: 500 kVA is given an open circuit test from which the iron loss is found to be 3.5k W. A short circuit test shows the copper loss to be 4.5kW with full-load current flowing. Calculate the all-day efficiency if the load cycle is as under: No load for 6 hours; 80 % load for 6 hours at 0.8 p.f.; 60% load for 8 hours at 0.75 p.f; 20% for 4 hours at 0.8 p.f.

: 500 kVA is given an open circuit test from which the iron loss is found to be 3.5k W. A short circuit test shows the copper loss to be 4.5kW with full-load current flowing. Calculate the all-day efficiency if the load cycle is as under: No load for 6 hours; 80 % load for 6 hours at 0.8 p.f.; 60% load for 8 hours at 0.75 p.f; 20% for 4 hours at 0.8 p.f.

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

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.53P

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