POWER SAVINGS AND APPLICATION GUIDE
For power factor correction and harmonic mitigation techniques
Normally, the term Power Factor (PF) is misunderstood. This often results in consumers paying penalties
due to poor Power Factor. In laymen’s terms, electric power is made up of two components, active
power (measured in kilo Watts / KW) that produces useful work, and reactive power (measured in Kilo
Volt Ampere Reactive / KVAR) necessary to generate magnetic fields for operation of inductive
electrical equipment. Although reactive power does not perform useful work, customers are billed for it
when it is supplied by their utility provider. Total power is measured in Kilo Volt Ampere / KVA,
which is the algebraic sum of KW and KVAR.
In an electrical system with substantial inductive loads such as induction motors, customers can lower
their electrical utility bills by supplying their own reactive energy from Capacitors. This has several
advantages, including improvement of cable, transformer and motor losses, voltage regulation, and
capacity relief of current carrying distribution equipment. This is depicted in figures 1 and 2.
Figure 1 Induction motor without power factor correction and utility supplying both reactive and active power
The ratio of useful power (measured in kilo Watts, KW) to total power (measured in kilo Volt Ampere,
KVA) is known as Power Factor. Thus, the objective of installing a capacitor bank is to increase such a
factor by supplying the reactive power when installed at or near inductive electrical equipment.
PF
=
KW
KVA
= cos (θ )
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Figure 2 Capacitor bank supplying the reactive power required by the loads
Figure 2 is a low voltage capacitor bank. A common reason for poor power factor is operating motors at
less than rated capacity. PF drops rapidly as illustrated in Figure 3 and outlined discussed within IEEE
Std 141-1993.
Figure 3 PF behavior at different load profiles with and without capacitor banks
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From figure 3 it can be seen that the power factor of a motor at full load is usually between 80 % and 90
% depending upon the motor type and speed. In general, induction motors do no operate at full load.
Despite the fact that PF varies from no load to full load conditions, the reactive power consumed by the
motor does not vary much. The table below summarizes typical examples of electrical loads which have
poor power factor.
Load
HVAC systems, pumps, cold headers, fans, blowers
Induction furnaces, weaving machines, spot welders
Automated machine tools, presses, industrial welders
Typical Power Factor
80% ~ 90%
60% ~ 80%
Less than 60%
SAVING MONEY
Usually, utilities include a penalty charge in their electrical rates for low power factor. This can be
alleviated when power capacitors are installed to supply the necessary reactive power (KVAR). By
supplying KVAR with PSS Capacitor banks at the load, the capacitor relieves the utility of the burden of
supplying the reactive current as much as 20 percent without a corresponding increase in the size of
transformers, conductors, etc. This makes the utility transmission/distribution system more efficient,
reducing cost for the utility as well as their customers. The utility will relieve the customers of the power
factor penalty when they maintain a proper power factor level.
BENEFITS OBTAINED FROM POWER FACTOR CORRECTION
Improving the PF of a system reduces the current flowing through it which provides many advantages.
Figure 4 shows the current and power triangles.
Figure 4 Current and power triangles
This illustrates that as the PF is corrected, the inductive or magnetizing current is reduced, in turn
minimizing power losses from the point of connection upstream to utility source.
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The main benefits from PF correction are:
a)
b)
c)
d)
e)
f)
Avoid utility penalties
Reduce transformer, cabling, and motor losses
Reduce total plant KVA for the same KW working power
Lower utility electrical billing
Improved voltage regulation
Minimize electrical distribution system investment costs
Avoid utility penalties
As PF correction reduces the magnetizing current through the power circuit, the utility company can
deliver electrical energy to other customers more efficiently.
Reduce transformers, cables, and motor losses
Since power losses are given by the Joule effect (I2 R), the per cent reduction of power losses (% RL) can
be expressed as follows:
 initial PF 

% RL = 100 − 100 ∗ 
 t arg et PF 
2
Where,
% RL =
Reduction in Losses in per cent
initial PF = Initial Power Factor before the application of capacitors
final PF = Target Power Factor, that is, the Power Factor value it is desired to achieve
For instance, if a poor PF of 57 % is improved to 98 %, the per cent reduction in power losses will be:
 0.57 
% RL = 100 − 100 ∗ 

 0.98 
2
= 66 %
Reduce total plant KVA for the same KW working power
Improving PF frees up actual KVA for additional loads that in turn allows system transformers operate
more efficiently.
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The freed up KVA can be calculated as follows:
KVA = KW

1

 actual PF
−

1

desired PF 
For example, if an 800KW plant with a poor power factor of 50 % has a total power of 1600 KVA, the
freed KVA obtained when power factor is improved to 80 % will be:
1 
 = 600 KVA
0 .8 
∴1600KVA − 600 KVA = 1000KVA will now be required.
KVA = 800
 1

 0 .5
−
Now, if the power factor was improved to unity the freed KVA would be:
KVA = 800
 1

 0 .5
−
1 
 = 800 KVA
1 .0 
Therefore, it is obvious that the better the PF, the more KVA capacity is released to serve the system.
Increase KW working power for the same KVA demand
Improving PF will release system capacity permitting the addition of more loads without overloading the
source transformers. By using the next formula, it is possible to verify what was stated above.
KW
= KVA ∗ PF
For example, a system with a poor PF of 50 % and a 500 KVA load profile operates at 250 KW capacity.
Upon improving the Power Factor to 90 %, the available real power improves to 450 KW.
Improved voltage Regulation
As PF correction helps to reduce line voltage drops, motors, conveyors and other equipment will have
more efficient performance.
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The voltage rise when capacitors are installed at the transformer is given by:
% VR =
KVAR ∗ % Z T
KVART
Where VR is the Voltage Rise in per cent and ZT is the transformer impedance % from nameplate. In
general, the lower the rated voltage of a motor, the higher the full load current, the lower the full load
speed and the lower the starting torque.
Less investment
In new installations, the reduction in size of distribution equipment (transformers, cables, switchgears,
etc) will imply less waste of money. The table below shows this idea.
KW
200
200
200
200
KVAR
0
67.24
150
300
KVA
200
211
250
360
PF
100
95
80
55
Conclusion
IDEAL
GOOD
POOR
BAD
Useful Power Factor Correction formulas:
KVA =
3 ∗V ∗ I
1000
=
HP ∗ 0.746
% Eff .
KW
XC
=
10 6
2 ∗π ∗ f ∗C
KVAR =
2 ∗ π ∗ f ∗ C ∗ (KV )
1000
2
Legend
V = Line to line voltage in Volts or kilo Volts (KV)
I = Line current in Amperes (Amps)
HP = Horse Power
C = Capacitance in micro Farads (µF)
XC = Capacitive Reactance in Ohms (Ω)
f = Fundamental Frequency in Hertz (60 Hz)
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Note: Sometimes equipment supplying reactive energy (KVAR) are operated either at a frequency lower
than the fundamental or at a voltage lower than the rated voltage. In that case, the following formula
could be useful:
2
KVAR E
f 
V 
= KVAR R ∗  A  ∗  A 
 fR 
 VR 
Where:
KVARE = Effective KVAR
KVARR = Rated KVAR
V A = Applied Voltage
VR = Rated Voltage
f A = Applied Frequency
f R = Rated Frequency
For example, if a 100KVAR, 240V rated capacitor bank is used in a 220V system, it will supply:
KVARE
 220 
= 100 ∗ 

 240 
2
≈ 84.03 KVAR
If the same equipment (designed to operate at 60 Hz) is used at a lower frequency (50 Hz), it will supply:
 220 
= 100 ∗ 

 240 
KVARE
2
 50 
∗   ≈ 70.03 KVAR
 60 
CALCULATING SIZE OF CAPACITOR BANK
When determining the right size of a capacitor bank all the potential benefits have to be analyzed in each
situation. For example, if there are penalty clauses imposed, it is recommended that the PF be raised to at
least 95%. If more loads are intended to be installed by using existing transformers, the degree of
correction will depend upon the amount of new loads required. Table A on following pages is usually
used as standard practice to calculate the required KVAR to be installed. It gives multipliers for KW to
get the capacitor KVAR needed to raise from initial to target corrected PF. All the values contained in the
table are obtained from the following formula:
KVARrequired
=
KW ∗ (tan ( ARC cos ( PPF ) ) − tan ( ARC cos (CPF ) ) )
Where:
PPF = Present Power Factor
CPF = Corrected Power Factor.
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Example (a)
The installed load of a power plant working at a PF of 55 % demands 150 KW on average. The PF is
desired to be raised at 95 %. Determine the required KVAR to perform this action.
Step1: Locate the original (actual) PF in Table A (rows).
Step2: Locate the desired PF (columns)
Step3: The point at which both values intercept is the correction factor (multiplier) that will be used for
further calculations. In this case it corresponds to 1.518.
Step4: The KVAR necessary to raise the PF to 95 % is obtained by multiplying the 150 KW by the factor
found in Table A. Therefore, 150 KW = 1.189 = 178.5 KVAR . Use 180 KVAR, this may
be higher due to capacitor value tolerances inherent to capacitor cells.
Example (b)
When a plant has a load of 650 KW and a total power requirement of 930 KVA, what is the size of
capacitor required to improve the actual PF to 98 %?
Step1: Calculate actual PF from the available data.
=
PFactual
KW
KVA
=
650
930
= 0.6989 ~ 0.7 = 70 %
Step2: Obtain the factor from the table A below (0.817)
Step3: Calculate the required KVAR in the same way
KVARC
= 0.817 ∗ 650 = 531.05 . Say, 535 KVARC
Example (c)
A group of inductive loads of 113 KW is connected to a 240V system. What is the size of capacitor is
required to correct the actual PF to 100% when the line current is 367 Amps?
Step 1: Calculate the actual PF.
KVA =
PFactual
3 ∗ kV ∗ I =
=
KW
KVA
=
3 ∗ 0.240 ∗ 367 = 152.55
113
152.55
= 0.7407 ~ 0.74 = 74 %
Step 2: Obtain the factor from the table A below (0.909)
Step 3: Calculate the required KVAR in the same way. KVARC = 0.909 ∗ 113 = 102.7 ~ 105 KVARC
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TABLE A
Another common way of calculating the required capacitor KVAR rating is by means of the following
tables according to IEEE Std 141-1986.
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10
POWER BILL SAVINGS
When electrical systems have poor PF, many of the utilities will impose a penalty due to increased
generation and transmission costs needed to provide and transport real and reactive power.
Here is a typical example:
Demand charge per month
First 50 KW
Next 100 KW
Next 150KW
Excess KW
$/KW
6.42
5.10
4.43
3.78
If the utility applies a PF penalty below 0.95, and the assumed method of charging the consumer is:
Billing demand
=
KWdemand ∗ 0.95
PFactual
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Actual load = 537 KW
Actual Power Requirement = 780 KVA
Actual PF = 0.6884
Actual Excess = 120 KW
Utility demand charges at actual PF = Billing demand
=
537 ∗ 0.95
0.6884
= 741.06 KW
Therefore, the charges will be:
50 KW x 6.42 $ / KW = $321
100 KW x 5.10 $ / KW = $510
150 KW x 4.43 $ / KW = $664.5
621.06 KW x 3.78 $ / KW = $2347.60
Total
= $3843.1 / month
After improving the actual PF to the minimum accepted value (0.95):
Billing demand
=
537 ∗ 0.95
0.95
= 537 KW
50 KW x 6.42 $USD / KW = $321
100 KW x 5.10 $USD / KW = $510
150 KW x 4.43 $USD / KW = $664.5
417 KW x 3.78 $USD / KW = $1576.26
Total
= $3071.76 / month
Total savings per month = $3843.1 - $3071.76 = $771.34
Total annual savings =
$771.34 X 12 = $9256.08
PAYBACK ANALYSIS
In order to correct the PF from 0.6884 to 0.95, from table 1 we see that 371.604 KVAR is required. Or
approximately 375 KVAR.
From PSS list price it is recommended to use: LAC 0375 480 05S N1
Prescribed equipment will perform automatic correction using 5 stages of 75 KVAR.
List price: $14,262
Payback: $14,262 / $9256.08 = 1.54 years
That is, one year and a half payback (based on list price of the capacitor bank)
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UNDERSTANDING HARMONICS
Power transmission and distribution systems are designed for operation with sinusoidal voltage and
current waveforms at a constant frequency. However, when non-linear loads (such as thyristor-based
drives, converters and arc furnaces) are connected to the system, excessive harmonic currents are
generated, causing both current and voltage distortion.
Harmonics and Standard Capacitors don’t mix
Harmonics are a natural by-product of non-linear loads such as drives (AC and DC), motor soft starters,
welders, UPS systems, robotics and other electronic loads. Harmonics are higher than 60 Hz currents and
voltages which are produced by these devices in response to the manner in which they draw current (see
figure 5). The amount of harmonic current rises as more and more non-linear loads are added. Harmonics
can cause serious problems to arise when standard fixed or automatic power factor correction capacitors
are added to the network.
150
100
50
0
-50
-100
-150
0
50
100
150
200
250
CH A
Ideal Sine Wave
150
100
50
0
-50
-100
-150
0
50
100
150
200
250
CH A
Distorted Sine Wave
Figure 5
There are two main concerns to consider when applying power factor correction capacitors. The first,
capacitors are a natural low impedance path for harmonic currents and will therefore absorb these
energies. The increased capacitor current results in higher element temperature which negatively impacts
the life of the capacitor. Also because capacitors reduce the network impedance, capacitor can actually
increase the level of harmonic current on the network. It’s important to remember that while capacitors
do not produce harmonic currents, they can magnify their effects. Furthermore, harmonic voltages
present on the network, create voltage stresses on the capacitor dielectric which can cause premature
dielectric breakdown and capacitor failure.
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The second, a potentially more serious concern, is network resonance. When capacitors are added to the
network, they set up a parallel resonance circuit between the capacitors and the system inductance’s (see
left figure below). Harmonic current components that are close to the parallel resonance point are
magnified (see right figure below). The magnified currents can cause serious problems such an excessive
voltage distortion, nuisance fuse operation, overvoltage trip of drives and insulation breakdown within
motors, transformers and conductors.
FIGURE 6
Harmonic filtering is the best way to eliminate this distortion from the power system, while at the same
time producing reactive power. Harmonic filters consist of capacitors connected in series with a reactor.
The capacitors produce reactive power at the filter’s fundamental frequency, and the circuit is designed to
achieve the required power factor correction. The inductance of the reactor is chosen so that the filter
forms a very low impedance series resonant circuit at the harmonic frequency. This ensures that a high
proportion of power system harmonic currents enter the filter.
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FIGURE 7
A typical harmonic filter consists of three series resonant circuits tuned to the most common harmonics
th
th
th
(5 , 7 and 11 harmonics). These filters are housed in steel cubicles. Each consists of a contactor,
reactor and capacitor. The unit is generally connected to the fuse feeders on the main distribution board.
Harmonic filters can be operated in the same way as automatic capacitor banks: they are controlled by
means of reactive power controller according to reactive power load curves.
PSS Harmonic filters are custom designed for each application using standard components. This ensures
that the best possible power factor correction and filtering characteristics are achieved for a reasonable
investment.
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