CHAPTER 9: EXTERNAL* INCOMPRESSIBLE
VISCOUS FLOWS
*(unbounded)
CHAPTER 9: EXTERNAL INCOMPRESSIBLE
VISCOUS FLOWS
= separated
bdy layer
thicker
adverse pressure
gradient
leads to separation
difficult to use theory
•Re = Ux/; Re = Uc/; …
•laminar and turbulent boundary layers
• displaced inviscid outer flow
• adverse pressure gradient and separation
Boundary Layer Provides Missing Link
Between Theory and Practice
Boundary layer, , where viscous stresses
(i.e. velocity gradient) are important we’ll define
as where u(x,y) = 0 to 0.99 U above boundary.
In August of 1904 Ludwig Prandtl, a 29-year old professor presented
a remarkable paper at the 3rd International Mathematical Congress in
Heidelberg. Although initially largely ignored, by the 1920s and 1930s
the powerful ideas of that paper helped create modern fluid dynamics
out of ancient hydraulics and 19th-century hydrodynamics.
(only 8 pages long, but arguably one of the most important
fluid-dynamics papers ever written)
• Prandtl assumed no slip condition
• Prandtl assumed thin boundary layer region where shear forces are
important because of large velocity gradient
• Prandtl assumed inviscid external flow
• Prandtl assumed boundary so thin that within it p/y  0; v << u
and /x << /y
• Prandtl outer flow drives boundary layer boundary layer can greatly
effect outer outer “inviscid” flow if separates
BOUNDARY LAYER HISTORY
- 1904 Prandtl
Fluid Motion with Very Small Friction
2-D boundary layer equations
- 1908 Blasius
The Boundary Layers in Fluids with Little Friction
Solution for laminar, 0-pressure gradient flow
- 1921 von Karman
Integral form of boundary layer equations
- 1924 Sir Horace Lamb
Hydrodynamics ~ one paragraph on bdy layers
- 1932 Sir Horace Lamb
Hydrodynamics ~ entire section on bdy layers
Theodore von Karman
INTERNAL
EXTERNAL
CAN BE
NEVER
NEVER
USUALLY - PLATE IS
EXCEPTION
THEORY
LAMINAR
PIPES, DUCTS,..
FLAT PLATE & ZERO
PRESSURE GRADIENT
GROWING
BOUNDARY
LAYER?
NOT WHEN
FULLY
DEVELOPED
PIPE/DUCT=N0
DIFFUSER=YES
ALWAYS
PIPE (EXAMPLE)
u(r)/Uc/l = (y/R)1/n
PLATE (EXAMPLE)
u(y)/Uo = (y/)1/n
FULLY
DEVELOPED?
WAKE?
ADVERSE
PRESSURE
GRADIENT
TURBULENT
EXPERIMENT
PLATE=MAYBE
BODIES=USUALLY
Note – throughout figures the
boundary layer thickness*,,
is greatly exaggerated!
(disturbance layer*)
Airline industry had to
develop flat face rivets.
Re = 20,000
Angle of attack = 6o
Symmetric Airfoil
16% thick
Flat Plate (no pressure gradient)
~ what is velocity profile?
~ wall shear stress/drag?
~ displacement of free stream?
~ laminar vs turbulent flow?
Immersed Bodies
~ wall shear stress/drag?
~ lift?
~ minimize wake
breath
FLAT PLATE – ZERO PRESSURE GRADIENT
Laminar Flow
/x ~ 5.0/Rex1/2
THEORY
Turbulent Flow
Rextransition > 500,000
u(y)/U = (y/)1/7
/x ~ 0.382/Rex1/5
EXPERIMENTAL
No simple theory
for Re < 1000;
(can’t assume
 is thin)
“At these Rex numbers
bdy layers so thin that
displacement effect on
outer inviscid layer is
small”
FLAT PLATE – ZERO PRESSURE GRADIENT
outside (x), U is constant so P is constant
u(x,y) is not constant, (x) is thin so
assume P inside (x) is impressed from the outside
ReL = 10,000 Visualization is by air bubbles see that boundary+ layer, ,
is thin and that outer free stream is displaced, *, very little.
+
Disturbance Thickness, (x) (pg 412); boundary layer thickness, (x) (pg 415)
FLAT PLATE – ZERO PRESSURE GRADIENT
Rex = Ux/
Assume Rextransition ~ 500,000
x
L
ReL = Ux/
ReL = 10,000 Visualization is by air bubbles see that boundary+ layer, ,
is thin and that outer free stream is displaced, *, very little.
+
Disturbance Thickness, (x) (pg 412); boundary layer thickness, (x) (pg 415)
SIMPLIFYING ASSUMPTIONS OFTEN MADE FOR
ENGINERING ANALYSIS OF BOUNDARY LAYER FLOWS
Development of laminar boundary layer
(0.01% salt water, free stream velocity 0.6 cm/s, thickness
of the plate 0.5 mm, hydrogen bubble method).
*
*
*
*
*
Rex  1000
breath
FLAT PLATE – ZERO PRESSURE GRADIENT: (x)
BOUNDARY OR DISTURBANCE LAYER
(x)
*


BOUNDARY OR DISTURBANCE LAYER
+
Boundary
Layer Thickness
(x)
Definition:
u(x,) = 0.99 of U=U=Ue
(within 1 % of U)
+Disturbance
 is at y location where u(x,y) = 0.99 U 
Because the change in u in the boundary layer
takes place asymptotically, there is some
indefiniteness in determining  exactly.
NOTE: boundary layer is
much thicker in turbulent flow.
Blasius showed theoretically for
laminar flow that /x = 5/(Rex)1/2
(Rex = Ux/)
  x1/2
Experimentally found*
for turbulent flow that
  x4/5
NOTE: velocity gradient at wall
(w =  du/dy) is significantly greater.
At same x: U/L > U /T
At same x: wL < wT
Note, boundary layer is not a streamline!
From theory (Blasius 1908, student of Prandtl):
= 5x/(Rex1/2) = 5x/(U/[x])1/2 = 51/2x1/2/U1/2
d/dx = 5 (/U)1/2 (½) x-1/2 = 2.5/(Rex)1/2
V/U = dy/dxstreamline = 0.84/(Rex1/2)
dy/dxstreamline  d/dx so  not streamline
U = U = Ue
Behavior of a fluid particle traveling along a streamline
through a boundary layer along a flat plate.
ASIDE ~ LAMINAR TO TURBULENT TRANSITION
LAMINAR TO TURBULENT TRANSITION
NOTE: Turbulence is not initiated at
Retr all along the width of the plate
Emmons spot ~ Rex = 200,000
Spots grow approximately linearly downstream at downstream
speed that is a fraction of the free stream velocity.
Emmons spot, Rex = 400,000
smoke in wind tunnel
x=0
Turbulent boundary layer is thicker and grows faster.
Transition not fixed but usually around Rex ~ 500,000
(2x105-3x106, MYO)
For air at standard conditions and U = 30 m/s, xtr ~ 0.24 m
Nevertheless: Treat transition as it
happens all along Rex = 500,000
Experimentally transition occurs around
Rex ~ 5 x 105
Water moving around 4 m/s past a ship,
transitions after about 0.14 m from the bow,
representing only about 0.1 % of total
length of 143 m long ship.
breath
FLAT PLATE – ZERO PRESSURE GRADIENT: *(x)
DISPLACEMENT THICKNESS

*
 (x)
DISPLACEMENT THICKNESS

Displacement Thickness
*
 (x)
Definition:

*
 = 0 (1 – u/U)dy
*
 is displacement of outer
streamlines due to boundary layer
Displacement thickness *
x=L
By definition, no flow passes through
streamline, so mass through 0 to h at x = 0
is the same as through 0 to h + * at x = L.
Uh = 0h+* udy = 0h+*  (U + u - U)dy
Uh = 0h+* Udy + 0h+* (u - U)dy
Uh = U(h + *) + 0h+*(u - U)dy
-U* = 0h+*(u -U)dy
* = 0
h+*(–

u/U + 1)dy  0 (1 – u/U)dy
displacement of outer
streamlines due to (x)


*  0 (1 – u/U)dy  0 (1 – u/U)dy
function of x!
breath
Displacement Thickness
*

Definition: * = 0 (1 – u/U)dy
U*w = 0  (U – u)dyw  0  (U – u)dyw
the deficit in mass flux through area w due to the
presence of the boundary layer.
=
mass flux passing through an area [*w] in the absence
of a boundary layer
Displacement Thickness *
Definition: * = 0 (1 – u/U)dy
U* = 0  (U – u)dy  0  (U – u[y])dy
* problem
Displacement Thickness, *, Problem
Suppose given velocity
profile:
u = 0 from y = 0 to 
u = Ue for y > .
Show that * = 
=
Displacement Thickness *
Suppose u = 0 from y = 0 to  and u = Ue for y > .
Show that * = 
* = 0 (1 – u/Ue)dy
= 0 (1 – 0/Ue)dy +  (1 – Ue/Ue)dy
*

=0

 dy
=
=
Laminar flow on flat plate in uniform free stream
Blasius = 5x/(Rex1/2)
Blasius* = 1.721x/(Rex)1/2
*(x) ~ 1/3 (x)
* = Distance that an equivalent inviscid flow
is displaced from a solid boundary as a consequence of
slow moving fluid in the boundary layer.
Displacement Thickness *
Definition: * = 0 (1 – u/Ue)dy
From the point of view of the flow outside the boundary
layer, * can be interpreted as the distance that the presence
of the boundary layer appears to “displace” the flow outward
(hence its name). To the external flow, this streamline
displacement also looks like a slight thickening of the body
shape.
*(x)
(x)
 and *are greatly magnified
breath
EXAMPLE

*
 (x)

PROBLEM: Find Ue as a function
*
of U, D and 
2-D DUCT
Ue(x)
Ue(x) = ?
U  f(y)
u(y)
Ue(x)
Continuity equation (per W):
{UD = -D2D/2 udy + -D2D/2 Uedy - -D2D/2 Uedy}
UD = -D2D/2 Uedy – {-D2D/2 (Ue-u)dy}
INLET OF DUCT
Continuity equation (per W):
UD = -D2D/2 Uedy – {-D2D/2 (Ue-u)dy}
UD =
D/2 U dy–{
(-D/2+) (U -u)dy+
D/2 (U -u)dy}



-D2
e
(-D/2)
e
(D/2- )
e
(used approximation that outside boundary layer u =Ue)
(note that Ue = function of x but not y)
INLET OF DUCT
*
*
Continuity equation (per W):
UD =
D/2 U dy – {
(-D/2+) (U -u)dy +
D/2 (U -u)dy}



-D2
e
(-D/2)
e
(D/2- )
e
UD = -D2D/2 Uedy – Ue{(-D/2) (-D/2+) (1-u/Ue)dy}
+ Ue{(D/2- ) D/2 (1-u/Ue)dy}
* = 0 (1 –u/Ue)dy  0 (1 – u/Ue)dy
INLET OF DUCT
*
*
Continuity equation (per W):
UD =
D/2 U dy–U {
(-D/2+) (1-u/U )dy+
D/2 (1-u/U )dy}



-D2
e
e (-D/2)
e
(D/2- )
e
-{ *
+
UD= UeD – 2Ue* = Ue[D – 2*]
Ue(x) = UD/[D – 2*(x)] QED
*}
from Continuity Equation
UD = Ue [D –
*
2 ]
We see that the free stream velocity in the duct
is given by the effective decrease in the cross
sectional area due to the growth of the
boundary layers, and this decrease in area is
measured by the displacement thickness!!!
FLAT PLATE – ZERO PRESSURE GRADIENT: (x)
MOMENTUM THICKNESS
Momentum Thickness

*

(x)
MOMENTUM THICKNESS
Momentum Thickness
 (x)
Definition:

 = 0 u/U(1 – u/U)dy
How define drag due to boundary layer?
Momentum Thickness 
Definition:  = 0 u/Ue(1 – u/Ue)dy
Ue2 w = 0  u(Ue – u)dyw
The momentum thickness, , is defined as the
thickness of a layer of fluid, with velocity Ue,
for which the momentum flux is equal to the
deficit of momentum flux through the boundary
layer. (Mass flux through  = 0  udyw)
U
[Flux of momentum deficit]
[(u)([U-u])]xdy
u
UU w
Summary:
= 0.99U (within 1 %)
*

= 0(1 – u/Ue)dy

 0(1 – u/Ue)dy
 =0 u/Ue(1 – u/Ue)dy

 0 u/Ue(1 – u/Ue)dy
*
 &  Easier to calculate form data
and more physical significance
but “can be” dependent on 
breath
Relate  to drag
Relate  to Drag
Control volume analysis of drag force
on a flat plate due to boundary shear
Relate  to Drag
PRESSURE IS UNIFORM - NO NET PRESSURE FORCE!
- NO DU/DX!
D
(NO BODY FORCES)
(STEADY)
0
Fx = -Don CV = (d/dt) udVol + u(V•n)dA
1.
2.
3.
4.
From (0,0) to (0,h): V•n = -Uo = -Ue; V = Ue
From (0,h) to (L,): streamline, V•n = 0 (no shear)
From (L, ) to (L,0): V•n = u(x=L,y)
From (L, 0) to (0,0): V•n = 0 (shear)
Fx = -D = u(V•n)dA
u(V•n)dy(w)
=  0h Uo(-Uo)wdy +  0 u(L,y) u(L,y)wdy
1.
2.
3.
4.
From (0,0) to (0,h): V•n = -Uo = -Ue; V = Ue
From (0,h) to (L,): no shear, V•n = 0
From (L, ) to (L,0): V•n = u(x=L,y); V = u(x,y)
From (L, 0) to (0,0): V•n = 0

-D = - Uo2wh +  0 u(L,y)2wdy
Not so useful because don’t
know h as a function of (x)

-D = - Uo2wh +  0 u(x,y)2wdy
Get rid of h by using conservation of mass.
   (V•n)dA= 0
h

- 0Uowdy +  0u(x,y)wdy = 0

Uoh = 0u(x,y)dy


-D = - Uow 0u(x,y)dy +  0 u(x,y)2wdy
-D = - Uow 0(x)u(y)dy +  0(x) u(x,y)2wdy
D = w [0(x)Uou(x,y)dy - u(x,y)2dy]
D = w [0(x) u(x,y) [Uo - u(x,y)]dy
D  w [0 u(x,y) [Uo - u(x,y)]dy
 Uo2w [0 [u(x,y)/Uo][1o – [u(x,y)/Uo]]dy
D(x) = Uo
2w
(x)
(First derived by Von Karman in 1921)
breath
Relate  to wall shear stress
D(x) = Uo
2w
(x)
D(x) = 0x wwdx = Uo2 w (x)
w(x) =  d/dx {Uo2 (x)}
No pressure gradient; dU/dx = 0
w(x)/ = Uo2 d/dx {(x)}
The shear stress at the wall is proportional to the rate of
change of momentum thickness with distance.
(Cf =w(x)/[(1/2) Uo2)])
NOTE
Class ~ derivation for constant pressure case
w(x)/ = Uo2 d/dx {(x)}
Book ~ derivation includes pressure gradient case
w(x)/ = d/dx { Uo(x) 2 (x)} *Uo(x)dUo(x)/dx
Uo  constant
dP/dx  0
book
w(x)/ = d/dx { Uo(x) 2 (x)} *Uo(x)dUo(x)/dx
Uo = constant; dP/dx = 0
me
w(x)/ = Uo2 d/dx {(x)}
Estimating (x), *(x), (x), Drag
breath
For laminar / turbulent / and pressure gradient flows
book
w(x)/ = d/dx { Ue(x) 2 (x)} *Ue(x)dUe(x)/dx
me
w(x)/ = Uo2 d/dx {(x)}
w(x)/ =
d/dx {(x)}
No pressure gradient; dU/dx = 0
2
Uo

w(x)/ = Uo2 d/dx {0 u/Ue(1 – u/Ue)dy
u(x)/U is assumed to be similar for all x
and is normally specified as a function of y/(x)
Defining  = y/
w(x)/=Uo2d/dx{ 01 u/Ue(1– u/Ue)d}
2
1
w(x)/=Uo d/dx{0 u/Ue(1– u/Ue)d}
 y/
Dimensionless velocity profile for a laminar boundary layer:
comparison with experiments by Liepmann, NACA Rept. 890,
1943. Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991
Turbulent bdy layer* ~ u/Ue = (y/)1/7; Ue = 1000 cm/s
y (cm)
turbulent bdy layer velocity profile
U = 10 m/s; x1=10cm
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
600
1.2
1
0.8
y/(x)
650
700
750
800
850
900
950
1000
1050
0.6
u (cm/sec)
0.4
turbulent bdy layer velocity profile
U = 10 m/s; x1=100cm
0.2
1
y (cm)
0.8
0
0.6
0.4
0.6
turbule nt bdy laye r v e locity profile
U = 10 m/s; x1=10cm
y (cm)
0.2
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
600
0
600
650
700
750
800
u
650
700
750
800
850
900
950
1000
0.65
900
950
1000
0.75
0.8
0.85
0.9
1050
(cm /se c)
850
0.7
1050
u (cm/sec)
* du/dy|wall = 
u(x)/U
0.95
1
1.05
Blasius developed an exact solution (but numerical integration
was necessary) for laminar flow with no pressure variation.
Blasius could theoretically predict boundary layer thickness (x),
velocity profile u(x,y)/U vs y/, and wall shear stress w(x).
Von Karman and Poulhausen derived
momentum integral equation
(approximation) which can be
used for both laminar (with and
without pressure gradient) and
turbulent flow
Von Karman and Polhausen method
(MOMENTUM INTEGRAL EQ. Section 9-4)
devised a simplified method by
satisfying only the boundary
conditions of the boundary layer flow
rather than satisfying Prandtl’s
differential equations for each and
every particle within the boundary layer.
breath
QUESTIONS
? In laminar flow along a plate, (x), (x), *(x) andw(x):
(a) Continually decreases
(b) Continually increases
(c) Stays the same
? In turbulent flow along a plate, (x), (x), *(x) andw(x):
(a) Continually decreases
(b) Continually increases
(c) Stays the same
?At transition from laminar to turbulent flow, (x), (x), *(x)
andw(x):
(a) Abruptly decreases
(b) Abruptly increases
(c) Stays the same
6:1 ellipsoid
wall
Turbulent
Laminar
forced
natural

Turbulent
Laminar
What is wrong with this figure?
What is wrong with this figure?
Are Antarctic Icebergs Towable Arctic News Record – Summer 1984; 36
Cf = 0.074/Re1/5 (Fox:Cf = 0.0594/Re1/5);
Area = 1 km long x 0.5 km wide
sea water = 1030 kg/m3; sea water = 1.5 x 10-6 m2/sec
Power available = 10 kW; Maximum speed = ?
P = UD
104 = U Cf½  U2A
= U [0.074 1/5/(U1/5L1/5)]( ½  U2A)
104 =
U14/5 [0.074(1.5x10-6)1/5/10001/5] x
[½ (1030)(1000)(500)]
U14/5 = 0.03054
U = 0.288 m/s
Rex ~ 3x108
80% of fresh water found in world
– Antarctic ice
Biggest problem not melting
but is = ?
THE END
extra “stuff”
breath
Derivation
Blasius developed an exact solution (but numerical integration
was necessary) for laminar flow with no pressure variation.
Blasius could theoretically predict boundary layer thickness (x),
velocity profile u(x,y)/U vs y/, and wall shear stress w(x).
Von Karman and Poulhausen derived
momentum integral equation
(approximation) which can be
used for both laminar (with and
without pressure gradient) and
turbulent flow
MOMENTUM INTEGRAL EQUATION
dP/dx is not a constant!
Deriving:
MOMENTUM INTEGRAL EQ
so can calculate (x), w.
u(x,y)
w
Surface Mass Flux Through Side ab
Surface Mass Flux Through Side cd
Surface Mass Flux Through Side bc
Surface Mass Flux Through Side bc
Apply x-component of momentum eq.
to differential control volume abcd
Assumption : (1) steady (3) no body forces
u
mf represents x-component of momentum flux
Fsx will be composed of shear force on boundary
and pressure forces on other sides of c.v.
Surface Momentum Flux Through Side ab
X-momentum
Flux =
cvuVdA
u
w
Surface Momentum Flux Through Side cd
X-momentum
Flux =
cvuVdA
u
Surface Momentum Flux Through Side bc
U=Ue=U
X-momentum
Flux =
cvuVdA
u
X-Momentum Flux Through Control Surface
a-b
c-d
c-d
b-c
IN SUMMARY
X-Momentum Equation
X-Force on Control Surface
Surface x-Force On Side bc
Note that p  f(y)
inside bdy layer
p(x)
w
w is unit width
into page
Surface x-Force On Side cd
p(x+dx)
w
w is unit width
into page
Surface x-Force On Side bc
p + ½ (dp/dx)dx
Note that since the velocity gradient goes
to zero at the top of the boundary layer,
then viscous shears go to zero there (bc).
pressure
Why w and
not w (bc)?
Surface x-Force On Side bc
p + ½ dp/dx along bc
In x-direction: [p + ½ (dp/dx)] wd
Surface x-Force On Side ad
-(w + ½ dw/dx]xdx)wdx
p(x)
Fx = Fab + Fcd + Fbc + Fad
Fx = pw -(p + [dp/dx]x dx) w( + d) + (p + ½ [dp/dx]xdx)wd
- (w + ½ (dw/dx)xdx)wdx
#
*
*
+
Fx = pw -(p w + p wd + [dp/dx]x dx) w + [dp/dx]x dx w d)
#
+
+ (p wd + ½ [dp/dx] dxwd) - ( + ½ (d /dx) dx)wdx
x
w
w
x
=
d  << 
dw << w
=
U
ad = 0
ab -cd
bc
Divide by wdx
dp/dx = -UdU/dx for inviscid flow outside bdy layer
 =  from 0 to  of dy
-(dp/dx) = -(0dy) (-UdU/dx) = (dU/dx)(0Udy)
p + ½ U2 + gz = constant outside boundary layer (“inviscid”)
dp/dx + UdU/dx = 0 outside bdy layer (dz = 0)
dp/dx = -UdU/dx for inviscid flow outside bdy layer
 =  from 0 to  of dy
Integration by parts
Multiply by U2/U2
Multiply by U/U
QED
FLAT PLATE – ZERO PRESSURE GRADIENT
LAMINAR FLOW
Blasius Theoretical Solution
Background
for u(x,y), (x), *(x), (x)
Blasius* (1908) developed an exact solution
for laminar flow over a flat plate with
no pressure variation. Blasius could
theoretically predict:
(x), *(x), (x),
velocity profile u(x,y)/U vs y/,
and wall shear stress w(x).
*(first graduate student of Prandtl)
Blasius* (1908) developed an exact
solution but required numerical integration,
not good if there is a pressure gradient
and not good for turbulent flow
hence we develop the momentum
integral equation.
Blasius Solution for Laminar, dP/dx=dUe/dx=0 Flow
2-D Governing Equations
Newton’s Law: F = ma
u(u/x) + v(u/y) = -(1/) p/x + (2u/x2 + 2u/y2)
u(v/x) + v(v/y) = -(1/) p/y + (2v/x2 + 2v/y2)
Cons. of Mass
u/x + v/y = 0
Boundary Conditions:
u = v = 0 at y = 0
Ue outside boundary layer
Blasius Solution for Laminar, dP/dx=dUe/dx=0 Flow
Prandtl made some simplifying assumptions, based on
the premise that the boundary layer is very thin (v and 
small compared to U and L) so 2-D equations over flat
plate more tractable:
v << u; d/dx << d/dy
Conservation of Mass
u/x + v/y = 0
Blasius Solution for Laminar, dP/dx=dUe/dx=0 Flow
Prandtl made some simplifying assumptions, based on
the premise that the boundary layer is very thin (v and 
small compared to U and L) so 2-D equations over flat
plate more tractable:
v << u; d/dx << d/dy
Conservation of Y-Momentum
u(v/x) + v(v/y) = -(1/) p/y + (2v/x2 + 2v/y2)
p/y  0
Pressure can vary only along boundary layer not through it!
Blasius Solution for Laminar, dP/dx=dUe/dx=0 Flow
Prandtl made some simplifying assumptions, based on
the premise that the boundary layer is very thin (v and 
small compared to U and L) so 2-D equations over flat
plate more tractable:
v << u; d/dx << d/dy
Conservation of X-Momentum
u(u/x) + v(u/y) = -(1/) p/x + (2u/x2 + 2u/y2)
p/y  0
u(u/x) + v(u/y) = -(1/) dp/dx + (2u/x2 + 2u/y2)
B.Eq.
u(u/x) + v(u/y)  UdU/dx + (1/)(/y)
Blasius Solution for Laminar, dP/dx=dUe/dx=0 Flow
With Prandtl’s simplifying assumptions plus no
pressure gradient for laminar flow,
left the 2-D mass and momentum
equations over flat plate “ripe” for Blasius:
u(u/x) + v(u/y) = (2u/y2)
u/x + v/y = 0
u = v = 0 at y = 0
u = Ue at y =  (outside boundary layer)
BOUNDARY LAYER EQUATIONS
Dimensional analysis = u(x,y)/Ue = f (Uex/, y/x)
By an ingenious coordinate transformation of
the boundary layer equation Blasius showed that
the dimensionless velocity distribution, u(x,y)/Ue,
was only a function of one composite dimensionless
variable, [y/x][Uex/]1/2.
u(x,y)/Ue = f([y/x][Uex/]1/2) = f()
 = y/
 y/(x)
Dimensionless velocity profile for a laminar boundary layer:
comparison with experiments by Liepmann, NACA Rept. 890,
1943. Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991
From Blasius solution it is found that:
= 5 (x/U)1/2 or /x = 5(Rex)-1/2
* = (x/U)1/2 or */x = 1.72(Rex)-1/2
 = (x/U)1/2 or /x = 0.664(Rex)-1/2
w = 0.332 U3/2 (/x)
u(x,y)/Ue =1 at y(Ue/[x])1/2 = 
u(x,y)/Ue = 0.99 at y(Ue/[x])1/2  5.0
u(x,y)