Page 126 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

STEADY STATE HEAT CONDUCTION IN ONE DIMENSION
118
and the heat generated will be equal to the heat lost at the surface, that is,
2
Gπr L =−k2πr o L dT (4.70)
o
dr
r o
Equation 4.68 can be rewritten as
1 d dT
k r + G = 0 (4.71)
r dr dr
The analytical solution for this problem is
2

T − T w r
= 1 − (4.72)
T c − T w r o
where T c is the temperature at r = 0 and is given by
Gr 2 o
T c = T w + (4.73)
4k
Let us now consider a finite element solution employing linear elements. The stiffness
matrix is (Equation 4.62 without convection)
2πk r i + r j 1 −1
[K] = (4.74)
l 2 −1 1
and the forcing vector is

T
{f}= G[N] 2πr dr (4.75)
r
per unit length.
In cylindrical coordinates, r may be expressed as

r = N i r i + N j r j (4.76)
Substituting the above equation into Equation 4.75 and integrating between r i and r j ,
we obtain
2πGl 2r i + r j
{f}= (4.77)
6 r i + 2r j
where l is the length of an element.

Example 4.3.2 Calculate the surface temperature in a circular solid cylinder of radius
3
25 mm with a volumetric heat generation of 35.3 MW/m . The external surface of the cylinder
is exposed to a liquid at a temperature of 20 C with a surface heat transfer coefficient of
◦
◦
4000 W/m 2 ◦ C. The thermal conductivity of the material is 21 W/m C.
Let us divide half of the region into four elements as shown in Figure 4.11, each of width
6.25 cm.