Extended Surfaces / Fins

8/14/2019 Extended Surfaces / Fins

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EXTENDED SURFACES / FINS

Convection: Heat transfer between a solid surface and a movingfluid is governed by the Newtons cooling law: q = hA(T s-T).Therefore, to increase the convective heat transfer, one can

Increase the temperature difference (T s-T) between thesurface and the fluid.

Increase the convection coefficient h. This can beaccomplished 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.


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Extended Surface Analysis

x

Tb

q kAdT dx x C = q q

dq

dx dx x dx x x

+ = +

dq h dA T T conv 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 dx

dx hP T T dx

x dx conv x x

S

C

= + = + +

+ =

+

( )

( ) ,2

2 0

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


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Extended Surface Analysis

(contd.)

d T dx

hPkA

T T

x T x T d dx

mhPkA

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 specifytwo 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


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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 xmL

q kAdT dx x hPkA mL M mL

b b

f C C

( ) cosh ( )cosh

( ) tanh tanh

= =

= = = =

0


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Temperature distribution for fins of different configurations

Case Tip Condition Temp. Distribution Fin heat transferA Convection heat

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

hmL

x Lmmk h x Lm

sinh)(cosh

)(sinh)()(cosh

+

+

M

mLmk hmL

mLmk hmL

osinh)(cosh

cosh)(sinh

+

+

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

x Lmcosh

)(cosh

mL M tanh0

C Given temperature:

(L)=

L mL

x Lm x Lmb

L

sinh

)(sinh)(sinh)( +

mL

mL M b

L

sinh

)(cosh0

D Infinitely long fin (L)=0

mxe M 0

bC bb

C

hPkA M T T

kAhPmT T

===

,)0(

, 2


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Example

An Aluminum pot is used to boil water as shown below. Thehandle 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 convectioncoefficient is 5 W/m 2 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/ m 2 C

x


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Example (contd...)

We can model the pot handle as an extended surface. Assumethat there is no heat transfer at the free end of the handle. Thecondition matches that specified in the fins Table, case B.h=5 W/ m 2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/mC, A C=Wt=0.00015(m 2), L=0.2(m)Therefore, m=(hP/kA C)1/2=3.138,M= (hPkA C)(T b-T)=0.111 b=0.111(100-25)=8.325(W)

T x T

T T

m L x

mLT x

T x x

b b

( ) cosh ( )

coshcosh[ . ( . )]

cosh( . * . ),

( ) . * cosh[ . ( . )]

-

= =

=

= +

25

100 253138 0 2

3138 0 2

25 62 32 3138 0 2


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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)=90.4 C. At the end T(0.2)=87.3 C.

Therefore, it should not be safe to touch the end of the handle


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Example (contd...)

The total heat transfer through the handle can be calculatedalso. q f =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 tovaporize 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/mC.Use the same parameter as before:

0281.0,47.122 / 1

===

=

C C

hPkA M kAhPm


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Example (contd...)

)](47.12cosh[3.1225)(

cosh )(cosh)(

x L xT

mL x LmT T T xT 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 atall. This example illustrates the important role played by the

thermal conductivity of the material in terms of conductive heattransfer.


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Fins-2

If the pot from previous lecture is made of other materials otherthan the aluminum, what will be the temperature distribution? Trystainless steel (k=15 W/m.K) and copper (385 W/m.K).

Recall: h=5W/m 2C, P=2W+2t=2(0.03+0.005)=0.07(m)AC=Wt=0.00015(m 2), L=0.2(m)Therefore, mss=(hP/kA C)1/2=12.47, mcu=2.46

M ss=(hPk ssAC) (T b-T)=0.028(100-25)=2.1(W)M cu= (hPk ssAC) 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( . )]

= =

=

= +

25100 25

12 0 212 0 2

25 12 3 12 0 2


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Fins-2 (contd....)

For copper,-T x T

T T m L xmL

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 0 225 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


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Fins-2 (contd...)

Inside the handle of the stainless steel pot, temperature drops

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


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Fins-2 (contd...)

How do we know the adiabatic tip assumption is good? Tryusing the convection heat transfer condition at the tip (case A infins table) We will use the aluminum pot as the example.h=5 W/m 2.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 ( / )sinhcosh[ . ( . )] . 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


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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 temperatureas expected since there is additional heat transfer at the tip.Note 2: There is no significant difference between these twosolutions, therefore, correct choice of boundary condition is notthat important here. However, sometimes correction might be

needed to compensate the effect of convective heat transfer atthe end. (especially for thick fins)


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Fins-2 (contd...)

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

Original fin length L

With convection

t

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

L

LC=L+t/2

t/2

Insulation


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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[ . ( . )]

-

= =

=

= +

25100 25

3138 0 20253138 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)


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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

T(0.2)=87.32 CTc(0.2)=87.09 C

Tcorr (0.2025)=87.05 C

slight improvement

over the uncorrectedsolution

Fins-2 (contd...)


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Correction Length

The correction length can be determined by using the formula:

Lc=L+(A c /P), where A c is the cross-sectional area and P is theperimeter of the fin at the tip.

Thin rectangular fin: A c=Wt, P=2(W+t)

2W, since t 2.

R(1)=0.762, means any increase beyond mL=1 will increase nomore than 23.8% of the fin heat transfer.

T t Di t ib ti


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Temperature Distribution

For an adiabatic tip fin case:cosh ( )

coshb

T T m L x R

T T mL

= =

Use m=5, and L=0.2as 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 ith a


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Correction Length for a Fin with aNon-adiabatic Tip

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

Thin rectangular fin: A c=Wt, P=2(W+t) 2W, since t


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Fin Design

Total heat loss: q f =Mtanh(mL) for anadiabatic fin, or q f =Mtanh(mL C) 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

b f b

t f

t f

m T T kA

T T q mL T T

R

R

=

= =

,C

resistance can be expressed as

( ) 1hPkA [tanh( )]

bt f

f

T T Rq mL

= =

Tb

T


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Fin EffectivenessHow effective a fin can enhance heat transfer is characterized by thefin effectiveness f : Ratio of fin heat transfer and the heat transferwithout 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 f f

C b C C

f

C C

q q mL kP mLq hA T T hA hA

kP k P

hA h A

= = = =

=

nsfer, 1.

However, 2 will be considered justifiable

If

Fin Effectiveness


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Fin Effectiveness(contd...)

To increase f , the fins 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 areusually placed on the gas side. (Why?)

f C C

kP k PhA h A

=


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


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Fin Effectiveness (contd...)

, ,

, ,

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

( ) ( ) /

It is a ratio of the thermal resistance due to convection tothe thermal resistance of a fin

f f b t f t h f

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.

ff


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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 ( )


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T(x)


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y ( )

Use an adiabatic rectangular fin as an example:

m a x

f

m a x

,,

( ) ta n ht a n h( ) ( )

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

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

, w h e r e1 /( )

f c b f

f b b

c

f f f f b

b b f t f

f f t f

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

m L m Lm Lh P

Lk A

q q h A T T

T T T T q R

h A R

= = =

= =

= = = =

,

, , b

1

T h e r m a l re s is ta n c e f o r a s in g le f in .

1A s c o m p a r e d to c o n v e c t iv e h e a t tr a n s f e r :

I n o r d e r to h a v e a l o w e r r e s is ta n c e a s th a t is r e q u ir e d to

e n h a n c e h e a t t r a n s f e r : o r A

f f

t bb

t b t f f f

h A

Rh A

R R A

=

=

> >

Oat is, to increase the effective area . t A

Th l R i C


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Thermal Resistance Concept

T1T

TbT2

T1 TTbT2

L1 t

R1=L 1 /(k 1A)

Rb=t/(k bA)

) /(1, Ot Ot hA R =

1 1

1 ,b t O

T T T T q R R R R

= =+ +

A=A b+NA b,f

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Extended Surfaces / Fins