Primary current and efficiency of 100 kVA Δ/Y transformer

Primary delta (phase voltage 11 kV), secondary star (phase voltage $400/\sqrt3$). $R_1, X_1, R_2, X_2$ are per-phase values. Load current 100 A per phase (= line current, star) at 0.85 pf lagging.

Turns ratio per phase

$$\begin{aligned} V_{2ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\ K &= \frac{V_{2ph}}{V_{1ph}} = \frac{230.94}{11000} = 0.020995 \end{aligned}$$

Primary line current

Neglecting no-load current:

$$\begin{aligned} I_{1ph} &= KI_{2ph} = 0.020995 \times 100 = 2.0995\ \text{A} \\ I_{1L} &= \sqrt3 \times 2.0995 = 3.636\ \text{A} \end{aligned}$$

Check: $I_{1L} = \frac{\sqrt3 \times 400 \times 100}{\sqrt3 \times 11000} = 3.636$ A.

Output power

$$P_{out} = 3V_{2ph}I_{2ph}\cos\phi = 3 \times 230.94 \times 100 \times 0.85 = 58890\ \text{W} = 58.89\ \text{kW}$$

Copper losses

$$\begin{aligned} P_{cu1} &= 3I_{1ph}^2R_1 = 3 \times 2.0995^2 \times 25 = 330.58\ \text{W} \\ P_{cu2} &= 3I_{2ph}^2R_2 = 3 \times 100^2 \times 0.04 = 1200\ \text{W} \\ P_{cu} &= 330.58 + 1200 = 1530.58\ \text{W} \end{aligned}$$

Efficiency

$$\begin{aligned} \text{Losses} &= 1000 + 1530.58 = 2530.58\ \text{W} \\ \eta &= \frac{58890}{58890 + 2530.58}\times 100 = 95.88\% \end{aligned}$$

Answer: Primary line current = 3.636 A; efficiency = 95.88%.