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′
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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.
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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
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