Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.121P
To determine

The steady flow rates in each pipe.

Expert Solution & Answer
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Answer to Problem 6.121P

Q1=0.66m3/sQ2=0.459m3/sQ3=0.153m3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.121P , additional homework tip  1

Fluid Mechanics, Chapter 6, Problem 6.121P , additional homework tip  2

Diameter of all pipes are 28cm and roughness =1mm

At three reservoir junction, if all flows are considered positive towards the junction, then

Q1+Q2+Q3=0(1)

Assuming the gauge pressure p1=p2=p3=0, the head loss Δh across each pipe is equal to,

Δh=f(LD)(V22g)=zhj

f - Friction factor

L - Length of the pipe

D - Diameter of the pipe

V -Velocity of the flow

z - Elevation

To solve this type of problems, we should guess the position of hj and solve above mentioned equation to find relevant velocities and flow rates, iterating until the flow rates satisfies the equation 1.

If the hj guessed was too high, then the sum of equation (1) will be negative, therefore we must reduce hj and vice versa.

Calculation:

First of all,

Guess ha=85m, therefore

(hf)1=z1ha=25m85m=60m

The roughness ratio for all pipes will be,

d=1×103m0.28m=3.571×103

To find Reynolds’s number,

Re=DVν Try V1=10m/s

Therefore,

(Re)1=DVν=(0.28m)(10m/s)1.01×106m2/s=2.77×106

The friction factor will be equal to,

1f1/21.8log(6.9Re+( /d 3.7 )1.11)1.8log(6.92.77× 106+( 3.571× 10 3 3.7 )1.11)=0.0276

Therefore, according to below equation

Δh=f(LD)(V22g)=zhj ( h f)1=f(LD)( V 2 2g)=(0.0276)( 95m 0.28m)( V 1 2 2( 9.81 ))=60mV1=11.212m/s

According to above result, the Reynolds’s number will be

(Re)1=DVν=(0.28m)(11.212m/s)1.01×106m2/s=3.11×106

To find the flow rate,

Q1=A1V1=[π(0.14)2(11.212m/s)]=0.69m3/s

The flow in pipe 1 is considered as flow away from the junction

Similarly for pipe 2,

(hf)2=z2ha=115m85m=30m

To find Reynolds’s number,

Fluid Mechanics, Chapter 6, Problem 6.121P , additional homework tip  3 Try V2=10m/s

Therefore,

(Re)2=DVν=(0.28m)(10m/s)1.01×106m2/s=2.77×106

The friction factor will be same.

( h f)2=f(LD)( V 2 2g)=(0.0276)( 125m 0.28m)( V 2 2 2( 9.81 ))=30mV1=6.92m/s

According to above result, the Reynolds’s number will be

(Re)2=DVν=(0.28m)(6.92m/s)1.01×106m2/s=1.92×106

To find the flow rate,

Q2=A2V2=[π(0.14)2(6.92m/s)]=0.426m3/s

The flow in pipe 2 is considered as flow towards the junction.

For pipe 3,

(hf)3=z3ha=85m85m=0

Q3=0

Therefore,

Q=Q2Q1=0.426m3/s0.69m3/s=0.264m3/s

Hence ha must be lower than, try ha=80m

(hf)1=z1ha=25m80m=55m

Try V1=10m/s

Therefore,

(Re)1=DVν=(0.28m)(10m/s)1.01×106m2/s=2.77×106

The friction factor will be same.

( h f)1=f(LD)( V 2 2g)=(0.0276)( 95m 0.28m)( V 1 2 2( 9.81 ))=55mV1=10.73m/s

According to above result, the Reynolds’s number will be

(Re)1=DVν=(0.28m)(10.73m/s)1.01×106m2/s=2.97×106

To find the flow rate,

Q1=A1V1=[π(0.14)2(10.73m/s)]=0.66m3/s

The flow in pipe 1 is considered as flow away from the junction

Similarly for pipe 2,

(hf)2=z2ha=115m80m=35m

To find Reynolds’s number,

Re=DVν Try V2=10m/s

Therefore,

(Re)2=DVν=(0.28m)(10m/s)1.01×106m2/s=2.77×106

The friction factor will be same.

( h f)2=f(LD)( V 2 2g)=(0.0276)( 125m 0.28m)( V 2 2 2( 9.81 ))=35mV1=7.46m/s

According to above result, the Reynolds’s number will be

(Re)2=DVν=(0.28m)(7.46m/s)1.01×106m2/s=2.07×106

To find the flow rate,

Q2=A2V2=[π(0.14)2(7.46m/s)]=0.459m3/s

The flow in pipe 2 is considered as flow towards the junction.

Similarly for pipe 3,

(hf)3=z3ha=85m80m=5m

To find Reynolds’s number,

Fluid Mechanics, Chapter 6, Problem 6.121P , additional homework tip  4 Try V3=1m/s

Therefore,

(Re)3=DVν=(0.28m)(1m/s)1.01×106m2/s=2.77×105

The friction factor will be equal to,

1f1/21.8log(6.9Re+( /d 3.7 )1.11)1.8log(6.92.77× 105+( 3.571× 10 3 3.7 )1.11)=0.0279

( h f)3=f(LD)( V 2 2g)=(0.0279)( 160m 0.28m)( V 3 2 2( 9.81 ))=5mV1=2.48m/s

According to above result, the Reynolds’s number will be

(Re)3=DVν=(0.28m)(2.48m/s)1.01×106m2/s=6.875×105

To find the flow rate,

Q3=A3V3=[π(0.14)2(2.48m/s)]=0.153m3/s

The flow in pipe 3 is considered as flow towards the junction.

Therefore the flow rate will be,

Q=(Q2+Q3)Q1=(0.153+0.459)0.66=0.048m3/s

By further iteration, we can get this close to zero, but for now, according to the obtained results, we can say,

Q1=0.66m3/sQ2=0.459m3/sQ3=0.153m3/s

Conclusion:

The flow rates in each pipe is equal to,

Q1=0.66m3/sQ2=0.459m3/sQ3=0.153m3/s.

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