Von Karman Momentum Integral Equation
The solution of the boundary layer equations, although more palatable in these days of computers, can still prove to be a formidable task in the general case of suction and when dp/dx is not zero. To circumvent these difficulties, a method, attributed to von Karman, is often used in which the
momentum theorem is applied to the gross characteristics of the boundary layer.
Consider Fig. 107. A control surface is formed by a differential segment of the boundary layer. The flux entering the left face of the control surface is
■a
The flux passing out through the right face can be written
The flux removed by suction is
Qs = v0 dx.
Because the flux in must equal the flux out, Q3, the flux in along the top, can be calculated from
or
The fluid along the top being drawn in has a velocity in the xdirection of Ux. Thus the total flux of momentum in the xdirection coming in is
The total momentum out is
The sum of the forces in the xdirection acting on the fluid around the control surface is
Z К = PS + (p + fx *)f Л – (p + fx *)(<5 + fx *) – •. *.
– * U."" 
where t0 is the shear stress at the wall. Equating ZF* = Momout — Momin and dropping higherorder terms produces the following:
However, from the definition of the displacement and momentum thicknesses
U dy = 11^(6 – 5*1
Substituting the above in (1019) and relating p and UrXj by Bernoulli’s equations, we obtain
uj* ^ + T – + u^0 = • (1020)
dx ax p
Equation (1020) is known as the von Karman momentum integral equation (modified for suction) and relates the displacement and momentum thickness to the wall shearing stress and the velocity U^. It applies equally well to laminar and turbulent boundary layers. For a Newtonian fluid such as air or water the wall shearing stress r0 for laminar flow is given by
(1021) where p is the coefficient of viscosity.
The advantage in using Eq. (1020) lies in the fact that the results do not depend to any great extent on the form of the velocity distribution u(y). We can assume a и(у) and by satisfying Eq. (1020) relating the gross characteristics arrive at reasonable values of these quantities.
For example, consider a flat plate with no suction; (1020) becomes
dO r0
dx pUi
Now assume an extremely crude profile, namely, a straightline variation from 0 at the wall to U„ at у = <3.
и у
u^ = s’
For this distribution 9/8 = 5, 8*/8 = and t0 = p(U^/8). Hence (1020) becomes
d5 6p
dx pUj
or integrating and substituting for в and 8* gives
These results agree surprisingly well with Blasius’ exact solution. Even the wall shearing stress is only about 13% lower than that calculated by the exact method.
KarmanPohlhausen Method
We represent the velocity profile by a fourthdegree polynomial in terms of the dimensionless distance from the wall, h = y/8.
j— = ah + bh2 + ch3 + dhA. (1022)
on
The constants a, b, c, and d are determined from the following boundary conditions. From Eq. (101)
at у = 0.
At у = ё, и = Uди/ду = 0, 82и/ду2 = 0. From these boundary conditions the constants are found to be
where
<52 «Л/, „
=—– j—> P =—–
v ax v
In terms of a, b, c, and d, the displacement and momentum thickness are given by
s* 
1 a b– 2 3 
c 
d 

5 
4 
5 

9 S ~ 
1 + 1 
a2 3 
ab У “ 
2ас + b2 5 
ad + be 2bd + c2 cd d2
3 7 ~4 ~ ~9
Equation (1020) can now be solved numerically, given UrJx). At each x, Rd, and the shape factors H or К are calculated and R6. compared with R6irjt from Figs. 104 and 105. In this manner the point of transition can be determined or the amount of suction required to delay transition can be calculated.