Page 118 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

STEADY STATE HEAT CONDUCTION IN ONE DIMENSION
110
In the case of a finite element formulation, we have to account for the heat generation
in the forcing vector such that

T N i GAL 1
{f} e = G[N] d
= G A dx = (4.34)

l N j 2 1
The heat generated is distributed equally between the two nodes ‘i’ and ‘j’. In all
linear elements, we observe that the heat generated, or any other type of load, is equally
distributed among the participating nodes.
Because of the symmetry of the problem, it is sufficient in this case if we take only
one half of the domain.
Example 4.2.2 Determine the temperature distribution in a plane wall of thickness 60 mm,
3
which has an internal heat source of 0.3 MW /m and the thermal conductivity of the mate-
◦
◦
rial is 21 W/m C. Assume that the surface temperature of the wall is 40 C.
Because of symmetry, we may consider only one half of the plane wall as shown in
Figure 4.7. Let us consider four elements, each of length 7.5 mm. Let the cross-sectional
2
area for heat flow, A = 1m .
The element stiffness matrix is
kA 1 −1 2800 −2800
[K] e = = (4.35)
L −1 1 −2800 2800
which is identical for every element and

GAL 1 1125
{f} e = = (4.36)
2 1 1125
which also is identical for all elements. Assembly gives
     
2800 −2800 0.0 0.0 0.0
 T 1  1125 
   
−2800 5600 5600 0.0 0.0    
     
T 2 2250
 
0.0 −2800 5600 −2800 0.0 2250 (4.37)
  T 3 =
     
  
 0.0 0.0 −2800 5600 −2800 T 4   
2250
   
0.0 0.0 0.0 −2800 2800    1125 
T 5
30 mm
1 2 3 4
12 3 45

x
x = 0
Figure 4.7 Finite element discretization