Loads - Scenario 1

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 = 26,794.87 * sin(18.19)
Q,sys = 8,364.53VAr

9.) Needed capacitor reactive power:

Q,sys = Q,load + Q,c
Q,c = Q,sys - Q,load
Q,c = 8,364.53VAr - 25,455.84VAr
Q,c = -17,091.31VAr
Q,c = -17.09kVAr
* Negative means "supplying".

10.) Capacitor complex power:

S<a = P,c + jQ,c
S<a = 0W - j17.09kVAr

11.) Capacitor reactance, scalar:

Q,c = V^2 / X,c
X,c = V^2 / Q,c
X,c = (240V)^2 / 17.09kVAr
X,c = 3.37 ohms

12.) Capacitance:

X,c = 1 / (2 pi f C)
C = 1 / (2 pi f X,c)

12.1.) Assuming 50 Hz freq,

C = 1 / (2 * pi * 50 * 3.37)

C = 944.54 uF @ 50 Hz

12.2.) Assuming 60 Hz freq,

C = 1 / (2 * pi * 60 * 3.37)
C = 787.12 uF @ 60 Hz

CONCLUSIONS

Capacitor size needed is 17.09 kiloVoltAmps-reactive, with a capacitance of 944.54 uF for a 50-Hz system, or 787.12 uF for a 60-Hz system.