Heat Transfer from Fins(Chapter 3)Zan Wu zan.wu@energy.lth.se Room: 5123

Fins

Fins/Extended surfaces

Why not called as convectors?

Radiators

Fins

Fan cooling is not sufficient for advanced microprocessors

Microfins

Microfin copper tube

Carbon nanotube microfinson a chip surface

Fin analysis

Two basic questionsØ What is the rate of heat dissipated by the fin?Ø What is the variation in the fin temperature from

the fin base to the fin tip?

Rectangular fin

2

2 0 (3 31)d Cdx Aϑ α

ϑλ

− = −

x dx

L

t1

Q1

.

tfb Z

Energy balance on the element from x to x + dx

A: area of a cross section normal to xC: perimeter of this section

)tt( f−=ϑSteady state1D

Cont’d

Boundary conditions:

Assume a long and thin fin, the heat transferred at the fin tip is negligible

)313(0AC

dxd2

2−=ϑ

λα

−ϑ

b2

bZZ2

ACm2

λα

=λ⋅α

≈λα

=

)tt(dxdt:Lx fLx −αʹ=λ−= =

0dxdt

Lx=⎟

⎠⎞

⎜⎝⎛

=

f111 tttt:0x −=ϑ=ϑ⇒==

x dx

L

t1

Q1

.

tfb Z

Rectangular fin

Solution:

cosh2

sinh2

mx mx

mx mx

e emx

e emx

−

−

+=

−=

1 2

3 4cosh sinh

mx mxC e C e

C mx C mx

ϑ −= + =

= +

Hyperbolic functions

At x = L ϑ = ϑ2

1 1

cosh ( ) (3 38)cosh

f

f

t t m L xt t mL

ϑϑ

− −= = −

−

2

1

1coshmL

ϑϑ

=

heat transfer from the fin?Q&

1 10

sinh ( )coshx

d mLQ A A mdx mL

CmA

ϑλ λ ϑ

αλ

=

⎛ ⎞= − = − ⋅ ⋅ −⎜ ⎟⎝ ⎠

= ⇒

&

1 1 1tanh 2 tanh (3 40)Q C A mL b Z mLα λ ϑ α λ ϑ= ⋅ = −g

Rectangular fin

Rectangular fin

α = 25 W/m2K, b = 2 cm, L = 10 cm

Rectangular fin

If the condition below is used, i.e., to consider heat loss from the fin tip

one has

and

and

LxLxdx

d=

=ϑαʹ=⎟

⎠⎞

⎜⎝⎛ ϑλ−

)413(mLsinh

mmLcosh

)xL(msinhm

)xL(mcosh

1−

λαʹ

+

−λαʹ

+−=

ϑϑ

)423(mLsinh

mmLcosh

11

2 −

λαʹ

+=

ϑϑ

)433(mLtanh

m1

mLtanhmAmQ 11 −

λαʹ

+

+λαʹ

ϑλ=!

Fins on Stegosaurus

Those plates absorb radiation from the sun or cool the blood?

Practical considerations

e.g.,Ø How to choose a fin material?ØHow to optimize fins?

Criterion for benefit

Fig. 3-13. Arrangement of rectangular fins

1

preferable if

0dQdL

>&

1 ( )Q function L=&L

.

Z

t1

b

1Q!

Fin effectiveness, fin efficiency

1

1

from the fin from the base area without the fin

QQ

η =&

&

1

1

from the finfrom a similar fin but with λ

QQ

ϕ == ∞

&&

Criterion: maximum heat flow at a given mass

M = ρ b L Z = ρ Z A1 A1 = b L, Z, ρ are given.

Find maximum for constant A1 = b⋅L.

C ≈ 2Z , A = b⋅Z

mLtanhACQ 11 ϑλα=!

b2

ACm2

λα

=λα

=

1Q!

!⎟⎠

⎞⎜⎝

⎛⋅

λα

⋅ϑ⋅αλ=bA

b2tanhZb2Q 1

11!

L

Zb

Optimal rectangular fin

Cont’d

Condition

1 0 gives optimum

after some algebra one obtains

21.419 (3 55)/ 2

dQdb

Lb b

λα

=

= −

&

M

Fin material selection

After some algebra one finds:

)523(bA

b2tanhZb2

mLtanhAmQ

11

11

−⎟⎠

⎞⎜⎝

⎛λα

ϑαλ=

=ϑλ=!

12 1.419Aub bαλ

= ⋅ =

1from the condition / 0dQ db =&

)a613(41

utanhu

Z1QA 233

3

1

11 −

λα⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛

ϑ=!

For an optimized rectangular fin

Cont’d

M = ρ b L Z = ρ Z A1 =

ρ/λ is the material parameter see Table 3-1.

Aluminum instead of Copper. ρ/λ Aluminum: 11.8; Copper: 23.0

Why not Magnesium? ρ/λ Magnesium: 10.2

λρ⋅

α⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛

ϑ= 232

3

1

1

41

utanhu

Z1Q!

Straight triangular fin

ϑ = t − tf

Heat balance ⇒

Solution:

K0 → ∞ as x → 0 ⇒ B = 0 because ϑ is finite for x = 0

x = L and ϑ = ϑ1 ⇒

)623(0bL2

x1

dxdx1

dxd2

2−=ϑ

λα

−ϑ

+ϑ

bL2

λα

=β

( ) )x2(BKx2AI 00 β+β=ϑ

( )L2AI01 β=ϑ

LxbZA ⋅=

δ

L

dx

x

b t1

tf1Q!

Bessel differential equation

I0 and K0 are the modified Bessel functions of order zero

Triangular fin

⇒

! )L2(IA

0

1β

ϑ=

)653()L2(I)x2(I

0

0

1−

β

β=

ϑϑ

Lx1 dx

dtAQ=⎭

⎬⎫

⎩⎨⎧ λ−=!!

)663()L2(I)L2(Ib2ZQ

0

111 −

ββ

αλϑ=⇒ !

Table 3.2 for numerical values of Bessel functions

Recap)383(

mLcosh)xL(mcosh

ft1tftt

1−

−=

−−

=ϑϑ

b22mλα

=

)403(mLtanh1Zb21Q −ϑαλ=!

mLtanhb2αλ

=η

mLmLtanh

=ϕ

21.419 (3 55)/ 2Lb b

λα

= −

)653()L2(0I)x2(0I

1−

ββ

=ϑϑ

bL2

λα

=β

11 1

0

(2 )2 (3-66)(2 )I LQ b ZI L

βαλ ϑ

β= ⋅ ⋅&

)L2(0I)L2(1I

b2

ββ

⋅αλ

=η

L)L2(0I/)L2(1I

βββ

=ϕ

21.309 (3 67)/ 2Lb b

λα

= −

Optimal fin: Maximum heat transfer at fixed fin mass

mL = 1.419 mL = 1.309

24

Circular or annular fins

Heat conducting area

A = 2πr b

Convective perimeter

C = 2 ⋅ 2πr = 4πr

r1 r2

b

How to calculate A and C for circular rod fins?

Fin efficiency for circular fins

How to use the fin efficiency in engineering calculations

s

flänsarareaoflänsad

QQQ

QQQ

finareaunfinned!!!

!!!

+=

+=

{( )

( )

fins b f

b b f fins

A t t

Q A t t Qλ

α

α ϕ=∞

−

= − + ⋅& &!

{ }( ) (3 71)b f b finsQ t t A Aα ϕ= − + ⋅ −&

Graphene

Anisotropic