Temperature Distribution-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

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Assignment of Chapter#

FINITE ELEMENT

METHOD

Submitted To

Problem-1.

Statement : Find the temperature distribution in the

stepped fin shown in Figure below using two finite

elements.

1- Idealization

Let the fin be idealized into two finite elements. If the temperatures at the nodes are taken as

unknowns there will be three unknowns, namely T 1

, T 2

and T

2- Interpolation (Temperature Distribution) Model

In each element e, the temperature is assumed to vary linearly as;

Where ‘a’ and ‘b’ are constants.

Inserting the boundary conditions

At

x=0 T(x) =

( )

1

e

T

x=l

e

T(x) =

( )

2

e

T

The constants are evaluated and hence equation (1) becomes,

3- Element Characteristic Matrices and Vectors

The governing Matrix equation is;

Where;

And K

(e)

is the characteristic matrix of element e given by;

T x ( )  a  bx ..............................(1)

( ) ( ) ( )

1 2 1 ( )

e e e

e

x

T x T T T

l

K T  P ......................................(3)

2

( )

1

e

e

K K





( ) ( )

( )

( ) ( )

e e

e

e e

hp l

K

l KA

Element matrices are assembled as;

Thus the governing finite element equation of the fin, equation (3) becomes;

5- Solution for Nodal Temperatures

The equation (6) has to be solved after applying boundary conditions which is;

T 1

=T 0

=

0

C (At Node 1)

So the first equation of (6) is replaced by;

T

1

=T

0

=

0

C

P

K

K

1

2

3

T

T

T

2

3

T

T

The remaining two equations are written in scalar form as;

Putting T 1

=

0

C in equation (7) we get;

Solving eq. (9) and (10) simultaneously we get;

T

2

=64.

0

C

And

T

3

=

0

C

Hence the nodal temperatures are;

T 2

=64.

0

C

And

T 3

=

0

C (Answer)

1 2 3

0.4 T  1.61 T  0.04 T 45.72..........................(7)

1 3

0.4 T  0.91 T 34.29........................................(8)

2 3

1.61 T  0.04 T 101.72........................................(9)

1 3

0.4 T  0.91 T 34.29...........................................(10)