Mid-term Exam1. Newtonian Fluid, Non-Newtonian Fluid, Ideal Fluid2. Viscosity, Kinematic Viscosity3. Viscosity4. Bulk Modulus of Elasticity5. Surface Tension, Capillarity 6. Vapor Pressure (Cavitation)7. Perfect Gas Law8. Pressure9. Pressure variation10. Monometer11. Pressure Variation and Center of Pressure12. Buoyancy13. Continuity Equation and Bernoulli Equation14. Force from Fluid

Viscosity (Absolute Viscosity)

Kinematic Viscosity

Physical Properties of Water

7. Capillarity

0. Newtonian Fluid

Shear Stress

Ideal Fluid = 0

Newtonian Fluid = constant

Non-Newtonian Fluid = not constant

Ideal Plastic = constant

> yield stress

du

dy

du

dy

1. Viscosity, kinematic viscosity

a. What is the viscosity of gasoline at 25 ºC ?

b. Determine the kinematic viscosity of benzen at 27 ºC ?

(Use the figure)

(Use the figure)

c. A fluid has a viscosity 0.6 Pa·s and a relative density of 0.7. Determine its kinematic viscosity.

1. Answer

a.42.8 10 [ ]Pa s

b.

Viscosity of gasoline at 25 ºC :

Viscosity of benzen at 27 ºC : 7 27.0 10 [ / ]m s

Kinematic Viscosity :

4 23

0.6[ ]8.6 10 [ / ]

0.7 1000[ / ]

Pa sm s

kg m

c.

2. Viscosity

Determine the viscosity of fluid between shaft and sleeve in the figure.

V = 0.1 m/s

200 mm

100 N

74 mm

0.07 mm

2. Answer

AUF

t

3

3 3 2

100[ ] (0.07 10 )[ ]

(74 10 ) (200 10 )[ ] 0.1[ / ]

1.5[ ]

Ft

AU

N m

m m s

Pa s

3

3 3 2

100[ ]

0.1[ / ]

0.07 10 [ ]

(74 10 ) (200 10 )[ ]

F N

U m s

t m

A m

3. Viscosity

A 25 mm diameter steel cylinder 300 mm long falls, because of its own gravity force at an uniform rate of 0.1 m/s inside a tube of slightly larger diameter. A castor-oil film of constant thickness is between the cylinder and the tube. Determine the clearance between the tube and the cylinder. The temperature is 38 ºC. Relative density of steel = 7.85.

3. Answer

AUF

t

233

1

3 3 2

25 10(300 10 ) 7.85 1000 [ ]

2

2.6 10 [ ]( )

0.1[ / ]

(25 10 ) (300 10 )[ ]

?[ ]

F ma g N

Pa s fromthe figure

U m s

A m

t m

3 31

233

4

(25 10 ) ( 300 10 ) 0.12.6 10

25 10( 300 10 ) 7.85 1000

2

2.1 10 [ ]

AUt

Fg

m

4. Viscosity

A piston of diameter 50.00 mm moves within a cylinder of 50.10 mm. Determine the percent decrease in force necessary to move the piston when the lubricant warms up from 0 to 120 ºC.　 (use viscosity of crude oil)

constant

89 % decrease

UAF

t

2 2

1 1

1 1 1 0.111 0.89F

F

4. Answer

Viscosity at 0ºC

1 = 1.8 10-2 2 = 2.0 × 10-3

Viscosity at 120ºC

5. Bulk Modulus of Elasticity

dp : Pressure change, ex. (p2 – p1)

dV : Volume change, ex. (V2 – V1)

Unit of K: [Pa]dp = 0.1 MPa

1 m3 of Water at 20 ºC

dV = ?

/

dpK

dV V

Determine the change of volume.(use the table)

5. Answer

6

9

5 3

0.1 10 1

2.2 101

4.5 10 [ ]22000

dp VdV

K

m

/

dpK

dV V Water at 20 ºC 2.20 GPa (from the table)

6. Bulk Modulus of Elasticity

For K = 2.2 GPa for bulk modulus of elasticity of water, what pressure is required to reduce its volume by 0.5 % ?

6. Answer

/

dpK

dV V

92.2 10 [ ]

/ 0.5 /100

?

K Pa

dV V

dp

9

7

/ 2.2 10 0.5 /100

1.1 10 [ ]

dp K dV V

Pa

7. Capillarity

Determine the capillary rise for distilled water at 40 ºC in a circular 6 mm diameter glass tube.(Use the figure)

7. Answer

From the figure,2.5 mm of capillarity rise

8. Vapor Pressure

A vertical cylinder 300 mm in diameter is fitted (at the top) with a tight but frictionless piston and is completely filled with water at 70 ºC. The outside of the piston is exposed to an atmospheric pressure of 100 kPa. Calculate the minimum force applied to the piston that will cause the water to boil ? (use the table to determine the vapor pressure)

8. Answer

d = 300mm

70 ºC

100 kPa

Force ?From the table, vapor pressure of waterat 70 ºC : 31.4 kPa

Deference between Atmosphere and Vapor Pressures: 100 – 31.4 [kPa]

The force required :

233

3

300 10(100 31.4) 10

2

4.85 10 [ ]

F p A

N

9. Cavitation

At What pressure can cavitation be expected at the inlet of a pump that is handling water at 20 ºC ?

Pump

9. Answer

From the table,cavitation is expected at 2.5 kPa of pressure

10. Perfect Gas Law

A gas with relative molecular mass of 44 is at a pressure of 0.9 MPa and a temperature of 20 ºC. Determine its density.

10. Answer

P = RT

61 30.9 10

1.6 10 [ / ]8312

(273 20)44

Pkg m

RT

11. Absolute Pressure

h = 380 mm

Atmospheric Pressure = ??

A B

Mercury (s = 13.6)

11. Answer

A Bp p

3

4

13.6 1000 9.80665 380 10

5.07 10 [ ]

A B HGp p gh

Pa

Pressure at A:

Pressure at B:

B HGp gh

Taking the points A and B,??Ap

12. Manometer

A

h2 = 3 mP0= 101.3 kPa

h1 = 380 mm

Mercury (s = 13.6)

Oil(s = 0.9)

PA (gage) = ??, PA (abs) = ??

12. AnswerA

h2 =

3 m

1

P0

2

h1 = 380 m

m

Mercury (s = 13.6)

Oil(s = 0.9)

PA (gage) = ??PA (abs) = ??

B C

B Cp p

Pressure at C:

1C HGp gh

Taking the points B and C,

Pressure at B:

1B C HGp p gh

Pressure at the point A:

2 1 2

1 2

3

4

9.80665 13.6 1000 380 10 0.9 1000 3

2.4 10 [ ]( )

A B oil HG oil

HG oil

p p gh gh gh

g h h

Pa Gage pressure

Absolute pressure at the point A:

, 0

4 3

5

2.4 10 101.3 10

1.3 10 [ ]( )

A abs Ap p p

Pa Absolute pressure

13. Manometer

Water

250 mm

Oil (s = 0.90)

PA

PB

1625 mm

500 mm

PA – PB = ??

13. Answer

Water

h2 = 250 mm

Oil (s = 0.90)

PA

PB

h 1 = 1

625

mm

h 3 = 5

00 m

m

PA – PB = ??

C D

2C D w oilp p gh

Pressures at A and B:

1 2

1 2 3

A C w

B D w

p p g h h

P p g h h h

Taking the points C and D,

Pressure difference at the points A and B:

1 2 1 2 3

1 2 1 2 3

2 3

2 3 2

3

9.80665 1000 0.250 0.500 0.9 1000 0.250

5.1 10

A B

C w D w

C D w

w oil w

w oil

p p

p g h h p g h h h

p p g h h h h h

gh gh

g h h gh

Pa

B

Oil (s = 0.90)

14. Manometer

A

150 mm

250mm150 mm

150 mm

200mm

Mercury (s=13.6)

Water

PA – PB = ??

14. Answer

B

Oil (s = 0.90)A

h1 = 150 mm

Mercury (s=13.6)

Water

C D

E F

G Hh2 = 150 mm

h 3 = 2

50 m

m

h 4 = 2

00 m

m

h 5 = 1

50 m

m

2 1

3

4

5

5 4 3 2 1

5 4 2 3 1

9.80665 1000 0.9 0.150 13.6 1000 0.200 0.150 1000 0.250 0.150

A C Hg w

C D

D E w

E F

F G Hg

G H

H B oil

A B oil Hg w Hg w

A B oil Hg w

p p gh gh

p p

p p gh

p p

p p gh

p p

p p gh

p p gh gh gh gh gh

p p g h h h h h

34.4 10 Pa

Water

15. Pressure variation (Inclined Surface)

22.2 kN

600

2.4 m

4.5 m

fixed

fixed

d = ?

Determine the depth d when the gate opens.

Answer:D >= 2.75m

15. Answer

Water

22.2 kN

= 600

b = 2.4 mh 1

= 4

.5 m

d = ?

Centroid

Center of Pressure

h 2

2

2

hy

y C

Moment balance on the plate:

2

2 22

32

22 2 2 2 2 2 2

2 2

1 1 2 2

2 2 2

2 22 2

32

3

2

sin

122

2 2 2 6 32 2

sin

2 2sin

3 3

2sin

9

2sin

9 sin

9sin 60

2

G G G

G

GC

C

c C w C C

w

w

w

w

d h

I k A k bh

bhI

k h bh h bh h h hy y

h hy

F h F h y

P A h y gy bh h y

h hg bh h

g bh

dg b

dgb

31 1

3 3

39

4 22.2 10 4.52 1000 9.80665 2.4

2.4

F h

d m

F 1

F 2

16. Pressure Variation

Water

d = ??

Atmospheric Pressure P0

1.3 m

1.2 m

G

y

b = 4 m

h = 3 m

Determine depth d when water gateopens automatically.

d > 6.45

16. Answer

Water

d = ??

a2 = 1.3 m

a1 = 1.2 m

G

y

h =

3 m

b = 4 m

2

2

1

1

1 2

2

1 1 2

1

2

1 2 1

122

2

,

122

2

2 2 12 2

GC

C

Equation for position of center of pressure

hk h

y y d ay h

d a

y d a a

Combining two equations above

hh

d a d a ah

d a

h h h hd a a d a

2 21 2

2 1 2

21 2

1 2

2

2

2 2 4 12 2

2 3 2

2

3.0 1.2 3.0 1.3 3.01.2 1.3

2 3 23.0

1.32

6.45

ha a hh h hd a a a

ha a hha a

dh

a

d

d

3 322 22

22

3

22 2 2

1

3 12

1212

2

hh

G hh

GG G G G

Conditions

y bhI y dA y b dy b

bhI h

I k A k bh kbh bh

hy d a

17. Buoyancy

350 kN

12 m

6 m

6 m

15 m

2.4 m

400 kN

d (draft) = ??

Water

17. Answer (1/2)

F2 = 350 kN

a1 = 12 m

a4 = 6 m

6 m

a2 = 15 m

a3 = 2.4 m

F1 = 400 kN

d (draft) = ??

Water

17. Answer (2/2)2

2 14 1 1 2

3

22 1 11

3 4

2 2 1 11 1

3 4

2 1

3

32

.

( )

2

( )0

2

( )4

2

( )2

2

(15 12) 750 1012 12 4

2 2.4 1000 9.80665 6

(15 12

w

w

w

Buoyancy equals gravity force

a a dga a d F F

a

a a F Fd a d

a ga

a a F Fa a

a gad

a a

a

2)

2 2.4

12512 144 2.5

12 13.269.80665 1.01, 20.21.25 1.25

0 1.01d d

3 3 31 2

1 2 13

1 4

41 1 2 1

3

2 1 41

3

22 1

4 1

400 10 350 10 750 10

( )

1

2

( )2

( )2

2

( )

2

w

w

w

w

w

Gravity Force

F F F N

Eqation between x and d

dx a a a

a

Buoyancy

F gV

g a x d a

dadg a a a a

a

a a d dag a

a

a a dga a d

3a

18. Bernoulli Eq. (Pitot Tube)

V1 1 = 300 mm 2 = 200 mm V2

Flow rate = 0.3 m3

h = ??

Mercury (s = 13.6)

water

18. Answer

1 2

2

22

2

22

2

2

1

Pr

,

4

2

4 1

2

1000 4 0.3 1

2 13.6 1000 1000 9.806653.1416 0.2

3.69 10

Hg w

wHg w

w

Hg w

essure Difference

p p gh

Combining two equations above

Qgh

Qh

g

Pa

1 1 2 2

2 2

1 21 2

2 2 22 22

2 21 1 2 2

1 2

1 1 2

222

1 2 22

,2 2

4

2

2 2

0,

4

2 2

w w

w

Continuity Equation

Q AV A V

A A

Q Q QV

A

Bernoulli Equation

V p V pgz gz

V z z

V Qp p

19. Bernoulli Eq. (Venturi Tube)

V1

d2 = 150 mm

Flow rate: Q = ???

Mercury (s = 13.6)h=480 mm

V2

d1 = 300 mm

z2

z1

1.5 m

oil = 0.800

(1)

(2)

19. Answer (1/2)

V1

d2 = 150 mm

Flow rate: Q = ???

Mercury (s = 13.6)h1=480 mm

V2

d1 = 300 mm

z2

z1

h2=1.5 m

oil = 0.800

(1)

(2)

(3)

19. Answer (2/2)

1 2 1 2

1 3 1 3 2 2

21 2

2 22 2

2 1

Pr int

,

,

1 1

2

2 2

Hg oil oil

Hg oil oil

Hg oil oil

oil

essure Diffrencebetween po s Aand B

p p gh gh

p p gh p p gh

Combining twoequations above

gh gh Q

d d

2

1 2 2 2 4 42 1

2 4 4

3

16 1 12 1

1000 13.6 16 1 12 1 9.80665 0.48

800 3.1416 0.150 0.300

0.224 /

Hg

oil

gh

Q gh gh ghd d

m s

1 1 2 2

2 2

1 21 2

2 21 1 2 2

1 2

2 21 22 1 2 1

22 21 2 2

2 2 221

2 21 2 2 2

22 1

,2 2

2 2

1

2

1

2

1 1

2

oil oil

oil

oil

oil

Continuity Equation

Q AV A V

d dA A

Bernoulli Equation

V p V pgz gz

p pV V g z z

p p AV V gh

A

p p A V

A A

22

21 2

22 22 1

1 1

2oil

gh

p p Qgh

A A

20. Force From Fluid

= ??F = ??

(1)

(2)

V1

V2

600

d1 = 300 mm

d2 = 200 mm

200 L/s

(Overhead View)

P1 = 150 kPa

Fluid : Water (Incompressible) = constant)

21. Force from Fluid

H = ??

V

V

Water Jet

1 = 150 mm

V

300

3 m

= 125 mm

How much is H at least to support the plate ?