Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.99P
To determine

(a)

The flow rate without diffuser.

Expert Solution
Check Mark

Answer to Problem 6.99P

Q=2.72×103m3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.99P , additional homework tip  1

The pipe entrance is sharp edged.

The energy equation can be written as below for the system without the diffuser,

z1z2=V222g+V122g(fLD+Kentrance)

But, we can say,

V1=V2

Therefore, the above equation can be rewritten as,

Δz=V122g(1+fLD+Kentrance)

Calculation:

According to the explanation given,

Assume,

f=0.02

Kentrance=0.5

Therefore,

Δz=V122g(1+fLD+K entrance)2m=V122( 9.81m/s)(1+( 0.02) 2m 0.03m+0.5)V1=3.721m/s

Calculate the Reynolds’s number,

Re=ρVdμ=(998kg/m3)(3.721m/s)(0.03m)0.001kg/ms=111406.74

Calculate the friction factor,

ζ=fcn(Re)ζ0.155Re1.75=f ( R e )22

0.155(111406.74)1.75=f ( 111406.74 )22f=0.017

Calculate the exact value of the velocity,

Δz=V122g(1+fLD+K entrance)2m=V122( 9.81m/s)(1+( 0.017) 2m 0.03m+0.5)V1=3.86m/s

Calculate the flow rate,

Q=VA=(3.86m/s)(π ( 0.015m )2)Q=2.72×103m3/s

Conclusion:

The flow rate is equal to 2.72×103m3/s without diffuser.

To determine

(b)

The flow rate with diffuser.

Expert Solution
Check Mark

Answer to Problem 6.99P

Q=3.19×103m3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.99P , additional homework tip  2

The pipe entrance is sharp edged.

The energy equation can be written as below for the system with the diffuser,

z1z2=V222g+V122g(fLD+Kentrance+KDiffuser)

Above equation can be rewritten as,

Δz=V222g+V122g(fLD+Kentrance+KDiffuser)

Calculation:

Calculate V2 in terms of V1

By equating the flow rates, we can say,

V2=(35)2V1=0.36V1

Assume,

f=0.02

Kentrance=0.5

KDiffuser=0.2

Therefore,

Δz=V222g+V122g(fLD+Kentrance+KDiffuser)

2m= ( 0.36 V 1 )22( 9.81m/ s 2 )+V122( 9.81m/ s 2 )(( 0.02) 2m 0.03m+0.5+0.2)V1=4.25m/s

Calculate the Reynolds’s number

Re=ρVdμ=(998kg/m3)(4.25m/s)(0.03m)0.001kg/ms=127245

Calculate the friction factor

ζ=fcn(Re)ζ0.155Re1.75=f ( R e )22

0.155(127245)1.75=f ( 127245 )22f=0.0164

Calculate the exact value of the velocity

Δz=V222g+V122g(1+fLD+Kentrance+KDiffuser)

2m= ( 0.36 V 1 )22( 9.81m/ s 2 )+V122( 9.81m/ s 2 )(( 0.0164) 2m 0.03m+0.5+0.2)V1=4.52m/s

Therefore,

Calculate the flow rate

Q=VA=(4.52m/s)(π ( 0.015m )2)Q=3.19×103m3/s

Conclusion:

The flow rate is equal to 3.19×103m3/s with diffuser.

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Chapter 6 Solutions

Fluid Mechanics

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