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Chapter 15Steady Heat Conduction

Critical Radius Of Insulation

3-83C In a cylindrical pipe or a spherical shell, the additional insulation increases the conductionresistance of insulation, but decreases the convection resistance of the surface because of the increase inthe outer surface area. Due to these opposite effects, a critical radius of insulation is defined as the outer

radius that provides maximum rate of heat transfer. For a cylindrical layer, it is defined as r k hcr = /

where k is the thermal conductivity of insulation and his the external convection heat transfer coefficient.

3-84CIt will decrease.

3-85C Yes, the measurements can be right. If the radius of insulation is less than critical radius ofinsulation of the pipe, the rate of heat loss will increase.

3-86Co.

3-87CFor a cylindrical pipe, the critical radius of insulation is defined as r k hcr = / . !n windy days, the

external convection heat transfer coefficient is greater compared to calm days. "herefore critical radius ofinsulation will be greater on calm days.

3-88 #n electric wire is tightly wrapped with a $%mm thick plastic cover. "he interface temperature andthe effect of doubling the thickness of the plastic cover on the interface temperature are to be determined.

Assumptions 1 &eat transfer is steady since there is no indication of any change with time. 2 &eattransfer is one%dimensional since there is thermal symmetry about the centerline and no variation in theaxial direction. 3 "hermal properties are constant. 4 "he thermal contact resistance at the interface isnegligible. 5 &eat transfer coefficient accounts for the radiation effects, if any.

Properties"he thermal conductivity of plastic cover is given to be k' (.$) */m+.

Analysis In steady operation, the rate of heat transfer from the wire is e-ual to the heat generated withinthe wire,

*(#$(010 ==== VIWQ e

"he total thermal resistance is

/*2()$.((34).(44$5.(

/*(34).(m$(0*/m.$).(06

$/6ln0

6

/ln0

/*44$5.(

m7m0$(0(.((28.*/m620

$$

plasticconvtotal

$6plastic

6

conv

=+=+=

=

=

=

=

==

RRR

kL

rrR

Ah

Roo

"hen the interface temperature becomes

C62.4=+=+=

= /*2()$.(0*(04(total$

total

6$ RQTTR

TTQ

"he critical radius of plastic insulation is

mm6).5m((56).(.*/m62

*/m.$).(6

==

==

h

krcr

Doubling the thickness of the plastic cover will increase the outer radius of the wire to 4 mm, which isless than the critical radius of insulation. "herefore, doubling the thickness of plastic cover will increasethe rate of heat loss and decrease the interface temperature.

4%54

Rconv

T6

Rplastic

T$

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3-89E#n electrical wire is covered with (.(6%in thick plastic insulation. It is to be determined if theplastic insulation on the wire will increase or decrease heat transfer from the wire.

Assumptions1&eat transfer from the wire is steady since there is no indication of any change with time.2 &eat transfer is one%dimensional since there is thermal symmetry about the centerline and no variationin the axial direction. 3 "hermal properties are constant. 4 "he thermal contact resistance at the interfaceis negligible.

Properties"he thermal conductivity of plastic cover is given to be k' (.(3) 9tu/hft+F.

Analysis "he critical radius of plastic insulation is

in(5$).(0in45.(ft(4.(F.9tu/h.ft).6

F9tu/h.ft.(3).(66

=>==

== r

h

krcr

:ince the outer radius of the wire with insulation is smaller than criticalradius of insulation, plastic insulation will increaseheat transfer from thewire.

3-90E #n electrical wire is covered with (.(6%in thick plastic insulation. 9y considering the effect ofthermal contact resistance, it is to be determined if the plastic insulation on the wire will increase ordecrease heat transfer from the wire.

Assumptions1&eat transfer from the wire is steady since there is no indication of any change with time.2 &eat transfer is one%dimensional since there is thermal symmetry about the centerline and no variationin the axial direction. 3 "hermal properties are constant

Properties"he thermal conductivity of plastic cover is given to be k' (.(3) 9tu/hft+F.

Analysis *ithout insulation, the total thermal resistance is 0per ft length of the wire

F/9tuh.2.$ft7ft0$0(.(4/$6F8.9tu/h.ft).60

$$6convtot =

===

ooAhRR

*ith insulation, the total thermal resistance is

F/9tuh.4(.$4(25.(4).(26.$6

F/9tuh.(25.(ft7ft0$0(.(4/$68

F/9tu.h.ft(($.(

F/9tuh.4).(ft$0F9tu/h.ft.(3).(06

(4.(/$64.(ln0

6

/ln0

F/9tuh.26.$6ft7ft0$0(.$64/$6F8.9tu/h.ft).60

$$

interfaceplasticconvtotal

6

interface

$6plastic

6conv

=++=++=

=

==

=

=

=

=

==

RRRR

A

hR

kL

rrR

AhR

c

c

oo

:ince the total thermal resistance decreases after insulation, plastic insulation will increase heat transfer

from the wire. "he thermal contact resistance appears to have negligible effect in this case.

4%52

*ireInsulation

*ireInsulation

Rplastic

Rinterface

Rconv

Ts

T

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3-91# spherical ball is covered with $%mm thick plastic insulation. It is to be determined if the plasticinsulation on the ball will increase or decrease heat transfer from it.

Assumptions1&eat transfer from the ball is steady since there is no indication of any change with time. 2&eat transfer is one%dimensional since there is thermal symmetry about the midpoint. 3 "hermal

properties are constant. 4 "he thermal contact resistance at the interface is negligible.

Properties"he thermal conductivity of plastic cover is given to be k' (.$4 */m+.

Analysis "he critical radius of plastic insulation for the spherical ball is

r k

hrcr = =

= = > =

6 6 ($4

6(( ($4 $4 6

0 ..

* / m.

* / m . m mm 0 3 mm

6

:ince the outer temperature of the ball with insulation is smaller than criticalradius of insulation, plastic insulation will increaseheat transfer from the wire.

4%5)

Insulation

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3-92

"GIVEN"D_1=0.005 "[m]""t_ins=1 [mm], parameter to be varied"k_ins=0.1 "[!m#$]"

%_ba&&=50 "[$]"%_in'init(=15 "[$]"

)_o=*0 "[!m+*#$]"

"N-/I/"D_*=D_1*t_ins$onvert2mm, m3_o=piD_*+*4_onv_o=12)_o_o34_ins=2r_*#r_1326pir_1r_*k_ins3r_1=D_1*r_*=D_**4_tota&=4_onv_o4_ins7_dot=2%_ba&%_in'init(34_tota&

tins8mm7 ; 8*7(.) (.(362

$.)65 (.$(4)

6.))4 (.$6)6

4.)3< (.$4 the thinner one will have higher effectiveness.

3-105C"he fin with the lower heat transfer coefficient will have the higher efficiency and the highereffectiveness.

3-106 # relation is to be obtained for the fin efficiency for a fin of constant cross%sectional area As ,

perimeter p, lengthL, and thermal conductivity kexposed to convection to a medium at T with a heat

transfer coefficient h. "he relation is to be simplified for circular fin of diameter Dand for a rectangularfin of thickness t.

Assumptions 1 "he fins are sufficiently long so that the temperature of the fin at the tip is nearly T . 2

&eat transfer from the fin tips is negligible.

Analysis"aking the temperature of the fin at the base to be Tb and using the heat transfer relation for a

long fin, fin efficiency for long fins can be expressed as

fin#ctual heat transfer rate from the fin

Ideal heat transfer rate from the fin

if the entire fin were at base temperature

=

=

= =

hpkA T T

hA T T

hpkA

hpL L

kA

ph

c b

fin b

c c0

0

$

"his relation can be simplified for a circular fin of diameter Dand rectangular fin of thickness t and width w to be

fin,circular

fin,rectangular

= = =

= =+

=

$ $ 2 $

6

$ $

6

$

6

$

6

6

L

kA

ph L

k D

D h L

kD

h

L

kA

ph L

k wt

w t h L

k wt

wh L

kt

h

c

c

0 /

0

0

0

0

4%5

heat chap03 083

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