LIFT AND THE STRENGTH OF DOUBLET IN ACCELERATION POTENTIAL

In the linearized thin-airfoil theory, the discontinuity of the pressure field across the airfoil can be represented by a layer of doublets, the strength of which varies with time. In order to find the relation between
the lift distribution and the oscillation mode, the lift force and the induced downwash velocity corresponding to a given distribution of doublets must be known. These relations will be derived in the present and the next sections.

First let us remark that, in the airfoil theory, the doublets are distributed over a surface (a line segment along the x axis in the two-dimensional theory). Consider, then, a line distribution of “doublets” along the x axis from x = — 1 to x = + 1. Let B(xa) be the strength per unit length of the doublet distribution at a point (x = x0, у = 0) on the airfoil. The total acceleration potential is then

ф(х, у, t) = B(x0) sin в dx0 (1)

where

Г = ]Pa ~ Xo)2 + ^ <2)

Let us evaluate the local lift distribution acting on a doublet element of length 2e situated at x = £, є being a small number compared with 1. Let L(x, t) be the lift per unit length acting bn the doublet layer at (ж, 0). Then the lift force on the element concerned is 2еЩ, t). Referring to Fig. 14.1,

Fig. 14.1. The region of integration.

let us take a curve C which is a narrow rectangle of height 2d and length 2e. According to Newton’s law, the rate of change of momentum of the fluid enclosed in C is equal to the external force acting on it. The latter consists of the pressure force acting on the boundaries of C and the lift force exerted by the doublet layer. Now, when the strength B(x0) is so chosen as to make the resulting flow field correspond to a physical prob­lem, the acceleration potential ф(х, у, t) must satisfy the kinematic bound­ary conditions on velocity and acceleration. In particular, the vertical acceleration 5ф/ду must be finite over the airfoil. Therefore, the rate of

(5)

(6)

6->0

The limit d -> 0 can be taken inside of the integral. Now

lim ф(х, (5, t) = lim eiwl Г B(x0) sin в dx0

6-е 0 6-+0 J-1

( ҐХ-В ҐХ + С Л1

= lim eiwt j + + (7)

d—+0 J — 1 Jx — € Jx+e)

When x — a:0| > e > 0, Нф2)(о>г’) is finite, but sin в tends to zero as <5 -» 0. Hence, the first and the last integrals in the bracelets of Eq. 7 vanish in the limit as long as d є. Applying the mean-value theorem to the middle integral in the bracelets of Eq. 7, we obtain

sin в H^wr’) dxn (8)

x—B

where 1 is some number between integral, notice that

Нф2)(г) = — — – z log z + (power series in г)

and

Substituting this result into Eq. 6, we obtain

ff+c 2i82a

2єЩ, t) = 2p0 е“(г+аЛ£) B(x +Xe)dx (- 1 < Я < 1)

«/| — e Cl)

Since Д(а; + Хє) is a continuous function, we obtain, in passing to the limit г -> 0

Щ, t) = eiat В(ф) (12)

CO

This formula relates the local strength of a continuously distributed doublet layer to the local lift force per unit length. It is seen that they are directly proportional.

The horizontal force acting on the distributed doublets with vertical axes is zero.