Solved Problems Samples in Fluid Flow.pdf

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University of Technology Solved Problems Samples in Fluid FlowChemical Eng. Dep. Instructor: Dr. Salah Salman IbrahimSecond Class

1

2011

A plate of size 60 cm x 60 cm slides over a plane inclined to the horizontal at an angle of30°. It is separated from the plane with a film of oil of thickness 1.5 mm. The plate weighs 25kg

and slides down with a velocity of 0.25 m/s. Calculate the dynamic viscosity of oil used aslubricant. What would be its kinematic viscosity if the specific gravity of oil is 0.95.

Problem (1)

Component of W along the plane =W cos(60) =W sin(30)Solution

= 25 (0.5) = 12.5 kgF = 12.5 kg (9.81 m/s 2) =122.625 Nτ = F/A = 122.625 N/(0.6 x0.6) m 2 = 340.625 Pa

poise s Pa

s

Pa dy du

44.20.044.2)0015.0 / 25.0(

625.340) / ( 1

==== −τ

µ

stoke s m m kg

s Pa5.21 / 00215.0

/ 950.044.2 2

3 ====

ρµ

ν

By Dr. Salah Salman

By dimensional analysis, obtain an expression for the drag force (F) on a partiallysubmerged body moving with a relative velocity (u) in a fluid; the other variables being thelinear dimension (L), surface roughness (e), fluid density (ρ), and gravitational acceleration (g).

Problem (2)

Drag force (F) N ≡ [MLT -2]Solution

Relative velocity (u) m/s ≡ [LT -1]Linear dimension (L) m ≡ [L] Surface roughness (e) m ≡ [L] Density ( ρ) kg/m 3 ≡ [ML -3]Acceleration of gravity (g) m/s 2 ≡ [L T -2]

F = k (u, L, e, ρ, g)f (F, u, L, e, ρ, g) = 0

n = 6, m = 3, ⇒ Π = n – m = 6 – 3 = 3

No. of repeating variables = m = 3The selected repeating variables is (u, L, ρ)

Π1 = ua1 L b1 ρc1 F --------------(1)

1.5m

w30°

60°30°


2/10


2

2011

Π2 = ua2 L b2 ρc2 e --------------(2)

Π3 = ua3 L b3 ρc3 g --------------(3)

For Π 1 equation (1)

[M 0 L0 T0] = [L T -1]a1 [L] b1[ML -3]c1[MLT -2] Now applied dimensional homogeneity

For M 0 = c1 + 1 ⇒ c1 = – 1

For T 0 = – a1 – 2 ⇒ a1 = – 2

For L 0 = a1 + b1 – 3c1+ 1 ⇒ b1 = – 2

Π1 = u-2 L -2 ρ-1 F

ρ 221 Lu

F =Π⇒


[M 0 L0 T0] = [L T -1]a2 [L] b2[ML -3]c2[L]

For M 0 = c2 ⇒ c2 = 0

For T 0 = – a2 ⇒ a2 = 0

For L 0 = a2 + b2 – 3c2+ 1 ⇒ b2 = – 1

Π2 = L-1 e

Le=Π⇒

2


[M 0 L0 T0] = [L T -1]a3 [L] b3[ML -3]c3[L T -2]

For M 0 = c3 ⇒ c3 = 0

For T 0 = – a3 – 2 ⇒ a3 = – 2

For L 0 = a3 + b3 – 3c3+ 1 ⇒ b3 = 1

Π3 = u-2 L g 23 u

g L=Π⇒

f 1 (Π 1, Π 2, Π 3) = 0 ⇒ f 1( ρ

22 Lu

F , Le

, 2ug L

) = 0

∴ ),( 222

u

g L Le

f LuF ρ =


3/10


3

2011

A conical vessel is connected to a U-tube having mercury and water as shown in theFigure. When the vessel is empty the manometer reads 0.25 m. find the reading inmanometer, when the vessel is full of water.

Problem (3)

P1 = P 2 Solution

P1 = (0.25 + H) ρw g + P o P2 = 0.25 ρm g +P o

⇒ (0.25 + H) ρw g + P o = 0.25 ρm g +P o

⇒ H = 0.25 ( ρm – ρw)/ ρw= 0.25 (12600 /1000) = 3.15 m

When the vessel is full of water, let the mercury level inthe left limp go down by (x) meter and the mercury levelin the right limp go to up by the same amount (x) meter.i.e. the reading manometer = (0.25 + 2x)

P1 = P 2 P1 = (0.25 + x +H + 3.5) ρw g + P o P2 = (0.25 + 2x) ρm g +P o

⇒ (0.25 + x +H + 3.5) ρw g + P o = (0.25 + 2x) ρm g +P o

⇒ 6.9 + x = (0.25 + 2x) ( ρm/ ρw) ⇒ x = 0.1431 mThe manometer reading = 0.25 + 2 (0.1431) = 0.536 m

A pump developing a pressure of 800 kPa is used to pump water through a 150 mmpipe, 300 m long to a reservoir 60 m higher. The flow rate obtained is 0.05 m 3/s. As a result ofcorrosion and scalling the effective absolute roughness of the pipe surface increases by afactor of 10 by what percentage is the flow ratereduced. μ= 1 mPa.s

Problem (4)

The total head of pump developing =(ΔP/ρg) =800,000/(1000 x 9.81)= 81.55 mH 2o

Solution

The head of potential energy = 60 m Neglecting the kinetic energy (same diameter)

02

2

=+−∆+∆+∆F hsg

W g

u z

gP η

α ρ

P1 P2

ρm mercury

25 cm

H

3.5 m

Po

300 m

1 .

.2

60 m

Datum line


4/10


4

2011

⇒ΔP/ρg + Δz +h F = 0

⇒ hF = – ΔP/ρg – Δz = 81.55 – 60 = 21.55 mu = Q/A = (0.05 m 3/s)/( π/4 0.15 2) =2.83 m/shFs = (– ΔP fs/ρg) = 4 f (L/d) (u

2/2g)

⇒ f = h Fs d 2g/(4Lu2) = (21.55) (0.15)(9.81)/(2 x 300 x 2.83 2) = 0.0066

Φ = 0.0033, Re = (1000 x 2.83 x 0.15)/0.001 = 4.23 x 105 From Figure (3.7) e/d = 0.003Due to corrosion and scalling the roughness increase by factor 10i.e. (e/d) new = 10 (e/d) old = 0.03The pump head that supplied is the same(– ΔP fs) = h Fs ρg = 21.55 (1 000) 9.81 = 211.41 kPa

ΦRe2 = (- ΔPfs/L)( ρd 3/4μ2) = [(211410)/( 300)][(1000)(0.15) 3/(4)(0.01) 2] = 6 x 10 8

From Figure (3.8) Re = 2.95 x 105 ⇒u = 1.97m/sThe percentage reduced in flow rate = (2.83 - 1.97)/ 2.83 x 100 % = 30.1 %.

It is required to pump cooling water from storage pond to a condenser in a processplant situated 10 m above the level of the pond. 200 m of 74.2 mm i.d. pipe is available andthe pump has the characteristics given below. The head loss in the condenser is equivalent to16 velocity heads based on the flow in the 74.2 mm pipe. If the friction factor Φ = 0.003,estimate the rate of flow and the power to be supplied to the pump assuming η = 0.5

Problem (5)

Q (m 3/s) 0.0028 0.0039 0.005 0.0056 0.0059Δh (m) 23.2 21.3 18.9 15.2 11.0

[Usually the pump done its exact duty and return the liquid at the suction pressure i.e thepump give the minimum limit of energy required to arrive the liquid to point 2 or d]

Solution

( ) ( ) ( )[ ] condenser F s F d F h h h g

u g P

z h +++++=α

ρ

2

2

gu

d L f h s d F 2

4)(2

=+ = 4(0.006)(200/0.0742)(u2/2g) = 3.3 u 2

gu

h condenser F 216)(

2

= = 0.815 u 2

u = Q/A = 321.26 Q


5/10


5

2011

⇒ Δh = 10 + (0.815 + 3.3)(321.26 Q) 2 = 10 + 2.2 x 10 5 Q 2

To draw the system curve

Q (m3

/s) 0.003 0.004 0.005 0.006Δh (m) 11.98 13.52 15.5 17.92

From Figure

Q = 0.0054 m 3/s

Δh = 16.4 m

Power required for pump =η

ρ g hQ = (0.0054)(16.4)(1000)(9.81)/0.5

= 17.375 kW

u = 1.25 m/s m gu

08.02

2

=⇒ ; Δz = 10 m; h f = 6.415 m

A Power-law liquid of density 961 kg/m 3 flows in steady state with an average velocity of1.523 m/s through a tube 2.67 m length with an inside diameter of 0.0762 m. For a pipeconsistency coefficient of 4.46 Pa.s n [or 4.46 (kg / m.s 2) sn], calculate the values of the

apparent viscosity for pipe flow ( μ a)P in Pa.s, the Reynolds number Re, and the pressure dropacross the tube for power- law indices n = 0.3, 0.7, 1.0, and 1.5 respectively.

Problem (6)

Solution

Apparent viscosity 1)8

()( −= n P a d u

Kp

= 4.46 (kg/m) s n-2 [8 (1.523)/0.0762] n-1sn-1

10

12

14

16

18

20

22

24

0.003 0.004 0.005 0.006 0.007

Q (m 3/s)

Δ h ( m ) Pump

System


6/10


6

2011

⇒ (μ a)P = 4.46 (159.9)n-1 (Pa.s)----------------------------(1)

-1n(159.9)4.46)(Re

ud ud

P a

ρµρ == = 961 (1.523)(0.0762) / 4.46 (159.9) n-1

⇒ Re = 25.006/ (159.9)n-1

----------------------------(2) –ΔPfs = 4 f (L/d) (ρu

2/2) = 4(16/Re) (2.67 / 0.0762)[961(1.523) 2/2] for laminar

⇒ –ΔPfs = 99950.56 (159.9)n-1 (Pa)--------------------------(3)

( η = 0.6)

n(μ a)P

Eq.(1)

Re

Eq.(2)

–ΔPfs

Eq.(3) (–ΔPfs) non-New / (– ΔPfs) New

Power

(W)

0.3 0.1278 872.44 2,865 0.0287 33

0.7 0.9732 114.6 21,809 0.218 252.5

1.0 4.46 25.006 99,950.56 1.0 1157

1.5 56.4 1.9776 1,263,890.541 12.7 14630

A (30cm x 15cm) Venturi meter is provided in a vertical pipe- line carrying oil of sp.gr. =0.9. The flow being upwards and the difference in elevations of throat section and entrancesection of the venture meter is 30 cm. The differential U-tube mercury manometer shows agauge deflection of 25 cm. Take Cd = 0.98 and calculate: -

Problem (7)

i- The discharge of oilii- The pressure difference between the entrance and throat

sections.

i-

Solution

22

21

2122

)(2

A A

A A g RC AuQ m d −

−==

ρρ

−=

44

22

15.03.0

])15.0(4 / [3.0900

81.9)12700)(25.0(298.0

π

= 0.1488 m 3/sii- Applying Bernoulli’s equation at points 1 and 2

2

222

1

211

22 z

gu

g P

z g

u g

P ++=++ρ

guu

z g P P

2

21

22

221

−+=

−ρ

u1 = 0.1488/( π/4 0.32) = 2.1 m/s, u 2 = 0.1488/( π/4 0.15

2) = 8.42 m/s

25cm

30cm

1

2


7/10


7

2011

⇒ P1 – P 2 = 900 (9.81) [0.3 + (8.422 – 2.1 2)/2(9.81)]

= 32.5675 kPa but P 1 – P 2 = 0.25 (13600–900)(9.81) = 31.1467 kPa% error = 4.36 %

A vacuum distillation plant operating at 7 kPa pressure at top has a boil-up rate of 0.125kg/s of xylene. Calculate the pressure drop along a 150 mm bore vapor pipe used to connectthe column to the condenser. And also calculate the maximum flow rate if L = 6 m, e = 0.0003m, Mwt = 106 kg/kmol, T = 338 K, μ = 0.01 mPa.s.

Problem (8)

042)(

ln 2

11

21

22

2

12

=+

−

+

G d L

P P P

P P

G φυ

Solution

G = 0.125 / [ π/4 (0.15) 2] = 7.074 kg/m 2.sP1 = 7 kPa, P 2 = Pressure at condenser

kg/kmol106 338K /kmol.K)(Pa.m8314 3

11 == Mwt RT

P υ

( )22 / / 68.26510 s m kg J ≡= Re = G d / μ = 7.074(0.15)/0.01 x 10 -3 = 1.06 x 10 5

, e/d = 0.002 ⇒ Φ = 0.003 (Figure 3.7)

015.06)003.0(4])107([10769.3107ln 23227

2

3

=+×−×+

×⇒ − P

P

( ) ( ) 723

2327

23

2322 10769.3

48.0) / 107ln(107

10769.3

48.0) / 107ln(107 −− ×

+×−×=⇒

×+×

−×=⇒ P P P

P

Solution by trial and error

P2 Assumed 5 x 103 6.8435 x 10 3 6.904 x 10 3 6.9057 x 10 3

P2 Calculated 6.8435 x 103 6.904 x 10 3 6.9057 x 10 3 6.9058 x 10 3

⇒ P2 = 6.9058 x 103 Pa

–Δ P = P1 – P2 = (7 – 6.9058) x 103

= 94.2Pa [(P1 – P2) / P 1] % = 0.665 % we can neglect the K.E. term in this problem

For maximum flow rate calculations11max11max / 1 / 1 υ P PG P P A m ww =⇒=

To estimate P w 081ln2

1

2

1 =+

−+

d L

P P

P P

ww

φ

Let X ≡ (P 1/Pw)2

LD

V

L

V


8/10


8

2011

⇒ ln(X) + 1 – X + 8 Φ L/d = 0 ⇒ X = 1.96 + ln(X)Solution by trial and error

X Assumed 1.2 2.14 2.72 2.96 3.074 3.086 3.087

X Calculated 2.14 2.72 2.96 3.074 3.086 3.087 3.087

⇒ X = 3.087 = (P 1/P w)2 ⇒ Pw = P1/(3.087)

0.5 = 3984 Pa

∴ This system did not reached maximum velocity (H.W. explain

⇒ Gmax = 3984 / (26510.68)0.5 = 24.47 kg/m 2.s

)

Calculate the theoretical power in Watt for a 0.1 m diameter, 6-blade flat blade turbineagitator running at 16 rev/s in a tank system without baffles and conforming to the standard

tank configuration. The liquid in the tank has a dynamic viscosity of 0.08 Pa.s, and a liquiddensity of 900 kg/m 3.

Problem (9)

(Re)m = ρ N DA2 / μ = (900) (16) (0.1) 2 / (0.08) = 1,800

Solution

From Power curve Figure (2) Φ = 2.2

The theoretical power for mixing P A = Φ [(Fr)m]y ρ N3 DA

5

β

m ylog(Re)−= 05638.0

40)1800log(1 −=

−=⇒ y

(Fr)m = N2 DA / g = (16)

2 (0.1) / 9.81 = 2.61

[(Fr)m]y = [2.61] -0.05638 = 0.9479

⇒PA = 2.2 (0.9479) (900) (16)3 (0.1) 5 = 76.88 W

In contact sulphuric acid plant the secondary converter is a tray type converter 2.3 mI.D. with the catalyst arranged in three layers, each 0.45 m thickness. The catalyst is in form ofcylindrical pellets 9.5 mm I.D. and 9.5 mm long. The void fraction is 0.35. The gas enters theconverter at 675 K and leaves at 720 K. Its inlet and outlet compositions are: -

Problem (10)

Gas SO 3 SO 2 O2 N2mol % In 6.6 1.7 10 81.7mol % Out 8.2 0.2 9.3 82.3

The gas flow rate is 0.68 kg/m 2.s. Calculate the pressure drop through the converter.Taken that the dynamic viscosity = 0.032 mPa.s.

Solution


9/10


9

2011

1.02

3

eR

4.0

eR

5

)1(

)(

′+

′=−

−=′=′ueS

e L

P J f ρ

and

)1(eR

e su−

=′µ

ρ

p p

p p

p p

P p p

P

P

d d

d d

L d

L d d

V A

S6

4 /

2 /

4 /

)()4 / (2

particleof Volume particleof areaSurface

3

22

2

2

=+

=+

===π

π

π

π

µ

ρ

)1(6)1(6eR

e

Gd

e

ud p p−

=−

=′⇒

2)(,

2,

)( outin avg

outin avg

avg

avg

gas

Mwt Mwt Mwt

T T T

RT

P Mwt +=

+== , ∑

==

n

iii Mwt x Mwt

1)(

(Mwt )in = 0.066(80) + 0.017(64) + 0.1(32) + 0.817(28) = 32.44 kg/kmol

(Mwt )out = 0.082(80) + 0.002(64) + 0.093(32) + 0.823(28) = 32.71 kg/kmol

(Mwt )aug = 32.58 kg/kmol; Tavg= 697.5 K

35

/ 569.0)5.697(8314

)1001325.1(58.32 m kg gas =

×=

76.5110032.0)35.01(6)105.9(68.0

)1(6eR 3

3

=×−

×=

−=′⇒ −

−

µe

Gd p366.0

)76.51(4.0

76.515

1.0 =+=′⇒

ρ

3

2

3

2 )1(6)1(e d

Ge Le

ueS L P

p

−′=

−′=−⇒ Pa65.3844

569.0)35.0(105.9)68.0)(35.01(6)45.03(366.0

33

2

=×

−×= −

What will be the settling velocity of a spherical steel particle, 0.4 mm diameter, in an oilof sp.gr 0.82 and viscosity 10 mPa.s? The spgr. of steel is 7.87.

Problem (11)

For a sphereSolution

GaC p D 34

(Re) 2 = , 2)(

µ

ρ p p gd Ga

−= = 36.29

34.48(Re) 2 =⇒ p DC 2.24(Re)22 =⇒

p DC

From Figure at 2.24(Re)2

2 = p

DC , (Re)p = 1.667

s m d

u p

p

p / 051.0)0004.0(820)1010(667.1(Re) 3 =

×==⇒

−

ρ

µ


10/10


10

2011

A mixture of gas and liquid flows through a tube of ID 0.02m at a steady total flow rateof 0.2 kg/s. Calculate the pressure gradient in the pipe using the Lockhart-Martinellicorrelation. Given that e = 0.0015 mm, µG = 0.01 and µL = 2 mPa.s, ρG = 60 and ρL = 1000kg/m 3, and the quality w =0.149.

Problem (12)

G = 0.2/(π/4 0.02 2) = 636.62 kg/m 2sSolution

ReL = [636.62 (1-0.149) x 0.02)/ 2 x 10-3 = 5.417 x 103 turbulent

ReG = [636.62 (0.149) x 0.02)/ 0.01 x 10-3 = 1.897 x 105 turbulent

5.01.09.01

−=

L

G

G

Ltt w

w X

ρ

ρ

µ

µ = 1.996

22 11

X X C

L ++=Φ , for tt C = 20 ⇒ ФL

2 = 11.27

ReL = 5.417 x 103 , e/d = 0.000075 ⇒ f = 0.009

L L

wGd

f d dP

ρ 2)1(1

422 −=

=

)1000(2)149.01(62.636

02.01

)009.0(422 −

= 264.157 Pa/m

L L

T d dP

d dP

Φ=

2 = 11.27(132.078) = 2977.05 Pa/m

A cylindrical tank 0.9 m ID and 2 m high open at top is filled with water to a depth of 1.5m. it is rotated about its vertical axis at N rpm. Determine the value of N which will raise water

level even with the brim.

Problem (12)

Solution

The water level even with brim P = P o , z = 2 mThe rise of liquid at wall y r = 2 – 1.5 = 0.5 mThe fall of liquid at wall y f = y r = 0.5 mzo = H - y f = 1 m, z = 2 m at r = R

22

2r

g z z o

ω +=

22

45.0)81.9(212 ω +=⇒

ω = 9.843 rad/s

π

π ω

2)843.9(60

602 =⇒= N N ≈ 94 rpm

R zo

ω

H

yr

yf

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