Q1// A 150 kVA, 2400/240V, single phase transformer has the following circuit model parameters: R, = 0.2 Q, R2= 2m 2, X, = 0.45 2, X2 = 0.45m N, and the iron loss is 2.427kw. Using the equivalent circuit referred to secondary, calculate: a) Voltage regulation. b) Efficiency at 2/3 full load at unity power factor. c) The kVA load for max. efficiency and value of max. efficiency at a power factor of 0-8 lagging. %3D %3D

Q1// A 150 kVA, 2400/240V, single phase transformer has the following circuit model parameters: R, = 0.2 Q, R2= 2m 2, X, = 0.45 2, X2 = 0.45m N, and the iron loss is 2.427kw. Using the equivalent circuit referred to secondary, calculate: a) Voltage regulation. b) Efficiency at 2/3 full load at unity power factor. c) The kVA load for max. efficiency and value of max. efficiency at a power factor of 0-8 lagging. %3D %3D

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 3.57P: A two-winding single-phase transformer rated 60kVA,240/1200V,60Hz, has an efficiency of 0.96 when...

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A two-winding single-phase transformer rated 60kVA,240/1200V,60Hz, has an efficiency of 0.96 when operated at rated load, 0.8 power factor lagging. This transformer is to be utilized as a 1440/1200-V step-down autotransformer in a power distribution system. (a) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the ratings as a two-winding transformer. Assume an ideal transformer. (b) Determine the efficiency of the autotransformer with the kVA loading of part (a) and 0.8 power factor leading.
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