Page 105 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

97
THE FINITE ELEMENT METHOD
The exact solution for this problem is
Q exact = kAm(T b − T a ) tanh (kml) = 1.593 W (3.291)
√
where m = hP/kA = 31.62. Therefore, the exact fin efficiency is
Q
(η f ) exact = = 88.48%. (3.292)
Q exact
(b) Let us consider a two-element solution of the same problem (3 nodes)
The length of the fin is divided equally into two elements, that is, l = 1.0 cm.
The stiffness matrix calculation is similar to the one for the single-element case, that is,

0.124 −0.118
[K 1 ] = [K 2 ] = (3.293)
−0.118 0.124
and the loading vectors are

0.15
{f 1 }= {f 2 }= (3.294)
0.15
On assembly we obtain

     
0.124 −0.118 0.0 0.15
T 1  
−0.118 0.124 + 0.124 −0.118 0.15 + 0.15 (3.295)
  T 2 =
0.0 −0.118 0.124    0.15 
T 3
◦
Now we have to incorporate the known value of base temperature, that is, T 1 = 100 C.
    
1.0 0.0 0.0 100.0

T 1  
0.0 0.248 −0.118 = 0.30 + 0.118(100) (3.296)
  T 2
0.0 −0.118 0.124    0.15 
T 3
Therefore, the two equations to be solved are
0.248T 2 − 0.118T 3 = 12.1

and
−0.118T 2 + 0.124T 3 = 0.15

◦
◦
Solving these equations, we get T 2 = 90.209 C, T 3 = 87.057 C.
Results, which have been generated using different number of elements are tabulated in
Tables 3.7 and 3.8.
As can be seen, the two-element solution is very good and is further improved with the
use of four elements. As a first idealization, even the one element solution is reasonably
good considering the small effort involved.