II. , The parameters of the equivalent circuit of a 150-kVA, 22000/400-V transformer are: R = 0.3 n X = 0.55 0 Re = 8 kn R2 = 2 x 10-3n X = 5.5 x 10-3n Xm = 1.5 ka (seen from 22000-V side) Calculate: 1) Open-circuit current, power and power factor when LV is excited at rated voltage 2) The voltage at which the HV should be excited to conduct a short-circuit test (LV shorted) with full load current flowing. What is the input power and its power factor?

II. , The parameters of the equivalent circuit of a 150-kVA, 22000/400-V transformer are: R = 0.3 n X = 0.55 0 Re = 8 kn R2 = 2 x 10-3n X = 5.5 x 10-3n Xm = 1.5 ka (seen from 22000-V side) Calculate: 1) Open-circuit current, power and power factor when LV is excited at rated voltage 2) The voltage at which the HV should be excited to conduct a short-circuit test (LV shorted) with full load current flowing. What is the input power and its power factor?

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.11P: For the transformer in Problem 3.10. The open-circuit test with 11.5 kV applied results in a power...

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A single-phase 100-kVA,2400/240-volt,60-Hz distribution transformer is used as a step-down transformer. The load, which is connected to the 240-volt secondary winding, absorbs 60 kVA at 0.8 power factor lagging and is at 230 volts. Assuming an ideal transformer, calculate the following: (a) primary voltage, (b) load impedance, (c) load impedance referred to the primary, and (d) the real and reactive power supplied to the primary winding.
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An ideal transformer has no real or reactive power loss. (a) True (b) False
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