Allday= Allday 240/P Allday 24 0/P+ (Pivon) 14 + (PW)24 A 800 KVA transformer During the day it is loaded as follows: For 4 hours For 4 hours For 4 hours 500 KW at 0.8 p.flag 700 KW at 0.9 p.flag 300 KW at 0.95 p.flag If the iron loss equal 7.5Kw and copper loss equal 14.2 KW, determine its all day efficiency. ANS. 95.86%

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I need the solution to this problem by an expert using the above-mentioned law.
Allday=
Allday 240/P
Allday 24 0/P+ (Pivon) 14 + (PW)24
A 800 KVA transformer During the day it is loaded as follows:
For 4 hours
For 4 hours
For 4 hours
500 KW at 0.8 p.flag
700 KW at 0.9 p.flag
300 KW at 0.95 p.flag
If the iron loss equal 7.5Kw and copper loss equal 14.2 KW, determine its all day
efficiency.
ANS. 95.86%
Transcribed Image Text:Allday= Allday 240/P Allday 24 0/P+ (Pivon) 14 + (PW)24 A 800 KVA transformer During the day it is loaded as follows: For 4 hours For 4 hours For 4 hours 500 KW at 0.8 p.flag 700 KW at 0.9 p.flag 300 KW at 0.95 p.flag If the iron loss equal 7.5Kw and copper loss equal 14.2 KW, determine its all day efficiency. ANS. 95.86%
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