Fluidmech in Pipe Velocity Shear Profile

8/21/2019 Fluidmech in Pipe Velocity Shear Profile

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FULLY DEVELOPED PIPE ANDCHANNEL FLOWSKUMAR DINKAR ANAND3 rd YEAR, MECHANICAL ENGG.IIT-KHARAGPURGUIDANCE : PROF. S CHAKRABORTYINDO-GERMAN WINTER ACADEMY-DECEMBER 2006: THE OUTLINE : Hydraulically developing flow through pipes and channels andevaluation of hydraulic entrance length. Hydraulically fully developed flows through pipes and channels . Hydraulically fully developed flow through non-circular ducts. Definition of Thermally fully developed flow and analysis of thermallyfully developed flow through pipe and channels. Analysis of the problem of Thermal Entrance: The Graetz Problem.Fully Developed Flows There are two types of fully developed flows :1.) Hydraulically Fully Developed Flow2.) Thermally Fully Developed FlowContdHydraulically Fully Developed FlowDefinition: As fluid enters any pipe or channel , boundary layers keep on growingtill they meet after some distance downstream from the entrance region. After this

distance velocity profile doesn't change, flow is said to be Fully Developed.Analysis of fluid flow before it is fully developed:Velocity in the core of the flow outside the boundary layer increases withincreasing distance from entrance. This is due to the fact that through any crosssection same amount of fluid flows, and boundary layer is growing.This meanshence0 > dx dU0 < dx dpWhere U=Free stream velocity in the core before flow is fully developedp= Free stream pressureContd

Schematic picture of internal flow through a pipe :Velocity Profile ,Using the boundary conditions :1.) At2.) At3.) AtWe get the velocity profile as :Contd2) ( cy by a y u + + =d = y U u =0 = ud = y

0 =

y

u2) ( ) ( 2) (d d y yUy u- =0 = yWhere Free stream velocity of entering flui


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Free stream core velocity insi

e the tubeCore velocity of fully

evelope

flowRa

ius of pipeNow from the principle of conservation of mass :Hence ,Cont

=U= U --+ =RRRUr

r ur

r R Uddp p p02

2 2 *r R y - == R=eU2) / ( 6 / 1 ) / ( 3 / 2 11R R UUd d + -=

22) / ( 6 / 1 ) / ( 3 / 2 1) / ( ) / ( 2R Ry yUud dd d+ --

=\ - + - =d dr r t0 02} ) / 1 ( ) / 1 ( / { dy U u


3/46

dxdUU dy U u U u UdxdwBounda

y Laye

momen

um in

eg

al equa

ion:Whe

e, Shea

s

ess a

wall,F

om Be

noulli's Equa

ion fo

f

ee s

eam flow

h

ough co

e:Using Navie

-S

okes equa

ion a

he wallCon

d

0 ==ywyu txpdxdU

U- =r1022==

yyuxpSolving fo

bounda

y laye

hickness :In

eg

a

e momen

um In

eg

al Equa

ionUsing

he bounda

y condi

ion a

Fo

de

e

mina

ion of En

ance Leng

h :pu

ing a

We ge

he exp

ession fo

En

ance Leng

h as:Con

d) ( d0 = d 0 = x) (eLR = deL x =


4/46

) (eLDeDLRe 03 . 0 =Analytical ex ression for Entrance Length :Hence it can be observe

that our ex ression for Entrance Length

iffersfrom the analytical ex ression

ue to the following reasons:1.) We have assume arabolic velocity rofile in the boun ary layer2.)We have not use

the Navier Stokes boun

ary equation at wall for velocity rofile

etermination3.) We are

oing boun

ary layer analysis which gives a roximate resultsCont ) (eLDeDLRe 06

. 0=2) / ( ) / ( 2 d d y yUu- =022=

=yyux

2) / ( 6 / 1 ) / ( 3 / 2 11R R U

Ud d + -=Schematic icture of internal flow through a channel:Velocity ProfileUsing the boun ary con itions :1.) At2.) At3.) At


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We get the velocity rofile as :Cont

2) ( cy by a y u + + =0 = y 0 = ud = y U u =d = y 0 =

y

u2) ( ) ( 2) (d d y yUy u- =Here , Distance between the arallel lates of channelWi th of the ChannelFree stream velocity of entering flui

Free stream velocity insi

e channelCore velocity of fully

evelo e

flowEntrance LengthHy

raulic DiameterCont

= D= W

=U= U=eU=eL=HD D

WWDPA H2244 = = =From the rinci le of conservation of mass:Hence when flow is fully evelo eCont

-

+ =d d ) 2 / (0 02 2 *DU y u y D U\) / ( 3 / 2 11


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D UUd -=\) / ( 3 / 2 1) / ( ) / ( 22Dy yUudd d--=) 2 / ( D = d= U U e 5 . 1From Boun

ary layer momentum integral equation :Where, Shear stress at wall,

From Bernoulli's Equation for free stream flow through core:Using Navier Stokes equation at the wallCont

- + - =d dr r t0 02} ) / 1 ( ) / 1 ( / { dy U udxdUU dy U u U u U

dxdw0 ==ywyu tx

pdxdUU- =r10


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22==yyu

xpSolving fo

bounda

y laye

hickness :In

eg

a

e momen

um In

eg

al Equa

ionUsing

he bounda

y condi

ion a

Fo

de

e

mina

ion of En

ance Leng

h :pu

ing a

We ge

he exp

ession fo

En

ance Leng

h as:ORCon

d) ( d0 = d 0 = x

) (eLeL x = R = d) (eLDeDLRe 025 . 0 =

HDHeDLRe 00625 . 0 =Analytical ex ression for Entrance Length :Hence it can be observe

that our ex ression for Entrance Length

iffersfrom the analytical ex ression ue to the following reasons:1.) We have assume

arabolic velocity rofile in the boun

ary layer2.) We have not use the Navier Stokes boun ary equation at wall forvelocity rofile

etermination

3.) We are

oing boun

ary layer analysis which gives a

roximate results.Cont

) (eLDeDLRe 05


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. 0=2) / ( ) / ( 2 d d y yUu- =022==yyux

) / ( 3 / 2 11

D UUd -=Analysis of fully

evelo e

flui

flow:Fully Develo e

Flow Through a Pi e: From Equation of continuity in cylin rical coor inates:for an incom ressible flui

flowing through a i eCont

0 ) (1=

+xurur rrHere, ra

ial velocityaxial velocityra

ius of i e

No flui

ro

erty varies with ,,at wall of the i ehence it is zero everywhere.Hence Equation of continuity re

uces to :Momentum Equation in ra ial coor inate:Cont

=ru= u


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= aq0 =ru , 0 =xu) (r u u =, 0 =rp ) (x p p =Momentum E uation in axial

irection :) (

r

ur

r

r

x

p =Solving above diffe

en

ial equa

ion in (

) using

he bounda

y condi

ions:1.) Axial veloci

y (u) is ze

o a

wall of pipe (

=R)2.) Veloci

y is fini

e a

he pipe cen

e

line (

=0).We ge

he fully developed veloci

y p

ofile:Con

d

-

- =2214 ar


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x auShea

S

ess Dis

ibu

ion :Shea

s

ess ,Maximum shea

s

ess a

wall ,

= =xp

d

du

x2 t

=xp a20tCon

dHence i

can be obse

ved

ha

Shea

s

ess dec

eases f

ommaximum

o ze

o a

pipecen

e

line and

hen inc

eases

o maximum again a

wall.Volume Flow Ra

e :

volume flow

a

e ,- = =


11/46

x

aur r Qa0482ppNow in a fully

evelo e

flow ressure gra

ient is constant ,Hence ,( )L

L x

ent exit- =

-=\L

aQp84

=Cont

Average Velocity :Average velocity ,- = = =

x aaQAQV p 82


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2Maximum Velocity :At the oint of maximum velocity ,0 =

r

uThis corres on

s to core of i e ,0 = rHence Vx

aU u ur2420 max=

- = = ==Con

dFully Developed Flow

h

ough Channel :F

om equa

ion of con

inui

y wi

hin

he en

ance leng

h : 0 =+

yvxuIn en

ance leng

h bounda

y laye

s g

owing ,0 xu0 v

It means flow is not parallel to walls in entrance regionCont

) (aE uation of Continuity for an incompressible flui in fully evelope region :0 =xu


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) (y u u =Momentum e uation in y

irection (transverse

irection) :0 =yp) (x p p =Momentum e uation in x

irection (along length of channel) :

=

22yuxpSolving above diffe

en

ial equa

ion in y using bounda

y condi

ions :u(y)=0 a

y=0 and y=aCon

dWe ge

he veloci

y p

ofile :

-


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=ayayxp au

222 Shea

S

ess Dis

ibu

ion :-

=

=21ayxpayuyx t Shea

S

ess ,

Maximum Shea

S

ess a

walls , - =x


15/46

a20tCon

dHence i

can be obse

ved

ha

Shea

s

ess dec

eases f

ommaximum

o ze

o a

cen

eof

he channel and inc

eases

o maximum again a

wall.Volume Flow Ra

e :

Volume flow

a

e pe

uni

wid

h of channel,30121axpudy Qa

- = =Con

dAve

age Veloci

y :2121

axpaQV

- = =Ave

age Veloci

y ,Maximum Veloci

y:A

he poin

of maximum veloci

y , 0 =yu


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This co

esponds

o cen

e of channel ,2ay =Hence , V axpu u23

812max= - = =

Con

dFully Developed Flow Th

ough Non-Ci

cula

Duc

s :) ( Ellip

ical C

oss Sec

ion :As flow is fully developed in

he ellip

ical sec

ion pipe :0 = =z yu uF

om equa

ion of con

inui

y fo

incomp

essible flow :, 0 =+

+zuyuxuzy

x0 =xu x) , ( z y u ux x= \


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Cont

12222=

+

bzayMomen

um Equa

ion in x-di

ec

ion :

+=2222z

uyuxpx xBounda

y condi

ion : on0 =x


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u12222=

+

bzaySolu

ion P

ocedu

e :Use ,Such

ha

non ze

o cons

an

s and

o be de

e

mined using :1c2c22

21) , ( ) , ( z c y c z y u z y ux x+ +=1.)2.) is cons

an

on

he wall .0 ) , (2=

z y u x) , ( z y u x Con

dUsing

he assumed veloci

y p

ofile and solving

he momen

umequa

ion using

wo s

a

ed condi

ions:21) , ( a c z y u x - =, along the wall


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Using La lace maximum criteria ( Maximum an

minimum of a functionsatisfying La lace equation lies on the boun

ary) :over entire

omain . ) , (21const a c z y u x = - =We get our velocity rofile as :

- -+

- =22222 22 2121) , (

bzayb ab ax

z y u xCon

dVolume

ic Flow Ra

e :Volume flow

a

e ,

dAbzayb ab axpdA z y u Q


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ion

ionx

- -+

- = =sec sec22222 22 2121) , (

\2 23 34 b ab axpQ+

- =pCont Thermally Fully Develo e

Flows :


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) ( Thermally fully

evelo e

flow through a i e :Cont

When flui

enters the tube with tube walls at a

ifferent tem eraturefrom the flui tem erature , thermal boun ary layer starts growing. After some

istance

ownstream (thermal entry length) thermally fully

evelo e

con

ition is eventually reache

:Thermally fully

evelo e

con

ition is

ifferent from Hy

raulicfully evelo e con ition ., 0 =xu, 0 xTfor hy

raulic fully

evelo e

flowat any ra ial location for thermally fully evelo eflow as convection heat transfer is occurring.Cont

Con

ition for Thermally Fully Develo e

Flow :Because of convective heat transfer , continuously changes

with axial coor

inate x .) (r TCon

ition for fully

evelo e

thermal flow is

efine

as :0) ( ) () , ( ) (=

--x T x Tx r T x Txm ssThis means although tem erature rofile changes with xBut the relative tem erature rofile

oes not change with x.) (r TCont

Here , = ) (x T s= ) (x T m Mean Tem eratureMean Tem erature ( ) is efine as: ) (x T mSurface Tem erature of the i evAc vmc mT

A uc


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Tc&=rThe

mal Ene

gy

anspo

ed by

he fluid as i

moves pas

anyc

oss sec

ion ,m vAc v

T c m TdA uc Ec&&= =rF

om New

on's Law of Cooling : ) (m s sT T h - =Since there is continuous heat transfer between flui

an

walls : 0 x

T mCont

0) ( ) () , ( ) (=-

-x T x Tx r T x Txm ssFrom the

efinition of thermally fully

evelo e

flow :Hence , ) () ( ) ( ) ( ) () , ( ) (0

0x fx T x TrTx T x Tx r T x Trm sr r


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r rm ss--=

--==Here is ra

ius of the i e . ) (0r

From Fourier's heat con

uction law at the wall an

Newton

s law of cooling:[ ] ) ( ) (00x T x T hrTkyTk m sr r

y ys- ==

- ==


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=Hence ,) (x fkhHere , Local convection heat transfer coefficient= h= k Coefficient of thermal con

uction (flui

)Hence, h is infinite in theBeginning (boun

ary layersjust buil ing u ), then ecaysex onentially to a constantvalue when flow is fully

evelo e

(thermally )an

thereafter remains constant.Cont

t f

x,f

hhx

Com

etition between Thermal an

Velocity boun

ary Layers :This com etition is ju

ge

by a

imensionless number , calle

Pran

tl number .=anPr

Where , Ki

em

tic frictio

coefficie

t (mome

tum diffusivity) = =rn= =pckra Therm l diffusivity

tPr

ddWhere , = d Velocity boun ary layer thickness=tdThermal boun ary layer thickness= n Positive ex onentCont If ,


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1 Pr1 PrIt means Velocity Boun

ary Layer grows faster than Thermalboun ary layer. Hence flow first hy raulically evelo e anthen thermally

evelo e

.If ,It means Thermal Boun

ary Layer grows faster than Velocityboun ary layer. Hence flow first thermally evelo e an thenhy

raulically

evelo e

.) ( Hence if, an

flow is sai

to be thermally

evelo e

it meansFlow is alrea

y hy

raulically

evelo e

.1 Pr) (Similarly if, an

flow is sai

to be hy

raulically

evelo e

itMeans flow is alrea

y thermally

evelo e

.1 PrCont

) ( Usually surface con

itions of i e fixe

by im osing con

itions :1.) Surface tem erature of i e is ma

e constant ,2.) Uniform surface heat flux ,. const T s =. const q s =Constant Surface Heat Flux :

From the

efinition of fully

evelo

e

thermal flow:0) ( ) () , ( ) (=--

x T x Tx r T x Txm ssdxdTT TT TdxdT

T TT TdxdTxTmm ss sm s


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s s--+--- =Cont

From Newton's Law of cooling :) (m s sT T h - =. const s =As, hence,

x

T

x

Tm s

=. const

x T

x

TxTm s= = =Hence using

efinition of thermally fully

evelope

flow

an

Newton's Law:sTmT. const s =s

Cont

Neglecting viscous issipation, energy e uation :

=


27/46

+

T

Tvx

TuaAssumi g the flow to be both hydr ulic lly d therm lly developed :, 0 =xu, 0 = vdxdTx

Tm=, 1 220

- =r

ru umHe ce e ergy equ tio reduces to :


28/46

-

=

2012 1

dxdT u

T

m maCo td,V u m =dxdTxTm=

I tegr ti g e ergy equ tio usi g bou d ry co ditio s :1.) Temper ture , is fi ite t ce tre , ) , ( x r T0 = r) (0x T Ts r r==


29/46

We get Temper ture profile :2.) Temper ture ,

-

+ - =204

02041161163 2) ( ) , (rrr

rdxdT r ux T x r Tm msaFrom defi itio of me temper ture ,vA


30/46

c vmc mTdA ucTc&=r

- =dxdT r ux T x Tm ms ma2

04811) ( ) (Co tdFrom the pri ciple of e ergy co serv tio :Pdx q dqs co v=dxm&m

Tm mdT T +x) (pv) ( ) ( pv d pv +r1= vspecificvolume,) ( pv T c d m dqm v conv

+= &Fo

an ideal gas, ,mRT pv = R c cv p+ =Pe

ime

e

,\ Pdx q dT c m dqs m p conv


31/46

= = &\ ( )m sp ps mT Tc mPhc mP q

dxdT- ==& &Co tdD P p =

=42Du mmpr &DHence combining

he equa

ions ob

ained by in

eg

a

ion of ene

gy

equa

ion in bounda

y laye

and conse

va

ion of ene

gy equa

ion :kD qdxdT

ux T x Ts m ms m- =

- = -48114811) ( ) (20


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a\\ ( ) ) ( ) (4811) ( ) ( x T x TkhDx T x Ts m s m- = -\ 36 . 41148= = =khDNu DHe ce Nusselt umber for fully developed flow through circul r pipeexposed to u iform he t flux o its surf ce is co st t ,i depe de t of

xi l loc tio ,Rey old's umber d Pr dtl umber .) (Co tdCo st t Surf ce Temper ture :

From the defi

itio

of fully developed therm

l flow :0) ( ) () , ( ) (=--

x T x Tx r T x Txm ssdxdTT TT TdxdT

T TT TdxdTxTmm ss sm s


33/46

s s--+--- =0 =

x

T sConstant surface temperature ,\--=

dxdTT TT TxTmm ssCo tdHe ce it c be see th t , depe ds o r di l coordi te.

Fully developed temper

ture profile for co

st

t w

ll temper

turehe ce differs from co st t surf ce he t flux co ditio .xT. co st T s =mTs

qCo tdNeglecti g viscous dissip tio , e ergy equ tio :


34/46

=+

T

TvxTuaAssumi g the flow to be both hydr ulic lly d therm lly developed :, 0 =xu

, 0 = v, 1 220

- =rru um,

V u m =--


35/46

=dxdTT TT TxTmm ssCo tdHe ce bou d ry l yer e ergy equ tio becomes :m ss m mT TT TrrdxdT urT

r r --

- =

20


36/46

12 1aAbove equ tio is solved usi g iter tive procedure :66 . 3 =DNuCo tdFully developed therm l flow through ch el :) ( Ch el w lls subjected to co st t he t flux :Here we co sider ch el with := =sq=sT= ) , ( x r T=mT= W ,

Depth of ch

el Width of ch

elHe t flux t the w llsTemper ture of fluid flowi g through ch elTemper ture t the w llMe Temper ture or Bulk Temper tureCo tdNeglecti g viscous dissip tio , e ergy equ tio :22yTyT

vxTu=+a

= = W P 2= = = = W

WPAD H 224 4Hydr ulic di meter


37/46

perimeterAssumi g the flow to be both Hydr ulic lly d therm lly developed :, 0 =xu, 0 = v. co stdxdTxTm= =

- - =

- =a

yayuayayxp a


38/46

um2 2262 Con

dHe

e, =mu Mean veloci

y , is defined as :x

p aAudAucAcmc- = =

rr122Now solving fo

bounda

y laye

ene

gy equa

ion :2 13246 126 c y cay

aydxdT uTm m+ +

- - =aCo st ts of i tegr tio obt i ed usi g :1.) t ( s temper ture profile issymmetric he ce h s extremev lue t ce tre.), 0 =dydT2


39/46

y =Co td2.) t the w ll ,

y y = = & 0sT T =sT c = 2Hence we obtain the temperature profile :

+ - - = -12 6 126324

y

y

ydxdT uT Tm msaFrom the defi itio of me temper ture,vA

c vmc mTdA ucTc&=r

- = -dxdT uT Tm ms ma


40/46

214017Co tdFrom co serv tio of e ergy method ( simil r to c se of pipe):( )m sp ps mT Tc mPhc mP qdxdT- ==& &He ce combi i g temper ture profile d co serv tio of e ergy :( )s mp

ms mT Tc mPh uT T - = -& a214017Usi g , ( ) , W u mmr = & , 2W P =

,pckra = D H 2 & =( )17140 2= =k

hNu

HDCo tdTherm l E tr ce : The Gr etz Problemu rx00T0


41/46

T0TwT0rProblem St teme t:Fluid i iti lly t u iform temper turee ters i to pipe t surf ce temper ture differe t th the fluid. Flow ssumed to be Hydr ulic lly developed .Co td...wTPr dtl umber of fluid is high , he ce therm l e tr ce st rts f rdow stre m.Flow lre dy hydr ulic lly developed. Pr dtl umber of fluid is high , he ce therm l e tr ce st rts f rdow stre m.Flow lre dy hydr ulic lly developed. Here , U iform temper ture of fluid before therm l e tr ce =0T=

wT U iform surf ce temper ture of w lls= ) , ( r x TFluid temper ture i therm l e tr ce regioAs the flow is hydr ulic lly fully developed :, 1 220

- =r

ru umNeglecti g viscous dissip tio , bou d ry l yer e ergy equ tio :


42/46

=

T

xT

uaCo tdBou d ry Co ditio s :1.) t2.) t, 0 x0T T =, 0 xwT x r T = ) , (0

Solutio

:Solutio do e with the help of o dime sio l v ri bles.,0*T TT TTww--= ,

0*rrr =Pr Re0*dxx =Here ,, Re

0nmu d=kc p nran= = Pr


43/46

He ce e ergy equ tio reduces to :( )

-=**** 2* **

*12rTrrr rxTCo tdBou d ry co ditio i terms of o dime sio l v ri bles :, 1 )

0 ,(* *= r T 0 ) , 1 (* *= x TSolvi g the e ergy equ tio usi g v ri ble sep r tio method :) ( ) ( ) , (* * * *x g r f x r T = Usi g ,the p rticul r solutioi e ergy equ tioWe obt i :( )

.122* **co stf r rf f rg


44/46

g= - =- + =lHence , ( )* 22 exp x C g l - =( ) 0 12* * 2 *= - + + f r r f f r lContdHence the particu ar so ution wi be :) ( ) 2 exp( ) , (* *2* **r f x C x r Tn n n nl - =

From the princip

e of

inearity and superposition : ==- =nnn n nr f x C x r T0* *2* * *

) ( ) 2 exp( ) , ( l1 ) 0 ( =nf0 ) 1 ( =nf, for simp icity, using the boundary condition0 ) , 1 (* *= x TUsing the other condition ,

=== =nnn nr f C r T0* * *


45/46

) ( 1 ) 0 , (To be determined using theory of orthogona

functions.,nCContd--=10*2* *10* * *) 1 () 1 (22

dr f r rdr f r rCnnnUsing theory of orthogona

functions :Now the rest of the prob em is numerica y so ved for Nusse t Number :( )( )-

-=-*2 2*22 exp ) 1 ( 22 exp ) 1 (x f Cx f C

Nun n n nn n nxl llcontd: KEY QUESTIONS : IF FLOW THROUGH A PIPE OR CHANNEL IS SAID TO BEHYDRAULICALLY FULLY DEVELOPED DOES THIS IMPLY


46/46

THERMALLY FULLY DEVELOPED AND VICE VERSA ???? IF TWO PLATES IN THE CHANNEL ARE MAINTAINED ATDIFFERENT TEMPERATURES THEN WHAT WILL BE THECRITEREA FOR THERMALLY FULLY DEVELOPED FLOW ????THANK YOU FOR YOUR COOPERATIONTHE END

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Fluidmech in Pipe Velocity Shear Profile