The problem of an oscillating airfoil in a two-dimensional supersonic flow can be solved in several different ways. In this section the Laplace – transformation method, in a form due to Stewartson,14 40 will be used. The theory will be limited to the linearized case, so that the airfoil must be infinitesimally thin, and executing harmonic oscillations of small ampli­tudes. The principle of superposition holds. It is sufficient to consider airfoils of zero thickness and zero camber, with stationary mean position. The fluid moves over it with an undisturbed velocity U at infinity. The x axis is taken in the direction of the free stream, and the origin of co­ordinates is taken at the leading edge of the airfoil. In the first-order theory the wing may be assumed to lie in the plane у = 0. The coordinate z, in the spanwise direction, does not appear in the problem. The flow is assumed to be irrotational, with a velocity potential Ux + Ф, and deviations in velocity components, pressure, and density are so small that squares and products of these deviations may be neglected in comparison with the first-order terms.

The equation of the velocity potential, referred to a frame of reference at rest relative to the fluid at infinity, is (§ 12.5):

УФ Э*Ф__1Э*Ф_

За;2 + dy2 a2 dt2 ~ ° ^

where a is the velocity of sound, which, in our order of approximation, is a constant. Transforming to axes moving with speed U in the negative x direction, so that x is replaced by x + Ut, Eq. 1 becomes (Eq. 2, p. 418):

The increment of pressure at any point due to the disturbance is given by the Eulerian equations of motion

3 u, 3tii 1 3

37 + Щ 5— = — – -— p

at oXj p dx{

and linearizing the result, one obtains

p0 being the density of the fluid at infinity. Hence a determination of Ф on the airfoil is sufficient to determine the pressure acting on it.

When the wing executes ;

a simple-harmonic motion, the time

t enters

as an exponential factor eimt

. Let

Then Eq. 2 becomes

Ф = Т(ж, у)еш


/32Y 32VF

32Y т 3’F

, = ^ + 2*»^ + (|юГР



where M is the Mach number Ufa and

p. = M2 – (7)

The boundary conditions must be formulated according to the following considerations:

1. The disturbances created at the leading edge propagate along a wedge which is called the Mach wedge. In front of the Mach wedge the disturbances cannot be felt, and the flow is uniform relative to the wing. Hence, one may put Ф = 0 for x < 0.

2. Inside the Mach wedge, the velocity of flow normal to the airfoil must conform to the actual motion of the airfoil. If the equation of the airfoil surface is specified by

У = Y(x, t) (8)

then the normal velocity of the flow on the airfoil must satisfy the following equation (§ 13.2):

Эф_ЭГ ЭГ Ъу Э t dx


In a simple-harmonic motion,

Y(x, t) == Z(x)eM


we have, on the airfoil (part of the plane у = 0),

ЭТ „ TdZ — — icoZ + І7 — dy dX


On the rest of the plane у = 0, the pressure must be continuous.

Equation 6 with the boundary condition 11 may be solved by Laplace transformation. Define

V = JSf’f’F} = JV8*¥(*) dx (12)

Then f satisfies

since Y and Э’Г/Эж vanish when x = becomes

where g(s) is the Laplace transform of the right-hand side of Eq. 11. Thus we require a solution of Eq. 13 so that -> 0 as y -> oo and such that Eq. 14 is satisfied. Let

y2 = /SV + 2Ms — + (—)2 (15)

a a }

The general solution of Eq. 13, if we take y. to be the branch on the right half plane, i. e., with Щу > 0, is

f = Ae-m + Bem

The constants A and В are determined by the boundary conditions. It is necessary to distinguish the solution on the upper and lower half-space. For the upper half-space (y > 0), the condition f -> 0 as у -> oo requires 5 = 0; and the condition 14 requires A — — g(s)/y. Similarly the solution for у < 0 can be determined. Hence,

where the symbol sgn у indicates a sign to be taken as positive on the positive side (y > 0) of у — 0 plane, and as negative on the negative side.

It is now necessary to find the inverse transform of Eq. 18. The inverse transform of g(s) is (Э’Г/Эу)г/=0 and is given by Eq. 11. From Table 10.1, we find

Пи™)} = – t=L= (19)

im MV

s + 7 p*j + ap

iMm(x — І)



The pressure change on the airfoil is, according to Eq. 3,

Since the pressure changes on the upper and lower surfaces of the airfoil are equal and opposite

Note that the integral is a function of the parameters M and Cl, or, alternatively, M and k.

The total lift on the airfoils is

L = 2bj^(x’) dx’ (32)

The moment (positive nose-up) on the airfoil about the leading edge is

The integration of Eqs. 32 and 33 has been discussed by von Borbely,14-29 Schwarz,14 39 Garrick and Rubinow.14-32

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