Page 61 - Introduction to Computational Fluid Dynamics

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1D HEAT CONDUCTION
Thus, A i and B i can be calculated by recurrence. The implementation steps are
as follows:
l
1. Prepare a i , b i , and c i for i = 2to N − 1 from knowledge of the T distribution.
i
2. From comparison of Equations 2.69 and 2.71, set A 2 = a 2 and B 2 = c 2 (because
b 2 = 0 via the boundary condition specification). Now evaluate A i and B i for
i = 3to N − 1 by recurrence using Equations 2.74 and 2.75.
3. Evaluate T i by backwards substitution using Equation 2.71, that is, from
i = N − 1 to 2. Note that since we prescribe boundary conditions such that
AE N−1 = 0, it follows that A N−1 = 0.
4. Evaluate fractional change as before and go to step 1 if the convergence criterion
is not satisfied.
The TDMA is essentially a forward elimination (implicit in the recurrence
relations) and backward substitution procedure in which temperatures at all i are
updated simultaneously in step 3. Hence, the TDMA is also called a line-by-line
procedure to contrast it with the point-by-point GS procedure introduced earlier.
Further, we note that if a i , b i , and c i were constants and not functions of T then
the TDMA would yield a solution in just one iteration whereas the point-by-point
procedure would require several iterations even when coefficients are constants.

2.8.3 Applications

To illustrate performance of the methods just described, we consider two steady-
state problems. 7

Problem 2 – Rectangular Fin [80]
A rectangular fin of length 2 cm, thickness 2 mm, and breadth 20 cm is attached
to a plane wall as shown in Figure 2.10. The wall temperature T w = 225 C
◦
◦
and ambient temperature T ∞ = 25 C. For the fin material, k = 45 W/m-K and
2
the operating h = 15 W/m -K. Determine the heat loss from the fin and its
effectiveness. Assume the tip heat loss to be negligible.

Solution
The exact solution to this problem is
coshm (L − x) √
T − T ∞
= , Q loss = hP k A (T w − T ∞ )tanh(mL),
coshmL
T w − T ∞
(2.76)
√
where m = hP /kA. In our problem, perimeter P = 2 × 20 = 40 cm, area A =
2
20 × 0.2 = 4cm , and L = 2 cm. Therefore, m = 18.257m −1 and Q loss = 23 W.

7 The USER files for these problems are given in Appendix B.