Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
Question
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Chapter 9, Problem 9.31P
To determine

(a)

The Mach number at point 2.

Expert Solution
Check Mark

Answer to Problem 9.31P

Ma2=0.938

Explanation of Solution

Given information:

At section 1,

p1=35lbf/in2T1=200°FV1=400ft/s

At section 2,

p2=18lbf/in2V2=1100ft/s

Speed of sound is defined as,

a=kRT

Where,

R - Gas constant

R=1716ft2/s2.°R

k - Specific heat capacity

The Mach number is defined as,

Ma=Va

Where,

V - Air velocity

For adiabatic flow, the stagnation temperature is defined as,

T0=T+V22Cp

Where,

Cp=6009ft2/s2.°R

Calculation:

Convert,

T1=200°F=659.67°R

Calculate the stagnation temperature,

T0=T1+V122Cp=659.67°R+( 400ft/s)22(6009ft2/s2.°R)=672.98°R

Calculate the temperature at point 2,

T0=T2+V222Cp

Substitute for known values,

672.98°R=T2+ ( 1100ft/s )22( 6009f t 2 / s 2 .°R)T2=572.29°R

Calculate the speed of sound,

a=kRT2=1.4(1716ft2/s2.°R)(572.29°R)=1172.54ft/s

Calculate the Mach number at section 2,

Ma2=V2a=1100ft/s1172.54ft/s=0.938

Conclusion:

The Mach number at section 2 is equal to Ma2=0.938.

To determine

(b)

Calculate Umax.

Expert Solution
Check Mark

Answer to Problem 9.31P

Umax=2843.92ft/s

Explanation of Solution

Given information:

At section 1,

p1=35lbf/in2T1=200°FV1=400ft/s

At section 2,

p2=18lbf/in2V2=1100ft/s

The maximum value of velocity is defined as,

Umax=2CpT0

Where,

T0 - Stagnation temperature

Cp=6009ft2/s2.°R Calculation:

According to sub-part a,

The stagnation temperature is equal to,

T0=672.98°R

Calculate the maximum value of velocity,

Umax=2CpT0=2(6009ft2/s2.°R)(672.98°R)=2843.92ft/s

Conclusion:

The maximum value of velocity is equal to Umax=2843.92ft/s.

To determine

(c)

Calculate p02p01.

Expert Solution
Check Mark

Answer to Problem 9.31P

p02p01=0.845

Explanation of Solution

Given information:

At section 1,

p1=35lbf/in2T1=200°FV1=400ft/s

At section 2,

p2=18lbf/in2V2=1100ft/s

The pressure ratio is defined as,

p0p=[1+12(k1)Ma2]k/k1

Where,

k=1.4

Speed of sound is defined as,

a=kRT

Where,

R - Gas constant

R=1716ft2/s2.°R

k - Specific heat capacity

The Mach number is defined as,

Ma=Va

Where,

V - Air velocity

Calculation:

Calculate the Mach number at section 1,

Ma1=V1a=V1kRT1=400ft/s1.4( 1716f t 2 / s 2 .°R)( 659.67°R)=0.318

Convert,

p1=35lbf/in2=5040lbf/ft2

Calculate the stagnation pressure at section 1,

p01p1=[1+12(k1)Ma12]k/k1

Substitute for known values,

p015040lbf/ft2=[1+12(1.41)( 0.318)2]1.4/1.41

Solve for stagnation pressure at section 1,

p01=5405.87lbf/ft2

Convert,

p2=18lbf/in2=2592lbf/ft2

Calculate the stagnation pressure at section 1,

p02p2=[1+12(k1)Ma22]k/k1

Substitute for known values,

p022592lbf/ft2=[1+12(1.41)( 0.938)2]1.4/1.41

Solve for stagnation pressure at section 2,

p02=4571.10lbf/ft2

Therefore,

p02p01=4571.10lbf/ft25405.87lbf/ft2=0.845

Conclusion:

The stagnation pressure ration is equal to p02p01=0.845.

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Chapter 9 Solutions

Fluid Mechanics

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