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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

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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.

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2edPU U U

U Vx y dx y

0U V

x y

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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

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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

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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

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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

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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

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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