Displacement Thickness The displacement thickness δ * is defined as: ∞
∞ ρ u u = − δ = ∫ 1 − d y 1 dy ∫ ρ u u e e 0 0 e *
c om om pr pr es ess ib ibl e flow
in co co mp mp re re ss ssi bl bl e flow
The displacement thickness has at least two useful interpretations: Interpretation #1 ∞
A
=∫ 0
y
u ue
dy ∞
u / (y) ue
A+B
(1)dy = ∫ (1) 0
A
B
u / ue
So, the difference is in area B.
⇒ δ * “represents” the decrease in mass flow due to viscous effects, i.e. lost visc = ρ e u e δ * m
Integral Boundary Layer Equations
Interpretation #2 am l in s t r e
e
∆y
ue
u(y)
y
y1
δ*(x)
x
Conservation of mass: y1 +∆ y
y1
∫ u dy = ∫ e
0
udy
0
y1
y1
∫ u dy = ∫ udy + ∆yu e
0
e
0 y1
⇒ ∆ yue = ∫ ( ue − u)dy 0
u ∆ y = ∫ 1 − dy ue 0 y1
Taking the limit of y1 → ∞ gives ∞
u ⇒ ∆ y = δ = ∫ 1 − dy ue 0 *
So, the external streamline is displaced by a distance δ * away from the body due to viscous effects.
⇒ Outer flow sees an “effective body”
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Integral Boundary Layer Equations
Karman’s Integral Momentum Equation This approach due to Karman leads to a useful approximate solution technique for boundary layer effects. It forms the basis of the boundary layer methods utilized in Prof. Drela’s XFOIL code. Basic idea: integrate b.l. equations in y to reduce to an ODE in x .
y
ue(x)
δ*(x)
x
Derivation:
− momentum
Add ( ρ u ) x continuity +
∂u ∂v ∂u due ∂u ∂2u ⇒ ρ u + + ρ u + v = ρ ue + µ 2 ∂ ∂ ∂ ∂ ∂y x y x y dx ( ρ u )×continuity
x − momentum
∂ (u 2 ) ∂ ∂ ∂u du ⇒ ρ + (uv) = ρ ue e + µ dx ∂y ∂y ∂ y ∂ τ
Now, we integrate from 0 to y1 : y1
∂(u 2 ) du y y ρ ∫ dy + ρ uv 0 = ρ ue e y1 + τ 0 ∂ x dx 0 1
1
Note: y1
ρ uv 0
y1
y1
∂v ∂u = ρ ue v( y1 ) = ρ ue ∫ dy = − ρ ue ∫ dy ∂ y ∂x 0 0
So, the equation becomes:
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Integral Boundary Layer Equations
y1
y1
∂(u 2 ) ∂u du y ρ ∫ dy − ρ ue ∫ dy = ρ ue e y1 + τ 0 ∂ x ∂x dx 0 0
1
After a little more manipulation this can be turned into (note we let y1 → ∞ also): τw =
d
2 * du ( ρ ueθ ) + ρ ueδ e dx dx
(1)
∞
ρ u
ρ u
ρ eue
∫ ρ u
where θ ≡ momentum thickness =
e e
0
∞
incompressible form =
u
u
ue
∫u o
1 −
e
1 −
dy
dy
Insight Integrate (1) from stagnation point along airfoil & then down the wake
y x x=0 at stag. point
∞
∫τ 0
But:
x 8 along wake
w
∞
* du dx = ( ρ ue θ ) + ρ ueδ e dx 0 dx 0 2
∞
∫
u e = 0 at stag. pt. ( x = 0) & −
dp
= ρ ue
due
dx dx Bernoulli
∞
∞
⇒ ρ ue2θ x = ∫ τ w dx + ∫ δ * →∞
0
drag (see Anderson Sec 2.6 for proof)
∞
∞
∫
∫
0
0
D′ = τ w dx + δ * N
friction drag
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0
dp dx
dp dx
dx
dx
form drag
4
Integral Boundary Layer Equations
Another common form of the integral momentum equation is derived below: τ w = τw 2 e e
