Chapter 14, Problem 14.1P

Repeat Problem 9.13 using

a. Newtonian theory

b. Modified newtonian theory

Compare these results with those obtained from exact shock-expansion theory (Problem 9.13). From this comparison, what comments can you make about the accuracy of newtonian and modified newtonian theories at low supersonic Mach numbers?

To determine

The value of lift and wave-drag coefficients using Newtonian theory.

The comparison between the results obtain from exact shock-expansion theory.

Answer to Problem 14.1P

The lift and wave-drag coefficient at $\alpha =5°$ is $C_L=0.01513$ and $\text{}{C}_D=0.001323$ respectively using Newtonian theory.

The lift and wave-drag coefficient at $\alpha =15°$ is $C_L=0.123$ and $C_D=0.0346$ respectively using Newtonian theory.

The lift and wave-drag coefficient at $\alpha =30°$ is $C_L=0.433$ and $C_D=0.25$respectively using Newtonian theory.

Explanation of Solution

Given:

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

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

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

Formula used:

The expression for the lift coefficient is given as,

  $C_L=\left(C_{P_3}-C_{P_2}\right)\cos \alpha$ .

The expression for the wave-drag coefficient is given as,

  $C_D=\left(C_{P_3}-C_{P_2}\right)\sin \alpha$

Calculation:

The value of lift and wave-drag coefficient at $\alpha =5°$ can be calculated as,

  $\begin{array}{c}{C}_{P_3}=2{\sin }^{2}5°\\ =0.01519\end{array}$

The lift coefficient can be calculated as,

  $\begin{array}{c}{C}_L=\left(C_{P_3}-C_{P_2}\right)\cos \alpha \\ =0.01513\end{array}$

The wave-drag coefficient can be calculated as,

  $\begin{array}{c}{C}_D=\left(C_{P_3}-C_{P_2}\right)\sin \alpha \\ =0.001323\end{array}$

The value of lift and wave-drag coefficient at $\alpha =15°$ can be calculated as,

  $\begin{array}{c}{C}_{P_3}=2{\sin }^{2}15°\\ =0.1339\end{array}$

The lift coefficient can be calculated as,

  $\begin{array}{c}{C}_L=\left(C_{P_3}-C_{P_2}\right)\cos \alpha \\ =0.129\end{array}$

The wave-drag coefficient can be calculated as,

  $\begin{array}{c}{C}_D=\left(C_{P_3}-C_{P_2}\right)\sin \alpha \\ =0.0346\end{array}$

The value of lift and wave-drag coefficient at $\alpha =30°$ can be calculated as,

  $\begin{array}{c}{C}_{P_3}=2{\sin }^{2}30°\\ =0.5\end{array}$

The lift coefficient can be calculated as,

  $\begin{array}{c}{C}_L=\left(C_{P_3}-C_{P_2}\right)\cos \alpha \\ =0.433\end{array}$

The wave drag coefficient can be calculated as,

  $\begin{array}{c}{C}_D=\left(C_{P_3}-C_{P_2}\right)\sin \alpha \\ =0.25\end{array}$

Conclusion:

Therefore, the lift and wave-drag coefficient at $\alpha =5°$ is $C_L=0.01513$ and $\text{}{C}_D=0.001323$ respectively using Newtonian theory.

Therefore, the lift and wave-drag coefficient at $\alpha =15°$ is $C_L=0.123$ and $C_D=0.0346$ respectively using Newtonian theory.

Therefore, the lift and wave-drag coefficient at $\alpha =30°$ is $C_L=0.433$ and $C_D=0.25$ respectively using Newtonian theory.

To determine

The value of lift and wave-drag coefficients using modified Newtonian theory.

The comparison between the results obtain from exact shock-expansion theory.

Answer to Problem 14.1P

The lift and wave-drag coefficient at $\alpha =5°$ is $C_L=0.0302$ and $C_D=0.002648$ respectively using modified Newtonian theory

The lift and wave-drag coefficient at $\alpha =15°$ is $C_L=0.25$ and $C_D=0.069$ respectively using modified Newtonian theory

The lift and wave-drag coefficient at $\alpha =30°$ is $C_L=0.86$ and $C_D=0.5$ respectively using modified Newtonian theory

Explanation of Solution

Given:

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

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

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

Formula used:

The expression for the lift coefficient is given as,

  $C_L=\left(C_{P_3}-C_{P_2}\right)\cos \alpha$

The expression for the wave-drag coefficient is given as,

  $C_D=\left(C_{P_3}-C_{P_2}\right)\sin \alpha$

Calculation:

The value of lift and wave-drag coefficient at $\alpha =5°$ can be calculated as,

  $\begin{array}{c}{C}_{P_3}=4{\sin }^{2}5°\\ =0.03\end{array}$

The lift coefficient can be calculated as,

  $\begin{array}{c}{C}_L=0.03\cos 5°\\ =0.0302\end{array}$

The wave drag coefficient can be calculated as,

  $\begin{array}{c}{C}_D=0.03\sin 5°\\ =0.002648\end{array}$

The value of lift and wave-drag coefficient at $\alpha =15°$ can be calculated as,

  $\begin{array}{c}{C}_{P_3}=4{\sin }^{2}15°\\ =0.26\end{array}$

The lift coefficient can be calculated as,

  $\begin{array}{c}{C}_L=0.26\cos 15°\\ =0.25\end{array}$

The wave drag coefficient can be calculated as,

  $\begin{array}{c}{C}_D=0.26\sin 15\\ =0.069\end{array}$

The value of lift and wave-drag coefficient at $\alpha =30°$ can be calculated as,

  $\begin{array}{c}{C}_{P_3}=4{\sin }^{2}30°\\ =1\end{array}$

The lift coefficient can be calculated as,

  $\begin{array}{c}{C}_L=1\cos \alpha \\ =0.86\end{array}$

The wave drag coefficient can be calculated as,

  $\begin{array}{c}{C}_D=1\sin \alpha \\ =0.5\end{array}$

On comparing values obtained using Newtonian and modified Newtonian theory with shock expansion theory (prob 9.13), we can see that the error % goes up to 90% for $\alpha =5°$ , 70% for $\alpha =15°$ and 60% for $\alpha =30°$ . This clarifies that at law supersonic mach number, Newtonian theory and modified Newtonian theory gives poor results.

Conclusion:

Therefore, the lift and wave-drag coefficient at $\alpha =5°$ is $C_L=0.0302$ and $C_D=0.002648$ respectively using modified Newtonian theory

Therefore, the lift and wave-drag coefficient at $\alpha =15°$ is $C_L=0.25$ and $C_D=0.069$ respectively using modified Newtonian theory.

Therefore, the lift and wave-drag coefficient at $\alpha =30°$ is $C_L=0.86$ and $C_D=0.5$ respectively using modified Newtonian theory.