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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc,clear
+printf('Example 3.3\n\n')
+
+//no load
+I=14 //input current
+V=230
+power_output_FL = 45*10^3
+power_input=V*I
+I_sh=2.55 //field current
+R_a=0.032 //armature resistance
+I_a=I-I_sh
+cu_loss_NL = I_a^2*R_a //no load copper loss
+brush_loss=2*I_a
+constant_loss= power_input - cu_loss_NL - brush_loss
+
+//full load
+
+//I=I_a+ 2.55
+//Motor input= Motor output + constant loss + brush loss + cu loss
+// solving for I_a , I_a^2 - 7125 I_a + 1487700.3 =0
+p=[1 -7125 1487700.3]
+roots(p)
+I_a=ans(2) //ignoring second root as its too large
+I=I_a+I_sh
+printf('Full load current is %.2f A\n',I)
+power_input=V*I
+eta=100*(power_output_FL/power_input)
+printf('Efficiency at full load is %.2f percent',eta)