

ANALYSIS

Find K

A$1 = 0!50

ReA =4]

0A:

=4× 3!0

0 × 5$12× 1!22× 10"5

= 7!51× 105

from Fig. 13.13:< = 0!625

Orice section area

I! = (0$4)× (5$12)2 = 0!136 ft2

Orice equation

!% =

µ]

<I!

¶2#

2

=

µ3

0!625× 0!136

¶2µ1!94

2

¶

= 1208 lbf$ ft2

Apply the manometer equation to determine the pressure di!erential across the manome-ter. The result is

!% = (waterM (6mercury " 1)1208 lbf$ ft2 =

¡62!37 lbf$ ft3

¢M (13!55" 1)

Solving the above equation gives the manometer deection (M)

M = 1!54 ft = 18!5 in

1176

PROBLEM 13.20

Situation: Water ows through a 6 inch orice in a 12 inch pipe. Assume & = 60!F' : = 1!22× 10"5 ft2$s.

Find: Discharge: ]

APPROACH

Calculate piezometric head. Then nd K and apply the orice equation.

ANALYSIS

Piezometric head

!M = (1!0)(13!55" 1) = 12!55 ft

Find parameters needed to use Fig. 13.13.

(A$1) = 0!50

(2)!M)045A$: = (2) × 12!55)045(0!5)$(1!22× 10"5)= 1!17× 106

Look up K on Fig. 13.13< = 0!625

Orice equation

] = <I0(2)!M)045

] = 0!625(0$4× 0!52)(64!4× 12!55)045 = 3.49 cfs

1177

PROBLEM 13.21

Situation: A rough orice is described in the problem statement.

Find: Applicability of gure 13.13

ANALYSIS

A rough pipe will have a greater maximum velocity at the center of the pipe relative tothe mean velocity than would a smooth pipe. Because more ow is concentrated nearthe center of the rough pipe less radial ow is required as the ow passes throughthe orice; therefore, there will be less contraction of the ow. Consequently thecoe"cient of contraction will be larger for the rough pipe. So, using < from Fig.13.13 would probably result in an estimated discharge that is too small.

1178

PROBLEM 13.22

Situation: Water ows through a 2.5 inch orice in a 5 inch pipe.Orice diameter is A = 2!5 in = 0!208 ft! Pipe diameter is 1 = 5 in = 0!417 ft!A piezometer measurement gives !M = 4 ft.

Find: Discharge: ]

Properties: Table A.5 (water at 60 !F): : = 1!22× 10"5 ft2$ s!

APPROACH

Find K using the upper horizontal scale on Fig. 13.13, and then apply the oriceequation.

ANALYSIS

Calculate value needed to apply Fig. 13.13

ReA<

=p2)!M

A

:

=p2× (32!2 ft$ s2)× (4 ft)

µ0!208 ft

1!22× 10"5 ft2$ s

¶

= 2! 736 × 105

For A$1 = 0!5' Fig. 13.3 gives

< % 0!63

Orice section area

I! =0

4× (2!5$12 ft)2

= 0!03409 ft2

Orice equation

] = <I!p2)!M

= 0!63×¡0!03409 ft2

¢p2× (32!2 ft$ s2)× (4 ft)

= 0!345 ft3$ s

] = 0!345 cfs

1179

PROBLEM 13.23

Situation: Kerosene at 20 !C ows through an orice. 1 = 3 cm, A$1 = 0!6'!% = 15 kPa

Find: Mean velocity in the pipe

Properties: Kerosene (20 !C) from Table A.4: # = 814 kg/m3' : = 2!37× 10"6 m2$s.

APPROACH

Find K using the upper horizontal scale on Fig. 13.13, and then apply the oriceequation to nd the discharge. Find the velocity in the pipe by using , = ]$I!

ANALYSIS

Calculate value needed to apply Fig. 13.13

ReA$< = (2!%$#)045(A$:)

= (2× 15× 103$814)045(0!6× 0!03$(2!37× 10"6))= 4!611× 104

From Fig. 13.13 for A$1 = 0!6< % 0!66

Orice section area

I! =0A2

4

=0 (0!6× 0!03m)2

4= 2! 545 × 10"4m2

Orice equation

] = <I0(2!%$#)045

= 0!66¡2! 545 × 10"4

¢(2× 15× 103$814)045

= 1! 020× 10"3m3$ s

Flow rate

,pipe =]

Ipipe

=4]

012

=4× (1! 020× 10"3m3$ s)

0 (0!03m)2

= 1! 443m

s

,pipe = 1!44 m$ s

1180

PROBLEM 13.24

Situation: Water at 20 !C ows in a pipe containing two orices, one that is horizontaland one that is vertical. For each orice, 1 = 30 cm and A = 10 cm. ] = 0!1 m3$s.

Find: (a) Pressure di!erential across each orice: !%) ' !%Z !(b) Deection for each mercury-water manometer: !M) ' !MZ

ANALYSIS

Find value needed to apply Fig. 13.13

4]$(0A:) = 4× 0!10$(0 × 0!10× 1!31× 10"6)= 9!7× 105

From Fig. 13.13 for A$1 = 0!333< = 0!60

Orice section area

I! = (0$4)(0!10)2

= 7!85× 10"3 m2

Orice equation

] = <I!p2)!M

Thus

!M = ]2$(<2I22)) = 0!12$(0!62 × (7!85× 10"3)2 × 2× 9!81)!M) = !MZ = 22!97 m"H2O

The deection across the manometers is

M) = MZ = 22!97$(6Hg " 6water) = 1.82 m

The deection will be the same on each manometer

Find !%

%= " %> = (!M = 9790× 22!97 = 224!9 kPa!%) = 225 kPa

For manometer F

((%9$() + R9)" ((%W$() + RW) = !M = 22!97 ftThus,

!%Z = %9 " %W = (!M" ((R9 " RW)= 9' 810(22!97" 0!3)

!%Z = 222 kPa

Because of the elevation di!erence for manometer F, !%) 6= !%Z

1181

PROBLEM 13.25

Situation: A pipe (1 = 30 cm) is terminated with an orice. The orice size isincreased from 15 to 20 cm with pressure drop (!% = 50 kPa) held constant.

Find: Percentage increase in discharge.

Assumptions: Large Reynolds number.

ANALYSIS

Find K valuesAssuming large Re' so K depends only on d/D. From Fig. 13.13

<15 = 0!62

<20 = 0!685

Orice equation

]15 = <15I15p2)!M

]15 = 0!62× (0$4)(0!15)2p2)!M

]15 = 0!01395(0$4)p2)!M

For the 20 cm orice

]20 = 0!685× (0$4)(0!20)2p2)!M

]20 = 0!0274(0$4)p2)!M

Thus the % increase is

(0!0274" 0!01395)$0!01395)× 100 = 96%

1182

PROBLEM 13.26

Situation: Water ows through the orice (vertical orientation) shown in the text-book. 1 = 50 cm, A = 10 cm, !% = 10 kPa, !R = 30 cm.

Find: Flow rate: ]

APPROACH

Find < and !M; then apply the orice equation to nd the discharge ]!

ANALYSIS

Piezometric head

!M = (%1$( + R1)" (%2$( + R2)= !%$( +!R

= 10' 000$9' 790 + 0!3

= 1!321 m of water

Find parameters needed to apply Fig. 13.13

A$1 = 10$50 = 0!20

ReA<

=p2)!M

A

:

=&2× 9!81× 1!321

0!1

10"6

= 5!091× 105

From Fig. 13.13< = 0!60

Orice equation

] = <I!p2)!M

= 0!60× (0$4)× (0!1)2&2× 9!81× 1!321

= 0.0240 m3/s

1183

PROBLEM 13.27

Situation: Flow through an orice is described in the problem statement.

Find: Show that the di!erence in piezometric pressure is given by the pressure dif-ference across the transducer.

ANALYSIS


%-I1 = %1 + (L1

%-I2 = %2 " (L2

so

%-I1 " %-I2 = %1 + (L1 " %2 + (L2= %1 " %2 + ((L1 + L2)

ButL1 + L2 = R1 " R2

or%-I1 " %-I2 = %1 " %2 + ((R1 " R2)

Thus,

%-I1 " %-I2 = (%1 + (R1)" (%2 + (R2)

1184

PROBLEM 13.28

Situation: Water (& = 50 !F, ] = 20 cfs) ows in the system shown in the textbook.b = 0!015!

Find: (a) Pressure change across the orice.(b)Power delivered to the ow by the pump.(c)Sketch the HGL and EGL.

APPROACH

Calculate pressure change by applying the orice equation. Then calculate thehead of the pump by applying the energy equation from section 1 to 2 (section 1 isthe upstream reservoir water surface, section 2 is the downstream reservoir surface).Then, apply the power equation.

ANALYSIS

Re = 4]$(0A:)

= 4× 20$(0 × 1× 1!41× 10"5) = 1!8× 106

Then for A$1 = 0!50, < = 0!625

Orice equation

] = <Ip2)!M or !M = (]$(<I))2$2)

where I = 0$4× 12! Then

!M = (20$(0!625× (0$4)))2$2)!M = 25!8 ft

!% = (!M = 62!4× 25!8 = 1,610 psf

Energy equation

%1$( + T1,21 $2) + R1 + M, = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + 10 + M, = 0 + 0 + 5 +X

ME

M, = "5 + , 2$2)(<G +<W + b2$1) + MEIorice

<G = 0!5; <W = 1!0

The orice head loss will be like that of an abrupt expansion:

MEI orice = (,V " ,pipe)2$(2))

Here, ,V is the jet velocity as the ow comes from the orice.

,V = ]$IV where IV = -1I0

1185

Assume-1 % 0!65 then ,V = 20$((0$4)× 12 × 0!65) = 39!2 ft/s

Also,, = ]$I, = 20$0 = 6!37 ft/s

ThenMEIorice = (39!2" 6!37)2$(2)) = 16!74 ft

Finally,

M, = "5 + (6!372$(2)))(0!5 + 1!0 + (0!015× 300$2)) + 16!74M, = 14!10 ft

* = ](M,$550

= 20× 62!4× 14!10$550

= 32.0 hp

The HGL and EGL are shown below:

1186

PROBLEM 13.29

Situation: Water ows (] = 0!03 m3/s) through an orice. Pipe diameter, 1 = 15cm. Manometer deection is 1 m-Hg.

Find: Orice size: A

APPROACH

Calculate!M! Then guess K and apply the orice equation. Check the guessed valueof < by calculating a value of Reynolds number and then comparing the calculatedvalue with the guessed value.

ANALYSIS

Piezometric head!M = 12!6× 1 = 12!6 m of water

Orice equation

I! = ]$(<p2)!M)

Guess < = 0!7' then

A2 = (4$0)]$(<p2)!M)

A2 = (4$0)× 0!03$h0!7p2) × 12!6

i= 3!47× 10"3 m2

A = 5!89 cm

A$1 = 0!39

ReA = 4× 0!03$(0 × 0!0589× 10"6) = 6!5× 105

< = 0!62

soA =

p(0!7$0!62)× 0!0589 = 0!0626 m

Recalculate < to nd that < = 0!62! Thus,

A = 6!26 cm

1187

PROBLEM 13.30

Situation: Gasoline (6 = 0!68) ows through an orice (A = 6 cm) in a pipe (1 = 10cm).!% = 50 kPa.

Find: Discharge: ]

Properties: : = 4× 10"7 m2/s (Fig. A-3)

Assumptions: & = 20!-!

ANALYSIS

Piezometric head

!M = !%$(

= 50' 000$(0!68× 9' 810)= 7!50 m

Find K using Fig. 13.13

A$1 = 0!60p2)!MA$: =

&2× 9!81× 7!50× 0!06$(4× 10"7) = 1!82× 106

< = 0!66

Orice equation

] = <I!p2)!M

= 0!66× (0$4)(0!06)2p2) × 7!50

] = 0!0226 m3/s

1188

PROBLEM 13.31

Situation: Water ows (] = 2 m3/s) through an orice in a pipe (1 = 1 m). !M = 6m-H2O.

Find: Orice size: A

APPROACH

Guess a value of <. Apply the orice equation to solve for orice diameter. Thencalculate Reynolds number and A$1 in order to nd a new value of <. Iterate untilthe value of < does not change.

ANALYSIS

Orice equation

] = <I!p2)!M

= <

µ0A2

4

¶p2)!M

Algebra

A =

·µ4]

0<

¶µ1

&2)!M

¶¸1*2

Guess < % 0!65

A =

·µ4× 20 · 0!65

¶µ1

&2× 9!81× 6

¶¸1*2

= 0!601m

Calculate values needed for Fig. 13.13

A

1=

0!601

1!0= 0!6

Re =4]

0A:

=4× 2

0 × 0!601× (1!14× 10"6)= 3!72× 106

From Fig. 13.13 with A$1 = 0!6 and Re = 3!72× 106, the value of < is

< = 0!65

Since this is the guessed value, there is no need to iterate.

A = 0!601m

1189

PROBLEM 13.32

Situation: Water ows (] = 3 m3/s) through an orice in a pipe (1 = 1!2 m).!% = 50 kPa.

Find: Orice size: A

Assumptions: < = 0!65; & = 20!-!

ANALYSIS

Piezometric head

!M = !%$(

= 50' 000$9790

= 5!11m

Orice equation

A2 = (4$0)× 3!0$(0!65&2× 9!81× 5!11) = 0!587

A = 0!766 m

Check <:

ReA = 4]$(0A:)

= 4× 3!0$(0 × 0!766× 10"6)= 5× 106

From Fig. 13.13 for A$1 = 0!766$1!2 = 0!64' < = 0!67Try again:

A =p(0!65$0!67)× 0!766 = 0!754

Check <: ReA = 5× 106 and A$1 = 0!63! From Fig. 13.13 < = 0!67 so

A =p(0!65$0!670)× 0!766 = 0.754 m

1190

PROBLEM 13.33

Situation: Water ows through a hemicircular orice as shown in the textbook.

Find: (a) Develop a formula for discharge.(b) Calculate ]!

APPROACH

Apply the ow rate equation, continuity principle, and the Bernoulli equation to solvefor ]!

ANALYSIS

Bernoulli equation%1 + #,

21 $2 = %2 + #,

22 $2


,1I1 = ,2I2; ,1 = ,2I2$I1

,2 =p2(%1 " %2)$%$

q1" (I22$I21)

Flow rate equation

] = I2,2·I2$

q1" (I22$I21)

¸p2!%$#

but I2 = -1I0 where I0 is the section area of the orice. Then

] =

·-1I0$

q1" (I22$I21)

¸p2!%$#

or orice equation

] = <I0p2!%$#

where < is the ow coe"cient. Assume < = 0!65; Also I = (0$8)× 0!302 = 0!0353m2! Then

] = 0!65× 0!0353p2× 80' 000$1' 000

] = 0!290 m3/s

1191

PROBLEM 13.34

Situation: Water (20 !C, ] = 0!75 m3$s) ows through a venturi meter (A = 30 cm)in a pipe (1 = 60 cm).

Find: Deection on a mercury manometer.

ANALYSIS

Reynolds number

ReA = 4× 0!75$(0 × 0!30× 1× 10"6)= 3!18× 106

For A$1 = 0!50' nd K from Fig. 13.13

< = 1!02

Venturi equation

!M = []$(<I5)]2 $(2))

=£!75$(1!02× (0$4)× 0!32)

¤2$(2× 9!81)

= 5!52 m H2O

Manometer equation

M66 = !M62B$

µ(66(62B

" 1¶

M = 5!52$12!6

M = 0!44 m

1192

PROBLEM 13.35

Situation: Water (] = 10 m3$s) ows through a venturi meter in a horizontal pipe(1 = 2 m). !% = 200 kPa.

Find: Venturi throat diameter.

Assumptions: & = 20!-!

ANALYSIS

Guess that < = 1!01' and then proceed with calculations

] = <I!$p2)!M

where !M = 200' 000 Pa/(9,790 N/m3) = 20!4 m. ThenVenturi equation

I5 = ]$(<p2)!M)

or

0A2$4 = ]$(<p2)!M)

A = (4]$(0<p2)!M))

1*2

A = (4× 10$(0 × 1!01p2) × 20!4))1*2 = 0!794 m

Calculate < and compare with the guessed value

Re = 4]$(0A:) = 1!6× 107

Also A$1 = 0!4 so from Fig. 13.13 < % 1!0! Try again:

A = (1!01$1!0)1*2 × 0!794 = 0.798 m

1193

PROBLEM 13.36

Situation: A venturi meter is described in the problem statement.

Find: Rate of ow: ]

ANALYSIS

Find K

!M = 4 ft and A$1 = 0!33

ReA$< = (1$3)&2× 32!2× 4$1!22× 10"5) = 4!4× 105

< = 0!97 (Estimated from Fig. 13.13)

Venturi equation

] = <Ip2)M

= 0!97(0$4× 0!3332)&2× 32!2× 4

] = 1!36 cfs

1194

PROBLEM 13.37

Situation: A venturi meter is described in the problem statement.

Find: Range that the venturi meter would read: !p

ANALYSIS

The answer is -10 psi G ! p G 0 so the correct choice is b) .

1195

PROBLEM 13.38

Situation: Water ows through a horizontal venturi meter. !% = 100 kPa,A = 1 m, 1 = 2 m.

Find: Discharge: ]

Properties: : = 10"6 m2$s.

ANALYSIS

!% = 100 kPa so !M = !%$( = 100' 000$9790 = 10!2 m

Find K

p2)!MA$: =

&2× 9!81× 10!2× 1$10"6

= 1!4× 107

Then < % 1!02 (extrapolated from Fig. 13.13).Venturi equation

] = <Ip2)!M

= 1!02× (0$4)× 12p2) × 10!2

= 11.3 m3/s

1196

PROBLEM 13.39

Situation: A poorly designed venturi meter is described in the problem statement.

Find: Correction factor: <

ANALYSIS

Because of the streamline curvature (concave toward wall) near the pressure tap, thepressure at point 2 will be less than the average pressure across the section. Therefore,]0 will be too large as determined by the formula. Thus, < G 1.

1197

PROBLEM 13.40

Situation: Water (50 !F) ows through a vertical venturi meter. !% = 6!2 psi, A = 6in., 1 = 12 in., : = 1!4× 10"5 ft2$s.

Find: Discharge: ]

ANALYSIS

!% = 6!20 psi = 6.2 × 144 psf

Thus!M = 6!20× 144$62!4 = 14!3 ft

Find K

ReA<

=p2)!M

A

:

=&2× 32!2× 14!3

µ6$12

1!4× 10"5

¶

= 10!8× 105

So < = 1!02.Venturi equation

] = <I5p2)!M

= 1!02× (0$4)× (6$12)2&2× 32!2× 14!3

] = 6!08 cfs

1198

PROBLEM 13.41

Situation: Gasoline (6 = 0!69) ows through a venturi meter. A di!erential pressuregage indicates !% = 45 kPa.A = 20 cm, 1 = 40 cm, 7 = 3× 10"4 N·s$m2!

Find: Discharge: ]

ANALYSIS

!M = 45' 000$(0!69× 9' 810) = 6!65 m: = 7$# = 3× 10"4$690 = 4!3× 10"7 m2/s

Then p2)!MA$: =

&2× 9!81× 6!65× 0!20$(4!3× 10"7) = 5!3× 106

From Fig. 13.13< = 1!02

Venturi equation

] = <Ip2)!M

= 1!02× (0$4)× (0!20)2&2× 9!81× 6!65

] = 0!366 m3/s

1199

PROBLEM 13.42

Situation: Water passes through a ow nozzle. !% = 8 kPa. A = 2 cm, A$1 = 0!5,: = 10"6 m2/s' # = 1000 kg/m3!

Find: Discharge: ]

APPROACH

Find K, and then apply the orice equation.

ANALYSIS

Find K

ReA$< = (2!%$#)045(A$:)

= ((2× 8× 103)$(1' 000))045(0!02$10"6)= 8!0× 104

From Fig. 13-13 with A$1 = 0!5; < = 0!99!Venturi equation

] = <I(2!%$#)045

= (0!99)(0$4)(0!022)(2× 8× 103$103)045

] = 0!00124 m3/s

1200

PROBLEM 13.43

Situation: Water ows through the annular venturi that is shown in the textbook.

Find: Discharge

Assumptions: -A = 0!98

ANALYSIS

From Eq. (13.5)

< = -A$p1" (I2$I1)2

= 0!98$&1" 0!752

< = 1!48

Venturi equation

I = 0!00147m2

] = <I(2)!M)045

] = (1!48)(0!00147)(2!0× 9!81× 1)045

] = 0!00964 m3/s

1201

PROBLEM 13.44

Situation: The problem statement describes a ow nozzle with A$1 = 1!3!

Find: Develop an expression for head loss.

APPROACH

Apply the sudden expansion head loss equation and the continuity principle.

ANALYSIS


,0I0 = ,VIV

,V = ,0I0$IV

= ,0 × (3$1)2 = 9,0


ME = (,V " ,0)2$2)

Then

ME = (9,0 " ,0)2$2)

= 64, 20 $2)

1202

PROBLEM 13.45

Situation: A vortex meter (1 cm shedding element) is used in a 5 cm diameter duct.For shedding on one side of the element, 65 = 0!2 and b = 50 Hz.

Find: Discharge: ]

APPROACH

Find velocity from the Strouhal number (6P = Q1$, ) ! Then, nd the dischargeusing the ow rate equation.

ANALYSIS

6P = Q1$,

, = Q1$6P

= (50)(0!01)$(0!2)

= 2!5 m/s

Flow rate equation

] = , I

= (2!5)(0$4)(0!052)

] = 0!0049 m3/s

1203

PROBLEM 13.46

Situation: A rotometer is described in the problem statement.

Find: Describe how the reading on the rotometer would be corrected for nonstandardconditions.

APPROACH

Apply equilibrium, drag force, and the ow rate equation.

ANALYSIS

The deection of the rotometer is a function of the drag on the rotating element.Equilibrium (drag force balances weight):

H9 = .

-9I#,2$2 = /)

Thus, =

p2)/$(#I-9)

Since all terms are constant except density

,$,std. = (#std.$#)045

applying the ow rate equation gives

] = , I

! ]$]std. = (#std.$#)045 (1)

Correct by calculating # for the actual conditions and then use Eq. (1) to correct ]!

1204

PROBLEM 13.47

Situation: A rotometer is calibrated for gas with #standard = 1!2 kg/m3' but is used

for # = 1!1 kg/m3.The rotometer indicates ] = 5 L$s.

Find: Actual gas ow rate (]) in liters per second.

APPROACH

Apply equilibrium, drag force, and the ow rate equation.

ANALYSIS

The deection of the rotometer is a function of the drag on the rotating element.Equilibrium of the drag force with the weight of the oat gives

H9 = .

-9I#, 2

2= /)

Use the above equation to derive a ratio of standard to nonstandard conditions gives

,

,std.=

r#std.#

also] = , I

Therefore]

]std.=

r#std.#

Thus

] = 5×r1!2

1!1

] = 5!22 L/s

1205

PROBLEM 13.48

Situation: One mode of operation of an ultrasonic ow meter involves the time for awave to travel between two measurement stations–additional details are provided inthe problem statement.

Find: (a) Derive an expression for the ow velocity.(b) Express the ow velocity as a function of 2' Z and P!(c) Calculate the ow velocity for the given data.

ANALYSIS

(a)

P1 = 2$(Z+ , )

P2 = 2$(Z" , )!P = P2 " P1

=2

Z" ,"

2

Z+ ,

=22,

Z2 " , 2(1)

Thus

(Z2 " , 2)!P = 22,

, 2!P+ 22, " Z2!P = 0, 2 + (22,$!P)" Z2 = 0

Solving for , :

[("22$!P)±p(22$!P)2 + 4Z2]$2 = ("2$!P)±

p(2$!P)2 + Z2

Selecting the positive value for the radical

, = (2$!P)["1 +p1 + (Z!P$2)2]

(b) From Eq. (1)

!P =22,

Z2for Z FF ,

, = 12!52E

(c)

, =(300)2(10× 10"3)

2× 20= 22.5 m/s

1206

PROBLEM 13.49

Situation: Water ows over a rectangular weir. 2 = 4 m; 4 = 0!20 m, * = 0!25 m.

Find: Discharge: ]

ANALYSIS

Flow coe"cient

< = 0!40 + 0!05

µ4

*

¶

= 0!40 + 0!05

µ0!20

0!25

¶

= 0!440

Rectangular weir equation

] = <p2)243*2

= 0!44×&2× 9!81× 4× (0!2)3*2

= 0!6973m3$ s

Thus] = 0!697 m3/s

1207

PROBLEM 13.50

Situation: Water ows over a 60! triangular weir. 4 = 0!35 m.

Find: Discharge: ]

ANALYSIS

Triangular weir equation

] = 0!179p2)45*2

] = 0!179&2× 9!81(0!35)5*2

] = 0!0575 m3/s

1208

PROBLEM 13.51

Situation: Two weirs (A and B) are described in the problem statement.

Find: Relationship between the ow rates: ]= and ]>

ANALYSIS

Correct choice is c) ]= G ]> because of the side contractions on I.

1209

PROBLEM 13.52

Situation: A rectangular weir is described in the problem statement.

Find: The height ratio: 41$42

ANALYSIS

Correct choice is b) (41$42) G 1 because < is larger for smaller height of weir asshown by Eq. (13-10); therefore, less head is required for the smaller * value.

1210

PROBLEM 13.53

Situation: A rectangular weir is being designed for ] = 4 m3$s, 2 = 3 m, Waterdepth upstream of weir is 2 m.

Find: Weir height: *

ANALYSIS

First guess 4$* = 0!60! Then

< = 0!40 + 0!05(0!60) = 0!43!

Rectangular weir equation (solve for 4)

4 = (]$(<p2)2))2*3

= (4$(0!43p(2)(9!81)(3))2*3 = 0!788 m

Iterate:

4$* = 0!788$(2" 0!788) = 0!65; < = 0!40 + !05(!65) = 0!433

4 = 4$(0!433p(2)(9!81(3))2*3 = 0!785 m

Thus:* = 2!0"4 = 2!00" 0!785 = 1.215 m

1211

PROBLEM 13.54

Situation: The head of the rectangular weir described in Prob. 13.53 is doubled.


ANALYSIS

Rectangular weir equation] = <

p2)243*2

Correct choice is c) .

1212

PROBLEM 13.55

Situation: A basin is draining over a rectangular weir. 2 = 2 ft, * = 2 ft. Initially,4 = 12 in.

Find: Time for the head to decrease from 4 = 1 ft to 0!167 ft (2 in).

ANALYSIS

With a head of 4 = 1 ft4

*=1

2= 0!5

thus

<# = 0!40 + 0!05 # 0!5= 0!425

With a head of 4 = 0!167 ft (2 in)

4

*=2$12

2= 0!0833

and

<( = 0!40 + 0!05 # 0!0833= 0!404

As a simplication, assume < is constant at

< = (!425 + !404) $2

= 0!415


] = 0!415p2)243*2

For a period of AP the volume of water leaving the basin is equal to I>A4 whereI> = 100 ft2 is the plan area of the basin. Also this volume is equal to ]AP.Equating these two volumes yields:

]AP = I>A4³0!415

p2)243*2

ÁP = I> A4

Separate variables

AP =I> A4¡

0!415&2)243*2

¢

=

¡100 ft2

¢A4³

0!415p2× (32!2 ft$ s2) (2 ft)43*2

´

=³15!01

&ft · s

´ A4

43*2

1213

Integrate

!5Z

0

AP = (15!01)

1Z

04167

A4

43*2

!P = ("15!01)µ2&4

¶1

04167

= ("15!01)µ2&1"

2&0!167

¶

= 43!44 s

!P = 43!4 s

1214

PROBLEM 13.56

Situation: A piping system and channel are described in the textbook. The channelempties over a rectangular weir.

Find: (a) Water surface elevation in the channel.(b) Discharge.

ANALYSIS


] = <p2)243*2

Assume 4 = 1$2 ft. Then < = 0!4 + 0!05(12$3) = 0!41, then

] = 0!41&64!4× 243*2

] = 6!5843*2 (1)

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + 100 = 0 + 0 + 3 +4 +X

ME (2)

Combined head lossX

ME = (, 2$2))(<G + b2$1 + 2<0 +<W)

= (, 2$2))(0!5 + b(100$(1$3)) + 2× 0!35 + 1)

Assume b = 0!02 (rst try). ThenX

ME = 8!2,2$2)

Eq. (2) then becomes97 = 4 + 8!2, 2$2) (3)

But , = ]$I so Eq. (3) is written as

97 = 4 + 8!2]2$(2)I2)

where

I2 = ((0$4)(1$3)2)2 = 0!00762 ft4

97 = 4 + 8!2]2$(2) × 0!00762)97 = 4 + 16!72]2 (4)

1215

Solve for ] and 4 between Eqs. (1) and (4)

97 = 4 + 16!72]2

97 = 4 + 16!72(6!5843*2)2

4 = 0!51 ft and ] = 2!397 ft3/s

Now check Re and bFlow rate equation

, = ]$I

= 27!5 ft/s

Reynolds number

Re = , 1$: = 27!5× (1$3)$(1!4× 10"5)Re = 6!5× 105

From Figs. 10.8 and Table 10.2 b = 0!017! Then Eq. (3) becomes

97 = 4 + 7!3, 2$2)

and Eq. (4) is97 = 4 + 14!88]2

Solve for 4 and ] again:

4 = 0!53 ft and ] = 2!54 ft3/s

1216

PROBLEM 13.57

Situation: Water ows into a tank at a rate ] = 0!1 m3/s. The tank has two outlets:a rectangular weir (* = 1 m, 2 = 1 m) on the side, and an orice (A = 10 cm) onthe bottom.

Find: Water depth in tank.

APPROACH

Apply the rectangular weir equation and the orice equation by guessing the head onthe orice and iterating.

ANALYSIS

Guess the head on the orice is 1.05 m.Orice equation

]orice = <I0p2)M; < % 0!595

]orice = 0!595× (0$4)× (0!10)2&2× 9!81× 1!05 = 0!0212 m3/s


]weir = <p2)243*2; 4weir = (]$(<

p2)2)2*3 where < % 0!405

4weir = ((0!10" 0!0212)$(0!405&2× 9!81× 1))2*3 = 0!124 m

Try again:

]orice = (1!124$1!05)1*2 × 0!0212 m3$s = 0!0219 m3$s4weir = ((0!10" 0!0219)$(0!405

&2× 9!81× 1))2*3 = 0.124 m

4weir is same as before, so iteration is complete. Depth of water in tank is 1.124 m

1217

PROBLEM 13.58

Situation: Weirs with sharp edges are described in the problem statement.

Find: (a) If the weir behave di!erently if the edges were not sharp.(b) Explain what might happen without the vent downstream and how it would a!ectthe ow and glow coe"cient.

ANALYSIS

(a) With a sharp edged weir, the ow will break free of the sharp edge and a denite(repeatable) ow pattern will be established. That assumes that the water surfacesboth above and below the nappe are under atmospheric pressure. However, if thetop of the weir was not sharp then the lower part of the ow may follow the roundedportion of the weir plate a slight distance downstream.

This would probably lessen the degree of contraction of the ow. With less contrac-tion, the ow coe"cient would be larger than given by Eq. (13.10).(b) If the weir is not ventilated below the Nappe, for example a weir that extends the

full width of a rectangular channel (as shown in Fig. 13.18), then as the water plungesinto the downstream pool air bubbles would be entrained in the ow and some of theair from under the Nappe would be carried downstream. Therefore, as the air underthe Nappe becomes evacuated, a pressure less than atmospheric would be establishedin that region. This would draw the Nappe downward and cause higher velocities tooccur near the weir crest. Therefore, greater ow would occur than indicated by useof Eqs. (13.9) and (13.10).

1218

PROBLEM 13.59

Situation: Water ows over a rectangular weir. 2 = 10 ft, * = 3 ft, and 4 = 1!8 ft.

Find: Discharge: ]

APPROACH

Apply the rectangular weir equation.

ANALYSIS

The ow coe"cient is

< = 0!40 + 0!05

µ4

*

¶

= 0!40 + 0!05

µ1!8

3!0

¶

= 0!43


] = <p2)243*2

= 0!43³&2 · 32!2

´10× 1!83*2

= 83!3 ft3$ s

1219

PROBLEM 13.60

Situation: Water (60 !F) ows into a reservoir through a venturi meter (< = 1,I! = 12 in2, !% = 10 psi). Water ows out of the reservoir over a 60! triangularweir.

Find: Head of weir: 4

ANALYSIS

Venturi equation

] = <I!p2!%$#

= 1× (12$144)p2× 10× 144$1!94

= 3!21 ft3/s


] = 0!179p2)45*2

3!21 = 0!179&64!445*2

4 = 1.38 ft = 16.5 in.

1220

PROBLEM 13.61

Situation: Water enters a tank through two pipes A and B. Water exits the tankthrough a rectangular weir.

Find: Is water level rising, falling or staying the same?

APPROACH

Calculate ]in and ]out and compare the values. Apply the rectangular weir equationto calculate ]out and the ow rate equation to calculate ]in.

ANALYSIS


]out = <(2))045243*2

< = 0!40 + 0!05(1$2) = 0!425

]out = 0!425(8!025)(2)(1)

= 6!821 cfs

Flow rate equation

]in = ,=I= + ,>I>

= 4(0$4)(12) + 4(0$4)(0!52)

= 0(1!25) = 3!927 cfs

]in G ]out; therefore, water level is falling

1221

PROBLEM 13.62

Situation: Water exits an upper reservoir across a rectangular weir (2$4F = 3' *$4F =2) and then into a lower reservoir. The water exits the lower reservoir through a 60!

triangular weir.

Find: Ratio of head for the rectangular weir to head for the triangular weir: 4F$4-


APPROACH

Apply continuity principle by equating the discharge in the two weirs.

ANALYSIS


] = (0!40 + !05(1$2))p2)(34F)4

145F (1)

Triangular weir equation] = 0!179

p2)4245

- (2)

Equate Eqs. (1) and (2)

(0!425p2)(3)4245

F = 0!179p2)4245

-

(4F$4- )245 = 0!179$(3× 0!425)

4F$4- = 0!456

1222

PROBLEM 13.63

Situation: For problem 13.62, the ow entering the upper reservoir is increased by50%.

Find: Describe what will happen, both qualitatively and quantitatively.

APPROACH

Apply the rectangular and triangular weir equations.

ANALYSIS

As soon as the ow is increased, the water level in the rst reservoir will start torise. It will continue to rise until the outow over the rectangular weir is equal tothe inow to the reservoir. The same process will occur in the second reservoir untilthe outow over the triangular weir is equal to the inow to the rst reservoir.

Calculations

Determine the increase in head on the rectangular weir with an increase in dischargeof 50%. Initial conditions: 4F$* = 0!5 so

< = 0!4 + !05× !5 = 0!425

Then]F# = 0!425

p2)24

3*2F# (1)

Assume<( = <# = 0!425 (rst try)

Then]F( = 0!425

p2)24

3*2F( (where ]F( = 1!5]#) (2)

Divide Eq. (2) by Eq. (1)

]F($]F# = (0!4252$0!4252)(4F($4F#)3*2

4F($4F# = (1!5)2*3 = 1!31

Check <# :< = 0!40 + !05× 0!5× 1!31 = 0!433

Recalculate 4F($4F#!

4F($4F# = ((0!425$0!433)× 1!5)2*3 = 1!29

The nal head on the rectangular weir will be 29% greater than the initial head . Nowdetermine the increase in head on the triangular weir with a 50% increase in discharge.

]-($]-# = (4-($4-#)5*2

or 4-($4-# = (]-($]-#)

= (1!5)2*5

= 1.18

The head on the triangular weir will be 18% greater with the 50% increase in discharge.

1223

PROBLEM 13.64

Situation: A rectangular weir (2 = 3 m) is situated in a canal.The water depth is 2 m and ] = 6 m3$s.

Find: Necessary weir height: *

APPROACH

Calculate the height by applying the rectangular weir equation by guessing < anditerating.

ANALYSIS

Rectangular weir equation] = <

p2)243*2

Assume < % 0!41 then

4 = (]$(0!41p2) × 3))2*3

4 = (6$(0!41×&2× 9!81× 3))2*3 = 1!10 m

Then

* = 2!0" 1!10 = 0!90 mand 4$* = 1!22

Check guessed < value:

< = 0!40 + 1!22× 0!05 = 0!461

Since this doesn’t match, recalculate 4:

4 = (6$(0!461×&2× 9!81× 3))2*3 = 0!986 m

So height of weir

* = 2!0" 0!986 = 1!01 m4$* = 0!976

Try again:

< = 0!40 + 0!976× 0!05 = 0!4494 = (6$(0!449×

&2× 9!81× 3))2*3 = 1!00 m

* = 2!00" 1!00 = 1.00 m

1224

PROBLEM 13.65

Situation: Water ows over a 60! triangular weir, 4 = 1!2 ft.

Find: Discharge: ]

APPROACH

Apply the triangular weir equation.

ANALYSIS

] = 0!179p2)45*2

] = 0!179p2× (32!2 ft$ s2)× (1!2 ft)5*2

] = 2!27 ft3/s

1225

PROBLEM 13.66

Situation: Water ows over a 45! triangular weir. ] = 10 cfm -A = 0!6!

Find: Head on the weir: 4

ANALYSIS

] = (8$15)-A(2))045 tan(K$2)45*2

] = (8$15)(0!60)(64!4)045 tan(22!5!)45*2

] = 1!06445*2

4 = (]$1!064)2*5

= (10$(60× 1!064))2*5

4 = 0!476 ft

1226

PROBLEM 13.67

Situation: A pump transports water from a well to a tank.The tank empties through a 60! triangular weir.Additional details are provided in the problem statement.

Find: Water level in the tank: M

Assumptions: b = 0!02

APPROACH

Apply the triangular weir equation to calculate M. Apply the ow rate equation andthe energy equation from well water surface to tank water surface to relate ] and M.

ANALYSIS

c&$1 = 0!001Energy equation

%1$( + T1,21 $2) + R1 + M, = %2$( + T2,

22 $2) + R2 +

XME

0 + 0 + 0 + M, = 0 + 0 + (2 + M) + (, 2$2))(<G + (b2$1) +<W)

Inserting parameter values

20 = (2 + M) + (, 2$2))(0!5 + (0!02× 2!5$0!05) + 1)18 = M+ 0!127, 2

, = ((18" M)$0!127)045

] = , I

= ((18" M)$0!127)045(0$4)(0!05)2 (10)

= 0!00551(18" M) (1)

Triangular weir equation] = 0!179

p2)4245

where 4 = M" 1! Then

] = 0!179p2)(M" 1)245 = 0!793(M" 1)245 (2)

To satisfy continuity, equate (1) and (2)

0!00551(18" M)045 = 0!793(M" 1)245

0!00695(18" M)045 = (M" 1)245

Solve for M:M = 1!24 m

Also, upon checking Re we nd our assumed b is OK.

1227

PROBLEM 13.68

Situation: A pitot tube is used to record data in subsonic ow. %5 = 140 kPa, % = 100kPa, &5 = 300 K.

Find: (a) Mach number: +(b) Velocity: ,

ANALYSIS

Use total pressure to nd the Mach number

%5$%1 = (1 +c " 12+2)

::"1

= (1 + 0!2+2)345 for air

(140$100) = (1 + 0!2+2)345

+ = 0!710

Total temperature

&5$& = 1 + 0!2+2

& = 300$1!10 = 273

Speed of sound

Z =&c"&

=p(1!4)(287)(273)

= 331 m/s

Mach number

, = +Z

= (0!71)(331)

, = 235 m/s

1228

PROBLEM 13.69

Situation: Eq. (13.13), the Rayleigh supersonic Pitot formula, can be used to calcu-late Mach number from data taken with a Pitot-static tube.

Find: Derive the Rayleigh supersonic Pitot formula.

ANALYSIS

The purpose of the algebraic manipulation is to express %1$%52 as a function of +1

only.

For convenience, express the group of variables below as

H = 1 + ((c " 1)$2)+2

X = c+2 " ((c " 1)$2)%1$%52 = (%1$%51)(%51$%52) = (%1$%51)(%1$%2)(H1$H2)

[*["1

From Eq. (12-38),%1$%2 = (1 + c+

22 )$(1 + c+

21 )

So%1$%52 = (%1$%51)((1 + c+

22 )$(1 + c+

21 ))(H1$H2)

[*["1

From Eq. (12-40), we have

(+1$+2) = ((1 + c+21 )$(1 + c+

22 ))(H2$H1)

1*2

Thus, we can write

(%1$%52) = (%1$%51)(+2$+1)(H1$H2)[+1*(2(["1))

But, from Eq. (12-41)+2 = (H1$X1)

1*2

Also, %1$%51 = 1$(H[*["11 )! So

%1$%52 = 1$(H[*["11 )(H

1*21 $X

1*21 )(1$+1)(H1$H2)

[+1*(2(["1))

= (X"1*21 $+1)H

"([+1)*2(["1)2

However,

H2 = 1 + ((c " 1)$2)+22 = 1 + ((c " 1)$2)(H1$X2) = (((c + 1)$2)+1)

2$X1

Substituting for H2 in expression for %1$%52 gives

%1$%52 = (1$+1)(X1*["11 )$((c + 1)$2+1)

[+1*["1

Multiplying numerator and denominator by (2$c + 1)1*["1 gives

%1$%52 ={[2c+2

1$(c + 1)]" (c " 1)$(c + 1)}1*(["1)

{[(c + 1)$2]+21}[*(["1)

1229

PROBLEM 13.70

Situation: A Pitot tube is used in supersonic airow. % = 54 kPa, %5 = 200 kPa,&5 = 350 K.

Find: (a) Mach number: +1

(b) Velocity: ,1

APPROACH

Apply the Rayleigh Pitot tube formula to calculate the Mach number. Then apply theMach number equation and the total temperature equation to calculate the velocity.

ANALYSIS

%1$%52 ={[2c+2

1$(c + 1)]" (c " 1)$(c + 1)}1*(["1)

{[(c + 1)$2]+21}[*(["1)

54$200 = (1!1667+21 " 0!1667)

245$(1!2+21 )345

and solving for +1 gives +1 = 1!79

&1 = &5$£1 + 0!5(c " 1)+2

1

¤

&1 = 350$(1 + 0!2(1!79)2)

= 213 K

Z1 =&c"&

=p(1!4)(287)(213)

= 293 m/s

,1 = +1Z1

= 1!79× 293,1 = 521 m/s

1230

PROBLEM 13.71

Situation: A venturi meter is used to measure ow of helium–additional details areprovided in the problem statement.%1 = 120 kPa %2 = 80 kPa c = 1!66 12$11 = 0!5, &1 = 17!- " = 2077 J/kg·K.


APPROACH

Apply the ideal gas law and Eq. 13.16 to solve for the density and velocity at section2. Then nd mass ow rate / = #2I2,2.

ANALYSIS

Ideal gas law

#1 = %1$("&1)

= 120× 103$(2' 077× 290)= 0!199 kg/m3

%1$#1 = 6!03× 105

Eq. (13.16)

,2 = ((5)(6!03× 105)(1" 0!666044)$(1" (0!666142 × 0!54)))1*2 = 686 m/s#2 = (%2$%1)

1*[#1 = (0!666)046#1 = 0!784#1 = 0!156 kg/m

3

Flow rate equation

/ = #2I2,2

= (0!156)(0$4× 0!0052)(686)

= 0.0021 kg/s

1231

PROBLEM 13.72

Situation: An orice is used to measure the ow of methane.%1 = 150 kPa, %2 = 110 kPa, & = 300 K' A = 0!8 cm, and A$1 = 0!5.


Properties: For methane: " = 518 J/kg*K, c = 1!31' and : = 1!6× 10"5 m2$s.

ANALYSIS

Ideal gas law

#1 =%1"&

=150× 103

518× 300= 0!965 kg/m3

Parameter on the upper scale of Fig. 13.13

2)!M = 2!%$#1= (2(30× 103))$0!965= 6!22× 104

ReA<

=p2)!M

µA

:

¶

=&6!22× 104

µ0!008

1!6× 10"5

¶

= 1!25× 105

From Fig. 13.13< = 0!62

S = 1" ((1$1!31)(1" (120$150))(0!41 + 0!35(0!4)4))= 0!936

Flow rate equation

/ = (0!63)(0!936)(0!785)(0!008)2p(2)(0!965)(30× 103)

= 0.00713 kg/s

1232

PROBLEM 13.73

Situation: Air ows through a 1 cm diameter orice in a 2 cm pipe. The pressurereadings for the orice are 150 kPa (upstream) and 100 kPa (downstream).

Properties: For air #(upstream) = 1!8 kg/m3' : = 1!8× 10"5 m2/s, c = 1!4!

Find: Mass ow rate

ANALYSIS

I0$I1 = (1$2)2 = 0!25; I0 = 7!85× 10"5 m2

Expansion factor:

S = 1" {(1$c)(1" (%2$%1))(0!41 + 0!35(I0$I1)2)}S = 1" {(1$1!4)(1" (100$150)(0!41 + 0!35(!25)2)}

= 0!897

/ = S I0<(2#1(%1 " %2))045

ReA$< = (2!%$#)045A$:

= (2× 50× 103$1!8)045(!01$(1!8× 10"5))= 236× 556= 1!31× 105

From Fig. 13.13 < = 0!63

/ = (0!897)(7!85× 10"5)(0!63)(2× 1!8× 50× 103)045

= 1.88×10"2 kg/s

1233

PROBLEM 13.74

Situation: Hydrogen (100 kPa, 15 !C) ows through an orice (A$1 = 0!5' < = 0!62)in a 2 cm pipe. The pressure drop across the orice is 1 kPa.

Find: Mass ow rate

ANALYSIS

A$1 = 0!50

A = 0!5× 0!02 m = 0!01 m

From Table A.2 for hydrogen (& = 15!- = 288<) : c = 1!41' and # = 0!0851 kg/m3.

I0 = (0$4)(0!01)2 = 7!85× 10"5 m2

/ = S I0<(2#1!%)

/ = (1)(7!85× 10"5)(0!62)(2(0!0851)(1000))045

/ = 6!35× 10"4 kg/s

1234

PROBLEM 13.75

Situation: Natural gas (50 psig, 70 !F) ows in a pipe.A hole (A = 0.2 in) leaks gas.%'5" = 14 psia

Find: Rate of mass ow out of the leak: /

Properties: For natural gas: c = 1!31' " = 3098 ft-lbf/slug !"!

Assumptions: The hole shape is like a truncated nozzle

ANALYSIS

Hole area

I =0A2

4=0 (0!2$12)2

4= 2!182× 10"4 ft2

Pressure and temperature conversions.

%5 = (50 + 14) = 64 psia = 9216 psfa

& = (460 + 70) = 530 !R

To determine if the ow is sonic or subsonic, calculate the critical pressure ratio

%&%5

=

µ2

c + 1

¶ ::"1

=

µ2

1!31 + 1

¶ 14311431"1

= 0!544

Compare this to the ratio of back pressure to total pressure:

%0%5

=14 psia64 psia

= 0!219

Since, %0$%5 G %&$%5, the exit owmust be sonic (choked). Calculate the critical mass owrate.

/ =%5I&&"&5

&c

µ2

c + 1

¶ (:+1)2(:"1)

=9216× 2!182× 10"4p

3098× 530)

&1!31

µ2

1!31 + 1

¶ (1431+1)2(1431"1)

= 0!00105 slug$ s

/ = 0!0338 lbm/s

1235

PROBLEM 13.76

Situation: Weirs are often subject to physical e!ects—additional details are providedin the problem statement.

Find: (a) List all physical e!ects not indicated in the text.(b) Explain how each might inuence the ow.

ANALYSIS

Some of the physical e!ects that might occur are:

a Abrasion might cause the weir crest to be rounded and this would undoubtedlyproduce greater ow than indicated by Eqs. 13.9 and 13.10 (see the answer toproblem 13.58)

b If solid objects such as oating sticks come down the canal and hit the weir theymay dent the weir plate. Such dents would be slanted in the downstreamdirection and may even cause that part of the weir crest to be lower than theoriginal crest. In either case these e!ects should cause the ow to be contractedless than before thus increasing the ow coe"cient.

c Another physical e!ect that might occur in an irrigation canal is that sedimentmight collect upstream of the weir plate. Such sediment accumulation wouldforce ow away from the bottom before reaching the weir plate. Therefore,with this condition less ow will be deected upward by the weir plate and lesscontraction of the ow would occur. With less contraction the ow coe"cientwould be increased. For all of the physical e!ects noted above ow would beincreased for a given head on the weir.

1236

PROBLEM 13.77

Situation: A constant head laboratory tank is described in the problem statement.

Find: Design a piece of equipment that could be used to determine the coe"cient ofcontraction for ow through an orice.

ANALYSIS

A jet to be studied can be produced by placing an orice in the side of a rectangulartank as shown below.

The plate orice could be machined from a brass plate so that the upstream edgeof the orice would be sharp. The diameter of the orice could be measured byinside calipers and a micrometer. The contracted jet could be measured by outsidecalipers and micrometer. Thus the coe"cient of contraction could be computed as-1 = (AV$A)

2. However, there may be more than desired error in measuring the waterjet diameter by means of a caliper. Another way to estimate AV is to solve for it fromIV where IV is obtained from IV = ]$,V! Then AV = (4IV$0)

12 ! The discharge, ],

could be measured by means of an accurate ow meter or by a weight measurementof the ow over a given time interval. The velocity at the vena contracta could befairly accurately determined by means of the Bernoulli equation. Measure the headon the orice and compute ,V from ,V =

&2)M where M is the head on the orice.

Because the ow leading up to the vena contracta is converging it will be virtuallyirrotational; therefore, the Bernoulli equation will be valid.

Another design decision that must be made is how to dispose of the discharge fromthe orice. The could be collected into a tank and then discharged into the labreservoir through one of the grated openings.

1237

PROBLEM 13.78

Situation: A laboratory setup is described in Prob. 13.77.

Find: Design test equipment to determine the resistance coe"cient, b , of a 2- indiameter pipe.

ANALYSIS

First, decisions have to be made regarding the physical setup. This should include:

a How to connect the 2 in. pipe to the water source.

b Providing means of discharging ow back into the lab reservoir. Probably have apipe discharging directly into reservoir through one of the grated openings.

c Locating control valves in the system

d Deciding a length of 2” pipe on which measurements will be made. It is desirableto have enough length of pipe to yield a measurable amount of head loss.

To measure the head loss, one can tap into the pipe at several points along thepipe (six or eight points should be su"cient). The di!erential pressure betweenthe upstream tap and downstream tap can rst be measured. Then measure thedi!erential pressure between the next tap and the downstream tap, etc., until thepressure di!erence between the downstream tap and all others has been completed.From all these measurements the slope of the hydraulic grade line could be computed.The discharge could be measured by weighing a sample of the ow for a period of time

and then computing the volume rate of ow. Or the discharge could be measuredby an electromagnetic ow meter if one is installed in the supply pipe.The diameter of the pipe should be measured by inside calipers and micrometer.

Even though one may have purchased 2 inch pipe, the nominal diameter is usuallynot the actual diameter. With this diameter one can calculate the cross-sectionalarea of the pipe. Then the mean velocity can be computed for each run: , = ]$I.Then for a given run, the resistance coe"cient, b , can be computed with Eq. (10.22).

Other things that should be considered in the design:

a) Make sure the pressure taps are far enough downstream of the control valve orany other pipe tting so that uniform ow is established in the section of pipewhere measurements are taken.

b) The di!erential pressure measurements could be made by either transducers ormanometers or some combination.

c) Appropriate valving and manifolding could be designed in the system so that onlyone pressure transducer or manometer is needed for all pressure measurements.

1238

d) The water temperature should be taken so that the specic weight of the watercan be found.

e) The design should include means of purging the tubing and manifolds associ-ated with the pressure di!erential measurements so that air bubbles can beeliminated from the measuring system. Air bubbles often produce erroneousreadings.

1239

PROBLEM 13.79

Situation: A laboratory setup is described in Prob. 13.77.

Find: Design test equipment to determining the loss coe"cients of 2- in gate andglobe valves.

ANALYSIS

Most of the design setup for this equipment will be the same as for Prob. (13.78)except that the valve to be tested would be placed about midway along the two inchpipe. Pressure taps should be included both upstream and downstream of the valveso that hydraulic grade lines can be established both upstream and downstream ofthe valve (see Fig. 10.15). Then as shown in Fig. (10.15) the head loss due to thevalve can be evaluated. The velocity used to evaluate<@ is the mean velocity in the2 in. pipe so it could be evaluated in the same manner as given in the solution forProb. (13.78).

1240

PROBLEM 13.80

Situation: A stagnation tube is used to measure air speed #air = 1!25 kg/m3, A = 2

mm, -, = 1!00Deection on an air-water manometer, M = 1 mm.The only uncertainty in the manometer reading is \7 = 0!1 mm.

Find: (a) Air Speed: ,(b) Uncertainty in air speed: \?

ANALYSIS

, =

µ2!%

#'#K-,

¶1*2

!% = M(%

Combining equations

, =

µ2(%M

#'#K-,

¶1*2=

µ(2)(9' 810)(0!001)

(1!25)(1!00)

¶1*2

, = 3!96 m/s

Uncertainty equation

\? =[,

[M\7

The derivative is[,

[M=

s2(%#'-,

1

2&M

Combining equations gives

\?,

=\72M

=0!1

2× 1!0= 0!05

So

\? = 0!05,

= 0!05× 3!96= 0.198 m/s

1241

PROBLEM 13.81

Situation: Water ows through a 6 in. orice situated in a 12 in. pipe. On a mercurymanometer, !M = 1 ft-Hg. The uncertainty values are \b = 0!03' \6 = 0!5 in.-Hg,\A = 0!05 in.

Find: (a) Discharge: ](b)Uncertainty in discharge: \X

APPROACH

Calculate discharge by rst calculating !M (apply piezometric head and manometerequation) and to apply the orice equation. Then apply the uncertainty equation.

ANALYSIS

Piezometric head

!M =

µ%1(%+ R1

¶"µ%2(%+ R2

¶

Manometer equation

%1 + (%R1 " (661 ft" (%(R2 " 1 ft) = %2%1 " %2(%

= "(R1 " R2) +µ(6D(%

¶1 ft" 1 ft

Combining equations

!M = (1!0 ft)µ(6D(%

" 1¶

= 1!0(13!55" 1) = 12!55 ft of water

Uncertainty equation for !M

\!7 =

µ0!5

12ft¶µ

(6D(%

" 1¶=

µ0!5

12

¶(13!55" 1)

= 0!523 ft of water

Orice equation

] = <0

4A2p2)!M

where < = 0!625 (from problem 13.20)

Thus, ] = 0!625×0

4× 0!52

&2× 32!2× 12!55

= 3.49 cfs

1242

Uncertainty equation applied to the discharge relationship

µ\X]

¶2=

ÃSXSb\b

]

!2+

ÃSXSA\A

]

!2+

ÃSXS!7\!7

]

!2

µ\X]

¶2=

µ\b<

¶2+

µ2\AA

¶2+

µ\!72!M

¶2

µ\X]

¶2=

µ!03

0!625

¶2+

µ2× 0!056

¶2+

µ!523

2× 12!55

¶2

\X]

= 0!055

\X = 0!055× 3!49 = 0.192 cfs

1243

PROBLEM 13.82

Situation: A rectangular weir (2 = 10 ft, * = 3 ft, 4 = 1!5 ft) is used to measuredischarge. The uncertainties are \[ = 5%' \6 = 3 in., \E = 1 in.

Find: (a) Discharge: ](b) Uncertainty in discharge: \X

APPROACH

Calculate < and apply the rectangular weir equation to nd discharge. Then applythe uncertainty equation.

ANALYSIS


< = 0!4 + 0!054

*= 0!4 + 0!05×

µ1!5

3!0

¶

= 0!425

] = <p2)243*2

= (0!425)&2× 32!2(10)(1!5)3*2

] = 62!7 cfs

Uncertainty equation

\2X =

µ[]

[<\b

¶2+

µ[]

[2\E

¶2+

µ[]

[4\6

¶2

µ\X]

¶2=

µ\b<

¶2+

µ\E2

¶2+

µ3

2×\64

¶2

= (!05)2 +

µ1$12

10

¶2+

µ3

2×3$12

1!5

¶2

= 0!2552

Thus, \X = 0!255]

= (0!255)(62!7)

\X = 16!0 cfs

1244

PROBLEM 13.83

Situation: Pitot tubes cannot measure low speed air velocities, because the pressuredi!erence between stagnation and static is too small. Additional details are providedin the problem statement.

Find: Develop ideas to measure air velocities from 1 to 10 ft$ s!

ANALYSIS

The are probably many di!erent approaches to this design problem. One idea is tosupport a thin strip of material in an airstream from a low friction bearing as shownin the gure.

bearing

rectangularstripwind

veloc ity

$

The drag force on the strip tends to rotate the strip and the angle of rotation willbe related to the ow velocity. Assume the strip has an area 6, a thickness i and amaterial density of #"! Also assume the length of the strip is 2! Assume that the forcenormal to the strip is given by the drag force associated with the velocity componentnormal to the surface and that the force acts at the mid point of the strip. Themoment produced by the ow velocity would be

+U/ = H92$2 = -96(#',20 cos

2 K$2)2$2

where K is the deection of the strip, #' is the air density and ,0 is the wind velocity.This moment is balanced by the moment due to the weight of the strip

+U/ =+)(2$2) sin K

Equating the two moments gives

+)(2$2) sin K = -96(#',20 cos

2 K$2)2$2

Solving for ,0 gives

, 20 =2+) sin K

-96#' cos2 K

,0 =

s2+) sin K

-96#' cos2 K

1245

But the mass of the strip can be equated to #"6i so the equation for velocity reducesto

,0 =

s2#"i) sin K

-9#' cos2 K

Assume the strip is a plastic material with a density of 800 kg/m3 and a thicknessof 1 mm. Also assume the drag coe"cient corresponds to a rectangle with an aspectratio of 10 which from Table 11.1 is 1.3. Assume also that a deection of 10o can bemeasured with reasonable accuracy. Assume also that the air density is 1.2 kg/m3!The wind velocity would be

,0 =

r2× 800× 0!001× 9!81× 0!174

1!3× 1!2× 0!9852= 1!3 m/s

This is close to the desired lower limit so is a reasonable start. The lower limit canbe extended by using a lighter material or possibly a wire frame with a thin lm ofmaterial. The relationship between velocity and angle of deection would be

Angle, deg

0 20 40 60 80 100

Velo

city

, m/s

02468

101214161820

This plot suggests that the upper range of 10 m/s could be reached with a deectionof about 70 degrees. The simple model used here is only an approximation for designpurposes. An actual instrument would have to be calibrated.

Other features to be considered would be a damping system for the bearing to handleow velocity uctuations and an accurate method to measure the deection. Thedesign calculations presented here show the concept is feasible. More detailed designconsiderations would then follow.

1246

PROBLEM 13.84

Situation: The volume ow rate of gas discharging from a small tube is less than aliter per minute.

Find: Devise a scheme to measure the ow rate.

ANALYSIS

One approach may be to use a very small venturi meter but instrumentation wouldbe di"cult (installing pressure taps, etc.). A better approach may be the use of somevolume displacement scheme. One idea may be to connect the ow to a exible bagimmersed in a water (or some liquid) bath as shown. As the gas enters the bag, thebag will expend displacing the liquid in the tank. The overow of the tank woulddischarge into a graduated cylinder to measure the displacement as a function of time.

overflow

graduatedcylinder

flexiblebag

valve

Features which must be considered are 1) the volume of the bag must be chosen suchthat pressure in the bag does not increase with increased displacement, 2) evaporationfrom the surface must be minimized and 3) a valve system has to be designed suchthat the ow can be diverted to the bag for a given time and then closed.

1247

PROBLEM 13.85

Situation: A owing uid.

Find: Design a scheme to measure the density of the uid by using a combination ofow meters.

ANALYSIS

The two ow meters must be selected such that one depends on the density of theuid and the other is independent of the uid density. One such combination wouldbe the venturi meter and the vortex meter as shown in the diagram.

! P

Venturi meter Vortex meter

The discharge in the venturi meter is given by the orice equation

] = <I!

s2!%

#

while the velocity measured by a vortex meter is

, =Q1

6P

where 1 is the size of the element. For a calibrated vortex ow meter one has

] = -b

where - is a calibration constant and b is the shedding frequency. The calibrationconstant is essentially independent of Reynolds number over a wide range of Reynoldsnumber. Thus we have

-b = <I!

s2!%

#

Solving for #

# =2!%(<I!)

2

(-b)2

The ow coe"cient does depend weakly on Reynolds number so there may be a sourceof error if < is not known exactly. If the viscosity of the uid is known, the Reynoldsnumber could be calculated and the above equation could be used for an iterativesolution.

1248

PROBLEM 14.1

Situation: A propeller is described in the problem statement.

Find: Thrust force.

ANALYSIS

From Fig. 14.2-- = 0!048!

Propeller thrust force equation

H- = --#14Q2

= 0!048× 1!05× 34 × (1' 400$60)2

H- = 2223N

1249

PROBLEM 14.2


Find: (a) Thrust.(b) Power.

APPROACH

Apply the propeller thrust force equation and the propeller power equation.

ANALYSIS

Reynolds number

Re = ,0$Q1

= (80' 000$3' 600)$((1' 400$60)× 3)= 0!317

From Fig. 14.2-- = 0!020


H = --#14Q2-

= 0!020× 1!05× 34 × (1' 400$60)2

H- = 926 N

From Fig. 14.2-, = 0!011

Propeller power equation

* = -,#Q315

= 0!011× 1!05× 35 × (1400$60)3

* = 35!7 kW

1250

PROBLEM 14.3


Find: (a) Thrust for ,0 = 25 mph.(b) Power for (a).(c) Thrust for ,0 = 0!

APPROACH

Apply the propeller thrust force equation and the propeller power equation. CalculateReynolds number to nd -- .

ANALYSIS

Reynolds number

Q = 1000$60 = 16!67 rev/sec

,0 = 25 mph = 36.65 fps

Advance ratio,0Q1

=36!65

16!67× 8= 0!27

Coe"cient of thrust and power (from Fig. 14.2)

-- = 0!023

-, = 0!011


H = --#14Q2-

= 0!023× 0!0024× 84 × 16!672

H- = 62!8 lbf


* = -,#Q315

= 0!011× 0!0024× 16!673 × 85

= 4372 ft-lb/sec

* = 7!95 hp

When the forward speed is 0 (,0 = 0) 'then the thrust coe"cient (Fig. 14.3) is

-- = 0!0475


H- = --#14Q2-

= 0!0475× 0!0024× 84 × 16!672

H- = 130 lbf

1251

PROBLEM 14.4


Find: Angular speed of propeller.

APPROACH

Use Fig 14.4 to nd the advance diameter ratio at maximum e"ciency.

ANALYSIS

,0 = 30 mph = 44 fps

From Fig. 14.3, ,0$(Q1) = 0!285

Q = 1$(0!285,0)

Q = 44$(0!285× 8)= 19!30 rps

8 = 1158 rpm

1252

PROBLEM 14.5


Find: (a) Thrust.(b) Power output.

APPROACH

Apply the propeller thrust force equation and the propeller power equation. Use Fig14.2 to nd -- and -\ at maximum e"ciency.

ANALYSIS

From Fig. 14.2

-- = 0!023

-, = 0!012


H- = --#14Q2

= 0!023× 0!0024× 64 × 25!732

H- = 47!4 lbf


* = -,#Q315

= 0!012× 0!0024× 65 × 25!733

= 3815 ft-lbf/s

* = 6!94 hp

1253

PROBLEM 14.6


Find: (a) Diameter of propeller.(b) Speed of aircraft.

APPROACH

Apply the Ideal gas law to get the density for the propeller thrust force equation tocalculate the diameter. Then apply the lift force equation to calculate the speed.

ANALYSIS

Ideal gas law

# = %$"&

= 60× 103$((287)(273))= 0!766 kg/m3


H- = --#Q214

H- = Drag = Lift$30 = (1' 200)(9!81)$(30) = 392 N

392 = (0!025)(0!766)(3' 000$60)214

1 = 1!69m

Lift force

2 = . = -E(1$2)#,20 6

2$(-E6) = (#, 20 $2)

= (1' 200)(9!81)$((0!40)(10)) = 2' 943

, 20 = (2' 943)(2)$(0!766) = 7' 684

,0 = 87!7m$ s

1254

PROBLEM 14.7


Find: Maximum allowable angular speed.

ANALYSIS

,tip = 0!9Z = 0!9× 335 = 301!5 m/s,tip = J3 = Q(20)3

Q = 301!5$(203) = 301!5$(01) rev/s

8 = 60× Q rpm

1 (m) 8 (rpm)2 2' 879

3 1' 919

4 1' 440

1255

PROBLEM 14.8


Find: Angular speed of propeller.

APPROACH

Use Fig 14.2 to nd the advance diameter ratio at maximum e"ciency.

ANALYSIS

Advance ratio (from Fig. 14.2)

,0$(Q1) = 0!285

Rotation speed

Q = ,0$(0!2851)

= (40' 000$3' 600)$(0!285× 2)= 19!5 rev/s

8 = 19!5× 608 = 1170 rpm

1256

PROBLEM 14.9


Find: (a) Thrust.(b) Power input.

APPROACH

Apply the propeller thrust force equation and the propeller power equation. Use Fig14.2 to nd -- and -\ at maximum e"ciency.

ANALYSIS

From Fig. 14.2,

-- = 0!023

-, = 0!012


H = --#14Q2-

= 0!023× 1!2× 24 × (19!5)2

H- = 168N


* = -,#Q315

= 0!012× 1!2× 25 × (19!5)3

* = 3!42 kW

1257

PROBLEM 14.10


Find: Initial acceleration.

APPROACH

Apply the propeller thrust force equation. Use Fig 14.2 to nd -- .

ANALYSIS

From Fig. 14.2-- = 0!048


H- = --#14Q2

= 0!048#14Q2

= 0!048× 1!1× 24 × (1' 000$60)2

= 235 N

Calculate acceleration

C = H$/

= 235$300

C = 0!782 m$ s2

1258

PROBLEM 14.11


Find: Discharge.

APPROACH

Apply discharge coe"cient. Calculate the head coe"cient to nd the correspondingdischarge coe"cient from Fig. 14.6.

ANALYSIS

Q = 1' 000$60

= 16!67 rev/s

Head coe"cient

-6 = !M)$12Q2

= 3× 9!81$((0!4)2 × (16!67)2)= 0!662

From Fig. 14.6, -X = ]$(Q13) = 0!625!Discharge coe"cient

] = 0!625× 16!67× (0!4)3

] = 0!667m3$ s

1259

PROBLEM 14.12


Find: (a) Discharge.(b) Power demand.

APPROACH

Apply discharge coe"cient and power coe"cient. Calculate the head coe"cient tond the corresponding discharge and power coe"cients from Fig. 14.6.

ANALYSIS

Angular velocity

Q = 690$60

= 11!5 rev/s

Head coe"cient

-6 = !M)$(Q212)

= 10× 9!81$((0!712)2(11!5)2)= 1!46

From Fig. 14.6,

-X = 0!40 and -, = 0!76

Discharge coe"cient

] = -XQ13

= 0!40× 11!5× 0!7123

] = 1!66 m3$ s

Power coe"cient

* = -,#15Q3

= 0!76× 1' 000× 0!7125 × 11!53

* = 211 kW

1260

PROBLEM 14.13


Find: (a) Discharge.(b) Power required.

APPROACH

Plot the system curve and the pump curve. Apply the energy equation from thereservoir surface to the center of the pipe at the outlet to solve the head of the pumpin terms of ]. Apply head coe"cient to solve for the head of the pump in terms of-6 . Apply discharge coe"cient to solve for -X in terms of ]—then use gure 14.6to nd the corresponding -6 . Find the power by using Fig. 14.7.

ANALYSIS

1 = 35!6 cm

Q = 11!5 rev/s

Energy equation from the reservoir surface to the center of the pipe at the outlet,

%1$( + ,21 $(2)) + R1 + M, = %2$( + ,

22 $(2)) + R2 +

XME

M, = 21!5" 20 + []2$(I22))](1 + b2$1 + cG + c0)2 = 64 m

Assume b = 0!014' 30$1 = 1! From Table 10-3, c0 = 0!35' cG = 0!1

M, = 1!5 + []2((0!014(64)$0!356) + 0!35 + 0!1 + 1)]$[2(9!81)(0$4)2(0!356)4]

= 1!5 + 20!42]2

-X = ]$(Q13) = ]$[(11!5)(0!356)3] = 1!93]

M, = -6Q212$) = -6(11!5)

2(0!356)2$9!81 = 1!71-6

](m3$s) -X -6 M,1 (m) M,2 (m)0.10 0.193 2.05 1.70 3.500.15 0.289 1.70 1.96 2.910.20 0.385 1.55 2.32 2.650.25 0.482 1.25 2.78 2.130.30 0.578 0.95 3.34 1.620.35 0.675 0.55 4.00 0.94

Then plotting the system curve and the pump curve, we obtain the operating condi-tion:

] = 0!22 m3/s

1261

From Fig. 14.7* = 6!5 kW

Q (m3 /s)

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

h p (m

)

0

1

2

3

4

5

pump curve

system curveoperating point

1262

PROBLEM 14.14


Find: (a) Discharge.(b) Power required.

APPROACH

Same solution procedure applies as in Prob. 14.13. To nd power, apply powercoe"cient (use gure 14.6 to nd the -\ that corresponds to the -X.

ANALYSIS

The system curve will be the same as in Prob. 14.13

-X = ]$[Q13] = ]$[15(0!356)3] = 1!48]

M, = -6Q212$) = -6(15)

2(0!356)2$9!81 = 2!91-6

] -X -6 M,0.20 0.296 1.65 4.790.25 0.370 1.55 4.510.30 0.444 1.35 3.920.35 0.518 1.15 3.34

Q (m2 /s)

0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.362

3

4

5

System curve

Pump curve

h(m

)p

Plotting the pump curve with the system curve gives the operating condition;

] = 0!32 m3$ s

-X = 1!48(0!32) = 0!474

Then from Fig. 14.6, -, = 0!70Power coe"cient

* = -,Q313#

= 0!70(15)3(0!356)51' 000

* = 13!5 kW

1263

PROBLEM 14.15


Find: (a) Discharge.(b) Head.(c) Power required.

APPROACH

Apply discharge, head, and power coe"cients. Use Fig. 14.6 to nd the discharge,power, and head coe"cients at maximum e"ciency.

ANALYSIS

From Fig. 14.6, -X = 0!64;-, = 0!60; and -6 = 0!75

1 = 1!67 ft

Q = 1' 100$60 = 18!33 rev/s

Discharge coe"cient

] = -XQ13

= 0!64× 18!33× 1!673

] = 54!6 cfs

Head coe"cient

!M = -6Q212$)

= 0!75× 18!332 × 1!672$32!2

!M = 21!8 ft

Power coe"cient

* = -,#15Q3

= 0!60× 1!94× 1!675 × 18!333

= 93' 116 ft-lbf/sec

* = 169!3 hp

1264

PROBLEM 14.16


Find: (a) Discharge.(b) Head.(c) Power required.

APPROACH

Apply discharge, head, and power coe"cients. Use Fig. 14.6 to nd the discharge,power, and head coe"cients at maximum e"ciency.

ANALYSIS

At maximum e"ciency, from Fig. 14.6, -X = 0!64; -, = 0!60; -6 = 0!75Discharge coe"cient

] = -XQ13

= 0!64× 45× 0!53

] = 3!60m3$ s

Head coe"cient

!M = -6Q212$)

= 0!75× 452 × 0!52$9!81

!M = 38!7m

Power coe"cient

* = -,#15Q3

= 0!60× 1' 000× 0!55 × 453

* = 1709 kW

1265

PROBLEM 14.17


Find: Plot the head-discharge curve.

APPROACH

Apply the discharge and head coe"cient equations at a series of coe"cients corre-sponding to each other from Fig. 14.6.

ANALYSIS

1 = 14$12 = 1!167 ft

Q = 1' 000$60 = 16!7 rev/s

Head coe"cient

!M = -6Q212$)

= -6(16!7)2(1!167)2$32!2

= 11!8-6 ft

Discharge coe"cient

] = -XQ13

= -X16!7(1!167)3

= 26!5-X cfs

-X -6 ](cfs) !M(ft)0.0 2.9 0 34.20.1 2.55 2.65 30.10.2 2.0 5.3 23.60.3 1.7 7.95 20.10.4 1.5 10.6 17.70.5 1.2 13.2 14.20.6 0.85 15.9 10.0

Discharge, cfs

0 2 4 6 8 10 12 14 16 185

10

15

20

25

30

35

40

Hea

d, ft

1266

PROBLEM 14.18



APPROACH

Apply the discharge and head coe"cient equations at a series of coe"cients corre-sponding to each other from Fig. 14.6.

ANALYSIS

1 = 60 cm = 0!60 m

8 = 690 rpm

Q = 11!5 rps

Head coe"cient

!M = -612Q2$)

= 4!853-6

Discharge coe"cient

] = -XQ13

= 2!484-X

-X -6 ](m3/s) M(m)0.0 2.90 0.0 14.10.1 2.55 0.248 12.40.2 2.00 0.497 9.70.3 1.70 0.745 8.30.4 1.50 0.994 7.30.5 1.20 1.242 5.80.6 0.85 1.490 4.2

Discharge, m3 /s

0 1 2

Hea

d, m

2

4

6

8

10

12

14

16

1267

PROBLEM 14.19


Find: (a)Head at maximum e"ciency.(b) Discharge at maximum e"ciency.

APPROACH

Apply discharge and head coe"cients. Use Fig. 14.10 to nd the discharge and headcoe"cients at maximum e"ciency.

ANALYSIS

1 = 0!371× 2 = 0!742 mQ = 2' 133!5$(2× 60) = 17!77 rps

From Fig. 14.10, at peak e"ciency -X = 0!121' -6 = 5!15!Head coe"cient

!M = -6Q212$)

= 5!15(17!77)2(0!742)2$9!81

!M = 91!3m

Discharge coe"cient

] = -XQ13

= 0!121(17!77)(0!742)3

] = 0!878 m3$ s

1268

PROBLEM 14.20

Situation: A fan is described in the problem statement.

Find: Power needed to operate fan.

APPROACH

Apply power coe"cient. Calculate the discharge coe"cient (apply the ow rateequation to nd ]) to nd the corresponding power coe"cient from Fig. 14.16.

ANALYSIS

Flow rate equation

] = , I

= (60)(0$4)(1!2)2

= 67!8 m3/s

Discharge coe"cient

-X = ]$(Q13)

= (67!8)$((1' 800$60)(2)3

= 0!282

From Fig. 14.16 -, = 2!6! ThenPower coe"cient

* = -,#15Q3

= (2!6)(1!2)(2)5(30)3)

* = 2!70MW

1269

PROBLEM 14.21


Find: Discharge through pipe.

APPROACH

Guess the pump head and iterate using Fig. 14.9 to get the corresponding ow rateand then Darcy-Weisbach equation to get the head for that ow rate (apply the owrate equation and Reynolds number to get the necessary parameters for the Darcy-Weisbach equation).

ANALYSIS

!R = 450" 366 = 84 mAssume !M = 90 m (F!R)' then from Fig. 14.9, ] = 0!24 m3/sFlow rate equation

, = ]$I

= 0!24$[(0$4)(0!36)2]

= 2!36 m/s; c&$1 = 0!00012

Assuming & = 20!-Reynolds number

Re = , 1$:

= 2!36(0!36)$10"6

= 8!5× 105

Frictional head loss (Darcy-Weisbach equation)from Fig. 10.8, b = 0!014

M( = (0!014(610)$0!36)((2!36)2$(2× 9!81)) = 6!73 m

M ! 84 + 6!7 = 90!7 m

from Fig. 14.9 ] = 0!23 m3/s;

, = 0!23$((0$4)(0!36)2) = 2!26 m/s

M( = [0!014(610)$0!36](2!26)2$(2× 9!81) = 6!18 m

so

!M = 84 + 6!2 = 90!2 m

, = 0!23$((0$4)(0!36)2) = 2!26 m/s

and from Fig. 14.9] = 0!225 m3/s

1270

PROBLEM 14.22


Find: Discharge.

APPROACH


ANALYSIS

1 = 0!371 m = 1!217 ft

Q = 1500$60 = 25 rps

Head coe"cient

!M = -6Q212$)

-6 = 150(32!2)$[(25)2(1!217)2]

= 5!217

from Fig. 14.10-X = 0!122

Discharge coe"cient

] = -XQ13

= 0!122(25)(1!217)3

] = 5!50 cfs

1271

PROBLEM 14.23


Find: Maximum possible head developed.

APPROACH

Apply head coe"cient.

ANALYSIS

Head coe"cient-6 = !4)$1

2Q2

Since -6 will be the same for the maximum head condition, then

!4 T Q2

or41I500 = 41I000 × (1' 500$1' 000)2

41I500 = 102× 2!25

41I500 = 229!5 ft

1272

PROBLEM 14.24


Find: Shuto! head.

APPROACH

Apply head coe"cient.

ANALYSIS

4 T Q2

so430$43546 = (30$35!6)

2

or

430 = 104× (30$35!6)2

430 = 73!8 m

1273

PROBLEM 14.25


Find: Discharge when head is 50 m.

APPROACH


ANALYSIS

Head coe"cient

-6 = !M)$(Q212)

= 50(9!81)$[(25)2(0!40)2]

= 4!91

from Fig. 14.10 -X = 0!136Discharge coe"cient

] = -XQ13

= 0!136(25)(0!40)3

] = 0!218 m3/s

1274

PROBLEM 14.26


Find: (a) Flow rate.(b) Pressure rise across pump.(c) Power required.

Properties: From table A.4 # = 814 kg/m3!

APPROACH

Apply the discharge, head, and power coe"cient equations. Use Fig. 14.10 to ndthe discharge, power, and head coe"cients at maximum e"ciency.

ANALYSIS

8 = 5' 000 rpm = 83!33 rps

From Fig. 14.10 at maximum e"ciency -X = 0!125; -6 = 5!15; -, = 0!69Discharge coe"cient

] = -XQ13

= (0!125)(83!33)(0!20)3

] = 0!0833 m3/s

Head coe"cient

!M = -612Q2$)

= (5!15)(0!20)2(83!33)2$9!81

!M = 145!8 m

Power coe"cient

* = -,#15Q3

= (0!69)(814)(0!20)5(83!33)3

* = 104!0 kW

1275

PROBLEM 14.27

Situation: A centrifugal pump with di!erent impeller diameters is described in theproblem statement.

Find: Plot ve performance curves for the di!erent diameters in terms of head anddischarge coe"cients.

APPROACH

Calculate the ve discharge coe"cients by applying the discharge coe"cient equation,and the ve head coe"cients by the applying head coe"cient equation.

ANALYSIS

Discharge coe"cient

-X = ]$Q13

The rotational speed is 1750/60=29.2 rps. The diameter for each impeller is 0.4167ft, 0.458 ft, 0.5 ft, 0.542 ft and 0.583 ft. One gallon per minute is 0.002228 ft3$s. Sofor each impeller, the conversion factor to get the discharge coe"cient is

5” gpm × 0!001055!5” gpm × 0!0007946” gpm × 0!0006106!5” gpm × 0!0004797” gpm × 0!000385

Head coe"cient

-6 =!4)

Q212

The conversion factors to get the head coe"cient are

5” ft × 0!21755!5” ft × 0!18006” ft × 0!15106!5” ft × 0!12857” ft × 0!1111

The performance in terms of the nondimensional parameters is shown on the graph.

1276

Discharge coefficient

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

Hea

d co

effic

ient

1

2

3

4

5

6

5 inch5.5 inch6 inch6.5 inch7 inch

1277

PROBLEM 14.28



APPROACH

Apply the head and discharge coe"cient equations at a series of coe"cients corre-sponding to each other from Fig. 14.10.

ANALYSIS

The rotational speed in rps is

Q = 500$60 = 8!33 rps

Discharge coe"cient

] = -XQ13

= -X(8!33)(1!523)

= 29!25-X (m3$s)

Head coe"cient

!M = -6Q212$)

= -6(8!332)(1!522)$9!81

= 16!34-6 (m)

-X ] -6 !M0 0 5!8 94!90!04 1!17 5!8 94!90!08 2!34 5!75 94!10!10 2!93 5!6 91!60!12 3!51 5!25 85!90!14 4!10 4!8 78!50!16 4!68 4!0 65!4

1278

Discharge, m 3/s

0 1 2 3 4 5

Hea

d, m

60

65

70

75

80

85

90

95

100

1279

PROBLEM 14.29

Situation: A pump is described in problem 14.13.

Find: (a) Suction specic speed.(b) Safety of operation with respect to cavitation.

APPROACH

Calculate the suction specic speed, and then compare that with the critical value of85' 000.

ANALYSIS

Suction specic speed

8&& = 8]1*2(8*64)3*4

8 = 690 rpm

8*64 ! 14!7 psi × 2!31 ft/psi" Mvap.press. ! 33 ft] = 0!22 m3$9× 264!2 gallons/s× 60 s/min = 3' 487 gpm8&& = 690× (3' 487)1*2$(33)3*4

8&& = 2' 960

8&& is much below 8,500; therefore, it is in a safe operating range.

1280

PROBLEM 14.30

Situation: A pump system is described in the problem statement.8 = 1' 500 rpm so Q = 25 rps; ] = 10 cfs; M = 30 ft

Find: Type of water pump.

APPROACH

Calculate the specic speed and use gure 14.14 to nd the pump range to which itcorresponds.

ANALYSIS

Specic speed

Q& = Qp]$[)3*4M3*4]

= (25)(10)1*2$[(32!2)3*4(30)3*4]

= 0!46

Then from Fig. 14.14, Q& F 0!60' so use a mixed ow pump.

1281

PROBLEM 14.31


Find: Type of pump.

APPROACH


ANALYSIS

Specic speed

Q = 25 rps

] = 0!30 m3/sec

M = 8 meters

Q& = Qp]$[)3*4M3*4]

= 25(0!3)1*2$[(9!81)3*4(8)3*4]

= 0!52

Then from Fig. 14.14, Q& G 0!60 so use a mixed ow pump.

1282

PROBLEM 14.32


Find: Type of pump.

APPROACH


ANALYSIS

Specic speed

8 = 1' 100 rpm = 18!33 rps

] = 0!4 m3/sec

M = 70 meters

Q& = Qp]$[)3*4M3*4]

= (18!33)(0!4)1*2$[(9!81)3*4(70)3*4]

= (18!33)(0!63)$[(5!54)(24!2)]

= 0!086

Then from Fig. 14.14, Q& G 0!23 so use a radial ow pump.

1283

PROBLEM 14.33

Situation: A pump is described in the problem statement.] = 5000 gpm

Find: Maximum speed.

APPROACH

Apply the suction specic speed equation setting the critical value for 8&& proposedby the Hydraulic Institute to 8500.

ANALYSIS

Suction specic speed8500 = 8]1*2$(8*64)3*4

The suction head is given as 5 ft. Then assuming that the atmospheric pressure is14.7 psia, and the vapor pressure is 0.256 psi, the net positive suction head (8*64)is

8*64 = 14!7 psi × 2!31 ft/psi+5 ft" Mvap.press. = 38!4 ft

Then

8 =8500× (8*64)3*4

]1*2

=8500× (38!4)3*4

50001*2

8 = 1850 rpm

1284

PROBLEM 14.34


Find: Type of pump

APPROACH


ANALYSIS

Specic speed

Q& = Qp]$()3*4M3*4)

Q = 10 rps

] = 1!0 m3$s

M = 3 + (1!5 + b2$1), 2$(2));

, = 1!27 m/s

Assume b = 0!01, so

M = 3 + (1!5 + 0!01× 20$1)(1!27)2$(2× 9!81)= 3!14 m

Then

Q& = 10×&1$(9!81× 3!14)3*4

= 0!76

From Fig. 14.14, use axial ow pump.

1285

PROBLEM 14.35

Situation: A blower for a wind tunnel is described in the problem statement.Max. air speed = 40 m/s; Area = 0!36 m2; Q = 2' 000$60 = 33!3 rps;

Find: (a) Diameter.(b) Power requirements for two blowers.

APPROACH

Apply the discharge and power coe"cient equations. Use Fig. 14.6 to nd thedischarge and head coe"cients at maximum e"ciency. Apply the ow rate equationto get the ] to calculate the diameter with discharge coe"cient.

ANALYSIS

Flow rate equation

] = , ×I= 40!0× 0!36= 14!4 m3/s

# = 1!2 kg/m3 at 20!-

From Fig. 14.6, at maximum e"ciency, -X = 0!63 and -, = 0!60Discharge coe"cient

13 = ]$(Q-X)

= 14!4$(33!3× 0!63)= 0!686 m3

1 = 0!882 m

Power coe"cient

* = -,#Q315

= 0!6(1!2)(33!3)3(0!882)5

* = 14!2 kW

1286

PROBLEM 14.36

Situation: A blower for air conditioning is described in the problem statement.Volume = 105 m3; time for discharge = 15 min = 900 sec

Find: (a) Diameter.(b) Power requirements.

APPROACH

Apply the discharge and power coe"cient equations. Use Fig. 14.6 to nd thedischarge and head coe"cients at maximum e"ciency. Apply the ow rate equationto get the ] to calculate the diameter with discharge coe"cient.

ANALYSIS

8 = 600 rpm = 10 rps

# = 1!22 kg/m3 at 60!F

] = (105 m3)$(900 sec) = 111.1 m3$sec

From Fig. 14.6, at maximum e"ciency, -X = 0!63; -, = 0!60

For two blowers operating in parallel, the discharge per blower will be one half so

] = 55!55 m3$sec

Discharge coe"cient

13 = ]$Q-X = (55!55)$[10× 0!63] = 8!815

1 = 2!066 m

Power coe"cient

* = -,#15Q3

= (0!6)(1!22)(2!066)5(10)3

* = 27!6 kW per blower

1287

PROBLEM 14.37

Situation: A centrifugal compressor is described in the problem statement.

Find: Shaft power to run compressor

Properties: From Table A.2 for methane " = 518 J/kg/K and c = 1!31!

ANALYSIS

*57 = (c$(c " 1))]%1[(%2$%1)(["1)*[ " 1]= (c /$(c " 1))"&1[(%2$%1)(["1)*[ " 1]= (1!31$0!31)(1)518(300)[(1!5)0431*1431 " 1]= 66!1 kW

*ref = *57$=

= 66!1$0!7

*ref = 94!4 kW

1288

PROBLEM 14.38

Situation: A compressor is described in the problem statement.

Find: Volume ow rate into the compressor.

APPROACH

Apply equation 14.17.

ANALYSIS

*57 = 12 kW× 0!6 = 7!2 kW*57 = (c$(c " 1))]%1[(%2$%1)(["1)*[ " 1]

= (1!3$0!3)]× 9× 104[(140$90)043*143 " 1]= 4!18× 104]

] = 7!2$41!8

] = 0!172 m3/s

1289

PROBLEM 14.39

Situation: A centrifugal compressor is described in the problem statement.

Find: The shaft power.

APPROACH

Apply equation 14.17.

ANALYSIS

*57 = %1]1LQ(%2$%1)

= /"&1LQ(%2$%1)

= 1× 287× 288× LQ4= 114!6 kW

*ref = 114!6$0!5

*ref = 229 kW

1290

PROBLEM 14.40

Situation: A turbine system is described in the problem statement.

Find: (a) Power produced.(b) Diameter of turbine wheel.

Assumptions: & = 10!-

APPROACH

Apply the energy equation from reservoir to turbine jet. Then apply the continuityprinciple and the power equation.

ANALYSIS

Energy equation

%1$( + ,21 $2) + R1 = %2$( + ,

22 $2) + R2 +

XME

0 + 0 + 650 = 0 + , 2jet$2) + 0 + (b2$1)(,2pipe$2))


,pipeIpipe = ,jetIjet

,pipe = ,jet(Ijet$Ipipe) = ,jet(0!16)2 = 0!026,jet

so(, 2jet$2))(1 + (b2$1)0!026

2) = 650

,jet = [(2× 9!81× 650)$(1 + (0!016× 10' 000)$1)0!0262)]1*2

= 107!3 m/s

Power equation

* = ](, 2jet=

= 107!3(0$4)(0!16)29' 810(107!3)20!85$(2× 9!81)* = 10!55 MW

,bucket = (1$2),jet

= 53!7 m/s = (1$2)J

1 = 53!7× 2$(360× (0$30))1 = 2!85 m

1291

PROBLEM 14.41

Situation: An impulse turbine is described in the problem statement.

Find: Referencing velocities to the bucket.

APPROACH


ANALYSIS

V-Vj B

V-Vj B

V-Vj B

Momentum principle

XHbucket on jet = #]["(,V " ,>)" (,V " ,>)]

Then XHon bucket = #,VIV2(,V " ,>)

assuming the combination of buckets to be intercepting ow at the rate of ,VIV! Then

* = H,> = 2#IV[,2V ,> " ,V,

2>]

For maximum power production, A*$A,> = 0' so

0 = 2#I(, 2V " ,V2,>)0 = ,V " 2,>

or,> = 1$2,V

1292

PROBLEM 14.42

Situation: A jet of water strikes the buckets of an impulse wheel—additional detailsare provided in the problem statement.

Find: (a) Jet force on the bucket.(b) Resolve the discrepancy with Eq. 14.20.

APPROACH


ANALYSIS

Consider the power developed from the force on a single bucket. Referencing velocitiesto the bucket givesMomentum principle

XHon bucket = #]rel. to bucket("(1$2),V " (1$2),V)

ThenHon bucket = #(,V " ,>)IV(,V)

but,V " ,> = 1$2,V

soHon bucket = 1$2#I,

2V

Then* = H,> = (1$2)#],

3V $2

The power is 1/2 that given by Eq. (14.20). The extra power comes from theoperation of more than a single bucket at a time so that the wheel as a whole turnsthe full discharge; whereas, a single bucket intercepts ow at a rate of 1$2 ,VIV!

1293

PROBLEM 14.43

Situation: A Francis turbine is described in the problem statement.

Find: (a) T1 for non-separating ow conditions .(b) Maximum attainable power.(c) Changes to increase power production.

ANALYSIS

Flow rate equation

,K1 = ^$(2031D)

= 126$(20 × 5× 1)= 4!01 m/s

J = 60× 20$60 = 20 rad/s

T1 = arc cot ((31J$,K1) + cot `1)

= arc cot ((5× 20$4!01) + 0!577)T1 = 6!78

!

T2 = arc tan (,K2$J32) = arc tan ((4!01× 5$3)$(3× 20)) = arc tan 0!355= 19!5!

Equation 14.24

* = #]J(31,1 cosT1 " 32,2 cosT2),1 = ,K1$ sinT1 = 4!01$0!118 = 39!97 m/s

,2 = ,K2$ sinT2 = 20!0 m/s

* = 998× 126× 20(5× 39!97× cos 6!78! " 3× 20!0× cos 19!5!)

* = 112 MW

Increase `2

1294

PROBLEM 14.44


Find: (a) T1 for non-separating ow conditions.(b) Power.(c) Torque.

ANALYSIS

,K1 = 3$(20 × 1!5× 0!3) = 1!061 m/s,K2 = 3$(20 × 1!2× 0!3) = 1!326 m/s;J = (60$60)20 = 209"1

T1 = arc cot ((31J$,K1) + cot `1) = arc cot ((1!5(20)$1!415) + cot 85!)

= arc cot (6!66 + 0!0875)

T1 = 8!250

,tan1 = 31J + ,K1 cot `1 = 1!5(20) + 1!061(0!0875) = 9!518 m/s

,tan2 = 32J + ,K2cot `2 = 1!2(20) + 1!326("3!732) = 2!591 m/s& = #](31,tan1 " 32,tan2)

= 1' 000(4)(1!5× 9!518" 1!2× 2!591)

& = 44' 671 N-m

Power = &J

= 44' 671× 20* = 280!7 kW

1295

PROBLEM 14.45


Find: T1 for non-separating ow conditions.

ANALYSIS

J = 120$60× 20 = 40 s"1

,K1 = 113$(20(2!5)0!9) = 7!99 m/s

T1 = arc cot ((31J$,K1) + cot `1)

= arc cot ((2!5(40)$7!99) + cot 45!)

= arc cot (3!93 + 1)

T1 = 11!280

1296

PROBLEM 14.46

Situation: A small hydroelectric project is described in the problem statement.

Find: (a) Power output.(b) Draw the HGL and EGL.

Assumptions: cG = 0!50; <W = 1!0; <0 = 0!2; <&$1 = 0!00016!

APPROACH

To get power apply the energy equation. Apply the ow rate equation to get , forthe head loss. Then apply the power equation.

ANALYSIS

Energy equation

%1$( + T1,21 $2) + R1 = %2$( + T2,

22 $2) + R2 +

XME + M5

0 + 0 + 3000 = 0 + 0 + 2600 +X

ME + M5X

ME = (, 2$2))(b(2$1) +<W +<G + 2<0)

Flow rate equation

, = ]$I = 8$((0$4)(1)2) = 10!19 ft/s;

Re = , 1$: = (10!19)(1)$(1!2× 10"5) = 8!5× 105

b = 0!0145XME = ((10!19)2$(64!4))[(0!0145)(1000$1) + 1!0 + 0!5 + 2× 0!2]

XME = 1!612(16!4) = 26!44 ft

M5 = 3000" 2600" 26!44 = 373!6 ft

Power equation

*in = (]M5$550

= (8)(62!4)(373!6)$550

*in = 339 hp

Power output from the turbine

*out = 339× g= 339× 0!8= 271!2 hp

1297

*out = 271 hp

Plot of HGL & EGLEGL

HGL

1298

PROBLEM 14.47

Situation: Pumps, with characteristics M,Ipump = 20[1 " (]$100)2] are connected inseries and parallel to operate a uid system with system curve M,'sys= 5 + 0!002]2!

Find: Operating point with a) one pump, b) two pumps connected in series and c)two pumps connected in parallel.

APPROACH

Equate the head provided by the pump and the head required by the system.

ANALYSIS

a) For one pump

20[1"µ]

100

¶2] = 5 + 0!002]2

20" 0!002]2 = 5 + 0!002]2

15 = 0!004]2

Q=61.2 gpm

b) For two pumps in series

2× 20[1"µ]

100

¶2] = 5 + 0!002]2

35 = 0!006]2

Q=76.4 gpm

c) For two pumps in parallel

20[1"µ

]

2× 100

¶2] = 5 + 0!002]2

20" 0!0005]2 = 5 + 0!002]2

15 = 0!0025]2

Q=77.4 gpm

1299

PROBLEM 14.48

Situation: Wind turbines are described in the problem statement.

Find: Width of wind turbine.

APPROACH

Apply the wind turbine maximum power equation.

ANALYSIS

Each windmill must produce 2 MW/20 = 100,000 W.Wind turbine maximum power

*max =16

54#, 3! I

In a 20 m/s wind with a density of 1.2 kg/m3' the capture area is

I =54

16

100000

1!2× 203= 35!16 m2

Consider the gure for the section of a circle.

$

R

H

The area of a sector is given by

I& =1

2K"2 "

1

2"4 cos(K$2)

where K is the angle subtended by the arc and 4 is the distance between the edgesof the arc. But

" =4

2 sin(K$2)so

I = 2I& =42

4[

K

sin2(K$2)" 2

cos(K$2)

sin(K$2)]

= 56!2× [K

sin2(K$2)" 2

cos(K$2)

sin(K$2)]

1300

Solving graphically gives K = 52!. The width of the windmill is

. = 4[1

sin(K$2)"

1

tan(K$2)]

Substituting in the numbers gives .=3.45 m.

1301

PROBLEM 14.49

Situation: A windmill is connected to a pump—additional details are provided in theproblem statement.

Find: Discharge of pump.

APPROACH

Apply the wind turbine maximum power equation to get * for the power equationto get ].

ANALYSIS

Wind turbine maximum power

* = (16$27$)(#I, 3$2)

= (16$27)(0!07$32!2)(0$4)(10)2(44)3$2

= 4' 309 ft-lbf/s

Power equation

0!80× * = (]M,

(0!80)(4' 309) = (]M,

3' 447 ft-lbf/s = (]M,

] = (3' 447)$((62!4)(10))

= 5!52 cfs = 331 cfm

] = 2476 gpm

1302

PROBLEM 14.50

Situation: A system is to supply water ow from a reservoir to an elevated tank—additional details are provided in problem 10.102.

Find: Design the system including the choice of pumps.

ANALYSIS

Assume that this system will be used on a daily basis; therefore, some safety shouldbe included in the design. That is, include more than one pump so that if onemalfunctions there will be at least another one or two to satisfy the demand. Also,periodic maintenance may be required; therefore, when one pump is down there shouldbe another one or two to provide service. The degree of required safety would dependon the service. For this problem, assume that three pumps will be used to supplythe maximum discharge of 1 m3/s. Then each pump should be designed to supply aow of water of 0.333 m3/s (5,278 gpm). Also assume, for the rst cut at the design,that the head loss from reservoir to pump will be no greater than 1 meter and thateach pump itself will be situated in a pump chamber at an elevation 1 m below thewater surface of the reservoir. Thus, the 8*64 will be approximately equal to theatmospheric pressure head, or 34 ft.Assume that the suction Specic speed will be limited to a value of 8,500:

8&& = 8' 500 = 8]1*2$(8*64)3*4

or 8]1*2 = 8' 500× (34)3*4 (1)

= 119' 681

Assume that 60 cycle A.C. motors will be used to drive the pumps and that thesewill be synchronous speed motors. Common synchronous speeds in rpm are: 1,200,1,800, 3,600; however, the normal speed will be about 97% of synchronous speed*.Therefore, assume we have speed choices of 1,160 rpm, 1,750 rpm and 3,500 rpm.Then from Eq. (1) we have the following maximum discharges for the di!erent speedsof operation:

8(rpm) ](gpm) ](m/s)1,160 10,645 0.6721,750 1,169 0.2953,500 1,169 0.074

Based upon the value of discharge given above, it is seen that a speed of 1,160 rpmis the choice to make if we use 3 pumps. The pumps should be completely free ofcavitation.Next, calculate the impeller diameter needed. From Fig. 14.10 for maximum e"-ciency -X ! 0!12 and -6 ! 5!2 or

0!12 = ^$Q13 (2)

and 5!2 = !4$(12Q2$)) (3)

1303

Then for 8 = 1' 160 rpm (Q = 19!33 rps) and ] = 0!333 m3/s we can solve for 1from Eq. (2).

13 = ]$(0!12 n)

= 0!333$(0!12× 19!33)= 0!144

or 1 = 0!524 m

Now with a 1 of 0.524 m the head produced will be

!4 = 5!212Q2$) (from Eq. (3))

= 5!2(0!524)2(19!33)2$(9!81)

= 54!4 m

With a head of 54.4 m determine the diameter of pipe required to produce a dischargeof 1 m3/s. From the solution to Prob. 10.100 (as an approximation to this problem),we have

M, = 50 m + (, 2$2))(2!28 + b2$1) m

Assume b = 0!012

2 = 400 m

so M, = 50 m + (, 2$2))(2!28 + 4!8$1) m

54 m = 50 + (, 2$2))(2!28 + 4!8$1) (4)

Equation (4) may be solved for 1 by an iteration process: Assume 1, then solve for, and then see if Eq. (4) is satised, etc. The iteration was done for 1 of 60 cm, 70cm and 80 cm and it was found that the closest match came with 1 = 70 cm. Nowcompute the required power for an assumed e"ciency of 92%.

* = ](M,$e!.

= 0!333× 9' 810× 54$0!92* = 192 kW

* = 257hp

In summary, 1 = 70 cm, 8 = 1,160 rpm,

] per pump = 0!333 m3/s, * = 192 kW

The above calculations yield a solution to the problem. That is, a pump and pipingsystem has been chosen that will produce the desired discharge. However, a trulyvalid design should include the economics of the problem. For example, the rst costof the pipe and equipment should be expressed in terms of cost per year based uponthe expected life of the equipment. Then the annual cost of power should be includedin the total cost. When this is done, the size of pipe becomes important (smallersize yields higher annual cost of power). Also, pump manufacturers have a multiple

1304

number of pump designs to choose from which is di!erent than for this problem. Wehad only one basic design although considerable variation was available with di!erentdiameters and speed.The design could also include details about how the piping for the pumps would becongured. Normally this would include 3 separate pipes coming from the reservoir,each going to a pump, and then the discharge pipes would all feed into the largerpipe that delivers water to the elevated tank. Also, there should be gate valves oneach side of a pump so it could be isolated for maintenance purposes, etc. Checkvalves would also be included in the system to prevent back ow through the pumpsin event of a power outage.

1305

PROBLEM 15.1

Situation: Water ows through a rectangular channel. ? = 4 in! , = 28 ft$ s!

Find: (a) Determine if the ow is subcritical or supercritical.(b) Calculate the alternate depth.

APPROACH

Check the Froude number, then apply the specic energy equation to calculate thealternative depth.

ANALYSIS

Froude number

H3 = ,$&)?

= 28&32!2× 0!333

= 8!55

The Froude number is greater than 1 so the ow is supercritical .Specic Energy Equation

N = ? + , 2$)

N = 0!333 + 282$(2× 32!2)= 12!51 ft

Let the alternate depth = ?2' then

N = ?2 +, 222)

= ?2 +]2

2) (?2 × 3)2

Solving for the alternative depth for N = 12!51 ft yields ?2 = 12!43 ft.

1306

PROBLEM 15.2

Situation: Water ows through a rectangular channel.] = 900 ft3$ s ? = 3 ft width = 16 ft!

Find: Determine if the ow is subcritical or supercritical.

APPROACH

Calculate average velocity by applying the ow rate equation. Then check the Froudenumber.

ANALYSIS

Flow rate equation

] = , I

900 = , × 18× 3, = 18!75

Froude number

H3 = ,$&)?

= 18!75$&32!2× 3)

= 4!09

The Froude number is greater than 1 so the ow is supercritical .

1307

PROBLEM 15.3

Situation: Water ows through a rectangular channel.] = 420 ft3$ s , = 9 ft$ s width = 18 ft!


APPROACH

Calculate ? by applying the ow rate equation. Then check the Froude number.

ANALYSIS

Flow rate equation

] = , I

420 = 9× 18× ?? = 2! 593 ft

Froude number

H3 =,&)?

=9 ft$ s

&32!2× 2! 593

H3 = 0!985

Since H3 G 1, the ow is subcritical

1308

PROBLEM 15.4

Situation: Water ows through a rectangular channel.] = 12m3$ s width = 3m!Three depths of ow are of interest: ? = 0!3' 1!0' and 2!0m!

Find:(a) For each specied depth:

(i) Calculate the Froude number.(ii) Determine if the ow is subcritical or supercritical.

(b) Calculate the critical depth

APPROACH

Calculate average velocities by applying the ow rate equation. Then check theFroude numbers. Then apply the critical depth equation.

ANALYSIS

Flow rate equation

] = , I

12 m3/s = , (3× ?),0430 = 12 m3/s $(3 m× 0!30 m) = 13!33 m/s;,140 = 12 m3$s $(3 m × 1 m) = 4 m/s,240 = 12 m3/s$(3 m× 2 m) = 2 m/s

Froude numbers

H3043 = 13!33 m/s$(9!81 m/s2 × 0!30 m)1*2 = 7.77 (supercritical)

H3140 = 4 m/s$(9!81 m/s2 × 1!0 m)1*2 = 1.27 (supercritical)

H3240 = 2 m/s /9.81 m/s2 × 1!0 m)1*2 = 0.452 (subcritical)

Critical depth equation

?1 = (^2$))1*3

= ((4 m2/s)2$(9!81 m/s2))1*3

= 1.18 m

1309

PROBLEM 15.5

Situation: Water ows through a rectangular channel.] = 12m3$ s width = 3m ? = 0!3m!

Find: (a) Alternate depth.(b) Specic energy.

APPROACH

Apply the ow rate equation to nd the average velocity. Then calculate specicenergy and alternate depth.

ANALYSIS

Flow rate equation

, =]

I

=12

3× 0!3= 13!33 m/s

Specic Energy Equation

N = ? + , 2$2)

= 0!30 + 9!06

= 9.36 m

Let the alternate depth = ?2' then

N = ?2 +, 222)

= ?2 +]2

2) (?2 × 3)2

Substitute numerical values

9!36 = ?2 +122

2× 9!81 (?2 × 3)2

Solving for ?2 gives the alternate depth.

? = 9!35 m

1310

PROBLEM 15.6

Situation: Water ows at the critical depth in a channel; , = 5m$ s!

Find: Depth of ow.

APPROACH

Calculate the critical depth by setting Froude number equal to 1.

ANALYSIS

Froude number

H31 =,&)?1

1 =5 m/sq

9!81 m/s2 × ?1

Critical depth

?1 =, 2

)

=(5 m/s)2

9!81 m/s2

?1 = 2.55 m

1311

PROBLEM 15.7

Situation: Water ows in a rectangular channel. ] = 320 cfs width = 12 ft!Bottom slope = 0.005 Q = 0!014!


APPROACH

Calculate ?, then calculate the average velocity by applying the ow rate equation.Then check the Froude number.

ANALYSIS

] =1!49

QI"2*361*2!

=1!49

QI(I$* )2*361*2!

=1!49

QD?(D?$(>+ 2?))2*361*2!

=1!49

Q12?(12?$(12 + 2?))2*361*2!

320 =1!49

0!01412?(12?$(12 + 2?))2*3(0!005)1*2

Solving for ? yields: ? = 2!45 ft.Flow rate equation

, = ]$I

= 320 ft3/s $(12 ft × 2!45 ft)= 10!88 ft/s

Froude number

H3 = ,$&)?

= 10!88$(32!2× 2!45)1*2

H3 = 1!22 supercritical

1312

PROBLEM 15.8

Situation: Water ows in a trapezoidal channel—additional details are provided in theproblem statement.


APPROACH

Calculate Froude number by rst applying the ow rate equation to nd averagevelocity and the hydraulic depth equation to nd the depth.

ANALYSIS

Flow rate equation

, =]

I

=10 m3/s

(3× 1 m2) + 12 m2

= 2!50 m/s

Calculate hydraulic depth

1 =I

&

=4m2

5m= 0!80m

Froude number

H3 =,&)1

=2!50

&9!81× 0!80

= 0!89

Since H3 G 1, the ow is subcritical

1313

PROBLEM 15.9

Situation: Water ows in a trapezoidal channel–additional details are provided inthe problem statement.

Find: The critical depth.

APPROACH

Calculate the critical depth by setting Froude number equal to 1, and simultaneouslysolving it along with the ow rate equation and the hydraulic depth equation.

ANALYSIS

For the critical ow condition, Froude number = 1!

,$p)1 = 1

or(,$

&1) =

&)

Flow rate equation

, = ]$I = 20$(3? + ?2)

1 = I$& = (3? + ?2)$(3 + 2?)

Combine equations

(20$(3? + ?2))$((3? + ?2)$(3 + 2?))045 =&9!81

Solve for ??1K = 1!40 m

1314

PROBLEM 15.10

Situation: Water ows in a rectangular channel—additional details are provided in theproblem statement.

Find: (a) Plot depth versus specic energy.(b) Calculate the alternate depth.(c) Calculate the sequent depths.

APPROACH

Apply the specic energy equation.

ANALYSIS

Specic Energy Equation for a rectangular channel.

N = ? + ^2$(2)?2)

For this problem

^ = ]$D = 18$6 = 3 m2$s

so

N = ? + 32$(2)?2)

= ? + 0!4587$?2

The calculated N versus ? is shown below

? (m) 0!25 0!3 0!4 0!5 0!6 0!7 0!8 0!9 1!0 1!1 1!4 2!0 4!0 7!0N (m) 7!59 5!4 3!27 2!33 1!87 1!64 1!52 1!47 1!46 1!48 1!63 2!11 4!03 7!01

The corresponding plot is

1315

The alternate depth to ? = 0!30 is ? = 5!38 mSequent depth:

?2 = (?1$2)(q1 + 8H 21 " 1)

H31 = ,$&)?1

= (3$0!3)$&9!81× 0!30

= 5!83

Hydraulic jump equation

?2 = (0!3$2)(&1 + 8× 5!832 " 1) = 2.33 m

1316

PROBLEM 15.11

Situation: A rectangular channel ends in a free outfall–additional details are providedin the problem statement.

Find: Discharge in the channel.

APPROACH

Calculate the critical depth by setting Froude number equal to 1, and simultaneouslysolve it along with the brink depth equation. Then apply the ow rate equation.

ANALYSIS

At the brink, the depth is 71% of the critical depth

Abrink % 0!71?1 (1)

Just before the brink where the ow is critical, H3 = 1

1 =,&)?1

=^p)?31

(2)


Abrink = 0!71

µ^2

)

¶ 13

Or

^ = )1*2µAbrink0!71

¶3*2

= (9!81)1*2µ0!35

0!71

¶3*2

= 1!084 m2/s

Discharge is

] = ^Y

=¡1!084 m2/s

¢(4m)

= 4.34m3/s

1317

PROBLEM 15.12

Situation: A rectangular channel ends in a free outfall–additional details are providedin the problem statement.


APPROACH


ANALYSIS


^ = (1!20× 32!21*3$0!71)3*2

^ = 12!47 m2/s

Then

] = 15× 12!47 = 187 cfs

1318

PROBLEM 15.13

Situation: A rectangular channel ends in a free outfall. ] = 500 cfs Width = 14 ft.

Find: Depth of water at the brink of the outfall.

APPROACH

Calculate the depth at the brink by setting Froude number equal to 1, and simulta-neously solve this equation along with the brink depth equation.

ANALYSIS

At the brink, the depth is 71% of the critical depth

Abrink % 0!71?1 (1)

Just before the brink where the ow is critical, H3 = 1

1 =,&)?1

=^p)?31

(2)


Abrink = 0!71

µ^2

)

¶ 13

where

^ =]

Y

=500 ft3$ s

14 ft= 35!71 ft2$ s

Thus

Abrink = 0!71

Ã¡35!71 ft2$ s

¢2

32!2 ft$ s2

! 13

Abrink = 2.42 ft

1319

PROBLEM 15.14

Situation: Water ows over a broad-crested weir–additional details are given in theproblem statement.

Find: Discharge of water.

APPROACH

Apply the Broad crested weir—Discharge equation.

ANALYSIS

To look up the discharge coe"cient, we need the parameter 66+\

4

4 + *= (1!5$3!5)

= 0!43

From Fig. 15.7 - = 0!89!Broad crested weir—Discharge equation

] = 0!385 -2p2)4145

] = 0!385(0!89)(10)&2× 32!2(1!5)145

] = 50!5 cfs

1320

PROBLEM 15.15

Situation: Water ows over a broad-crested weir.(b) The weir height is * = 2m The height of water above the weir is 4 = 0!6m!(c) The length of the weir is 2 = 5m!

Find: Discharge.

APPROACH


ANALYSIS

To look up the discharge coe"cient, we need the parameter 66+\

4

4 + *=

0!6

0!6 + 2= 0!23

From Fig. 15.7- % 0!865

Broad crested weir—Discharge equation

] = 0!385 -2p2)43*2

= (0!385) (0!865)(5)&2× 9!81(0!60)145

] = 3!43 m3/s

1321

PROBLEM 15.16

Situation: Water ows over a broad-crested weir.Additional details are given in the problem statement.

Find: The water surface elevation in the reservoir upstream.

APPROACH


ANALYSIS

From Fig. 15.7, - % 0!85Broad crested weir—Discharge equation

] = 0!385 -2p2)43*2

25 = 0!385(0!85)(10)&2× 9!8143*2

Solve for 4

(4)3*2 = 1!725

4 = 1!438m

Water surface elevation

Elev. = 100m + 1!438m

= 101.4 m

1322

PROBLEM 15.17

Situation: Water ows over a broad-crested weir.Additional details are given in the problem statement.

Find: The water surface elevation in the upstream reservoir.

APPROACH


ANALYSIS

From Fig. 15.7, - % 0!85Broad crested weir—Discharge equation

] = 0!385- 2p2)43*2

1' 200 = 0!385(0!85)(40)&64!443*2

4 = 5!07 ft

Water surface elevation = 305.1 ft

1323

PROBLEM 15.18

Situation: Water ows in a rectangular channel.Two situations are of interest: an upstep and a downstep.Additional details are provided in the problem statement.

Find: (a) Change in depth and water surface elevation for the upstep.(b) Change in depth and water surface elevation for the downstep.(c) Maximum size of upstep so that no change in upstream depth occurs.

APPROACH

Apply the specic energy equation and check the Froude number.

ANALYSIS

Specic Energy Equation for the upstep

N1 = ?1 + ,21 $2)

= 3 + 32$(2× 9!81)= 3!46 m

Froude number

H31 = ,1$&)?1

= 3$&9!81× 3

= 0!55 (subcritical)

Then

N2 = N1 "!Rstep = 3!46" 0!30 = 3.16 m


?2 + ^2$(2)?22) = 3!16 m

?2 + 92$(2)?22) = 3!16

?2 + 4!13$?22 = 3!16

Solving for ?2 yields?2 = 2!49m

Then

!? = ?2 " ?1= 2!49" 3!00= -0.51 m

1324

So water surface drops 0!21 m!For a downstep

N2 = N1 +!Rstep

= 3!46 + 0!3 = 3.76 m

?2 + 4!13$?22 = 3!76

Solving for ?2 gives?2 = 3!40/

Then

!? = ?2 " ?1= 3!40" 3= 0.40 m

Water surface elevation change = +0!10 mMax. upward step before altering upstream conditions:

?1 = ?2 =3p^2$) = 3

p92$9!81 = 2!02

N1 = !Rstep +N2

where

N2 = 1!5?1 = 1!5× 2!02 = 3!03 m

Maximum size of step

Rstep = N1 "N2 = 3!46" 3!03 = 0.43 m

1325

PROBLEM 15.19

Situation: Water ows in a rectangular channel.Two situations are of interest: an upstep and a downstep.Additional details are provided in the problem statement.

Find: (a) Change in depth and water surface elevation for the upstep.(b) Change in depth and water surface elevation for the downstep.(c) Maximum size of upstep so that no change in upstream depth occurs.

APPROACH

Apply the specic energy equation by rst calculating Froude number and criticaldepth.

ANALYSIS

For the upstep

N2 = N1 " 0!60,1 = 2 m/s

Froude number

H31 = ,1$&)?1

= 2$&9!81× 3

= 0!369


N2 = 3 + (22$(2× 9!81))" 0!60 = 2!60 m?2 + ^

2$(2)?22) = 2!60

where ^ = 2× 3 = 6 m3/s/m. Then

?2 + 62$(2× 9!81× ?22) = 2!60

?2 + 1!83$?22 = 2!60

Solving, one gets ?2 = 2!24 m. Then

1326

!? = ?2 " ?1 = 2!34" 3!00 = -0.76 m

Water surface drops 0.16 mFor downward step of 15 cm we have

N2 = (3 + (22$(2× 9!81)) + 0!15 = 3!35 m?2 + 6

2$(2× 9!81× ?22) = 3!35

?2 + 1!83$?22 = 3!35

Solving: ?2 = 3!17 m or

?2 " ?1 = 3!17" 3!00 = +0.17 m

Water surface rises 0.02 mThe maximum upstep possible before a!ecting upstream water surface levels is for?2 = ?1Critical depth equation

?1 =3p^2$) = 1!54 m

Then

N1 = !Rstep +N2Icrit

!Rstep = N1 "N2Icrit = 3!20" (?1 + , 21 $2)) = 3!20" 1!5× 1!54!Rstep = +0.89 m

1327

PROBLEM 15.20

Situation: Water ows over an upstep–additional details are provided in the problemstatement.

Find: Maximum value of !R to permit a unit ow rate of 6 m2$s.

ANALYSIS


?1 = (^2$))1*3

= (62$9!81)04333

= 1!542 m

where ?1 is depth allowed over the hump for the given conditions.Specic Energy Equation

N1 = N2

,1 = ^$?1 = 6$3 = 2 m/s

,2 = 6$1!542 = 3!891 m/s

, 21 $2) + ?1 = , 22 $2) + ?2 +!R

22$2) + 3 = (3!8912$(2× 9!81)) + 1!542 +!R!R = 3!204" 0!772" 1!542 = 0.89 m

1328

PROBLEM 15.21

Situation: A rectangular channel has a gradual contraction in width–additionaldetails are provided in the problem statement.

Find: (a) Change in depth.(b) Change in water surface elevation.(c) Greatest contraction allowable so that upstream conditions are not altered.

ANALYSIS

Froude number

H31 = ,1$&)?1

= 3$&9!81× 3



N1 = N2

= ?1 + ,21 $2)

= 3 + 32$2× 9!81 = 3!46 m^2 = ]$D2 = 27$2!6 = 10!4 m3/s/m

Then

?2 + ^2$(2)?22)

= ?2 + (10!4)2$(2× 9!81× ?22) = 3!46

?2 + 5!50$?22 = 3!46

Solving: ?2 = 2!71 m.

!Rwater surface = !? = ?2 " ?1 = 2!71" 3!00 = 0.29 m

Max. contraction without altering the upstream depth will occur with ?2 = ?1

N2 = 1!5?1 = 3!45; ?1 = 2!31 m

Then

, 21 $2) = ?1$2 = 2!31$2 or ,1 = 4!76 m/s

]1 = ]2 = 27 = D2?1,1

D2 = 27$(2!31× 4!76) = 2!46 m

The width for max. contraction = 2.46 m

1329

PROBLEM 15.22

Situation: Ships streaming up a channel cause a problem due to a phenomena called“ship squat.” Additional details are provided in the problem statement.

Find: The change in elevation or “ship squat” of a fully loaded supertanker.

APPROACH

Apply the specic energy equation from a section in the channel upstream of the shipto a section where the ship is located. Then apply the ow rate equation and solvefor ?2.

ANALYSIS


N1 = N2

, 21 $2) + ?1 = , 22 $2) + ?2

I1 = 35× 200 = 7' 000 m2

,1 = 5× 0!515 = 2!575 m/s2!5752$(2× 9!81) + 35 = (]$I2)

2$(2× 9!81) + ?2 (1)

where ] = ,1I1 = 2!575× 7' 000 m3/sI2 = 200 m × ?2 " 29× 63

Flow rate equation

] = ,1I1 (2)

= 2!575× 7' 000 m3/sI2 = 200 m × ?2 " 29× 63 (3)

Substituting Eqs. (2) and (3) into Eq. (1) and solving for ?2 yields ?2 = 34!70 m.Therefore, the ship squat is

?1 " ?2 = 35!0" 34!7= 0.30 m

1330

PROBLEM 15.23

Situation: A rectangular channel has a small reach that is roughened with angleirons–additional details are provided in the problem statement.

Find: Determine the depth of water downstream of angle irons.

APPROACH

Apply the momentum principle for a unit width.

ANALYSIS

Momentum principle

XH$ =

X/!,! "

X/#,#

(?21$2" (?22$2" 2000 = "#, 21 ?1 + #,

22 ?2

Let ,1 = ^$?1 and ,2 = ^$?2 and divide by (

?21$2" ?22$2" 2000$( = "^21?1$()?

21) + ^

22?2$()?

22)

1$2" ?22$2" 3!205 = (+(20)2$32!2)("1 + 1$?2)

Solving for ?2 yields: ?2 = 1!43 ft

1331

PROBLEM 15.24

Situation: Water ows out of a reservoir into a steep rectangular channel–additionaldetails are provided in the problem statement.

Find: Discharge.

Assumptions: Negligible velocity in the reservoir and negligible energy loss. Thenthe channel entrance will act like a broad crested weir.

APPROACH


ANALYSIS


] = 0!545&)243*2

where 2 = 4 m and 4 = 3 m. Then

] = 0!545&9!81× 4× 33*2

= 35.5 m2/s

1332

PROBLEM 15.25

Situation: A small wave is produced in a pond.Pond depth = 8 in.

Find: Speed of the wave.

APPROACH

Apply the wave celerity equation.

ANALYSIS

Wave celerity

, =&)?

=q32!2 ft2$ s× 8$12 ft

= 4.63 ft/s

1333

PROBLEM 15.26

Situation: A small wave travels in a pool of water.Depth of water is constant.Wave speed = 1!5m$ s!

Find: Depth of water.

APPROACH


ANALYSIS

Wave celerity

, =&)?

1!5 =p9!81?

? = 0!23 m

1334

PROBLEM 15.27

Situation: As ocean waves approach a sloping beach, they curve so that they arealigned parallel to the beach.

Find: Explain the observed phenomena.

APPROACH


ANALYSIS

As the waves travel into shallower water their speed is decreased.Wave celerity

, =&)?

Therefore, the wave in shallow water lags that in deeper water. Thus, the wave creststend to become parallel to the shoreline.

1335

PROBLEM 15.28

Situation: A ba#ed ramp is used to dissipate energy in an open channel–additionaldetails are provided in the problem statement.

Find: (a) Head that is lost.(b) Power that is dissipated.(c) Horizontal component of the force exerted by ramp on the water.

Assumptions: The kinetic energy correction factors are % 1!0! E positive in thedirection of ow.

APPROACH

Let the upstream section (where ? = 3 ft) be section 1 and the downstream section(? = 2 ft) be section 2. Solve for the velocities at 1 and 2 using the ow rate equation.Then apply the energy equation and power equation. Determine the force of rampby writing the momentum equation between section 1 and 2. Let H$ be the force ofthe ramp on the water.

ANALYSIS

Flow rate equation

, = ]$I

,1 = 18$3

= 6 ft/s

,2 = 18$2

= 9 ft/s

Energy equation

?1 + T1,21 $2) + R1 = ?2 + T2,

22 $2) + R2 + ME

3 + 62$(2× 32!2) + 2 = 2 + 92$(2× 32!2) + MEME = 2!30 ft

Power equation

* = ](ME$550

= 18× 62!4× 2!3$550

* = 4.70 horsepower

1336

Momentum principle

XH$ = #^(,2$ " ,1$)

(?21$2" (?22$2 + H$ = 1!94× 18(9" 6)

(62!4$2)(33 " 22) + H$ = 104!8

H$ = "51!2 lbf

The ramp exerts a force of 51.2 lbf opposite to the direction of ow.

1337

PROBLEM 15.29

Situation: Water ows out a reservoir, down a spillway and then forms a hydraulicjump near the base of the spillway.Flow rate is ^ = 2!5m3$ s per m of width.Additional details are provided in the problem statement.

Find: Depth downstream of hydraulic jump.

APPROACH

Apply the specic energy equation to calculate ?1. Then calculate Froude numberin order to apply the Hydraulic jump equation.

ANALYSIS

Specic Energy

?0 + ^2$(2)?20) = ?1 + ^

2$(2)?21)

5 + 2!52$(2(9!81)52) = ?1 + 2!52$(2(9!81)?21)

?1 = 0!258 9 m

Froude number

H31 =^p)?31

=2!5p

9!81(0!258 9)3

= 6! 059


?2 = (?1$2)

µq1 + 8H 21 " 1

¶

= (0!258 9$2)³p

1 + 8(6! 0592)" 1´

= 2.09 m

1338

PROBLEM 15.30

Situation: Water ows out a sluice gate–additional details are provided in the prob-lem statement.

Find: (a) Determine if a hydraulic jump can exist.(b) If the hydraulic jump can exist, calculate the depth downstream of the jump.

APPROACH

Calculate Froude number, then apply the hydraulic jump equation.

ANALYSIS

Calculate Froude number

H3 =,&)?

=^p)?3

=3!6m2$ s

&9!81× 0!33m2$ s

= 7!00

Thus, a hydraulic jump can occur.Hydraulic jump equation

?2 = (?1$2)

µq1 + 8H 21 " 1

¶

= (0!3$2)³&1 + 8× 72 " 1

´

?2 = 2. 82 m

1339

PROBLEM 15.31

Situation: A dam and spillway are described in the problem statement.

Find: Depth of ow on the apron just downstream of the hydraulic jump.

Assumptions: ,0 is negligible; kinetic energy correction factors are negligible.

APPROACH

First develop the expression for ?1 and ,theor.!Begin by applying the energy equationfrom the upstream pool to ?1. Then nd ^ by applying the rectangular weir equation.Then solve for the depth of ow on the apron by applying the hydraulic jump equation.

ANALYSIS

Energy equation

T0,20 $2) + R0 = T1,

21 $2) + R1

0 + 100 = , 2theor.$2) + ?1 (1)

But

,theor = ,act$0!95 (2)

and

,act. = ^$?1 (3)

Consider a unit width of spillway. ThenRectangular weir equation

^ = ]$2 = <p2)4145

= 0!5p2)(5145)

^ = 44!86 cfs/ft (4)

Solving Eqs. (1), (2), (3), and (4) yields

?1 = 0!59 ft

1340

and,act. = 76!03bP$9

Froude number

H31 = ,$&)?1

= 76!03$p(32!2)(0!59)

= 17!44


?2 = (?1$2)((1 + 8H321)045 " 1)

= (0!59$2)((1 + 8(17!442))045 " 1)

?2 = 14.3 ft

1341

PROBLEM 15.32

Situation: A hydraulic jump is described in the problem statement.

Find: Depth upstream of the hydraulic jump.

APPROACH

Apply the hydraulic jump equation.

ANALYSIS


?2 = (?1$2)((1 + 8H321)045 " 1)

where Froude number

H321 = ,21 $()?1) = ^

2$()?31)

Then

?2 = (?1$2)((1 + 8^2$()?31))

045 " 1)?2 " ?1 = (?1$2)[((1 + 8^

2$()?31))045 " 1" 2]

However

?2 " ?1 = 14!0 ft (given)

^ = 65 ft2$s

Therefore

14!0 ft = (?1$2)[((1 + 8× 652$()?31))045 " 1)" 2]

?1=1.08 ft

1342

PROBLEM 15.33

Situation: An obstruction in a channel causes a hydraulic jump.On the upstream side of the jump: ,1 = 8m$ s ?1 = 0!40m!

Find: Depth of ow downstream of the jump.

APPROACH

Calculate the upstream Froude number. Then apply the Hydraulic jump equationto nd the downstream depth.

ANALYSIS

Froude number

H31 =,&)?1

=8

&9!81× 0!4

= 4! 039


?2 =?12

·q1 + 8H321 " 1

¸

=0!40

2

h&1 + 8× 4! 0392 " 1

i

?2 = 2.09 m

1343

PROBLEM 15.34

Situation: A hydraulic jump is described in the problem statement.( = 9' 810 N/m2' D3 = 5 m, ?1 = 40 cm= 0!40 m.

Find: Depth of ow downstream of jump.

ANALYSIS

Check H3 upstream to see if the ow is really supercritical ow. Then apply themomentum principle.

H3 = ,$()1)045

1 = I$&

= (D? + ?2)$(D + 2?)

13 = 044 = (5× 0!4 + 0!42)$(5 + 2× 0!4)= 0!372 m

Then

H31 = 10 m/s/((9!81 m/s2)(0!372))045

H31 = 5!23

so ow is supercritical and a jump will form. Applying the momentum equation (Eq.15.23):

%1I1 + #],1 = %2I2 + #],2 (1)

Evaluate %1 by considering the hydrostatic forces on the trapezoidal section dividedinto rectangular plus triangular areas as shown below:

Then

%1I1 = %=I= + %>I> + %)I)

= ((?1$3)(?21$2) + ((?1$2) D?1 + ((?1$3)(?

21$2)

= ((?31$6) + (D(?21$2) + ((?

31$6)

= ((?31$3) + (D(?21$2)

%1I1 = (((?31$3) +D(?21$2))

Also

1344

#],1 = #]]$I1 = #]2$I1

Equation (1) is then written as

(((?31$3) + (D(?21$2)) + #]

2$I1 = (((?32$3) +D(?

22$2)) + #]

2$(D?2 + ?22)

Flow rate equation

] = ,1I1 = 21!6 m3/s

I1 = 5× 0!4 + 0!42 = 2!16 m2

Solving for ?2 yields: ?2 = 2!45 m

1345

PROBLEM 15.35

Situation: A hydraulic jump occurs in a wide rectangular channel.The upstream depth is ?1 = 0!5 ft!The downstream depth is ?2 = 10 ft!

Find: Discharge per foot of width of channel.

APPROACH

Apply the Hydraulic jump equation to solve for the Froude number. Next, use thevalue of the Froude number to solve for the discharge ^.

ANALYSIS


?2 =?12

·q1 + 8H321 " 1

¸

12 =0!5

2

·q1 + 8× H321 " 1

¸

Solve the above equation for Froude number.

H31 = 14!49

Froude number

H31 =^p)?31

14!49 =^

&32!2× 0!53

Solve the above equation for ^

^ = 29!07 ft2/s

1346

PROBLEM 15.36

Situation: A rectangular channel has three di!erent reaches–additional details areprovided in the problem statement.

Find: (a) Calculate the critical depth and normal depth in reach 1.(b) Classify the ow in each reach (subcritical, critical or supercritical).(c) For each reach, determine if a hydraulic jump can occur.

APPROACH

Apply the critical depth equation. Determine jump height and location by applyingthe hydraulic jump equation.

ANALYSIS


?1 = (^2$))1*3

^ = 500$20 = 25 cfs/ft

?1 = (252$32!2)1*3 = 2.69 ft

Solving for ?:I1yields 1!86 ft.Thus one concludes that the normal depth in each reach is

• Supercritical in reach 1

• Subcritical in reach 2

• Critical in reach 3

If reach 2 is long then the ow would be near normal depth in reach 2. Thus, theow would probably go from supercritical ow in reach 1 to subcritical in reach 2. Ingoing from sub to supercritical a hydraulic jump would form.


?2 = (?1$2)((1 + 8H321)045 " 1)

H31 = ,1$()?1)045 = (25$1!86)$(32!2× 1!86)045 = 1!737

?2 = (1!86$2)((1 + 8× 1!7372)045 " 1) = 3.73 ft

Because ?2 is less than the normal depth in reach 2 the jump will probably occur inreach 1. The water surface prole could occur as shown below.

1347

1348

PROBLEM 15.37

Situation: Water ows out a sluice gate and then over a free overfall–additionaldetails are provided in the problem statement.

Find: (a) Determine if a hydraulic jump will form.(i) If a jump forms, locate the position.(ii)If a jump does not form, sketch the full prole and label each part.

(b) Sketch the EGL

APPROACH

Check Froude numbers. Then determine ?1 for a ?2 of 1.1 m by applying the hydraulicjump equation.

ANALYSIS

Froude number

H31 = ,1$&)?1 = 10$

&9!81× 0!10 = 10!1 (supercritical)

,2 = ^$?2 = (0!10 m) (10 m/s)/(1.1 m) = 0!91 m/s

H32 = ,2$()?2)045 = 0!91$(9!81× 1!1)045 = 0!277

A hydraulic jump will form because ow goes from supercritical to subcritical.Hydraulic jump equation

?1 = (?2$2)((1 + 8H322)045 " 1)

= (1!1$2)((1 + 8× !2772)045 " 1)?1 = 0!14 m

Therefore the jump will start at about the 29 m distance downstream of the sluicegate. Prole and energy grade line:

1349

PROBLEM 15.38

Situation: Water ows out a sluice gate and then over a free overfall–additionaldetails are provided in the problem statement.

Find: Estimate the shear stress on the bottom of the channel 0.5 m downstream ofthe sluice gate.

Assumptions: The ow can be idealized as boundary layer ow over a at plate,where the leading edge of the plate is located at the sluice gate.

APPROACH

Apply the local shear stress equation.

ANALYSIS

Reynolds number

Re$ =, E

:

=10× 0!510"6

= 5× 106

Since Re$ F 500' 000, the boundary layer would be turbulent. The most appropriatecorrelation is given by Eq. (9.52a):

Z( =0!455

ln2 (0!06Re$)

=0!455

ln2 (0!06× 5× 106)= 0!00286

Local shear stress

B ! = Z(#, 2

2

= 0!002861000× 102

2= 143Pa

Therefore, the correct choice is (d) B F 40 N/m2

1350

PROBLEM 15.39

Situation: Water ows out of sluice gate and then through a hydraulic jump–additional details are provided in the problem statement.

Find: Horsepower lost in hydraulic jump.

Assumptions: Negligible energy loss for ow under the sluice gate.

APPROACH

Apply the Bernoulli equation from a location upstream of the sluice gate to a locationdownstream. Then, calculate the Froude number and apply the equations that governa hydraulic jump. Calculate the power using * = ](ME$550, where the number"550" is a unit conversion.

ANALYSIS

Bernoulli equation

?0 + ,20 $2) = ?1 + ,

21 $2)

65 + neglig. = 1 + , 21 $2)

,1 =&64× 64!4 = 64!2 ft/s

Froude number

H31 = ,1$&)?1

= 64!2$&32!2× 1

= 11!3

Hydraulic jump equations

?2 = (?1$2)(q1 + 8H 21 " 1)

= (1$2(&1 + 8× 11!32 " 1)

= 15!5 ft

ME = (?2 " ?1)3$(4?1?2)= (14!51)3$(4× 1× 15!51)= 49.2 ft

Power equation

* =](ME550

=(64!2× 1× 5)× 62!4× 49!2

550

= 1793 horsepower

1351

PROBLEM 15.40

Situation: A ume is to be designed. This ume will be used to verify the hydraulicjump relationships given in Section 15.2.

Find: Basic specications of the ume.

ANALYSIS

For this experiment, it is necessary to produce supercritical ow in the ume andthen force this ow to become subcritical. The supercritical ow could be producedby means of a sluice gate as shown in Prob. 15.39 and the jump could be forced bymeans of another sluice gate farther down the ume. Therefore, one needs to includein the design an upstream chamber that will include a sluice gate from which the highvelocity ow will be discharged.

The relevant equation for the hydraulic jump is Eq. (15.28). To verify this equation?1' ?2 and ,1 can be measured or deduced by some other means. A fairly accuratemeasurement of ?2 can be made by means of a point gage or piezometer. The depth?1 could also be measured in the same way; however, the degree of accuracy of thismeasurement will be less than for ?2 because ?1 is much smaller than ?2. Perhaps amore accurate measure of ?1 would be to get an accurate reading of the gate openingof the sluice gate and apply a coe"cient of contraction to that reading to get ?1. The-) for a sluice gate could be obtained from the literature.

The velocity, ,1, which will be needed to compute HK1, can probably be best calculatedby the Bernoulli equation knowing the depth of ow in the chamber upstream of thesluice gate. Therefore, a measurement of that depth must be made.

Note that for use of ,1 and ?1 just downstream of the sluice gate, the hydraulicjump will have to start very close to the sluice gate because the depth will increasedownstream due to the channel resistance. The jump location may be changed byoperation of the downstream sluice gate.

COMMENTS Other things that could or should be considered in the design:

A. Choose maximum design discharge. This will be no more than 5 cfs (see Prob.13.77).

B. Choose reasonable size of chamber upstream of sluice gate. A 10 ft depth wouldbe ample for a good experiment.

C. Choose width, height and length of ume.

D. Work out details of sluice gates and their controls.

1352

PROBLEM 15.41

Situation: Water ows in a rectangular channel.A sill installed on the bottom of the channel forces a hydraulic jump to occur.Additional details are provided in the problem statement.

Find: Estimate the height of hydraulic jump (the height is the change in elevation ofthe water surface).

Assumptions: Q = 0!012!

APPROACH

Calculate Froude number in order to apply the Hydraulic jump equation.

ANALYSIS

, = (1$Q)"2*361*20

" = I$* = (0!4× 10)$(2× 0!4 + 10) = 0!370 m, = (1$0!012)(0!370)2*3 × (0!04)1*2 = 8!59 m/s

Froude number

H31 = ,$&)?1

= 8!59$&9!81× 0!40

= 4!34 (supercritical)


?2 = (?1$2)(q1 + 8× H 21 " 1)

= (0!40$2)(p1 + 8× (4!34)2 " 1)

?2 = 2.26 m

1353

PROBLEM 15.42

Situation: Water ows in a rectangular channel.A sill installed on the bottom of the channel forces a hydraulic jump to occur.Additional details are provided in the problem statement.

Find: (a) Estimate the shear force associated with the jump.(b) Calculate the ratio H&$H6 , where H& is shear force and H6 is the net hydrostaticforce acting on the jump.

Assumptions: (a) The shear stress will be the average of B 01 (associated with uniformow approaching the jump), and B 02 (associated with uniform ow leaving the jump).(b) The ow may be idealized as normal ow in a channel.

APPROACH

Apply the local shear stress equation 10.21 and calculate the Reynolds numbers.Then nd ,2 by applying the same solution procedure from problem 15.41. Thenestimate the total shear force by using an average shear stress.

ANALYSIS

Local shear stress

B 0 = b#,2$8

where b = b(Re' c&$4")

"G1 = ,1(4"1)$: "G2 = ,2 × (4"2)$:

From solution to Prob. 15.41

,2 = ,1 × 0!4$2!26 = 1!52 m/s

"G1 = 8!59× (4× 0!37)$10"6 "2 = I$* = (2!26× 10)$(2× 2!26 + 10) = 1!31 m"G1 = 1!3× 107 "G2 = 1!52× (4× 1!56)$10"6 = 9!5× 106

Assume c& = 3× 10"3 m

c&$4"1 = 3× 10"3$(4× 0!37) c&$4"2 = 3× 10"3$(4× 1!56)c&$4"1 = 2× 10"3 c&$4"2 = 4!8× 10"4

From Fig. 10-8, b1 = 0!024 and b2 = 0!017! Then

B 01 = 0!024× 1' 000× (6!87)2$8 B 02 = 0!017× 1' 000× (1!52)2$8B 01 = 142 N/m

2 B 02 = 4!9 N/m2

B avg = (142 + 4!9)$2 = 73 N/m2

1354

Then

H& = B avgI& = B avg*2

where 2 ! 6?2' * ! D + (?1 + ?2)! Then

H2 ! 73(10 + (0!40 + 2!26))(6× 2!26)= 10' 790 N

H6 = (($2)(?22 " ?21)D

= (9' 810$2)((2!26)2 " (0!40)2)× 10= 242' 680 N

Thus

H&$H6 = 10' 790$242' 680

= 0.044

COMMENTS

The above estimate is probably inuenced too much by B 01 because shear stress willnot be linearly distributed. A better estimate might be to assume a linear distributionof velocity with an average b and then integrate B 0AI from one end to the other.

1355

PROBLEM 15.43

Situation: Water ows out of a sluice gate–additional details are provided in theproblem statement.

Find: (a) Determine the type of water surface prole that occurs downstream of thesluice gate.(b) Calculate the shear stress on bottom of the channel at a horizontal distance of0.5 m downstream from the sluice gate.

Assumptions: The ow can be idealized as a boundary layer ow over a at plate,with the leading edge of the boundary layer located at the sluice gate.

APPROACH

Apply the hydraulic jump equation by rst calculating ^ applying the ow rate equa-tion. Then apply the local shear stress equation.

ANALYSIS

Flow rate equation

^ = 0!40× 10

= 4!0m2

s


?1 = 3p^2$)

= 3p(4!0)2$9!81

= 1!18m

Then we have ? G ?: G ?1; therefore, the water surface prole will be an 63!Reynolds number

Re$ ! , × 0!5$:Re$ = 10× 0!5$10"6

= 5× 106

The local shear stress coe"cient is

Z( =0!455

ln2 (0!06Re$)

=0!455

ln2 (0!06× 5× 106)= 0!00286

1356

Local shear stress

B 0 = Z(#, 202

= 0!00286998× 102

2

= 143 N/m2

1357

PROBLEM 15.44

Situation: Water ows in a rectangular channel–additional details are provided inthe problem statement.

Find: Classify the water surface prole asa.) S1b.) S2c.) M1d.) M2

ANALYSIS

?: = 2 ft

?1 = (^2$))1*3 = (102$32!2)1*3 = 1!46 ft.

? F ?: F ?1

From Fig. 15-16 the prole is M1. Thus, the correct choice is c.

1358

PROBLEM 15.45

Situation: Water ows in a rectangular channel–additional details are provided inthe problem statement.

Find: Classify the water surface prole asa.) M2b.) S2c.) H1d.) A2

ANALYSIS

The correct choice is d).

1359

PROBLEM 15.46

Situation: The problem statement shows a partial sketch of a water-surface prole.

Find: (a) Sketch the missing part of the water prole.(b) Identify the various types of proles.

APPROACH

Check the Froude number at points 1 and 2. Apply the Broad crested weir—Dischargeequation to calculate ?2 for the second Froude number.

ANALYSIS

Froude number

H31 = ^$p)?3

= (5$3)$p9!81(0!3)3

= 3!24 F 1(supercritical)


] = (0!40 + 0!054$* )2p2)43*2

5 = (0!40 + 0!054$1!6)× 3p2(9!81)43*2

Solving by iteration gives 4 = 0!917 m. Depth upstream of weir = 0!917+1!6 = 2!52m

H2 = (5$3)$p9!81(2!52)3 = 0!133 G 1 (subcritical)

Therefore a hydraulic jump forms.Hydraulic jump equation

?2 = (?1$2)(q1 + 8H 21 " 1)

?2 = (0!3$2)(p1 + 8(3!24)2 " 1)

?2 = 1!23m

1360

PROBLEM 15.47

Situation: A rectangular channel ends with a free overfall–additional details areprovided in the problem statement.

Find: Determine the classication of the water surface just before the brink of theoverfall.

ANALYSIS

The prole might be an + prole or an 6 prole depending upon whether the slopeis mild or steep. However, if it is a steep slope the ow would be uniform right tothe brink. Check to see if + or 6 slope. assume Q = 0!012

] = (1!49$Q)I"046676045

I"2*3 = ]$((1!49$Q)(6045));

= 120$((1!49$0!012)(0!0001)045)

(>?)(>?$(10 + 2?))4667 = 96!6

With > = 10 ft we can solve for ? to obtain ? = 5!2 ft.Flow rate equation

, = ]$I

= 120$32

= 2!31 ft/s

Froude number

H3 = ,$&)?

= 2!31$(p32!2× 5!2)


Therefore, the water surface prole will be an M2.

1361

PROBLEM 15.48

Situation: Water ows out a sluice gate and thorough a rectangular channel.A weir will be added to the channel.Additional details are provided in the problem statement.

Find: (a) Determine if a hydraulic jump will occur.(b) If a jump form, calculate the location.(c) Label any water surface proles that may be classied.

ANALYSIS


] = <p2)243*2

where < = 0!40 + 0!054$* . By trial and error (rst assume < then solve for 4,etc.) solve for 4 yield 4 = 2!06 ft.Flow rate equation

, = ]$I

= 108$(4!06× 10)= 2!66 ft/s

Froude number

H3 = ,$&)?

= 2!66$(32!2× (4!06))045


The Froude number just downstream of the sluice gate will be determined:Flow rate equation

, = ]$I

= 108$(10× 0!40)= 27 ft/s

Froude number

H3 = ,$&)?

= 27$&32!2× 0!40

= 7!52 (supercritical)

Because the ow is supercritical just downstream of the sluice gate and subcriticalupstream of the weir a jump will form someplace between these two sections.

Now determine the approximate location of the jump. Let ?2 = depth downstreamof the jump and assume it is approximately equal to the depth upstream of the weir

1362

(? ! 4!06 ft). By trial and error (applying the hydraulic jump equation 15.25))it can be easily shown that a depth of 0.40 ft is required to produce the given ?2.Thus the jump will start immediately downstream of the sluice gate and it will beapproximately 25 ft long. Actually, because of the channel resistance ?2 will besomewhat greater than ?2 = 4!06 ft; therefore, the jump may be submerged againstthe sluice gate and the water surface prole will probably appear as shown below.

1363

PROBLEM 15.49

Situation: A rectangular channel is described in the problem statement.

Find: (a) Sketch all possible water-surface proles.(b) Label each part of the water-surface prole with its classication.

APPROACH

Apply the critical depth equation to determine if a hydraulic jump will form.

ANALYSIS


?1 = 3p^2$)

= 3p202$32!2

= 2!32 ft

Thus the slopes in parts 1 and 3 are steep.

If part 2 is very long, then a depth greater than critical will be forced in part 2 (thepart with adverse slope). In that case a hydraulic jump will be formed and it mayoccur on part 2 or it may occur on part 1. The other possibility is for no jump toform on the adverse part. These three possibilities are both shown below.

1364

PROBLEM 15.50

Situation: Water ow through a sluice gate and down a rectangular channel is de-scribed in the problem statement.

Find: Sketch the water surface prole until a depth of 60 cm. is reached.

ANALYSIS

Froude number

H31 = ^$p)?3

= 3$p9!81(0!2)3 = 10!71

H32 = 3$p9!81(0!6)3 = 2!06

Therefore the prole is a continuous 43 prole.

? ? , , N !N 6( !E E0.2 15 11.6678 0

0.25 12.5 6.2710 0.1593 39.40.3 10 5.3968 39.4

0.35 8.75 2.1298 0.0557 38.20.4 7.5 3.2670 77.6

0.45 6.75 0.9321 0.0258 36.10.5 6.0 2.3349 113.7

0.55 5.5 0.4607 0.0140 32.90.6 5.0 1.8742 146.6

1365

PROBLEM 15.51

Situation: A horizontal channel ends in a free outfall–additional details are providedin the problem statement.

Find: Water depth 300 m upstream of the outfall.

APPROACH

Apply the critical depth equation. Then carry out a step solution for the proleupstream from the brink.

ANALYSIS

^ = ]$D

= 12$4 = 3 m3/s/m

?1 = 3p^2$)

= 0!972 m (This depth occurs near brink.)

Reynolds number

Re ! , × 4"$: ! 3× 1$10"6 ! 3× 106

c&$4" ! 0!3× 10"3$4 ! 0!000075b ! 0!010

See solution table below.

1366

Section number upstream of yc

Depth y,m

Velocity atsection V,m/s

Mean Velocity in reach

(V1+V0)/2 V2Hydraulic Radius

R=A/P,m

Mean Hydraulic Radius

Rm=(R1+R2)/2sf=fV2

mean/8gRmean

"x=((y2+V22/2g)-

(y1+V1/2g))/Sf

Distance upstream from brink x,m

1 at y=yc 0.972 3.086 0.654 3.9m

3.073 9.443 0.656 1.834 x 10-3 0.1m 4.0m2 0.980 3.060 0.658

3.045 9.272 0.660 1.790 x 10-3 0.4m 4.4m3 0.990 3.030 0.662

2.986 8.916 0.669 1.698 x 10-3 1.7m 6.1m4 1.020 2.941 0.675

2.886 8.327 0.684 1.551 x 10-3 4.7m 10.9m5 1.060 2.830 0.693

2.779 7.721 0.701 1.403 x 10-3 7.7m 18.6m6 1.100 2.727 0.710

2.613 6.828 0.730 1.192 x 10-3 33.2m 51.8m7 1.200 2.500 0.750

2.404 5.779 0.769 9.576 x 10-4 55.3m 107.1m8 1.300 2.308 0.788

2.225 4.951 0.806 7.83 x 10-4 80.0m 187.1m9 1.400 2.143 0.824

2.072 4.291 0.841 6.501 x 10-4 107.4m 294.5m10 1.500 2.000 0.857The depth 300 m upstream is approximately 1.51 m

Solution Table for Problem 15.51

1367

PROBLEM 15.52

Situation: Water ows through a sluice gate, down a channel and across a hydraulicjump.Additional details are provided in the problem statement.

Find:(a) Determine the water-surface prole classication

i) Upstream of the jump.ii) Downstream of the jump.

(b) Determine how the addition of ba#e block will e!ect the jump.

ANALYSIS

Upstream of the jump, the prole will be an 43.Downstream of the jump, the prole will be an 42.The ba#e blocks will cause the depth upstream of I to increase; therefore, the jumpwill move towards the sluice gate.

1368

PROBLEM 15.53

Situation: Water ows out of a reservoir, down a spillway and then over an outfall.Additional details are provided in the problem statement.


Assumptions: ,1 = 0 and T2 = 1!0!

APPROACH

Apply the energy equation from the reservoir, (1), to the entrance section (2) and setthe Froude number equal to 1 (critical ow) to solve for ?1 and ,1. Then calculatethe discharge by applying the ow rate equation.

ANALYSIS

The channel is steep; therefore, critical depth will occur just inside the channel en-trance.Energy equation

?1 + T1,21 $2) = ?2 + T2,

22 $2)

Then

2 = ?2 + ,22 $2)

Froude number

, 22 $2) = , 21 $2) (1)

= 0!5?1

The energy equation becomes?1 = ?1 + 0!5?1

Let ?1 = 2m and solve for ?1

?1 = 2m$1!5 = 1!33 m

From Eq. (1)

, 21 $) = ?1

= 1!33

or ,1 = 3!62 m/s

1369

Flow rate equation

] = ,1I2

= 3!62× 1!33× 4] = 19!2 m3/s

1370

PROBLEM 15.54

Situation: Water ows out a reservoir and down a channel.

Find: (a) Estimate the discharge.(b) Describe a procedure for calculating the discharge if the channel length was 100m.

Assumptions: Uniform ow is established in the channel except near the downstreamend. Q = 0!012 .

APPROACH

Apply the energy equation from the reservoir to a section near the upstream endof the channel to solve for , . Then apply the ow rate equation to calculate thedischarge.

ANALYSIS

(a) Energy equation

2!5 ! , 2: $2) + ?: (1)

Also

,: = (1$Q)"2*361*2

, 2: $2) = (1$Q2)"4*36$2) (2)

where

" = I$* = 3!5?:$(2?: + 3!5) (3)

Then combining Eqs. (1), (2) and (3) we have

2!5 = ((1$Q2)((3!5?:$(2?: + 3!5))4*36$2)) + ?: (4)

Assuming Q = 0!012 and solving Eq. (4) for ?: yields: ?: = 2!16 m; also solving (2)yields ,: = 2!58 m/s. Then

] = , I

= 2!58× 3!5× 2!15] = 19!5 m3/s

(b) With only a 100 m-long channel, uniform ow will not become established in thechannel; therefore, a trial-and-error solution is required. Critical depth will occurjust upstream of the brink, so assume a value of ?1, then calculate ] and calculatethe water surface prole back to the reservoir. Repeat the process for di!erent valuesof ?1 until a match between the reservoir water surface elevation and the computedprole is achieved.

1371

PROBLEM 15.55

Situation: During ood ow, water ows out of a reservoir.

Find: Calculate the water surface prole upstream from the dam until the depth issix meters.

APPROACH

Apply the critical depth equation. Then carry out a step solution for the proleupstream from the dam.

ANALYSIS

^ = 10 m3/s/m

?1 = 3p^2$)

= 3p102$9!81

= 2!17 m

? ? , , N !N 6(×104 !E E elev.52.17 0.1917 52.170 0 52.17

51.08 0.1958 2.168 0.00287 -5,42950 0.20 50.002 5,430 52.17

45 0.2222 9.999 0.00419 -25,02440 0.25 40.003 -30,450 52.18

35 0.2857 9.997 0.00892 -25,04830 0.333 30.006 -55,550 52.22

25 0.400 9.993 0.02447 -25,14620 0.50 20.013 -80,650 52.26

15 0.6667 9.962 0.11326 -25,63110 1.00 10.051 -106,280 52.51

9 1.1111 1.971 0.5244 -5,6718 1.25 8.080 -111,950 52.78

7 1.4286 1.938 1.1145 -6,7166 1.667 6.142 -118,670 53.47

1372

PROBLEM 15.56

Situation: Water ows in a wide rectangular concrete channel.Additional details are provided in the problem statement.

Find: Determine the water surface prole from section 1 to section 2.

Assumptions: Q = 0!015' < = 0!42' c& = 0!001 ft so c&$4" = 0!00034!

APPROACH

Determine whether the uniform ow in the channel is super or subcritical. Determine?: and then see if for this ?: the Froude number is greater or less than unity. Thenapply the hydraulic jump equation to get ?2. Then apply the Rectangular weirequation to nd the head on the weir. A rough estimate for the distance to where thejump will occur may be found by applying Eq. (15.35) with a single step computation.A more accurate calculation would include several steps.

ANALYSIS

Froude number

] = (1!49$Q)I"2*361*2

12 = (1!49$0!015)× ? × ?2*3 × (0!04)1*2

?: = 0!739 ft and , = ]$?: = 16!23 ft/s

H = ,$&)?: = 3!33

Solving for ?: gives ?: = 0!739 ft and

, = ]$?: = 16!23 ft/s

Therefore, uniform ow in the channel is supercritical and one can surmise that ahydraulic jump will occur upstream of the weir. One can check this by determiningwhat the sequent depth is. If it is less than the weir height plus head on the weirheight plus head on the weir then the jump will occur.

Now nd sequent depth:

?2 = (?1$2)(q1 + 8H 21 " 1)

= (0!739$2)(&1 + 8× 3!332 " 1)

?2=3.13 ft


] = <p2)243*2

12 = 0!42&64!4× 1×43*2

4 = 2!33 ft

4$* = 2!33$3 = 0!78

1373

so

< = 0!40 + 0!05× 0!78

A better estimate is4 = 2!26 ft < = 0!44

Then depth upstream of weir = 3 + 2!26 = 5!56 ft. Therefore, it is proved that ajump will occur.

The single-step calculation is given below:

!E = ((?1 " ?2) + (, 21 " ,22 ))$2)$(6( " 60)

where ?1 = 3!13 ft; ,1 = ^$?1 = 12$3!13 = 3!83 ft/s; , 21 = 14!67 ft2/s2 and ?2 = 5!56

ft; ,2 = 2!16 ft/s.

, 22 = 4!67 ft2$s2

6( = b, 2avg$(8)"avg)

,avg = 3!00 ft/s

"avg = 4!34 ft

Assuming c& = 0!001 ft so c&$4" = 0!00034!

Re = , × 4"$: = ((3!83 + 2!16)$2)× 4× 4!34$(1!22× 10"5) = 4!33× 106

Thenb = 0!015

and

6( = 0!015× 3!02$(8× 32!2× 4!34) = 0!000121

!E = ((3!13" 5!56) + (14!67" 4!67)$(64!4))$(0!000121" 0!04) = 57.0 ft

Thus, the water surface prole is shown below:

1374

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