Heat Loss in Fins

8/12/2019 Heat Loss in Fins

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SME 3033 FINITE ELEMENT METHOD

Heat Loss in Thin Fins(Initial notes are designed by Dr. Nazri Kamsah)


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What is Fin?

Fin is an extended surface, added onto a surface of a structure to

enhance the rate of heat transfer from the structure.Example: The fins fitted around a motorcycle engine block.

Quiz: Can you think of other practical examples?


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Thin Rectangular Fin

We will develop finite element formulation to model and analyze heat

transfer process in thin rectangular fins. The objective is to determine:a) Temperature distribution in the fin; and b) Total heat dissipated.

Thin rectangular fins used in heat sink design for microelectronic cooling.


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Heat Transfer in Thin Fins

Temperature gradient is assumed to exists in onedirection only. Thus, heat

transfer through thin fin can be treated as one-dimensional.Note: Heat flows through the fin by conduction and is dissipated to ambient

air by convection. The two heat transfer modes occur simultaneously.


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Performing an energy balance across an elemental section of the thin

rectangular fin yields thegoverning equation for heat transfer troughthe fin, given by

0

Q

dx

dTk

dx

d

where Qrepresents the internalheat generation, in W/m3.

Note:

In general, thermal conductivity, k

varies along thex-direction.

For isotropic, homogeneous

material, kis uniform and has the

same value in all directions.

(7-1)

The Governing Equation

TTA

ph

dxA

TThdxPQ

c

c


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Finite Element Modeling

Since heat transfer through thin fin is assumed one-dimensional, we will

model the fin using one-dimensional heat transfer elements.The fin is discretized into threeelements, as shown.

A single element will be in a localcoordinate system.

Globalcoordinate

Localcoordinate


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Element Conductivity Matrix

A one-dimensional heat transfer element is used to model thin fins, which is

similar to the element used to model the plane wall (Chapter 6).

Therefore the element conductivity matrixfor the plane wall can be used

for the thin fin, so that

21 1 W

1 1 m K

e eeT Tfin wall

e

kk k

l

where kis thermal conductivity of the fin material (W/m K) and leis the elementlength.

(7-4)


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Exercise 1

A circular rod of diameterD= 5 mm and lengthL= 190 mm has its base

maintained at Tb

= 100C. It is made of copper with thermal conductivity k=

398 W/m K. The surface of the rod is exposed to ambient air at temperature

T= 25C, with convection heat transfer coefficient, h= 100 W/m2K. Model

the rod using twoelements. Assemble theglobal: a) conductivity matrix;

b) heat transfer matrix due to convection. Assume thickness, t 0.8D.

A protruding rod, which can be treated as apinfin.


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Element Heat Rate Vector

In thin fins, the heat rate vector is contributed also by the heat loss by

convection from the fins.

Using the Galerkins approach, it can be shown that the element heat

rate vectorcan be expressed as

21 W

1 m

e ehl T

r t

where,

T= ambient temperature (inK);

h= convective heat transfer coefficient( in W/m2K);

t= thickness of the fin (in m);

le= element length (in m).

(7-6)


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System of Linear Equations

Thesystem of linear equations (SLEs) for a single element, can be

written in a condensed matrix form as,

e e e eT Tk h T r

where [kT]e= element conductivity matrix (due to conduction);

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

{T}e= nodal temperature vector (unknown);

{r

}e= element heat rate vector (due to convection).

(7-7)

Note:

If the fin is insulated, then the magnitude of both [hT]e and {r

}ewill be zero.


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Exercise 2

Reconsider Exercise 7-1. a) Assemble theglobalheat rate vector for the

rod; b) Write the global system of linear equation for the problem.



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Thermal Boundary Conditions

Temperature at the base of the fin, Tbis usually assumed to be the same

as the temperature of the structure.

The tipof the fin can be assumed as adiabaticor insulated. Therefore, the heat

flux q= 0. Hence, the boundary conditions for the fin are,

at 0, i.e. specified temperature

0 at , i.e. specified heat flux

b oT T x

q x L

Global coordinate0bT T


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Exercise 3

Reconsider Exercise 7-2. a) Apply the boundary conditions; b) Solve the

modified global system of linear equations for the unknown temperature

distribution in the rod.



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The heat dissipated,He, from a single element can be estimated using the

Newtons law of cooling,given by

e avg sH h T T A where

The totalheat dissipated from the fin,

1

n

e

e

H H

Heat Dissipated From an Element

h= heat transfer coefficient (W/m2K),Tavg= average temperature in the element (K),T= ambient temperature (K),

As= surface are of the fin (m2),

w= width of the fin (m),le

= length of the element (m).

(7-8)

(7-9)


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Exercise 4

Reconsider Exercise 7-3. Knowing the temperature distribution, estimate the

total amount of heat dissipated from the rod to the ambient air. Note that for

a circular-shape (or pin) fin, the area for heat transfer,As= DL.

A protruding rod, which can be treated as a pin fin.


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Example 1

A metallic fin is 0.1 cm thick,10 cm long, and 1 m width, has a thermal

conductivity, k= 360 W/mo

C. It extends from a plane wall whose surfacetemperature is 235oC. The convective heat transfer coefficient, h= 9

W/m2oC. Determine: a) Temperature distribution in the fin, b) Total heat

dissipated from the fin to ambient air, at 20oC.

Solution

Model the fin using threeone-dimensional elements. Assume the tip of the fin isinsulated.


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1 2 3

2

1 1360

1 13.33 10T T Tk k k

1100

1210

0121

0011

1033.3

3602T

K

1. Write the elementconductivity matrices

Since the fin is made up of homogeneous material and that all elements

have the same length, the conductivity matrix for all elements will be thesame, i.e.

2. Assembleglobal conductivity matrix, we get

1 2 3 4 Connectivity with the

global nodes.


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1

1

001.0

201033.392

321rrr

1

2

2

1

001.0

201033.392

R

5. Write the elementheat rate vector

Since all elements have the same length and thickness, the heat rate

vector for the elements will be the same, i.e.

6. Assembleglobal heat rate vector, we get

1

2

3

4

Connectivity with the

global nodes.


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7. Write theglobalsystem of linear equations, in the form

1

22

2

3

4

2

1 1 0 0 2 1 0 0

1 2 1 0 1 4 1 0360 9 3.33 10

0 1 2 1 0 1 4 13.33 10 3 0.001

0 0 1 1 0 0 1 2

129 3.33 10 20

20.001

1

T

T

T

T

T TK H T R

We have,


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8. Impose the thermal boundary conditions.

1

22

2

3

4

2

1 1 0 0 2 1 0 0

1 2 1 0 1 4 1 0360 9 3.33 10

0 1 2 1 0 1 4 13.33 10 3 0.001

0 0 1 1 0 0 1 2

129 3.33 10 20

20.001

1

T

T

T

T

Given, T1= 235 C. Using the elimination method, we delete the 1strow

and column of the global SLEs, and modify the right side term accordingly.We have,


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22

32

4

2

2

2 1 0 4 1 0360 9 3.33 10

1 2 1 1 4 13.33 10 3 0.001

0 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 0

3 0.0010

9. Write the modified system of linear equations

Imposing the boundary condition, the system of linear equations is

reduced to,


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2

o

3

4

211.7

197.0 C

192.2

T

T

T

2

1

2

2

2

3

235 211.79 20 2 1 3.33 10 121.9 W

2

211.7 197

9 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 modifiedsystem of linear equations

11. Compute the heat loss from each element

Solving the modified system of linear equations yields the unknown

global nodal temperatures,


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12. Compute the totalheat loss

The total heat loss from the fin is,

1 2 3

121.9 110.5 104.7

337.1 W

H H H H

H

H

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Heat Loss in Fins

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