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Since height and weight of gird is same

h = k = 0.8m

The nodes are shown in Figure below:

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 u_{1}:

u_{3} + u_{2} + 75 + 50 – 4u_{1} = 0 ———-(1)

-4u_{1} + u_{2} + u_{3} = -125

For u_{2}:

u_{4} + u_{3} + 100 + 50 – 4u_{2} = 0 ———-(2)

u_{1} – 5u_{2} + u_{4} = -150

for u_{3}:

u_{4} + u_{1} + 300 + 75 – 4u_{3} = 0 ———-(3)

u_{1} + u_{4} – 4u_{3} = -375

For u_{4}:

u_{2} + u_{3} + 300 + 100 – 4u_{4} = 0 ———-(4)

u_{2} + u_{3} – 4u_{4} = -400

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

-4u_{1 } + u_{2} + u_{3} = -125

u_{1} – 4u_{2} + u_{4} = -150

u_{1} + u_{4} – 4u_{3} = -375

u_{2} + u_{3} – 4u_{4} = -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 u_{2} = 0, u_{3} = 0, u_{4} = 0, we get

Values →Iteration ↓ |
u_{1} |
u_{2} |
u_{3} |
u_{4} |

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,

u_{1} = 96.873

u_{2} = 103.124

u_{3} = 159.374

u_{4} = 165.625

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