16100lectre14 cg
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This document presents the solution to determining the vortex strengths (Γ1, Γ2, Γ3) for a problem involving 3 point vortices. It sets up a system of equations relating the vortex strengths to the velocities at 3 control points, but notes this system has infinite solutions. It then explains that applying the Kutta condition, which sets the velocity around one point vortex to zero, allows determining a unique solution by setting Γ3 = 0 and solving the reduced system.
16100lectre14 cg
- 1. Solution • 0 = ⋅ n u at control pt #1: The velocity at control pt #1 is the sum of the freestream + 3 point vortices’ velocities at that point: 1 2 3 1 2 2 2 2 2 2 u V i i i j π π π ∞ Γ Γ Γ = + − + The normal at control pt #1 is: 1 1 2 1 1 0 2 2 2 2 n i u n V π π ∞ = − Γ Γ ⇒ ⋅ = − − + = Rearranging: 1 2 V π π ∞ Γ Γ − = − (1) • 0 = ⋅ n u at control pt #2: Now, following the same procedure for control pt #2: =10 miles V = 100 mph 8 Γ1 Γ2 Γ3 x y
- 2. Solution 16.100 2002 2 2 2 3 2 2 1 3 2 2 0 2 n i j V u n π π ∞ = + Γ Γ ⋅ = − + = 2 3 2 V π π ∞ Γ Γ − = (2) • 0 = ⋅ n u at control pt #3: 3 1 3 3 3 1 3 2 2 0 2 n i j V u n π π ∞ = − Γ Γ ⋅ = + − = 1 3 2 V π π ∞ Γ Γ ⇒ − + = (3) Final System of Equations Combine the numbered equations: 1 2 3 vortex strengths (unknowns) Influence matrix 1 1 0 1 1 0 2 1 1 0 2 i V n V V V π π π π π π ∞ ∞ ∞ ∞ − − Γ − Γ = Γ − i The problem with these equations is that they have infinitely many solutions. One clue is that the determinant of the matrix is zero. In particular we can add a constant strength to any solution because: 0 0 0 influence 0 matrix Γ Γ = Γ
- 3. Solution 16.100 2002 3 ⇒ Given a solution 0 0 0 Γ Γ Γ , then 1 0 2 0 3 0 Γ Γ Γ + Γ Γ Γ is also a solution where 0 Γ is arbitrary. So, how do we resolve this? Answer: the Kutta condition! 2 2 . . . . 1 1 2 2 t e upper t e lower p V p V ρ ρ + = + ⇒ 0 upper lower V V = ≠ What’s the Kutta condition for the windy city problem: Kutta: 3 0 Γ = ⇒ no flow around node 3! So, we can now solve our system of equations starting with 0 3 = Γ 1 2 3 2 2 0 V V π π ∞ ∞ Γ = − ⇒ Γ = Γ = V 8 Γ1 Γ2 Γ3