boundary layer theory_2
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CE F312 Hydraulics Engineering 1
STANDARD BOUNDARY LAYER THICKNESS
� That distance from the boundary surface (or
plate) upto which the velocity reaches 99% of the
velocity of the main stream.
y = δ where u = 0.99U
Somewhat an easy to understand
but arbitrary definition as the limit
of boundary layer is not easily
defined13
99% THICKNESS
U
U is the free-stream velocity
δ(x)
x
y
δ(x) is the boundary layer thickness when u(y) ==0.99U
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CE F312 Hydraulics Engineering 2
� For greater accuracy the boundary layer
thickness is defined in terms of certain
mathematical expressions which are the
measures of the effect of boundary layer on the
flow.
� Displacement thickness (δ*)
� Momentum thickness (θ)
� Energy thickness (δE)
THICKNESS OF BOUNDARY LAYER
15
DISPLACEMENT THICKNESS (δ*)
� The boundary layer retards the fluid, so that the mass flux is less
than it would be in the absence of the boundary layer.
� Thus the displacement thickness is the distance by which the
boundary surface would have to be moved so that the actual
discharge deficit would be same as that of an ideal (or
frictionless) fluid past the displaced boundary.
Amount of fluid
being displaced
outward
δ*
Equal
Areas
U-u
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CE F312 Hydraulics Engineering 3
DISPLACEMENT THICKNESS (δ*)
∫∞
−=
0
*dy
U
u1δ
Uδq *ρ=
The areas under each curve are defined as being equal:
( )∫∞
−=0
dyuUq ρ
and
Equating these gives the equation for the displacement
thickness:
The loss due to boundary layer
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MOMENTUM THICKNESS (θ)
� The distance from the actual boundary surface
such that the momentum flux corresponding to
the main stream velocity V through this distance
is equal to the deficiency or loss in momentum
due to the boundary layer formation.
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CE F312 Hydraulics Engineering 4
MOMENTUM THICKNESS (θ)
θρUm 2=•
∫∞
−=
0
dyU
u1
U
uθ
In the boundary layer, the fluid loses momentum, so imagining an
equivalent layer of lost momentum:
( )∫∞•
−=0
dyuUρum and
Equating these gives the equation for the momentum thickness:
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ENERGY THICKNESS (δE)
� The distance from the actual boundary surface
such that the energy flux corresponding to the
main stream velocity U through this distance (δE)
is equal to the deficiency or loss of energy due to
the boundary layer formation
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CE F312 Hydraulics Engineering 5
ENERGY THICKNESS (δE)
3 2 2
0
1 1( )
2 2EU U u udy
∞
ρδ = ρ −∫
2
2
0
1E
u udy
U U
∞ δ = −
∫
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SOLVE!� Air flowing into a 2-ft-square duct with a uniform velocity of 10 ft/s
forms a boundary layer on the walls as shown in Fig. The fluid withinthe core region (outside the boundary layers) flows as if it were inviscid.From advanced calculations it is determined that for this flow theboundary layer displacement thickness is given by
where δ* and x are in feet.� Determine the velocity U = U(x) of the air within the duct but outside of
the boundary layer.
( ) 2/1* 0070.0 x=δ
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CE F312 Hydraulics Engineering 6
SOLUTION
� The volume flow rate across any section of the duct is equal to
that at the entrance (i.e., Q1=Q2). That is
� According to the definition of the displacement thickness, the
flowrate across section (2) is the same as that for a uniform flow
with velocity U through a duct whose walls have been moved
inward by δ*
( ) ∫==×=)(
3 /ft ft ft/s 2
2
1140210 udAsAU
( ) ( )
( )ft/s
.
.*/ft
/
/
)(
3
221
2212
2
007001
10
00700142240
xU
xUUudAs
−=⇒
−=−== ∫ δ
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PROBLEM
� Assuming that the shear stress distribution in a
laminar boundary layer is such that
Calculate the displacement and momentum
thickness of this boundary layer in terms of δ.
−=δ
ττy
10
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CE F312 Hydraulics Engineering 7
BOUNDARY LAYER EQUATIONS
� Equations of continuity and motion for the
steady flow of an incompressible, inviscid fluid in
2-D without body forces are:
1v v pu vx y y
∂ ∂ ∂+ = −
∂ ∂ ρ ∂
CONTINUITY EQUATION
EULER’S EQUATION IN X DIR.
EULER’S EQUATION IN Y DIR.
0u v
x y
∂ ∂+ =
∂ ∂
1u u pu vx y x
∂ ∂ ∂+ = −
∂ ∂ ρ ∂
Chapter 6, Sec 6.6
Chapter 7, Sec 7.3
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� If a viscous fluid is considered, eqn. of continuity
remains unchanged
� Additional terms due to viscous stresses will be
introduced in the eqns. of motion
BOUNDARY LAYER EQUATIONS
τ + dτ
τ
δy
x
y
δx
Edge of
boundary layer
( ) x y xy
∂ττ + ∂τ δ = τ + δ δ ∂
Net force = x yy
∂τδ δ
∂26
CE F312 Hydraulics Engineering 8
1 1u u pu vx y x y
∂ ∂ ∂ ∂τ+ = − +
∂ ∂ ρ ∂ ρ ∂
BOUNDARY LAYER EQUATIONS
u
y
∂τ = µ
∂
� Simplification……..� Since the boundary layer is thin, it is expected that the component
of velocity normal to the plate is much smaller than the parallel tothe plate and that the rate of change of any parameter across theboundary layer should be much greater than that along the flowdirection. That is
Again2
2
1u u p uu vx y x y
∂ ∂ ∂ ∂+ = − + υ
∂ ∂ ρ ∂ ∂
v << u and yx ∂∂
<<∂∂
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� Boundary conditions
� y = 0, u = 0, v = 0
� y = ∞, u = U,
10
p
y
∂− =ρ ∂
0u v
x y
∂ ∂+ =
∂ ∂2
2
1u u p uu vx y x y
∂ ∂ ∂ ∂+ = − + υ
∂ ∂ ρ ∂ ∂
Equations that govern the flow in the
steady, 2-D laminar boundary layer on
a flat plate
PRANDTL’S BOUNDARY PRANDTL’S BOUNDARY PRANDTL’S BOUNDARY PRANDTL’S BOUNDARY
LAYER RQUATIONSLAYER RQUATIONSLAYER RQUATIONSLAYER RQUATIONS
Although the mathematical problem is well-
posed, no one has obtained an analytical
solution to these equations for flow past any
shaped body!
BOUNDARY LAYER EQUATIONS
0=∂∂y
uSolution ? ……still extremely
difficult to obtain..
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