Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.28P
To determine

(a)

The Mach number at point 1.

Expert Solution
Check Mark

Answer to Problem 9.28P

Ma1=0.5

Explanation of Solution

Given information:

At section 1,

p1=250kPaT1=125°CV1=200m/s

At section 2,

A2=0.25m2Ma2=2.0

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

Calculation:

Calculate the speed of sound,

For ideal gas,

k=1.4

Convert,

T1=125°C=398K

a=kRT=(1.4)(287m2/s2.K)(398K)=399.89m/s

Calculate the Mach number,

Ma1=Va=200m/s399.89m/s=0.5

Conclusion:

The Mach number at section1 is equal to Ma1=0.5.

To determine

(b)

The temperature at point 2.

Expert Solution
Check Mark

Answer to Problem 9.28P

T2=232.16K

Explanation of Solution

Given information:

At section 1,

p1=250kPaT1=125°CV1=200m/s

At section 2,

A2=0.25m2Ma2=2.0

The temperature ratio can be defined in terms of Mach number as,

(T0T)=[1+12(k1)Ma2]

For ideal gas,

k=1.4

Calculation:

Calculate the stagnation temperature,

(T0T1)=[1+12(k1)Ma12]

For point 1,

According to sub-part a,

Ma1=0.5

Substitute for known values,

(T0398K)=[1+12(1.41)(0.5)2]

Therefore,

T0=417.9K

Find the temperature at section 2,

(T0T2)=[1+12(k1)Ma22]

Substitute for known values,

(417.9KT2)=[1+12(1.41)(2.0)2]

Therefore,

T2=232.16K

Conclusion:

The temperature at section 2 is equal to T2=232.16K.

To determine

(c)

To calculate:

The velocity at point 2.

Expert Solution
Check Mark

Answer to Problem 9.28P

V2=610.84m/s

Explanation of Solution

Given information:

At section 1,

p1=250kPaT1=125°CV1=200m/s

At section 2,

A2=0.25m2Ma2=2.0

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

Calculation:

According to sub-part a,

T2=232.16K

Calculate the speed of sound,

a=kRT=(1.4)(287m2/s2.K)(232.16K)=305.42m/s

Calculate the velocity at point 2,

V2=Ma2a=(2.0)(305.42m/s)=610.84m/s

Conclusion:

The velocity at section 2 is equal to V2=610.84m/s.

To determine

(d)

To calculate:

The mass flow.

Expert Solution
Check Mark

Answer to Problem 9.28P

m=86.74kg/s

Explanation of Solution

Given information:

At section 1,

p1=250kPaT1=125°CV1=200m/s

At section 2,

A2=0.25m2Ma2=2.0

The density at section 2 is defined as,

ρ2=p2RT2

The pressure ration is defined as,

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

The mass flow is defined as,

m=ρ2A2V2

Where,

A2 - Area at section 2,

For ideal gas,

R=287m2/s2.Kk=1.4

Calculation:

Calculate the stagnation pressure,

p0250kPa=[1+12(1.41)( 0.5)2]1.4/1.41

Therefore,

p0=296.55kPa

Calculate the pressure at section 2,

296.55kPap2=[1+12(1.41)( 2.0)2]1.4/1.41

Therefore,

p2=37.9kPa

Calculate the density at section 2,

ρ2=p2RT2=37900Pa(287m2/s2.K)(232.16K)=0.568kg/m3

Calculate the mass flow,

m=ρ2A2V2=(0.568kg/m3)(0.25m2)(610.84m/s)=86.74kg/s

Conclusion:

The mass flow is equal to m=86.74kg/s.

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