ME3322
Fluids and Heat Transfer
The Blasius solution of the momentum equation for the Laminar Boundary Layer:
According to the boundary layer assumptions of this analysis, we have:
• Incompressible or non-compressed flow,  = constant.
• Flow is laminar, no unsteadiness (particularly turbulence)
•
Newtonian fluid so that we can write:    u y .
•
Flat plate, p y = 0, no streamline curvature.

y
x
•
•
•
Boundary layer flow, only diffusion in cross-stream direction.
Steady state and steady flow.
No significant streamwise acceleration (indirectly goes with the statement "flat plate," which distinguishes this
from a "wedge flow").
By continuity:
u
v
x  y  0
From the streamwise momentum equation:
u u x  v u y  
1 dp
2u
  2 , where the pressure gradient term is zero. We can come to this
 dx
y
conclusion by writing the momentum equation for y >  :
If
du 
u
du 
1 dp
.
dx    dx
dp
dx = 0, then dx = 0. We can solve this set with the boundary conditions:
u  u  as y  
•
•
u = 0 @ y = 0 for all x
• u = u  @ x = 0 for all y > 0
The boundary conditions suggest that the velocity profiles may be self-similar, allowing us to reduce the number of dependent
variables. Let's look into this further:
Morgan's theorem - If a well-posed problem is invariant under a one-parameter group of transformations of the
independent and dependent variables, the number of independent variables can be reduced by one. Try the following
one-parameter transformations (stretching factors):
u  A u , v  B v , x  P x , y  Q y , and u   C u  . Then, for example,
 A  u
x   P   x , etc. Substituting into the boundary layer equations:
Continuity:
Momentum:
u
u
 BP 
 v
 
0
 x  AQ   y
u u
 P
 BP 
v u
 
 

 x  AQ 
 y  AQ2
 2
 u
2

y

With boundary conditions:
u  0 , v  0 at y  0 and x  0
 A u
u  C
,
as y   and x  0 and at x  0, y  0
If the problem is to remain invariant under the transformation:
BP
AQ  1 ,
which gives
vx
uy
invariant
Laminar Boundary Layer Flow - Page 1
P
AQ2
which fixes P
 1,
C A 1,
AQ 2
and makes  x
uy 2
u
u
which gives
invariant
invariant
The above terms are invariant. We have transformed the problem in new variables without any of the old variables "left over"
and without changing the variables (no left over stretching factors). The transformation was successful. But, any combination
of the new variables is also invariant. Lets look for a more suitable form. Looking for a form where y is singled out:
 uy 2

 x

 u
y
u

 u  y x  x

vx
u
Re x which is the canonical (simplest) form.
 u
v



Singling out v: uy y 
u
 x  u 
v
ux

u

Re x and singling out u:
u
. Now that we have
u
this grouping, assign new variable names to these quantities:
u
v
u
ux
  y  ; f1  
; and f 2  
u
u
x



Now look at the boundary conditions:
u = 0 @ y = 0, x > 0 transforms to
f1 0   0
v = 0 @ y = 0, x > 0 transforms to
f 2 0   0
u  u  when y   transforms to
f1    1 and,
f1    1 .
u = u  when x = 0, y > 0 to
Note that the last two boundary conditions collapsed into one. This suggests that we can reduce the number of
dependent variables.
We now have two dependent variables f1   and f 2   and one independent variable,  .
The original variables have been replaced by the new -- they remained invariant.
We could reduce f1   and f 2   to one dependent variable if we introduce the stream function,  , such that;
u 


y and v   x .
The continuity eqn. becomes:
 2
xy 
 2
. Thus, it is always satisfied.
yx  0
The momentum equation becomes (the subscripts indicate partial differentiation):
 y  yx   x  yy   yyy ,
with boundary conditions:
u and v  0 at y  0 ;
 x   y  0 at y  0 x  0
u  u as y   ;
 y  u  as y   x  0
u  u at x  0
 y  u  at x  0 y  0
Laminar Boundary Layer Flow - Page 2
Now let's apply Morgan's theorem to reduce that number of independent variables. In doing so, we find the
invariants:
y
x

and
. In canonical form:  
x
uy
y
Re x and f  
 

 Re x
.
Now, writing the equations in terms of  and f  :
  f  Rex ,
u  f   u  , so;
u
 y  f u 
 yy  f u  u 
 yyy  f 
2
u
 xy   f  
y u
2 x
 x   f 
yu
2x 
f  u

x
2
x
 
u  f  
x
u
x
Substituting into the momentum equation gives:
1
ff   0
2
f  
The Blasius Equation (Continuity and Momentum)
With boundary conditions:
u=0
@
y=0

f    1
f 0  0
x > 0;
u  u  as y   , x  0
u  u
f    1 Note that the last two collapse into one.
at x  0 , y  0
Also, from the momentum equation:

2u
y
2
 u
u
u
v
x
y
at the wall, u = v = 0
 2u
y 2
f 0   0 . Or f 0  is a constant (constant shear stress).
0
f 0  
From the Blasius equation at the wall,
This requires that:
Thus,
f 0 f 0   0 but,
1
f 0 f 0   0
2


f 0    u 

y

0
1
u u
for all x.
x
f 0  could not be zero (otherwise the shear stress would be zero for all x).



Therefore, it must be that f 0  0 . Since f 0      ,
 xu 
 0

We can find the velocity profile if we can solve the Blasius Eqn.
f  

 
 0 =0.

1
ff   0 f  0  0 , f    1 , and f 0  0 .
2
One way to solve this is to nodalize the range of
 
 into discrete  , write the equations (below) for each and
solve them simultaneously, matching the f    1 boundary condition at the outermost node. Another way is the

"shoot and correct method." Here, one first guesses the value of the wall shear stress f  0 , then integrates the
following discretized set of equations stepwise from the wall to the edge of the boundary layer:
1
1
f    ff 
f    ff 
2
2
f   f 
f  f 
Laminar Boundary Layer Flow - Page 3
The edge of the boundary layer is where
 
 is sufficiently large that f" and f   0 . If this solution doesn't hit the

required f    1 , guess another value of shear stress ( f  0 ) and try again. The solution to this equation is
tabulated as Table 5-1 , below (Though published in many texts, this table is taken from Convective Heat Transfer
by L. Burmeister, 2nd ed., John Wiley Press).
From this solution, we can process the quantities we need:
u
v
cf
u
Re x 
u
2 

 

0



x
 f    f  y
0
u 
2


f  

f 0  
0
Re x ,
1
f f  ,
2
3
2
2
x u
u 
u
1

  u  u


u
  1
u
0

2
Re x
 
2 f 0  
 2
y
2

 y 

f    Re x 
 x 

2
x
2
3
Re x
f 0 
,
2 f   ,
f 0  ,
Re x


 dy ,


x


 dy , and

x


1
Re x
1
Re x




f    1 f    d ,
0

 1 f  d .
0
Using the tabulated results of Table 5-1.
1
f f   0.86
2
f"(0) = 0.332


0
0
 f 1 f d  0.664
 1 f d  1.72
   0.664,     1.72,     at u  0.99u    4.6
Let's look at some typical values. Air at room temperature and pressure,   2 x10
Let
u   15 m sec ,
4.6
 0.017 ,  99 = 2 mm (  1/16 inch at x = 10 cm)
x 
Re x
v
0.86
,v  0.05m / sec
u 
Re x 
u
m2
sec .
x = 10 cm (4 inches), Rex = 75,000
99
Re  
5
  180
Probably ready for transition, if it were in a disturbed flow (called "engineering flows").
Laminar Boundary Layer Flow - Page 4
Laminar Boundary Layer Flow - Page 5