Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.129P
To determine

The flow rate in each pipe.

Expert Solution & Answer
Check Mark

Answer to Problem 6.129P

Q1=0.0388m3/sQ2=0.0223m3/sQ3=0.0241m3/sQ4=0.0365m3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.129P , additional homework tip  1

Fluid Mechanics, Chapter 6, Problem 6.129P , additional homework tip  2

All pipes are made of cast iron.

The diameter of all pipes are 8cm and the length id equal to 45m

The fluid is water at 20°C

The basic rules of pipe networks are,

“The net pressure change in a closed loop must be equal to zero”

“The net flow into a junction must be equal to zero”

“All pressure changes must satisfy moody and minor loss friction co-relation”

The head loss hf can be defined as,

hf=8LfQ2π2gd5=KQ2

In above equation,

L - Length of pipe

f - Friction factor

Q - Flow rate

d - Diameter

K - Loss co efficient

The total head loss is equal to zero in a closed loop.

The pressure difference ΔP can be defined as,

ΔP=ρghf

Where, ρ - density

Calculation:

To solve this problem,

Assume pressure at junction,

pa=500kPa

But we know that,

hf=ΔPρg=p1paρg=(950× 103Pa500× 103Pa)9790N/m3=45.96m

We can define head loss as,

hf=8LfQ2π2gd5

In above equation,

Assume,

f1=0.02

Therefore,

45.96m=8( 45m)( 0.02)Q12π2( 9.81m/ s 2 ) ( 0.08m )5Q1=0.045m3/s

Similarly, find the flow rate for all pipes

For pipe 2,

hf=ΔPρg=p2paρg=(350× 103Pa500× 103Pa)9790N/m3=15.32m

15.32m=8( 45m)( 0.02)Q22π2( 9.81m/ s 2 ) ( 0.08m )5Q2=0.0259m3/s

For pipe 3,

( 675× 10 3 Pa500× 10 3 Pa)9790N/m3=8( 45m)( 0.02)Q32π2( 9.81m/ s 2 ) ( 0.08m )5Q3=0.028m3/s

For pipe 4,

( 100× 10 3 Pa500× 10 3 Pa)9790N/m3=8( 45m)( 0.02)Q42π2( 9.81m/ s 2 ) ( 0.08m )5Q4=0.0424m3/s

To find the exact value of the flow rate,

Calculate the roughness ratio

εd=0.26mm80mm=3.25×103

Calculate the Reynolds’s number for each pipe

Re=ρVdμ=4ρQπμd

For pipe 1,

Re1=4ρQ1πμd=4(998kg/m3)(0.045m3/s)π(0.001kg/ms)(0.08m)=714764.85

For pipe 2,

Re2=4ρQ2πμd=4(998kg/m3)(0.0259m3/s)π(0.001kg/ms)(0.08m)=411386.88

For pipe 3,

Re3=4ρQ3πμd=4(998kg/m3)(0.028m3/s)π(0.001kg/ms)(0.08m)=444742.57

For pipe 4,

Re4=4ρQ4πμd=4(998kg/m3)(0.0424m3/s)π(0.001kg/ms)(0.08m)=673467.32

Calculate the friction factor for all pipes,

For pipe 1,

1f1 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1.8log( 6.9 714764.85+ ( 3.25× 10 3 3.7 ) 1.11)f10.0269

For pipe 2,

1f2 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1.8log( 6.9 411386.88+ ( 3.25× 10 3 3.7 ) 1.11)f20.027

For pipe 3,

1f3 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1.8log( 6.9 444742.57+ ( 3.25× 10 3 3.7 ) 1.11)f30.027

For pipe 4,

1f4 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1.8log( 6.9 673467.32+ ( 3.25× 10 3 3.7 ) 1.11)f40.0269

Friction factors are almost same for all four pipes.

Now, calculate the exact flow rates in each pie,

For pipe 1,

45.96m=8( 45m)( 0.0269)Q12π2( 9.81m/ s 2 ) ( 0.08m )5Q1=0.0388m3/s

For pipe 2,

15.32m=8( 45m)( 0.027)Q22π2( 9.81m/ s 2 ) ( 0.08m )5Q2=0.0223m3/s

For pipe 3,

( 675× 10 3 Pa500× 10 3 Pa)9790N/m3=8( 45m)( 0.027)Q32π2( 9.81m/ s 2 ) ( 0.08m )5Q3=0.0241m3/s

For pipe 4,

( 100× 10 3 Pa500× 10 3 Pa)9790N/m3=8( 45m)( 0.0269)Q42π2( 9.81m/ s 2 ) ( 0.08m )5Q4=0.0365m3/s

Net junction flow is equal to zero, therefore,

(0.0388m3/s)(0.0223m3/s)+(0.0241m3/s)(0.0365m3/s)=+0.0041

The assumed pa is little low. By assuming pa=515kPa we can get more accurate answers.

But for now, we can say that,

Q1=0.0388m3/sQ2=0.0223m3/sQ3=0.0241m3/sQ4=0.0365m3/s

Conclusion:

The assumed pa is little low. By assuming pa=515kPa we can get more accurate answers.

But for now, we can say that,

Q1=0.0388m3/sQ2=0.0223m3/sQ3=0.0241m3/sQ4=0.0365m3/s.

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