Question

A 100-km long, three-phase, 50-Hz transmission line has resistance/phase/ km = 0.1 Ω ; reactance/phase/km = 0.5 Ω ; susceptance/phase/km = 10 × 10⁻⁶ siemens.

If the line supplies a load of 20 MW at 0.9 p.f. lagging at 66 kV at the receiving end, calculate by nominal ‘p’ method, the regulation and efficiency of the line. Neglect leakage.

Answer 1

 For a 100 km line resistance/phase = 0.1 × 100 = 10 Ω ; inductive reactance/phase = 0.5 × 100 = 50 Ω Capacitive susceptance/phase = 10 × 10⁻⁴ × 100 = 10 × 10⁻⁴ siemens Susceptance at each end = 5 × 10⁻⁴ siemens

(ii) line input = √3 × 76 × 103 × 177.6 × 0.905 = 21.5 × 10⁶ Ω = 21.15 MW

∴ transmission η = 20 × 100/21.15 = 94.56%