Question

QUESTION 5:

A) Given that V_an = 230  ̶ 10^o for positive sequence voltage Y connection load, Find the phase voltages and line voltages then draw a phasor diagram to show all voltages.

B) Consider a 300 kVA single phase load with pf = 0.85 lagging and it is required to correct the pf to 0.90 leading, Find the reactive power (Qc) added after using the parallel capacitors and the value of capacitor added if V = 380 kV and f = 50 Hz.

Answer 1

Solution :

A ) The solution is attached below

B) Let the true power be P Watts

and Apparent power (Given) S = 300 KVA

Given p.f.1 = 0.85 lagging

p.f.2 = 0.90 lagging

f = 50 Hz

V = 380 KV

To find Q_C  and the value of capacitor added for power factor correction.

Power Factor = True power / apparent Power

=> p.f.1 = True power (P)/Apparent power(S)

=> 0.85 = P(W)/(300*1000)(VA) {1000 is multiplied to convert KVA into VA}

=> P = 0.85 * 300 *1000 (W)

=> P = 255000 W or 255 KW

Now as

p.f.1 = 0.85 lag = => = 31.788

p.f.2 = 0.90 lag = => = 25.841

The required capacitor KVAR i.e. Q_C to improve the power factor will be

Q_c

=> Qc = 255 (tan(31.788)- tan(25.841))

=> Qc = 34.535 KVAR

i.e. reactive power Qc added after parallel capacitor addition is 34.535 KVAR

now the value of capaciotr required C will be calculated by the formula

Q_C   = 2 * pi * f * C* V^2

34.535 = 2 *3.14 *50 *C *380 * 380

=> C = 0.7616 microfarad,