I need a code for problem 10. ,so reference problem 9. Thank you!!! The concept

Question

I need a code for problem 10. ,so reference problem 9.

Thank you!!!

The concept of thermal resistance described in Problem 8 can be used to nd the temperature distribution in the at square plate shown in Figure P9(a). The plate's edges are insulated so that no heat can escape, except at two points where the edge temperature is heated to Ta and Tb, respectively. The temperature varies through the plate, so no single point can describe the plate's temperature. One way to estimate the temperature distribution is to imagine that the plate consists of four subsquares and to compute the temperature in each subsquare. Let R be the thermal resistance of the material between the centers of adjacent subsquares. Then we can think of the problem as a network of electric resistors, as shown in part (b) of the gure. Let qij be the heat ow rate between the points whose temperatures are Ti and Tj. If Ta and Tb remain constant for some time, then the heat energy stored in each subsquare is constant also, and the heat ow rate between each subsquare is constant. Under these conditions, conservation of energy says that the heat ow into a subsquare equals the heat ow out. Applying this principle to each subsquare gives the following equations. qa1 = q12 + q13 q12 = q24 q13 = q34 q34 + q24 = 4b Substituting q = (Ti - Tj)/R. we nd that R can be canceled out of every equation, and they can be rearranged as follows: T1 = 1/3(Ta + T2 + T3) T2 = 1/2 (T1 + T4) T3 =1/2 (T1 + T4) T4 =1/3 (T2 + T3 + T5) Use the averaging principle developed in Problem 9 to nd the temperature distribution of the plate shown in Figure P10. using the 3 times 3 grid and the given values Ta = 150 degree C and Tb = 20 degree C. Figure P10

Explanation / Answer

close all clear all clc Ta = 150; Tb = 20; %writing the equating for temperature in matrix form A= [ 3 -1 0 -1 0 0 0 0 0; -1 3 -1 0 -1 0 0 0 0; 0 -1 2 0 0 -1 0 0 0; -1 0 0 3 -1 0 -1 0 0; 0 -1 0 -1 4 -1 0 -1 0; 0 0 -1 0 -1 3 0 0 -1; 0 0 0 -1 0 -1 2 0 0; 0 0 0 0 -1 0 -1 3 -1; 0 0 0 0 0 -1 0 -1 3 ]; B = [Ta/3 0 0 0 0 0 0 0 Tb/3]; T = inv(A)*B' %please rate first

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