500 kVA transformer efficiency full/half load

The efficiency is found from output ÷ (output + iron loss + copper loss). The power factor is not given, so unity power factor is assumed (the result at 0.8 lagging is also shown).

Rated currents

• Primary: I₁ = 500 000 / 6600 = 75.76 A
• Secondary: I₂ = 500 000 / 400 = 1250 A

Copper loss at full load

• Primary: I₁²R₁ = (75.76)² × 0.4 = 2295.7 W
• Secondary: I₂²R₂ = (1250)² × 0.001 = 1562.5 W
• Total P_cu(fl) = 3858.2 W ≈ 3.858 kW
• Iron loss P_i = 3000 W (constant, same at all loads)

(a) Full load (unity pf)

Output   = 500 kVA x 1.0           = 500 000 W
Losses   = 3000 + 3858.2           = 6858.2 W
Input    = 500 000 + 6858.2        = 506 858.2 W
η        = 500 000 / 506 858.2     = 0.9865

Efficiency at full load = 98.65 %

(b) Half load (unity pf)

Copper loss varies with (load)², so P_cu = (0.5)² × 3858.2 = 964.5 W.

Output   = 250 000 W
Losses   = 3000 + 964.5            = 3964.5 W
η        = 250 000 / 253 964.5     = 0.9844

Efficiency at half load = 98.44 %

Answer: η(full load) ≈ 98.65 %, η(half load) ≈ 98.44 % at unity pf.