Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.8P

Consider a Mach 4 airflow at a pressure of 1 atm. We wish to slow this flow to subsonic speed through a system of shock waves with as small a loss in total pressure as possible. Compare the loss in total pressure for the following three shock systems:

a. A single normal shock wave

b. An oblique shock with a deflection angle of 25 . 3 ° , followed by a normal shock.

c. An oblique shock with a deflection angle of 25 . 3 ° , followed by a second oblique shock of deflection angle of 2 0 ° , followed by a normal shock.

From the results of (a), (b), and (c), what can you induce about the efficiency of the various shock systems?

(a)

Expert Solution
Check Mark
To determine

The comparison in total pressure loss for the single normal shock wave.

Answer to Problem 9.8P

The loss in pressure is 130.87atm

Explanation of Solution

Given:

The Mach number is M1=4 .

The pressure is p1=1atm .

Formula used:

The expression for p0,1 is given as,

  p0,1=p1[1+γ12M12]γγ1

The expression for p0,2 from normal shock table at M1=4 is given as,

  p0,2=0.1388p0,1

The expression for loss in pressure is given as,

  loss=p0,1p0,2

Calculation:

The pressure p0,1 can be calculated as,

  p0,1=p1[1+ γ12M12]γ γ1p0,1=1×[1+ 1.412×42] 1.4 1.41p0,1=151.83atm

The pressure p0,2 can be calculated as,

  p0,2=0.1388p0,1p0,2=0.1388×151.83atmp0,2=20.953atm

The loss in pressure can be calculated as,

  loss=p0,1p0,2loss=151.83atm20.953atmloss=130.87atm

Conclusion:

Therefore, the loss in pressure is 130.87atm .

(b)

Expert Solution
Check Mark
To determine

The comparison in pressure for an oblique shock with a deflection angle of 25.3° .

Answer to Problem 9.8P

The loss in pressure is 105.58atm .

Explanation of Solution

Given:

The Mach number is M1=4 .

The pressure is p1=1atm .

The deflection angle of oblique shock wave is θ=25.3° .

Formula used:

The expression for Mn1 is given as,

  Mn1=M1sinβ

The expression for M2 is given as,

  M2=Mn2sin(βθ)

The expression for p0,3 is given as,

  p0,3=p0,3p0,2p0,2p0,1p0,1

The expression for loss in pressure is given as,

  loss=p0,1p0,3

Calculation:

From θβM diagram at θ=25.3° value of angle β=38.5° .

  Fundamentals of Aerodynamics, Chapter 9, Problem 9.8P , additional homework tip  1

    Figure (1)

The Mach number Mn1 can be calculated as,

  Mn1=M1sinβMn1=4×sin(38.5)Mn1=2.49

The pressure ratio for Mach number Mn1=2.49 from appendix B is given as,

  p0,2p0,1=0.5030

From appendix B Mn2=0.514

The Mach number M2 can be calculated as,

  M2=M n2sin( βθ)M2=0.514sin( 38.525.3)M2=2.25

The pressure ratio for Mach number M2=2.25 from appendix B is given as,

  p0,3p0,2=0.6055

The pressure p0,3 can be calculated as,

  p0,3=p 0,3p 0,2p 0,2p 0,1p0,1p0,3=0.5030×0.6055×151.83atmp0,3=46.242atm

The pressure loss can be calculated as,

  loss=p0,1p0,3loss=151.83atm46.242atmloss=105.58atm

Conclusion:

Therefore, the loss in pressure is 105.58atm .

(c)

Expert Solution
Check Mark
To determine

The comparison in pressure for the an oblique shock with a deflection angle of 25.3° followed by a second oblique shock of deflection angle of 20° , followed by normal shock.

Answer to Problem 9.8P

The loss in pressure is 88.28atm .

Explanation of Solution

Given:

The Mach number is M1=4 .

The pressure is p1=1atm .

The deflection angle of second oblique shock wave is θ=20° .

Formula used:

The expression for the Mach number Mn2 can be given as,

  Mn2=M2sinβ2

The expression for Mach number M3 is given as,

  M3=Mn,3sin(β2θ)

The expression for the pressure p0,4 is given as,

  p0,4=p0,4p0,3p0,3p0,2p0,2p0,1p0,1

The expression for loss in pressure is given as,

  loss=p0,1p0,3

Calculation:

From θβM diagram for M2=2.25 and θ=20° the value of angle β2=47.3 .

  Fundamentals of Aerodynamics, Chapter 9, Problem 9.8P , additional homework tip  2

    Figure (2)

The Mach number Mn2 can be calculated as,

  Mn2=M2sinβ2Mn2=2.25×sin47.3°Mn2=1.653

The pressure ratio for Mach number from appendix B is given as,

  p0,3p0,2=0.8740

Refer to appendix B Mn3=0.655

The Mach number M3 can be calculated as,

  M3=M n,3sin( β 2 θ)M3=0655sin( 47.3°20°)M3=1.428

The pressure ratio for Mach number M3=1.428 from appendix B is given as,

  p0,4p0,3=0.9501

The pressure p0,4 can be calculated as,

  p0,4=p 0,4p 0,3p 0,3p 0,2p 0,2p 0,1p0,1p0,4=0.9501×0.8740×0.5030×151.83atmp0,4=63.41atm

The pressure loss can be calculated as,

  loss=p0,1p0,4loss=151.83atm63.41atmloss=88.42atm

From a, b and c it is clear that the most efficient way to decrease supersonic flow to subsonic flow is through a combination of supersonic diffuser and then normal shock wave at the end.

Conclusion:

Therefore, the loss in pressure is 88.28atm .

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