# Document ```EXTENDED SURFACES - FINS
The 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
Q&amp; =aS (T − T )
(3.63)
s
s
s
∞
• “Extending” S increases Q&amp; for a fixed T
s
Rewrite (3.63)
s
s
T =T +
s
∞
Q&amp;
s
aS
s
• “Extending” surface at fixed Q&amp; s lowers Ts
Based 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
(a) constant area (b) variable area
straight fin
straight fin
Fig. 3.12
(c) pin fin
(d) annular fin
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
r
Distribution in Fins
• Heat flow through a fin is
axial and lateral (2-D)
δ
T∞
x
T
h,TT∞
α,
∞
Fig. 3.13
• Temperature distribution is
2-D
r
δ
• Note temperature profile,
Fig. 3.13
T∞
x
T
h,TT∞
α,
∞
Fig. 3.13
The Fin Approximation
Neglect temperature variation in the lateral direction r
and assume uniform temperature at any section:
T ≈ T(x)
• Criterion for justifying this approximation:
Biot number = Bi = αδ /λ &lt;&lt; 1
Bi &lt; 0.1
(3.64)
• Meaning of the Biot number:
δ/λ internal resistance
Bi =
=
1/α external resistance
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 &lt; 0.1: T ≈ T(x)
• Conservation of energy for the element for steady state
and no energy generation:
E&amp; in = E&amp; out
Energy in by conduction = Q&amp; x
Energy out by conduction = Q&amp; x + (dQ&amp; x /dx)dx
(3.65)
Energy out by convection = dQ&amp;
conv
= Q&amp;
(a)
x
Eout = Q&amp; x +
dQ&amp; x
dx +dQ&amp; conv
dx
(b)
(a) and (b) into (3.65)
d Q&amp;
x
dx
dx + dQ&amp;
conv
=0
dS=P.dx
dA
s
C
P
qx
&amp;c
dq
dQ
conv
qx +
dqx
dx
dx
(b)
(c)
Fourier's law and Newton’s law:
dT
&amp;
Q = − λA
x
dx
(d)
dQ&amp;
conv
= α (T − T )P.dx
∞
(e)
α is a surface average heat transfer coefficient
A = cross-sectional conduction area
dS = surface area of element through which heat is
convected
(d) and (e) into (c)
d ⎡ dT ⎤
λA ⎥dx − α(T − T∞ )Pdx = 0
⎢
dx ⎣ dx ⎦
(f)
Assume: constant λ
d 2T αP
− (T − T∞ ) = 0
2 λA
dx
(3.66)
Or assuming T∞ constant and replacing T with T − T∞
d 2 (T - T ) αP
∞ −
(T − T∞ ) = 0
2
λA
dx
(3.66)
• Equation (3.66) is the heat equation for fins
Boundary Conditions
Two B.C. are needed
Determination of Fin Heat Transfer Rate Q&amp; o
• Conservation of energy applied to a fin at steady state:
qQ&amp; fo = conduction at the base
= convection at the surface
• Two methods to determine Q&amp; o
(1) Conduction at the base: Fourier's law
d (T − T∞ )
&amp;
Qo = − λA
dx
x =0
(3.68)
(2) Convection at the fin surface: Newton's law
Q&amp; o = ∫ α[T(x) − T∞ ] P.dx
S
(3.69)
• Modify (3.69) to include heat exchange by radiation
• Introducing a new parameter
fin parameter m
αP
mL =
L
λA
Rewrite (3.66)
d 2 (T - T ) 2
∞ −m (T − T ) = 0
∞
dx 2
(3.70)
`
• Equation (3.73) is valid for:
(2) Constant λ
(3) No energy generation
(5) Bi &lt;&lt; 1
(6) Constant fin area
(7) Constant ambient temperature T∞
Solution
d 2 (T - T ) 2
∞ −m (T − T ) = 0
∞
dx 2
(3.70)
Solution to (3.70) is
T(x) − T∞ = C1e mx + C 2 e − mx
(3.71)
C1 and C are integration constants. They depend on:
`
• Location of the origin
• Direction of coordinate axis x
`
• Two boundary conditions
Special Cases
Consider 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
•
•
d 2 (T - T ) 2
∞ −m (T − T ) = 0
Fin equation: (3.70)
∞
dx 2
Temperature solution: (3.71)
• Objective: To determine:
T(x) − T∞ = C1e mx + C 2 e − mx
(1) The temperature distribution in the fin T ≈ T(x)
(2) The heat transfer rate (tepeln&yacute; tokQ&amp; o )
Case (i): Semi-infinite fin with specified
temperature at the base
α,h,
TT∞
0
To
• The base is at temperature To
• Ambient temperature is T∞
• Fin equation: (3.70)
• Solution: (3.71)
x
∞
α,
h, T∞
∞
Ac
A
Fig. 3.16
T(x) − T∞ = C1e mx + C 2 e − mx
• B.C. are:
CP
(3.71)
T(0) =To , T(0) − T∞ = T0 − T∞
(a)
T(∞) =T∞ , T( ∞ ) − T∞ = 0
(b)
B.C. (b)
0 = C1 ⋅ ∞ + C 2 ⋅ 0,
B.C. (a)
∴ C1 = 0
C 2 = T0 − T∞
T(x) − T∞ = (To − T∞ )exp( −mx)
Θ=
T(x) − T
T −T
o
∞
∞
= exp( −mx)
(3.72)
• Fin heat transfer Q&amp; o :
d (T − T∞ )
&amp;
Qo = − λA
dx
x =0
= λAm(T0 − T∞ )
Q&amp; o = λAα P (T0 − T∞ )
(3.73)
Q&amp; o
= (T0 − T∞ )
λAα P
Obviously,
no dependence on the fin length
Case (ii): Finite length fin with specified
temperature at the base and insulated tip
Same as Case (i) except
the tip is insulated.
∂T
&amp;
Q x = L = − λS
∂x
=0
x=L
0
To
x
α,T
h, T∞
∞
x=L
h, T∞∞
α,T
Fig. 3.17
=0
General solution:
C
ht
Tip B.C.
d (T − T∞ )
dx
P
T(x) − T∞ = C1e mx + C 2 e − mx
Ac
• Two B.C. - fin base (specified temperature) and tip
zero heat transfer rate give C1 and C2 ,
sinhx e x − e − x
ex + e−x
ex − e−x
= x
; tghx =
; coshx =
sinhx =
coshx e + e − x
2
2
Solution:
T ( x ) − T∞ cosh m( L - x )
=
Θ=
cosh mL
To − T∞
Tip temperature (x=L):
Similarly heat transfer rate
(tepeln&yacute; tok):
TL − T∞
1
=
ΘL ≡
T0 − T∞ cosh mL
d (T − T∞ )
&amp;
Qo = − λA
dx
sinh mL
&amp;
Qo = λAm(T0 − T∞ )
cosh ml
Q&amp; o
= tgh (mL )
λAα P (T0 − T∞ )
x =0
Compare with the semi-infinite fin:
Q&amp; o
= (T0 − T∞ )
λAα P
ΘL ≡
TL − T∞
= 0,014
T0 − T∞
TL − T∞ = 0,014.(T0 − T∞ )
Case (iii): Finite length fin with specified
temperature at the base and heat
transfer at the tip
0
To
x
α,T
h, T∞
∞
α,T∞
P
ht
h, T∞∞
α,T
Ac
Fig. 3.17
T ( x ) − T∞ cosh[m ( L − x )] + (α L / mλ )sinh[m ( L − x )]
=
Θ≡
T0 − T∞
cosh mL + (α L / mλ )sinh mL
C
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
Lc = L + ΔLc
• The correction increment ΔLc = t/2 fin half width
• Error negligible if
αt
≤ 0,0625
λ
Fin Effectiveness (efektivnost) εf
Fin Efficiency (&uacute;činnost) ηf
Fin performance is described by two parameters:
Fin Effectiveness εf : Measures heat transfer
Q&amp; o
Q&amp; o
&amp; o = λAm(T0 − T∞ ) sinh mL
Q
=
Defined as ε f =
cosh ml
&amp;
αA(To − T∞ ) Qbase
Ratio of heat transfered by the fin and heat transfered
from the original area A of the fin Qbase = αA(To − T∞ )
A
For a semi-infinite fin:
εf =
λAαP (To − T∞ )
λP
=
αA(To − T∞ )
αA
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
εf must be &gt;2, otherwise don’t use the fin.
Fin Efficiency η f : Compares heat transfer from a fin
with the maximum heat that the fin
can transfer. Defined as
Q&amp; o
(3.89)
ηf =
Q&amp; o,max
Q&amp; o,max = heat transfer from fin if its entire surface is at
the base temperature
Q&amp; o,max = αS f (To − T∞ )
S f = total fin surface area
Introduce the above into (3.89)
Q&amp; o
ηf =
αS f (To − T∞ )
(3.90)
Total Fin Efficiency (&uacute;činnost) ηtot
ηTOT =
Q&amp; TOT
Q&amp; TOT,max
Q&amp; TOT
=
αSTOT (To − T∞ )
Sf
So
Total finned surface (including surface between
individual fins) STOT = S f + S o
STOT
Total heat transfer rate from the total finned
surface of an actual finned surface
Q&amp; TOT
Q&amp; TOT = αS f η f (To − T∞ ) + αSo (To − T∞ )
Q&amp; TOT,max
When the entire finned surface is at the
base temperature
(3.91)
After some rearrangement of (3.91)
Sf
⎡
⎤
&amp;
(
QTOT = αSTOT ⎢1 −
1 − η f )⎥(To − T∞ )
⎣ STOT
⎦
Taking the definition of the total efficiency
Q&amp; TOT
ηTOT =
⇒ Q&amp; TOT = αSTOT ηTOT (To − T∞ )
αSTOT (To − T∞ )
we come to the definition of a total efficiency
ηTOT = 1 −
Sf
STOT
(1 − η f )
Practical procedure how to calculate heat transfer rate
from a finned surface
1.
Specify efficiency of an individual fin ηf using graphs
2.
STOT
Calculate total surface of fins
and= S f + So
“free” space between fins
3.
η
TO
4.
Calculate total
S f efficiency
(
ηTOT = 1 −
1 −η f )
STOT
Calculate maximum heat transfer rate
Q&amp; TOT,max = αS TOT (To − T∞ )
5.
Calculate actual heat transfer rate
Q&amp; TOT = ηTOT Q&amp; TOT,max
So
Sf
Graphs of fin efficiency
Fin efficiency for three types of straight fins
Graphs of fin efficiency
Fin efficiency for annular fins
```