(14 15) boundary layer theory

8/12/2019 (14 15) Boundary Layer Theory

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SOLUTION OF BOUNDARY LAYER

Associate Professor

IIT Delhi

E-mail: [email protected]

P.Talukdar/Mech-IITD


2/25

Boundary layer ApproximationApplying Newtons 2nd law in the

y-direction, we get y-momentum

equation2

x

P

y

u

y

uv

x

uu

2

2

=

+

X momentum:

yx

v

y

v

vx

v

u 2 =

+

0P=

)x(PP,Thus =dx

dP

x

P,Hence =

y

For a flate plate, since u = U= constant and v = 0 outside

the boundary layer, X-0

x

P=

momen um equa on g ves

Therefore,forflowoveraflatplate,thepressureremainsconstantovertheentire


plate(bothinsideandoutsidetheboundarylayer).


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Boundary layer over a flat plate



4/25

Thecontinuityandmomentumequationswerefirstsolvedin1908bytheGerman

Blasius (1883 1970) was a

German Fluid Dynamic

Engineer. He was one of theeng neer . as us,as u en o .Prandtl.

Thiswasdonebytransformingthetwo

first students of Prandtl.

Born 9August1883

partialdifferentialequationsintoasingleordinarydifferentialequationbyintroducinganewindependentvariable,

er n, ermanyDied

24April1970(aged86)Hamburg,WestGermany .

Thefindingofsuchavariable,assumingitexists,ismoreofanartthanscience,and

Fields Fluidmechanicsand

mechanicalengineeringAlmamater UniversityofGttingen

itrequirestohaveagoodinsightoftheproblem.

Doctoraladvisor

LudwigPrandtl



5/25

The shape of the velocity profile remains the same along the plate.

Blasius reasoned that the nondimensional velocity profile u/u should

remain unchanged when plotted against the nondimensional distance

,

given x.

That is, althou h both and u at a iven var with x, the velocit u ata fixed y/ remains constant

Blasius was also aware from the work

of Stokes that is proportional to

x

u



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Scale Analysis

Dividing by x to express the result in dimensionless form gives



7/25

Similarity VariableThe significant variable is y/, and weassume that the velocity may be expressed as a

function of this variable. We then have

=

u

x/ydefineWe

/y,makesThis

Here, is called the similarity variable, and g()is the function we seek as a solution



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Variable TransformationA stream function was defined such that: y

=

x

v

=

x

uy

= udyd =

( ) = fuxux

udyd

=

= d)(g)(fwhere

( ) ( )

=

= uxuf

uxuf

=

=

=

=

=

=

=

fdfu1

fudfx

uv

ddfu

xu

ddf

uxu

yyu


dx2xu2xdux


9/25

Differentiating the previous equation with respect to x and y

3222

2

2

fduufduu

,d

fd

x2

u

x

u

=

322

dx

y

,

dx

uy

=

=

simplifying, we obtain

0d

fd

fd

fd

2 2

2

3

3

=+

This is a third-order nonlinear differential equation.

.



10/25

Blasius Solution0

d

fdf

d

fd2

2

2

3

3

=

+

The value of corresponding to u/u = 0.992is 5.0

xu5

xuy

=

=


Significance

of u, , xxRe

x.

xu

.=

=


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12/25

The shear stress at the wall can be

eterm ne rom:

2

2

wfdu

uu

=

=00y ==

2u332.0u

x

wRex

u. ==

21x2

w

2

wx,f Re664.0

2u2VC

=

=

=

Note that unlike the boundary layer

thickness, wall shear stress and the skin


friction coefficient decrease along the plate

as x-1/2

.


13/25

Ener E uationIntroduce a non-dimensional temperature s

TT

T)y,x(T)y,x(

=

Substitution gives an energy equation of the form:2

2

yyv

xu

=+

Temperature profiles for flow over an isothermal flat plate are similar like the

velocity profiles.

Thus, we expect a similarity solution for temperature to exist.

Further, the thickness of the thermal boundary layer is proportional to u/x

Using the chain rule and substituting the u and v expressions into the energy equationgives

22


2 ydydf

dx2xddu

=

+


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Using the chain rule and substituting the u and v expressions into theenergy equation gives

22ddf

dfu1ddfu

=

+

Compare

df/d is replaced by

fdfd 230d

fPrd

22

=

+

For Pr = 1 dd 23 d

1d

dfand0d

df

== ==( ) ( ) 100 == and

Thusweconcludethatthevelocityandthermalboundarylayerscoincide,and

forsteady,incompressible,laminarflowofafluidwithconstantpropertiesandPr=1overanisothermalflatplate


The value of the temperature gradient at the surface (Pr =1) ??

332.0d

fddd

2

2

=

=


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This eq. is solved for numerous values of Prandtl numbers.

For Pr > 0.6, the nondimensional temperature gradient at0d

fPrd

22

=

+

Pr>0.6the surface is found to be proportional to Pr

31Pr332.0d

=

0=

s

s

TT

T)y,x(T)y,x(

=

x

uy

y

=

=

The temperature gradient at the surface is

( ) ( )

u

ydTT

yTT

y

31

0y0

s

0y

s

0y ==

==

=

=


xr. s

=


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This solution is given by Pohlhausen


17/25

The local convection coefficient can be expressed as:

uy

Tk

q 310ys

= u

y

T

0y

=

=

And the local Nusselt number becomes

( ) ( ) x.

TTTT ssx

x

r. s

21x

31xx RePr332.0

k

xhNu == Pr > 0.6

Solving the thermal boundary layer equation numerically for the temperature

profile for different Prandtl numbers, and using the definition of the thermal

boundar la er it is determined that

31

t

Pr

P.Talukdar/Mech-IITD x3131t

RePr

x0.5

Pr

=

=


18/25

-0

vu=

+

yx

puuu 2

xyyx 2

22 TTTT

=

22p yxy

vx

uc

s

s*

2

*****

TT

TTTand

V

pp,

V

vv,

V

uu,

L

yy,

L

xx

=

=====



19/25

** 0yx ** =

+

**2**

Continuity:

*2*L** dxyRey

v

x

u

=

+

Momentum:

2*

*2

L*

**

*

**

y

T

PrRe

1

y

Tv

x

Tu

=

+

Energy:

With the boundary conditions:

******** ====

( ) ( ) ( ) 1,xT,00,xT,1y,0T,,,,,,,

****** ===



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21/25

Functional forms of Friction and

onvect on oe c ents

**2**

** dpu1uu

*2*L** dx

y

Reyx

=

,

can be expressed as L**

1* Re,y,xfu =

T en t e s ear stress at t e sur ace ecomes

( )L*2**

0

s Re,xfL

V

y

u

L

V

y

u

*

=

=

==

( ) ( ) ( )L*3L*2L*222s

x,f Re,xfRe,xf2

Re,xfLV

C ==

=

=



22/25

*2**

The solution for T* can be expressed as

2*L**

yPrRey

vx

u

=

+

Energy:

Pr,Re,y,xgT L**

1* =

Using the definition of T*, the convection heat transfer coefficient becomes

0y

**

0y

**

s

s

s

0y

**yT

LyT

)TT(LTTh

==

==

=

=

usse num er:Pr),Re,x(gyT

kNu L

*2

0y

**x *

====

Note that the Nusselt number is equivalent to the

dimensionless temperature gradient at the


,

dimensionless heat transfer coefficient


23/25


24/25

WhenPr=1(approximately the case

* * =2*

*2

L*

**

*

** u

Re

1

y

uv

x

uu

=

+

. .

plate)

2*

*2

L*

*

**

*

*

yT

Re1

yTv

xTu

=+

*ThL

0y*x

*yk =

x*

*

s NuL

Vu

L

Vu =

=

= ** Tu

0yy ==

x2

x

2

sx,f Nu

2Nu

L

V

C =

=

=

0y*

0y*

** yy == =

xL

x,f NuRe

C = (Pr=1)x

x,fSt

2

C= (Pr=1)


PrRe

Nu

Vc

hSt

Lp

=

=


25/25

Clinton-Colburn Analogy

Also called modified Reynolds analogy

21

Colburn j-factor

xx,f e.= xx er.u =

Taking the ratio between Cf,x and Nux

Valid for

31x

xx,f PrNu

2

ReC = H

32

p

xx,f jPrVc

h

2

=

Although this relation is developed using relations

for laminar flow over a flat plate (for which

P*/x* = 0), experimental studies show that it is

0.6

(14 15) boundary layer theory

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