Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.55P
To determine

(a)

To compute: the pressure just downstream of this shock.

Expert Solution
Check Mark

Answer to Problem 9.55P

p2=260.91kPa

Explanation of Solution

Given information:

Stagnation pressure is equal to 450kPa

Throat area is 12cm2

Shock is at A1=20cm2

The pressure ratio is defined as,

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

Where,

k=1.4

Calculation:

Calculate the area ratio,

A1A=20cm212cm2=1.667

According to the table B.1 which represents the isentropic flow of perfect gas

AA=1.5553Ma=1.90AA=1.6875Ma=2.0

By interpolation,

Ma1=(1.6671.5553)(1.68751.5553)(2.01.9)+1.9=1.9845

Calculate the pressure at point 1,

p0p1=[1+12( k1)Ma12]k/k1450kPap1=[1+12( 1.41) ( 1.9845 )2]1.4/1.41p1=58.91kPa

According to the table B.2 which represents the normal shock relations for a perfect gas,

Ma=1.90p2p1=4.0450Ma=2.0p2p1=4.5

By interpolation,

p2p1=(1.98451.90)(2.01.9)(4.54.0450)+4.0450=4.429

Therefore, the pressure p2 just downstream of shock,

p2p1=4.429p2=4.429(58.91kPa)=260.91kPa

Conclusion:

The pressure just downstream of shock is equal to p2=260.91kPa.

To determine

(b)

To estimate: p3.

Expert Solution
Check Mark

Answer to Problem 9.55P

p3=314.22kPa

Explanation of Solution

Given information:

Stagnation pressure is equal to 450kPa

Throat area is 12cm2

Shock is at A1=20cm2

The pressure ratio is defined as,

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

Where,

k=1.4

Calculation:

According to sub-part a,

We have found,

Ma1=1.9845

According to the table B.2 which represents the normal shock relations for a perfect gas,

Ma=1.90A2A1=1.3032Ma=2.0A2A1=1.3872

By interpolation,

A2A1=(1.98451.90)(2.01.9)(1.38721.3032)+1.3032=1.3742

Therefore,

A2=1.3742(12cm2)=16.49cm2

At A3=30cm2

A3A2=30cm216.49cm2=1.82

Therefore, calculate the relevant Mach number Ma3

According to table B.1

Ma3=1.821.59012.03511.5901(0.40.3)+0.3=0.35

According to the table B.2 which represents the normal shock relations for a perfect gas,

Ma=1.9p 0 2 p 0 1 =0.7674Ma=2.0p 0 2 p 0 1 =0.7209

By interpolation,

p02p01=(1.98451.9)(2.01.9)(0.76740.7209)+0.7209=0.76

Therefore,

p02=0.76(450kPa)=342kPa

Calculate the pressure at point 3,

p 0 2 p3=[1+12( k1)Ma32]k/k1342kPap3=[1+12( 1.41) [ 0.35]2]1.4/1.41p3=314.22kPa

Conclusion:

The pressure at point 3 is equal to p3=314.22kPa.

To determine

(c)

To estimate: throat area.

Expert Solution
Check Mark

Answer to Problem 9.55P

A2=16.49cm2

Explanation of Solution

Given information:

Stagnation pressure is equal to 450kPa

Throat area is 12cm2

Shock is at A1=20cm2

According to sub-part a,

We have found,

Ma1=1.9845

Therefore, by using table B.2, we can find the relevant area ratio.

Calculation:

According to the table B.2 which represents the normal shock relations for a perfect gas,

Ma=1.90A2A1=1.3032Ma=2.0A2A1=1.3872

By interpolation,

A2A1=(1.98451.90)(2.01.9)(1.38721.3032)+1.3032=1.3742

Therefore,

A2=1.3742(12cm2)=16.49cm2

Conclusion:

The throat area is equal to A2=16.49cm2.

To determine

(d)

To estimate: Ma3.

Expert Solution
Check Mark

Answer to Problem 9.55P

Ma3=0.35

Explanation of Solution

Given information:

Stagnation pressure is equal to 450kPa

Throat area is 12cm2

Shock is at A1=20cm2

According to sub-part a,

We have found,

Ma1=1.9845

Therefore, by using table B.2 and B.1 we can able to estimate the Mach number at point 3

Calculation:

According to the table B.2 which represents the normal shock relations for a perfect gas,

Ma=1.90A2A1=1.3032Ma=2.0A2A1=1.3872

By interpolation,

A2A1=(1.98451.90)(2.01.9)(1.38721.3032)+1.3032=1.3742

Therefore,

A2=1.3742(12cm2)=16.49cm2

At A3=30cm2

A3A2=30cm216.49cm2=1.82

Therefore, calculate the relevant Mach number Ma3

According to table B.1

AA=1.5901Ma=0.4AA=2.0351Ma=0.3

Ma3=1.821.59012.03511.5901(0.40.3)+0.3=0.35

Conclusion:

The Mach number at point 3 is equal to Ma3=0.35.

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