EXTENDED SURFACES - FINSThe Function of Fins• Increase heat transfer rate for a fixed surface

temperature, or

• Lower surface temperature for a fixed heat transfer rate

Newton's law of cooling

(3.63))T(T aSQ sss ∞−=&

Rewrite (3.63)

s

ss aS

QTT

&+=

∞

sQ&• “Extending” Ss for a fixed Tsincreases

• “Extending” surface at fixed lowers sQ& sTBased on presentations by professor Latif M. Jiji, The City College of New York

• A surface is “extended” by adding fins • Examples of fins:

• Thin rods on the condenser in back of refrigerator.• Honeycomb surface of a car radiator.• Corrugated surface of a motorcycle engine.• Coolers of PC boards.

Types of Fins

finstraight areaconstant (a) finpin(c)

finstraight areavariable(b) finannular (d)

3.12Fig.

Fin terminology and types

• Fin base (pata žebra)• Fin tip (konec žebra)• Straight fin: Fig. 3.12 (a) and (b). • Variable cross-sectional area fin: Fig. 3.12 (b), (c) and

(d). • Spine or a pin fin: Fig. 3.12 (c).• Annular or cylindrical: Fig. 3.12 (d).

Heat Transfer and Temperature Distribution in Fins• Heat flow through a fin is

axial and lateral (2-D)x

∞Th,

δT

∞Tr

3.13 Fig.∞

Tα,

• Note temperature profile, Fig. 3.13

• Temperature distribution is 2-D

The Fin ApproximationNeglect temperature variation in the lateral direction r and assume uniform temperature at any section:

T ≈ T(x)

• Criterion for justifying this approximation:(3.64)Biot number = Bi = αδ /λ << 1

Bi < 0.1

x

∞Th,

δT

∞Tr

3.13 Fig.∞

Tα,

• Meaning of the Biot number:

resistance externalresistance internal

1/αδ/λBi ==

The Fin Heat Equation• Objective: Determine the heat transfer rate from a fin.

Need the temperature distribution• Procedure: Formulate the fin heat equation:

Conservation of energy for element dx.

• Select an origin and a coordinate axis x

• Assume Bi < 0.1: T ≈ T(x)

• Conservation of energy for the element for steady state and no energy generation:

(3.65)outin EE && =

Energy in by conduction = xQ&

Energy out by conduction = /dx)dxQ(dQ xx&& +

Energy out by convection = convQd &

Neglect radiation:

(a) and (b) into (3.65)

0Qddxxd

Qdconv

x =+ &&

(c)

Fourier's law and Newton’s law:

dxdTAλQx −=& (d)

(a)xQ&=

(b)convx

xout QddxdxQd

QE &&

& ++=

)(b

CsdA

cdq

xq dxdxdqq x

x +

dS=P.dx

convQd &

P

(d) and (e) into (c)

A = cross-sectional conduction area

= surface area of element through which heat isdSconvected

0Pdx)Tα(TdxdxdTλA

dxd

=−−⎥⎦⎤

⎢⎣⎡

∞ (f)

Assume: constant λ

(3.66)0)T(TλAαP

dxTd2

2=−−

∞

(e))P.dxT(TαQd conv ∞−=&

α is a surface average heat transfer coefficient

• Equation (3.66) is the heat equation for fins

0)T(TλAαP

dx

)T-(Td2

2

=−−∞

∞ (3.66)

∞−TTOr assuming constant and replacing T with

∞T

Boundary Conditions

Two B.C. are needed

Determination of Fin Heat Transfer Rate oQ&

• Conservation of energy applied to a fin at steady state:

(1) Conduction at the base: Fourier's law

(3.68)( )

0xo dxTTdλAQ =∞−

−=&

base theat conduction =fqsurface theat convection =

oQ&

• Two methods to determine oQ&

(2) Convection at the fin surface: Newton's law

(3.69)∫ −= ∞S

o P.dx]Tα[T(x)Q&

• Modify (3.69) to include heat exchange by radiation

• Introducing a new parameter fin parameter m L

λAPαmL =

Rewrite (3.66)

0)T(Tmdx

)T-(Td 22

2

=−−∞

∞ (3.70)

` • Equation (3.73) is valid for:(1) Steady state(2) Constant λ(3) No energy generation(4) No radiation(5) Bi << 1

(6) Constant fin area (7) Constant ambient temperature ∞T

Solution

Solution to (3.70) is

(3.71) mx2

mx1 eCeCTT(x) −

∞+=−

C1 and C are integration constants. They depend on:

0)T(Tmdx

)T-(Td 22

2

=−−∞

∞ (3.70)

• Location of the origin` • Direction of coordinate axis x`

• Two boundary conditions

Special CasesConsider 3 cases of constant area fins(i) Specified temperature at base, semi-infinite fin(ii) Finite fin, specified temperature at the base,

insulated tip(iii)Finite fin, specified temperature at the base, heat

transfer at the tip

• Fin equation: (3.70)

• Temperature solution: (3.71)

• Objective: To determine:(1) The temperature distribution in the fin T ≈ T(x) (2) The heat transfer rate (tepelný tok ) oQ&

0)T(Tmdx

)T-(Td 22

2

=−−∞

∞

mx2

mx1 eCeCTT(x) −

∞+=−

• The base is at temperature oT• Ambient temperature is ∞T• Fin equation: (3.70)• Solution: (3.71)

• B.C. are:

Case (i): Semi-infinite fin with specified temperature at the base

(3.71) mx2

mx1 eCeCTT(x) −

∞+=−

0 xC∞Th,

∞Th,oT cA

3.16 Fig.

∞Tα,

∞Tα,

P

A

T(0) = ,To (a) ∞∞−=− TTTT(0) 0

(b) T(∞) = ,T∞ 0T)T( =−∞

∞

B.C. (b)

∴ C1 = 0 0,CC0 21 ⋅+∞⋅=

B.C. (a)

mx)exp(TTTT(x)

Θo

−=−

−=

∞

∞ (3.72)

∞−= TTC 02

mx))exp(T(TTT(x) o −−=−∞∞

• Fin heat transfer oQ& :

(3.73)

( ) )TλAm(Tdx

TTdλAQ 00xo ∞=∞ −=

−−=&

( )∞−= TTPαλAQ 0o&

Obviously, no dependence on the fin length

( )∞−= TTPαλA

Q0

o&

Case (ii): Finite length fin with specified temperature at the base and insulated tip

Same as Case (i) except the tip is insulated.

mx2

mx1 eCeCTT(x) −

∞+=−General solution:

( ) 0dx

TTdLx =

−=

∞

0=∂∂

−==

=Lx

Lx xTλSQ& cA

Cth0

oT

∞Th,x

∞Th,

3.17 Fig.

∞α,T

∞α,T

P

Tip B.C.

About hyperbolic functions:

xx

xxxxxx

eeee

coshxsinhxtghx ;

2eecoshx ;

2eesinhx −

−−−

+−

==+

=−

=

• Two B.C. - fin base (specified temperature) and tip zero heat transfer rate give C1 and C2,

mLxLm

TTTxTΘ

o cosh )-(cosh )(

=−−

=∞

∞Solution:

mlmL)TλAm(TQ 0o cosh

sinh∞−=&

Similarly heat transfer rate(tepelný tok):

( ))(tgh

0mL

TTPAQo =

− ∞αλ&

( )0xo dx

TTdλAQ =∞−

−=&

Tip temperature (x=L):mLTT

TTΘ LL cosh

10

=−−

≡∞

∞

Compare with the semi-infinite fin: ( )∞−= TTPαλA

Q0

o&

)(0,014. 0 ∞∞ −=− TTTTL

014,00

=−−

≡∞

∞TTTTΘ L

L

Case (iii): Finite length fin with specified temperature at the base and heattransfer at the tip

cA

Cth0

oT

∞Th,x

∞Th,

3.17 Fig.

∞α,T

∞α,T

P

∞α,T

[ ] ( ) [ ]( ) mLmmL

xLmmxLmTT

TxTΘL

Lsinh/cosh

)(sinh/)(cosh)(0 λα

λα+

−+−=

−−

≡∞

∞

Corrected Length Lc

• Fins with insulated tips have simpler solutions than fins with convection at the tip

• Simplified model: Assume insulated tip and compensate by increasing the length by ΔLc

• The corrected length Lc is cc LLL Δ+=

• The correction increment ΔLc = t/2 fin half width

0,0625λαt

≤• Error negligible if

Fin Effectiveness (efektivnost) εfFin Efficiency (účinnost) ηf

Fin performance is described by two parameters:

Fin Effectiveness εfenhancement due to fin addition.

: Measures heat transfer

Defined as

A

Ratio of heat transfered by the fin and heat transferedfrom the original area A of the fin ( )∞−= TTαAQ obase

( ) base

o

o

of Q

QTTαA

Qε&

&&=

−=

∞mlmL)TλAm(TQ 0o cosh

sinh∞−=&

εf must be >2, otherwise don’t use the fin.

( )( ) αA

λPTTαA

TTλAαPεo

of =

−−

=∞

∞

For a semi-infinite fin:

How to increase fin effectivness:• high conductivity material λ• high perimeter to area ratio P/A – thin fin and close

each to other• use fin where heat transfer coefficient α is low

fS = total fin surface area

Introduce the above into (3.89)

Fin Efficiency fηwith the maximum heat that the fin can transfer. Defined as

: Compares heat transfer from a fin

(3.89) o,max

of Q

Qη&

&=

the base temperature= heat transfer from fin if its entire surface is ato,maxQ&

( )∞−= TTαSQ ofo,max&

(3.90) ( )∞−

=TTαS

Qηof

of

&

Total Fin Efficiency (účinnost) ηtot

( )∞−==

TTαSQ

QQη

oTOT

TOT

TOT,max

TOTTOT

&

&

&

TOTQ& Total heat transfer rate from the total finned surface of an actual finned surface

When the entire finned surface is at the base temperature

TOT,maxQ&

TOTS Total finned surface (including surface between individual fins) ofTOT SSS +=

fS

oS

( ) ( )∞∞ −+−= TTSTTSQ oooffTOT αηα& (3.91)

( ) ( )∞−⎥⎦

⎤⎢⎣

⎡−−= TT

SS

SQ ofTOT

fTOTTOT ηα 11&

After some rearrangement of (3.91)

( )fTOT

fTOT S

Sηη −−= 11

Taking the definition of the total efficiency

( )( )∞

∞−=⇒

−= TTηαSQ

TTαSQη oTOTTOTTOT

oTOT

TOTTOT &

&

we come to the definition of a total efficiency

Practical procedure how to calculate heat transfer rate from a finned surface

fS

oS

ofTOT SSS +=

1. Specify efficiency of an individual fin ηf using graphs

2. Calculate total surface of fins and “free” space between fins

3. Calculate total efficiencyTOη ( )f

TOT

fTOT S

Sηη −−= 11

4. Calculate maximum heat transfer rate )T(TαSQ oTOTTOT,max ∞−=&

5. Calculate actual heat transfer rate

TOT,maxTOTTOT QηQ && =

Fin efficiency for three types of straight fins

Graphs of fin efficiency

Fin efficiency for annular fins

Graphs of fin efficiency