University:
Massachusetts Institute of Technology
Course:
16.100 | Aerodynamics
Academic year:

2020
Views:

498

Pages:

3
Author:

izgledaoyp

Kutta Condition Thought Experiment1 Suppose we model the flow around an airfoil using a potential flow approach. L' 8 V 8 p Question: Answer: We know the following: u = ∇φ L′ = ρV∞ Γ D′ = 0 ω=0 u in = 0 Bernoulli applies How many potential flow solutions are possible? Infinitely many! For example: Flow #1 Both of these flows have circulation which are not all equal Γ1 ≠ Γ 2 ⇒ L1′ ≠ L2′ 1 Anderson, Sec. 4.5 Flow #2 Kutta Condition Another difference can be observed at the trailing edge: Flow #1 Flow leaves t.e. smoothly. Velocity is not infinite. Flow #2 Flow turning around a sharp corner has infinite velocity at corner for potential flow. As a result of this and the physical evidence, Kutta hypothesized: In a physical flow (i.e. having viscous effects), the flow will smoothly leave a sharp trailing edge. -Kutta Condition ⇒ Flow #1 is physically correct! Let’s look at Flow #1 a little more closely: Finite angle T.E. (α te > 0) Vlower αt.e. T.E. Vupper Upper and lower surface velocities must still be tangent to their respective surfaces. This implies 2 different velocities at TE. 16.100 2002 2 Kutta Condition Only realistic option: Vlower = Vupper = 0 for finite angle T.E. Note: from Bernoulli, this implies pt .e. = p∞ + 1 1 ρV∞2 − ρ Vt .2e. 2 2 =0 1 ρV∞2 2 ⇒ TE is a stagnation point with pt .e. ≡ total pressure pt .e. = p∞ + Cusped TE (α te = 0) Vupper T.E. Vlower In this case, velocities from upper and lower surface are aligned. In order for the pressure at the TE to be unique: Vupper = Vlower 16.100 2002 3