Heat & Mass Transfer Week 06 Instructor: Mr. Adnan Qamar ...€¦ · Heat & Mass Transfer Week_06...

Heat & Mass Transfer

Week_06

Instructor: Mr. Adnan Qamar

Mechanical Engineering Department

1

One Dimensional Steady State Heat Conduction

Extended Surfaces-Fins

2


3


4


5


6

Common types of fin configurations are shown if Fig below:


7


8


Types of Fins

1. Variable Cross Sectional Area Fins

❖ Conical Fins

❖ Triangular Fins

2. Constant Cross Sectional Area Fins

❖ Rectangular Fins

❖ Cylindrical Fins

9

Variable Cross Sectional Area Fins

Let us consider a variable cross sectional area fin as shown in Fig. below;

10


Let,

A= Cross sectional area of the fin

a= Surface area of the fin

da=Surface area of the small element

L=Total length of the fine

K=Thermal conductivity of the fin material

To= Temperature of the fin base

Tα=Surrounding temperature of the fin

h=convective heat transfer coefficient

11


Since for the steady state system;

12

0=−−conveccondcond

outoutinQQQ

0=−− + convecoutdxxx QQQ

( ) ( ) 0=−−

+− TThdadxQx

xQQ xx

( ) ( ) 0=−−

−− TThdadxQ

xQQ xxx

( ) ( )

( ) ( )

( ) ( ) 0

0

0

=−−

=−−

−

−

=−−

−

TThdadxdx

dTxkA

x

TThdadxdx

dTxkA

x

TThdadxQx

x


Equation (1) is known as differential equation for a fin of variable cross section.

13

( ) ( )

( )

( ) )1(0

0

0

→=−

=−

=−−

kdx

hda

dx

dxA

x

hdadxdx

dxkA

x

TThdadxdx

dTxkA

x

Constant Cross Sectional Area Fins

Let us consider a constant cross sectional area fin as shown in Fig. below;

14


For a constant cross sectional area equation (1) can be written as;

Here;

A= Cross sectional area = Z × t

a = Surface area = P × L

For small element = da = P×dx

15

( )

( )

)2(0

0

)1(0

2

2

→=−

=−

→=−

kAdx

hda

dx

d

kdx

hda

dx

dxA

dx

d

kdx

hda

dx

dxA

x


By putting all values in equation (2), we will get;

Using D-Operator method to solve the equation (3), let; d/dx=D

16

)3(000 2

2

2

2

2

2

2

→=−=−=−

mdx

d

kA

hP

dx

d

kAdx

hPdx

dx

d

( )0))((

00 2222

=+−=

=−==−=

mDmD

mDmD


17

011 =− mD

011 =− mdx

d

11

mdx

d=

mdxd

=1

1

Cmx +=1lnmxeC11 =

022 =+ mD

022 =+ mdx

d

22

m

dx

d−=

mdxd

−=2

2

Dmx +−=2ln

mxeC −= 22


From the above two values of “θ” we have;

The equation (4) gives the exact solution and by using this equation we can find the

value of “θ” for any value of “x”, where θ=T-Tα

The general solution for equation (4) can also be written in form of hyperbolic

function. Since we know that;

18

)5(

)4(

21

21

→+=−

→+=

−

−

mxmx

mxmx

eCeCTT

eCeC

)(2

sinh aee

mxmxmx

→−

=−

)(2

cosh bee

mxmxmx

→+

=−


By adding equation (a) and (b), we will have;

By subtracting equation (a) and (b), we will have;

By putting the resultant equations in equation (4), we will have;

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)(2

sinh aee

mxmxmx

→−

=−

)(2

cosh bee

mxmxmx

→+

=−

mxemxmx =+ coshsinh

mxemxmx −=− sinhcosh

)sinh(cosh)cosh(sinh 21 mxmxCmxmxC −++=

( ) ( ) mxCCmxCC coshsinh 2121 ++−=


Here are some special cases for fins of constant cross sectional area with variousboundary conditions depending upon the various physical conditions.

CASE 1 The fin is very long, and the temperature at the end of the fin is essentially

that of the surrounding fluid.

CASE 2 The fin is of finite length, specified cross sectional area and loses heat by

convection from its end.

CASE 3 The fin is of finite length, negligible cross sectional area and insulated

tip. so that dT/dx = 0 at x = L.

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By putting the resultant equations in equation (4), we will have;

)sinh(cosh)cosh(sinh 21 mxmxCmxmxC −++=

( ) ( ) mxCCmxCC coshsinh 2121 ++−=

)6(coshsinh 21 →+= mxCmxC


Since we know that from the exact solution;

By applying boundary conditions;

At x=0; T=Tο; θ=θο

At x=α; T=Tα; θ=0

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CASE 1 The fin is very long, and the temperature at the end of the fin is essentially

that of the surrounding fluid.

)7(0 21

2

2

2

→+==− −mxmx eCeCmdx

d

)(0

)(

21

0

2

0

10

beCeC

aeCeC

→+=

→+=

−


By solving equation (a) and (b) we have;

By putting all values in equation (7), we will have;

22

)(0

)(

21

0

2

0

10

beCeC

aeCeC

→+=

→+=

−

02

1 0

=

=

C

C

)8(

)0(

00

00

→=−

−=

=+=

−

−

−−

mxmx

mxmxmx

eTT

TTe

eee


23


Since we know that from the exact solution form equation (6);



At x=L; Qcond = Qconv -kAdT/dx = hA θ

24

CASE 2 The fin is of finite length, specified cross sectional area and loses heat by

convection from its end.



By applying boundary conditions 1 in equation (9);

Differentiating equation (9) and the by applying boundary 2;

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( ) ( ) )(0cos0sin 02

0

2

0

10 aCCC →=+=

LxLx mxCmmxCmdx

dmxCmmxCm

dx

d==

+=+= sinhcoshsinhcosh 2121

LxLx mxCmxChAmxCmmxCmkA ==+=+− )coshsinh()sinhcosh( 2121

LxLx mxCmxChmxCmmxCmk ==+=+− )coshsinh()sinhcosh( 2121


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LxLx mxCmxChmxCmmxCmk ==+=+− )coshsinh()sinhcosh( 2121

)coshsinh()sinhcosh( 2121 mLCmLChmLCmmLCmk +=+−

)coshsinh()sinhcosh( 2121 mLCmLChmLCmLCmk +=+−

)coshsinh()sinhcosh( 2121 mLCmLCmk

hmLCmLC +=+−

)coshsinh()sinhcosh( 0101 mLmLCmk

hmLmLC +=+−

mLmk

hmLmL

mk

hCmLC coshsinhsinhcosh 0011 +=−−

mLmk

hmLmL

mk

hmLC coshsinh)sinh(cosh 0

01

+=+−


By putting all values in equation (9) we will have,

27

mLmk

hmLmL

mk

hmLC coshsinh)sinh(cosh 0

01

+=+−

)(

sinhcosh

coshsinh 00

1 b

mLmk

hmL

mLmk

hmL

C →

+

+

−=

mxmx

mLmk

hmL

mLmk

hmL

coshsinh

sinhcosh

coshsinh

0

00

+

+

+

−=


28

mxmx

mLmk

hmL

mLmk

hmL

coshsinh

sinhcosh

coshsinh

0

00

+

+

+

−=

mxmx

mLmk

hmL

mLmk

hmL

coshsinh

sinhcosh

coshsinh

0

+

+

+

−=

)10(coshsinh

sinhcosh

coshsinh

0

→+

+

+

−=−

−mxmx

mLmk

hmL

mLmk

hmL

TT

TT


29

mxmx

mLmk

hmL

mLmk

hmL

TT

TTcoshsinh

sinhcosh

coshsinh

0

+

+

+

−=−

−

mLmk

hmL

mxmLmk

hmxmLmxmL

mk

hmxmL

TT

TT

sinhcosh

coshsinhcoshcoshsinhcoshsinhsinh

0 +

++−−

=−

−

)11(

sinhcosh

)sinhcoshcosh(sinh)sinhsinhcosh(cosh

0

→

+

−+−

=−

−

mLmk

hmL

mxmLmxmLmk

hmxmLmxmL

TT

TT


Since we know that from the exact solution form equation (6);



At x=L; dT/dx = 0; θ=0

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CASE 3 The fin is of finite length, negligible cross sectional area and insulated tip, so

that dT/dx = 0 at x = L..



By applying boundary conditions 1 in equation (11);

Differentiating equation (12) and the by applying boundary 2;

31


( ) ( ) )(0cos0sin 02

0

2

0

10 aCCC →=+=

0sinhcosh

0sinhcosh

sinhcosh

01

21

21

=+

=+=

+=

==

mLmmLCm

mxCmmxCmdx

d

mxCmmxCmdx

d

LxLx


By putting all values in equation (12), we have,

32

)(cosh

sinh0sinhcosh 0

101 bmL

mLCmLmmLCm →

−==+

)13(cosh

)(cosh

cosh

coshcoshsinhsinh

coshsinhcosh

sinh

coshsinhcosh

sinh

0

0

0

0

→−

=−

−

+−=

+−=

+−

=

mL

xLm

TT

TT

mL

mxmLmxmL

mxmxmL

mL

mxmxmL

mL


33


34


35

th

k

ttZ

tZ

A

P

f

=

+=

2

2)(2


36

Conduction Heat Transfer-Class Problems

Example 6.1:

37


38


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Example 6.2:

40


41


42


43


Example 6.3:

44


45


46


Example 6.4:

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48


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Home Work (Heat Transfer, J.P. Holman)All Examples and Problems are from J.P. Holeman 10th Edition

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Chapter 01 Chapter 02

Examples Examples

1.1,1.2,1.3,1.4,1.5,1.6 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7,

2.8, 2.9, 2.10, 2.11, 2.12

Exercise Problems Exercise Problems

1.3, 1.16, 1.23, 1.29, 1.31, 1.36,

1.37, 1.40, 1.41, 1.43

2.2, 2.11, 2.15, 2.19, 2.21, 2.26,

2.30, 2.39, 2.51, 2.60, 2.66, 2.88,

2.112, 2.125, 2.137, 2.138

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Heat & Mass Transfer Week 06 Instructor: Mr. Adnan Qamar ...€¦ · Heat & Mass Transfer Week_06...

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