Fluid Mechanics, 8 Ed
Fluid Mechanics, 8 Ed
8th Edition
ISBN: 9789385965494
Author: Frank White
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 6, Problem 6.115P
To determine

(a)

The flow rate from reservoir 1 if valve C is closed.

Expert Solution
Check Mark

Answer to Problem 6.115P

Q=0.0152m3/s

Explanation of Solution

Given information:

Fluid Mechanics, 8 Ed, Chapter 6, Problem 6.115P , additional homework tip  1

All pipes are made of cast iron and the diameter is equal to 8cm

If the valve C is closed, the energy equation for the above system can be written as,

z1z2=hfA+hmA+hfB+hmB

Since, the flow is series and the diameter is same for all pipes, the flow rate and the velocity will be equal in both pipe A and pipe B.

Calculation:

Assume, the water at 20°C will have,

ρ=998kg/m3

μ=0.001kg/ms

The loss co efficient will be equal to,

Ksharpentrance=0.5Klinejunction=0.9KBranchjunction=1.3KSubmergedexit=1.0

According to the explanation, the energy equation will be,

z1z2=hfA+h m A +hfB+h m B ΔZ=V22g(f L A D+K sharpentrance+K linejunction+f L B D+K Submergedexit)25m=V22( 9.81m/ s 2 )(f 100m 0.08m+0.5+0.9+f 50m 0.08m+1.0)

Assume,

f=0.02

Therefore,

25m=V22( 9.81m/ s 2 )(( 0.02) 100m 0.08m+0.5+0.9+( 0.02) 50m 0.08m+1.0)V=3.5m/s

Calculate the Reynolds’s number

Red=(ρVd)μ=(998kg/m3)(3.5m/s)(0.08m)0.001kg/ms=279440

Calculate the roughness ratio

εd=0.26mm80mm=3.25×103

Calculate the exact value of friction factor

1f 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1f 1/21.8log( 6.9 279440+ ( 3.25× 10 3 3.7 ) 1.11)f=0.0272

Now, calculate the exact value of the velocity

z1z2=hfA+h m A +hfB+h m B ΔZ=V22g(f L A D+K sharpentrance+K linejunction+f L B D+K Submergedexit)25m=V22( 9.81m/ s 2 )(( 0.0272) 100m 0.08m+0.5+0.9+( 0.0272) 50m 0.08m+1.0)V=3.03m/s

Calculate the flow rate

Q=VA=(3.03m/s)(π( 0.04m)2)=0.0152m3/s

Conclusion:

The flow rate is equal to 0.0152m3/s.

To determine

(b)

The flow arte from reservoir 1 if valve C is open.

Expert Solution
Check Mark

Answer to Problem 6.115P

QA=0.0174m3/sQB=0.00955m3/sQC=0.00784m3/s

Explanation of Solution

Given information:

Fluid Mechanics, 8 Ed, Chapter 6, Problem 6.115P , additional homework tip  2

All pipes are made of cast iron and the diameter is equal to 8cm

If the valve C is opened, the energy equation for the above system can be written as,

z1z2=hfA+hm AB+hfB+hmB=hfA+hm AC+hfB+hmC

Since, the flow is parallel in pipe B and C, and the diameter is same for all pipes,

QA=QB+QC

VA=VB+VC

Calculation:

Assume, the water at 20°C will have,

ρ=998kg/m3

μ=0.001kg/ms

The loss co-efficient will be equal to,

Ksharpentrance=0.5Klinejunction=0.9KBranchjunction=1.3KSubmergedexit=1.0

According to the explanation, the energy equation will be,

z1z2=hfA+h m AB +hfB+h m B =hfA+h m AC +hfB+h m C ΔZ=VA22g(f L A D+K sharpentrance+K linejunction)+VB22g(f L B D+K Submergedexit)25m=VA22( 9.81m/s)(f 100m 0.08m+0.5+0.9)+VB22( 9.81m/s)(f 50m 0.08m+1.0)

Similarly,

ΔZ=VA22g(f L A D+K sharpentrance+K branchjunction)+VC22g(f L C D+K Submergedexit)25m=VA22( 9.81m/s)(f 100m 0.08m+0.5+1.3)+VC22( 9.81m/s)(f 70m 0.08m+1.0)

But, we also know

VA=VB+VC

Assume the friction factor is same for all pipe and it is equal to,

f=0.02

25m=VA22( 9.81m/s)(( 0.027) 100m 0.08m+0.5+0.9)+VB22( 9.81m/s)(( 0.027) 50m 0.08m+1.0)25m=1.791VA2+0.911VB2(1)

Similarly,

25m=VA22( 9.81m/s)(( 0.027) 100m 0.08m+0.5+1.3)+VC22( 9.81m/s)(( 0.027) 70m 0.08m+1.0)25m=1.811VA2+1.255VC2(2)

VA=VB+VC(3)

By solving we get,

VA=3.48m/sVB=1.91m/sVC=1.57m/s

Now find the Reynolds’s number and friction factor separately for each pipe

ReA=( ρVd)μ=( 998kg/ m 3 )( 3.48m/s)( 0.08m)0.001kg/ms=277843.2ReB=( ρVd)μ=( 998kg/ m 3 )( 1.91m/s)( 0.08m)0.001kg/ms=152494.4ReC=( ρVd)μ=( 998kg/ m 3 )( 1.57m/s)( 0.08m)0.001kg/ms=125348.8

1fA 1/21.8log( 6.9 277843.2+ ( 3.25× 10 3 3.7 ) 1.11)fA0.02721fB 1/21.8log( 6.9 152494.4+ ( 3.25× 10 3 3.7 ) 1.11)fB0.02751fC 1/21.8log( 6.9 125348.8+ ( 3.25× 10 3 3.7 ) 1.11)fC0.0277

25m=VA22( 9.81m/s)(( 0.0272) 100m 0.08m+0.5+0.9)+VB22( 9.81m/s)(( 0.0275) 50m 0.08m+1.0)25m=1.804VA2+0.927VB2(1)

25m=VA22( 9.81m/s)(( 0.0272) 100m 0.08m+0.5+1.3)+VC22( 9.81m/s)(( 0.0277) 70m 0.08m+1.0)25m=1.824VA2+1.283VC2(2)

According to the equations,

VA=3.46m/sVB=1.9m/sVC=1.56m/s

The relevant flow rate,

QA=VAA=(3.46m/s)(π ( 0.04m )2)=0.0174m3/sQB=VBA=(1.9m/s)(π ( 0.04m )2)=0.00955m3/sQC=VCA=(1.56m/s)(π ( 0.04m )2)=0.00784m3/s

Conclusion:

The flow rate in pipe A is equal to 0.0174m3/s

The flow arte in pipe B is equal to 0.00955m3/s

The flow arte in pipe C is equal to 0.00784m3/s.

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