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FLUID MECHANICS
Lecture 7 Exact solutions
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• To present solutions for a few representative laminar boundary layers where the boundary conditions enable exact analytical solutions to be obtained.
Scope of Lecture
3
Solving the boundary-layer equations
• The boundary layer equations
• Solution strategies:– Similarity solutions: assuming “self-similar” velocity profiles. The solutions are
exact but such exact results can only be obtained in a limited number of cases.– Approximate solutions: e.g. using momentum integral equ’n with assumed
velocity profile (e.g. 2nd Yr Fluids)– Numerical solutions: finite-volume; finite element,…
• CFD adopts numerical solutions; similarity solutions useful for checking accuracy.
2
2edPU U U
U Vx y dx y
0U V
x y
4
SIMILARITY SOLUTIONS
• Incompressible, isothermal laminar boundary layer over a flat plate at zero incidence (Blasius (1908))
• Partial differential equations for U and V with x and y as independent variables.
• For a similarity solution to exist, we must be able to express the differential equations AND the boundary conditions in terms of a single dependent variable and a single independent variable.
0U V
x y
2
2
U U UU V
x y y
0edP
dx.U const
5
U/U∞
U/U∞=F(
• Condition of similarity
is likely to be a function of x and y
• Strategy: to replace x and y by
• Similarity variable
• How to find before solving the equation?
THE SIMILARITY VARIABLE
( )U
FU
y
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THE SIMILARITY VARIABLE
• To find the order of magnitude of From previous lecture
• So
• Similarity variable:
[ ] /
y y
O x U
x
xO
Re][
)(][Re2
2
L
OL xL
[ ]x
OU
Rex
U x
2
2[Re ] ( )x
xO
7
NON-DIMENSIONAL STREAM FUNCTION
• For incompressible 2D flows stream function exists.
• Strategy: To replace U and V with • Define a non-dimensional stream function f
such that f is a function of only
Uy
V
x
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• Find the order of magnitude of stream function
• Non-dimensional stream function
Uy
[ ] [ ]O U
fU x
[ ] [ ]O U x
U x f
[ ]x
OU
NON-DIMENSIONAL STREAM FUNCTION
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Substituting
so as to replace U, V by f
• Convert the boundary layer equations
SIMILARITY SOLUTIONS
0U V
x y
2
2
U U UU V
x y y
Automatically satisfied by
Uy
V
x
U x f Replace
by derivatives WRT
2,,
yyx
/
y y
x U
023
3
2
2
d
fd
d
fdf
0'''2'' fffor
Let
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Some details of analysis
• Thus with:
• we find:
;
• So finally:
• with b.c.’s
( )xU f y U x
( )U
U xU f U fy y x
1( )
2
UV f f
x x
UU UU f
y y x
2 0ff f
0 : 0; : 1f f f
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• Substituting into the boundary layer equation2
2
U U UU V
x y y
3rd order ordinary differential equation
' 0U
fU
0fU x
@ wall 0, hence
@ wall U=0, hence
' 1U
fU
SIMILARITY SOLUTIONS
@ y= , U=U , hence
023
3
2
2
d
fd
d
fdf
0'''2'' fffor
12
VELOCITY PROFILES
5x
U
0.992U
U
5,U
yx
At
Boundary layer thickness:
The tabulated Blasius velocity profile can be found in many textbooks
Uy
x
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VELOCITY PROFILES
V is much smaller than U.
Uy
x
U xV
U
Uy
x
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'0 0
1 1 1.7208U y x
dy f dU U
' '
0 0
1 1 0.664U U y x
dy f f dU U U
x
x
Re
7208.1
x
x
Re
664.0
x
x
Re
5 /
y
x U
U
y x
• Boundary layer thickness
• Displacement thickness
• Momentum thickness
• */=0.344, /=0.133
DISPLACEMENT AND MOMENTUN THICKNESS
*, all grow with x1/2
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U =10m/s, =17x10-6 m2/s
DISPLACEMENT AND MOMENTUN THICKNESS
• Typical distribution of , * and
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Other Useful Results
• Shape factor:
• Wall shear stress:
,Re
7208.1
x
x
x
x
Re
664.0 59.2
H
''
2 2
0.6642 (0)
1 1 Re2 2
wf
xw
UC f
y U xU U
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Short Problem
• The boundary layer over a thin aircraft wing can be treated as that over a flat plate. The speed of the aircraft is 100m/s and the chord length of the wing is 0.5m. At an altitude of 4000m, the density of air is 0.819kg/m3 and the kinematic viscosity is 20x10-6m2/s, Assuming the flow over the wing is 2D and incompressible, – Calculate the boundary-layer thickness at the trailing edge– Estimate the surface friction stress at the trailing edge – Will the boundary layer thickness and friction stress
upstream be higher or lower compared to those at the trailing edge?
– What will be the boundary-layer thickness and the surface friction stress at the same chord location if the speed of the aircraft is doubled?
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SOLUTIONS
• The Reynolds number at x=0.5m:
• The boundary layer thickness:
• Friction stress at trailing edge of the wing
mx
Ux
x
x
0016.0105.2
5.05
Re
5
105.21020
5.0100Re
6
66
2
6
2
2
2
/72.1105.2
664.0100819.05.0
Re
664.05.0
5.0
mN
u
Cu
x
fw
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SOLUTIONS
• Will the boundary layer thickness and friction stress upstream higher or lower compared to those at the trailing edge? – is smaller upstream as is proportional to x1/2
– w is higher upstream as is thinner.
• If the speed of the aircraft is doubled, Rex will be doubled since
– will be smaller as Re increases.
– w will be higher as the increase in U∞ has a greater effect than the increase in Rex.
Rex
U x
x
w uRe
664.05.0 2
x
x
Re
5
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Plane stagnation flow
• Flows with pressure gradients can be self-similar … but it has to be a pressure gradient compatible with self-similarity. See Schlichting and other “advanced” textbooks on fluid mechanics for examples.
• Stagnation flow provides one such example where and potential flow)• Note by Euler equ’n.
• The boundary layer equation thus becomes:
• or• Here primes denote diff’n wrt
0 /eU U x L1 e e
e
dP dUU
dx dx
2 20
2
UU U x UU V
x y L L y
0 /eV U y L
21 0f ff f 0U
yL
21
Stagnation flow results
• Note that the boundary layer has a constant thickness!
• However, the mass within the boundary layer increases continuously since the velocity rises linearly with distance from the stagnation point.
• Unlike the flat-plate boundary layer, the shear stress decreases continuously from the wall.
• For this flow i.e. less than for the zero-pressure-gradient boundary layer.
* / 2.21H
22
Asymptotic Suction Flow• Sometimes it may be desirable to withdraw fluid through
the wall
• If the suction is uniform a point is reached where the boundary layer no longer grows with distance downstream and no further change with x occurs.
• Thus the continuity equation is just V/y = 0, i.e V= Vw
• The x-momentum equation becomes:
• …which is readily integrated to give
• A question for you: What is the skin friction coefficient?
wV V
2
2w
dU d UV
dy dy
1 exp( )wyVU
U