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# Heat Map Modeling Using Resistive Network

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2D steady state and dynamic temperature distribution modeling with heat sources and sinks using both thermal resistance and capacitance.

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### Heat Map Modeling Using Resistive Network

1. 1. Steven Surgnier<br />Advanced Numerical Methods<br />Final Project – 4/3/2010<br />Find the steady-state temperature distribution on a rectangular printed circuit board 8 cm×10 cm with two power sources as shown in the figure below. <br />Assume: <br />(a) Thermal resistance per square is R□=0.5 K/W <br />(b) Thermal conductance per square centimeter between PC board and ambient is GTH=2.5 W/(K×cm2) <br />(c) The radiation is proportional to area A and it is the function of the absolute temperature Prad=3×10-10×T4 W/cm2 <br />(d) Both initial and radiator temperatures are 10C <br />Power dissipation P1=2W and P2=5W as shown in the figure below. Please use 81×101 uniform mesh. <br />(1) Analyze the temperature distribution in the printed circuit board. Plot the temperature distribution <br />(2) Consider the thermal capacitance of each node is C=0.001 W×s/K. Plot the temperature distribution after 1, 2, 3, 4 and 5 minutes <br />-34290021209000In order to secure accurate solution please use very small time increment (like 0.1ms) <br /><ul><li>Steady State
2. 2. clear all; format compact;n=101; m=81; T=zeros(n,m)+10+273;P=zeros(n,m); Ta=10+273;A = 1/100; % unit surface area cm^2K=A*3e-10; G=A*2.5; R=0.5;P(21:46,16:41)=5; % 5 Watt sourceP(71:91,41:61)=2; % 2 Watt sourcefor ite=1:500 for j=2:n-1 for i=2:m-1 t=T(j-1,i)+T(j+1,i)+T(j,i-1)+T(j,i+1); T(j,i)= (t/R + Ta*G - K*T(j,i)^4 + P(j,i))/(4/R + G); end; end; T(1:56,61:81)=20+273; % radiator temp %Boundary conditions: T(n,:)=T(n-1,:); T(1,:)=T(2,:); T(:,m)=T(:,m-1); T(:,1)=T(:,2);end;figure(8); mesh(T-273); axis ([0 81 0 101 10 130]);zlabel('Temp (C)'); title('Steady State'); view(43,28);
3. 3. Transient Response
4. 4. clear all; format compact; clc;n=101; m=81; T=zeros(n,m)+10+273;T1=zeros(n,m); T2=zeros(n,m); T3=zeros(n,m);T4=zeros(n,m); T5=zeros(n,m);P=zeros(n,m); SS=zeros(n,m);A = 1/100; % unit surface area cm^2Ta=10+273; K=A*3e-10; G=A*2.5; R=0.5; C=0.001; P(21:46,16:41)=5; % 5 Watt sourceP(71:91,41:61)=2; % 2 Watt sourceTs=70e-3; % 70 ms simulation timedt=100e-6; % time stepk=Ts/dt;for x=1:k for j=2:n-1 for i=2:m-1 t=T(j-1,i)+T(j+1,i)+T(j,i-1)+T(j,i+1); % cm^2 area Tx=T(j,i); p = (t-4*Tx)/R + (Ta-Tx)*G - K*Tx^4 + P(j,i); T(j,i)=Tx+dt*(p)/C; end; end; T(1:56,61:81)=20+273; % radiator temp% Boundary conditions: T(n,:)=T(n-1,:); T(1,:)=T(2,:); T(:,m)=T(:,m-1); T(:,1)=T(:,2); if (x==1e-3/dt) T1=T; end; % after 1 ms if (x==5e-3/dt) T2=T; end; % after 5 ms if (x==10e-3/dt) T3=T; end; % after 10 ms if (x==20e-3/dt) T4=T; end; % after 20 ms if (x==40e-3/dt) T5=T; end; % after 50 msend;figure(1); mesh(T1-273); axis ([0 81 0 101 10 130]); zlabel('Temp (C)'); title('After 1 ms'); view(43,28);figure(2); mesh(T2-273); axis ([0 81 0 101 10 130]); zlabel('Temp (C)'); title('After 5 ms'); view(43,28);figure(3); mesh(T3-273); axis ([0 81 0 101 10 130]); zlabel('Temp (C)'); title('After 10 ms'); view(43,28);figure(4); mesh(T4-273); axis ([0 81 0 101 10 130]); zlabel('Temp (C)'); title('After 20 ms'); view(43,28);figure(5); mesh(T5-273); axis ([0 81 0 101 10 130]); zlabel('Temp (C)'); title('After 40 ms'); view(43,28);figure(6); mesh(T-273); axis ([0 81 0 101 10 130]); zlabel('Temp (C)'); title('After 70 ms'); view(43,28);
5. 5. -4000526320753427095273685034080455346700-590555346700341757031750-4953031750