ADI Method

Use the ADI method to solve for the temperature of the plate in Examples 29.1 and 29.2. At t = 0, assume that the temperature of the plate is zero and the boundary temperatures are instantaneously brought to the levels shown in Fig. 29.4. Employ a time step of 10 s. Recall from Example 30.1 that the coefficient of thermal diffusivity for aluminum is $k=0.835 cm ^{2} / s$ .

Question

ADI Method

Use the ADI method to solve for the temperature of the plate in Examples 29.1 and 29.2. At t = 0, assume that the temperature of the plate is zero and the boundary temperatures are instantaneously brought to the levels shown in Fig. 29.4. Employ a time step of 10 s. Recall from Example 30.1 that the coefficient of thermal diffusivity for aluminum is [latex]k=0.835 cm ^{2} / s[/latex].

Accepted Answer

A value of Δx = 10 cm was employed to characterize the 40 × 40-cm plate
from Examples 29.1 and 29.2. Therefore, [latex]\lambda=0.835(10) /(10)^{2}=0.0835[/latex]. For the first step to t = 5 (Fig. 30.11a), Eq. (30.20)

[latex]-\lambda T_{i, j-1}^{l+1 / 2}+2(1+\lambda) T_{i, j}^{l+1 / 2}-\lambda T_{i, j+1}^{l+1 / 2}=\lambda T_{i-1, j}^{l}+2(1-\lambda) T_{i, j}^{l}+\lambda T_{i+1, j}^{l}[/latex] (30.20)

is applied to nodes (1, 1), (1, 2), and (1, 3) to yield the following tridiagonal equations:

[latex]\left[\begin{array}{ccc}2.167 & -0.0835 & \-0.0835 & 2.167 &-0.0835 \& -0.0835 & 2.167\end{array} ight]\left\{\begin{array}{l}T_{1,1} \T_{1,2} \T_{1,3}\end{array} ight\}=\left\{\begin{array}{r}6.2625 \6.2625 \14.6125\end{array} ight\}[/latex]

which can be solved for

[latex]T_{1,1}=3.01597 \quad T_{1,2}=3.2708 \quad T_{1,3}=6.8692[/latex]

In a similar fashion, tridiagonal equations can be developed and solved for

[latex]T_{2,1}=0.1274 \quad T_{2,2}=0.2900 \quad T_{2,3}=4.1291[/latex]

and

[latex]T_{3,1}=2.0181 \quad T_{3,2}=2.2477 \quad T_{3,3}=6.0256[/latex]

For the second step to t = 10 (Fig. 30.11b), Eq. (30.22) is applied to nodes (1, 1), (2, 1), and (3, 1) to yield

[latex]\left[\begin{array}{ccc}2.167 & -0.0835 & \-0.0835 & 2.167 &-0.0835 \& -0.0835 & 2.167\end{array} ight]\left\{\begin{array}{l}T_{1,1} \T_{2,1} \T_{3,1}\end{array} ight\}=\left\{\begin{array}{c}12.0639 \0.2577 \8.0619\end{array} ight\}[/latex]

which can be solved for

[latex]T_{1,1}=5.5855 \quad T_{2,1}=0.4782 \quad T_{3,1}=3.7388[/latex]

Tridiagonal equations for the other rows can be developed and solved for

[latex]T_{1,2}=6.1683 \quad T_{2,2}=0.8238 \quad T_{3,2}=4.2359[/latex]

and

[latex]T_{1,3}=13.1120 \quad T_{2,3}=8.3207 \quad T_{3,3}=11.3606[/latex]

The computation can be repeated, and the results for t 5 100, 200, and 300 s are depicted in Fig. 30.12a through c, respectively. As expected, the temperature of the plate rises. After a suffi cient time elapses, the temperature will approach the steady-state distribution of Fig. 29.5.

Question 30.5: ADI Method Use the ADI method to solve for the temperature o...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions