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

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 kAdT

dxx C q qdq

dxdxx dx x

x

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 qdq

dxdx hdA T T

kAd T

dxdx hP T T dx

x dx conv xx

S

C

( )

( ) ,2

20

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

Extended Surface Analysis

(contd….) d T

dx

hP

kAT T

x T x T

d

dxm

hP

kAD 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 & - m

The 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) = T

The 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

-

The fin heat transfer rate is

These results and other solutions using different end conditions are

tabulated in Table 3.4 in HT textbook, p. 118.

T x T

T T

m L x

mL

q kAdT

dxx hPkA mL M mL

b b

f C C

( ) cosh ( )

cosh

( ) tanh tanh

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

xLm

cosh

)(cosh

mLM tanh0

C Given temperature: (L)=L

mL

xLmxLmb

L

sinh

)(sinh)(sinh)(

mL

mLM b

L

sinh

)(cosh

0

D Infinitely long fin (L)=0

mxe M 0

bCbb

C

hPkAMTT

kA

hPmTT

,)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. h=5 W/ m2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/m C, 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 T

T T

m L x

mL

T x

T x x

b b

( ) cosh ( )

cosh

cosh[ . ( . )]

cosh( . * . ),

( ) . * cosh[ . ( . )]

-

25

100 25

3138 0 2

3138 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, temperature drops off very quickly. At the midpointT(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

Example (contd...)

The total heat transfer through the handle can be calculated also. qf=Mtanh(mL)=8.325*tanh(3.138*0.2)=4.632(W)Very small amount: latent heat of evaporation for water: 2257 kJ/kg. Therefore, the amount of heat loss is just enough to vaporize 0.007 kg of water in one hour.

If a stainless steel handle is used instead, what will happen:For a stainless steel, the thermal conductivity k=15 W/m°C.Use the same parameter as before:

0281.0,47.122/1

C

C

hPkAMkA

hPm

Example (contd...)

)](47.12cosh[3.1225)(

cosh

)(cosh)(

xLxT

mL

xLm

TT

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 played by the thermal conductivity of the material in terms of conductive heat transfer.

Fins-2

If the pot from previous lecture is made of other materials other than the aluminum, what will be the temperature distribution? Try stainless steel (k=15 W/m.K) and copper (385 W/m.K).Recall: h=5W/m2C, P=2W+2t=2(0.03+0.005)=0.07(m) AC=Wt=0.00015(m2), L=0.2(m)Therefore, mss=(hP/kAC)1/2=12.47, mcu=2.46Mss=(hPkssAC) (Tb-T)=0.028(100-25)=2.1(W) Mcu= (hPkssAC) b=0.142(100-25)=10.66(W)

For stainless steel, -T x T

T T

m L x

mL

T x

T x x

ss

b b

ss

ss

( ) cosh ( )

cosh

cosh[ .47( . )]

cosh( .47 * . ),

( ) . * cosh[ .47( . )]

25

100 25

12 0 2

12 0 2

25 12 3 12 0 2

Fins-2 (contd....)

For copper, -T x T

T T

m L x

mL

T x

T x x

cu

b b

cu

cu

( ) cosh ( )

cosh

cosh[ .46( . )]

cosh( .46 * . ),

( ) . * cosh[ .46( . )]

25

100 25

2 0 2

2 0 2

25 66 76 2 0 2

0 0.04 0.08 0.12 0.16 0.275

80

85

90

95

100

T( )x

T ss( )x

T cu( )x

x

copper

aluminum

stainless steel

Fins-2 (contd...)

Inside the handle of the stainless steel pot, temperature drops

quickly. Temperature at the end of the handle is 37.3C. This is

because the stainless steel has low thermal conductivity and heat

can not penetrate easily into the handle.

Copper has the highest k and, correspondingly, the temperature

inside the copper handle distributes more uniformly. Heat easily

transfers into the copper handle.

Question? Which material is most suitable to be used in a heat

sink?

Fins-2 (contd...)

How do we know the adiabatic tip assumption is good? Try using the convection heat transfer condition at the tip (case A in fins table) We will use the aluminum pot as the example.h=5 W/m2.K, k=237 W/m.K, m=3.138, M=8.325W

T x T

T T

m L x h mk m L x

mL h mk mL

x x

T x x x

b b

( ) cosh[ ( )] ( / )sinh[ ( )]

cosh ( / )sinh

cosh[ . ( . )] . sinh[ . ( . )]

cosh( . ) . sinh( . )

( ) . {cosh( . . ) . sinh( . . )}

-

3138 0 2 0 00672 3138 0 2

0 6276 0 00672 0 6276

25 62 09 0 6276 3138 0 00672 0 6276 3138

Long equation

Fins-2 (contd….)

0 0.04 0.08 0.12 0.16 0.285

88.75

92.5

96.25

100

T( )x

T c( )x

x

T: adiabatic tip

Tc: convective tip

T(0.2)=87.32 °CTc(0.2)=87.09 °C

Note 1: Convective tip case has a slightly lower tip temperature as expected since there is additional heat transfer at the tip.Note 2: There is no significant difference between these two solutions, therefore, correct choice of boundary condition is not that important here. However, sometimes correction might be needed to compensate the effect of convective heat transfer at the end. (especially for thick fins)

Fins-2 (contd...)

In some situations, it might be necessary to include the convective heat transfer at the tip. However, one would like to avoid using the long equation as described in case A, fins table. The alternative is to use case B instead and accounts for the convective heat transfer at the tip by extending the fin length L to LC=L+(t/2).

Original fin length L

With convection

t

Then apply the adiabatic condition at the tip of the extended fin as shown above.

L

LC=L+t/2

t/2

Insulation

Fins-2 (contd...)

T x T

T T

m L x

mL

T x

T x x

corr

b b

c

c

corr

corr

( ) cosh ( )

cosh

cosh[ . ( . )]

cosh( . * . ),

( ) . * cosh[ . ( . )]

-

25

100 25

3138 0 2025

3138 0 2025

25 62 05 3138 0 2025

Use the same example: aluminum pot handle, m=3.138, the

length will need to be corrected to

LC=l+(t/2)=0.2+0.0025=0.2025(m)

0 0.04 0.08 0.12 0.16 0.285

88.75

92.5

96.25

100

T( )x

T c( )x

T corr( )x

x

T(0.2)=87.32 °C Tc(0.2)=87.09 °CTcorr(0.2025)=87.05 °C

slight improvement over the uncorrected solution

Fins-2 (contd...)

Correction Length

The correction length can be determined by using the formula: Lc=L+(Ac/P), where Ac is the cross-sectional area and P is the perimeter of the fin at the tip.

Thin rectangular fin: Ac=Wt, P=2(W+t)2W, since t < W Lc=L+(Ac/P)=L+(Wt/2W)=L+(t/2)

Cylindrical fin: Ac=(/4)D2, P= D, Lc=L+(Ac/P)=L+(D/4)

Square fin: Ac=W2, P=4W, Lc=L+(Ac/P)=L+(W2/4W)=L+(W/4)

Optimal Length of a Fin

In general, the longer the fin, the higher the heat transfer. However, a long fin means more material and increased size and cost. Question: how do we determine the optimal fin length? Use the rectangular fin as an example:

tanh , for an adiabatic tip fin

( ) , for an infinitely long fin

Their ratio: R(mL)= tanh( )

f

f

f

f

q M mL

q M

qmL

q

0 1 2 3 40

0.2

0.4

0.6

0.8

1

R( )mL

mL

Note: heat transfer increases with mL as expected. Initially the rate of change is large and slows down drastically when mL> 2.

R(1)=0.762, means any increase beyond mL=1 will increase no more than 23.8% of the fin heat transfer.

Temperature Distribution

For an adiabatic tip fin case:

cosh ( )

coshb

T T m L xR

T T mL

Use m=5, and L=0.2 as an example:

0 0.05 0.1 0.15 0.20.6

0.8

11

0.648054

R ( )x

0.20 x

High T, good fin heat transfer Low T, poor fin heat transfer

Correction Length for a Fin with a Non-adiabatic Tip

The correction length can be determined by using the formula: Lc=L+(Ac/P), where Ac is the cross-sectional area and P is the perimeter of the fin at the tip.

Thin rectangular fin: Ac=Wt, P=2(W+t)2W, since t < W Lc=L+(Ac/P)=L+(Wt/2W)=L+(t/2)

Cylindrical fin: Ac=(/4)D2, P= D, Lc=L+(Ac/P)=L+(D/4)

Square fin: Ac=W2, P=4W, Lc=L+(Ac/P)=L+(W2/4W)=L+(W/4)

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 T

R

R

,

C

resistance can be expressed as

( ) 1

hPkA [tanh( )]b

t ff

T TR

q mL

Tb

T

Fin EffectivenessHow 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 kPmL

q hA T T hA hA

kP k P

hA h A

nsfer, 1.

However, 2 will be considered justifiable

If <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 P

hA 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 R

q hA T T T T R R

. In order to enhance heat transfer,

the fin's resistance should be lower than that of the resistance

due only to convection.

Fin Efficiency

Define Fin efficiency:

where represents an idealized situation such that the fin is made up

of material with infinite thermal conductivity. Therefore, the fin should

be at the same temperature as the temperature of the base.

f

q

q

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:

max

f

max

,,

( ) tanhtanh

( ) ( )

tanh tanh (see Table 3.5 for of common fins)

The fin heat transfer: ( )

, where 1/( )

f c bf

f b b

c

f f f f b

b bf t f

f f t f

q hPkA T T mLM mL

q hA T T hPL T T

mL mL

mLhPL

kA

q q hA T T

T T T Tq R

hA R

,

, , b

1

Thermal resistance for a single fin.

1As compared to convective heat transfer:

In order to have a lower resistance as that is required to

enhance heat transfer: or A

f f

t bb

t b t f f f

hA

RhA

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 T

h A NA N A T T h A NA T T

NAhA T T hA T T

A

NA

A

Heat Transfer from a Fin Array

,,

,

,

1( ) where

Compare to heat transfer without fins

1( ) ( )( )

where is the base area (unexposed) for the fin

To 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 R

R hA

q hA T T h A NA T ThA

A

A A

Oat is, to increase the effective area .tA

Thermal Resistance Concept

T1T

TbT2

T1 TTbT2

L1 t

R1=L1/(k1A)

Rb=t/(kbA)

)/(1, OtOt hAR

1 1

1 ,b t O

T T T Tq

R R R R

A=Ab+NAb,f