This document discusses heat transfer through extended surfaces called fins. It begins by introducing fins and explaining that they are used to increase the surface area for heat transfer. It then derives the governing differential equation for one-dimensional, steady-state heat conduction through a fin. The document explores several boundary conditions and derives equations for the temperature distribution, heat transfer rate, and efficiency of fins with different boundary conditions, including infinitely long fins, fins with an insulated tip, and fins with a prescribed tip temperature. It concludes by discussing fin effectiveness and the factors that influence it.

1. Heat Transfer through Extended Surfaces
Parag Chaware
Department of Mechanical Engineering
Cummins College of Engineering, Pune
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2. Introduction
Figure: Fins
The rate of heat transfer from a
surface at a temperature Ts to
the surrounding medium at T∞
is given by Newton‘s law of
cooling.
Increasing h may require the
installation of a pump or fan.
The alternative is to increase
the surface area by attaching
extended surfaces called fins
made of highly conductive
materials such as aluminum.
e.g. Radiator, Cylinder head of
IC engine
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3. Heat transfer through fin I
(Heat Conducted into element) = (Heat Conducted out of the element)
+ (Rate of heat convection from the element)
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4. Heat transfer through fin II
Q̇∆x = Q̇x+∆x + Q̇conv (1)
where,
Q̇conv = h(p∆x)(T − T∞) (2)
Dividing by ∆x we get,
Q̇x+∆x − Q̇∆x
∆x
+ hp(T − T∞) = 0 (3)
taking limit ∆x → 0
d(Q̇cond )
dx
+ hp(T − T∞) = 0 (4)
From Fourier‘s law
Q̇cond = −kAc
dT
dx
(5)
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5. Heat transfer through fin III
d
dx
−kAc
dT
dx
+ hp(T − T∞) = 0 (6)
substituting eq. (5) in eq. (6) and assuming (T − T∞) = θ
d2T
dx2
− m2
θ = 0 (7)
m =
r
hp
kAc
(8)
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6. Heat transfer through fin IV
Equation 7 is a linear, homogeneous, second-order differential equation
with constant coefficients. Therefore, the general solution of the
differential equation is
θ(x) = C1emx
+ C2e−mx
(9)
where C1 and C2 are arbitrary constants whose values are to be determined
from the boundary conditions at the base and at the tip of the fin.
Figure: Boundary Conditions for fin
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7. Heat transfer through fin V
θ(x) = C1emx
+ C2e−mx
(10)
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8. Infinitely Long fin I
For a sufficiently long fin of uniform cross section (Ac = constant), the
temperature of the fin at the fin tip will approach the environment
temperature T∞.
So BC‘s are θ(L) = T − T∞ = 0 as L → ∞
So BC‘s are;
At x = 0 θ = θ0
At x = ∞ θ = 0
First BC gives C1 + C2 = θ0
Since C2e−mx is zero
C1emx
= C2e−mx
(11)
Possible when C1 → 0
θ(x) = C2e−mx
(12)
Therefore
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9. Infinitely Long fin II
C2 = θ0
applying BC’s at base i.e. θ(0) = T0 − T∞
T(x) = T∞ + (T0 − T∞)e−mx
(13)
and
Qfin =
p
hpkAc(T0 − T∞) (14)
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10. Insulated Tip I
Heat transfer from the fin is proportional to its surface area, and the
surface area of the fin tip is usually a negligible fraction of the total fin
area. Therefore;
BC1
dθ
dx

14. x=L
= 0 (15)
BC2
θ(0) = T0 − T∞ (16)
The temperature distribution is
θ
θ0
=
coshm(L − x)
coshmL
(17)
The heat transfer form fin is
Qfin =
p
hPkAc(T0 − T∞)tanhmL (18)
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15. Prescribed temperature
This is a condition when the temperature at the tip is known
θ
θ0
=
(θL/θ0)sinhmx + sinhm(L − x)
sinhmL
(19)
Qfin =
p
hPkAc
coshmL − (θL/θ0)
sinhmL
(20)
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16. Fin Efficiency I
Figure: Fin temperature distribution
In the limiting case of zero
thermal resistance or infinite
thermal conductivity (k∞), the
temperature of the fin will be
uniform at the base value of T0
In reality, however, the
temperature of the fin will drop
along the fin, and thus the heat
transfer from the fin will be less
because of the decreasing
temperature difference
(T(x) − T0) toward the fin tip,
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17. Fin Efficiency II
ηf =
Actual heat transfer rate from the fin
Ideal heat transfer rate from the fin if the entire
fin were at base temperature
(21)
ηf long =
√
hpkAc(T0 − T∞)
hAf (T0 − T∞)
=
1
mL
(22)
(23)
(24)
ηf Insulated Tip =
√
hpkAc(T0 − T∞)tanhmL
hAf (T0 − T∞)
=
tanhmL
mL
(25)
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18. Fin Effectiveness I
The performance of fins expressed in terms of the fin effectiveness (εf )
εf =
˙
Qfin
˙
QWithout fin
(26)
εf =
Heat transfer rate from
the fin of base area Ab
Heat transfer rate from
the surface of area Ab
(27)
εLong fin =
Qfin
QNo Fin
=
√
hpkAc(T0 − T∞)
hAc(T0 − T∞)
=
r
kp
hAc
(28)
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19. Fin Effectiveness II
k should be as high as possible, (copper, aluminum, iron). Aluminum
is preferred: low cost and weight, resistance to corrosion.
p/Ac should be as high as possible. (Thin plate fins and slender pin
fins)
Most effective in applications where h is low. (Use of fins justified if
when the medium is gas and heat transfer is by natural convection).
Therefor the fins are provided on gas side rather than liquid side.
ε = 0 Fin is not contributing the heat transfer
ε 0 Fin act as insulation (if low k material is used)
ε 0 Enhancing heat transfer (use of fins justified if fin 2)
Use of Fin
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20. Figure: Temperature distribution and heat loss for fins of uniform cross sectiona
a
Fundamentals of Heat Transfer by Incropera Dewitt
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