Heat Loss in Thin Fins - Universiti Teknologi Malaysiataminmn/SME 3033 Finite Element...

SME 3033 FINITE ELEMENT METHOD

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


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?


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.


Heat Transfer in Thin Fins

Temperature gradient is assumed to exists in one direction 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.


Performing an energy balance across an elemental section of the thin

rectangular fin yields the governing equation for heat transfer trough

the fin, given by

0

Q

dx

dTk

dx

d

where Q represents the internal

heat generation, in W/m3.

Note:

In general, thermal conductivity, k

varies along the x-direction.

For isotropic, homogeneous

material, k is uniform and has the

same value in all directions.

(7-1)

The Governing Equation

TTA

ph

dxA

TThdxPQ

c

c


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 three elements, as shown.

A single element will be in a local coordinate system.

Global coordinate

Local coordinate


The temperature varies from T1 at node 1 to T2 at node 2 in an element.

To predict temperature at any point between the two nodes we need to

establish a temperature function.

For simplicity, it is assumed that the temperature varies linearly within the

element. Therefore, we establish a linear temperature function in the

form,

eTNTNTNT 2211

1

2

1;

2

1

2

N

N

Temperature Function

where N1 and N2 are linear shape functions given by

(7-2)

(7-3)


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 matrix for the plane wall can be used

for the thin fin, so that

2

1 1 W

1 1 m K

e e eT Tfin wall

e

kk k

l

where k is thermal conductivity of the fin material (W/m K) and le is the element

length.

(7-4)


Element Heat Transfer Matrix

In thin fins, heat is dissipated primarily by convection heat transfer.

The effect of conduction through the fin is represented by the conductivity

matrix [kT] just established.

To include the effect of convection heat transfer, we need to establish a

heat transfer matrix, [hT].

It can be shown that the element heat transfer matrix is given by,

2

2 1 W

1 23 m K

e eT

hlh

t

where,

h = heat transfer coefficient (W/m2K)

le = element length (m)

t = thickness of the fin (m)

(7-5)


Exercise 1

A circular rod of diameter D = 5 mm and length L = 190 mm has it’s 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 two elements. Assemble the global: a) conductivity matrix;

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

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


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 Galerkin’s approach, it can be shown that the element heat

rate vector can be expressed as

2

1 W

1 m

e ehl Tr

t

where,

T = ambient temperature (in K);

h = convective heat transfer coefficient

( in W/m2K);

t = thickness of the fin (in m);

le = element length (in m).

(7-6)


System of Linear Equations

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

written in a condensed matrix form as,

e e e e

T 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}e will be zero.


Exercise 2

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

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



Thermal Boundary Conditions

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

as the temperature of the structure.

The tip of the fin can be assumed as adiabatic or 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 coordinate 0bT T


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.



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

Newton’s law of cooling, given by

e avg sH h T T A

where

The total heat 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)


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.



Example 1

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

conductivity, k = 360 W/moC. It extends from a plane wall whose surface

temperature 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 three one-dimensional elements. Assume the tip of the fin is

insulated.


1 2 3

2

1 1360

1 13.33 10T T Tk k k

1100

1210

0121

0011

1033.3

3602TK

1. Write the element conductivity 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 the

same, i.e.

2. Assemble global conductivity matrix, we get

1 2 3 4 Connectivity with the

global nodes.


21

12

001.03

1033.39 2321

TTT hhh

2100

1410

0141

0012

001.03

1033.39 2

TH

3. Write the element heat transfer matrices.

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

matrix for all elements will be the same, i.e.

4. Assemble global heat transfer matrix

1 2 3 4 Connectivity with the

global nodes.


1

1

001.0

201033.39 2321

rrr

1

2

2

1

001.0

201033.39 2

R

5. Write the element heat 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. Assemble global heat rate vector, we get

1

2

3

4

Connectivity with the

global nodes.


7. Write the global system 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

1

29 3.33 10 20

20.001

1

T

T

T

T

T TK H T R

We have,


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

1

29 3.33 10 20

20.001

1

T

T

T

T

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

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

We have,


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

21

9 3.33 10 2350

3 0.0010

9. Write the modified system of linear equations

Imposing the boundary condition, the system of linear equations is

reduced to,


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 1979 20 2 1 3.33 10 110.5 W

2

197 192.29 20 2 1 3.33 10 104.7 W

2

H

H

H

10. Solve the modified system of linear equations

11. Compute the heat loss from each element

Solving the modified system of linear equations yields the unknown

global nodal temperatures,


12. Compute the total heat 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 Thin Fins - Universiti Teknologi Malaysiataminmn/SME 3033 Finite Element...

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