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Unformatted text preview: ied. Determine the magnitude of the primary current. 3‐ The following data were obtained from the tests on a 12‐kVA, 480/120‐V 60 Hz two winding transformer: Open circuit test: Voltage=120 V, Current=1.71 A and power=72 W Short circuit test: Voltage=73 V, Current=25A and power=937.5 W a) Find the equivalent circuit parameters and sketch the equivalent circuit of the transformer viewed from the low voltage side and high voltage side. b) Find the efficiency of the transformer when the power factor at full load is 0.866 lagging c) What is the maximum efficiency of the transformer 4‐ Consider a single‐phase 10 KVA, 2200/220 V , 60 Hz transformer with an impedance of Zeq=10.4+j31.3 ohm referred to the high voltage side. Considering the rated values of the transformer as the base values: a) Draw per unit equivalent circuit of the transformer b) Per unit value of copper losses at full load c) Voltage regulation of the transformer if the transformer is delivering 75% of its rated power at rated voltage with a power factor of 0.6 lagging. 5‐ Three single‐phase, 50‐kVA 2400:240‐V transformers, each has the equivalent impedance of 1.42+j1.82 Ω (referred to high voltage side), are connected Y‐Y to form a three‐phase 150‐kVA bank to step down the voltage at the load end of a feeder whose impedance is 0.15+j1.00 Ω/phase. On the secondary sides, the transformers supply a balanced three‐phase load through a feeder whose impedance is 0.0005+j0.002Ω/phase. The load line‐line voltage is 416V and draws rated 150 kVA power at a power factor of 0.80 lagging. a) Determine the single‐phase equivalent circuit. b) Calculate the sending end voltage of the 2400‐V feeder. c) Calculate the transformer voltage regulation....
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This note was uploaded on 12/02/2012 for the course ECSE 2110 taught by Professor Parsa during the Spring '12 term at Rensselaer Polytechnic Institute.
- Spring '12