Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
Question
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Chapter 9, Problem 9.33P
To determine

(a)

To calculate:

The Mach number at point 1.

Expert Solution
Check Mark

Answer to Problem 9.33P

Ma1=1.467

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

p0=300kPaT0=500K

At section 1,

A1=0.2m2V1=550m/s

Speed of sound is defined as,

a=kRT

Where,

R - Gas constant

R=287m2/s2.K

k - Specific heat capacity

The Mach number is defined as,

Ma=Va

Where,

V - Air velocity

The stagnation temperature is defined as,

T0=T+V22Cp

Where,

Cp=1005m2/s2.K

Calculation:

Calculate the temperature at point 1,

T0=T1+V122Cp

Substitute for known values,

500K=T1+( 550m/s)22(1005m2/s2.K)

Solve for temperature T1,

T1=349.5K

Calculate the speed of sound,

a=kRT1=1.4(287m2/s2.K)(349.5K)=374.74m/s

Calculate the Mach number at section 1,

Ma1=V1a=550m/s374.74m/s=1.467

Conclusion:

The Mach number at section 1 is equal to Ma1=1.467.

To determine

(b)

To calculate:

The Temperature at point 1.

Expert Solution
Check Mark

Answer to Problem 9.33P

T1=349.5K

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

p0=300kPaT0=500K

At section 1,

A1=0.2m2V1=550m/s

The stagnation temperature is defined as,

T0=T+V22Cp

Where,

Cp=1005m2/s2.K

Calculation:

Calculate the temperature at point 1,

T0=T1+V122Cp

Substitute for known values,

500K=T1+( 550m/s)22(1005m2/s2.K)

Solve for temperature T1,

T1=349.5K

Conclusion:

The temperature at section 1 is equal to T1=349.5K.

To determine

(c)

To calculate:

The pressure at point 1.

Expert Solution
Check Mark

Answer to Problem 9.33P

p1=85.7kPa

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

p0=300kPaT0=500K

At section 1,

A1=0.2m2V1=550m/s

The pressure ratio is defined as,

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

Where,

k=1.4

Calculation:

Calculate the pressure at point 1,

p0p1=[1+12(k1)Ma12]k/k1

According to sub-part a,

Ma1=1.467

Therefore,

300kPap1=[1+12(1.41)( 1.467)2]1.4/1.41

Solve for pressure at point 1,

p1=85.7kPa

Conclusion:

The pressure at section 1 is equal to p1=85.7kPa.

To determine

(d)

To calculate:

The mass flow.

Expert Solution
Check Mark

Answer to Problem 9.33P

m=93.982kg/s

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

p0=300kPaT0=500K

At section 1,

A1=0.2m2V1=550m/s

The density at section 1 is defined as,

ρ1=p1RT1

The pressure ratio is defined as,

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

The mass flow is defined as,

m=ρ1A1V1

Where,

A1 - Area at section 1,

For ideal gas,

R=287m2/s2.Kk=1.4

Calculation:

Calculate the density at section1,

ρ1=p1RT1=85700Pa(287m2/s2.K)(349.5K)=0.854kg/m3

Calculate the mass flow,

m=ρ1A1V1=(0.854kg/m3)(0.2m2)(550m/s)=93.982kg/s

Conclusion:

The mass flow is equal to m=93.982kg/s.

To determine

(e)

Calculate A.

Expert Solution
Check Mark

Answer to Problem 9.33P

A=0.173m2

The flow is choked

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

p0=300kPaT0=500K

At section 1,

A1=0.2m2V1=550m/s

For perfect gas, where k=1.4

Area change is defined as,

AA=1Ma( 1+0.2M a 2 )31.728

Calculation:

Calculate A

A1A=1Ma1( 1+0.2M a 1 2 )31.728

According to subpart a,

Ma1=1.467

Therefore,

0.2m2A=11.467 ( 1+0.2 ( 1.467 ) 2 )31.728A=0.173m2

Conclusion:

The flow is choked, because to produce supersonic flow in the duct

A=0.173m2.

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Chapter 9 Solutions

Fluid Mechanics

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