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).
Cp=M∞2−12θ+[2(M∞2−1)2(γ+1)M∞4−4M∞2+4]θ2 (10.23a)
or
Cp=C1θ+C2θ2 (10.23b)
Since panel 1 is parallel to the free stream, Cp1=0 as before. For panel 2 ,
Cp2=22−12(−20π/180)+2(22−1)2(1.4+1)(2)4−4(2)2+4(18020π)2=−0.4031+0.1787=−0.2244
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,
Cp3=0.4031+0.1787=0.5818
For panel 4,
Cp4=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:
Cl=(γ/2)p∞M∞2c∑pcosθ(0.5c/cosδw) (10.24a)
where δw 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
Cl=2cosδw1ΣCpcosθ (10.24b)
where the signs assigned to the Cp terms account for the direction of the force. Thus,
Cl=2cos10∘1(−Cp2cos20∘+Cp3cos20∘)=0.3846
Similarly, we can calculate the section wave-drag coefficient:
Cd=(γ/2)p∞M∞2c∑psinθ(0.5c/cosδw) (10.25a)
or
Cd=2cosδw1ΣCpsinθ (10.25b)
Applying this relation to the airfoil section of Fig 10.5, the section wave-drag coefficient for α=10∘ is
Cd=2cos10∘1(Cp3sin20∘−Cp2sin20∘)=0.1400
Let us now calculate the moment coefficient with respect to the midchord of the airfoil section (i.e., relative to x=0.5c ). 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,
Cm0.5c=21ρ∞U∞2c2m0.5c=(−p1+p2+p3−p4)21ρ∞U∞2c2c2/8+(p1−p2−p3+p4)21ρ∞U∞2c2(c2/8)tan2δw(10.26)
Note that, as usual, a nose-up pitching moment is considered positive. Also note that we have accounted for terms proportional to tan2δw. Since the pitching moment due to a uniform pressure acting on any closed surface is zero, equation (10.26) can be written as
Cm0.5c=(−Cp1+Cp2+Cp3−Cp4)81+(Cp1−Cp2−Cp3+Cp4)8tan2δw(10.27)
The reader is referred to equations (5.11)
M0=∬px dx dy (5.11)
through (5.15)
CM0=q∞ScM0 (5.15)
for a review of the technique. Thus,
Cm0.5c=0.04329