Question

A Delta-Y connected bank of three identical 150-kVA, 2400-V:120-V, 60-Hz transformers is supplied at its high-voltage terminals through a feeder whose impedance is 6.4 ohm + j154m ohm per phase. The voltage at the sending end of the feeder is held constant at 2400 V line to line. The results of a single-phase short-circuit test on one of the transformers with its low-voltage terminals short-circuited are VH = 131 V IH = 62.5 A P = 1335 W

Calculate the series impedance of this three-phase transformer bank as referred to its high-voltage terminal.

Answer 1

The series impedance of the three-phase transformer bank can be calculated by referring to the high-voltage terminal. Using the per-unit values of voltage and current obtained from the short circuit test results, the series impedance is found to be 21.84 times the rated impedance of one transformer.

Step-by-step explanation:

The series impedance of the three-phase transformer bank can be calculated by referring to the high-voltage terminal. To do this, we need to calculate the per-unit values of voltage and current based on the short circuit test results.

First, we calculate the per-unit voltage, which is given by the formula:

Vpu = Vtest / Vbase, where Vtest is the test voltage and Vbase is the base voltage. In this case, the test voltage is 131 V and the base voltage is 120 V. So, Vpu = 131 / 120 = 1.092.

Next, we calculate the per-unit current, which is given by the formula:

Ipu = Itest / Ibase, where Itest is the test current and Ibase is the base current. In this case, the test current is 62.5 A and the base current is the rated current of the transformer, which is 150 kVA / 120 V = 1250 A. So, Ipu = 62.5 / 1250 = 0.05.

Finally, we calculate the per-unit impedance, which is given by the formula:

Zpu = Vpu / Ipu = 1.092 / 0.05 = 21.84.

Therefore, the series impedance of the three-phase transformer bank as referred to its high-voltage terminal is 21.84 times the rated impedance of one transformer.