Forces

The force contribution per unit span is given by

d~F’ = -2pU2 (c-(x)(rtdl)- + C+ (x)(tdl}+) (4.43)

where (itdl) = (dz, – dx) and (itdl)+= (-dz+, dx), see Fig. 4.7.

The lift corresponds to the z-component of the force, i. e. dL’ = 1 pU2 (c~ (x) – C+ (x}) dx ^ dCi = (c~ (x) – C+ (x)) у (4.44)

With the help of Ackeret formulae, one finds

2 c, !, dx a

Ci = (a – f -(x) + f + (x) – a)— = 4- (4.45)

P 0 c P

where we accounted for f ±(0) = f ±(c) = 0. The result is remarkable in that, contrary to what happens in subsonic flow, camber does not contribute to lift.

The drag is the x – component with

dD’ = -2pU2 (є-(x)dz – – C+ (x)dz+) ^

dz, dz+ dx

dCd = – C-(x)— – C + (x)— – (4.46)

Again, using Ackeret formulae, one can evaluate an inviscid drag which is called the “wave drag” with coefficient given by

Cd = 2 ^ ((f ‘-(x) – a)2 + (f’+ (x) – a)2) ^2

= 4 + (f -(x))2 + (f +(x))2 (4.47)

в в 0 c

Let (Cd)a=0 be the integral term in the above equation. Introducing camber and thickness, it can be rewritten as

(4.48)

The result shows that in supersonic flow, camber, thickness and incidence, all contribute to wave drag.

Application:

2

• drag coefficient for a flat plate: Cd = 4 5

• drag coefficient for a biconvex profile of thickness |: Cd = 45+35 (§)2

• drag coefficient for a parabolic plate of camber d: Cd = 4 5+35 (4- )2

Note, that in linearized supersonic flow theory, the flow at the leading edge of the profile is still governed by a wave-type hyperbolic equation and the upper and lower surfaces do not communicate. Hence, the flow does not go around the leading edge as it does in subsonic flow, and there does not exist a suction force.