Loads - Scenario 1
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SITUATION An induction motor, rated 240 Volts, draws a current of 150 Amps and brings the system power factor down to 70.7%. What capacitor size is needed to raise system power factor to 95%? ANALYSIS 1.) Complex power equations, vector: S-> = P-> + Q-> S<a = P + jQ S<a = S cos(a) + j S sin(a) 2.) System power factor angle: P = S cos(a) pf = P/S = cos(a) = 0.707 a = arccos(0.707) = 45deg 3.) Motor apparent power, scalar: S = Vi = 240V * 150A = 36kVA 4.) Motor real and reactive power: P,load = S cos(a) P,load = 36k cos(45) = 25,455.84W Q,load = S sin(a) Q,load = 36k sin(45) = 25,455.84VAr 5.) System power factor angle after capacitor correction: pf = cos(a) = 0.95 a = arccos(0.95) = 18.19deg 6.) Real power, unaffected: P,load = 25,455.84W 7.) Desired system apparent power: P = S cos(a) 25,455.84 = S cos(18.19) S = 25,455.84 / cos(18.19) S = 26,794.87VA 8.) Desired system reactive power: Q,sys = S sin(a) Q,sys = 26,794.