A plate of dimension 18cm x 18cm is subjected to temperatures as follows: left side at \(100^{0}c\), right side at \(200^{0}c\). Upper part at \(50^{0}c\), and lower at \(150^{0}c\). If square grid length of 6cm x 6cm is assumed, what will be the temperature at the interior nodes?

This answer is restricted. Please login to view the answer of this question.

Login Now

Solution:

Since height and weight of gird is same

h = k = 0.8m

The nodes are shown in Figure below:

- Hamro CSIT

Now to get the temperature at the interior nodes we have to write Equation for all the combinations of i and j, i = 1, ……., m; j = 1, ……., n-1

For u1:

u3 + u2 + 75 + 50 – 4u1 = 0    ———-(1)

-4u1 + u2 + u3 = -125

For u2:

u4 + u3 + 100 + 50 – 4u2 = 0   ———-(2)

u1 – 5u2 + u4 = -150

for u3:

u4 + u1 + 300 + 75 – 4u3 = 0   ———-(3)

u1 + u4 – 4u3 = -375

For u4:

u2 + u3 + 300 + 100 – 4u4 = 0   ———-(4)

u2 + u3 – 4u4 = -400

Equations (1) to (4) represent a set of four simultaneously linear equations, which is given below:

-4u + u2 + u3 = -125

u1 – 4u2 + u4 = -150

u1 + u4 – 4u3 = -375

u2 + u3 – 4u4 = -400

Now,

\(u_1 = \frac{u_2 + u_3 + 125}{4}\)

\(u_2 = \frac{u_1 + u_4 + 150}{4}\)

\(u_3 = \frac{u_1 + u_4 + 375}{4}\)

\(u_4 = \frac{u_2 + u_3 + 400}{4}\)

Solving above equations by using Gauss-Seidel method with initial guess u2 = 0, u3 = 0, u4 = 0, we get

Values →
Iteration ↓
u1 u2 u3 u4
1 31.25 45.313 101.563 136.719
2 67.969 88.672 144.922 158.398
3 89.648 99.512 155.762 163.818
4 95.068 102.222 158.472 165.173
5 96.423 102.899 159.149 165.512
6 96.762 103.069 159.319 165.597
7 96.847 103.111 159.361 165.618
8 96.868 103.121 159.371 165.623
9 96.873 103.124 159.374 165.625

Thus,

u1 = 96.873

u2 = 103.124

u3 = 159.374

u4 = 165.625

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .