GATE 2018 EE Q.51: solution and answer

GATE 2018 · EEQ.51 · 2 marks · NAT

A 3-phase 900 kVA, 3 kV / √3 kV (Δ/Y), 50 Hz transformer has primary (high voltage side) resistance per phase of 0.3 Ω and secondary (low voltage side) resistance per phase of 0.02 Ω. Iron loss of the transformer is 10 kW. The full load % efficiency of the transformer operated at unity power factor is ________ (up to 2 decimal places).

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Answer and solution

ANSWER

Full-load % efficiency at unity power factor = 97.36%.

Official answer: 97.20 to 97.55Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Identify winding phase voltages: primary is Δ-connected at line voltage 3 kV, so per-phase voltage = 3 kV = 3000 V; secondary is Y-connected at line voltage √3 kV, so per-phase voltage = √3 kV / √3 = 1 kV = 1000 V.
Full-load primary phase current: $I_{1} = \frac{900000}{3 \times 3000} = 100 A$ .
Full-load secondary phase current: $I_{2} = \frac{900000}{3 \times 1000} = 300 A$ .
Consistency check: phase turns ratio $= 3000/1000 = 3$ , and $I_{2} / I_{1} = 300/100 = 3$ — matches, confirming the current directions and side assignment are correct.
Total full-load copper loss (sum over all three phases, both windings): $P_{c u} = 3 (I_{1}^{2} R_{1} + I_{2}^{2} R_{2}) = 3 (10 0^{2} \times 0.3 + 30 0^{2} \times 0.02) W$
$P_{c u} = 3 (3000 + 1800) = 14400 W = 14.4 kW$ .
Total full-load losses $= P_{c u} + P_{i r o n} = 14.4 + 10 = 24.4 kW$ .
Full-load output at unity power factor: $P_{o u t} = 900 \times 1 = 900 kW$ .
$η = \frac{P _{o u t}}{P _{o u t} + total losses} = \frac{900}{900 + 24.4} = \frac{900}{924.4} = 0.97360$
% efficiency $= 97.36%$ .

KEY IDEA

For a Δ/Y three-phase transformer, phase voltage equals line voltage on the Δ side but line voltage/√3 on the Y side; full-load phase currents on each side then follow from rated kVA = 3 × V_phase × I_phase for that side. Total copper loss is the sum of I²R loss in all three phases of both windings, and efficiency at any power factor is (output)/(output + copper loss + iron loss), with output = rated kVA × cosφ at full load.

WATCH OUT FOR

Using the line voltage as the phase voltage on the Y (secondary) side (forgetting to divide by √3), or computing copper loss for only one phase instead of multiplying by 3, both of which give a wrong current/loss and shift the efficiency outside the key range.

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GATE 2018 EE Q.51: solution and answer

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