500 kVA delta/star transformer efficiency

The copper loss is found from the phase currents of the delta (HV) and star (LV) windings. The resistances are given per phase, so the loss is 3 × I_ph² × R. The load is taken at 0.8 pf lagging.

Rated phase currents

• HV (delta): phase voltage = 33 kV, so I_ph1 = 500 000 / (3 × 33 000) = 5.051 A
• LV (star): I_ph2 = I_L = 500 000 / (√3 × 11 000) = 26.24 A

Full-load copper loss

• HV: 3 × (5.051)² × 35 = 2678.3 W
• LV: 3 × (26.24)² × 0.876 = 1809.9 W
• P_cu(fl) = 4488.2 W
• Iron loss P_i = 3050 W

Full load at 0.8 pf

Output = 500 000 x 0.8             = 400 000 W
Losses = 3050 + 4488.2             = 7538.2 W
η      = 400 000 / 407 538.2       = 0.9815

η(full load) = 98.15 %

Half load at 0.8 pf

P_cu = (1/2)² × 4488.2 = 1122.1 W

Output = 200 000 W
Losses = 3050 + 1122.1             = 4172.1 W
η      = 200 000 / 204 172.1       = 0.9796

η(half load) = 97.96 %

Answer: 98.15 % at full load and 97.96 % at half load.