Question 10.2: Let us calculate the pressure coefficient on each panel of t......
Let us calculate the pressure coefficient on each panel of the airfoil in Example 10.1 using Busemann’s theory. We will use equation (10.23).
C_p=\frac{2 \theta}{\sqrt{M_{\infty}^2-1}}+\left[\frac{(\gamma+1) M_{\infty}^4-4 M_{\infty}^2+4}{2\left(M_{\infty}^2-1\right)^2}\right] \theta^2 (10.23a)
or
C_p=C_1 \theta+C_2 \theta^2 (10.23b)
Since panel 1 is parallel to the free stream, C_{p 1}=0 as before. For panel 2 ,
\begin{aligned} C_{p 2} & =\frac{2(-20 \pi / 180)}{\sqrt{2^2-1}}+\frac{(1.4+1)(2)^4-4(2)^2+4}{2\left(2^2-1\right)^2}\left(\frac{20 \pi}{180}\right)^2 \\ & =-0.4031+0.1787 \\ & =-0.2244 \end{aligned}
As noted, the airfoil in this sample problem is relatively thick, and therefore the turning angles are quite large. As a result, the differences between linear theory and higher-order approximations are significant but not unexpected. For panel 3,
\begin{aligned} C_{p 3} & =0.4031+0.1787 \\ & =0.5818 \end{aligned}
For panel 4,
C_{p 4}=0
since the flow along surface 4 is parallel to the free stream.
Having determined the pressures acting on the individual facets of the double-wedge airfoil, let us now determine the section lift coefficient:
C_{l}=\frac{\sum p \cos \theta\left(0.5 c / \cos \delta_{w}\right)}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c} (10.24a)
where \delta_{w} is the half-angle of the double-wedge configuration. We can use the fact that the net force in any direction due to a constant pressure acting on a closed surface is zero to get
C_{l}=\frac{1}{2 \cos \delta_{w}} \Sigma C_{p} \cos \theta (10.24b)
where the signs assigned to the C_{p} terms account for the direction of the force. Thus,
\begin{aligned} C_{l} & =\frac{1}{2 \cos 10^{\circ}}\left(-C_{p 2} \cos 20^{\circ}+C_{p 3} \cos 20^{\circ}\right) \\ & =0.3846 \end{aligned}
Similarly, we can calculate the section wave-drag coefficient:
C_{d}=\frac{\sum p \sin \theta\left(0.5 c / \cos \delta_{w}\right)}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c} (10.25a)
or
C_{d}=\frac{1}{2 \cos \delta_{w}} \Sigma C_{p} \sin \theta (10.25b)
Applying this relation to the airfoil section of Fig 10.5, the section wave-drag coefficient for \alpha=10^{\circ} is
\begin{aligned} C_{d} & =\frac{1}{2 \cos 10^{\circ}}\left(C_{p 3} \sin 20^{\circ}-C_{p 2} \sin 20^{\circ}\right) \\ & =0.1400 \end{aligned}
Let us now calculate the moment coefficient with respect to the midchord of the airfoil section (i.e., relative to x=0.5 c ). As we have seen, the theoretical solutions for linearized flow show that the midchord point is the aerodynamic center for a thin airfoil in a supersonic flow. Since the pressure is constant on each of the facets of the double-wedge airfoil of Fig. 10.5 (i.e., in each numbered region), the force acting on a given facet will be normal to the surface and will act at the midpoint of the panel. Thus,
\begin{aligned} C_{m_{0.5 c}}=\frac{m_{0.5 c}}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c^{2}}=\left(-p_{1}\right. & \left.+p_{2}+p_{3}-p_{4}\right) \frac{c^{2} / 8}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c^{2}} \\ & +\left(p_{1}-p_{2}-p_{3}+p_{4}\right) \frac{\left(c^{2} / 8\right) \tan ^{2} \delta_{w}}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c^{2}} & (10.26) \end{aligned}
Note that, as usual, a nose-up pitching moment is considered positive. Also note that we have accounted for terms proportional to \tan ^{2} \delta_{w}. Since the pitching moment due to a uniform pressure acting on any closed surface is zero, equation (10.26) can be written as
\begin{aligned} C_{m_{0.5 c}}=\left(-C_{p 1}+C_{p 2}\right. & \left.+C_{p 3}-C_{p 4}\right) \frac{1}{8} \\ & +\left(C_{p 1}-C_{p 2}-C_{p 3}+C_{p 4}\right) \frac{\tan ^{2} \delta_{w}}{8} & (10.27) \end{aligned}
The reader is referred to equations (5.11)
M_0=\oiint p x d x d y (5.11)
through (5.15)
C_{M_0}=\frac{M_0}{q_{\infty} S c} (5.15)
for a review of the technique. Thus,
C_{m_{0.5 c}}=0.04329
