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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Calculating the specific electric loading
+clc;
+disp('Example 6.9, Page No. = 6.13')
+// Given Data
+Pc = 1000;// Core loss (in W)
+R = 0.025;// Armature resistance (in ohm)
+l = 230;// Specific loss dissipation (in W per degree celsius per meter square)
+a = 2;// Since a=z for lap winding
+Z = 270;// Number of conductors
+L = 0.25;// Core length (in meter)
+D = 0.25;// Armature diameter (in meter)
+T = 40;// Temperature rise (degree celsius)
+// Calculation of the specific electric loading
+c = 1/l;// Cooling co-efficient
+S = %pi*D*L;// Dissipation surface (in meter square)
+Q = S*T/c;// Maximum allowable pwer dissipation from armature surface
+Ia = ((Q-Pc)/R)^(1/2);// Armature current (in Ampere)
+Iz = Ia/a;// Current in each conductor (in A)
+ac = Iz*Z/(%pi*D);// Specific electric loading
+disp(ac,'Specific electric loading (ampere conductors per meter)=');
+//in book answer is 31000 (ampere conductors per meter). The answers vary due to round off error