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99

Heat Transfer from Extended Surfaces

UNIT 6 HEAT TRANSFER FROM EXTENDED SURFACES

Structure

6.1 Introduction

Objectives

6.2 Extended Surfaces

6.3 A General Conduction Analysis 6.4 Fins of Uniform Cross-sectional Area

6.4.1 Case A 6.4.2 Case B 6.4.3 Case C 6.4.4 Case D

6.5 Fin Performance

6.6 Fins of Non-uniform Cross-sectional Area 6.7 Overall Surface Efficiency

6.8 Summary 6.9 Key Words 6.10 Answers to SAQs

6.1 INTRODUCTION

The term extended surface is commonly used to depict an important special case involving heat transfer by conduction within a solid and heat transfer by convection (and/or radiation) from the boundaries of the solid. Until now, we have considered heat transfer from the boundaries of a solid to be in the same direction as heat transfer by conduction in the solid. In contrast, for an extended surface, the direction of heat transfer from the boundaries is perpendicular to the principal direction of heat transfer in the solid.

Objectives

After studying this unit, you should be able to

 understand the wide application of extended surfaces in different heat transfer devices,

 distinguish different types of extended surfaces,

 formulate the heat transfer equations with fins,

 appreciate need of an extended surface,

 evaluate heat transfer and temperature distribution for a system with fins, and

 solve some problems on extended surfaces.

6.2 EXTENDED SURFACES

Consider a strut that connects two walls at different temperatures and across which there is fluid flow (Figure 6.1). With T1 > T2, temperature gradients in the x-direction sustain heat transfer by conduction in the strut. However, with T1 > T2 > T there is concurrent

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100

Conduction heat transfer by convection to the fluid, causing qx, and hence the magnitude of the temperature gradient, dT

dx , to decrease with increasing x.

Figure 6.1 : Combined Conduction and Convection in a Strut

Although there are many different situations that involve such combined

conduction-convection effects, the most frequent application is one in which an extended surface is used specifically to enhance heat transfer between a solid and an adjoining fluid. Such an extended surface is termed a fin.

Consider the plane wall of Figure 6.2(a). If Ts is fixed, there are two ways in which the heat transfer rate may be increased. The convection coefficient h could be increased by increasing the fluid velocity, and/or the fluid temperature T could be reduced. However, there are many situations for which increasing h to the maximum

(a) (b)

Figure 6.2 : Use of Fins to Enhance Heat Transfer from a Plane Wall, (a) Bare Surface, (b) Finned Surface

possible value is either insufficient to obtain the desired heat transfer rate or the associated costs are prohibitive. Such costs are related to the blower or pump power requirements needed to increase h through increased fluid motion. Moreover, the second option of reducing T is often impractical. Examining Figure 6.2(b), however, we see that there exists a third option. That is, the heat transfer rate may be increased by increasing the surface area across which the convection occurs. This may be done by employing fins that extend from the wall into the surrounding fluid. The thermal conductivity of the fin material has a strong effect on the temperature distribution along the fin and therefore influences the degree to which the heat transfer rate is enhanced.

Ideally, the fin material should have a large thermal conductivity to minimize temperature variations from its base to its tip. In the limit of infinite thermal conductivity, the entire fin would be at the temperature of the base surface, thereby providing the maximum possible heat transfer enhancement.

A

T

h T

T A

H T

Q = hA

(T -T) T2

qx, 2

T

h Fluid

T1

qx, 1

qconv

a

O L

T(x)

T1 > T2 >T

T1 T2

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101

Heat Transfer from Extended Surfaces

There are many engineering applications of fin such as cooling of engine heads, radiators, condenser coils of refrigerators, lawn mowers, electric generators, electric power transformers, etc. Various types of fins are used for different applications. Some configurations of fins are given in Figure 6.3.

(a) (b)

(c) (d)

Figure 6.3 : Fin Configurations (a) Straight Fin of Uniform Cross Section, (b) Straight Fin of Non-Uniform Cross Section, (c) Annular Fin and (d) Pin Fin.

A straight fin is any extended surface that is attached to a plane wall. It may be of uniform cross-sectional area, or its cross-sectional area may vary with the distance x from the wall (Figures 6.3(a) (b)). An annular fin is one that is circumferentially attached to a cylinder, and its cross section varies with radius from the wall of the cylinder (Figure 6.3(c)). The foregoing fin types have rectangular cross sections, whose area may be expressed as a product of the fin thickness t and the width w for straight fins or the circumference 2r for annular fins. In contrast, a pin fin, or spine, is an extended surface of circular cross section. Pin fins may also be of uniform or non-uniform cross section (Figure 6.3(d)). In any application, selection of a particular fin configuration may depend on space, weight, manufacturing, and cost considerations, as well as on the extent to which the fins reduce the surface convection coefficient and increase the pressure drop associated with flow over the fins.

SAQ 1

(a) What is a fin?

(b) Give some examples of fins.

(c) Why it is necessary to attach fins in a heat exchanger?

x

W t

x

r

x

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102

Conduction

6.3 A GENERAL CONDUCTION ANALYSIS

Consider the extended surface of Figure 6.4. The analysis is simplified if certain assumptions are made. We choose to assume one-dimensional conditions in the longitudinal (x) direction, even though conduction within the fin is actually two-dimensional.

Figure 6.4 : Energy Balance for an Extended Surface

The rate at which energy is convected to the fluid from any point on the fin surface must be balanced by the rate at which energy reaches that point due to conduction in the transverse (y, z) direction. However, in practice the fin is thin and temperature changes in the longitudinal direction are much larger than those in the transverse direction.

Hence, we may assume one-dimensional conduction in the x-direction. We will consider steady-state conditions and also assume that the thermal conductivity is constant, that radiation from the surface is negligible, that heat generation effects are absent, and that the convection heat transfer coefficient h is uniform over the surface.

Applying the conservation of energy to the section in Figure 6.4.

conv

x x dx

q q dq . . . (6.1)

From Fourier’s law we know that

x c

q k A dt

  dx . . . (6.2)

where Ac is the cross-sectional area, which may vary with x. Since the conduction heat rate at x + dx may be expressed as

x dx x x

q q dq dx

  dx . . . (6.3)

It follows that

x dx c c

dT d dT

q kA k A dx

dx dx dx

     . . . (6.4)

The convection heat transfer rate may be expressed as

conv s ( )

dq h d A T T . . . (6.5) where dAs is the surface area of the differential element. Substituting the foregoing rate equations into the energy balance, Eq. (6.1), we obtain

( ) 0

c s

d A dT h dA T T

dx dx k dx

    

 

  . . . (6.6)

z y x

X qx

dAs

dx

Ac (x) qx-dx

dq

x

Convection

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103

Heat Transfer from Extended Surfaces

or 22 1 1

( ) 0

c s

c c

dA dA

d T dT h

T T A dx dx A k dx

dx

   

     

    . . . (6.7)

Eq. (6.7) provides a general form of the energy equation for an extended surface. Its solution for appropriate boundary conditions provides the temperature distribution, which may be used with Eq. (6.2) to calculate the conduction rate at any x.

6.4 FINS OF UNIFORM CROSS-SECTIONAL AREA

To solve Eq. (6.7) it is necessary to be more specific about the geometry. We begin with the simplest case of straight rectangular and pin fins of uniform cross section (Figure 6.5). Each fin is attached to a base surface of temperature T (0) = Tb and extends into a fluid of temperature T.

(a) Straight Fin (b) Thin Rod Figure 6.5 : Straight Fin of Uniform Cross-section (a) Rectangular, and (b) Pin Fin

For the prescribed fins, Ac is a constant and As = Px where As is the surface area measured from the base to x and P is the fin perimeter. Accordingly, with dAc 0

dx  and dAs

dx P, Eq. (6.7) reduces to

2

2 ( ) 0

c

d T hP T T

dx  kA   . . . (6.8)

To simplify the form of the equation, we transform the dependent variable by defining an excess temperature  as

( )x T x( ) T

   . . . (6.9)

where, since T is a constant, d dT dx dx

 . Substituting Eq. (6.9) into Eq. (6.8), we then obtain

2 2

2 0

d m

dx

    . . . (6.10)

where 2

c

m hP

 kA . . . (6.11)

Eq. (6.10) is a linear, homogeneous, second-order differential equation with constant coefficients. Its general solution is of the form

1 2

( )x C emx C emx

   . . . (6.12)

By substitution it may readily be verified that Eq. (6.12) is indeed a solution to Eq. (6.10).

x Q

QO Qx

TO

T Q

l

T1

L h

Qx + dx dx

qconv

D x L Th

t h h

qf

AC

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104

Conduction To evaluate the constants C1 and C2 of Eq. (6.12), it is necessary to specify appropriate boundary conditions. One such condition may be specified in terms of the temperature at the base of the fin (x = 0)

(0) Tb T b

     . . . (6.13)

The second condition, specified at the fin tip (x = L), may correspond to one of four different physical situations. These are described below as Case A, Case B, Case C and Case D.

6.4.1 Case A

The first condition considers convection heat transfer from the fin tip. Applying an energy balance to a control surface about this tip (Figure 6.6), we obtain

[ ( ) ]

c c

x L

h A T L T k A dT

dx

   . . . (6.14)

or ( )

x L

h L k d dx

    . . . (6.15)

Figure 6.6 : Conduction and Convection in a Fin of Uniform Cross Section

That is, the rate at which energy is transferred to the fluid by convection from the tip must equal the rate at which energy reaches the tip by conduction through the fin.

Substituting Eq. (6.12) into Eqs. (6.13) and (6.15), we obtain, respectively

1 2

b C C

   . . . (6.16)

and h C e( 1 mL C e2 mL)km C e( 2 mL C e1 mL) . . . (6.17) Solving for C1 and C2, we get,

cosh ( ) sinh ( )

cosh sinh

b

m L x h m L x

mk

mL h mL

mk

 

   

   

   

. . . (6.18)

The temperature distribution ( vs x) is shown schematically in Figure 6.7.

Figure 6.7 : Temperature Distribution along the Length of a Fin with Uniform Cross-section qconv

Fluid, T

hAc [T(L) - T] qb = qf

Tb

kAcdT x L

dx

b

θ (x)

θO L

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105

Heat Transfer from Extended Surfaces

It may be noted that the magnitude of the temperature gradient decreases with increasing x. This trend is a consequence of the reduction in the conduction heat transfer qx (x) with increasing x due to continuous convection losses from the fin surface.

We are particularly interested in the amount of heat transferred from the entire fin. From Figure 6.7 it is evident that the fin heat transfer rate qf may be evaluated in two

alternative ways, both of which involve use of the temperature distribution. The simpler procedure, and the one that we will use, involves applying Fourier’s law at the fin base.

That is,

0 0

f b c c

x x

dT d

q q k A k A

dx dx

      . . . (6.19)

Hence, knowing the temperature distribution, ( ), x qf may be evaluated, giving

sinh cosh

cosh sinh

f c b

mL h mL

q h P k A mk

mL h mL

mk

 

  

 

 

  

. . . (6.20)

However, conservation of energy dictates that the rate at which heat is transferred by convection from the fin must equal the rate at which it is conducted through the base of the fin. Accordingly, the alternative formulation for qf is

[ ( ) ]

f

f A s

q 

h T x T d A . . . (6.21) ( )

f

f A s

q 

h x d A . . . (6.22)

where Af is the total, including the tip, fin surface area. Substitution of Eq. (6.18) into Eq. (6.22) would yield Eq. (6.20).

6.4.2 Case B

The second tip condition corresponds to the assumption that the convective heat loss from the fin tip is negligible, in which case the tip may be treated as adiabatic and

0

x L

d dx

  . . . (6.23)

Substituting from Eq. (6.12) and dividing by m, we obtain

1 mL 2 mL 0

C e C e  . . . (6.24)

Using this expression with Eq. (6.16) to solve for C1 and C2 and substituting the results into Eq. (6.12), we obtain

cosh ( )

b cosh

m L x mL

  

 . . . (6.25)

Using this temperature distribution with Eq. (6.19), the fin transfer rate is then

f c b tanh

q  h P k A  mL . . . (6.26)

6.4.3 Case C

In the same manner, we can obtain the fin temperature distribution and heat transfer rate for Case C, where the temperature is prescribed at the fin tip. That is the boundary condition is ( ) L  L, and the resulting expressions are of the form

sinh sinh ( )

sinh

L b b

mx m L x

mL

   

 

   

 . . . (6.27)

cosh sinh

L

f c b b

mL q h P k A

mL

 

   . . . (6.28)

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106

Conduction 6.4.4 Case D

Fin of infinite length

The very long fin, case as L  , L  0 and it is easily verified that

mx b

e

 

 . . . (6.29)

f c b

q  h P k A  . . . (6.30)

SAQ 2

Derive an expression for heat transfer and temperature distribution for a rectangular fin with its tip at adiabatic condition.

6.5 FIN PERFORMANCE

Fins are used to increase the heat transfer from a surface by increasing the effective surface area. However, the fin itself represents a conduction resistance to heat transfer from the original surface. For this reason, there is no assurance that the heat transfer rate will be increased through the use of fins. An assessment of this matter may be made by evaluating the fin effectiveness f.

It is defined as the ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin. Therefore,

, f f

c b b

q

  h A

 . . . (6.31)

where Ac, b is the fin cross-sectional area at the base. In any rational design the value of f

should be as large as possible, and in general, the use of fins may rarely be justified unless f  2.

Subject to any one of the four tip conditions that have been considered, the effectiveness for a fin of uniform cross section may be obtained by dividing the appropriate expression for qf by h Ac b,b. Although the installation of fins will alter the surface convection coefficient, this effect is commonly neglected. Hence, assuming the convection

coefficient of the finned surface to be equivalent to that of the unfinned base, it follows that, for the infinite fin approximation (Case D), the result is

1 2 F

c

kP hA

 

   

  . . . (6.32)

Several important trends may be inferred from this result.

Obviously, fin effectiveness is enhanced by the choice of a material of high thermal conductivity. Aluminum alloys and copper come to mind. However, although copper is superior from the standpoint of thermal conductivity, aluminum alloys are the more common choice because of additional benefits related to lower cost and weight. Fin effectiveness is also enhanced by increasing the ratio of the perimeter to the cross- sectional area. For this reason the use of thin, but closely related fins is preferred, with the provision that the fin gap not be reduced to a value for which flow between the fins is severely impeded, thereby reducing the convection coefficient.

Eq. (6.32) also suggests that the use of fins can be better justified under conditions for which the convection coefficient h is small. Hence, it is evident that the need for fins is stronger when the fluid is a gas rather then a liquid and when the surface heat transfer is by free convection. If fins are to be used on a surface separating a gas and a liquid, they

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107

Heat Transfer from Extended Surfaces

are generally placed on the gas side, which is the side of lower convection coefficient. A common example is the tubing in an automobile radiator. Fins are applied to the outer tube surface, over which there is flow of ambient air (small h), and not to the inner surface, through which there is flow of water (large h). Note that, if f > 2 is used as a criterion to justify the Implementation of fins, Eq. (6.32) yields the requirement that

4

c

kP h A

 

 

  .

Eq. (6.32) provides an upper limit to f which is reached as L approaches infinity.

However, it is certainly not necessary to use very long fins to achieve near maximum heat transfer enhancement.

Fin performance may also be quantified in terms of a thermal resistance. Treating the difference between the base and fluid temperatures as the driving potential, a fin resistance may be defined as

, b

t f f

R q

  . . . (6.33)

The result is extremely useful, particularly when representing a finned surface by a thermal circuit. Note that, according to the fin tip condition, an appropriate expression for qf may be obtained from Eqs. (6.20), (6.26), (6.28) and (6.30).

Dividing Eq. (6.33) into the expression for the thermal resistance due to convection at the exposed base,

,

,

1

t b

c b

R  h A . . . (6.34)

and substituting from Eq. (6.31), it follows that

, , t b

f t f

R

  R . . . (6.35)

Hence the fin effectiveness may be interpreted as a ratio of thermal resistances, and to increase f it is necessary to reduce the conduction/convection resistance of the fin. If the fin is to enhance heat transfer, its resistance must not exceed that of the exposed base.

Another measure of fin thermal performance is provided by the fin efficiency f. The maximum driving potential for convection is the temperature difference between the base (x = 0) and the fluid,  b Tb T. Hence, the maximum rate at which a fin could dissipate energy is the rate that would exist if the entire fin surface were at the base temperature. However, since any fin is characterised by a finite conduction resistance, a temperature gradient must exist along the fin and the above condition is an idealisation.

A logical definition of fin efficiency is, therefore,

max

f f

f

f b

q q

q h A

  

 . . . (6.36)

where Af is the surface area of the fin. For a straight fin of uniform cross-section and an adiabatic tip

tanh tanh

f

b

M mL mL

h P L mL

  

 . . . (6.37)

From Eq. (6.37), it is observed that f approaches its maximum and minimum values of 1 and 0, respectively, as L approaches 0 and .

As discussed earlier, Eq. (6.20) gives a cumbersome expression for heat transfer from a straight rectangular fin with an active fin tip. An approximate prediction for the heat transfer for the same fin may be obtained by using the adiabatic tip boundary condition.

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108

Conduction

In such a case, the length of the fin in Eq. (6.37) is corrected to

c 2t

L L  

     for a rectangular fin. The corrected value of length can be taken as

c D4

L L  

     in case of pin fin.

This correction is based on assuming equivalence between heat transfer from the actual fin with tip convection and heat transfer from a longer, hypothetical fin with an adiabatic tip.

Hence, with tip convection, the fin heat rate may be approximated as

f tanh c

q M mL . . . (6.38)

The corresponding efficiency is tanh c

f

c

mL

  mL . . . (6.39)

Errors associated with the approximation are negligible if ht k

 

 

  or 0.0625

2 hD

k

  

 

  .

If the width of a rectangular fin is much larger then its thickness, wt, the perimeter may be approximated as P = 2w, and

1 1

2 2 2

c c c

c

hP h

mL L L

kA kt

   

    . . . (6.40)

Multiplying numerator and denominator by L1/ 2c and introducing a corrected fin profile area, Ap L tc , it follows that

1 2 3 2

2

c c

p

mL h L

kA

 

   . . . (6.41)

Hence, as shown in Figures (6.8) and (6.9), the efficiency of a rectangular fin with tip convection may be represented as a function of

1 3 2 2

c

p

L h k A

 

 

 

  .

Figure 6.8 : Efficiency of Straight Fins ( Rectangular, Triangular and Parabolic Profiles)

O 0.5 1.0 1.5 2.0 2.5

20 40 60 80 100

Lc = L+1/2

Ap = Lc t L 1/2

L 1/2 Lc = L Ap = L1/2

Lc = L Ap = L1/3

1/2 x L y

y y-x

y-x2

f (%)

Lc3/2 (h/kAp)1/2

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109

Heat Transfer from Extended Surfaces

Figure 6.9 : Efficiency of Annular Fins of Rectangular Profile

SAQ 3

(a) Define fin efficiency?

(b) What is fin effectiveness?

(c) What are the parameters to consider for evaluating the fin effectiveness?

Discuss their effect on performance of a fin.

(d) When is the use of fin is not justified?

6.6 FINS OF NON-UNIFORM CROSS-SECTIONAL AREA

Analysis of the thermal behavior is more complicated for the case of fins with non- uniform cross section. In such cases the second term of Eq. (6.7) must be retained, and the solutions are no longer in the form of simple exponential or hyperbolic functions.

For example, consider the annular fin shown Figure 6.10. It is assumed that the fin thickness is uniform (t is independent of r), the cross-sectional area, Ac  2 r t, varies with r.

Figure 6.10 : Annular Fin

Replacing x by r in Eq. (6.7) and expressing the surface area as As  2 (r2 r12), the general form of the fin equation reduces to

2 2

1 2 ( ) 0

d T dT h T T

r dr kt

dr     . . . (6.42)

or, with 2 2h

m  kt and   T T,

f (%)

Lc3/2 (h/kAp)1/2 r2c = r2+h/2

Lc = L+h2 Ap = Lct t

L r1

r2

1= r2c/r1

2 3 5

0.5 1.0 1.5 2.0 2.5

0 20 40 60 80 110

r2

r1

s Th

T- h

(12)

110

Conduction 2

2 2

1 0

d d

r dr m dr

     . . . (6.43)

The foregoing expression is a modified Bessel equation of order zero, and its general solution is of the form

1 0 2 0

( )r C I (mr) C K (mr)

   . . . (6.44)

where I0 and K0 are modified, zero-order Bessel functions of the first and second kinds, respectively. If the temperature at the base of the fin is prescribed, ( )r1  b, and an adiabatic tip is presumed,

2

0

r

d dr

  , C1 and C2 may be evaluated to yield a temperature distribution of the form

0 1 2 0 1 2

0 1 1 2 0 1 1 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

b

I mr K mr K mr I mr I mr K mr K mr I mr

  

  . . . (6.45)

where 1 [ 0 ( )]

( )

( ) d I mr

I mr  d mr and 1 [ 0( )]

( )

( ) d I mr

K mr d mr

  are modified, first-order Bessel functions of the first and second kinds, respectively. The Bessel functions are tabulated at the end of the unit (Appendix-I).

With the fin heat transfer rate expressed as

1 1

, (2 1 )

f c b

r r r r

dT d

q k A k r t

dr dr

      . . . (6.46)

It follows that

1 1 1 2 1 1 1 2

1

0 1 1 2 0 1 1 2

( ) ( ) ( ) ( )

2 ( ) ( ) ( ) ( )

f b

K mr I mr I mr K mr q k r t m

K mr I mr I mr K mr

   

 . . . (6.47)

From which the fin efficiency becomes

1 1 1 1 2 1 1 1 2

2 2 2 2

0 1 1 2 0 1 1 2

2 1 2 1

2 ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 ( ) ( )

f f

b

q r K mr I mr I mr K mr

K mr I mr I mr K mr

h r r m r r

   

     . . . (6.48)

This result may be applied for an active (convective) tip, if the tip radius r2 is replaced by a corrected radius of the form 2 2

c 2t

r r     . Results are represented graphically in Figure 6.9.

Knowledge of the thermal efficiency of a fin may be used to evaluate the fin resistance, where, from Eqs. (6.33) and (6.36), it follows that

, 1

t f

f f

R h A

 . . . (6.49)

Although results for the fins of uniform thickness or diameter were obtained by assuming an adiabatic tip, the effects of convection may be treated by using a corrected length (Eq. (6.39)) or radius (Eq. (6.48)). The triangular and parabolic fins are of non- uniform thickness, which reduces to zero at the fin tip. The volume of a straight fin is simply the product of its width and profile area, V w Ap.

Fin design is often motivated by a desire to minimise the fin material and/or related manufacturing costs required to achieve a prescribed cooling effectiveness. Hence, a straight triangular fin is attractive because, for equivalent heat transfer, it requires much

(13)

111

Heat Transfer from Extended Surfaces

less volume (fin material) than a rectangular profile. In this regard, heat dissipation per unit volume,

f

q V

  

  is largest for a parabolic profile. However, since

f

q V

  

  for the parabolic profile is only slightly larger than that for a triangular profile, its use can rarely be justified in view of its larger manufacturing costs. The annular fin of rectangular profile is commonly used to enhance heat transfer to or from circular tubes.

6.7 OVERALL SURFACE EFFICIENCY

In contrast to the fin efficiency f, which characterises the performance of a single fin, the overall surface efficiency 0 characterises an array of fins and the base surface to which they are attached. Representative arrays are shown in Figure 6.10, where S designates the fin pitch. In each case the overall efficiency is defined as

0 max

t t

t b

q q

q h A

  

 . . . (6.50)

where qt is the total heat rate from the surface area At associated with both the fins and the exposed portion of the base (often termed the prime surface). If there are N fins in the array, each of surface area Af, and the area of the prime surface is designated as Ab, the total surface area is

t f b

A  NA  A . . . (6.51)

The maximum possible heat rate would result if the entire fin surface, as well as the exposed base, were maintained at Tb.

The total rate of heat transfer by convection from the fins and the prime (unfinned) surface may be expressed as

t f f b b b

q  N h A   h A  . . . (6.52)

where the convection coefficient h is assumed to be equivalent for the finned and prime surfaces and f is the efficiency of a single fin. Hence,

( ) 1 f (1 )

t f f t f b t f b

t

q h N A A NA h A NA

A

 

 

           

  . . . (6.53)

Substituting Eq. (6.53) into Eq. (6.50), it follows that

0 1 f (1 f)

t

NA

   A   . . . (6.54)

From knowledge of 0, Eq. (6.50) may be used to calculate the total heat rate for a fin array.

Recalling the definition of the fin thermal resistance, Eq. (6.33) and Eq. (6.50) may be used to infer an expression for the thermal resistance of a fin array. That is,

, 0

0 t b

t t

R q

q h A

  

 . . . (6.55)

where Rt, 0 is an effective resistance that accounts for parallel heat flow paths by

conduction/convection in the fins and by convection from the prime surface. Figure 6.11 illustrates the thermal circuits corresponding to the parallel paths and their representation in terms of an effective resistance.

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112

Conduction

Figure 6.11 : Fin Array and Thermal Circuit with Fins Integral to the Base

If fins are machined as an integral part of the wall from which they extend, there is no contact resistance at their base. However, more commonly, fins are manufactured separately and are attached to the wall by a metallurgical or adhesive joint. Alternatively, the attachment may involve a press fit, for which the fins are forced into slots machined on the wall material. In such cases (Figure 6.12), there is a thermal contact resistance, Rt, c, which may adversely influence overall thermal performance.

Figure 6.12 : Fin Array and Thermal Circuit with Fins Attached to the Base

An effective circuit resistance may again be obtained, where, with the contact resistance,

, ( )

( ) b 1

t o c

t o c t

R q hA

  

 . . . (6.56)

It is readily shown that the corresponding overall surface efficiency is

( )

1

1 f 1 f

o c

t

NA

A C

  

     

  . . . (6.57)

where 1 ,

,

1 f f t c

c b

C h A R

A

  

     . . . (6.58) In manufacturing, care must be taken to render Rt c, Rt f, .

Example 6.1

12 (Twelve) number of fins each having thermal conductivity k = 75 W/mK and 0.75 mm thickness protrude 25 mm from a cylindrical surface of 50 mm diameter and 1 m length placed in an atmosphere of 40oC. If the cylindrical surface is maintained at 150oC and the heat transfer coefficient is h = 23 W/m2 . K, calculate

Tb T

Nqf

qb

(fhAf)-1 q1

T

Tb

(NfhAf)-1 RfNAc b

Th Tb

qb

q1

m

Tb

q1

qb

Tb

Th

(NfhAf)-2 Nqf

qb

[h(A2-NAf)]-2

(fhAf)-1 q1

T

Tb

q1

Tb

T

(15)

113

Heat Transfer from Extended Surfaces

(a) the rate of heat transfer,

(b) the percentage increase in heat transfer due to fins, (c) the temperature at the centre of the fins, and (d) fin efficiency and fin effectiveness.

Solution

Refer to the following figure : The perimeter of one fin

2 ( ) 2 (1 0.75 10 )3 2 m P Lb   

3 2

0.75 10 1 0.00075 m A bL    

1 1

2 23 2 2 1

28.6 m 0.00075 75

m hP kA

  

 

      28.6 0.025 0.715

ml  

tanhmltanh 0.715 0.61

Heat transfer from one fin

tanh 28.6 75 0.00075 (150 40) 0.61 108

o o

Q m k A ml       W

Heat transfer from 12 fins

12 o 12 108 1296

Q Q    W

Heat transfer from uniform portion of the surface

0

Qs h A

23 [(   0.05 l) (12 0.00075 l)] (150 40) 23 0.148 110 = 374.44   W

Total heat transfer from the cylindrical surface 1296 374.44 1670.44

T s

Q QQ    W Ans. (a)

If the cylinder were without fins, the heat transfer would have been

23 ( 0.05 ) 110 397.4

Q     l  W

Increase in heat transfer due to fins 1670.44 397.44

100 320.3%

397.44

    Ans. (b)

cosh (1 )

o cosh

m x

ml

  

l = 25mm

L = 1m

b = 0.75

TW = 150oC T= 40oc

R = 23 w/m2k

(16)

114

Conduction

or, 40 cosh (28.6 0.0125) 1.062

0.845

150 40 cosh 0.715 1.257

Tc     

Centre temperature of the fins, Tc = 133oC Ans. (c) If tip loss is neglected

fin tanh 0.61 0.853

0.715 ml

  ml  

or, fin 85.3% Ans. (d)

Fin effectiveness

1670.44 4.2 397.44

   Ans. (e) As fin effectiveness is more than 2, use of fins is justified.

Example 6.2

One end of a long rod is inserted into a furnace while the other projects into ambient air. Under steady state the temperature of the rod is measured at two points 75 mm apart and found to be 125oC and 88.5oC, respectively, while the ambient temperature is 20oC. If the rod is 25 mm min diameter and h = 23.36 W/m2K, find the thermal conductivity of the rod material.

Solution

The temperature distribution is given by

1 mx 2 mx

C e C e

  

where

1

hP 2

m kA

 

   At x0,  o At x   , 0

 C2 = 0

1 mx

C e

 

Let, l be the distance between the two points where temperatures are measured.

Then,   1 oemx1 and   2 0em x(1l)

1 1

1

2 .

mx ml

mx ml

e e

e e

  

or, 125 20 105

88.5 20 68.5 eml

  

 ml = 0.427

or,

1

1 2 1

2 2

2

0.427

5.696 2

0.075

4

hP h d h

m kA k d kd

 

  

   

         

 

2 4 h 32.44

m  kd 

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115

Heat Transfer from Extended Surfaces

4 23.36

115.2 W/m.K 0.025 32.44

k  

Example 6.3

A bar of square cross section connects two metallic structures. One structure is maintained at a temperature 200oC and the other is maintained at 50oC. The bar, 20 mm  20 mm, is 100 mm long and is made of mild steel (k = 0.06 kW/m . K).

The surroundings are at 20oC and the heat transfer coefficient between the bar and the surroundings is 0.01 kW/m2 . K.

Derive an expression for the temperature distribution along the bar and hence calculate the total heat flow rate from the bar to the surroundings.

Solution

1

Q kAdT

  dx

2 1 1

Q Q dQ dx

  dx

2

1 2 d dT d T2 ( )

Q Q kA dx kA dx h P dx T T

dx dx dx

 

       

Let,   T T

2

2 0

d hP

dx kA

   

or, (D2 m2) 0 where

1

hP 2

m kA

 

  

The general solution is

1 mx 2 mx

C e C e

  

At x0,   1 C1C2 180 Co At x    1, 2 C e1 ml C e2 ml 30 Co

1 1 1

2 0.01 4 0.02 2 4 2

0.06 0.02 0.02 0.12

m hP kA

   

   

       

T = 20oc

Q1 Q2

x

x

K = 0.06 kw/m.k.

w/m2k

l = 0.1 m h = 0.01 kw/m2 k

T2 = 50oc T1 = 200oc

20mm 20mm

x-x

(18)

116

Conduction ml = 0.577

ml 1.78 e 

0.561 eml

1 2

30C 1.78C 0.561

1 2

3.173C C 53.476 From Eq. (2)

1 2 180.00

C C  On subtraction,

2.173C1 126.524 C1 =  58.22 From Eq. (2)

C2 = 238.22 The temperature distribution is

  58.22e5.77x 238.22e5.77x

5.77 5.77

1 0

0

( 58.22 x 238.22 x)x 5.77

x

Q kA d kA e e

dx

 

        

0.06 4 10  4(58.22e5.77x 238.22e5.77x)x05.7741.0W

2

x l

Q kA d

dx

 

   

 0.06 4 10  4 ( 58.22 5.77 1.78  238.22 5.77 0.561)  32.8W Heat flow rate from the bar to the surroundings

1 2 41.0 32.8 8.2

QQ Q    W Example 6.4

Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m . K. The base temperature is 100oC and the fluid is at 20oC. The heat transfer coefficient between the fin and the fluid is

30 W/m2 . K.

Solution

The extended length

20 40 2 20.95 cm 0.21 m

c 84 L l A

P

      

40 2 80 cm2

A  

0.21 0.84 0.1764 m2

s c

A L P  

(19)

117

Heat Transfer from Extended Surfaces 1 1

2 30 0.84 2 1

4.583 m 150 0.008

m hP kA

  

 

     

0 0 tanh

Q  m k A ml

4.583 150 0.008 80 tanh (4.583 0.21) 328.0       W Also, fin tanh tanh (4.583 0.21)

0.775 (4.583 0.21)

ml ml

    

0 fin s ( 0 )

Q   A  h T T

0.775 30 0.1764 (100    20) 328 W Example 6.5

Derive the condition for the maximum heat flow for a given weight of a rectangular fin with minimum weight.

Solution

In some applications such as aircraft, heat transfer from engine should be

maximum with minimum weight of the engine. For a given weight, the maximum heat transfer is desirable.

Weight of one fin     b l L where  is the density of the material.

Let, A1 b l= area of fin cross-section normal to L.

The length of the fin L is fixed at a given dimension. Here, we have to change the other two dimensions (i.e. b and l) so as to obtain maximum heat flow for a given area A1 (Figure 6.3(a)).

We have

1

2h 2

m kb

 

    A b L  and A1 b l If the tip loss is neglected,

1

2 0 0

( ) tan tanh

Q h p k A  ml m k A ml

1 1

2 2 1

0

2 2

tanh A

h h

k b L

kb kb b

 

    

     

1 1 1

2 1

2 2

0 3

2

(2 ) tanh 2h A

hk b L

k b

 

   

     

For a given area A1, Q will be maximum when

1 1 1

2 1 1

2 2

0 3

2

2 1

0 (2 ) tanh

2 A

dQ h

hk L b

db k

b

 

 

 

     

(20)

118

Conduction

12

1 1 5

2 2 2

0 1 1

2 2 1

3 2

1 2 3

(2 ) 2

cosh 2

hk L b h A b

A k h

k b

   

       

 

 

 

  

 

 

Putting

1

2h 2

u kb

 

   

and A1

b ml

We get,

1 1 2

1 2 2

2

tanh 3 2 1

2 cosh 0

2

u A h

b k u

b

 

   

or

1 1 2

2

2 1

tan 3 0

cosh

A h

u b kb u

 

   

or cosh sinhu u3u0

or sinh 3

2 u  u

or 2 2 6

2

u u

e e

u

or

2 3 4 5

4 8 16 32

1 2 . . .

2 6 24 120

u u u u

 u 

     

 

 

 

2 3 4 5

4 8 16 32

1 2 . . . 12

2 6 24 120

u u u u

u u

 

       

or 4 16 3 64 5 12

6 120

u u

u   u

or u4 5u2 150

or

1

2 5 (25 60)2

2 2.1

u     

or uml1.419

Now,

1

2 2

1.419 h l

kb

  

 

  or

1

1 2 2

1.419 2

k

b hb

 

  

This is the condition for the maximum heat flow for a given weight of fin, giving the optimum ratio of fin height to half the fin thickness.

Exercise 6.1

Show that the Fin Efficiency for a Rectangular Fin is given by

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119

Heat Transfer from Extended Surfaces 1

2 2

fin 1

2 2

tanh 2

2

c

c

hl kb hl kb

 

 

 

 

 

 

 

 

 

where

c 2 l b

l   = corrected length

or c A

l l

  P Exercise 6.2

Show that the Total Heat Transfer from a Finned Wall is given by

 

0 ( fin) fin

Q h A l   A

where A = Total area of finned and unfinned surfaces, Afin = Area of the finned surface, and

fin = Fin efficiency.

and  0 T0T

Exercise 6.3

A very long rod 5 mm in diameter has one end maintained at 100oC. The surface of the rod is exposed to ambient air at 25oC with a convection heat transfer coefficient of 100 W/m2 . K.

(a) Determine the temperature distribution along rods constructed from pure copper, 2024 aluminium alloy, and type AISI 316 stainless steel.

What are the corresponding heat losses from the rods?

(b) Estimate how long the rods must be for the assumption of infinite length to yield an accurate estimate of the heat loss.

[Hint : Evaluate the properties at mean temperature, i.e.

62.5 Co

2 Tb T

T    , kcu 398 W/m.K, 180 W/m.K

kal  and for stainless steel, kss 14 W/m.K] Exercise 6.4

A long circular aluminium rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.

(a) If the diameter of the rod is tripled, by how much would the rate of heat removal change?

(b) If a copper rod of the same diameter is used in place of the aluminium rod, by how much would the rate of heat removal change?

Exercise 6.5

Consider two long, slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at 100oC, while the surface of the rods are exposed to ambient air at 20oC. By traversing the length of each rod with a thermocouple, it was observed that the temperatures of the rods were equal at the positions xA = 0.15 m and xB = 0.075 m, where x is measured

(22)

120

Conduction from the base surface. If thermal conductivity of the rod A is known to be kA = 70 W/m . K, determine the thermal conductivity for the rod B.

6.8 SUMMARY

Extended surfaces are used in many heat transfer appliances to increase the heat transfer rate from the device. There are different types of extended surfaces, also know as fins, such as rectangular, triangular, annular, pin fin etc. I the present unit, heat transfer equation through a fin is formulated and solutions are presented for different possible conditions. Performance indices of fin are also discussed. Condition under which use of fin is acceptable is also described. The solution for heat transfer rate and temperature profile with and without fins are illustrated with examples.

6.9 KEY WORDS

Fin : Enhance heat transfer between a solid and an

adjoining fluid.

Fin Effectiveness : Ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin.

Fin Efficiency : It is a measure of the thermal performance of a fin.

Adiabatic : No heat transfer.

6.10 ANSWERS TO SAQs

Please refer the relevant preceding text in this unit for answers to SAQs.

Appendix-I

Bessel function of First Kind

x J0 (x) J1 (x)

0.0 0.1 0.2 0.3 0.4

1.0000 0.9975 0.9900 0.9776 0.9604

0.0000 0.0499 0.0995 0.1483 0.1960 0.5

0.6 0.7 0.8 0.9

0.9385 0.9120 0.8812 0.8463 0.8075

0.2423 0.2867 0.3290 0.3688 0.4059 1.0

1.1 1.2 1.3 1.4

0.7652 0.7196 0.6711 0.6201 0.5669

0.4400 0.4709 0.4983 0.5220 0.5419 1.5

1.6 1.7 1.8

0.5118 0.4554 0.3980 0.3400

0.5579 0.5699 0.5778 0.5815

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121

Heat Transfer from Extended Surfaces

1.9 0.2818 0.5812

2.0 2.1 2.2 2.3 2.4

0.2239 0.1666 0.1104 0.0525 0.0025

0.5767 0.5683 0.5560 0.5399 0.5202

Modified Bessel Function of First and Second Kinds x e– x I0 (x) e– x I1 (x) ex K0 (x) ex K1 (x)

0.0 1.0000 0.0000  

0.2 0.8269 0.0823 2.1407 5.8334

0.4 0.6974 0.1368 1.6627 3.2587

0.6 0.5993 0.1722 1.4167 2.3739

0.8 0.5241 0.1945 1.2582 1.9179

1.0 0.4657 0.2079 1.1445 1.6361

1.2 0.4198 0.2152 1.0575 1.4429

1.4 0.3831 0.2185 0.9881 1.3010

1.6 0.3533 0.2190 0.9309 1.1919

1.8 0.3289 0.2177 0.8828 1.1048

2.0 0.3085 0.2153 0.8416 1.0335

2.2 0.2913 0.2121 0.8056 0.9738

2.4 0.2766 0.2085 0.7740 0.9229

2.6 0.2639 0.2046 0.7459 0.8790

2.8 0.2528 0.2007 0.7206 0.8405

3.0 0.2430 0.1968 0.6978 0.8066

3.2 0.2343 0.1930 0.6770 0.7763

3.4 0.2264 0.1892 0.6579 0.7491

3.6 0.2193 0.1856 0.6404 0.7245

3.8 0.2129 0.1821 0.6243 0.7021

4.0 0.2070 0.1787 0.6093 0.6816

4.2 0.2016 0.1755 0.5953 0.6627

4.4 0.1966 0.1724 0.5823 0.6453

4.6 0.1919 0.1695 0.5701 0.6292

4.8 0.1876 0.1667 0.5586 0.6142

5.0 0.1835 0.1640 0.5478 0.6003

5.2 0.1797 0.1614 0.5376 0.5872

5.4 0.1762 0.1589 0.5279 0.5749

5.6 0.1728 0.1565 0.5188 0.5633

5.8 0.1696 0.1542 0.5101 0.5525

6.0 0.1666 0.1520 0.5019 0.5422

6.4 0.1611 0.1479 0.4865 0.5232

6.8 0.1561 0.1441 0.4724 0.5060

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122

Conduction 7.2 0.1515 0.1405 0.4595 0.4905

7.6 0.1473 0.1372 0.4476 0.4762

8.0 0.1434 0.1341 0.4366 0.4631

8.4 0.1398 0.1312 0.4264 0.4511

8.8 0.1365 0.1285 0.4168 0.4399

9.2 0.1334 0.1260 0.4079 0.4295

9.6 0.1305 0.1235 0.3995 0.4198

10.0 0.1278 0.1213 0.3916 0.4108

1 1 1

( ) ( ) 2 ( )

x x n

I x I x I x

   x

 

REFERENCES

J. P. Holman, (2002), Heat Transfer, 9th Edition, Tata McGraw-Hill, New Delhi.

M. N. Ozisik, (1985), Heat Transfer A Basic Approach, McGraw-Hill International Edition.

F. P. Incropera and D. P. Dewitt, (2004), Fundamentals of Heat and Mass Transfer, 5th Edition , John Wley and Sons.

P. K. Nag, (2002), Heat Transfer, Tata McGraw-Hill Publishing Company Limited, New Delhi.

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Heat Transfer from Extended Surfaces

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124

Conduction

CONDUCTION

Present block is devoted to conduction heat transfer. The entire block consists of three units.

Unit 4 introduces the governing equations of conduction heat transfer. Application of Fourier’s law to steady and unsteady state conduction has been analysed. The governing equation of heat conduction is presented in rectangular, cylindrical as well as spherical coordinates. Heat transfer and determination of temperature distribution for plane wall, cylinder and sphere under steady state condition are discussed. Transient heat transfer concept is introduced with heat generation number, Biot number and Fourier number. A detail analysis of transient heat transfer by lumped capacitance method is described.

Unit 5 describes various methods for solving heat conduction equations. Analytical and graphical methods are described. Limitation of these methods are highlighted and it is emphasized that numerical methods is used in complicated geometries, boundary conditions and variable thermal properties. Finite difference method is described both for the steady and transient conduction problems.

Unit 6 describes application of extended surfaces for heat transfer devices. Different types of fins and their applications are cited in this unit. Formulation and solution are done for extended surfaces. Performance parameters such as fin efficiency and effectiveness are discussed in detail.

The units are supported with solved problems. All the three units include some unsolved problems and SAQ’s which will help you understanding the heat conduction mechanism.

References

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