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authorpriyanka2015-06-24 15:03:17 +0530
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+//Chapter 21, Problem 19
+clc
+k=400000 //transformer rating
+v1=5000 //primary current
+v2=320 //secondary current
+r1=0.5 //resistance in ohm
+r2=0.001 //resistance in ohm
+lfe=2500 //iron loss
+pf=0.85 //power factor
+i1=k/v1 //primary current
+i2=k/v2 //secondary current
+lcu=(i1^2*r1)+(i2^2*r2) //total copper loss
+lt=lcu+lfe //total loss
+pt=k*pf //total output power
+pi=pt+lt //input power
+n=(1-(lt/pi))*100 //efficiency
+lc=lcu*(1/2)^2 //total copper loss at half load
+lh=lc+lfe //total loss at half loss
+ph0=(1/2)*pt //output power at half load
+phi=(ph0+lh) //input power at half load
+n1=(1-(lh/phi))*100 //efficiency
+printf("(a) Efficiency on full load = %.3f percent\n\n",n)
+printf("(b) Efficiency at half load = %.3f percent\n\n",n1)