A 500-kVA, 3-phase, 50-Hz transformer has a voltage ratio (line voltage) of 33/11-kV and is delta/star connected. The resistances per phase are: high voltage 35 Ω, low voltage 0.876 Ω and the iron loss is 3050W. Calculate the value of efficiency at full-load and one-half of full-load respectively (a) at unity p.f. and (b) 0.8 p.f.
Transformation ratio
Per phase
Secondary phase current
Full-load condition
Full load total Cu loss,
= 4,490 W;
Iron loss = 3,050 W
Total full-load losses
= 4,490 +3,050 = 7,540 W;
Output at unity p.f. = 500 kW
∴ F.L. efficiency = 0.9854 or 98.54%;
Output at 0.8 p.f.= 400 kW
∴ Efficiency = 0.982 or 98.2%
Half-load condition
Output at unity p.f.= 250 kW
Cu losses = 1,222 W
Total losses = 3,050 + 1,122 = 4,172 W
∴ = 0.9835 = 98.35%
Output at 0.8 p.f. = 200 kW
= 0.98 or 98%
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