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+//Problem 20.18: A 400 kVA transformer has a primary winding resistance of 0.5 ohm and a secondary winding resistance of 0.001 ohm . The iron loss is 2.5 kW and the primary and secondary voltages are 5 kV and 320 V respectively. If the power factor of the load is 0.85, determine the efficiency of the transformer (a) on full load, and (b) on half load.
+
+//initializing the variables:
+S = 400000; // in VA
+R1 = 0.5; // in Ohm
+R2 = 0.001; // in Ohm
+V1 = 5000; // in Volts
+V2 = 320; // in Volts
+Pi = 2500; // in Watt
+pf = 0.85; // power factor
+
+//calculation:
+//Rating = 400 kVA = V1*I1 = V2*I2
+//Hence primary current
+I1 = S/V1
+//secondary current
+I2 = S/V2
+//Total copper loss = I1*I1*R1 + I2*I2*R2,
+Pcf = I1*I1*R1 + I2*I2*R2
+//On full load, total loss = copper loss + iron loss
+Plf = Pcf + Pi
+// full-load power output = V2*I2*pf
+Pof = S*pf
+//Input power at full-load = output power at full-load + losses
+PIf = Pof + Plf
+//Efficiency = output power/input power = (input power—losses)/input power
+//Efficiency = 1 - losses/input power
+efff = (1-(Plf/PIf))*100
+
+//Half full-load power output = V*I*pf/2
+Poh = S*pf/2
+//Copper loss (or I*I*R loss) is proportional to current squared
+//Hence the copper loss at half full-load is
+Pch = Pcf/(2*2)
+//Iron loss = 2500 W (constant)
+//Total losses
+Plh = Pch + Pi
+//Input power at half full-load = output power at half full-load + losses
+PIh = Poh + Plh
+//efficiency
+effh = (1-(Plh/PIh))*100
+
+printf("\n\n Result \n\n")
+printf("\n (a)the transformer efficiency at full load is %.2f percent", efff)
+printf("\n (b)the transformer efficiency at half full load is %.2f percent", effh)
+\ No newline at end of file