Quick Visit to Bernoulli Land
Although we have seen the Bernoulli equation and seen it derived before, this
next note shows its derivation for an uncompressible & inviscid flow. The
derivation follows that of Kuethe &Chow most closely (I like it better than
Anderson).1
Start from inviscid, incompressible momentum equation
∂u
1
+ u ⋅ ∇u = − ∇ p
∂t
ρ
There is a vector calculus identity:
1 2
u • ∇u = ∇  u  − u × ( ∇ × u )
2

ω ,vorticity
⇒
∂u
1 2 1
+ ∇  u  + ∇p = u × ω
∂t
2
 ρ
From here, we can make the final re-arrangement:
1
∂u
2

∇  p + ρ u  = ρu × ω − ρ
2
∂t


Two common applications:
1. Steady irrotational flow
∂u
ω=0
=0
∂t
Irrotational
Steady
1
2

⇒ ∇ p + ρ u  = 0
2


⇒ p+
1
1
2
ρ u = const. for entire flow
2
Kuethe and Chow, 5th Ed. Sec 3.3-3.5
Quick Visit to Bernoulli Land
2. Steady but rotational flow
∂u
ω≠0
=0
∂t
Rotational
Steady
1
2

⇒ ∇  p + ρ u  = ρu × ω
2


This is a vector equation. If we dot product this into the streamwise
direction:
u
s ≡ ← streamwise direction
u
1
2

⇒ s • ∇  p + ρ u  = ρ s • (u × ω )
2


=0, ( u ×ω ) ⊥u
⇒
1
d 
2
p+ ρ u =0
2
ds 

⇒ p+
1
2
ρ u = const. along streamline
2
Vortex Panel Methods2
Step#1: Replace airfoil surface with panels
m
i+1
i
Original airfoil
i-1
4
3 2
m+1
1
m-panels (m+1 nodes)
Step #2: Distribute singularities on each panel with unknown strengths
In our case we will use vortices distributed such that their strength varies linearly
from node to node:
Recall a point vortex at the origin is:
φ=−
Γ
Γ
 y
θ =−
tan −1  
2π
2π
x
2
Kuethe and Chow, 5th Ed. Sec. 5.10
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Quick Visit to Bernoulli Land
A point vortex at xˆ , yˆ is:
φ=−
Γ
 y − yˆ 
tan −1 

2π
 x − xˆ 
Next, consider an arbitrary panel:
(xj,yj)
sj
j
Vortex of strength
γ(sj)ds
Sj
expanded
view
γ(sj)
dsj
Midpoints will
be (xj,yj)
j+1 (xj+1,yj+1)
At any s j , we will place a vortex with strength γ ( s j ) ds :
⇒ dφ ( x, y ) = −
γ ( s j )ds −1  y − yˆ j 
tan 
 x − xˆ j 
2π


where
xˆ j ≡ x j + ( x j +1 − x j )
sj
Sj
yˆ j ≡ y j + ( y j +1 − y j )
sj
Sj
Thus, the potential at any ( x, y ) due to the entire panel j is:
Sj
φ j ( x, y ) ≡ − ∫
0
γ (sj )
 y − yˆ j 
ds
tan −1 
 x − xˆ j 
2π


We will assume linear varying γ on each panel:
γ ( s j ) = γ j + (γ j +1 − γ j )
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sj
Sj
3
Quick Visit to Bernoulli Land
With this type of panel, we have m+1 unknowns = γ 1, γ 2, γ 3...γ m −1,γ m, γ m +1 , so we
need m+1 equations.
Step#3: Enforce Flow Tangency at Panel Midpoints
The next step is to enforce some approximation of the boundary conditions at the
airfoil surface. To do this, we will enforce flow tangency at the midpoint of each
panel.
Panel method lingo: control point is a location where u • n = o is enforced.
To do this, we need to find the potential and the velocity at each control point.
The potential has the following form:
 freestream 
 individual panel 
+ ∑ 
φ =


potential
 potential  # panels 

Suppose freestream has angle α :
m Sj
φ ( x, y ) = V∞ ( x cos α + y sin α ) − ∑ ∫
γ (sj )
j =1 0
The required boundary condition is
 y − yˆ j 
ds
tan −1 
 x − xˆ j 
2π


∂φ
( xi , yi ) = 0 for all i = 1 → m
∂ni
So, let’s carry this out a little further:
m S j γ (s ) 
∂φ
j  ∂
xi , yi ) = V∞ ( cos α i + sin α j ) • ni − ∑ ∫
(

∂n j
j =0 0 2π  ∂ni
component of freestream normal

to surface of panel i
And recall γ (s j ) = γ j + (γ j +1 − γ j )
 −1  y − yˆ j   
 tan 
   ds = 0

 x − xˆ j   xi , yi 
normal velocity due to panel j
at control point of panel i
sj
.
Sj
We can re-write these integrals in a compact notation:
γ ( s j )  ∂
∫0 2π  ∂ni

Sj

 −1  y − yˆ j   
 tan 
   ds = Cn1ij γ j + Cn 2ij γ j +1

 x − xˆ j   xi , yi 
Cn 1ij = Influence of panel j due to
xi , y i
node j on control point of panel i
i.e. Cn1ij γ j = normal velocity from panel j due to node j on control point of panel i .
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Quick Visit to Bernoulli Land
Cn 2ij = Influence of panel
j due to node j + 1 on control point at panel i
⇒ Total normal velocity at control point of panel i due to panel j = Cn1ij γ j + Cn 2ij γ j +1
So, let’s look at the control point normal velocity
So, for panel i , flow tangency looks like:
∑ (C
m
j =1
n1 ij
)
γ j + Cn 2 γ j +1 = V∞ ( cos α i + sin α j )ini , for all i = 1 → m
ij
V∞ ni
We can write this as a set of m equations for m+1 unknowns.
Question: What can we do for one more equation?
Step#4: Apply Kutta condition
We need to relate Kutta condition to the unknown vortex strengths γ j . To do this,
consider a portion of a vortex panel.
γ
ds
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Quick Visit to Bernoulli Land
Put a contour about differential element ds
U2
dn
γds
V1
V2
U1
ds
Γ = − ∫ u ⋅ ds
Recall: ⇒ γ ds = − [V2dn − U 2ds − V1dn + U1ds ]
= (V1 − V2 ) dn − (U1 − U 2 ) ds
Now let dn & ds → 0 :
dn → O , γ ds = − (U1 − U 2 ) ds
γ = U 2 − U1 , or
γ = U top − U bottom , in general
So, since the Kutta condition requires U top = U bottom at TE:
γ t .e. = 0, Kutta condition
For the vortex panel method, this means:
γ 1 + γ m +1 = 0
Step#5: Set-up System of Equations & Solve
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Quick Visit to Bernoulli Land
 I11
I
 21
 I31

 I41
 I51

 I61
I
 71
 I81
1

uin = 0@i = 1
uin = 0@i = 2
uin = 0@i = m
Kutta
I12
I22
I32
0
I33
0
0
V∞ n1 
I19  γ 1 
 
 γ 
V∞n 2 
  2
. 
 γ 3 
 
 
.
. 
 
 .  = − . 
 
 
.
 
 
 
 
 
 
V∞n8 
 γ 8 
 



1  γ 9 
0 
γ
I
V∞n
Where I ij = total influence of node j at control point i
I 37 = C n137 + C n 236
For example:
6
5
7
Cn236
4
8
Cn137
3
9
2
Nodes
Control points
1
The problem thus reduces to Ax = b , or, using our notation
Iγ = V∞n ,
which we solve to find the vector of γ ’s!
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Quick Visit to Bernoulli Land
Step #6: Post-processing
The final step is to post-process the results to find the pressures and the lift
acting on the airfoil.
L′ = ρV∞ Γ = ρV∞
∫
γ ds
airfoil
So, for our method, this reduces to:
L′ =
m
1
ρV∞ ∑ (γ i + γ i +1 )Si
2
i =1
⇒ Cl =
m
γ
γ S
= ∑  i + i +1  i
1
ρV∞2c i =1  V∞ V∞  c
2
L′
Vortex Panel Method Summary
In practice, the vortex panel method used for airfoil flows is a little different than
the strategy used in the windy city problem. Here’s a summary:
Step #1: Replace airfoil surface with panels
m
4
3 2
m+1
1
Note: the trailing edge is double-numbered ⇒ m + 1 points, m panels .
Step #2: Distribute vortex singularities with linear strength variables on each
panel
γ(sj)
sj
γ ( s j ) = γ j + (γ j +1 − γ j )
Sj
γ(j)
γ
γ(j+1)
j
γds
sj
ds
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Sj
j+1
Quick Visit to Bernoulli Land
This means we have m+1 unknowns:
γ 1, γ 2, γ 3, ............γ m, γ m +1
Step #3: Enforce flow tangency at panel midpoints
u • n = 0 at midpoint of every panel
⇒ m equations
Step#4: Apply Kutta condition
Kutta condition becomes:
γ t .e. = 0 ⇒ γ 1 + γ m+1 = 0
⇒ m + 1 equations & m + 1 unknowns
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