Practical power factor correction
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:
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:
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:
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 180o 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:
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).
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