100° T1 T2 60° 40° T3 T4 0° (i) Using the Gauss elimination...

100° T1 T2 60° 40° T3 T4 0° (i) Using the Gauss elimination or Gauss-Jordan elimination methods, determine the temperature of the interior nodes [Hint: first write down the nodal equations using the finite difference method. Now you have four equations. Afterwards, make the augmented matrix and apply either the Gauss elimination or Gauss-Jordan elimination methods. This way you will get the solutions for all interior nodes.] (ii) Using the Jacobi iteration method, determine the temperature of the interior nodes (perform only three iterations)

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00:01 Now, according to the given quotient, i can write t1, t3, t3, this is t4, this is 40 degree.

00:30 That implies i can write here that is t1, t3, t3, t4.

00:54 So, in the right side, this is 40 degree, this is 40 degree, this is 40 degree, this is 0 degree, this is 0 degree, this is 0 degree, 100 degree, this is 100 degree and this is 100 degree.

01:13 Now, at first using finite differentiation i can write.

01:21 So, finite differentiation, finite difference method we can write t1 equal to 1 by 4 60 degree plus 100 degree plus t3 plus t2.

01:53 So, that imply i can write 4t1 minus t2 minus t3 equal to 160.

02:06 This is equation number 1.

02:10 So, again i can write also i can write t2 equal to 1 by 4 100 degree plus 40 degree plus t1 plus t4.

02:34 So that imply 4t2 minus t1 minus t4 equal to 140.

02:46 Second equations.

02:50 Also i can write t3 equal to 1 by 4, 60 plus 0 plus t1 plus t4.

03:05 So, that implies 4t3 equal to or 4t3 minus t1 minus t4 equal to 60.

03:20 This is relation 3.

03:24 Also, i can write t4 equal to 1 by 4, 40 plus 0 plus t2 plus t3.

03:44 That implies 4t4 minus t2 minus t3 equal to 40.

03:55 This is relation number.

03:56 Now, so from 1 forming matrices, so from 1 we can form the matrices that is matrix a equal to 4 minus 1 minus 1, 0.

04:23 Minus 1, 4, 0, minus 1, minus 1, 0, 4, minus 1, 0, minus 1, minus 1 and b equal to also i can write 1 6 0 1 4 0 60 40 only the column so we know that a x a x equal to b that imply x equal to to matrix ab.

05:16 Therefore we can write that is t1, t2, t3, t4 single column that equal to i can write here 4 minus 1 minus 1 0 1 60 minus 1 4 0 minus 1 1 40 minus 1 0 4 minus 1 60 0 0 minus 1 minus 1 4 40.

06:12 Now so we have to do the operations at first c1 equal to c2 minus c1 and c3 equal to that is c4 by 4 plus c3.

06:34 So, t1, t2, t3, t4 that equal to so i can write here 4, 0, minus 1, 0, 1, 6, 0...

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