Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 11, Problem 11.31P
To determine

(a)

The best efficiency point and the maximum efficiency for given parameters.

Expert Solution
Check Mark

Answer to Problem 11.31P

The required maximum efficiency of 85% occurs at Q 2000gal/min

Explanation of Solution

Given Information:

Fluid Mechanics, Chapter 11, Problem 11.31P , additional homework tip  1

Temperature(T)=20oC

Formula used:

η=γQHP

Calculation:

To solve this problem, first we will convert the given data into efficiency, like at Q=400gal/min.

We know that, η=γQHP ;

Put the values in the above equation;

(62.4lbf/ft3)(400/448.8ft3/s)(115ft)(36×550ftlbf/s)

0.32

32%

After conversion the table look like:

Q gal/min 0 400 800 1200 1600 2000 2400
η % 0 32% 55% 70% 80% 85% 82%

Therefore the maximum efficiency of 85% occurs at Q 2000gal/min.

To determine

(b)

The most effective flow rate, resulting head and brake horsepower.

Expert Solution
Check Mark

Answer to Problem 11.31P

The required values are:

Flow rate = 24000gal/min

Head = 729ft

BHP = 5180hp

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 11, Problem 11.31P , additional homework tip  2

Temperature(T)=20oCNr=50%increased

Formula used:

CQ*=Q1n1D13 = Q2n2D23

CH*=gH1n12D12=gH2n22D22

CP*=P1ρn13D15=P2ρn23D25

Calculation:

Since we don’t know the values of CQ*, CH* and CP*, yet we can set them equal for conditions 1 and 2;

CQ*=Q1n1D13 = Q2n2D23

Put the values in the above equation;

Q2(1.5n1)(2 D 1)3

And, Q2=12Q1 ;

12(2000gpm)

24000gal/min

CH*=gH1n12D12=gH2n22D22=gH2(1.5 n 1)2(2 D 1)2

orH2=9H1=9(81ft)=729ft

CP*=P1ρn13D15=P2ρn23D25P2ρ( 1.5n1 )3(2 D 1)5

P2=108P1=108(48hp)

5180hp.

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Chapter 11 Solutions

Fluid Mechanics

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