// Estimating the hot spot temperature clc; disp('Example 4.7, Page No. = 4.14') // Given Data l = 1;// Length of mean turn in meter Sf = 0.56;// Space Factor p = 120;// Total loss in the coil in Watt pi = 8;// Thermal resistivity in ohm*meter A = 100*50;// Area of cross-section in mm square t = 50*10^(-3);// Thickness of coil in meter // Calculation of the temperature of the hot spot pe = pi*(1-Sf^(1/2));// Effective thermal resistivity in ohm*meter V = A*l*10^(-6);// Volume of coil(in meter cube) q = p/V;// Heat dissipated in Watt per meter cube T0 =q*pe*t*t/8;// Assuming equal inward and outward heat flows disp(T0,'Temperature of the hot spot (degree celsius)='); //in book answers is 15 degree celsius. The answers vary due to round off error