A Primer on Power Factor Correction
By Gonzalo Sandoval and Cesar Chavez Soria, Inelap
Jun 1, 2004 12:00 PM
Improving power factor can reduce system and conductor losses, boost voltage levels, and free up
capacity
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Power factor correction is a
frequently misunderstood
topic. Improper techniques
can result in overcorrection, undercorrection, and/or harmonic
resonance, so it can be
helpful to understand the
process for determining the
correct methods of sizing
capacitors for various
applications. It's also important for calculating the values of system and conductor losses, power
factor improvement, voltage boost, and freed-up system capacity (kVA) you can expect to realize from
their installation.
The power triangle and its components. The term “factor” implies a proportional relationship
between two quantities. In this case, power factor is the proportion of active power (P) to apparent
power (S). A power triangle is used to represent the proportion and calculate the reactive power,
using the Pythagorean Theorem (Fig. 1 below).
Active power, also known as working power, is the energy
converted into useful work. Apparent power, on the other hand, is
the total energy consumed by a load or delivered by the utility. So
power factor is the proportion of power converted into useful work
to the total power consumed by the loads or delivered by the
power source. The power not converted into useful work is called
reactive power (Q). You need this power to generate the magnetic
field in inductors, motors, and transformers.
Fig. 1. Power factor is the
Suppose you have a load with a power factor of 0.95. What does
proportion of active power (p) to
this mean? Basically, it means that the load consumes 95% of the
apparent power (s), as shown in
apparent power and converts it into work. But how much is
this power triangle.
reactive power? The answer, using the power triangle and the
Pythagorean Theorem is calculated as follows:
S2=P2+Q2
Q2=S2-P2
Q = sq rt (S2 - P2)
Q = sq rt (1002 - 952) = 31.225
If PF is corrected to 1.0, then reactive power=0 and apparent power=SQRT (952+02)=95. This
demonstrates an actual reduction of energy consumption and peak demand of 5% (100kVA95kVA=5kVA).
Ideal power factor is unity (1.00), which means that the load is using 100% of the power to perform
actual work.
Why reactive power is undesirable. Although you need reactive power for all the magnetic devices
found in any electrical system, it's nevertheless undesirable because it causes a low power factor. A
low power factor means a higher apparent power, which translates into excessively high current flows
and inefficient use of electrical power. These currents cause elevated losses in transmission lines,
excess voltage drop, and poor voltage regulation. Additionally, the installation must have sufficient
capacity to conduct both the active and the reactive power. Many utilities apply a penalty to users with
a low power factor as a way to get reimbursed for supplying the total apparent power.
The most common method for improving power factor is to add capacitors banks to the system.
Capacitors are attractive because they're economical and easy to maintain. Not only that, they have
no moving parts, unlike some other
devices used for the same purpose.
When you add a capacitor bank to your
system, the capacitor supplies the
reactive power needed by the load. If you
size and select the capacitor bank to
compensate to a unity power factor, it
can supply all the reactive power needed
by the load, and no reactive power is
demanded from the utility. If you design
the capacitor bank to improve the power
factor to a quantity less than 1.0, the
reactive power supplied by the bank will
be its rated kVARs (or MVARs), while the
rest of the reactive power needed by the
load will be supplied by the utility.
Fig. 2. Prior to the installation of a capacitor bank, all the
reactive power (noted as “Q1” in Fig. 2b) of the facility is
supplied by the utility, so the apparent power is high,
because both the active and the reactive power has to be
supplied by the utility (noted as “S1”). The added capacitor
bank “Qcap” in Fig. 2a supplies reactive power to the load,
so the facility doesn’t have to draw this reactive power
from the utility, but rather only the difference (Q1 2 Qcap).
A low demand of reactive power translates into a low
consumption of apparent power to the utility, thus
How power factor correction releases releasing capacity in the system, as shown by the power
system capacity. Consider Fig. 2. Prior triangle of Fig. 2c.
to the installation of a capacitor bank, all
the reactive power of the facility (Q1 in Fig. 2b) is supplied by the utility, so the apparent power (S1
in Fig. 2b) is high because both the active and the reactive power have to be supplied by the utility.
Equipment like transformers, switchgear, and cables must be large enough to handle the total
apparent power, but this results in higher equipment costs when the power factor is low.
After installation of power factor correction capacitors, the capacitor bank (Qcap in Fig. 2a) supplies
reactive power to the load, so the facility doesn't have to draw this reactive power from the utility, but
rather only the difference (Q1-Qcap). A low demand of reactive power translates into a low
consumption of apparent power to the utility, which releases capacity in the system, as shown by the
power triangle of Fig. 2c.
Calculate the release of system capacity by using the following equation:
%Ir = (1 - pf1/pf2) x 100
R = P (1/pf1 - 1/pf2)
where R is the release of system capacity, P is the active power of the load, pf1 is the power factor
without capacitor banks, and pf2 is the target power factor.
When power factor is low, the potential capacity to be released for the facility is greater than for high
power factor.
Fig. 3 shows the release of power in perunit of the active power in a facility. It
demonstrates that the maximum release
occurs in those cases where the existing
power factor is low and the target power
factor is high, such as near or equal to
one.
For example, suppose an industrial facility
has a 1MVA transformer and its load has
a power factor of 0.82 and 800kW of
power consumption. The transformer is
therefore loaded at 975kVA (97.5% of its
total capacity).
S=P÷pf=800kW÷0.82=975kVA
If you connect a capacitor bank in the
secondary of the transformer to improve
the power factor to 0.98, the released
capacity in the transformer is:
Fig. 3. This graph of the release of the power system in
per-unit of active power in a facility demonstrates that you
get the maximum release in those cases where the
existing power factor is low and the target power factor is
high, such as near or equal to one.
R = 800 (1/0.82 - 1/0.98) = 159kVA
This means you have reduced the transformer load to 816kVA, which is 81.6% of its total capacity.
The reduction of apparent power translates into reduced current because both are directly
proportional. Before the use of the capacitor bank, the current has two components, active and
reactive. With the addition of the bank, power factor can be corrected to unity and the current will
have only the active component.
Calculate the percentage of current reduction by using the following equation:
This equation shows that the reduction obtained is a function of the power factor before and after the
improvement.
Fig. 4 shows the percent of current
reduction in the circuit feeders when the
power factor is improved. The greatest
current reductions occur in those systems
with a low power factor when the
improvement is near or equal to one.
How power factor correction reduces
conductor losses. The losses in
conductors are equal to the product I2R,
so the improvement of the power factor
also accomplishes a reduction in the
system losses. Transformer losses are
also reduced because the current through
its windings is reduced by the capacitor
bank. The losses are also smaller in the
circuit that carries the current from the
secondary to the main panel if the
capacitor bank is installed close to the
load.
Fig. 4. As shown in this graph of the percent of current
reduction in circuit feeders when power factor is improved,
the greatest current reductions from those systems occur
when the original power factor is low and is improved to
(or near) unity.
The curves in Fig. 5 show the reduction
in feeder losses when you improve power
factor. This graph shows that the greatest
reduction of losses in feeders occurs when
the original power factor in the facility is
low and is improved to (or near) unity.
With the reduction of current, voltage
drop in the facility also decreases because
both are directly proportional. If the
capacitor bank has sufficient capacity, the
voltage could rise above its nominal
value. Calculate the voltage rise in a
central compensation scheme, neglecting
short circuit resistance of the windings,
Fig. 5. The curves show the reduction in feeder losses
by using the following equation:
when you improve power factor. The greatest reduction of
losses occurs when the original power factor in the facility
V2 = [(ST/ZPU)/(ST/ZPU - Qcap)] x V'1
is low and is improved to (or near) unity.
where ST is the nominal apparent power of the transformer, Zpu is the per-unit transformer
impedance, Qcap is the rated reactive power of the capacitor bank at its rated voltage, V'1 is the
voltage before the power factor improvement, and V2 is the expected voltage after the power factor
improvement.
Suppose a facility has a 750kVA transformer with an impedance of 6.3%. Although its nominal
secondary voltage is 220V, the voltage with load decreases to 215V. If you install a 75kVAR capacitor
bank on the secondary side of the transformer, the voltage will be:
V2 = (750/0.063)/(750/0.063-75) x 215 = 216.3V
The corresponding reduction in current means that the apparent power demanded from the utility will
be lower than without the capacitor bank. This reduction of apparent power translates directly into
energy savings because the power consumption from the utility feed is lower. If the utility penalizes
you for a low power factor, then power factor correction will not only eliminate the penalty, it might
also help you get a refund if the corrected power factor is greater than the minimum value required by
the utility.
Determining the capacitor kVAR requirement. Calculate the capacity, in kVAR or MVAR (P must
be kW or MW), of the capacitor bank needed to improve power factor from pf1 (actual power factor) to
pf2 (target power factor) by using the following equation:
Qcap = P x [(sq rt (1-pf12)/pf1) - (sq rt (1-pf22)/pf2)]
When active power is constant, you can use this equation to calculate the reactive power of the
capacitor bank. But when active power isn't constant, you must consider other factors. You should
consider the average value of the active power (P) as well as the average power factor in the facility.
Using these two values, you can calculate the capacitor bank for the average operating condition. You
should also consider the worst case operating conditions (highest active power and lowest power
factor).
Calculating the kVAR requirement based on maximum active power. Looking at the preceding
equation (hereafter referred to as Equation A), you can see that either of two factors can cause the
calculated value of reactive power of the capacitor bank to be less than the value required:
•
•
The active power (P) is higher than the average value used in Equation A.
The power factor (pf1) is lower than the average value used in Equation A.
Taking this into account, you need to re-calculate the reactive power requirement of the capacitor
bank using the maximum active power in the system and the power factor measured under this
operating condition. Equation A can now be expressed as:
Qcap = Pmax [(sq rt (1 - pf12-Pmax)/pf1-Pmax) - (sq rt (1 - pf22)/pf2)]
(hereafter referred to as Equation B) where Pmax is the maximum active power in the facility and pf1is the power factor in the facility when the active power is Pmax.
Pmax
If the reactive power requirement for the capacitor bank, as calculated using Equation B, is greater
than the average value previously calculated using Equation A, then the capacitor bank sized for the
average value won't be sufficient for compensating the reactive power of the load when the active
power reaches its maximum value. As a result, the power factor in the facility won't reach the target
value. In this case, you should select the capacitor based on the maximum active power and the
actual power factor under that operating condition (Equation B).
Calculating the kVAR requirement based on minimum power factor. The next consideration is
to calculate the capacitor bank needed when the power factor is minimum. Do so by using the
following equation:
Qcap = Ppf1min [(sq rt (1 - pf12-min)/pf1min) - (sq rt (1 - pf22)/pf2)]
(hereafter referred to as Equation C) where pf1min is the minimum power factor measured in the facility
and Ppf1min is the active power when the power factor is pf1min.
If the two previously calculated values (average and maximum active power conditions) are less than
the value calculated using Equation C, the capacitor bank kVAR previously determined using Equation
A or Equation B won't be sufficient to compensate for the reactive power of the load when the power
factor reaches its minimum value, and the power factor in the facility won't reach the target value. In
this case, you should select the capacitor based on the minimum power factor as calculated in
Equation C.
Note: The best value of capacitance will be the greater of all calculations above (Equation A, Equation
B, and Equation C), because the capacitor bank will have the capacity for compensating for the
maximum active power condition as well as minimum power factor condition. Automatic capacitor
banks can ensure high power factor under widely varying operating conditions.
Once you've calculated the capacitor bank size in kVAR or MVAR, you should perform a voltage and
current spectrum analysis to check for the presence of harmonics in the facility because they can
cause severe damage to capacitors. When harmonics are present, you should use only capacitors
equipped with capacitor protection reactors.
Beware of power system resonance. From the load point of view, the capacitor and the
transformer form a parallel resonant circuit, while the same elements form a series resonant circuit
from the source point of view. At the resonant frequency, the admittance of a parallel circuit is very
low, so a small current in the load side of the circuit results in very large currents in both paths
(inductive and capacitive) when the frequency of the current is equal to the resonant frequency. The
impedance of a series circuit at its resonant frequency is very low, so a small voltage on the primary
side of the transformer results in very large currents in the capacitor bank and the transformer when
the frequency of the voltage is equal to the resonant frequency.
To avoid resonance, you should perform an electrical survey prior to installation of the capacitor bank.
Harmonic currents on the load side can excite the parallel circuit, while the harmonic voltages in the
primary side of the transformer can excite the series circuit. Both situations will result in very high
currents in the transformer and the capacitor bank, as well as very distorted voltages in the secondary
side of the transformer.
To perform the survey correctly, use a power quality monitor that can record the trend of electrical
parameters, such as voltage, current, active, reactive and apparent power, and power factor, with a
time stamp. It should also be able to monitor and record the waveform, spectrum analysis, and
harmonic distortion levels.
With this kind of instrument, it's relatively easy to determine the reactive power rating of the capacitor
bank necessary for power factor improvement. The trend graphs will allow you to calculate the
reactive power requirement for maximum active power and minimum power factor conditions, thus
helping you to determine the best value of the bank. The waveform and spectrum analysis will provide
an indication of any harmonic components present in the voltage and current.
Determining the resonant frequency of the power system will require system modeling and analysis
based on an equivalent electrical circuit. You can do this by using simulation software or by doing the
calculations by hand. If the system is small and relatively simple, you should be able to do a hand
calculation of the resonant point fairly easily. However, you'll need large simulation software on large
and complex systems. Once you've developed the equivalent electrical circuit, then you can perform a
frequency scan to determine the resonant point of the system.
The power monitor, and its associated software, must be able to calculate total harmonic distortion
(THD) for voltage according to the definition in IEEE Std. 519 - 1992 (distortion based on the nominal
system voltage) and calculate total demand distortion (TDD) defined in the same standard (distortion
based on the rms value of the fundamental current component at maximum demand). These
capabilities are a must.
Sandoval is an electrical engineer and Chavez is manager of electrical engineering at Inelap in
Naucalpan, Mexico. John Houdek, electrical engineer and president of Allied Industrial Marketing in
Milwaukee, provided editing assistance on this article.