500 kVA 3-phase transformer efficiency

Copper loss is found from the phase currents and phase resistances of both windings. The HV side is delta (phase current = line current $/\sqrt{3}$) and the LV side is star (phase current = line current).

Phase currents at full load

HV (delta), line voltage = phase voltage = 33 kV:

$$I_{ph1} = \frac{500000}{3 \times 33000} = 5.051\ \text{A}$$

LV (star), line current = phase current:

$$I_{ph2} = \frac{500000}{\sqrt{3}\times 11000} = 26.243\ \text{A}$$

Full-load copper loss

$$\begin{aligned} P_{cu} &= 3\left(I_{ph1}^2R_1 + I_{ph2}^2R_2\right) \\ &= 3\left[(5.051)^2(35) + (26.243)^2(0.876)\right] \\ &= 3\,[892.8 + 603.3] = 4488.2\ \text{W} \end{aligned}$$

Iron loss $P_i = 3050$ W.

Efficiency at full load (0.8 p.f.)

$$\eta = \frac{500000\times 0.8}{500000\times 0.8 + 3050 + 4488.2}\times 100 = \frac{400000}{407538.2}\times 100 = 98.15\%$$

Efficiency at half load (0.8 p.f.)

Copper loss $= (0.5)^2 \times 4488.2 = 1122.1$ W.

$$\eta = \frac{200000}{200000 + 3050 + 1122.1}\times 100 = \frac{200000}{204172.1}\times 100 = 97.96\%$$

Answer: $\eta_{\text{full load}} = 98.15\%$ and $\eta_{\text{half load}} = 97.96\%$ at 0.8 p.f.