% All day n = Output energy in kWh during a day. Input energy in kWh during a day × 100 Output energy in kWh during a day × 100 Output energy+ Energy spent for total losses Q15-A 1000 KVA transformer During the day it is loaded as follows: For 10 hours 600 KW at 0.8 p.f lag For 8 hours For 6 hours 450 KW at 0.9 p.f lag 225 KW at 0.9 p.f lag If the iron loss equal 1.8Kw and total loss equal 10 KW, determine its all day efficiency

% All day n = Output energy in kWh during a day. Input energy in kWh during a day × 100 Output energy in kWh during a day × 100 Output energy+ Energy spent for total losses Q15-A 1000 KVA transformer During the day it is loaded as follows: For 10 hours 600 KW at 0.8 p.f lag For 8 hours For 6 hours 450 KW at 0.9 p.f lag 225 KW at 0.9 p.f lag If the iron loss equal 1.8Kw and total loss equal 10 KW, determine its all day efficiency

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

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.1P: Given the complex numbers A1=630 and A2=4+j5, (a) convert A1 to rectangular form: (b) convert A2 to...

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