laminar boundary layer
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Laminar Boundary layer
Unit#2
(Ref 8.6.2, 8.6.3 Potter)
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Boundary layer
• Types of boundary layer - At Rex < Recr Laminar layer
- At Rex > Recr Turbulent layer
Critical Reynolds number, Recr = 500,000
Boundary layer properties is found by either
1. Rigorous solution of Navier Stokes equation or (Blasius)
or 2. Simple Von Karman momentum integral method
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Von Karman Integral Method .
It is valid for both laminar and turbulent flow
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Von Karman Integral Equation for
• Force balance on fluid element from previous pagegives the Momentum eqn. in flow direction when dP/dx
=0
• Mass balance gives
• Neglecting pressure gradient
where θ is the momentum thickness
[ ]
∫ ∫
∂
∂−
∂
∂=−−
−−=+++−++
δ δ
ρ ρ τ δ
τ δ δ δ δ
0 0
2
0
0
)(
))(()
2
(
xU udydx x
dydxu x
dxdp
M M M dxd dp pd dp
p p topinout
δ ρ
δ
Ud dydxu x
mmm inout top =∂
∂=−= ∫
0
∫∫∫ −=−=
δ δ δ
ρ ρ ρ τ dyuU ud
dyud
dyU d
)(2
∞∞dxdxdx
00
0
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Laminar velocity profile• Assume laminar profile
u = ay + b y2+ cy3
• Use boundary conditions y=0, u=0; u=U,
(du/ d y)= 0 at y =δ (d2u/ d y2)= 0 at y = 0
• It gives laminar velocity
profile
• (2)3
3
2
1
2
3
δ δ
y y
U
u−=
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Find Boundary layer thickness of
Laminar BL using cubic velocity profile
• Substitute velocity profile (2) in Von Karman eqn (1)
• From the definition of shear stress
• Equating above two one gets
.
dx
d U dy
y y y y
dx
d U dyuU u
dx
d δ ρ
δ δ δ δ ρ ρ τ
δ δ 2
0 3
3
3
32
0
0 139.022
31
22
3)( =
+−
−=−= ∫∫
δ
µ µ τ
2
3
0
0
U
dy
du
y
=
=
=
x x Re
65.4
=
δ
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Displacement thickness & skin friction (Appx)
• Displacement thickness is defined as
• Skin friction on the wall is
found by using value of δ
• From the definition of local skin friction coefficient, C fx
• Average skin friction coefficient
∫
−=
δ
δ 0
* 1 dyU
u
x
fxU
C Re
646.022
0 == ρ
τ
L
L L
x
fx f
L
dxC
L
C
Re
29.1
Re
646.011
0 0
=== ∫ ∫
x
U U
dy
du
yRe
323.0
2
3 2
0
0
ρ
δ µ µ τ ==
= =
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Exact solution of Blasius• Blasius through more rigorous solution of BL eqns.
found values of BL thickness, displacement thickness
& local friction coefficient.
•He found average skin friction coefficient for the entire
length L
• Use these equations for solving problems. Shear stress
;Re
664.0;
Re
72.1;
Re
5 *
5.0
x
x
x x
C x x
=== δ δ
L
f C Re
33.1
=
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Laminar boundary layer
[Douglas p-393]
• Oil with a free stream velocity of 3 m/sflows over a thin plate of 1.25 m wide and 2
m long. Determine the boundary layer
thickness and the shear stress at mid-lengthand calculate the total, double sided
resistance of the plate (density 860 kg/m3,kinematic viscosity 10-5 m2/s)
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Problem
• The floor of the wind tunnel of a model of FrenchTGV moves with the speed corresponding to that
of the main flow. This prevents a boundary layer
from building up on the floor as the fluid reachesthe turbo-train and more closely resembles the
actual flow relative to a moving train. Find what
would be boundary layer if floor was not moving.Air velocity is 6 m/s and train is 2.5 m from the
leading edge. Take viscosity 1.55x10-5 m2/s
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Model testing of world’s fastest train