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20 kVA transformer iron loss, efficiency

Short answer

Iron loss Pi=250²/595.2=105 W. At half load and 0.8 pf: output 8000 W, copper loss 80 W, efficiency=8000/(8000+105+80)=97.7%.

A 20 kVA, 250/2500 V, 50 Hz, single phase transformer has Ro = 595.2 Ω, Xo = 187.26 Ω, Ro1 = 0.05 Ω and Xo1 = 0.14 Ω. Calculate (i) iron loss of the transformer (ii) efficiency of the transformer at half full load with 0.8 pf lagging.

Given: 20 kVA, 250/2500 V, 50 Hz. Ro = 595.2 Ω, Xo = 187.26 Ω (the no-load branch, referred to the 250 V primary); Ro1 = 0.05 Ω, Xo1 = 0.14 Ω (series, referred to primary).

(i) Iron loss

The iron loss is the power consumed in Ro at the rated primary voltage 250 V:

Pi = V1²/Ro = 250²/595.2 = 105.0 W

(ii) Efficiency at half load, 0.8 pf lagging

Rated primary current  I1 = 20 000/250 = 80 A
Half-load current      = 40 A
Cu loss = I² Ro1 = 40² × 0.05 = 80 W
Output  = 0.5 × 20 000 × 0.8 = 8000 W
η = 8000/(8000 + 105 + 80) = 8000/8185 = 97.74 %

Check: full-load Cu loss = 80² × 0.05 = 320 W, and at half load it is 0.25 × 320 = 80 W.

Answer: (i) iron loss = 105 W; (ii) η ≈ 97.7 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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