Trefftz Plane Analysis of Induced Drag
Consider an inviscid, incompressible potential flow around a body (say a wing).
We define a control volume surrounding the body as follows
Trefftz plane
ST (part of S )
body
z
Sbody
y
8
S
8
x, V
Upstream flow is V∞ and is in x − direction. Thus, drag is the force in
x − direction. Apply integral momentum in x to find induced drag.
KK K
K
∫∫ ρuu ⋅ ndS = − ∫∫ pndS
S body + S ∞
S body + S ∞
K K
First, on the body u ⋅ n = 0 , so:
KK K
K
∫∫ ρuu ⋅ ndS = − ∫∫ pndS
S body + S ∞
S∞
Next, also on the body,
−
K
∫∫ pndS =
force of body acting on fluid
S body
We are interested in the exact opposite, i.e. the force acting on the body. In x ,
this is the drag, in z this is the lift, and in y this is a yaw or side force:
Trefftz Plane Analysis of Induced Drag
⇒
−
K
K K
K
p
n
dS
D
i
Y
j
L
k
=
−
−
−
∫∫
S body
⇒
K
K K
K
KK K
Di + Yj + Lk = − ∫∫ pndS − ∫∫ ρ u u ⋅ ndS
S∞
S∞
Now, let’s pull out the drag:
K K
KK K
D = − ∫∫ pn ⋅ i dS − ∫∫ ρu u ⋅ n dS
S∞
S∞
The next piece is to apply Bernoulli to eliminate the pressure:
p = p∞ +
1
1
ρV∞2 − ρ (u 2 + v 2 + w 2 )
2
2
⇒
1
1
K K

K K
D = − ∫∫  p ∞ + ρV∞2 − ρ (u 2 + v 2 + w 2 )n ⋅ i dS − ∫∫ ρu u ⋅ n dS
2
2

S∞ 
S∞
But,
∫∫ ( p∞ +
S∞
1
1
K K
ρV∞2 )n ⋅ idS = ( p∞ + ρV∞2 )
2
2
K K
n
∫∫ ⋅ idS
S∞
= 0 for a closed
surface
1
K K
K K
D = ∫∫ ρ (u 2 + v 2 + w 2 )n ⋅ i dS − ∫∫ ρu u ⋅ n dS
2
S∞
S∞
Next, we divide the velocity into a freestream and a perturbation:
u = V∞ + uˆ
v=
vˆ
w=
wˆ
where uˆ, vˆ, wˆ are perturbation velocities (not necessarily small).
Substitution gives:
D=
1
K K
K K
ρ ∫∫ (V∞2 + 2V∞uˆ + uˆ 2 + vˆ 2 + wˆ 2 )n ⋅ idS − ρ ∫∫ (V∞ + uˆ )u ⋅ ndS
2 S∞
S
∞
But, we note that
K K
K K
ρ ∫∫V∞ u ⋅ n dS = ρV∞ ∫∫ u ⋅ n dS = 0 from conservation of mass
S∞
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S∞
2
Trefftz Plane Analysis of Induced Drag
⇒
1
K K
K K
K K
D = ρV∞ ∫∫ uˆ n ⋅ i dS + ρ ∫∫ (uˆ 2 + vˆ 2 + wˆ 2 )n ⋅ i dS − ρ ∫∫ uˆ u ⋅ n dS
2 S∞
S∞
S∞
If we take the control volume boundary far away from the wing, then the velocity
perturbations go to zero except downstream. Downstream the presence of
trailing vortices will create non-zero perturbations (more on this in a bit).
So, uˆ, vˆ, wˆ → 0 except on S T .
⇒
D = ρV∞
1
ˆ + ρ ∫∫ (uˆ
∫∫ udS
2
ST
⇒
D=
2
ST
+ vˆ 2 + wˆ 2 )dS − ρ ∫∫ uˆ ( V∞ + uˆ)dS
ST
1
ρ ∫∫ (vˆ 2 + wˆ 2 − uˆ 2 )dS
2 ST
The final step is to note that far downstream the x − velocity perturbation must die
away (in inviscid flow). The reason is that the trailing vortices, which far
downstream must be in the x − direction, cannot induce an x − component of
velocity.
So, this brings us to the final answer
D=
1
ρ ∫∫ (vˆ 2 + wˆ 2 )dS
2 ST
In other words, the induced drag is the kinetic energy which is transferred into the
crossflow (i.e. the trailing vortices)!
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