Part II $(70$ pt$)$ In a steady$-$state heat transfer problem $($Figure $2),$ the temperatures at $10$ nodal

points $(T_1,T_2,T_3,$dots,$T_{10})$ can be calculated using the following equations $(R$ : thermal resistance

and $c$

Figure $2.$ Steady$-$state heat transfer $($thermal resistance model$)$

Node $1$: $\frac{T_1-T_{,1}}{R_{,1}}+\frac{T_1-T_2}{R_{1-2}}=q_1$

Node $10$: $\frac{T_{10}-T_{,2}}{R_{,2}}+\frac{T_{10}-T_9}{R_{9-10}}=q_{10}$

Node $q_1=q_{10}=5T_{,1}=270,T_{,2}=310R_{,1}=1\mathrm{.}5,R_{,2}=1\mathrm{.}0R_{1-2}=R_{2-3}=R_{3-4}=0\mathrm{.}2$;$R_{4-5}=R_{5-6}=R_{6-7}=0\mathrm{.}4$;$R_{7-8}=R_{8-9}=R_{9-10}=0\mathrm{.}6AT=b.(T^T=[T_1{T}_2{T}_{3}dots{T}_{10}])A1-10={10}^{-8}{T}_i(T_i=T_i^{(k+1)}-T_i^{(k)})(>1)1$

Boundary Temperatures: $T_{,1}=270,T_{,2}=310$

Thermal resistance: $(Boundaries)R_{,1}=1\mathrm{.}5,R_{,2}=1\mathrm{.}0$

$(Conduction)R_{1-2}=R_{2-3}=R_{3-4}=0\mathrm{.}2$;$R_{4-5}=R_{5-6}=R_{6-7}=0\mathrm{.}4$;$R_{7-8}=R_{8-9}=R_{9-10}=0\mathrm{.}6$

Present your answers using $at$ least eight decimal places.

$(a)$ Express the system $of$ the linear equations $in$ the form $ofAT=b.(T^T=[T_1{T}_2{T}_{3}dots{T}_{10}])$

Find the coefficient matrix A and the nonhomogeneous vector $b.(using$ 'for', $\text{'}if\text{'}$

statements the code can $be$ much reduced$).$

$(b)$ Find the temperatures $at$ nodes $1-10,$ solving the system $of$ the linear equations $by$ the

Gauss elimination method with MATLAB code.

Now, solve the system $of$ linear equations $by$ iterative methods using MATLAB code. Use

the convergence criterion $={10}^{-8}.$ The maximum $absolute$ value $of{T}_i$ will $be$ compared with

$(T_i=T_i^{(k+1)}-T_i^{(k)}).$

$(c)$ Find the temperatures using Jacobi iteration and the number $of$ iterations.

$(d)$ Find the temperatures using Gauss$-$Seidel iteration and the number $of$ iterations.

$(e)$ Find the temperatures using the successive$-$over$-$relaxation $(SOR)$ method. Test $at$

least $5$ relaxation factors, $(>1)$ and find $$ that can lead $to$ the least number $of$

iterations.