Currents and efficiency of 132/11 kV transformer at 4 MW

Primary is star (line current = phase current), secondary is delta (phase voltage = line voltage = 11 kV, line current = $\sqrt3$ × phase current). Winding resistances are taken as per-phase values.

Load apparent power:

$$S = \frac{P}{\cos\phi} = \frac{4 \times 10^6}{0.8} = 5\ \text{MVA}$$

(i) Secondary line and phase currents

$$\begin{aligned} I_{2L} &= \frac{S}{\sqrt3\,V_{2L}} = \frac{5\times10^6}{\sqrt3 \times 11000} = 262.43\ \text{A} \\ I_{2ph} &= \frac{I_{2L}}{\sqrt3} = \frac{262.43}{\sqrt3} = 151.52\ \text{A} \end{aligned}$$

(ii) Primary line current

Neglecting no-load current, input VA = output VA:

$$I_{1L} = I_{1ph} = \frac{5\times10^6}{\sqrt3 \times 132000} = 21.87\ \text{A}$$

Check through the turns ratio: $V_{1ph} = 132000/\sqrt3 = 76210$ V, $K = 11000/76210 = 0.1443$, $I_{1ph} = K I_{2ph} = 0.1443 \times 151.52 = 21.87$ A.

(iii) Efficiency

Copper losses (three phases):

$$\begin{aligned} P_{cu1} &= 3I_{1ph}^2R_1 = 3 \times 21.87^2 \times 10 = 14348\ \text{W} \\ P_{cu2} &= 3I_{2ph}^2R_2 = 3 \times 151.52^2 \times 0.02 = 1377\ \text{W} \\ P_{cu} &= 14348 + 1377 = 15725\ \text{W} = 15.73\ \text{kW} \end{aligned}$$

Total losses $= P_i + P_{cu} = 20 + 15.73 = 35.73$ kW.

$$\eta = \frac{4000}{4000 + 35.73}\times 100 = 99.11\%$$

(The leakage reactances do not cause power loss, so they are not needed for efficiency.)

Answer: (i) $I_{2L} = 262.43$ A, $I_{2ph} = 151.52$ A; (ii) $I_{1L} = 21.87$ A; (iii) $\eta = 99.11\%$.