BLayer Flows 13c

8/10/2019 BLayer Flows 13c

1/14

The Falkner-Skan Flows

The similarity solution for an external stream velocity proportional

to xm. The way in which acceleration (favorable pressure gradient)

or deceleration (adverse pressure gradient) of the external stream

affects the velocity profile inside the boundary layer, skin friction

and heat transfer characteristics can be shown by a family of

solutions given by Falkner-Skan.

Special Case: m = 1 stagnation point flow

U(x) = cxm

m/(1+m)

m > 0U(x)

m < 0

Acceleration Deceleration

U(x)


2/14

2

2

2

2

2

2

0

:Equations

y

CD

y

Cv

x

Cu

y

T

y

Tv

x

Tu

y

u

dx

dUU

y

uv

x

uu

y

v

x

u

AB

m

s

s

cxxU

CCTTxUu

CCTTvu

)(where

,),(:flowpotential

:layerboundarytheOutside

.and,0,0

:surfacetheAt

:conditionsBoundary

Falkner-Skan Flow with Heat Transfer


3/14

x

U

yand

x

U

y

m

xU

U

x

Uy

x

Uy

xx

xUy

fxUyx

2

22

1

2

1

2

1

variablesimilaritytheiswhere

)(),(

:functionStream

SolutionSimilarity


4/14

mffd

dfm

x

Uv

d

fd

x

U

yd

fdU

y

u

d

fd

x

UU

yd

fdU

y

u

d

fdUm

xd

dfU

xd

fdU

d

dfU

x

u

d

dfUu

2

1

2

1

2

1

2

1

2

1

2

1

3

322

3

3

2

2

2

2

2

2

2

2

2

2


5/14

2

2

2

2

2

2

2

2

,

,21

21

,

,2

1

2

1

)(and)(

With

:onsdistributiionconcentratandeTemperatur

d

d

x

U

yd

d

x

U

y

ddm

xx

d

d

x

U

yd

d

x

U

y

d

dm

xx

CC

CC

TT

TT

s

s

s

s


6/14

0)1(2

:

0)1Pr(2

:

012

1

:

2

fmSc

ionConcentrat

fm

EquationEnergy

mfmffmf

EquationBlasius

EquationMomentum

.1and1

,1

:),(

b.layertheOutside

00

0,0

:)0(surfacetheAt

:conditionsBoundary

f

y

and

ff

Reduced Problem


7/14

Pr10

010

110

2

0.3

0.6

0.9

1.2

1.5

1.8

2.1

2.4

f"(0)'(0)0.5704 Pr

2/5

Skin friction and heat transfer coefficient for

stagnation point flow (m = 1)

5/2Pr5704.0)0(and

23259.1)0(

f


8/14

2/12/1

0

,

2/12/1

5/2

Re

93036.4

Re

)0(4)(

1

ReWith:Lover)(ofvalueAverage

Re

146518.2

Re

)0(2

ReWith:frictionSkin

frictionskinstress,shearWallPr5704.0)0(and23259.1)0(1For

:flowpointstagnation

forsimulationnumericalofresultstheFrom

LL

L

faverf

Lf

xx

f

x

fdxxC

L

C

ULxC

fC

Ux

fm


9/14

2/14.0

4.0

0

2/14.04.0

RePr14.1and

Pr14.1Re)(1

overaveragedtcoefficientransferheattheiswhere

lengthofplateflat

ofsectionthefornumberofvalueaveragedThe

RePr57.0andPr57.0

:transferHeat

Laver

L

L

L

aver

aver

aver

L

xx

k

LhNu

L

k

dxxhLh

Lh

k

Lh

Nu

L

Nu

k

hxNu

x

Ukh


10/14

2/14.0,

4.0

0

,

,

,

2/14.04.0

Re14.1and

14.1Re)(1

overaveragedtcoefficientransfermasstheiswhere

lengthofplateflat

ofsectionthefornumberofvalueaveragedThe

Re57.0and57.0

transferMass

LAB

avermL

LAB

L

maverm

averm

AB

averm

L

xm

xm

ScD

LhSh

ScL

D

dxxhLh

Lh

D

Lh

Sh

L

Sh

Sck

xhShSc

x

Ukh


11/14

Turbulent flat-plate Boundary layer

71

71

y

U

u

The oneseventh profile, used by many researchers

Will not permit evaluation of shearing stress at the wall using

du/dyat y= 0

URewhere,ReU0.0233

2w

41

We use the result from pipe flow:

51

51

x2

21

wf

x Re

0.0594

U

Cand,

Re

0.382

x

And so we obtain:


12/14

Drag force cont.

For laminar flowover a plate ( parallel to flow):

Lx

L

0x

f

ReL

x

UL

L

x

xURe

anddxw AusingandRe

0.664C

We obtain:

L0

L

L

0 L

D

Re

1.328

L

xd

L

x

Re

0.664

ReLx

dxw0.664

wL

1C

21

21

L


13/14

Drag force cont.

For turbulent flowover a plate ( parallel to flow):

10Re10for

Relog

0.455C

10Re10*5forRe0.07425C

:RelationsEmpirical

9L

7

58.2L

D

7L

5

L

D5

1


14/14

3/15/4

7L

53/15/43/1

3/2

PrRe037.0

:tcoefficientransferheatAveraged

)10Re105(PrRe0296.0PrRe

2

1

)60Pr5.0(2

1Pr

:AnalogyColburnReynolds

PrRe

number;Stanton

:numberNusseltandtcoefficientransferHeat

LL

xxfx

f

x

x

Nu

CNu

CSt

Nu

cU

hSt

Heat Transfer coefficient in turbulent flow over a flat plate

BLayer Flows 13c

Documents

Transcript of BLayer Flows 13c