Boundary Layers Thickness-Displacement-Momentum
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Boundary Layers
As a fluid flows over a body, the no-slip condition ensures that the fluid next to the boundary is subject to large shear. A pipe is enclosed, so the fluid is fully bounded, but in an open flow at what distance away from the boundary can we begin to ignore this shear?
There are three main definitions of boundary layer thickness:1. 99% thickness2. Displacement thickness3. Momentum thickness
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99% ThicknessU
U is the free-stream velocity
(x)
x
y
(x) is the boundary layer thickness when u(y) ==0.99U
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Displacement thickness
There is a reduction in the flow rate due to the presence of the boundary layer
This is equivalent to having a theoretical boundary layer with zero flow
y
u
y
uU
U
d
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Displacement thicknessThe areas under each curve are defined as being equal:
0
dyuUq and Uδq d
0d dy
Uu1δ
Equating these gives the equation for the displacementthickness:
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Momentum thicknessIn the boundary layer, the fluid loses momentum, so imagining an equivalent layer of lost momentum:
0
dyuUρum and m2δρUm
0m dy
Uu1
Uuδ
Equating these gives the equation for the momentumthickness:
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Laminar boundary layer growth + d
dy
x
y
Boundary layer => Inertia is of the same magnitude as Viscosity
a) Inertia Force: a particle entering the b.l. will be slowed from a velocity U to near zero in time, t. giving force FI U/t. But u=x/t => t l/U where U is the characteristic velocity and l the characteristic length in the x direction.
Hence FI U2/l
b) Viscous force: F /y 2u/y2 U/2
since U is the characteristic velocity and the characteristic length in the y direction
(x)
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Laminar boundary layer growthComparing these gives:
ρUμδ l
So the boundary layer grows according to l
Alternatively, dividing through by l, the non-dimensionalisedboundary layer growth is given by:
lRl1δ
Note the new Reynolds numbercharacteristic velocity and characteristic length
υU
μρU llRl
)(ρUμ5δ Blasiusl
U2/l U/2
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Boundary layer growth
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Length Reynolds Number
μρUlRl
l
U
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Flow at a pipe entry
l
Ud
δ
If the b.l. meet while the flow is still laminar the flow in the pipe will be laminar
If the b.l. goes turbulent before they meet, then the flow in the pipe will be turbulent
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Length Reynolds number and Pipe Reynolds number
The critical Reynolds number for flow along a surface is Rl=3.2*105
In a pipe, the Reynolds number is given by
dumRe
Considering a pipe as two boundary layers meeting, d=2a=2and from above
ρUμ5δ l
lRll 10μ
ρU10ρUμ10.
μρURe
The mean velocity in the pipe, um, is comparable to the free-stream velocity, U
If Rl=3.2*105 then Re=5657
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Boundary layer equations for laminar flow
These may be derived by solving the Navier-Stokes equationsin 2d.
0
yv
xu
dtdu
yu
xu
ρμ
xp
ρ1
2
2
2
2
Continuity MomentumU
Assume:1. The b.l. is very thin compared to the length2. Steady state
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Boundary layer equations for laminar flow
yuv
xuu
yu
ρμ
xp
ρ1
2
2
This gives Prandtl’s b.l. equation:
rate of change of u with x is small compared to y
Blasius produced a perfect solution of these equations validfor 0<x<3.2*105, and demonstrated the shape of the boundarylayer profile
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Blasius Solution
0
5
0 1
u/U
y'
y' f' (or u/U) f'' 0 0 0.3321 0.330 0.3322 0.630 0.3233 0.846 0.2674 0.956 0.1615 0.992 0.0646 0.999 0.0027 1.000 0.000
lμρUy'y
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Laminar skin frictionThe shear stress at the surface can be found by evaluatingthe velocity gradient at the surface
00 y
uμτ
The friction drag force along the surface is then found byintegrating over the length
dxyubμF
0y0f
l
where b is the breadth of the surface
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Laminar skin frictionFrom the Balsius solution, the gradient of the velocityprofile at y=0 yields the result:
0.5x0 R
xU0.332μτ
The shear force can be obtained by integration along the surface
0.5
00f R0.664UbdxτbF l
l
μ The frictional drag coefficient can then be calculated
21
R33.1ρAUFC 2
21
ff
l
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Force and momentum in fluid mechanics - refresher
Newton’s laws still apply. Consider a stream tube:
u1,A1
q1=u1A1
u2,A2
q2=u2A2
mass entering in time, δt, is ρu1A1δt
momentum entering in time, δt, is m1 = (ρu1A1δt)u1
momentum leaving in time, δt, is m2 = (ρu2A2δt)u2
Impulse = momentum change, F = (m2 – m1)/ δt = ρ(u22A2-u1
2A1)
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The von Karman Integral Equation (VKI)
Boundary Layer
A
B
C
D
Flow enters on AB and BC, and leaves on CD
1 2
2 - 1
x
U
u1(y)u2(y)
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VKIThe momentum change between entering and leaving the control volume is equal to the shear force on the surface:
122
δ
0
21
δ
0
220 δδρUdyρudyρuxτ
12
(CD) (AB) (BC)
By conservation of fluid mass, any fluid entering the control volume must also leave, therefore
12
01
0212 )(
dyudyuU
12 δ
01
21
δ
02
220 dyUuuρdyUuuρxτ
Force on fluid
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VKI
δ
0
20 dyUuuρddxτ
As x 0, the two integrals on the right become closer andthe equation may be written as a differential:
δ
0
20 dy
Uu1
Uu
dxdρUτ
The integral is the definition of the momentum thickness, so
dxdδρUτ m2
0 dxdUUρδd if U(x)
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Turbulent boundary layersThe assumption is made that the flat plate approximates to the behaviour in a pipe. The free stream velocity, U, corresponds to the velocity at the centre, and the boundary layer thickness, , corresponds to the radius, R.
1/7 Power LawFrom experiments, one possibility for the shape of the boundary layer profile is
71
δy
Uu
and measurements of the shear profile give41
UδυU0.0225τ 2
0
ρ
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Turbulent boundary layersPutting the expression for the 1/7 power law into the equations for displacement and momentum thickness
δ727δ,
8δδ md
=99%
d
m
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Turbulent boundary layers
dxdδρUτ m2
0 becomesdxdδρU
727τ 2
0
Equating this to the experimental value of shear stress:41
Uδυ0.0225
dd
727
x
Integrating gives:51
υUx0.37xδ
The turbulent boundary grows as x4/5, faster than the laminar boundary layer.
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Turbulent boundary layersMomentum thickness
51
υUx0.036xδ
727δm
To find the total force, first find the shear stressdx
dδρUτ m20
then integrate over the plate length
m2
0
m2
00 δρUdx
dxdδρUdxτF
ll
f
For a plate of length, l, and width b, 51
υUbU0.036F 2
llρf51
0.074RC lf )10R10*5( 75 l
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Logarithmic boundary layerFrom the mixing length hypothesis it can be shown thatthe profile is logarithmic, but the experimental valuesare different from those in a pipe
υyVln85.556.5
Vu *
*
and the friction coefficient llf R
ARlog455.0C 58.2
(A is a correction constant if part of the b.l. is laminar)
)10R0( 9 l
ritrit
58.2rit
RR
1.328Rlog455.0
ccc
A
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Quadratic
0
1
0 1
u/U
y/
Blasius (exact)
Quadratic approximation to the laminar boundary layer
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Quadratic approximation to the laminar boundary layer
2
δy
δy2
Uu
Remember - boundary layer theory is only applicable insidethe boundary layer.
This is sometimes written with =y/ and F()=u/U as
2η2ηηF
It provides a good approximation to the shape of the laminar boundary layer and to the shear stress at the surface
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Turbulent Boundary Layer
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Laminar Sub-Layer