Lect - 4 Steady conduction One Dimensional.pptx

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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5/12/16 Slide 2

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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5/12/16 Slide )

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



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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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5/12/16 Slide 6

.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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5/12/16 Slide A

-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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5/12/16 Slide 17

-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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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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5/12/16 Slide 13


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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5/12/16 Slide 16


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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5/12/16 Slide 1>


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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&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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5/12/16 Slide 2>

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


+lane $all

=lectrical analo%y

O"erall heat transfer throu%h a plane $all


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5/12/16 Slide 2?


.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$


*ollo$ cylinder $ith con"ection boundaries

=lectrical analo%y

*ollo$ cylinder $ith con"ectionboundaries


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5/12/16 Slide 2A


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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5/12/16 Slide )7

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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5/12/16 Slide )2

&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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5/12/16 Slide ))


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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5/12/16 Slide )3


.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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.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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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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5/12/16 Slide )?


.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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5/12/16 Slide )A


.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


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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5/12/16 Slide 31

Case 2@ .he end of the fin is insulated


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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Lect - 4 Steady conduction One Dimensional.pptx

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Transcript of Lect - 4 Steady conduction One Dimensional.pptx