chap 6 heat loss in fins

8/11/2019 Chap 6 Heat Loss in Fins

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6-2 Thin Rectangular Fins

We will consider only thinrectangular fin.

The heat transfer through the fincan be treated as one-dimensional, since temperaturegradient exists in the x-directiononly, thus, dT/dx 0.

Temperature gradient along thewidth, dT/dz , and across thethickness, dT/dy , are negligible.

L = fins lengthw = fins widtht = fins thickness

The governing equation for heattransfer trough the fin,

0=+

Q

dxdT

k dxd

where Q = internal heat generation (W/m 3).

6-3 Finite Element ModelingSince heat transfer through the thin fin can be considered one-dimensional, we can model the fin one-dimensional elements.

Suppose the fin is discretized into three elements, as shown.

Consider a single element,

Global coordinate

Local coordinate


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Within an element, temperature varies from T 1 at local node 1 to T 2at local node 2.

To predict the temperature at any point between the two nodes, weneed to establish a temperature function.

For simplicity, we will assume that the temperature varies withinthe element in a linear fashion. Therefore, we establish a lineartemperature function in the form,

( ) [ ]{ }eT N T N T N T =+= 2211

1 21 1 and

2 2 N N += =

6-4 Temperature Function

where

are the linear shape functions.

6-5 Element Conductivity Matrix

Recall that we use one-dimensional element to model thin fins,which is similar to the case of a plane wall (Chapter 5).

Therefore the conductivity matrixconductivity matrix for the plane wall element can alsobe used for the fin element, i.e.

[ ] 2o1 1 W m C1 1e e

T e

k k l

= where k is thermal conductivity of the fin material (W/m. oC) and l e isthe element length.


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6-6 Element Heat Transfer Matrix

Recall that in thin fins, heat transfer by convection also takes placetogether with conduction. To include this effect in our analysis, weneed to establish a heat transfer matrixheat transfer matrix, [ hT ].

It can be shown that,

[ ] 2o2 1 W

m C1 23e e

T

hl h

t

where,h = heat transfer coefficient (W/m 2o C)

l e = element length (m)t = thickness of the fin (m)

6-7 Element Heat Rate Vector

For heat transfer in thin fins, the heat rate is contributed by the heatloss by convection.

Using the Galerkins approach, it can also be shown that the heatrate vector for a given element is given by,

{ } 21 W

m1e ehl T r

t

where

T = ambient temperatureh = convective heat transfer coefficient

t = thickness of the fin

l e = element length


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6-8 System of Linear Equations

For a single fin element, the system of linear equations (SLEs) canbe written in a condensed matrix form as,

[ ] [ ]( ){ } { }e e e eT T k h T r + =where

[k T ]e = element conductivity matrix (due to conduction)

[hT ]e = element heat transfer matrix (due to convection)

{T }e = nodal temperature vector

{r }e

= element heat rate vector

6-9 Thermal Bondary Conditions

It is usually assumed that temperature at the base of the fin is thesame as that of the structure, denoted by T o.

The tip of the fin can be assumed to be adiabatic or insulated,thus, heat flux q = 0 at the tip.

Therefore, the boundary conditions for fin problems are

at 0, i.e. specified temperature

0 at , i.e. specified heat fluxoT T x

q x L

= == =

Global coordinate


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The heat loss, H e, from each fin element can be estimated usingthe NewtonNewton s laws law of cooling, given by

( ) ( ) ( )2e avg s avg e H h T T A h T T w l = = where

The totaltotal heat loss from the entire fin can be obtained using,

1

n

ee

H H =

=

6-10 Heat Loss Through the Fin

h = convective heat transfer coefficient,T avg = average temperature within the element,T = ambient temperature,

A s = surface are of the fin,w = width of the fin,

le = length of the element.

Example 6-1

A metallic fin, 0.1 cm thick,10 cm long, 1 m width and with thermalconductivity k = 360 W/m oC, extends from a plane wall whosetemperature is 235 oC. Determine the temperature distribution withinthe fin and amount of heat loss from the fin to the ambient air, whichis at 20 oC. Take h = 9 W/m 2oC.

SolutionWe will model the fin using three one-dimensional elements andassume that the tip of the fin is insulated.


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[ ]( ) [ ]( ) [ ]( )1 2 3 21 13601 13.33 10T T T

k k k = = =

[ ]

=

1100

1210

0121

0011

1033.3360

2T K

1. Element conductivity matrices

Since the fin is made up of homogeneous material and that allelements have the same length, the element conductivity matrixfor the elements will be the same, i.e.

2. Assemble global conductivity matrix

1 2 3 4 Global nodes

[ ]( ) [ ]( ) [ ]( )

===

21

12

001.031033.39 2321

T T T hhh

[ ]

=

2100

1410

0141

0012

001.031033.39 2

T H

3. Element heat transfer matrices

Since all elements have the same length and thickness, theelement heat transfer matrix for the elements will be the same,

4. Assemble global heat transfer matrix

1 2 3 4 Global nodes


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{ }( ) { }( ) { }( ) ===

1

1

001.0201033.39 2321 r r r

{ } =

12

2

1

001.0

201033.39 2 R

5. Element heat rate vector

Since all elements have the same length and thickness, theelement heat rate vector for the elements will be the same, i.e.

6. Assemble global heat rate vector

1

2

34

Global nodes

7. Write the system of linear equations

1

2 22

3

4

2

1 1 0 0 2 1 0 0

1 2 1 0 1 4 1 0360 9 3.33 100 1 2 1 0 1 4 13.33 10 3 0.001

0 0 1 1 0 0 1 2

1

29 3.33 10 2020.001

1

T

T T

T

+

=

[ ] [ ]( ){ } { }T T K H T R+ =We get,


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8. Impose the thermal boundary conditions.

1

22

23

4

2

1 1 0 0 2 1 0 0

1 2 1 0 1 4 1 0360 9 3.33 100 1 2 1 0 1 4 13.33 10 3 0.001

0 0 1 1 0 0 1 2

1

29 3.33 10 2020.001

1

T

T

T

T

+

=

We have T 1 = 235 C. Using the Gauss eliminationGauss elimination method, wedelete the 1 st row and column of the SLEs, and modify the rightside term accordingly.

22

32

4

2

2

2 1 0 4 1 0360 9 3.33 10

1 2 1 1 4 13.33 10 3 0.0010 1 1 0 1 2

29 3.33 10 20

20.001

1

1360 235

03.33 10

0

T

T

T

+

=

2

19 3.33 10 235

03 0.001

0

9. Write the reduced system of linear equations

After imposing the boundary condition, the system of linearequations are reduced to,


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2o

3

4

211.7

197.0 C

192.2

T

T

T

=

21

22

23

235 211.79 20 2 1 3.33 10 121.9 W

2

211.7 1979 20 2 1 3.33 10 110.5 W2

197 192.29 20 2 1 3.33 10 104.7 W

2

H

H

H

+ = =

+ = = + = =

10. Solve the reduced SLEs

Solving the reduced SLEs yields

11. Heat loss from each element

12. Total heat loss

The total heat loss from the fin is given by

1 2 3

121.9 110.5 104.7

337.1 W

H H H H

H

H

= + += + +

=


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End of Chapter 6

chap 6 heat loss in fins

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