Page 139 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

STEADY STATE HEAT CONDUCTION IN MULTI-DIMENSIONS
The elemental [K] matrices for elements 1, 3, 5 and 7 can be written as (refer to
Equation 5.8)
 2 2  131
b + c b 1 b 2 + c 1 c 2 b 1 b 4 + c 1 c 4
tk 1 1
2
[K] 1 = [K] 3 = [K] 5 = [K] 7 =  b 1 b 2 + c 1 c 2 b + c 2 2 b 2 b 4 + c 2 c 4  (5.12)
2
4A 2 2
b 1 b 4 + c 1 c 4 b 2 b 4 + c 2 c 4 b + c
4 4
where the area of the elements can be written as
& &
1.00.00.0
& &
& & 2
2A = det 1.00.50.0 = 0.25 m (5.13)
&
&
& &
& 1.00.00.5 &
Substituting the area into Equation 5.12, we get the final form of the elemental
equation as
 
2.0 −1.0 −1.0
tk
[K] 1 = [K] 3 = [K] 5 = [K] 7 =  −1.0 1.0 0.0  (5.14)
2
−1.0 0.0 1.0
Similarly, we can calculate the elemental [K] matrices for elements 2, 4, 6 and 8 as
 
1.0 −1.0 0.0
tk
[K] 2 = [K] 4 = [K] 6 = [K] 8 =  −1.0 2.0 −1.0  (5.15)
2
0.0 −1.0 1.0
The assembled equations are (see Appendix C)
2.0 −1.0 0.0 −1.0 0.0 0.0 0.0 0.0 0.0 T 1 0.0
     
 
−1.0 4.0 −1.0 0.0 −2.0 0.0 0.0 0.0 0.0 T 2   


0.0
 
     
 0.0 −1.0 2.0 0.0 0.0 −1.0 0.0 0.0 0.0   T 3   
 0.0 
 
     
 −1.0 0.0 0.0 4.0 −2.0 0.0 −1.0 0.0 0.0   T 4   


tk     0.0 
T 5
 0.0 −2.0 0.0 −2.0 8.0 −2.0 0.0 −2.0 0.0    = 0.0 (5.16)
2      
T
 0.0 0.0 −1.0 0.0 −2.0 4.0 0.0 0.0 −1.0     
0.0
   6   
T


 0.0 0.0 0.0 −1.0 0.0 0.0 2.0 −1.0 0.0     
0.0
   7  

0.0 0.0 0.0 0.0 −2.0 0.0 −1.0 4.0 −1.0 T 8 0.0

     
 


0.0 0.0 0.0 0.0 0.0 −1.0 0.0 −1.0 2.0 T 9 0.0
The only unknown quantity in the above equation is T 5 , which can be calculated from
the equation corresponding to the fifth node, that is, from
8T 5 = 2T 2 + 2T 4 + 2T 6 + 2T 8 (5.17)
◦
◦
◦
Substituting T 2 = T 4 = T 6 = 100 C and T 8 = 500 C, we get T 5 = 200 C
The analytical solution to this problem is given by (Holman 1989)
nπy

∞ n+1 sinh
2 (−1) + 1 nπx w
T (x, y) = (T top − T side ) sin + T side (5.18)
π n w nπH
n=1 sinh
w