Page 98 - Basics of Fluid Mechanics and Introduction to Computational Fluid Dynamics

Dynamics of Inviscid Fluids 83

Aj
®; = In rpjdsj.
Qn
J
Obviously, the potential at P due to all the panels is the sum

Suppose now that P is the control point, that is the midpoint of the
i-th panel. Then we have

© (23, yi) = > a J nrsdsy

j=1 j
while the normal component of the velocity at (xj, y;) is

d
Un = ® (24, Yi) ’
dn,
n(n,;) being the outward unit normal vector to the 7-th panel. Because

for 7 = 1, rj; = 0 at the control point and, when the derivative is carried
out, rj; appears in the denominator (thus creating a singular point), it
would be useful to evaluate directly the contribution of the 2-th panel
to this derivative calculated at (z;,y;). Since it is about a source which
acts only on a half-circumference (the other half-circumference does not
interfere due to the rigid wall), its strength will be Mi and this is the
looked for contribution to the normal component of the velocity. Hence

= yp 3 | a (Inr;;) ds; .

ti i
Taking into account that the normal component of the free-stream
velocity Woo at the same point (2;,4;) 18 Yoon = Woo'Ni = Voo c08 Bi, 8;
being the angle between V,, and nj, the slip-condition will be oon +

Un = 0, which means

;
ttle fe (In rig) ds + Yoo Cos Bi = 0.
<
ay j
Applying this approach to all the panels, the above equalities with
a = 1,2,...,n, represent a linear algebraic system with n unknowns

Ai, A2,+--; An, Which can be solved by conventional numerical methods.