robertplant22
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- Feb 1, 2012
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Why are the copper losses multiplied by 4 and not by 2 in the solution to this problem? The question is as follows:
A power transformer has the following test data at rated voltage:
% of Nameplate kVA Total Losses (W)
0 460
50 2,370
The total transformer loses (W) at 100% of nameplate kVA are most nearly?
In this question I understand that the Core losses are Pcore= 460W. I also understand that to find the Copper losses, Pcu,50% at 50% loading you need to subtract the Core losses (Pcore= 460W) from the Total losses (PT,50% = 2,370W); therefore:
Pcu,50%= PT,50% - Pcore= 2,370W - 460W = 1,910W
Here is the part I dont understand:
Pcu,100%= (4)(Pcu,50%) + Pcore = (4)(1,910W) + 460W = 8,100W
A power transformer has the following test data at rated voltage:
% of Nameplate kVA Total Losses (W)
0 460
50 2,370
The total transformer loses (W) at 100% of nameplate kVA are most nearly?
In this question I understand that the Core losses are Pcore= 460W. I also understand that to find the Copper losses, Pcu,50% at 50% loading you need to subtract the Core losses (Pcore= 460W) from the Total losses (PT,50% = 2,370W); therefore:
Pcu,50%= PT,50% - Pcore= 2,370W - 460W = 1,910W
Here is the part I dont understand:
Pcu,100%= (4)(Pcu,50%) + Pcore = (4)(1,910W) + 460W = 8,100W
