Question

A 3-phase, 20 km line delivers a load of 10 MW at 11 kV having a lagging p.f. of 0.707 at the receiving end. The line has a resistance of 0.02 Ω/km phase and an inductive reactance of 0.07 Ω/km/phase. Calculate the regulation and efficiency of the line. If, now, the receivingend p.f. is raised to 0.9 by using static capacitors, calculate the new value of regulation and efficiency

Answer 1

(i) When p.f. = 0.707 (lag)

Line current = 10 × 10⁶ /√3 × 11,000 × 0.707 = 743 A

V_R per phase = 11,000/√3 = 6,352 V

Total resistance/phase for 20 km = 20 × 0.02 = 0.4 W 

Total reactance/phase for 20 km = 20 × 0.07 = 1.4 W

∴ Total impedance/phase = (0.4 + j 1.4) Ω

If V_R is taken as the reference vector, then drop per phase 

= 743 (0.707 − j 0.707) (0.4 + j 1.4) = (945 + j 525)

∴ V_S = 6,352 + 945 + j 525 = 7,297 + j 525 

or V_S = √ ( 7297² + 525² ) = 7,315

∴ % regulation = 7,315 - 6,352 / 6,352 x 100 = 15.1 %

Total line loss = 3I² R = 3 × 7432 × 0.4 = 662 kW

Total output = 10 + 0.662 = 10.662 MW 

∴ η = 10 × 100/10.662 = 94%

(ii) When p.f. = 0.9 (lag)

Line current = 10⁷ /√3 × 11,000 × 0.9 = 583 A 

Drop/phase = 583 (0.9 − j 0.435) (0.4 + j 1.4) = 565 + j 633

V_S = 6,352 + (565 + j 633) = 6,917 + j 633

∴ V_S = √( 6,917² + 633² ) = 6,947 V

Total line loss = 3 × 5832 × 0.4 = 408 kW ; Total output = 10.408 MW

∴ η = 10 × 100/10.408 = 96.1%