Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.125P
To determine

The flow rate in each pipe and the direction of it.

Expert Solution & Answer
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Answer to Problem 6.125P

Q1=6.78×103m3/s(towards)Q2=1.4×103m3/s(away)Q3=5.81×103m3/s(away)

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.125P

Pressures in each pipe is equal to,

p1=200kPap2=160kPap3=100kPa

The fluid is water at 20°C

The pipe is made of cast iron.

The pressure drop ΔP is defined as,

ΔP=f(LD)(ρV22)

In the above equation,

f - Friction factor

L - Length of pipe

D - Diameter of pipe

ρ - Density of the fluid

V - Velocity of the flow

The pressure drop in each pipe can be defined as below,

p1pB=f1( L 1 D 1 )( ρ V 1 2 2)p2pB=f2( L 2 D 2 )( ρ V 2 2 2)pBp3=f1( L 3 D 3 )( ρ V 3 2 2)

Consider flow in pipe (1) is towards B and the flow in pipe (3) is away from B and the flow in pipe (2) cannot be confirmed.

Calculation:

For pipe 1,

Assume,

V1=0.5m/s

Calculate the roughness ratio,

εD1=0.26mm120mm=2.16×103

Calculate the Reynolds’s number,

Re=ρVD1μ=(998kg/m3)(0.5m/s)(0.12m)0.001kg/ms=59880

Find the friction factor,

1f 1/2 =1.8log( 6.9 R e + ( ε/d 3.7 ) 1.11)=1.8log( 6.9 59880+ ( 2.16× 10 3 3.7 ) 1.11)=0.026

Calculate the pressure at point B,

p1pB=f1( L 1 D 1 )( ρ V 1 2 2)200×103PapB=(0.026)( 800m 0.12m)( ( 998kg/ m 3 ) ( 0.5m/s ) 2 2)pB=178.37kPa

Calculate the flow rate,

Q=VA=(0.5m/s)(π( 0.06m)2)=5.654×103

For pipe 2,

Assume,

V2=0.5m/s

Calculate the roughness ratio,

εD2=0.26mm80mm=3.25×103

Calculate the Reynolds’s number,

Re=ρVD2μ=(998kg/m3)(0.5m/s)(0.08m)0.001kg/ms=39920

Find the friction factor,

1f 1/2 =1.8log( 6.9 R e + ( ε/d 3.7 ) 1.11)=1.8log( 6.9 39920+ ( 3.25× 10 3 3.7 ) 1.11)=0.0294

Calculate the pressure at point B,

p2pB=f1( L 1 D 1 )( ρ V 1 2 2)160×103PapB=(0.0294)( 600m 0.08m)( ( 998kg/ m 3 ) ( 0.5m/s ) 2 2)pB=259kPa

Calculate the flow rate,

Q=VA=(0.5m/s)(π( 0.04m)2)=2.513×103

For pipe 3,

Assume,

V3=0.5m/s

Calculate the roughness ratio,

εD3=0.26mm100mm=2.6×103

Calculate the Reynolds’s number,

Re=ρVD3μ=(998kg/m3)(0.5m/s)(0.1m)0.001kg/ms=49900

Find the friction factor,

1f 1/2 =1.8log( 6.9 R e + ( ε/d 3.7 ) 1.11)=1.8log( 6.9 49900+ ( 2.6× 10 3 3.7 ) 1.11)=0.0276

Calculate the pressure at point B,

p2pB=f1( L 1 D 1 )( ρ V 1 2 2)100×103PapB=(0.0276)( 900m 0.1m)( ( 998kg/ m 3 ) ( 0.5m/s ) 2 2)pB=69kPa

Calculate the flow rate,

Q=VA=(0.5m/s)(π( 0.05m)2)=3.92×103

Calculate the average value of pB

pB=(178.37kPa+259kPa+69kPa)3=168.79

The friction factor will not differ a lot; we can use the same friction factor to find relevant velocities,

For pipe (1)

200×103Pa168.79×103Pa=(0.026)( 800m 0.12m)( ( 998kg/ m 3 ) ( V 1 ) 2 2)V1=0.6m/s

The flow rate is equal to,

Q1=V1A=(0.6m/s)(π( 0.06m)2)=6.78×103m3/s

The flow in pipe (1) is considered as a flow towards junction B.

For pipe (2)

160×103Pa168.79×103Pa=(0.0294)( 600m 0.08m)( ( 998kg/ m 3 ) ( V 2 ) 2 2)V2=0.28m/s

The flow rate is equal to,

Q2=V2A=(0.28m/s)(π( 0.04m)2)=1.4×103m3/s

The flow in pipe (2) is considered as a flow away from junction B.

For pipe (3)

100×103Pa168.79×103Pa=(0.0276)( 900m 0.1m)( ( 998kg/ m 3 ) ( V 3 ) 2 2)V2=0.74m/s

The flow rate is equal to,

Q3=V3A=(0.74m/s)(π( 0.05m)2)=5.81×103m3/s

The flow in pipe (3) is considered as a flow away from junction B.

Q=(Q1( Q 2 + Q 3 ))=(6.78× 10 3m3/s( 1.4× 10 3 m 3 /s+5.81× 10 3 m 3 /s))=0.00043

By further iteration, we an get more accurate answers.

Conclusion:

The flow rates in each pipe is equal to,

Q1=6.78×103m3/s(towards)Q2=1.4×103m3/s(away)Q3=5.81×103m3/s(away).

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