Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.73P
To determine

(a)

Determine whether the nozzle is choked.

Expert Solution
Check Mark

Answer to Problem 9.73P

The nozzle is choked since the flow is supersonic.

Explanation of Solution

Given Information:

Section pressure, P1=101kPa

Section temperature, T1=300k

Velocity, V1=868m/s

Area of throat, At=3cm2

Section 1 Mach number is given by,

Ma1=V1C1

=V1kRT1

=8681.4×287×300

Ma1=2.5

The nozzle is choked since the flow is supersonic

Conclusion:

The nozzle is choked since the flow is supersonic.

To determine

(b)

Determine the area of section one.

Expert Solution
Check Mark

Answer to Problem 9.73P

The area of section one is A1=7.92cm2.

Explanation of Solution

Given Information:

Section pressure, P1=101kPa

Section temperature, T1=300k

Velocity, V1=868m/s

Area of throat, At=3cm2

Section 1 area be A1,

We know,

A1At=1Ma1( ( 1+0.2Ma1 2 ) 31.728)

A13=12.5(( 1+0.2 (2.5) 2 )31.728)

A13=2.64

A1=7.92cm2

Conclusion:

The area of section one is A1=7.92cm2.

To determine

(c)

Determine the mass flow. Also, determine the new P1, T1 and V1 if there is change in gas properties at the section one

Expert Solution
Check Mark

Answer to Problem 9.73P

The mass flow is m=0.805kg/s . The new V1=926m/s, T1=248.44K and P1=52.2kPa.

Explanation of Solution

Given Information:

Section pressure, P1=101kPa

Section temperature, T1=300k

Velocity, V1=868m/s

Area of throat, At=3cm2

Let m = mass flow

If the process is isentropic then,

P0P1=(1+k12Ma12)kk1

P0101=(1+1.412( 2.5)2)1.41.41

P0=101(1+0.2( 2.5)2)3.5

P0=1725.68kPa

Similarly, we know,

T0T1=(1+k12Ma12)

T0300=(1+1.412(2.5)2)

T0=300(1+0.2(2.5)2)

T0=675k

We know,

m=mmax

m=0.6847P0AtRT0

m=0.6874×1725.68×103×3104287×675

m=0.805kg/s

Therefore, the mass flow is m=0.805kg/s

Throat area reduced to 2 cm2

We know,

A1At=1Ma1(( 1+0.2M a 1 2 )31.728)

7.922=1Ma1(( 1+0.2M a 1 2 )31.728)

3.96=1Ma1(( 1+0.2M a 1 2 )31.728)

(1+0.2Ma12)3=7.057×Ma1

(1+0.2Ma12)=(7.057×Ma1)1/3

By trial and error method,

Ma1=2.93

We know,

P0P1=(1+k12Ma12)kk1

1725.68P1=(1+1.412( 2.93)2)1.41.41

1725.68P1=(1+0.2( 2.93)2)3.5

P1=1725.68(1+0.2 ( 2.93 ) 2)3.5

P1=52.19kPa

P1=52.2kPa

We also know,

T0T1=(1+k12Ma12)

675T1=(1+0.2(2.93)2)

T1=675(1+0.2( 2.93)2)

T1=248.44K

V1=Ma×kRT

V1=2.93×1.4×287×248.44

V1=926m/s

Conclusion:

The mass flow is m=0.805kg/s . The new V1=926m/s, T1=248.44K and P1=52.2kPa.

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Chapter 9 Solutions

Fluid Mechanics

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