### Loads - Scenario 5

SITUATION It is desired to improve the power factor of a 230V, 60 Hz, 3-phase system from 0.7 to 0.92 lagging. A wye-connected capacitor bank with capacitance of 10 microfarad per phase is to be used. What is the kW of the load? ANALYSIS 1.) Capacitor reactance, 1ph: 1.1.) Scalar: X,c,1ph = 1 / (2 pi f C) X,c,1ph = 1 / (2 pi 60 * 10uF) X,c,1ph = 265.2582 ohms 1.2.) Vector: X,c,1ph-> = 265.2582 ohms / j X,c,1ph-> = -j 265.2582 ohms X,c,1ph-> = 265.2582 ohms < (-90 deg) 2.) Capacitor power, vector: 2.1) Real power The capacitor is purely a reactive device. Since there is no resistance, no power dissipation happens. Therefore, capacitor real power does not exist. P,c-> = 0 Watts 2.2.) Reactive power, 3ph: Q,c,1ph-> = V,c,LN-> (i,c,1ph->)* Q,c,3ph-> = 3 Q,c,1ph-> Q,c,3ph-> = 3 V,c,LN-> (i,c,1ph->)* -- Conjugate (i->)* means "to reverse angle sign". Q,c,3ph-> = 3 V,c,LN-> (V,c,LN-> / X,c,1ph