EXTENDED SURFACES / FINS

Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling law: q = hA(Ts -T

). Therefore, to increase the convective heat transfer, one can

Increase the temperature difference (Ts -T

) between the surface and the fluid.

Increase the convection coefficient h. This can be accomplished by increasing the fluid flow over the surface since h is a function of the flow velocity and the higher the velocity, the higher the h. Example: a cooling fan.

Increase the contact surface area A. Example: a heat sink with fins.

Extended Surface Analysis

x

Tb

q kA dTdxx C q q

dqdx

dxx dx xx

dq h dA T Tconv S ( )( ), where dA is the surface area of the elementS

AC is the cross-sectional area

Energy Balance:

if k, A are all constants.

x

C

q q dq qdqdx

dx hdA T T

kA d Tdx

dx hP T T dx

x dx conv xx

S

C

( )

( ) ,2

2 0

P: the fin perimeterAc : the fin cross-sectional area

Extended Surface Analysis

(contd….)d Tdx

hPkA

T T

x T x T

ddx

m hPkA

D m

x C e C e

C

C

mx mx

2

2

2

22 2 2

1 2

0

0 0

( ) ,

( ) ( ) ,

, , ( )

( ),

A second - order, ordinary differential equation

Define a new variable = so that

where m

Characteristics equation with two real roots: + m & - mThe general solution is of the form

To evaluate the two constants C and C we need to specify two boundary conditions: The first one is obvious: the base temperature is known as T(0) = TThe second condition will depend on the end condition of the tip

2

1 2

b

Extended Surface Analysis (contd...)

For example: assume the tip is insulated and no heat transferd/dx(x=L)=0

The temperature distribution is given by

)cosh()(cosh)(

mLxLm

TTTxT

bb

The fin heat transfer rate is given by

mLMmLhPkAxdxdTkAq CCf tantanh)0(

Temperature distribution for fins of different configurations

Case Tip Condition Temp. Distribution Fin heat transfer A Convection heat

transfer: h(L)=-k(d/dx)x=L mLmk

hmL

xLmmkhxLm

sinh)(cosh

)(sinh)()(cosh

MmLmk

hmL

mLmkhmL

osinh)(cosh

cosh)(sinh

B Adiabatic (d/dx)x=L=0 mL

xLmcosh

)(cosh mLM tanh0

C Given temperature: (L)=L

mL

xLmxLmb

L

sinh

)(sinh)(sinh)(

mL

mLM b

L

sinh

)(cosh0

D Infinitely long fin (L)=0

mxe M 0

bCbb

C

hPkAMTT

kAhPmTT

,)0(

, 2

Example

An Aluminum pot is used to boil water as shown below. The handle of the pot is 20-cm long, 3-cm wide, and 0.5-cm thick. The pot is exposed to room air at 25C, and the convection coefficient is 5 W/m2 C. Question: can you touch the handle when the water is boiling? (k for aluminum is 237 W/m C)

100 C

T

= 25 Ch = 5 W/ m2 C

x

Example (contd...)

We can model the pot handle as an extended surface. Assume that there is no heat transfer at the free end of the handle. The condition matches that specified in the fins Table, case B. Use the following data:h=5 W/ m2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/mC, AC =Wt=0.00015(m2), L=0.2(m)Therefore, m=(hP/kAC )1/2=3.138, M=(hPkAC )(Tb -T

)=0.111b =0.111(100-25)=8.325(W)T x TT T

m L xmL

T x

T x x

b b

( ) cosh ( )cosh

cosh[ . ( . )]cosh( . * . )

,

( ) . * cosh[ . ( . )]

-

25100 25

3138 0 23138 0 2

25 62 32 3138 0 2

Example (contd…)Plot the temperature distribution along the pot handle

0 0.05 0.1 0.15 0.285

90

95

100

T( )x

x

As shown in the figure, temperature drops off but not very steeply. This is because k of aluminium is fairly high. At the midpoint, T(0.1)=90.4C. At the end T(0.2)=87.3C. Therefore, it should not be safe to touch the end of the handle.The end condition is insulated, hence the gradient is zero.

Example (contd...)The total heat transfer through the handle can be calculated also. qf =Mtanh(mL)=8.325tanh[(3.138)(0.2)]=4.632 W

If a stainless steel handle is used instead, what will happen?For a stainless steel, the thermal conductivity k=15 W/m°C, which is much less compared to aluminium.Using the same process parameter as before:

0281.0,47.122/1

C

C

hPkAMkAhPm

Example (contd...)

)](47.12cosh[3.1225)(cosh

)(cosh)(

xLxTmL

xLmTT

TxT

b

0 0.05 0.1 0.15 0.20

25

50

75

100

T x( )

x

Temperature at the handle (x=0.2 m) is only 37.3 °C, not hot at all. This example illustrates the important role of the thermal conductivity of the material in the temperature distribution in a fin.

Fin Design

Total heat loss: qf =Mtanh(mL) for an adiabatic fin, or qf =Mtanh(mLC ) if there is convective heat transfer at the tip

C C

C,

,

hPwhere = , and M= hPkA hPkA ( )

Use the thermal resistance concept:( )hPkA tanh( )( )

where is the thermal resistance of the fin.For a fin with an adiabatic tip, the fin

b bc

bf b

t f

t f

m T TkA

T Tq mL T TR

R

,C

resistance can be expressed as( ) 1

hPkA [tanh( )]b

t ff

T TRq mL

Tb

T

Fin Effectiveness

How effective a fin can enhance heat transfer is characterized by the fin effectiveness f : Ratio of fin heat transfer and the heat transfer without the fin. For an adiabatic fin:

ChPkA tanh( )tanh( )

( )If the fin is long enough, mL>2, tanh(mL) 1, it can be considered an infinite fin (case D of table3.4)

In order to enhance heat tra

f ff

C b C C

fC C

q q mL kP mLq hA T T hA hA

kP k PhA h A

nsfer, 1.

However, 2 will be considered justifiableIf <1 then we have an insulator instead of a heat fin

f

f

f

Fin Effectiveness (contd...)

To increase f , the fin’s material should have higher thermal

conductivity, k.

It seems to be counterintuitive that the lower convection

coefficient, h, the higher f . But it is not because if h is very high,

it is not necessary to enhance heat transfer by adding heat fins.

Therefore, heat fins are more effective if h is low. Observation: If

fins are to be used on surfaces separating gas and liquid. Fins are

usually placed on the gas side. (Why?)

fC C

kP k PhA h A

P/AC should be as high as possible. Use a square fin with

a dimension of W by W as an example: P=4W, AC=W2,

P/AC=(4/W). The smaller W, the higher the P/AC, and the

higher f.

Conclusion: It is preferred to use thin and closely spaced

(to increase the total number) fins.

Fin Effectiveness (contd...)

Fin Effectiveness (contd...)

, ,

, ,

The effectiveness of a fin can also be characterized as( ) /

( ) ( ) /It is a ratio of the thermal resistance due to convection to the thermal resistance of a fin

f f b t f t hf

C b b t h t f

q q T T R Rq hA T T T T R R

. In order to enhance heat transfer,the fin's resistance should be lower than that of the resistancedue only to convection.

Fin Efficiency

Define Fin efficiency:

where represents an idealized situation such that the fin is made upof material with infinite thermal conductivity. Therefore, the fin shouldbe at the same temperature as the temperature of the base.

f

qq

q

q hA T T

f

f b

max

max

max ( )

T(x)<Tb for heat transferto take place

Total fin heat transfer qf

Real situation Ideal situation

For infinite kT(x)=Tb, the heat transferis maximum

Ideal heat transfer qmax

Tb x x

Fin Efficiency (contd…)

Fin Efficiency (cont.)

Use an adiabatic rectangular fin as an example:

m a x

f

m a x

,,

( ) ta n hta n h( ) ( )

ta n h ta n h (se e T a b le 3 .5 fo r o f c o m m o n f in s )

T h e f in h e a t tra n s fe r : ( )

, w h e re 1 /( )

f c bf

f b b

c

f f f f b

b bf t f

f f t f

q h P k A T T m LM m Lq h A T T h P L T T

m L m Lm Lh P L

k Aq q h A T T

T T T Tq Rh A R

,

, , b

1

T h e rm a l re s is ta n c e fo r a s in g le f in .1A s c o m p a re d to c o n v e c tiv e h e a t tr a n s fe r :

In o rd e r to h a v e a lo w e r re s is ta n c e a s th a t is r e q u ire d to e n h a n c e h e a t tra n s fe r : o r A

f f

t bb

t b t f f f

h A

Rh A

R R A

Overall Fin Efficiency

Overall fin efficiency for an array of fins:

Define terms: Ab : base area exposed to coolant

Af : surface area of a single fin

At : total area including base area and total

finned surface, At =Ab +NAf

N: total number of fins

qb

qf

Overall Fin Efficiency (contd…)

( ) ( )[( ) ]( ) [ (1 )]( )

[1 (1 )]( ) ( )

Define overall fin efficiency: 1 (1 )

t b f b b f f b

t f f f b t f f b

ft f b O t b

t

fO f

t

q q Nq hA T T N hA T Th A NA N A T T h A NA T T

NAhA T T hA T T

ANAA

Heat Transfer from a Fin Array

,,

,

,

1( ) where

Compare to heat transfer without fins1( ) ( )( )

where is the base area (unexposed) for the finTo enhance heat transfer Th

bt t O b t O

t O t O

b b b f b

b f

t O

T Tq hA T T RR hA

q hA T T h A NA T ThA

AA A

Oat is, to increase the effective area .tA

Thermal Resistance Concept

T1T

TbT2

T1 TTbT2

L1 t

R1 =L1 /(k1 A)

Rb =t/(kb A)

)/(1, OtOt hAR

1 1

1 ,b t O

T T T TqR R R R

A=Ab +NAb,f