Heat Transfer Chapter 3

7/27/2019 Heat Transfer Chapter 3

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Chapter 3:Steady Heat Conduction

Yoav PelesDepartment of Mechanical, Aerospace and Nuclear Engineering

Rensselaer Polytechnic Institute

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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ObjectivesWhen you finish studying this chapter, you should be able to:

Understand the concept of thermal resistance and its limitations,and develop thermal resistance networks for practical heatconduction problems,

Solve steady conduction problems that involve multilayerrectangular, cylindrical, or spherical geometries,

Develop an intuitive understanding of thermal contact resistance,and circumstances under which it may be significant,

Identify applications in which insulation may actually increaseheat transfer,

Analyze finned surfaces, and assess how efficiently andeffectively fins enhance heat transfer, and

Solve multidimensional practical heat conduction problems

using conduction shape factors.


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3.1 Steady Heat Conduction in Plane Walls

1) Considerable temperature

difference between the inner

and the outer surfaces of thewall (significant temperature

gradient in thex direction).

2) The wall surface is nearly

isothermal.

Steady one-dimensional modelingapproach is

justified.


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Assuming heat transfer is the only energy interaction

and there is no heat generation, the energy balancecan be expressed as

or

The rate of heat transfer through the

wall must be constant ( ).

P0

0wallin out dE

Q Qdt

=

= =

Rate of

heat transfer

into the wall

Rate of

heat transfer

out of the wall

Rate of change

of the energy

of the wall

- =

Zero for steady

operation

, constantcond wallQ =

(3-1)


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For the control volume (CV):

[] - [] + [] = []


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Then Fouriers law of heat conduction for the wall

can be expressed as

Remembering that the rate of conduction heat transfer

and the wall areaA are constant it follows

dT/dx=constant

the temperature through the wall varies linearly withx.

Integrating the above equation and rearranging yields

, (W)cond walldT

Q kAdx

= (3-2)

(3-3)1 2, (W)cond wallT T

Q kA L

=


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Thermal Resistance Concept-

Conduction Resistance

Equation 33 for heat conduction through a

plane wall can be rearranged as

WhereRwall

is the conduction resistance

expressed as

(3-4)1 2

, (W)cond wallwall

T TQ

R

=

(3-5)( C/W)wallL

R

kA

= D


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Analogy () to Electrical Current Flow

Eq. 3-5 is analogous to the relation forelectric currentflowI, expressed as

Heat Transfer Electrical current flow

Rate of heat transfer Electric currentThermal resistance Electrical resistanceTemperature difference Voltage difference

(3-6)1 2

eR

V VI

=


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Convection Resistance Thermal resistance can also be applied to convection

processes.

Newtons law of cooling for convection heat transfer

rate ( ) can be rearranged as

Rconv is the convection resistance

( )conv s sQ hA T T =

(3-7)(W)s

conv

conv

T TQ

R

=

(3-8)1

( C/W)conv

s

RhA

= D


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Radiation Resistance The rate of radiation heat transfer between a surface and

the surrounding

(3-9)

( )4 4 ( ) (W)s surrrad s s surr rad s s surr rad

T TQ A T T h A T T

R

= = =

(3-10)1

( /W)radrad s

R Kh A

=

( )( )2 2 2(W/m K)( )

radrad s surr s surr

s s surr

Qh T T T T

A T T= = + +

(3-11)


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Thermal Resistance Concept

- Radiation and Convection Resistance A surface exposed to the surrounding might involves

convection and radiation simultaneously.

The convection and radiation resistances are parallel to

each other.

When TsurrT, the radiationeffect can properly be

accounted for by replacing h

in the convection resistance

relation by

hcombined= hconv+hrad(W/m2

K) (3-12)


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Thermal Resistance Network

consider steady one-dimensional heat transfer through aplane wall that is exposed to convection on both sides.

Rate ofheat convection

into the wall

Rate ofheat conduction

through the wall

Rate ofheat convection

from the wall

= =

(3-13)

( )

( )

1 ,1 1

1 2

2 2 ,2

=

Q h A T T

T TkAL

h A T T

=

=


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Rearranging and adding

1 2 wallT T Q R =

,1 ,2

1 2

1 1( C/W)total conv wall conv

LR R R R

h A kA h A= + + = + + D

(3-15),1 ,2 (W)

total

T TQ

R

=

(3-16)

where

,1 ,2T T = ,1 ,2( )conv wall convQ R R R+ +

totalQ R=

,1 1 ,1convT T Q R =

2 ,2 ,2convT T Q R

+

=


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Multilayer Plane Walls

The rate of steady heat transfer through a two-layercomposite wall can be expressed through Eq. 3-15 where

the total thermal resistance is

1 2,1 ,1 ,2 ,2

1 1 2 2

1 1total conv wall wall conv L LR R R R Rh A k A k A h A= + + + = + + +

(3-22)


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3.2 Thermal Contact Resistance

In reality surfaces have some roughness.

When two surfaces are pressed against each other, thepeaks form good material contact but the valleys form

voids filled with air.

As a result, an interface containsnumerous air gaps of varying sizes

that act as insulationbecause of thelow thermal conductivity of air.

Thus, an interface offers some

resistance to heat transfer, whichis termed the thermal contactresistance,Rc.


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The value of thermal contact resistance

depends on the surface roughness,

material properties,

temperature andpressure at the interface,

type of fluid trapped at the interface.

Thermal contact resistance is observed todecrease with decreasing surface roughnessand increasing interface pressure.

The thermal contact resistance can beminimized by applying a thermally conductingliquid called a thermal grease.


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3.3 Generalized Thermal Resistance Network

The thermal resistance concept can be used to solve

steady heat transfer problems that involve parallel

layers or combined series-parallel arrangements. The total heat transfer of two parallel layers

( )1 2 1 21 2 1 21 2 1 2

1 1T T T T Q Q Q T T R R R

= + = + = +

(3-29)

1

totalR

1 2

1 2 1 2

1 1 1=total

total

R RR

R R R R

= +

+

(3-31)


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Combined Series-Parallel Arrangement

The total rate of heat transfer

Through the composite system

where

31 21 2 3

1 1 2 2 3 3 3

1; ; ; conv

LL LR R R R

k A k A k A hA= = = =

(3-32)1

total

T TQR

=

12 3

1 23

1 2

total conv

conv

R R R R

R RR R

R R

= + +

= + +

+

(3-33)

(3-34)


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3.4 Heat Conduction in Cylinders

Consider the long cylindrical layer

Assumptions:

the two surfaces of the cylindricallayer are maintained at constant

temperatures T1 and T2,

no heat generation, constant thermal conductivity,

one-dimensional heat conduction.

Fouriers law of heat conduction

, (W)cond cyldT

Q kAdr

= (3-35)


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Separating the variables and integrating from r=r1,

where T(r1)=T1, to r=r2, where T(r2)=T2

SubstitutingA =2rL and performing the integrations

give

Since the heat transfer rate is constant

, (W)cond cyldT

Q kA

dr

= (3-35)

2 2

1 1

,

r T

cond cyl

r r T T

Q dr kdT A= =

=

(3-36)

( )1 2

,

2 1

2ln /

cond cyl

T TQ Lk

r r

= (3-37)

1 2

,cond cylcyl

T TQ

R

= (3-38)


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Thermal Resistance with Convection

Steady one-dimensional heat transfer through acylindrical or spherical layer that is exposed to

convection on both sides

where

(3-32),1 ,2

total

T TQ

R

=

( )

( )

( )

,1 ,2

2 1

1 1 2 2

ln /1 1

2 2 2

total conv cyl convR R R R

r r

r L h Lk r L h

= + + =

= + +(3-43)


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Multilayered Cylinders

Steady heat transfer

through multilayered

cylindrical or sphericalshells can be handled

just like multilayered plane.

The steady heat transfer rate through a three-layered

composite cylinder of lengthL with convection on both

sides is expressed by Eq. 3-32 where:

( )

( ) ( ) ( )

( )

,1 ,1 ,3 ,3 ,2

2 1 3 2 4 3

1 1 1 2 3 2 2

ln / ln / ln /1 1

2 2 2 2 2

total conv cyl cyl cyl convR R R R R R

r r r r r r

r L h Lk Lk Lk r L h

= + + + + =

= + + + +

(3-46)


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3.5 Critical Radius of Insulation ()

Adding more insulation to a wall or to the attic

always decreases heat transfer. Adding insulation to a cylindrical pipe or a spherical

shell, however, is a different matter.

Adding insulation increases the conduction resistanceof the insulation layer but decreases the convection

resistance of the surface because of the increase in the

outer surface area for convection. The heat transferfrom the pipe may increase or

decrease, depending on which effect dominates.


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A cylindrical pipe of outer radius r1

whose outer surface temperature T1 is

maintained constant.

The pipe is covered with an insulator(kand r2).

Convection heat transfer at T

and h.

The rate of heat transfer from the insulated pipe to the

surrounding air can be expressed as

( )( )

1 1

2 1

2

ln / 1

2 2

ins conv

T T T T Q r rR R

Lk h r L

= =+

+

(3-37)


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2

0dQ

dr

=

, (m)cr cylinderk

rh

=

The variation of the heat transfer rate with the outer

radius of the insulation r2 is shownin the figure.

The value ofr2 at which

reaches a maximum isdetermined by

Performing the differentiation

and solving forr2 yields

Thus, insulating the pipe may actually increase therate of heat transfer instead of decreasing it.

(3-50)

Q


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3.6 Heat Transfer from Finned Surfaces

Newtons law of cooling

Two ways to increase the rate of heat transfer: increasing the heat transfer coefficient,

increase the surface areafins

Fins are the topic of this section.

( )conv s sQ hA T T =


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Fin Equation

Under steady conditions, the energy balance on thisvolume element can be expressed as


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where

Substituting and dividing by x, we obtain

Rate of heat

conduction into

the element at x

Rate ofheat

conductionfrom the

element at x+x

Rate of heat

convection from

the element

= +

, ,cond x cond x x convQ Q Q+= +

( ) ( )convQ h p x T T =

( ), , 0cond x x cond xQ Q

hp T T

x

+

+ =

(3-52)


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Taking the limit as x 0 gives

From Fouriers law of heat conduction we have

Substitution of Eq. 3-54 into Eq. 353 gives

( ) 0conddQ

hp T T

dx

+ =

(3-53)

cond c

dTQ kA dx=

(3-54)

( ) 0cd dT

kA hp T T dx dx

=

(3-55)

For constant cross section and constant thermal conductivity


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For constant cross section and constant thermal conductivity

Where

Equation 356 is a linear, homogeneous, second-order differential

equation with constant coefficients.

The general solution of Eq. 356 is

C1 and C2 are constants whose values are to be determined from the

boundary conditions at the base and at the tip of the fin.

22

2 0d

mdx

= (3-56)

;

c

hpT T m

kA

= =

1 2( ) mx mxx C e C e

= + (3-58)


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Boundary Conditions

Several boundary conditions are typically employed at thefin tip :

At the fin base

Specified temperatureboundary condition, expressedas: (0)= b= Tb-T

( ) I fi it l L Fi (T T)


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(a) Infinitely Long Fin (Tfin tip = T)

For a sufficiently long fin,the temperature at the fin

tip approaches the

ambient temperature

Boundary condition:

(L) = T(L) - T

= 0


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Whenx , so does emx

C1=0

@x = 0: emx = 1 C2= b

The temperature distribution:

heat transferfrom the entire fin

/( ) cx hp kAmx

b

T x Te eT T

= = (3-60)

( )0

c c b

x

dTQ kA hpkA T T

dx

=

= = (3-61)


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(b) Adiabatic Tip

Boundary condition at fin tip:

After some manipulations, the temperature

distribution:

heat transferfrom the entire fin

0x L

d

dx

= = (3-63)

( )cosh( )coshb

m L xT x T

T T mL

=

(3-64)

( )0

tanhc c bx

dTQ kA hpkA T T mL

dx

=

= = (3-65)


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(c & d) Convection (or Combined Convection

and Radiation) from Fin Tip

A practical way of accounting for the heat loss from

the fin tip is to replace thefin length L in the relationfor the insulated tip case by a corrected length

defined as

Lc=L+Ac/p (3-66) For rectangular and cylindrical

finsLc is

Lc,rectangular=L+t/2

Lc,cylindrical=L+D/4

Fin Efficiency


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Fin Efficiency

To maximize the heat transfer from a fin thetemperature of the fin should be uniform (maximized)

at the base value ofTb

In reality, the temperature drops along the fin, and thusthe heat transfer from the fin is less

To account for the effect we define

a fin efficiency

or

(3-69)

,max

fin

fin

fin

Q

Q

= =

Actual heat transfer rate from the fin

Ideal heat transfer rate from the finif the entire fin were at base temperature

,max

( )fin fin fin fin fin b

Q Q hA T T = =


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Fin Efficiency

For constant cross section of very long fins:

For constant cross section with adiabatic tip:

( )

( ), ,max

1 1fin c b clong fin

fin fin b

Q hpkA T T kA

Q hA T T L hp mL

= = = =

(3-70)

( )( ), ,maxtanh

tanh

fin c badiabatic fin

fin fin b

Q hpkA T T aLQ hA T T

mL

mL

= =

=

(3-71)


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Fin Effectiveness

The performance of the fins is judged on the basis of the

enhancement in heat transfer relative to the no-fin case.

The performance of fins is expressedin terms of thefin effectiveness fin

defined as

( )fin fin

fin

no fin b b

Q Q

Q hA T T

= = =

Heat transfer rate

from the surface

ofarea Ab

Heat transfer rate

from the fin ofbasearea Ab

(3-72)


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Remarks regarding fin effectiveness The thermal conductivity kof the fin material should

be as high as possible. It is no coincidence that fins

are made from metals. The ratio of theperimeterto the cross-sectional area

of the finp/Ac should be as high as possible.

The use of fins is most effective in applicationsinvolving a low convection heat transfer coefficient.

The use of fins is more easily justified when themedium is agas instead of a liquid and the heattransfer is by natural convection instead of by forced

convection.


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Overall Effectiveness

An overall effectiveness for a

finned surface is defined as the

ratio of the total heat transferfrom the finned surface to the

heat transfer from the same

surface if there were no fins.

( )

,

fin

fin overall

no fin

unfin fin fin

no fin

Q

Q

h A A

hA

=

+=

(3-76)


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Proper Length of a Fin

An important step in the design of a fin is the

determination of the appropriate length of the fin once

the fin material and the fin cross section are specified.

The temperature drops along

the fin exponentially and

asymptotically approaches the

ambient temperature at somelength.

3 7 Heat Transfer in Common Configurations


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3.7 Heat Transfer in Common Configurations

Many problems encountered in practice are two- orthree-dimensional and involve rather complicatedgeometries for which no simple solutions are

available. An important class of heat transfer problems for

which simple solutions are obtained encompassesthose involving two surfaces maintained at constant

temperatures T1 and T2.

The steady rate of heat transfer between these twosurfaces is expressed as

Q=Sk(T1=T2) (3-79)

Sis the conduction shape factor, which has the

dimension oflength.

T bl 3 7


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Table 3-7


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Heat Transfer Chapter 3