% 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

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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