Derivation of Von Karman Integral Momentum Equation
  u   v 

0
x
y
Multiply this by edge velocity ue(x). Integrate by parts.
We start with continuity:
u
 u    uu e 

 u e
x
x
x
 v    vue 
ue

y
y
 uu e 
u
 vue 
Result :
 u e 
0
x
x
y
ue
(1)
Subtract from this the u-momentum equation:
 u 2
uv  p  xy



x
y
x
y


(2)
Use also the Euler equation: VdV+dp=0 which gives at boundary layer edge
u
p
 e ue e 
0
x x
(3)
Result after some minor rearranging:
u 
 
u  u  
2
vue  vu   e u e e 1  u
  e u e 1  

x 
x 
 eue
 u e   e u e  y


   xy
y

(4)
Integrate equation (4) from y=0 to y=.


u
 
u  u

2 

u
dy   vue  vu 0   e u e e
 e e  1  
x 
u e   e u e 
x
0


u
0 1   e u e



dy    xy dy
y

0
Use definitions of displacement thickness and momentum thickness. Second term goes to zero,
since v=- at wall, and u=ue at the edge. We get:
u

 e u e2   e u e e  *   wall
x
x
(5)
Equation (5) works for laminar and turbulent, compressible and incompressible flows.
Expand first term. Simplify this for incompressible flows, by assuming e is constant.
We use ordinary derivatives from now on, since there are no derivatives with respect to y left.
We get:
u
u
d
 e u e2
 2  e u e e  H e u e e   wall
dx
x
x


Divide thru’ by  e u e2 We get
Cf
d
1 du e
 (2  H )

dx
u e dx
2
(6)