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+// Calculating the loss that will pass through copper bar to iron
+clc;
+disp('Example 4.1, Page No. = 4.3')
+// Given Data
+D = 12;// Diameter of copper bar in mm
+t = 1.5;// Thickness of micanite tube in mm
+p = 8;// Resistivity of macanite tube in ohm*meter
+T = 25;// Temperature difference in degree celsius
+L = 0.2;// Length of copper bar
+// Calculation of loss.that will pass through copper bar to iron
+S = %pi*(D+t)*10^(-3)*L;//Area of insulation in the path of heat flow
+R =( p*t*10^(-3))/S;// Thermal resistance of micanite tube
+Q_con= T/R;// Heat Dissipated
+disp(Q_con,'Heat Dissipated(W)=');
+//in book answer is 17.67 W. The answers vary due to round off error