Page 123 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

STEADY STATE HEAT CONDUCTION IN ONE DIMENSION
Now, on substituting the above relation into the first and third nodal equations, we have

K 21 K 23 K 12 115
K 11 − K 12 T 1 + K 13 − K 12 T 3 = f 1 − f 2 (4.53)
K 22 K 22 K 22
for the first node, and

K 21 K 23 K 32
K 31 − K 32 T 1 + K 33 − K 32 T 3 = f 3 − f 2 (4.54)
K 22 K 22 K 22
for the second node. Now the matrix form of the equation can be rewritten as
   
K 21 K 23
K 11 − K 12 K 13 − K 12  K 12 
 f 1 − f 2

 
 K 22 K 22  T 1 K 22
 = (4.55)


K 21 K 23 T 3
 
 K 32 
K 31 − K 32 K 33 − K 32 f 3 − f 2 


K 22 K 22 K 22
Note that the number of equations have been reduced, which leads to a small decrease
in computational cost. However, extending this procedure to multi-dimensional problems
is difficult and therefore not widely used.
Example 4.2.5 Repeat Example 4.2.4 using the static condensation procedure.
Substituting all relevant values into Equation 4.55 and applying the boundary condition
(T 3 = 40 C) leads to the following:
◦

700.00.0 T 1 4499.89 + 700(40)
= (4.56)
0.01 T 3 40.0
◦
The solution to the above equation results in T 1 = 46.43 C, which is identical to the
exact solution.
4.3 Radial Heat Flow in a Cylinder
Many problems in industry, such as heat exchangers, crude oil transport, and so on, involve
the flow of hot fluids in very long pipes that have uniform boundary conditions along the
circumference, both inside and outside as shown in Figure 4.10. In such problems, the
heat transfer mainly takes place along the radial direction apart from the end effects. The
governing differential equation for heat flow in cylindrical geometries is

1 d dT
rk = 0 (4.57)
r dr dr
The boundary conditions are as follows:

At r = r i ,T = T w
dT
and at r = r o , −k = h(T o − T a ) (4.58)
dr