Question 10.1: Let us use the linear theory to calculate the lift coefficie......
Let us use the linear theory to calculate the lift coefficient, the wave-drag coefficient, and the pitching-moment coefficient for the airfoil section whose geometry is illustrated in Fig 10.5. For purposes of discussion, the flow field has been divided into numbered regions, which correspond to each of the facets of the double-wedge airfoil, as shown. In each region the flow properties are such that the static pressure and the Mach number are constant, although they differ from region to region. We seek the lift coefficient, the drag coefficient, and the pitching-moment coefficient per unit span of the airfoil given the free-stream flow conditions, the angle of attack, and the geometry of the airfoil neglecting the effect of the viscous boundary layer. The only forces acting on the airfoil are the pressure forces. Therefore, once we have determined the static pressure in each region, we can then integrate to find the resultant forces and moments.
Let us now evaluate the various geometric parameters required for the linearized theory:
\begin{aligned} & z_u(x)= \begin{cases}x \tan 10^{\circ} & \text { for } 0 \leq x \leq \frac{c}{2} \\ (c-x) \tan 10^{\circ} & \text { for } \frac{c}{2} \leq x \leq c\end{cases} \\ & z_l(x)= \begin{cases}-x \tan 10^{\circ} & \text { for } 0 \leq x \leq \frac{c}{2} \\ -(c-x) \tan 10^{\circ} & \text { for } \frac{c}{2} \leq x \leq c\end{cases} \end{aligned}
Furthermore,
\overline{\sigma_l^2}=\int_0^1 \sigma_l^2 d\left(\frac{x}{c}\right)=\delta_w^2
and
\overline{\sigma_u^2}=\int_0^1 \sigma_u^2 d\left(\frac{x}{c}\right)=\delta_w^2
We can use equation (10.8)
C_l=\frac{4 \alpha}{\sqrt{M_{\infty}^2-1}} (10.8)
to calculate the section lift coefficient for a 10^{\circ} angle of attack at M_{\infty}=2.0 :
C_l=\frac{4(10 \pi / 180)}{\sqrt{2^2-1}}=0.4031
Similarly, the drag coefficient can be calculated using equation (10.16):
C_d=\frac{d}{q_{\infty} c}=\frac{4 \alpha^2}{\sqrt{M_{\infty}^2-1}}+\frac{2}{\sqrt{M_{\infty}^2-1}}\left(\overline{\sigma_u^2}+\overline{\sigma_l^2}\right) (10.16)
C_d=\frac{4(10 \pi / 180)^2}{\sqrt{2^2-1}}+\frac{2}{\sqrt{2^2-1}}\left[\left(\frac{10}{57.296}\right)^2+\left(\frac{10}{57.296}\right)^2\right]
Therefore,
C_d=0.1407
The lift/drag ratio is
\frac{l}{d}=\frac{C_l}{C_d}=\frac{0.4031}{0.1407}=2.865
Note that in order to clearly illustrate the calculation procedures, this airfoil section is much thicker (i.e., t=0.176 c ) than typical supersonic airfoil sections for which t \approx 0.05 c (see Table 5.1). The result is a relatively low lift/drag ratio.
Similarly, equation (10.22)
C_{m_{x 0}}=\frac{-4 \alpha}{\sqrt{M_{\infty}^2-1}}\left(\frac{1}{2}-\frac{x_0}{c}\right)+\frac{4}{\sqrt{M_{\infty}^2-1}} \int_0^1 \frac{d z_c}{d x} \frac{x-x_0}{c} d\left(\frac{x}{c}\right) (10.22)
for the pitching moment coefficient gives
C_{m_{x 0}}=\frac{-4 \alpha}{\sqrt{M_{\infty}^2-1}}\left(\frac{1}{2}-\frac{x_0}{c}\right)
since the mean-camber coordinate x_c is everywhere zero. At midchord, we have
C_{m_{0.5 c}}=0
This is not a surprising result, since equation (10.22) indicates that the moment about the aerodynamic center of a symmetric (zero camber) thin airfoil in supersonic flow vanishes.