Answer to Question 7

A 250 kVA transformer has a full load copper loss of 3.0 kW and an iron loss of 2.0 kW. It has a power factor of 0.80. Calculate: (a) The output apparent power at which the efficiency of the transformer is at a maximum; (b) The maximum percentage efficiency.

(a) Let x be the fraction of the load at which the efficiency is at a maximum: Copper loss = 3.0 kW × x^{2} Copper loss = iron loss = 2.0 kW x^{2} = 2.0 kW ÷ 3.0 kW = 0.667 x = 0.8165
The output apparent power S = 0.8165 × 250 kVA = 204.1 kVA
(b) Now work out the useful power: P = 0.80 × 204.1 kVA = 163.3 kW
Total loss = 2.0 kW + 2.0 kW = 4.0 kW
Input power = 163.3 kW + 4.0 kW = 167.3 kW
Efficiency = 1  (4.0 kW ÷ 167.3 kW) = 0.9755 = 97.55 %
