Technically Speaking
No. 0011
Industrial Products Division
Capacitors for Power Factor Correction
4-20-2012
1. Basic Power
In any power system, there are three components that effect connected equipment: real, reactive, and
active power. Active power is the most commonly understood, and is measured in watts (W) or kilowatts
(kW). Real power is given in volt-amps (kVA), and represents the power that is actually doing work. Reactive
power (kVAR) represents the portion of power flow that is temporarily stored in the form of electric or
magnetic fields, due to inductive and capacitive network elements, and returned to power source. The power
factor (PF) of an AC electrical system is given as a percentage of the actual power drawn. That percentage
represents how much of the total power being drawn is real power, and how much is reactive power.
2. Why Correct Power Factor
The goal of power factor correction is to get the percentage of power used as close to 100% active power
as possible. There are several reasons for trying to improve power factor, the main reason being that utilities
can charge fees if power factor is below a certain value. Also, if there is a lot of reactive current traveling on
the distribution system, it may cause voltage drops which can cause electrical equipment to abruptly shut down
or overheat. In order to prevent these situations, the industry often tries to correct the power factor to
approximately 95%. This correction is achieved by connecting capacitors in parallel with the connected motor
circuits. This configuration results in a capacitive current that leads the lagging current from the supply. This
leading current cancels the lagging current, and brings the percentage of active power closer to 100%, usually
between 92% and 95%.
3. Power Factor Correction Capacitors
The standard power factor correction system includes three capacitors in a bank, which are rated by the
manufacturer using the total kVAR of the system, rather than the capacitance of the individual capacitors. The
system also includes current limiting fuses, as well as discharge resistors that will reduce voltage to 50 Volts
or less in one minute on motors rated at a voltage below 2400V. On machines with a voltage higher than
2400V, the resistors can drop the voltage within 5 minutes, which is in compliance with the NEC. These
capacitors are depicted below in Figure 1, showing the three units in parallel, with 2 current limiting fuses.
TS0011
Reference D.S. 10.3.7
Technically Speaking
[3] Power Factor
[2] Current
Correction Capacitors
Limiting Fuses
Figure 1
Figure 2 below shows the connection of these capacitors to the motor leads. The capacitor
leads are simply connected directly in parallel with the main motor leads.
Figure 2
TS0011
Reference D.S. 10.3.7
Technically Speaking
4. Equations for Selecting Capacitors
A standard power factor correction system is selected based on the no load kVA of the motor. To
determine the correct kVAR value of the capacitor to be used for PF correction, the following calculations can
be applied. All of the data required to perform these calculations is provided in the standard software package,
i.e. the motor datasheets. It is therefore advisable to perform all calculations and use the lower calculated
kVAR for capacitor selection.
Equation 1:
0.9
1000
Where:
1.73
1
= No load current at rated voltage.
= Rated Voltage.
= Maximum expected percent overspeed.
An industry standard capacitor is selected with a value equal to or less than the kVAR calculated above. The
goal is to select a capacitor that is as close to this calculated kVAR as possible without exceeding it. This
ensures the best power factor correction without risking damage to the motor or anything else connected to the
capacitor.
Example 1:
For a motor that runs at 4000V, has a no load current of 30A, and will be running at 15% overspeed, a kVAR
value would be calculated as:
0.9 30
1000
4000 1.73
1 .15
186840
1322.5
141.278
From this value, a capacitor would be selected with a kVAR of 125 because the next higher capacitor is rated
at 150 kVAR, which would exceed the calculated value.
Some other basic calculations are listed below, including how to calculate line power factor, as well as
an alternate method for calculating a kVAR for the capacitors that are to be used.
TS0011
Reference D.S. 10.3.7
Technically Speaking
Equation 2:
To determine the value of a capacitor in kVAR that will correct power factor to 95%:
tan cos
1
tan cos 1 . 95
Where:
=
Input power in kilowatts with rated output horsepower or rated output power of a
generator.
This value is calculated as:
746
1000
.
. = Full load efficiency in per unit
notation (from datasheet)
= Uncorrected full load power factor in per
unit notation
If a power factor of 95% is requested, the value of the necessary capacitor can be calculated using the above
equations. This calculation will give a different kVAR value from that calculated in Equation 1. The lesser of
the two values is always taken to ensure that the motor is not damaged by the addition of the capacitors in the
circuit. For this calculation, the same procedure is used as above, i.e. an in industry standard capacitor with
kVAR value less than or equal to the calculated value. Also, the value of 0.95 in the above equation can be
substituted with another value to correct to a different power factor. For example, to correct to a power factor
of 85%, one could write:
tan cos
1
tan cos 1 . 85
Example 2:
To correct the power factor on a motor that runs at 1500 HP, 4000V, 91% power factor, and 95.6% efficiency,
up above 95%, plug these values into equation 2, and write:
tan cos 1. 91
. 4556
1500
1000
tan cos 1 . 95
.3287
746
.4556 .3287
.956
1119000
.1269
956
148.5
TS0011
Reference D.S. 10.3.7
Technically Speaking
For this power factor correction, use a 125 kVAR capacitor because the next higher rating is 150 kVAR and
that exceeds the calculated value. The difference between Equation 2 and Equation 1 is that Equation 2 is
based on the horsepower, efficiency, and uncorrected power factor of the motor. However, Equation 1 gives a
kVAR value equal to 90% of the no load kVA of the motor.
Equation 3:
To determine the line PF of a system with known capacitive reactance:
PFL
Cos Tan‐1
Tan
.
‐
.746
Where:
and PF is the uncorrected PF of the system.
Example 3:
Given a machine that is 1500 HP, using a 125 kVAR capacitor, has an uncorrected power factor of 91% and
which is 95.6% efficient, write:
PFL
Cos Tan‐1
Tan
‐
125 .956
1500 .746
Cos Tan‐1 . 45 ‐ . 11
Cos . 34
.94
PFL
Giving a corrected line power factor of 94%
5. Precautions
The kVAR calculated in all of the above examples is the total three phase kVAR for the motor.
Additionally, precautions should be taken when connecting capacitors to motors that are multispeed, wye-delta
start, or used with autotransformers. In cases like these, any fast switching will parallel the incoming line
voltage with whatever voltage was stored in the capacitors. The resulting transient voltage is unpredictable and
can generate transient currents and torques that could be high enough to damage the machine. When
connecting power factor correction capacitors in conjunction with other equipment, the manufacturer of said
equipment should be consulted to be sure that the equipment is compatible with the capacitors.
TS0011
Reference D.S. 10.3.7