THE VORTEX PANEL METHOD y j m α V 4 3 2 panel 1 a) Approimate the contour of the airfoil by an inscribed polygon with m sides, called panels. Number the panels clockwise with panel #1 starting on the lower (pressure) side, at the trailing edge and panel #m ending on the trailing edge from the upper (suction) side. b) Assume that each panel represents a planar vorte sheet with linearly varying strength and such that the end strength of each panel is the same as the starting strength of the net panel: Eception: γ 1 γ m+1 γ(s j ) = γ j + s j (γ j+1 γ j ) γ j+1 γ j+2 j + 1 panel j panel γ(s j ) s j γ j Page 1 of 6
Instead of the dimensional strength γ (in velocity units), it is more convenient to use the dimensionless strength γ, defined as γ = γ 2πV Then, γ (s j ) = γ j + s j (γ j+1 γ j) c) The only unknowns of this problem are the end strengths γ j, j = 1, 2,..., m + 1. They can be found by solving m + 1 equations consisting of i) m-equations, epressing the no penetration condition at the m mid-points of the panels (called control points ) ii) one equation, epressing the Kutta condition. Once γ j are found, the velocity can be found by superposition and the pressure can be computed from Bernoulli s equation. d) Apply the Kutta condition at the trailing edge. Note that strict application of K-c would require that Instead, apply the weaker condition (V u ) t.e. = (V l ) t.e. = 0 (V u V l ) t.e. = 0 which implies that γ 1 + γ m+1 = 0 (A) e) Apply the no penetration condition at the m control points (midpanels). Notice that all panels are assumed to be parts of the same streamline. The equipotential lines are normal to streamlines and, therefore, to panels. Let φ be the velocity potential. Then, φ = V, which implies that V ni = φ Page 2 of 6
n i panel i where V ni is the velocity component normal to the panel i, and φ/ is the derivative of φ in the direction of the normal vector, n i. Therefore, the n-equation of no penetration becomes φ( i, y i ) = 0, i = 1, 2,..., m (B) f) The velocity potential φ( i, y i ) at a control point can be found by superposition of the potential due to the uniform stream and the potentials due to the m vorte panels, as φ( i, y i ) = V ( i cos α+y i sin α)+ m j=1 panel j γ(s j ) 2π tan y i y j i j ds j The remaining work is only algebraic and trigonometric manipulations. (C) (X i,y i ) θ i ( i,y i ) panel i (X i+1,y i+1 ) (X j+1,y j+1 ) ( j,y j ) θ j ds j r ij tan 1 yi y j i j panel j r ij = ( i j ) 2 +(y i y j ) 2 s j (X j,y j ) Page 3 of 6
g) Notice that i δ i δn i = sin θ i y i δy i δn i = cos θ i n i θ i δn θ δy ( i,y i ) δ Then, by chain rule, φ = φ i + φ y i = φ sin θ i + φ cos θ i i y i i y i In epression (C), everything has already been epressed in terms of i, y i, so it is straightforward to apply the chain rule to the various derivatives. Reminder Then i y i t tan 1 [f(t)] = [ tan 1 y ] i y j = i j [ tan 1 y ] i y j = i j 1 + 1 df 1 + f 2 (t) dt 1 + 1 ( ) 2 yi y j i j 1 ( ) 2 yi y j i j y i y j ( i j ) 2 1 i j h) The integral in eq. (C) is w.r.t. the j panel, therefore i, y i are constant during integration. Because j, y j, s j vary, it is necessary to epress all of them in terms of one variable, e.g. s j. j = X j s j cos θ j Page 4 of 6
y j = Y j + s j sin θ j Also γ(s j ) has been epressed in terms of s j. Then the integral for panel j becomes Sj Sj γ j 0 f 1j (s j )ds j + γ j+1 which can be computed numerically. 0 f 2j (s j )ds j i) Following various manipulations (see tetbook for some details) one gets the system of m + 1 linear, algebraic equations Anij γ j = sin(θ i α), i = 1, 2,..., m γ 1 + γ m+1 = 0 where the constants A nij depend only on the coordinates of the end points of panels i and j and can be computed immediately once the panels have been selected. Notice that the effect of panel i on itself has also been taken into account. The system of the above equations is solved numerically to provide the end strengths. γ j, j = 1, 2,..., m + 1 j) The velocity at the control points is tangential to the panel and can be found as V i = V ti = φ = φ i + φ y i t i i t i y i t i or V i = φ i cos θ i + φ y i sin θ i n i δ θ δy θ i ( i,y i ) t i δ t Page 5 of 6
Instead of the dimensional velocity V i, the program computes a dimensionless velocity as i = V i = 1 φ V V t i Following various manipulations, this is found as i = m+1 j=1 = A tij γ j + cos(θ i α), i = 1, 2,..., m where the coefficients A tij j alone. depend on the end coordinates of panels i and k) The pressure at the control points can be found from Bernoulli s eq. as C Pi = P i P 1 2 ρv 2 = 1 2 i l) The lift can be found in two ways: i) By integrating all vorte strengths and using the Kutta-Joukowsky theorem. ii) By integrating the surface pressure. Page 6 of 6