Efficiencies of 120 kVA Y/Y transformer

Given: $S = 120$ kVA, $P_i = 1600$ W, maximum efficiency at $x = \frac34$ full load.

Full-load copper loss

At maximum efficiency, copper loss = iron loss:

$$\begin{aligned} x^2P_{cu,FL} &= P_i \\ P_{cu,FL} &= \frac{1600}{(0.75)^2} = 2844.44\ \text{W} \end{aligned}$$

Efficiency at ¾ full load (maximum efficiency)

Power factor not stated for this part, so unity pf is assumed:

$$\begin{aligned} \text{Output} &= 0.75 \times 120 \times 1 = 90\ \text{kW} \\ \text{Losses} &= 1600 + 1600 = 3200\ \text{W} \\ \eta_{max} &= \frac{90000}{90000 + 3200}\times 100 = 96.57\% \end{aligned}$$

(i) Full load, 0.8 pf

$$\begin{aligned} \text{Output} &= 120 \times 0.8 = 96\ \text{kW} \\ \text{Losses} &= 1600 + 2844.44 = 4444.44\ \text{W} \\ \eta &= \frac{96000}{96000 + 4444.44}\times 100 = 95.58\% \end{aligned}$$

(ii) Half load, unity pf

$$\begin{aligned} \text{Output} &= 0.5 \times 120 \times 1 = 60\ \text{kW} \\ P_{cu} &= 0.5^2 \times 2844.44 = 711.11\ \text{W} \\ \eta &= \frac{60000}{60000 + 1600 + 711.11}\times 100 = 96.29\% \end{aligned}$$

The Y/Y connection and voltage rating do not affect the answer because efficiency depends only on total power and losses.

Answer: $\eta_{max}$ (¾ FL, upf) = 96.57%; (i) 95.58%; (ii) 96.29%.