Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.103P
To determine

Estimate the flow of water at 20° C.

Expert Solution & Answer
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Answer to Problem 6.103P

Q=0.11ft3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.103P

Both reservoirs are connected by cast iron pipes with sharp edged entrance and exit. Reservoir 1 is 45ft higher than reservoir 2. The fluid is water at 20°C

The volumetric flow rate Q can be defined as below,

Q=VA

In above equation,

V - Velocity of the flow

A -Area

The steady flow equation between the reservoir 1 & 2 will be,

p1ρg+V122g+z1=p2ρg+V222g+z2+hf+hmhp

We know that, p1=p2 and V1=V20, therefore the above equation can be rewritten as,

hp=z2z1+hf+hm

The Reynolds’s number is defined as,

Re=DVν

D - Diameter of the pipe

V - Velocity of the flow

ν - Kinematic viscosity

The friction factor can be determined by,

1f1/21.8log(6.9Re+( /d 3.7 )1.11)

Assume that the water at 20°C will have,

ρ=1.94slug/ft3

μ=2.09×105slug/fts

Assume that the wrought iron will have a roughness of,

ε=0.00015ft

Calculation:

Assume small pipe as (a) and the large pipe as (b).

First of all, the roughness ratio will be,

For pipe (a)

εda=0.00015ft112ft=1.8×103

For pipe (b)

εdb=0.00015ft212ft=9×104

Loss co-efficient for pipe (a) will be,

Ksharpentrance=0.5Ksuddenexpansion=0.56

Similarly for pipe (b),

Ksubmergedexit=1.0

By equating the volumetric flow for both pipes,

Qa=QbVa( π d a 2 4)=Vb( π d b 2 4)Va( π d a 2 4)=Vb( π ( 2 d a ) 2 4)Va=4Vb

According to the steady flow equation between reservoir (1) & (2),

45ft=Va22(32.2ft/s2)(fa( 20ft)1 12ft+(0.5+0.56))+Vb22(32.2ft/s2)(fb( 20ft)2 12ft+(1.0))

We know that,

Vb=Va4

Therefore,

45ft=Va22( 32.2ft/ s 2 )(240fa+1.06)+ ( V a 4 )22( 32.2ft/ s 2 )(120fb+( 1.0))=[ V a 22( 32.2ft/ s 2 )](( 240 f a +1.06)+( 120 f b +( 1.0 ) 16 ))

To find Va the friction factor needs to be known, therefore assume fa=fb=0.02

45ft=[ V a 22( 32.2ft/ s 2 )](( 240(0.02)+1.06)+( 120(0.02)+( 1.0 ) 16 ))Va=21.845ft/s

To find Reynolds’s number,

Re=DVν=1 12ft(21.845ft/s)1.09× 10 5=167010.70

To find friction factor,

1f 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1.8log( 6.9 167010.7+ ( 1.8× 10 3 3.7 ) 1.11)0.0238

Similarly for pipe (b),

Vb=Va4=21.845ft/s4=5.46125ft/s

To find Reynolds’s number,

Re=DVν=2 12ft(5.46125ft/s)1.09× 10 5=83505.35

To find the friction factor,

1f 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1.8log( 6.9 83505.35+ ( 9× 10 4 3.7 ) 1.11)0.022

Therefore, to find the absolute value of Va

45ft=[ V a 22( 32.2ft/ s 2 )](( 240(0.0238)+1.06)+( 120(0.022)+( 1.0 ) 16 ))Va=20.347ft/s

To calculate the flow of water,

Q=VA=(20.347ft/s)(π ( 0.5 12 )2)=0.11ft3/s

Conclusion:

The flow of water is equal to Q=0.11ft3/s.

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