Blasius Solution of Boundary Layer Flow over a Flat Plate

vx  U 

Consider the problem of flow over a thin, flat plate where upstream of the plate the flow
is a simple streaming flow as shown. Starting from the leading edge of the plate, a
boundary layer will develop on either side of the plate, and the above figure shows the
result on only one side. We wish to determine the shape of the boundary layer and the
velocity profiles inside this boundary layer. Specifically, we will study the classical
analysis due to Blasius.
In this problem, the boundary layer equations which we discussed in the previous lectures
take the form.
Continuity:
v x v y

0
x
y
X – momentum balance:
vx  vy  0
vx  U
at
vx
v x
v
dU
2v
 vy x  U
  2x
x
y
dx
y
y  0,
x0
as y   for all
x  0.
Idea due to Blasius: Seek a similarity solution! Let   y / x1/ 2 , where  is as yet
unspecified. Then, postulate a function f  η  , so that
df
 Uf ;
d

Uf 
 Uf 

;
x
2x
Uf 
 2 v x Uf 
 1/ 2 ;
 2 ;
x
y 2
 x
vx  U
v x
x
v x
y
161
We can determine what vy should be from continuity equation. From continuity,
v x

dy   Uf  dy
x
2x
O

y
y
v y  x, y    

  Uf 
O
 y
d
2x 

 U 
  1/ 2   f d
 2x  0
 vy 
U
 f   f 
2x1/ 2
where we’ve assumed that f  0  0. The above choice of v y ensures that continuity is
satisfied. Substituting these expressions for velocities and the derivatives into the xmomentum balance reduces to
  
1
  2U  f   2 f  f   0,


0
(240)
f  0  0
comes from
vy  0
at
y0
f   0  0
comes from
vx  0
at
y0
f      1 comes from
v x  U as
(241)
y
Choose    / U , so that
y

   Re x1/ 2 
x

where Re x 
(242)
Ux
 equation (1) then becomes,

1
f   ff   0,
2
0    .
(243)
Blasius solved numerically Eq. (243) subject to the boundary conditions (241), and
determined f   , from which he obtained f    . This then yielded the velocity
information.
162
y

   Re x1/ 2 
x

Note: V in the figure is the same as U in the notes. Similarly, R in the figure refers to
Rex in the text. The data in the figures were obtained by Nikuradse at different
conditions.
It is clear from the figure that f    ~ 0.99 when  5. Thus we can identify  5 as a
measure of boundary layer thickness. If the boundary layer thickness is given by
y    x  , then   5 corresponds to
  x   5x Rex1/ 2 .
The shear stress at the wall (on the wall by the fluid),   x  , is given by
163
(244)
x  
v x
y
y 0

U
U 1/ 2
f   0  
Re x f   0  
1/ 2
x
x
Numerically, f   0  is found to be 0.332, so
  0.332
U 1/ 2
Re x .
x
(245)
Note that    at x  0. This is an integrable singularity as the average shear
stress over a plate of length L,  , is given by
1
U 1/ 2
UL
   dx  0.664
Re L , where Re L 

L0
L

L
(246)
Clearly,  is finite for all L > 0.
Another classic problem in the study of boundary layers is flow past a wedge, which is
illustrated in the figure below.
Flow past a wedge. Figure (a) shows the definition of the wedge angle,  . The other
figures indicate various cases corresponding to different choices of the wedge angle.
Source: Deen, page 350.
164
It can be shown that, in the vicinity of the tip, the inviscid flow solution, namely
the outer solution, has the form
vox  x, y  0   cx m  U  x 
(247)
where the superscript “o” denotes outer solution; the orientation of the x-axis is along the
wedge and the y-axis is perpendicular to the x-axis (and pointing into the flowing fluid),
and m is related to the wedge angle (see figure in previous page for definition of the
wedge angle) according to
m   /  2 
(248)
The similarity solution for this problem differs from that for the flow over a flat plate we
just considered in two ways:
a 
 b
U
dU
is now not zero.
dx
vx  U  x  as y  , while previously vx  U (which was a
constant independent of x).
The analysis of the boundary layer flow in this wedge example proceeds in the same
manner as in Blasius solution.
Let
vx  U  x 
df
,
d
 1  m  y 1/ 2

Re x

 2  x
Ux x
Re x 

1/ 2
where
and
The x-momentum balance then reduces to
2
f   ff    1   f     0


(249)
which is called Falkner-Skan equation. Solutions for various values of  are shown in
the figure below.
165
Numerical solutions to the Falkner-Skan equation for various wedge angles. Source:
Deen, page 352.
dv x
d
becomes zero at   0. For values of  smaller than -0.199, v x assumes negative values
for a range of η values close to   0. This indicates flow reversal, as sketched in the
figure below.
Note that negative  's correspond to flow in divergent ducts. At   0.199,
As the outer flow (which is inviscid and irrotational) satisfies the Bernoulli’s equation,
we can write
166
1 dP
dU
 x, y  0  U  c2mx 2m1 .
 dx
dx
(250)
dP
dP
 0 implies that
 x, y  0   0 . Physically,
dx
dx
the fluid is decelerating as x increases. Thus, we conclude that an adverse pressure
gradient, namely one which causes fluid deceleration, can induce flow reversal. When a
flow reversal occurs at some point on the solid surface, we say that the boundary layer
separation has occurred there.
Therefore, when   0 , m  0 and
When flow separation occurs, the site of the boundary layer grows rapidly and the
original assumption that the boundary layer can be treated as an “inner” region falls apart.
For example in the problem of flow past a cylinder at high Re, the variation of dynamic
pressure with  was found earlier to be given by:
P  
U 2
2

2  
, so that
1

sin

R 
dP
U 2

sin   / R  cos   / R  .
d
R
Thus, in the region 0   
R dP
,
 0 and no boundary layer separation is possible. At
2 d
dP
 0 and flow separation can occur. When flow separation occurs a
d
wake develops behind the object, as illustrated in the figure below (see page 169). Figure
(a) in page 169 shows the streamlines corresponding to the potential flow theory for flow
past a cylinder, while figure (b) illustrates a steady wake which has developed behind the
cylinder.
the rear side,
Application of boundary layer theory to this flow problem reveals that the wall shear
stress. w , is a function of θ . It is zero at   180 (see figure in page 159 of the notes;
this position corresponds to the forward stagnation point marked in figure a in page 169).
The shear stress increases as we decrease  to 120 , reaching a maximum there. It then
decreases, becoming zero at  ~ 81 and negative as we move towards the rear stagnation
point at   0 . A negative value for wall shear stress indicates flow reversal. Thus the
onset of flow reversal occurs at the point with zero wall shear stress.
167
Boundary layer theory is not useful for treating wakes. However, this theory
provides us with an insight into why wakes arise. This theory is rather accurate as long
as flow reversal has not occurred. Even when a wake is present behind the cylinder, the
boundary layer theory provides us with a good solution near the forward stagnation point.
When wakes form, the flow is no longer irrotational (on a macroscopic length
scale), flow in the wake can contribute significantly to viscous dissipation, which implies
that the formation of wakes (dramatically) increases drag.
Although we will not get into a detailed discussion of the wake structure, it is
interesting to examine the flow patterns obtained at different Reynolds numbers. The
plates below, taken from Batchelor’s book, An Introduction to Fluid Dynamics, show
streamlines for flow past a cylinder at different Reynolds numbers (marked by the
symbol R in the figures). R = 2aU/, where a is the radius of the cylinder.
Plate 1 (see page 170 of this notes) shows the flow at a low Reynolds number
case. Even when R is as small as 3.64, there appears to be a tiny wake near the rear
stagnation point. When R is 9.1, this wake is clearly visible. As R increases, the size of
the wake increases steadily. In all these panels, the wake is steady and it remains attached
to the cylinder. However, as R increases, the wake loses stability and it begins to oscillate
at first, and gives way to periodic shedding of the wakes at higher R values, as illustrated
in Plate 2 (see page 171 of this notes). When the wake manifests a time-dependent
behavior, the drag force on the cylinder varies in a periodic manner; in addition, there is
also a time-dependent lift force which acts in the plane of the paper in these photographs,
but perpendicular to the main fluid flow direction. The time-average value of the lift
force is zero. At even higher Reynolds number values, the wake shedding becomes
aperiodic and the wake is said to be turbulent. This is illustrated in the schematic diagram
(taken from BSL) shown below (see page 169).
A similar sequence occurs in flow past a sphere as well.
168
Schematic of flow patterns at different Reynolds numbers. Flow past a cylinder. Source:
Bird, Stewert and Lightfoot, page 100.
169
170
171