The Indirect Problem (Case of Equal Potentials): P-G Transformation – Version II

In the indirect problem, the requirement is to find a transformation, for the profile, by which we can obtain a body in incompressible flow with exactly the same pressure distribution, as in the compressible flow.

For two-dimensional or planar bodies, the pressure coefficient Cp is given by Equation (9.73a) as:

C„ = —2 .

p

and the perturbation velocity component, u, is:

дф

dx

But in this case, Cp = CPlnc; therefore, from the above expressions for Cp and u, we have:

Cp Cpinc, U uinc, ф ф1пе.

For this situation the transformation Equation (9.79) gives:

К = 1. (9.88)

From Equation (9.83b) with K2 = 1, we get:

dz dx	dzin dxin
(9.89)

Equation (9.89) is the relation between the geometries of the actual profile in compressible flow and the transformed profile in the incompressible flow, resulting in same pressure distribution around them.

From Equation (9.89), we see that in a compressible flow, the body must be thinner by the factor ■Jl — M^ than the body in incompressible flow as shown in Figure 9.11. Also, the angle of attack in compressible flow must be smaller by the same factor than in incompressible flow.

From the above relation for Cp, we have:

Cp CL

Cpinc CLinc

a

V cP

V cP

(b)

That is, the lift coefficient and pitching moment coefficient are also the same in both the incompressible and compressible flows. But, because of decreased a in compressible flow:

dCL 1 (dCL}

da 1 – M2 da lnc’

This is so because of the fact that the disturbances introduced in a compressible flow are larger than those in an incompressible flow and, therefore, we must reduce a and the geometry by that amount (the difference in the magnitude of disturbance in a compressible and an incompressible flow). In other words, because of Equation (9.79) (z = K1 zlnc), every dimension in the z-direction must be reduced and so the angle of attack a should also be transformed.