lect - 4 steady conduction one dimensional.pptx
TRANSCRIPT
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8/17/2019 Lect - 4 Steady conduction One Dimensional.pptx
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Steady Conduction One DimensionalDr. Senthilmurugan S. Department of Chemical Engineering IIT Guwahati - Part 4
Slabs, cylinders and spheres; Critical
thickness of insulation
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Ob!ecti"es
#nderstand the concept of thermal resistance and its
limitations, and de"elop thermal resistance net$orks for
practical heat conduction problems
Sol"e steady conduction problems that in"ol"e multilayer
rectan%ular, cylindrical, or spherical %eometries
De"elop an intuiti"e understandin% of thermal contactresistance, and circumstances under $hich it may be
si%nificant
&dentify applications in $hich insulation may actually
increase heat transfer
'naly(e finned surfaces, and assess ho$ efficiently and
effecti"ely fins enhance heat transfer
Sol"e multidimensional practical heat conduction problems
usin% conduction shape factors
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Contents
Steady state conduction heat transfer one dimension
Steady *eat Conduction in +lane alls
-enerali(ed .hermal esistance 0et$orks
.hermal Contact esistance
*eat conduction in cylinders and spheres
.he O"erall *eat.ransfer Coefficient
Critical .hickness of &nsulation
Steady *eat Conduction $ith *eatSource Systems
Cylinder $ith *eat Sources
ConductionCon"ection Systems
ins
.hermal Contact esistance
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Steady State *eat Conduction in +lane alls
ate of heat transfer into the $all ate
of heat transfer out of the $all 4 ate of
chan%e of the ener%y of the $all
.herefore, the rate of heat transfer into
the $all must be eual to the rate of
heat transfer out of it &n other $ords,
the rate of heat transfer throu%h the $allmust be constant
7
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Steady State *eat Conduction in +lane alls
&f the system is in a steady state, ie, if
the temperature does not chan%e $ith
time, then the problem is a simple one,
and $e need only inte%rate ourier8s la$
of heat conduction euation and
substitute the appropriate "alues tosol"e for the desired uantity
hen k is constant
One Dimension 9 Constant 'rea
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.hermal esistance Concept
.he rate of heat conduction throu%h a
plane $all is proportional to the a"era%e
thermal conducti"ity, the $all area, and
the temperature difference, but is
in"ersely proportional to the $all
thickness
heat conduction throu%h a plane $all
can be rearran%ed as
.he thermal resistance of the $all
a%ainst heat conduction or simply the
conduction resistance of the $all
.he thermal resistance of a medium
depends on the %eometry and the
thermal properties of the medium .his euation for heat transfer is
analo%ous to the relation for electric
current flo$ &, e:pressed as Ohm8s la$
in electriccircuit theory<
Similarity $ith =lectric resistance
electrical conducti"ity
Electric resistance
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'nalo%y bet$een thermal and electrical resistanceconcepts
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Onedimensional heat transfer throu%h a composite$all and electrical analo%
.1
.2.)
.3
'ssumption@ esistance for
$all contact is ne%li%ible
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-enerali(ed thermal resistance net$orks
.he thermal resistance concept or the
electrical analo%y can also be used to
sol"e steady heat transfer problems that
in"ol"e parallel layers or combined
seriesparallel arran%ements
4 1B2
.hermal resistance net$ork
for t$o parallel layers
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-enerali(ed thermal resistance net$orks
.he total rate of heat transfer throu%h
this composite system can a%ain be
e:pressed as
.he result obtained is some$hat
appro:imate, since the surfaces of thethird layer are probably not isothermal,
and heat transfer bet$een the first t$o
layers is likely to occur
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-enerali(ed .hermal esistance 0et$orks
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-enerali(ed .hermal esistance 0et$orks ethod
.$o assumptions in sol"in% comple:
multidimensional heat transfer problems
by treatin% them as onedimensional
usin% the thermal resistance net$ork are
1
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.hermal esistance Concept
0e$ton8s la$ of coolin% for con"ection
heat transfer rate can be rearran%ed as
Con"ection resistance
ith con"ection heat transfer
* , , then
't surfaces $here boilin% and condensation
occur
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.hermal esistance Concept
hen the $all is surrounded by a %as, the radiation effects .he rate of radiation
heat transfer bet$een a surface of emissi"ity and area 's at temperature .s and the
surroundin% surfaces at some a"era%e temperature .surr can be e:pressed as
adiation resistance
adiation heat transfer coefficient
.emperature must be in kel"in in the e"aluation of hrad
ith radiation heat transfer
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.hermal contact resistance
.he temperature drop at plane 2, the
contact plane bet$een the t$o
materials, is said to be the result of a
thermal contact resistance
.hermalcontactresistance
effect physical situation
.emperature profile
$here the uantity 1 /hc A is called the thermal
contact resistance and hc is called the contact
coefficient
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.hermal contact resistance
0o real surface is perfectly smooth, and
the actual surface rou%hness is belie"ed
to play a central role in determinin% the
contact resistance
.here are t$o principal contributions to
the heat transfer at the !oint@ .he solidtosolid conduction at the
spots of contact
.he conduction throu%h entrapped
%ases in the "oid spaces created by
the contact
.he second factor is belie"ed to
represent the ma!or resistance to heat
flo$, because the thermal conducti"ity of
the %as is uite small in comparison to
that of the solids
.heory
Actual (imperfect) thermal contact
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.hermal contact resistance
Desi%natin% the contact area by Ac and
the "oid area by Av , $e may $rite for the
heat flo$ across the !oint
$here Lg is the thickness of the "oid
space and k f is the thermal conducti"ity
of the fluid $hich fills the "oid space.he total crosssectional area of the
bars is A Sol"in% for hc , the contact
coefficient, $e obtain
.heory
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.hermal Contact esistance
&n most instances, air is the fluid fillin% the
"oid space and kf is small compared $ith k ' and kE &f the contact area is small, the
ma!or thermal resistance results from the
"oid space
.he main problem $ith this simple theory is
that it is e:tremely difficult to determine
effecti"e "alues of 'c, '", and F% forsurfaces in contact
.hermal contact resistance can be reduced
markedly, perhaps as much as >5 percent,
by the use of a Gthermal %reaseH like GDo$
)37H
rom the physical model, $e may
tentati"ely conclude@
.he contact resistance should increase
$ith a decrease in the ambient %as
pressure $hen the pressure is decreased
belo$ the "alue $here the mean free path
of the molecules is lar%e compared $ith a
characteristic dimension of the "oid
space, since the effecti"e thermal
conductance of the entrapped %as $ill be
decreased for this condition
.he contact resistance should be
decreased for an increase in the !oint
pressure since this results in a
deformation of the hi%h spots of the
contact surfaces, thereby creatin% a
%reater contact area bet$een the solids
.heory
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.hermal Contact esistanceeasurement .echniue
.he thermal contact resistance can be
determined from euation %i"en belo$ by
measurin% the temperature drop at the
interface and di"idin% it by the heat flu:
under steady conditions
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.hermal contact conductance of some metal surfacesin air from "arious sources<
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*eat conduction in cylinders
Consider a lon% cylinder of inside radius
r i , outside radius r o, and len%th L,
e e:pose this cylinder to a
temperature differential T i IT o
or a cylinder $ith len%th "ery lar%e
compared to diameter, it may beassumed that the heat flo$s only in a
radial direction, so that the only space
coordinate needed to specify the system
is r
'%ain, ourier8s la$ is used by insertin%
the proper area relation .he area for
heat flo$ in the cylindrical system is
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*eat conduction in cylinders
ourier8s la$
ith the boundary conditions
.he solution to ourier8s la$ =uation is
.hermal resistance
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Onedimensional heat flo$ throu%h multiple cylindricalsections and electrical analo%
.he thermalresistance concept may be used for multiplelayer cylindrical $alls !ust
as it $as used for plane $alls
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ourier8s la$
ith the boundary conditions
.he solution to ourier8s la$ =uation is
*eat conduction in Spheres
r i
r o
D i i
D i i
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System Flu !esistanceDri"ingforce
=lectrical & ∆J
CartesianConduction
∆.
Cylindrical
Conduction ∆.Conductionthrou%h sphere
∆.
Con"ection ∆.
adiation ∆.
System Flu !esistanceDri"ingforce
=lectrical & ∆J
CartesianConduction
∆.
Cylindrical
Conduction ∆.Conductionthrou%h sphere
∆.
Con"ection ∆.
adiation ∆.
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*eat Conduction in Composite Cylinder
0et *eat transfer rate
0et *eat transfer rate amon% the indi"idual
layer is eual
Onedimensional heat flo$ throu%h
multiple cylindrical sections
=lectrical analo%y
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O"er all *eat .ransfer Coefficient
Consider the plane $all sho$n in i%ure,
e:posed to a hot fluid A on one side and
a cooler fluid B on the other side .he
heat transfer is e:pressed by
O"er all *eat .ransfer Coefficient
+lane $all
=lectrical analo%y
O"erall heat transfer throu%h a plane $all
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O"er all *eat .ransfer Coefficient
.he o"erall heat transfer by combined
conduction and con"ection is freuently
e:pressed in terms of an o"erall heat
transfer coefficient U,
$here ' is some suitable area for the
heat flo$
O"er all *eat .ransfer Coefficient
*ollo$ cylinder $ith con"ection boundaries
=lectrical analo%y
*ollo$ cylinder $ith con"ectionboundaries
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O"er all *eat .ransfer Coefficient
0ote that the area for con"ection is not the same for both fluids in this case, these
areas dependin% on the inside tube diameter and $all thickness .he o"erall heat
transfer $ould be e:pressed by
Jariable heat transfer area system
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Critical &nsulation .hickness
e kno$ that addin% more insulation to a$all al$ays decreases heat transfer .he
thicker the insulation, the lo$er the heat
transfer rate .his is e:pected, since the
heat transfer area A is constant, and addin%
insulation al$ays increases the thermal
resistance of the $all without increasingthe con"ection resistance
'ddin% insulation to a cylindrical pipe or a
spherical shell, ho$e"er, is a different
matter .he additional insulation increases
the conduction resistance of the insulation
layer but decreases the con"ectionresistance of the surface because of the
increase in the outer surface area for
con"ection .he heat transfer from the pipe
may increase or decrease, dependin% on
$hich effect dominates
.hermal heat transfer
An insulated cylindricalpipe
exposed to convectionfrom the outer surface
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Critical &nsulation .hickness
.he rate of heat transfer from theinsulated pipe to the surroundin% air can
be e:pressed as
Cylinder
con"
ins
total
total4con"Bins
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&mportance of Critical &nsulation .hickness
Should $e al$ays check and make surethat the outer radius of insulation
sufficiently e:ceeds the critical radius
before $e install any insulationK
+robably not, as e:plained here
.he "alue of the critical radius r cr is thelar%est $hen k is lar%e and h is small
Fo$est "alue of h encountered in
practice is L 5 /m2MN for the case of
natural con"ection of %ases, and k L
775 /mMN for common insulator
.he lar%est "alue of the critical radius$e are likely to encounter is
Do $e need to be concerned about the criticalradius of insulation $hen insulatin% hot$ater
pipes or e"en hot$ater tanks
.his "alue $ould be e"en smaller $hen the
radiation effects are considered .he critical
radius $ould be much less in forced con"ection,
often less than 1 mm, because of much lar%er h
"alues associated $ith forced con"ection 1
mm in "ery small considerin% practice
.herefore, $e can insulate hot$ater or steam
pipes freely $ithout $orryin% about the
possibility of increasin% the heat transfer by
insulatin% the pipes
Eut, the radius of electric $ires may be smallerthan the critical radius .herefore, the plastic
electrical insulation may actually enhance the
heat transfer from electric $ires and thus keep
their steady operatin% temperatures at lo$er
and thus safer le"els
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Steady *eat Conduction $ith *eatSource Systems
Consider the plane $all $ith uniformlydistributed heat sources sho$n in
i%ure .he thickness of the $all in the x
direction is 2L
'ssumptions@ &t is assumed that the
dimensions in the other directions aresufficiently lar%e that the heat flo$ may
be considered as one dimensional
.he heat %enerated per unit "olume is ,
and $e assume that the thermal
conducti"ity does not "ary $ith
temperature +ractical 'pplications@ +assin% a current
throu%h an electrically conductin%
material 0uclear fuel rod, nuclear
$eapons
One dimensional +lane all
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Steady *eat Conduction $ith *eatSource Systems
.he differential euation that %o"ernsthe heat flo$ is
EC1@ &nterface temperature
EC2 @ .hermal Symmetry
Solution
Substitutin% EC2 C147
Substitutin% EC1
Substitute =uation 2< in 1<
One dimensional +lane all
1<
2<
)<
3<
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Steady *eat Conduction $ith *eatSource SystemsOne dimensional Cylinder
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Steady *eat Conduction $ith *eatSource Systems
.he differential euation that %o"erns
the heat flo$ for cylindrical coordinates
is
EC1@ &nterface temperature
EC2 @ .hermal Symmetry
Solution
One dimensional Cylinder
Substitutin% EC2
C147Substitutin% EC1
1<
2<
)<
3<
Substitute =uation 2< in 1<
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Steady *eat Conduction $ith *eatSource Systems
Deri"e an e:pression for thetemperature distribution in a hollo$
cylinder $ith heat sources that "ary
accordin% to the linear relation
ith the %eneration rate per unit"olume at r 4r i .he inside and outside
temperatures are T 4T i at r 4r i and T 4T o
at r 4r o
Deri"e an e:pression for thetemperature distribution in a hollo$
sphere $ith heat sources that "ary
accordin% to the linear relation
ith the %eneration rate per unit"olume at r 4r i .he inside and outside
temperatures are T 4T i at r 4r i and T 4T o
at r 4r o
'ssi%nment problem *ollo$ Cylinder and Sphere
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ConductionCon"ection Systems
.he heat that is conducted throu%h abody must freuently be remo"ed or
deli"ered< by some con"ection process
=ner%y in left face4ener%y out ri%ht face
B ener%y lost by con"ection
Con"ection *eat loss 0e$ton8s Fa$ ofCoolin%<
&0
( )
2
2
co
x
x dx
x dx
nv p
dT q kAdx
dT dT d T q kA kA dx
dx dx dx
q hA T T
++
∞
= −
= − = − + ÷
= −
co x x dx nvq q q+− =
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ConductionCon"ection Systems
.he ener%y balance yields
Fet
.o sol"e second order differential
euation 2 EC are must
EC1 @ :47 4
.he other boundary condition EC2<depends on the physical situation
Se"eral cases may be considered@
Case 1@ .he fin is "ery lon%, and the
temperature at the end of the fin is
essentially that of the surroundin% fluid Case 2@ .he end of the fin is insulated
so that d./d:47 at :4F adiabatic in tip
Case )@ .he fin is of finite len%th and
loses heat by con"ection from its end
Case 3@ Specified temperature at fin tip
&0
( )2
2 0
phAd T T T
dx kA ∞− − =
2
2 0 phAd
dx kA
θ θ − =
2
2
2 0
d m
dx
θ θ − =
2 phA
mkA
=
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Solution for Case 1@ Jery Fon% &0
.he ener%y balance yields
EC1 @ 4 at :47
EC2 @ P 47 at : 4
Solution for differential euation
.emperature at the end of the fin is essentially that of the surroundin% fluid
2
2
2 0
d m
dx
θ θ − =
2 phA
mkA
=
1 2
mx mxC e C eθ −= +
0 1 21: BC C C θ = +
22 : 0 BC C =
1
0 0
mxT T C e
T T
θ
θ
−∞
∞
−= =
−
0 0
L
c
x
c
dT q kA hP dx
dx
hPKA
θ
θ
=
= − =
=
∫
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Case 2@ .he end of the fin is insulated
.he ener%y balance yields
EC1 @ 4 at :47
EC2 @ at : 4 F Solution for differential euation
Sol"in% for the constants C1 and C2, $e
obtain
'diabatic in tip
'pplications.o pro"ide safe temperature
ran%e end of fins
2
2
2 0
d m
dx
θ θ − =
2 phA
mkA
=
1 2
mx mxC e C eθ −= + 0 1 21: BC C C θ = +
( )1 22 : 0
mx mxd
BC m C e C edx
θ −
= = − +
( )2 2
0 0
cosh
1 1 cosh
mx mx
mL mL
m L xT T e e
T T e e mL
θ
θ
−∞
−∞
− − = = + =− + +
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Case )@ Con"ection at =nd.he fin is of finite len%th and loses heat by con"ection from its end
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Case 3@Specified temperature at fin tip
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5/12/16 Slide 33
in .emperature .emperature profile
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