Monday, January 9, 2012

Power Factor Improvement by Capacitor

Capacitor is commonly used to improve the power factor (pf, Cos Phi) on alternating current (AC). This method is quite popular, and it is applied in fluorescent lamps, motors, transformers, etc.. The schematic diagram can be seen in the figure below. Instruments needed are Power meter (P) and Ammeter (A). Capacitor (C) have a certain value, not too big, and not too small, so that the power factor will be close to 1 (one, ideal) and economical electric power consumption. The position of the capacitor should be as close as possible to the load.



To check voltage and current waveform  simulation chart at a specific power factor is discussed in separate articles. The definition of: efficiency, power factor,watts, voltamperes; are also discussed in a separate article.


Sample calculation below is contained in the attached Excel sheet which has formulas to calculate power factor correction by capacitor. Before the capacitor installed to improve power factor, firts we need to know how much power factor which occurs to a certain power load (real power).

Suppose the data obtained as follows:
Frequency: 60 Hz
Voltage: 220 volts
Current: 1 amperes
Real power: 160 watts

Supplied power by power source:

apparent power = 220 volts x 1 amperes = 220 voltamperes

Power factor = pf = real power / apparent power = 160/220 = 0.73

Then calculate reactive power by Pythagoras formula:

Reactive power = ((apparent power)2 - (real power)2) 1 / 2
            Reactive power = (2202 - 1602) 1 / 2 = 151 var (reactive voltamperes)

Based on reactive power data above, then the reactive load can be obtained:

Reactive load = Voltage2 / reactive power = 2202/151 = 320.5 W

To calculate the amount of capacity, we use Faraday's formula:

Capacity = 1 / (2p x frequency x reactive load) = 1 / (2p x 60 x 320.5) = 8.27 μf

If we pick and install parallel 8 μf capacitor to correct power factor, then the value for the capacitive reactance by this capacitor is:

xc = 1 / (2 p x 60 x 8 x 10-6) = 331.44 W

Current flowing to capacitive reactance is:

ac = 220 / 331.44 = 0.66 amperes

For reactive power for the this capacitive reactance, please note that the vector is opposite to reactive power before, so it has a negative sign (-) :

pc = 220 x 0.66 = - 146.03 var

The new or total reactive power is : Q1 = 151 - 146.03 = 4.97 var

Corrected apparent power is :

S1 = ((real power)2 + (new reactive power))1 / 2 = (1602 + 4.97)1/2 = 160.08 va

Therefore the improved power factor is:

pf1 = 160 / 160.08 = 0.9995

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