Heat transfer from extended surfaces (or fins)

Lectures on Heat Transfer --
HEAT TRANSFER FROM EXTENDED
SURFACES (FINS)
by
Dr. M. ThirumaleshwarDr. M. Thirumaleshwar
formerly:
Professor, Dept. of Mechanical Engineering,
St. Joseph Engg. College, Vamanjoor,
Mangalore
India

Preface
• This file contains slides on Heat Transfer
from Extended Surfaces (FINS).
• The slides were prepared while teaching
Heat Transfer course to the M.Tech.
students in Mechanical Engineering Dept.
of St. Joseph Engineering College,
Vamanjoor, Mangalore, India, during Sept.
– Dec. 2010.
Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful
to teachers, students, researchers and
professionals working in this field.
• For students, it should be particularly
useful to study, quickly review the subject,useful to study, quickly review the subject,
and to prepare for the examinations.
•
Aug. 2016 3MT/SJEC/M.Tech.

References
• 1. M. Thirumaleshwar: Fundamentals of Heat &
Mass Transfer, Pearson Edu., 2006
• https://books.google.co.in/books?id=b2238B-
AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false
• 2. Cengel Y. A. Heat Transfer: A Practical
Approach, 2nd Ed. McGraw Hill Co., 2003
Aug. 2016 MT/SJEC/M.Tech. 4
Approach, 2nd Ed. McGraw Hill Co., 2003
• 3. Cengel, Y. A. and Ghajar, A. J., Heat and
Mass Transfer - Fundamentals and Applications,
5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.
• 4. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.
• 5. M. Thirumaleshwar: Software Solutions to• 5. M. Thirumaleshwar: Software Solutions to
Problems on Heat Transfer – CONDUCTION-
Part-II, Bookboon, 2013
• http://bookboon.com/en/software-solutions-problems-on-heat-
transfer-cii-ebook

Extended surfaces- Fins:
Outline:
Governing differential eqn – different boundary
conditions – temp. distribution and heat transfer
rate for: infinitely long fin, fin with insulated end,
fin losing heat from its end, and fin with specified
temperatures at its ends – performance of fins -
‘fin efficiency’ and ‘fin effectiveness’ – fins of
non-uniform cross-section- thermal resistance
and total surface efficiency of fins – estimation of
error in temperature measurement - Problems

• Fins are generally used to enhance the heat
transfer from a given surface.
• Addition of fins can increase the heat transfer
from the surface by several folds.
• Typical application areas of Fins are
HEAT TRANSFER FROM EXTENDED
SURFACES (FINS)
• Typical application areas of Fins are:
• Radiators for automobiles
• Air-cooling of cylinder heads of Internal
Combustion engines (e.g. scooters, motor
cycles, aircraft engines etc.), air compressors
etc.

• Economizers of steam power plants
• Heat exchangers of a wide variety, used
in different industries
• Cooling of electric motors, transformers
Fins – Applications (contd.)
• Cooling of electric motors, transformers
etc.
• Cooling of electronic equipments, chips,
I.C. boards etc.

Fins of uniform cross-section
(rectangular or circular) – Governing
differential equation:
Qconv
To
t Y
Ac
h, Ta
L
dx
Qx Qx+dx
w
X
X
Z
Fig. 6.2(a) Rectangular fin of uniform crosssection

• Assumptions:
• Steady state conduction,
with no heat generation
in the fin
• Thickness ‘t’ is small
• Thickness ‘t’ is small
compared to length L
and width w, i.e. one-
dimensional conduction
in the x-direction only.
• Thermal conductivity, k
of the fin material is
constant.

• Assumptions (contd.):
• Isotropic (i.e. const. k in all
directions) and homogeneous
(i.e. const. density) material.
• Uniform heat transfer coeff. h,
over the entire length of fin.
• Uniform heat transfer coeff. h,
over the entire length of fin.
• No bond resistance in the
joint between the fin and the
base wall, and
• Negligible radiation effect.

• Consider an elemental section
of thickness dx at a distance x
from the base as shown. Let us
write an energy balance for this
element:
element:
• Energy going into the element
by conduction =
(Energy leaving the element by
conduction + Energy leaving the
surface of the element by
convection)
i.e. Qx = Qx+dx + Qconv ……(a)

• We have:

• Substituting the terms in eqn (a) and simplifying:

• Eqn. (6.1) is the governing differential equation for the fin
of uniform cross-section considered.
• Eqn. (6.1) is a second order, linear, ordinary differential
equation. Its general solution is given by calculus theory,
in two equivalent forms:
in two equivalent forms:

• To calculate the set of constants C1 and
C2, or A and B, we need two boundary
conditions:
• One of the B.C’s is that the temperature of
the fin at its base.
• i.e. B. C. (i): at x = 0, T = To
• Regarding the second boundary condition,
there are several possibilities:

• Case (i): Infinitely long fin,
• Case (ii): Fin insulated at its end (i.e.
negligible heat loss from the end of the
fin),
fin),
• Case (iii): Fin losing heat from its end by
convection, and
• Case (iv): Fin with specified temperature
at its end.

Few useful relations for hyperbolic
functions:

Case 1: Infinitely long fin:
• The governing differential
eqn., as already derived, is:
And, we shall choose for its
solution for temperature
distribution eqn. (6.2, a), i.e.
θ x( ) C1 exp m x.( ). C2 exp m x.( ). ........(6.2, a)

• C1 and C2 are obtained from the B.C’s:
• B.C. (i): at x = o, T = To
• B.C. (ii): as x , T Ta, the ambient temperature.
• We get:
at x = 0, θ x( ) T o T a θ o
at x = ∞, θ x( ) T a T a 0
at x = ∞, θ x( ) T a T a 0
Therefore:
Substituting C1 and C2 back in eqn. (6.2, a), we get:
θ x( ) θ o exp m x.( ).

• Eqn.(6.3) gives the temperature distribution in an
infinitely long fin of uniform cross-section, along the
i.e.
θ x( )
θ o
exp m x.( )
i.e.
T x( ) T a
T o T a
exp m x.( ) ..........(6.3)
infinitely long fin of uniform cross-section, along the
length.
• To determine the heat transfer rate:
Q fin k A c
. d
dx
T x( )
x 0
. k A c
. d
dx
θ x( )
x 0
. .......(c)
i.e. Q fin k A c
. d
dx
θ o e
m x.
.
x 0
.

• Substituting for m:
i.e. Q fin k A c
. m. θ o
. ........(6.4)
Q fin k A c
. h P.
k A c
.
. θ o
.
i.e. Q fin h P. k. A c
. θ o
. h P. k. A c
. T o T a
. ......(6.5)
Eqn. (6.4) or (6.5) gives the heat transfer rate through the fin.

Case 2: Fin of finite length with
insulated end:
• To determine the
temperature distribution:
• The governing
differential eqn:
To
t
Ac
h, Ta
(dT/dx)x=L = 0
Q
d
2
θ 2.
L
X
Fig. 6.4(a) Fin of finite length, end insulated
(dT/dx)x=L = 0
d
2
θ
dx
2
m
2
θ. 0 ............(6.1)
And, we shall choose for its solution for
temperature distribution, eqn. (6.2, b),
i.e.
θ x( ) A cosh m x.( ). B sinh m x.( ). ..........(6.2, b)

• Constants A and B are obtained from the B’C’s:
• B.C.(i): at x = 0, θ x( ) T o T a θ o
B.C.(ii): at x = L,
dT
dx
dθ
dx
0 since the end is insulated.
From B.C. (i) and eq. (6.2,b): A θ o
Temp. profile
T
To
From B.C. (ii) and eq. (6.2,b): dθ
dx x L
0
i.e. A m. sinh m L.( ). B m. cosh m L.( ). 0 ...using relations in Table 6.1
i.e. B θ o
sinh m L.( )
cosh m L.( )
.We get:
X
Fig. 6.4(b) Temp. profile for fin insulated at its end
Ta
TL

• Substituting for A and B in eqn. (6.2, b) and simplifying:
θ x( ) θ o cosh m x.( ). θ o
sinh m L.( )
cosh m L.( )
. sinh m x.( ).
i.e.
θ x( )
θ o
cosh m L.( ) cosh m x.( ). sinh m L.( ) sinh m x.( ).
cosh m L.( )
i.e.
θ x( ) cosh m L x( ).( )
.....(6.6)......using relation no. ( n ) from Table 6.1
i.e.
θ o cosh m L.( )
.....(6.6)......using relation no. ( n ) from Table 6.1
i.e.
T x( ) T a
T o T a
cosh m L x( ).( )
cosh m L.( )
.......(6.7)
Eqn. (6.6) or (6.7) gives the temperature distribution in
the fin with negligible heat transfer from its end.

• Temperature at the end of the fin:
• This is easily determined by putting x = L in eqn. (6.6) or (6.7):
θ L( )
θ o
1
cosh m L.( )
.....(6.6, a)......since cosh(0) = 1
i.e.
T L T a
T o T a
1
cosh m L.( )
.....(6.7, a)
and,
or, T L
T o T a
cosh m L.( )
T a ......(6.7, b)
Eqn. (6.7,b) gives the temperature at the end of a
fin (i.e. at x = L, when the end of the fin is insulated.

• To determine the heat transfer rate:
Q fin k A c
. θ o
. m sinh m L x( ).( ).
cosh m L.( ) x 0
.
i.e. Q fin k A c
. m. θ o
. tanh m L.( ). .......(6.8)
i.e. Q h P. k. A. θ. tanh m L.( ). ......(6.9)
. θ o
. tanh m L.( ). ......(6.9)
Eqn. (6.8) or (6.9) gives the heat transfer rate from the fin,
insulated at its end.
Comparing eqn. (6.8) with that obtained for heat transfer
from an infinitely long fin viz. eqn. (6.4), we see that a fin
with insulated end becomes equivalent to an infinitely long
fin when tanh(m.L) = 1.

Table 6.2
Values of tanh(X) for different values of X
X
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
tanh X( )
0
0.19738
0.37995
0.53705
0.66404
0.76159
0.83365
0.88535
0.92167
0.94681
0.96403
0.97574
tanh X( )
1
cosh X( )
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X versus tanh (X) and {1/cosh(X)}
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
4
4.2
4.4
4.6
4.8
5
0.97574
0.98367
0.98903
0.99263
0.99505
0.99668
0.99777
0.99851
0.999
0.99933
0.99955
0.9997
0.9998
0.99986
0.99991
X
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
0
0.1
tanh(X)
1/cosh(X)

• It is observed from the Table that when
(m.L) for the insulated-end fin reaches a
value of about 2.8, heat transfer rate
becomes about 99% of that obtained for
an infinitely long fin.
• And, beyond a value of (m.L) more than 5,
the fin with insulated end can be
the fin with insulated end can be
considered as infinitely long.
• Therefore, from the heat transfer point of
view, there is no great advantage in
having a fin with (m.L) greater than 2.8 or
3.

Case 3: Fin of finite length losing
heat from its end by convection:
• Here, heat conducted to the
tip of the fin must be equal to
the heat convected away from
the tip to the ambient i.e.
dT
k A c
. dT
dx x L
. h A c
. T L T a
.
i.e. k
dT
dx x L
. h θ L
.

• To determine the temperature distribution:
• The governing differential eqn., as already derived, is
given by eqn. (6.1), viz.
d
2
θ
dx
2
m
2
θ. 0 ............(6.1)
Temp. profile
X
Ta
TL
To
Convection
And, we shall choose for its solution for temperature
distribution, eqn. (6.2, b), i.e.
B.C.(i): at x = 0, θ x( ) T o T a θ o
X
Fig. 6.5(b) Temp. profile for fin losing heat at its end

• B.C. (ii): at x = L,
heat conducted to the end = heat convected from the end
i.e. k A c
. dθ x( )
dx x L
. h A c
. θ L( ). where θ(L) = TL - Ta
i.e. k
dθ x( )
dx x L
. h θ L( ). 0
x L
i.e. k A m. sinh m L.( ). B m. cosh m L.( ).( ). h A cosh m L.( ). B sinh m L.( ).( ). 0
...using relations in Table (6.1)
A θ oWe get: B
θ o sinh m L.( )
h
m k.
cosh m L.( )..
cosh m L.( )
h
m k.
sinh m L.( ).
i.e.

• Substituting for A and B in eqn. 6.2(b) and proceeding as earlier, we
get::
i.e.
θ x( )
θ o
cosh m L x( ).( )
h
m k.
sinh m L x( ).( ).
cosh m L.( )
h
m k.
sinh m L.( ).
..(6.10).....using relations (n) and (p)
from Table (6.1)
And, the heat transfer rate:
tanh m L.( )
h
i.e. Q fin k A c
. m. θ o
.
tanh m L.( )
h
m k.
1
h
m k.
tanh m L.( ).
. ......(6.11)

• In practice, even when the fin is losing heat from its tip, it
is easier to use eqn. (6.8) or (6.9) obtained for a fin with
insulated tip, but with a ‘corrected length’, Lc rather than
the actual length, L, to include the effect of convection at
the tip.
• In that case, only to evaluate Q, L is replaced by a
corrected length Lc, in eqn. (6.8) or (6.9), as follows:
For rectangular fins: L c L
t
2
where 't'is the thickness of fin
For cylindrical (round) fins: L c L
r
2
where 'r'is the radius of the cylindrical fin

Case 4: Fin of finite length with
specified temperatures at its ends:
T1
L
t
h, Ta
Q1
T2
Q2
Qconv
Temp. profile
X
Fig. 6.6. Fin of finite length, with specified temp. at two ends
and the temperature profile along the length
X
T1
T2

To determine the temperature distribution:
• The governing differential eqn., as already derived, is
given by eqn. (6.1), viz.
d
2
θ
dx
2
m
2
θ. 0 ............(6.1)
And, we shall choose for its solution for temperature
Constants A and B are obtained from the B’C’s:
B.C.(i): at x = 0: T T1 i.e. θ θ 1
B.C.(ii): at x = : T T2 i.e. θ θ 2

• We get: A θ 1
Substituting for A and B in eqn. (6.2,b):
Eqn. (6.12) gives the temperature distribution along the
To determine the heat transfer rate:
Total heat transfer rate from the fin is determined by integrating the
convection heat transfer over the length of the fin:
Eqn. (6.12) gives the temperature distribution along the
length of fin, when its two ends are maintained at two
specified temperatures.

Simplifying, we get:
Eqn. (6.13) gives the heat transfer rate for a fin with
specified temperatures at its both ends.
To find the min. temperature in the fin:
Differentiate eqn. (6.12) w.r.t. x and equate to zero;
solving it, we get xmin , the position where min.
temperature occurs. Then, substitute this value of xmin
back in eqn. (6.12) to get the value of Tmin

• Special case: When both the ends of fin are at the
same temperature:
• Now, T1 = T2 (i.e. ), and obviously, the min. temperature
will occur at the centre i.e. at x = L/2.
• Then, substituting and x = L/2 in eqn. (6.12), we get for min.
temperature:
θ 1 θ 2
θ 1 θ 2
θ x( )
θ 1 sinh m L x( ).( ). θ 2 sinh m x.( ).
sinh m L.( )
......(6.12)
Therefore, θ min
θ 1 sinh m L
L
2
.. θ 1 sinh m
L
2
..
sinh m L.( )
i.e. θ min
2 θ 1
. sinh
m L.
2
.
sinh m L.( )
.......(6.14)
Remember: θ min T min T a

Performance of fins:
• 1. Fin efficiency: fin efficiency is defined as the
amount of heat actually transferred by a given fin to
the ideal amount of heat that would be transferred if
the entire fin were at its base temperature. i.e.
η f
Q fin
Q max
....(6.16)
(a) For an infinitely long fin:
For an infinitely long fin, actual heat transferred is given by eqn. (6.5):
. θ o
. h P. k. A c
. T o T a
. ......(6.5)

• To calculate Qmax, if the entire fin surface were to be at a
temperature of To, the convective heat transfer from the
surface would be:
Q max h P. L. T o T a
. .........eqn. (A)
Dividing eqn. (6.5) by eqn. (A):

• For a fin with insulated end:
For the case of a fin with an insulated end, we get
actual heat transferred Qfin from eqn. (6.7):
and, fin efficiency is given by:
. .
η f
h P. k. A c
. T o T a
. tanh m L.( ).
h P. L. T o T a
.
i.e. η f
tanh m L.( )
h P.
k A c
.
L.
i.e. η f
tanh m L.( )
m L.
....(6.18)...fin efficiency for a fin with insulated end

• It is instructive to represent eqn. (6.18) in graphical form:
Let X = m.L
X 0.1 0.2, 5.. ....let (m.L) vary from (m.L) =0.1 to 5 with an increment of 0.1
The graph looks as follows:
0.8
0.9
1
Fin efficiency graph - insulated end
Note: X = m.L
tanh X( )
X
X
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8

Fins of non-uniform cross-section:
t
w
L
r1
w
L L
t
r2
t
(a) Straight rectangular fin (b) Straight triangular fin (c) Circular fin of rectangular section
Fig. 6.8 Typical fins: (a) and (d) of uniform cross-section , and
(b), (c) and (e): of nonuniform cross-section
L
D D
L
(e) Pin fin, conical section(d) Pin fin, circular section

Table 6.4
Fin efficiency for a few fin shapes
• Ac = area of crosssection, Af = total fin surface area, Lc = corrected
length, P = perimeter of fin section, h = heat tr. coeff., m = {h.P /
(k.Ac)}

Note: In the above Table:
Io = modified zero order Bessel function of first kind
Ko = modified zero order Bessel function of second kind
I1 = modified first order Bessel function of first kind
K1 = modified first order Bessel function of second kind

Fin efficiency graphs:
• As can be seen from the above Table, expressions for
fin efficiency of fins of non-uniform cross-sections are
rather complicated and involve Bessel functions.
• In practice, to find out the heat transfer from such fins,
we use ‘Fin efficiency charts’.
• Once the fin efficiency is obtained from these graphs,
• Once the fin efficiency is obtained from these graphs,
actual heat transferred from the fin is calculated using
the definition of fin efficiency, viz.
η f
Q fin
Q max
....(6.16)
Q max h P. L. T o T a
.
where

• Fig. gives fin efficiency
values for fins of
rectangular and
triangular sections.
• It may be noted that
graph for rectangular
fins is also valid for
cylindrical, pin fins
Note: On x-axis use: (m.L c) for rectangular fins and, (m.L) for triangular fins
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
Fin efficy. for rect. & triangular fins
FinEfficiency
cylindrical, pin fins
since eqn. for fin
efficiency is the same
(see Table 6.4); of
course, m and Lc will
be different for pin fins.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
0.3
0.35
0.4
0.45
0.5
0.55
rectangular fins
triangular fins
mLc or mL

Fin effectiveness,ε f
. = (heat transfer rate with fin)/ (heat
transfer rate without fin)
ε f
i.e. ε f
Q fin
h A c
. T o T a
.
.....(6.20)
Obviously, should be as large
as possible.
Use of fins is hardly justified
unless fin effectiveness is
greater than about 2.
ε f

• Consider an infinitely long fin:
• Then, we have:
ε f
h P.
k A c
.
T o T a
.
h A c
. T o T a
.
....fin effectiveness for very long fin
i.e. ε f
k P.
h A c
.
......(6.21)
c
Following significant conclusions may be derived from
this equation:
(i)Thermal conductivity, k should be as high as possible; that is
why we see that generally, fins are made of copper or Aluminium.
Of course, Aluminium is the preferred material from cost and
weight considerations.

(ii) Large ratio of perimeter to area of cross-section is
desirable; that means, thin, closely spaced fins are
preferable. However, fins should not be too close as to
impede the flow of fluid by convection.
(iii) Fins are justified when heat transfer coeff. h is small,
i.e. generally on the gas side of a heat exchanger rather
than on the liquid side. e.g. the car radiator has fins on
the outside of the tubes across which air flows.
the outside of the tubes across which air flows.
Requirement that , gives us the criterion:ε f 2>
k P.
h A c
.
4> ....(6.22)

Thermal resistance of a fin:
• Convective thermal resistance of the base area is:
R b
1
h A c
.
...(6.24,a)....convective thermal resistance of base area
When fin is attached, we compute a thermal resistance for
the fin, from the usual definition vizc
R fin
∆T
Q
T o T a
Q
.....(6.24,b).....thermal resistance of fin
R fin
Q fin Q fin
Values of Qfin depend on the conditions at the tip of the fin and
may be obtained from Table 6.3.
Dividing eqn. (6.24,a) by eqn. (6.24,b), we get:
R b
R fin
Q fin
h A c
. T o T a
.
ε f ..(6.25)....from the definition of εf, in eqn.(6.20)

Total surface efficiency (or, overall surface
efficiency, or area-weighted fin efficiency)
• Total heat exchange area (At) may be considered as
made up of two areas:
the base or ‘prime’ surface area, Ap, on which there
are no fins, and the total fin surface area (N.Af) ,
where
where
N is the total no. of fins, and Af is the surface area of
each fin.
• Now, the ‘prime’ surface (or, un-finned surface) is
100% effective; but, all the fin surface area provided
is not 100% effective, since there is always a
temperature gradient along the fin.

• From the definition of fin efficiency, we know
that effective area of the fin surface is f . Af .
• Therefore, considering the total area of the
array, viz. (Ap + N.Af), we can define an ‘total
or overall surface efficiency’, , such that:η t

Eqn. (6.26) gives the value of overall or total surface
efficiency (or, area –weighted fin efficiency) for a fin array.
In other words, effective heat transfer area of the array is:
where At is the total area of the prime surface plus all
the fin area.
t
the fin area.
Concept of ‘overall surface efficiency’ is very useful in
calculating the heat transfer rates in heat exchangers
where fins may be provided on one or both sides of the
wall.

Application of fin theory for error estimation in
temperature measurement:
• Temperature of a
fluid flowing in a
pipe is generally
measured with a
thermometer
thermometer
placed in a
‘thermo-well’
welded radially or
obliquely to the
pipe wall.

• If the temperature of a hot
fluid flowing in the pipe is
Ta, obviously, the
temperature indicated by
the thermometer, TL will
not be equal to Ta , but
less than T , because of
a
less than Ta , because of
heat loss along the wall of
the thermo-well from its tip
to the root (and, vice-
versa, for a cold fluid
flowing in the pipe).

• Considering the thermo-well to be a fin protruding from the pipe wall,
with an insulated tip (i.e. no heat transfer from its tip), we can write,
from eqn. (6.7):
T x( ) T a
T o T a
cosh m L x( ).( )
cosh m L.( )
.......(6.7)
At the tip, i.e. at x = L, T(x) = TL.
Substituting x = L in eqn. (6.7):
T L T a
T o T a
1
cosh m L.( )
.....(a)
And, the error in thermometer reading is given by:
T L T a
T o T a
cosh m L.( )
.....(b)

• From eqn. (b), we observe that to reduce the temperature error, we
should have the factor 1/cosh(m.L) as small as possible. To achieve
this, looking at the graph of 1/cosh (m.L) vs. (m.L) given earlier, it is
clear that (m.L) should be as large as possible. i.e.
h P.
k A c
.
L. must be as large as possible.
This leads to the following conclusions:
This leads to the following conclusions:
• value of heat transfer coeff., h should be large
• value of thermal conductivity, k should be small
• thermo-well should be long and thin-walled. (thermo-well may be
placed obliquely inside the pipe, to make it long).

• Again, for the thermo-well, treated as a fin, we have:
i.e. fin parameter, 'm'does not depend upon thermo-
well pocket diameter, when the wall thickness is very
small compared to its diameter.

Ref: Incropera et al, 6th ed., p.176

It may be observed that:
(i) higher the thermal conductivity, higher is
(i) higher the thermal conductivity, higher is
the steady state temperature attained at a
given location.
(ii) to attain the same temp. on the rod,
longer length is required for a material of
lower thermal conductivity.
(iii) it is verified that fin can be considered
as 'infinitely long'if the lengths are 2.437,
1.768 and 0.927 m for copper, Aluminium
and steel,respectively. i e. the end
temperature becomes equal to the ambient
temp.

Heat transfer from extended surfaces (or fins)

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