Heat Transfer from Extended SurfacesHeat Transfer Enhancement by

Fins

Bare surface Finned surface

Typical finned-tube heat exchangers

Straight fin of uniform cross

section

Straight fin of nonuniform cross section

Annular fin Pin fin

Equation for Extended Surfaces

x

T∞, h

Ac(x)dx

dx

Tb

cond,inq cond,outq

conv,outq

cond,in cond,out conv,outq q q

dAs(x)

dx

Ac(x)dAs(x)

cond,inq cond,outq

conv,outq

cond,in cond,out conv,outq q q

x

( ) ( )c c

dT d dTkA x kA x dx

dx dx dx

( ) ( )shdA x T x T

cond,inq ( )x c

dTq kA x

dx

cond,outq x dxq

conv,outq

T∞, h

T(x)

convdq

When k = constant,

( ) ( ) ( ) 0c s

d dTkA x dx hdA x T x T

dx dx

( ) ( ) 0sc

dAd dTkA x h T x T

dx dx dx

( ) ( ) 0sc

dAd dT hA x T x T

dx dx k dx

Fins of Uniform Cross-Sectional Area

Ac(x) = constant,

and dAs = Pdx

P: fin perimeter

x

Ac

Tb

L

dAsdx

,T h

( ) ( ) 0sc

dAd dT hA x T x T

dx dx k dx

2

2( ) 0

c

d T hPT x T

dx kA

P

boundary conditionsat x =

0:

excess temperature : (x) = T(x) - T∞

2

2( ) 0

c

d T hPT x T

dx kA

22

20

dm

dx

where 2

c

hPm

kA

dxx

Tb

, T h

(0) bT T (0) (0)T T bT T b

L

T(x)

1 2( ) mx mxx C e C e 1 2sinh( ) cosh( )D mx D mx

at x = L: 3 cases1) very long fin (L → ∞):

2) convection tip:

3) negligible heat loss: adiabatic tip

dxx

Tb, T h

L( ) ( )x T x T

( )T L T ( ) 0T T

( )L

dTk h T L T

dx ( ) 0L

d hL

dx k

0L

dTk

dx 0

L

d

dx

T(x)

Temperature distribution

2) convection tip:

1) long fin:

3) adiabatic tip:

( ) mx

b

xe

cosh ( ) / sinh ( )( )

cosh / sinhb

m L x h mk m L xx

mL h mk mL

( ) cosh ( )

coshb

x m L x

mL

Total heat loss by the fin

1) long fin:

3) adiabatic tip:

2) convection tip:

,

0

f c b

x

dTq kA

dx

or ( )f

f Aq h T x T dA

x

Ac

dAs

Tb

L

dx

P, T h

f c bq hPkA M

sinh / cosh

cosh / sinhf c b

mL h mk mLq hPkA

mL h mk mL

tanh tanhf c bq hPkA mL M mL

Find: 1) Temperature distribution T(x) and heat loss qf when the fin is constr

ucted from: a) pure copper, b) 2024 aluminum alloy, and c) type AISI 316 stainless steel.

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

Assumption:very long fin

Example 3.9

, , 5mmk L D

C100bT 2

25 C100 W/m K

Th

air

1) For a very long fin

heat loss:

conductivity at2

bT TT

Copper: k = 398 W/m.KAluminum alloy: k = 180 W/m.KStainless steel: k = 14 W/m.K

Copper: 8.3 W

Aluminum alloy: 5.6 W

Stainless steel: 1.6 W

qf

( ) mx

b

xe

( )

b

T

T

T

T

x

( ) mx

bT T T eT x

, , 5mmk L D

C100bT 2

25 C100 W/m K

Th Air

c

hPm

kA

4h

kD

2

,4c

DP D A

cf bhPkAq

2)

1/ 22.65

2.65 ckAL

m hP

To get an accuracy over 99%

For the adiabatic condition at the tip

qf = MtanhmL (long fin: qf = M)

Copper: 0.19 m

Aluminum alloy: 0.13 m

Stainless steel: 0.04 m

tanh 0.99mL 2.65mL or

Fins of Nonuniform Cross-Sectional Area : Annular Fin

T∞, h2r t

cond,in cond,out conv,outq q q

bT

dr

1rr

cond,inqcond,outq

conv,outq

c 2A rt

T∞, h

1r

2rrdr

t

conv,outq

cond,outq cond,inq

2r

dTq k rt

dr

cond,outq 2 2r dr

dT d dTq k rt k rt dr

dr dr dr

cond,inq

conv,outq conv 2 2 ( )dq h rdr T r T

T(r)

c 2A rt

bT

2 2sdA rdr

When k = const,

boundary conditions: when an adiabatic tip is presumed

2 4 ( ) 0d dT

k rt h r T r Tdr dr

2( ) 0

d dT hr r T r T

dr dr kt

in terms of excess temperature

( ) ( )r T r T

2 0d d

r m rdr dr

where 2 2hm

kt

1( ) ,br 2

0r r

d

dr

( ) ( ) 0sc

dAd dT hA x T x T

dx dx k dx

2r tdr

1rr

c 2A rt

2 2sdA rdr

42 ( ) 0

d dT h rdrrt T x T

dr dr k dr

Particular solutions

: real : real : zero or integer

J andand

(or )J Y

nJ nY

: imaginary : fractional

: zero or integer

I and

and

(or )I K

nI nK

Solutions to generalized Bessel equations

2 0, 2 0d d

r rdr dr

General Solution:

/ 1/( ) ,r r Z r

where (1 ) /( 2), 2 /( 2), / (1 ) / 2

present case:

boundary conditions:

2 0d d

r m rdr dr

2 0, 2 0d d

r rdr dr

2 21, m

Thus,

, 0, mi

( )r Z mr 1 0 2 0( ) ( )r C I mr C K mr

1 1 0 1 2 0 1( ) ( ) br C I mr C K mr

2 2 2

0 01 2

( ) ( )0

r r r r r r

dI mr dK mrdC C

dr dr dr

/ 1/( )r r Z r

1( ) ( ),n nn n

dx I x x I x

dx

1( ) ( )n nn n

dx K x x K x

dx

2 2

0 01 2

( ) ( )0

r r r r

dI mr dK mrC C

dr dr

01

( )( ),

dI mrmI mr

dx 0

1

( )( )

dK mrmK mr

dx

1 1 2 2 1 2( ) ( ) 0C I mr C K mr

1 21

0 1 1 2 0 1 1 2

( ),

( ) ( ) ( ) ( )bK mr

CI mr K mr K mr I mr

1 22

0 1 1 2 0 1 1 2

( )

( ) ( ) ( ) ( )bI mr

CI mr K mr K mr I mr

1 2 0 1 2 0

0 1 1 2 0 1 1 2

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )b

K mr I mr I mr K mrr

I mr K mr K mr I mr

1 0 2 0( ) ( )r C I mr C K mr

Heat loss from the fin

1

,f c b

r r

dTq kA

dr

1

12r r

dk r t

dr

1 1 1 2 1 1 1 21

0 1 1 2 0 1 1 2

( ) ( ) ( ) ( )2

( ) ( ) ( ) ( )f b

K mr I mr I mr K mrq kr t

K mr I mr I mr K mr

T∞, h

1r

2rrdr

t

T(r)

c 2A rtbT

1 2 0 1 2 0

0 1 1 2 0 1 1 2

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )b

K mr I mr I mr K mrr

I mr K mr K mr I mr

Fin Performance

• fin effectiveness

• fin resistance

• fin efficiency

heat loss without fin

Tb

T∞, h

Ac,b: fin cross-sectional area at the base

Ac,b

1. Fin effectiveness

fin effectiveness:

design criteria:

, ,b c b b c b bq hA T T hA

,

ff

c b b

q

hA

(rule of a thumb)

2f

Assume hs are the same for with or without fin.

Ex) long straight fin with uniform cross- sectional area

In order to get high fin performance

• installation of fins at the lower h side

• large k material

• thin shape

f c bq hPkA

,

c bf

c b b

hPkA

hA

c

kP

hA

provides upper limit of f, which is reached as L approaches infinity.

Practically qf for the adiabatic tip reaches 9

8% of heat transfer when mL = 2.3.

Thus, the fin length longer than L = 2.3/m is not effective.

f

c

kP

hA

f c bq hPkA long fin

c

hPm

kAtanh ,f c bq hPkA mLadia. tip

2. Fin resistance

b: driving potential

Thermal resistance due to convection at the exposed base: Rt,b

,

bf

t f

qR

,t fb

f

Rq

,b

t bb

Rq

,

b

c b bhA

,

1

c bhA

Tb

T∞, h

Ac,b

Ex) straight fin of uniform cross-sectional area with an adiabatic tip

3. Fin efficiency

qmax: heat loss when the whole fin is assumed at Tb

max

ff

q

q f

f b

q

hA

tanhc b

b

hPkA mL

hPL

ff

f b

q

hA

tanh mL

mL

x

Ac

dAs

Tb

L

dx

P, T h

For an active tip, the above relation can be used with fin length correction.rectangular fin:

pin fin:

tanhtanh , c

cf

c

f

mq M m

Lm

LL

Errors can be negligible if

or /ht k / 2 0.0625hD k

2c

tL L

4c

DL L

cf mL

Tb

x, T h

L

t D

When w >> t, P ~ 2w,

Lc

wt

Corrected fin profile area

cA wt

1/ 2

c cc

hPmL L

kA

1/ 22

c

h wL

kwt

1/ 22

c

hL

kt

pp c

c

AA L t t

L

1/ 22

c c

hmL L

kt

1/ 2

2 cc

p

hLL

kA

1/ 2

3 / 22c

p

hL

kA

1/ 23 / 2 2 /f c pf L h kA

Ap

Efficiency of straight fins (rectangular, triangular, and parabolic

profile)

Efficiency of annular fins of rectangular profile

Overall Surface Efficiency

overall efficiency of array of fins:

At : area of fins + exposed portion of the base

single fin efficiency:

ff

f b

q

hA

t f bA NA A

max

to

q

q t

t b

q

hA

f f b b bN hA hA

tqf b bNq hA f

ff b

q

hA

b f f t fh N A A NA

1f ft b f

t t

NA NAhA

A A

1 1ft b f

t

NAhA

A

1 1ft

t b

o f

t

q

hA

NA

A

t f bA NA A

In the case of press fit: thermal contact resistance

Tb T∞Tc

(

x

(

a

))

m

tc

co

q

q ( )t c

t b

q

hA

( ) ( )t c f c b bq Nq hA

( ), ,

bf c

t c t f

qR R

( ),

b cf c

t c

T Tq

R

( ) ( ) ,

, ,/b c

f c f c c bt c c b

T Tq q A

R A

, , ,/t c t c c bR R A

,t cR ,t fR( )f cq

bT cT T

, , fbt f f

f f b

qR

q hA

,

1t f

f f

RhA

( ), ,/ 1/

bf c

t c c b f f

qR A hA

( ) ( )t c f c b bq Nq hA

, ,/ 1/b

b bt c c b f f

NhA

R A hA

, ,1 /f f b

b bf f t c c b

N hAhA

hA R A

, ,1 /f f

b bf f t c c b

N Ah A

hA R A

let , , 11 /f f t c c bhA R A C

( )

1

f ft c b b

N Aq h A

C

1

f f t ft b

t t

N A A NAhA

C A A

1

1 1f ft b

t

NAhA

A C

(

x

(

a

))

m

tc

co

q

q ( )t c

t b

q

hA

1

1 1f f

t

NA

A C

Find: Increase in heat transfer, q = qt – qwo, associated

with using fins

Example 3.10

.0 15mH

2300K50 W/m K

Th

air

S

500 KbT

6mmt

1 25mmr 20mmL

2 45mmr

Engine cylinder

Cross-section

(2024 T6 Al alloy)

Annual fins

1 1ft f

tt bA

NAh

Aq

tA f bNA A

1) Heat transfer rate

12fNA r H Nt

fA 2 22 12 cr r

2cr 2 2

tr

2 2 22 0.048 0.025 0.0105 mfA

25 0.0105 2 0.025 0.15 5 0.006 0.0716 mtA

0.0060.045 0.048 m

2

1 25mmr 20mmL

2 45mmr

H =

0.1

5 m

t = 6

mm

500 KbT

2300K50 W/m K

Th

f: known

To get f, use Fig. 3.19.

Parameters:

2

1

0.0481.92

0.025cr

r

1/ 2

1/ 2 3 / 23 / 24

50/ 0.023 0.15

186 1.38 10c pL h kA

cL2

tL

0.0060.02 0.023 m

2

pA cL t 4 20.023 0.006 1.38 10 m

1 25mmr 20mmL

2 45mmr

H =

0.1

5 m

t = 6

mm

2024 T6 Alk = 186 at 400 K

2

1

,cr

r 1/ 23 / 2 /c pL h kA

1/ 23 / 2 / 0.15c pL h kA

2

1

1.92cr

r

0.95f

0.15

0.95

Without fins

wo 454 Wt qq q

5 0.010550 0.0716 1 1 0.95 500 300 690 W

0.0716

1 1ft f b

ttq

NAhA

A

woq 12 bh r H 236 W

The amount of increase in heat transfer

wo

6902.67

236tq

q

1 25mmr 20mmL

2 45mmr H

= 0

.15

m

t = 6

mm

500 KbT

2300K50 W/m K

Th

Comments:

Fixed fin thickness : 6 mm, Minimum fin gap : 4 mm Nmax= H/S =15

Fixed fin gap : 4 mm, Minimum fin thickness : 2 mm Nmax= H/S =25

Example 3.11Hydrogen-air Proton Exchange Membrane (PEM) fuel cell

Known: 1) Dimensions of a fuel cell and finned heat sink2) Fuel cell operating temperature3) Rate of thermal energy generation: 11.25 W

4) Power production: P = 9 W

5) Relationship between the convection coefficient and the air channel dimensions

50 mm 50 mm 6 mm

10 mm gaps50 mm 26 mm

56.4 CcT

9.4 m/s, 25 CV T

Without finned heat sink

With finned heat sink

air1.78 f fh k L a L a 3/ 1000 W/(m s)f fP C

Assumptions: 1) Steady-state conditions2) Negligible heat transfer from the edges of the fuel

cell, as well as from the front and back faces of the finned heat sink

3) 1D heat transfer through the heat sink4) Adiabatic fin tips5) Negligible radiation when the heat sink is in place.

Find: 1) Net power of the fuel cell-fan system for no heat sink, Pnet = P – Pf

2) # of fins N needed to reduce the fan power consumption by 50%

Volumetric flow rate of cooling air: ,cf VA c cA W H t

f cV W H t 9.4m/s 0.05m (0.026m 0.006m) 3 39.4 10 m /s

net

fPP C 3 -3 39.0 W 1000 W/(m /s) 9.4 10 m /s

9.0 W 9.4 W 0.4 W

Fan power consumption:

3/ 1000 W/(m s)f fP C

9.4 m/s, 25 CV T

The fan consumes more power than is generated by the fuel cell, and the system cannot produce net power.

P = 9 W

fP C net fPP P 1.

Fuel cell

3/ 1000 W/(m s)f fP C

Aluminum fined sink: k = 200 W/m.K

air1.78 f fh k L a L a

+ finned sink base(conduction)

+ exposed base of finned side(convection)

Lc = 50 mm

+ fins(conv+cond)

11 1equiv , , ( )t b t f NR R R

= 11.25 W

2. To reduce the fan power consumption by 50%, 3 34.7 10 m /sf

9.4 W / 2 4.7 WfP

, ,base equiv

c

t c t

T Tq

R R R

Thermal circuit

3 2, 10 m K/Wt cR

contact joint(contact)

3 2, 10 m K/Wt cR

Aluminum fined sink: k = 200 W/m.K

Lc = 50 mm

,

,c b

t c

T Tq

R

q Aq, /

c b

t c

T T

R A

,

c b

t c

T T

R

3 2(10 m K/W)/(2 0.05m 0.05m)=0.2K/W

, , /t c t cR R A , / 2t c c cR L W : 2 sides of the heat sink assembly

,basetR /L kA /(2 )b c ct kL W

(0.02m)/(2 200W/m K 0.05m 0.05m) = 0.002K/W

, :t cR

,base :tR

N, h are needed to evaluate Rt,b

1

(2 0.05m 0.001m) 0.05 mNh

,t bR1

hA

1

2 c f ch W LNt

1

0.005 0.0005 Nh

, :t bR

3 2, 10 m K/Wt cR

Aluminum fined sink: k = 200 W/m.K

Lc = 50 mm

air1.78 f fh k L a L a

, ( ) :t f NR, ( )

b

t f N

T T

R

( )f Nq fNq

,

b

t f

T TN

R

,, ( )

t ft f N

R

NR

For a single fin,

, /t f b fR q

For a fin with an insulated fin tip,

tanhf c b fq PkA mh L

,

1,

tanhb

t ff c f

Rq PkA mh L

3 2, 10 m K/Wt cR

Aluminum fined sink: k = 200 W/m.K

Lc = 50 mm

c

mhP

kA

2 c fL t

c fL t

2

0.102m10.2

200W/m K 0.00005mc

Pm

kA

h hh

Lc = 50 mm

Aluminum fined sink: k = 200 W/m.K

2 (0.05m 0.001m) 0.102m P

cA 20.05m 0.001m =0.00005m

2

1

0.102m 200W/m K 0.00005m tanh( 10.2 0.008 m)h h

For N fins,

Aluminum fined sink: k = 200 W/m.K

N, h are also needed to evaluate Rt,f

,

1

tanht f

c f

RPkA Lh m

,, ( )

t ft f N

R

NR

1

0.00102 tanh(0.008 10.2 )hN h

11 1equiv , , ( )t b t f NR R R

The equivalent fin resistance, Reqiuv

tot , ,ba use eq iv

11.25 Wc c

t c t

T T T Tq

R R R R

56.4 CcT

25 CT

equiv , ,basec

t c t

T TR R R

q

56.4 C 25 C(0.2 0.02)K/W 2.59K/W

11.25W

The total thermal resistance, Rtot

tot , ,base equivt c tR R R R

1 1, , ( )

12.59K/W

t b t f NR R

,

1,

0.005 0.0005t b NR

h

, ( )

1

0.00102 tanh(0.008 10.2 )t f N

hR

N h

air1.78 f fk Lh a L a

2 0.05m 0.001m 0.1 0.001N N

N N

a

2 c fW N

N

t

air1.78 f fk Lh a L a

Properties: Table A.4. air( 300K)T

air 0.0263W/m Kk

0.1 0.001 0.1 0.0011.78 0.0263 0.08 0.08

N N

Nh

N

1 1, , ( )

12.59K/W

t b t f NR R

,

1

0.005 0.0005t b h NR

, ( )

1

0.00102 tanh(0.008 10.2 )t f N

hR

N h

For N =

22, 2 -10.0035 m, 19.1W/m K, 13.9ma h m

, ( ) ,2.94K/W, 13.5K/Wt f N t bR R

equiv tot2.41K/W, 2.61K/WR R

54.4 CcT

For N =

20,

For N =

24,

58.9 CcT

50.7 CcT

11 on top and 11 on the bottom,

net 9.0W 4.7W 4.3WfP P P

( 56.4 C)cT