power factor correction

Practical power factor correction







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When the need arises to correct for poor power factor in an AC power
system, you probably won't have the luxury of knowing the load's exact
inductance in henrys to use for your calculations. You may be fortunate
enough to have an instrument called a power factor meter
to tell you what the power factor is (a number between 0 and 1), and
the apparent power (which can be figured by taking a voltmeter reading
in volts and multiplying by an ammeter reading in amps). In less
favorable circumstances you may have to use an oscilloscope to compare
voltage and current waveforms, measuring phase shift in degrees and calculating power factor by the cosine of that phase shift.


Most likely, you will have access to a wattmeter for measuring true
power, whose reading you can compare against a calculation of apparent
power (from multiplying total voltage and total current measurements).
From the values of true and apparent power, you can determine reactive
power and power factor. Let's do an example problem to see how this
works: (Figure below)





Wattmeter reads true power; product of voltmeter and ammeter readings yields appearant power.


First, we need to calculate the apparent power in kVA. We can do this by multiplying load voltage by load current:




As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which
tells us that the power factor in this circuit is rather poor
(substantially less than 1). Now, we figure the power factor of this
load by dividing the true power by the apparent power:




Using this value for power factor, we can draw a power triangle, and
from that determine the reactive power of this load: (Figure below)





Reactive power may be calculated from true power and appearant power.


To determine the unknown (reactive power) triangle quantity, we use the
Pythagorean Theorem “backwards,” given the length of the hypotenuse
(apparent power) and the length of the adjacent side (true power):




If this load is an electric motor, or most any other industrial AC
load, it will have a lagging (inductive) power factor, which means that
we'll have to correct for it with a capacitor
of appropriate size, wired in parallel. Now that we know the amount of
reactive power (1.754 kVAR), we can calculate the size of capacitor
needed to counteract its effects:




Rounding this answer off to 80 µF, we can place that size of capacitor in the circuit and calculate the results: (Figure below)





Parallel capacitor corrects lagging (inductive) load.


An 80 µF capacitor will have a capacitive reactance of 33.157 Ω,
giving a current of 7.238 amps, and a corresponding reactive power of
1.737 kVAR (for the capacitor only). Since the capacitor's current is 180^o
out of phase from the the load's inductive contribution to current
draw, the capacitor's reactive power will directly subtract from the
load's reactive power, resulting in:




This correction, of course, will not change the amount of true power
consumed by the load, but it will result in a substantial reduction of
apparent power, and of the total current drawn from the 240 Volt
source: (Figure below)





Power triangle before and after capacitor correction.


The new apparent power can be found from the true and new reactive
power values, using the standard form of the Pythagorean Theorem:




This gives a corrected power factor of (1.5kW / 1.5009 kVA), or
0.99994, and a new total current of (1.50009 kVA / 240 Volts), or 6.25
amps, a substantial improvement over the uncorrected value of 9.615
amps! This lower total current will translate to less heat losses in
the circuit wiring, meaning greater system efficiency (less power
wasted).