Heat transfer from extended surfaces (or fins)

Lectures on Heat Transfer --HEAT TRANSFER FROM EXTENDED

SURFACES (FINS)

by

Dr. M. ThirumaleshwarDr. M. Thirumaleshwarformerly:

Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,

MangaloreIndia

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

Ach, Ta


L

dx

Qx Qx+dx w

X

X

Z

Fig. 6.2(a) Rectangular fin of uniform crosssection


differential equation:• 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.


differential equation:• 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.


differential equation:• 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)


differential equation:• We have:



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



differential equation:



differential equation:• 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 itssolution for temperaturedistribution 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 oexp m x.( )

i.e.T x( ) T a

T o T aexp 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

dxT x( )

x 0

. k A c. d

dxθ x( )

x 0

. .......(c)

i.e. Q fin k A c. d

dxθ 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

Ach, Ta

(dT/dx)x=L = 0

Q

d2 θ 2 .


L

X

Fig. 6.4(a) Fin of finite length, end insulated

(dT/dx)x=L = 0d2 θ

dx2m2 θ. 0 ............(6.1)

And, we shall choose for its solution fortemperature 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

dθdx

0 since the end is insulated.

From B.C. (i) and eq. (6.2,b): A θ oTemp. profile

T

To


From B.C. (ii) and eq. (6.2,b): dθdx

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 osinh m L.( )

cosh m L.( ).We get:

X

Fig. 6.4(b) Temp. profile for fin insulated at its end

TaTL

• Substituting for A and B in eqn. (6.2, b) and simplifying:

θ x( ) θ o cosh m x.( ). θ osinh 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 inthe 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) = 1i.e.

T L T a

T o T a

1

cosh m L.( ).....(6.7, a)and,


or, T LT o T a

cosh m L.( )T a ......(6.7, b)

Eqn. (6.7,b) gives the temperature at the end of afin (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

.


. tanh m L.( ). .......(6.8)

i.e. Q h P. k. A. θ. tanh m L.( ). ......(6.9)



. 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

00.20.40.60.81

1.21.41.61.82

2.2

tanh X( )

00.197380.379950.537050.664040.761590.833650.885350.921670.946810.964030.97574

tanh X( )

1

cosh X( )

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1X versus tanh (X) and {1/cosh(X)}


2.22.42.62.83

3.23.43.63.84

4.24.44.64.85

0.975740.983670.989030.992630.995050.996680.997770.998510.999

0.999330.999550.99970.99980.999860.99991

X0 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. kdT

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.

d2 θ

dx2m2 θ. 0 ............(6.1)

Temp. profile

X

Ta

TL

To

Convection


And, we shall choose for its solution for temperaturedistribution, 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. kdθ 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



.tanh m L.( )

h

m k.

1h

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 Lt

2where 't' is the thickness of fin

For cylindrical (round) fins: L c Lr

2where '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

T2Q2

Qconv


Temp. profile

X

Fig. 6.6. Fin of finite length, with specified temp. at two endsand 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.

d2 θ

dx2m2 θ. 0 ............(6.1)

And, we shall choose for its solution for temperaturedistribution, eqn. (6.2, b), i.e.


distribution, eqn. (6.2, b), i.e.


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 theconvection heat transfer over the length of the fin:

Eqn. (6.12) gives the temperature distribution along thelength of fin, when its two ends are maintained at twospecified temperatures.

Simplifying, we get:

Eqn. (6.13) gives the heat transfer rate for a fin withspecified 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 xminback 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 LL

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 fQ 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):


. 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 fh P. k. A c

. T o T a. tanh m L.( ).

h P. L. T o T a.

i.e. η ftanh m L.( )

h P.

k A c.

L.

i.e. η ftanh 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

1Fin efficiency graph - insulated end

Note: X = m.L


tanh X( )

X

X0 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 6.4Fin 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 kindI1 = 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 fQ 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

1Fin efficy. for rect. & triangular fins

Fin

Eff

icie

ncy


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 30.3

0.35

0.4

0.45

0.5

0.55

rectangular finstriangular fins

mLc or mL



Fin effectiveness, ε f. = (heat transfer rate with fin)/ (heattransfer rate without fin)

ε f

i.e. ε fQ fin

h A c. T o T a

......(6.20)


Obviously, should be as largeas possible.Use of fins is hardly justifiedunless fin effectiveness isgreater 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 fk P.

h A c.

......(6.21)


c

Following significant conclusions may be derived fromthis equation:

(i)Thermal conductivity, k should be as high as possible; that iswhy we see that generally, fins are made of copper or Aluminium.Of course, Aluminium is the preferred material from cost andweight 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 ratherthan on the liquid side. e.g. the car radiator has fins onthe 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 b1

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 viz c

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 andmay 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


whereN 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 allthe fin area.


tthe 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


aless 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 aT 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, Aluminiumand steel,respectively. i e. the end temperature becomes equal to the ambient temp.






