500 kVA transformer efficiency, full and half load

Efficiency is output divided by input. Output = $x \times$ kVA $\times$ p.f., and losses = iron loss + $x^2 \times$ full-load copper loss, where $x$ is the fraction of full load.

The power factor is not given, so unity is assumed first (values for 0.8 lagging are added at the end).

Full-load currents

$$I_1 = \frac{500000}{6600} = 75.76\ \text{A}, \qquad I_2 = \frac{500000}{400} = 1250\ \text{A}$$

Full-load copper loss

$$P_{cu} = I_1^2R_1 + I_2^2R_2 = (75.76)^2(0.4) + (1250)^2(0.001) = 2295.7 + 1562.5 = 3858.2\ \text{W}$$

Iron loss $P_i = 3000$ W (constant at all loads).

(a) Full load (unity p.f.)

$$\eta = \frac{500000}{500000 + 3000 + 3858.2} \times 100 = \frac{500000}{506858.2}\times 100 = 98.65\%$$

(b) Half load (unity p.f.)

Copper loss at half load $= (0.5)^2 \times 3858.2 = 964.5$ W.

$$\eta = \frac{250000}{250000 + 3000 + 964.5} \times 100 = \frac{250000}{253964.5}\times 100 = 98.44\%$$

Answer: efficiency at full load $= 98.65\%$ and at half load $= 98.44\%$ (unity p.f.). At 0.8 p.f. lagging the values are 98.31% (full load) and 98.06% (half load).