Currents and efficiency of 132/11 kV transformer at 5 MW

Reading of the data: the load is 5 MW in total (balanced, three-phase) at 0.8 pf lag, supplied at 11 kV line voltage from the delta secondary; the 132 kV primary is star-connected. Winding R and X are per phase. Copper loss is found from winding currents; the voltage drop is neglected when finding currents.

Secondary currents (delta, 11 kV)

$$\begin{aligned} I_{L2} &= \frac{P}{\sqrt3 V_L\cos\phi} = \frac{5\times10^6}{\sqrt3\times11000\times0.8} = 328.04\ \text{A} \\ I_{ph2} &= \frac{I_{L2}}{\sqrt3} = \frac{328.04}{\sqrt3} = 189.39\ \text{A} \end{aligned}$$

Primary current (star, 132 kV)

Phase voltages: primary $132/\sqrt3 = 76.21$ kV, secondary 11 kV, so the turns ratio per phase is $76.21/11 = 6.928$.

$$I_{L1} = I_{ph1} = \frac{189.39}{6.928} = 27.34\ \text{A}$$

Check: $5\times10^6/(\sqrt3\times132000\times0.8) = 27.34$ A.

Losses

$$\begin{aligned} P_{cu1} &= 3 I_{ph1}^2 R_1 = 3\times27.34^2\times10 = 22419\ \text{W} \\ P_{cu2} &= 3 I_{ph2}^2 R_2 = 3\times189.39^2\times0.02 = 2152\ \text{W} \\ P_{cu} &= 22419 + 2152 = 24571\ \text{W} \\ P_{loss} &= 24571 + 20000 = 44571\ \text{W} \end{aligned}$$

(The leakage reactances cause voltage drop but no loss, so they do not enter the efficiency.)

Efficiency

$$\eta = \frac{5000}{5000 + 44.571}\times100 = 99.12\%$$

Answer: secondary phase current = 189.4 A, secondary line current = 328.0 A, primary line current = 27.34 A, efficiency = 99.12%.