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FLUID MECHANICS

Lecture 7 Exact solutions

2

• 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.

2

2

edPU U UU V

x y dx yρ ρ µ

∂ ∂ ∂+ = − +

∂ ∂ ∂

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 yν

∂ ∂ ∂+ =

∂ ∂ ∂

U=V=0 at y=0;

U U ∞→ as y ∞→

0edP

dx=.U const∞ =

5

U/U∞U/U∞=F(η)

• Condition of similarity

� η is likely to be a function of x and y

• Strategy: to replace x

and y by η

• Similarity variable

• How to find δ before solving the equation?

THE SIMILARITY VARIABLE

( )U

FU

η∞

=

δη

y=

6

THE SIMILARITY VARIABLE

• To find the order of magnitude of δ

• Similarity variable:[ ] /

y y

O x Uη

δ ν ∞

= =

=

x

xO

Re][δ

)1

(][Re2δ

OL = )(][Re2

2

δ

LOL =

L

δδ →

xL →

[ ]x

OU

νδ

∞

=

Re x

U x

ν∞=

2

2[Re ] ( )

x

xO

δ=

7

NON-DIMENSIONAL STREAM FUNCTION

• For incompressible 2D flows stream function ψ exists.

• Strategy: To replace U and V with ψ

• Define a non-dimensional stream function fsuch that f is a function of η only.

Uy

ψ∂=

∂V

x

ψ∂= −

∂

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• Find the order of magnitude of stream function ψ

• Non-dimensional stream function

Uy

ψ∂=

∂

[ ] [ ]O Uψ δ∞=

fU x

ψ

ν ∞

=

[ ] [ ]O U xψ ν ∞=

U x fψ ν ∞=

[ ]x

OU

νδ

∞

=

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 yν

∂ ∂ ∂+ =

∂ ∂ ∂

Automatically satisfied by ψ

Uy

ψ∂=

∂V

x

ψ∂= −

∂

U x fψ ν ∞=Replace

by derivatives WRT η

2,,

yyx ∂

∂

∂

∂

∂

∂

/

y y

x Uη

δ ν ∞

= =

023

3

2

2

=+ηη 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 fψ ν η∞≡ y U xη ν∞≡

( )U

U xU f U fy y x

ψ ψ ην η

η ν∞

∞ ∞

∂ ∂ ∂′ ′= = = =

∂ ∂ ∂

1( )

2

UV f f

x x

νψη∞∂

′= − = −∂

UU UU f

y y x

η

η ν∞

∞

∂ ∂ ∂′′= =

∂ ∂ ∂

2 0ff f′′ ′′′+ =

0 : 0; : 1f f fη η′ ′= = = → ∞ →

11

• Substituting into the boundary layer equation2

2

U U UU V

x y yν

∂ ∂ ∂+ =

∂ ∂ ∂

3rd order ordinary

differential equation

' 0U

fU∞

= =

0fU x

ψ

ν ∞

= =@ wall ψ =0, hence

@ wall U=0, hence

' 1U

fU∞

= =

SIMILARITY SOLUTIONS

@ y= ∞, U=U∞, hence

023

3

2

2

=+ηη d

fd

d

fdf

0'''2'' =+ fffor

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VELOCITY PROFILES

x

x

Re

5=5

x

U

νδ

∞

= =

0.992U

U∞

=5,U

yx

ην

∞= =At

Boundary layer thickness:

The tabulated Blasius velocity profile can be found

in many textbooks

Uy

xη

ν∞=

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VELOCITY PROFILES

V is much smaller than U.

Uy

xη

ν∞=

U xV

U ν∞

∞

Uy

xη

ν∞=

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( )'

0 0

1 1 1.7208U y x

dy f dU U

νδ η

η

∞ ∞∗

∞ ∞

∂= − = − =

∂ ∫ ∫

( )' '

0 0

1 1 0.664U U y x

dy f f dU U U

νθ η

η

∞ ∞

∞ ∞ ∞

∂= − = − =

∂ ∫ ∫

x

x

Re

7208.1=∗δ

x

x

Re

664.0=θ

x

x

Re

5=δ /

y

x Uη

ν ∞

= U

y x

η

ν∞∂

=∂

• Boundary layer thickness

• Displacement thickness

• Momentum thickness

• δ*/δ =0.344, θ/δ =0.133

DISPLACEMENT AND MOMENTUN THICKNESS

� δ, δ*, θ all grow with x1/2

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U∞ =10m/s, ν=17x10-6 m2/s

0

0.002

0.004

0.006

0.008

0.01

0 0.5 1 1.5 2

x (m)

δδ δδ, δδ δδ*,

θ(

θ(

θ(

θ(m

m)) )) δ

δ∗

θ

DISPLACEMENT AND MOMENTUN THICKNESS

• Typical distribution of δ, δ* and θ

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Other Useful Results

• Shape factor:

• Wall shear stress:

,Re

7208.1

x

x====∗∗∗∗δδδδ

x

x

Re

664.0====θθθθ 59.2========

∗∗∗∗

θθθθ

δδδδH

''

2 2

0.6642 (0)

1 1 Re2 2

wf

xw

UC f

y U xU U

τ µ ν

ρ ρ ∞∞ ∞

∂= = = =

∂

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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 kinematicviscosity 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

6

6

====××××

××××========

××××====××××

××××========

−−−−

δδδδ

νννν

2

6

2

2

2

/72.1105.2

664.0100819.05.0

Re

664.05.0

5.0

mN

u

Cu

x

fw

====××××

××××××××××××====

====

====

∞∞∞∞

∞∞∞∞

ρρρρ

ρρρρττττ

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SOLUTIONS

• Will the boundary layer thickness and friction

stress upstream higher or lower compared to

those at the trailing edge?

– δ is smaller upstream as δ is proportional to x1/2

– τw is higher upstream as δ is thinner.

• If the speed of the aircraft is doubled, Rex will be

doubled since

– δ will be smaller as Re increases.

– τw will be higher as the increase in u∞

has a greater effect than the increase in Rex.

νννν

Uxx ====Re

x

w uRe

664.05.0

2

∞∞∞∞==== ρρρρττττ

x

x

Re

5====δδδδ

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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 L=

1 e ee

dP dUU

dx dxρ− =

2 2

0

2

UU U x UU V

x y L L yν

∂ ∂ ∂+ = +

∂ ∂ ∂

0 /eV U y L= −

21 0f ff f′′′ ′′ ′+ + − =0U

yL

ην

≡

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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 δ θ≡ =

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 dyν− =

1 exp( )wyVU

U ν∞

= − −