Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
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Chapter 12, Problem 12.1P

Using the results of linearized theory, calculate the lift and wave-drag coefficients for an infinitely thin flat plate in a Mach 2.6 freestream at angles of attack of

(a) α = 5 ° (b) α = 15 ° (c) α = 30 °

Compare these approximate results with those from the exact shock- expansion theory obtained in Problem 9.13. What can you conclude about the accuracy of linearized theory in this case?

(a)

Expert Solution
Check Mark
To determine

The lift and wave drag coefficients and comparison of accuracy.

Answer to Problem 12.1P

The lift and drag coefficients are 0.1454_ and 0.012692_ .

Explanation of Solution

Given:

The angle of attack is α1=5° .

The Mach number is M=2.6 .

Formula used:

The expression for lift is given as,

  Cl=4αM21

The expression for drag is given as,

  Cd=4α2M21

Calculation:

The coefficient of lift can be calculated as,

  Cl=4α1 M 2 1=4×5°×π 180°rad ( 2.6 ) 2 1=0.1454

Refer to problem 9.13, the error can be calculated as,

  E%=0.14750.14540.1475×100=1.69%

The coefficient of drag can be calculated as,

  Cd=4α12 M 2 1=4× ( 5°× π 180° rad )2 ( 2.6 ) 2 1=0.012692

Refer to problem 9.13, the error can be calculated as,

  E%=0.01290.01260.0129×100=2.209%

Conclusion:

Therefore, the lift and drag coefficients are 0.1454_ and 0.012692_ .

(b)

Expert Solution
Check Mark
To determine

The lift and wave drag coefficients and comparison of accuracy.

Answer to Problem 12.1P

The lift and drag coefficients are 0.4363_ and 0.1142_ .

Explanation of Solution

Given:

The angle of attack is α2=15° .

The Mach number is M=2.6 .

Formula used:

The expression for lift is given as,

  Cl=4αM21

The expression for drag is given as,

  Cd=4α2M21

Calculation:

The coefficient of lift can be calculated as,

  Cl=4α2 M 2 1=4×15°×π 180°rad ( 2.6 ) 2 1=0.4363

Refer to problem 9.13, the error can be calculated as,

  E%=0.4490.43630.449×100=1.4%

The coefficient of drag can be calculated as,

  Cd=4α22 M 2 1=4× ( 15°× π 180° rad )2 ( 2.6 ) 2 1=0.1142

Refer to problem 9.13, the error can be calculated as,

  E%=0.1170.11420.117×100=2.39%

Conclusion:

The lift and drag coefficients in first case are 0.4363_ and 0.1142_ .

(c)

Expert Solution
Check Mark
To determine

The lift and wave drag coefficients and comparison of accuracy.

Answer to Problem 12.1P

The lift and drag coefficients are 0.8756_ and 0.4600_ .

Explanation of Solution

Given:

The angle of attack is α3=30°

The Mach number is M=2.6

Formula used:

The expression for lift is given as,

  Cl=4αM21

The expression for drag is given as,

  Cd=4α2M21

Calculation:

The coefficient of lift can be calculated as,

  Cl=4α3 M 2 1=4×30°×π 180°rad ( 2.6 ) 2 1=0.8756

Refer to problem 9.13, the error can be calculated as,

  E%=1.210.87561.21×100=27.63%

The coefficient of drag can be calculated as,

  Cd=4α32 M 2 1=4× ( 30°× π 180° rad )2 ( 2.6 ) 2 1=0.4600

Refer to problem 9.13, the error can be calculated as,

  E%=0.690.46000.69×100=33.33%

Conclusion:

Therefore, the lift and drag coefficients are 0.8756_ and 0.4600_ .

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