Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.30P
`
To determine

The exit mass flow rate as a function of pressure, temperature, gas constant, specific heat and nozzle diameter.

Expert Solution & Answer
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Answer to Problem 5.30P

Required dimensionless function m=PoD2m.To12R12f(CpR)

Explanation of Solution

Given information:

Mass flow rate =mTank pressure = PoTemperature = ToGas constant =RSpecific heat = CpNozzle diameter = D

ṁ is a function of P0, To,R,cp and D.

ṁ = f (ṁ, P0, T0, R, Cp, D)

or g(ṁ, P0, T0, R, Cp, D) = 0

Dimensions of ṁ = [MT-1]

Dimensions of P = [ML-1 T-2]

Dimensions of R = [L2 T-2 θ- 1]

Dimensions of D = [L]

Dimensions of CP = [L2 T-2 θ- 1]

Dimensions of TO = [θ]

Total number of variables m = 6

Number of fundamental variables n = 4 [MLTθ]

So, Number of (terms = m-n = 2(π12)

Calculation:

Taking (P0, T0, R, D) as repeating variables: -

π1=Pa1Tob1Rc1Dd1m.[M0L0T0θ0]=[ML 1T 2]a1[θ]b1[L2T 2θ 1]c1[L]d1[MT1][M0L0T0θ0]=[M a 1+1L a 1 +2 c 1 + d 1T 2 a 1 2 c 11θ b 1 c 1]

comparing powers of [MLTθ]

a1+1 = 0

-a1+2c1+d1 = 0

-2a1-2c1-1 = 0

b1-c1 = 0

solving above equations

a1 = -1

b1 = ½

c1 = ½

d1 = -2

so first π-term :-

π1=Po1To12R12D2m.π1=To 1 2 R 1 2 m.PoD2

Second π-term :-

π2=Pa2Tob2Rc2Dd2Cp[M0L0T0θ0]=[ML 1T 2]a2[θ]b2[L2T 2θ 1]c 2 [L]d2[L2T2θ1][M0L0T0θ0]=[M a 2L a 2 +2 c 2 + d 2+2T 2 a 2 2 c 22θ b 2 c 21]

comparing powers of [MLTθ]

a2 = 0

-a2+2c2+d2+2 = 0

-2a2-2c2-2 = 0

b2-c2-1 = 0

solving above equations

a2 = 0

b2 = 0

c2 = -1

d2 = 0

so second π-term :-

π2=P0To0R1D0Cpπ2=CpR

Thus π1 = f(π2)

π1=f(π2)To 1 2 R 1 2 m.PoD2=f( C p R)m.=PoD2m.To 1 2 R 1 2 f( C p R)

Conclusion:

The required dimensionless function is

m.=PoD2m.To12R12f(CpR).

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Chapter 5 Solutions

Fluid Mechanics

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