fluidmech in pipe velocity shear profile
TRANSCRIPT
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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
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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 =
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) (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 ,- = =
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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
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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 :
=
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+
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 :
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-
=
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==
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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
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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
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= = &\ ( )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
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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 :
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=+
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 =--
-
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=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
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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
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perimeterAssumi g the flow to be both Hydr ulic lly d therm lly developed :, 0 =xu, 0 = v. co stdxdTxTm= =
- - =
- =a
yayuayayxp a
-
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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
-
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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
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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
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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 :
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=
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
-
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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
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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* * *
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) ( 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
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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