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) $\alpha =5°$ (b) $\alpha =15°$ (c) $\alpha =30°$

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?

To determine

The lift and wave drag coefficients and comparison of accuracy.

Answer to Problem 12.1P

The lift and drag coefficients are $\underset{\_}{0.1454}$ and $\underset{\_}{0.012692}$ .

Explanation of Solution

Given:

The angle of attack is ${\alpha }_1=5°$ .

The Mach number is $M_{\infty }=2.6$ .

Formula used:

The expression for lift is given as,

  $C_l=\frac{4\alpha }{\sqrt{M_{\infty }^2-1}}$

The expression for drag is given as,

  $C_d=\frac{4{\alpha }^2}{\sqrt{M_{\infty }^2-1}}$

Calculation:

The coefficient of lift can be calculated as,

  $\begin{array}{c}{C}_l=\frac{4{\alpha }_1}{\sqrt{M_{\infty }^2-1}}\\ =\frac{4\times 5°\times \frac{\pi }{180°}\text{rad}}{\sqrt{{\left(2.6\right)}^2-1}}\\ =0.1454\end{array}$

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

  $\begin{array}{c}E\%=\frac{0.1475-0.1454}{0.1475}\times 100\\ =1.69\%\end{array}$

The coefficient of drag can be calculated as,

  $\begin{array}{c}{C}_d=\frac{4{\alpha }_1^2}{\sqrt{M_{\infty }^2-1}}\\ =\frac{4\times {\left(5°\times \frac{\pi }{180°}\text{rad}\right)}^2}{\sqrt{{\left(2.6\right)}^2-1}}\\ =0.012692\end{array}$

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

  $\begin{array}{c}E\%=\frac{0.0129-0.0126}{0.0129}\times 100\\ =2.209\%\end{array}$

Conclusion:

Therefore, the lift and drag coefficients are $\underset{\_}{0.1454}$ and $\underset{\_}{0.012692}$ .

To determine

The lift and wave drag coefficients and comparison of accuracy.

Answer to Problem 12.1P

The lift and drag coefficients are $\underset{\_}{0.4363}$ and $\underset{\_}{0.1142}$ .

Explanation of Solution

Given:

The angle of attack is ${\alpha }_2=15°$ .

The Mach number is $M_{\infty }=2.6$ .

Formula used:

The expression for lift is given as,

  $C_l=\frac{4\alpha }{\sqrt{M_{\infty }^2-1}}$

The expression for drag is given as,

  $C_d=\frac{4{\alpha }^2}{\sqrt{M_{\infty }^2-1}}$

Calculation:

The coefficient of lift can be calculated as,

  $\begin{array}{c}{C}_l=\frac{4{\alpha }_2}{\sqrt{M_{\infty }^2-1}}\\ =\frac{4\times 15°\times \frac{\pi }{180°}\text{rad}}{\sqrt{{\left(2.6\right)}^2-1}}\\ =0.4363\end{array}$

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

  $\begin{array}{c}E\%=\frac{0.449-0.4363}{0.449}\times 100\\ =1.4\%\end{array}$

The coefficient of drag can be calculated as,

  $\begin{array}{c}{C}_d=\frac{4{\alpha }_2^2}{\sqrt{M_{\infty }^2-1}}\\ =\frac{4\times {\left(15°\times \frac{\pi }{180°}\text{rad}\right)}^2}{\sqrt{{\left(2.6\right)}^2-1}}\\ =0.1142\end{array}$

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

  $\begin{array}{c}E\%=\frac{0.117-0.1142}{0.117}\times 100\\ =2.39\%\end{array}$

Conclusion:

The lift and drag coefficients in first case are $\underset{\_}{0.4363}$ and $\underset{\_}{0.1142}$ .

To determine

The lift and wave drag coefficients and comparison of accuracy.

Answer to Problem 12.1P

The lift and drag coefficients are $\underset{\_}{0.8756}$ and $\underset{\_}{0.4600}$ .

Explanation of Solution

Given:

The angle of attack is ${\alpha }_3=30°$

The Mach number is $M_{\infty }=2.6$

Formula used:

The expression for lift is given as,

  $C_l=\frac{4\alpha }{\sqrt{M_{\infty }^2-1}}$

The expression for drag is given as,

  $C_d=\frac{4{\alpha }^2}{\sqrt{M_{\infty }^2-1}}$

Calculation:

The coefficient of lift can be calculated as,

  $\begin{array}{c}{C}_l=\frac{4{\alpha }_3}{\sqrt{M_{\infty }^2-1}}\\ =\frac{4\times 30°\times \frac{\pi }{180°}\text{rad}}{\sqrt{{\left(2.6\right)}^2-1}}\\ =0.8756\end{array}$

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

  $\begin{array}{c}E\%=\frac{1.21-0.8756}{1.21}\times 100\\ =27.63\%\end{array}$

The coefficient of drag can be calculated as,

  $\begin{array}{c}{C}_d=\frac{4{\alpha }_3^2}{\sqrt{M_{\infty }^2-1}}\\ =\frac{4\times {\left(30°\times \frac{\pi }{180°}\text{rad}\right)}^2}{\sqrt{{\left(2.6\right)}^2-1}}\\ =0.4600\end{array}$

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

  $\begin{array}{c}E\%=\frac{0.69-0.4600}{0.69}\times 100\\ =33.33\%\end{array}$

Conclusion:

Therefore, the lift and drag coefficients are $\underset{\_}{0.8756}$ and $\underset{\_}{0.4600}$ .