Question

A 50 KVA, TRANSFORMER WITH A TRANSFORMATION RATIO OF 20 IS TESTED FOR EFFICIENCY AND REGULATION BY PERFORMING OPEN CIRCUIT AND SHORT CIRCUIT TESTS. ON OPEN CIRCUIT TEST, THE AMMETER, VOLTMETER AND WATTMETER READINGS ARE 0.6 A, 230 VOLTS, 300 WATTS RESPECTIVELY. ON A SHORT CIRCUIT TEST, THE AMMETER, VOLTMETER AND WATTMETER READINGS ARE 9.87 A, 150 V, 600 WATTS RESPECTIVELY. CALCULATE THE EFFICIENCY OF THE TRANSFORMER IF IT OPERATES AT 20% OVERLOAD AND 85% POWER FACTOR.

Answer 1

Given data;Transformer rating = 50 KVA,Transformation ratio = 20Open circuit test results;Ammeter, I0 = 0.6 AVoltmeter, V0 = 230 VWattmeter, W0 = 300 WShort circuit test results;Ammeter, Isc = 9.87 AVoltmeter, Vsc = 150 VWattmeter, Wsc = 600 WEfficiency of the transformer;Efficiency of transformer = output power / input powerWe know that, the efficiency of transformer on full load is given by;Efficiency, η = output power / input powerOn full load, output power = input powerFrom the short circuit test, the copper losses = Wsc = 600 W and this is at normal voltage (230 V)From the open circuit test, the iron losses = W0 = 300 W and this is at normal voltage (230 V)Now, the full load current can be calculated as follows;Full load current, Ifl = Transformer rating / (√3 * LV)Where;LV = low voltage side voltage = normal voltage / transformation ratio = 230 V / 20 = 11.5 VTherefore;Ifl = 50 × 10^3 / (√3 × 11.5 V)Ifl = 2295.94 AAt 20% overload, the current drawn, I2 = 1.2 × 2295.94 A = 2755.12 AAt 85% power factor;True power = apparent power × power factor = 50 × 10^3 × 0.85 = 42,500 WApparent power, S = √3 × LV × Ifl = √3 × 11.5 × 2295.94S = 71,059.92 VAThe full load efficiency is calculated using the formula;η = output power / input power = (output power) / (output power + copper losses + iron losses)On full load, output power = input power, therefore;ηfl = output power / input power = output power / (output power + copper losses + iron losses)ηfl = 50 × 10^3 / (50 × 10^3 + 300 + 600)ηfl = 96.61%At 20% overload, the efficiency of transformer can be calculated as follows;η2 = (true power / apparent power) / [1 + (copper losses / (apparent power)^2)]From the data given;Copper losses = Wsc = 600 WApparent power, S = 71,059.92 VACurrent, I2 = 2755.12 APower factor = 0.85Therefore;η2 = (42,500 / 71,059.92) / [1 + (600 / (71,059.92)^2)]η2 = 0.6698 or 66.98%Therefore, the efficiency of transformer when it operates at 20% overload and 85% power factor is 66.98%.