Page 104 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

THE FINITE ELEMENT METHOD
96
Note that T a is in Cas h is expressed in W/m C.
◦
2◦
Since only one element is employed, no assemblage of element contribution is necessary.
Thus, the simultaneous equation system may be written as

0.068 −0.056 T 1 0.30
= (3.284)
−0.056 0.068 T 2 0.30
◦
We now incorporate the known base temperature of 100 C at node 1. It is done in such
a way that the symmetry of the [K] matrix is retained. This is essential if a symmetric matrix
solution procedure is employed in the solution of the simultaneous equations. The following
steps give a typical implementation procedure for the temperature boundary condition:
(i) The diagonal element of the first row is assigned a value of 1 and the remaining
elements on that row are zero.
(ii) Replace the first row value of the loading vector f by the known value of T 1 , that is,
100.

(iii) In order to retain the symmetry, the first term of the second row in the [K] matrix is
transferred to the right-hand side and replaced with a zero value as given below:

1.0 0.0 T 1 100.0
= (3.285)
0.00.068 T 2 0.30 + 0.056(100.0)
The equation to be solved is
0.068T 2 = 0.3 + 0.056(100) (3.286)

◦
◦
Therefore, the solution is T 1 = 100 C and T 2 = 86.765 C.
Heat dissipated is
kA
Q = (T 1 − T 2 ) = 0.7941 W (3.287)
l
The above answer is very approximate. However, a more accurate value can be deter-
mined by using the following convection condition, that is,

M
T 1 + T 2

Q = hP l − T a = 1.64 W (3.288)
2
e=1
where M is the total number of elements. The maximum theoretically possible heat transfer
is
M

Q max = hP l (T 1 − T a ) = 1.8 W (3.289)
e=1
The efficiency is defined as
Q 1.64
η f = = = 91.11% (3.290)
Q max 1.80