Today, we will continue
explaining the technical part for Power
Factor Correction Capacitors Sizing Calculations. We will explain the following
topics:
 What causes low power factor?
 Bad impacts of low power factor,
 Benefits of Power Factor correction.
1 What causes
Low Power Factor?

There are many causes for getting a low power factor for some
equipments as follows:

1.A Equipment Design
A low power factor can be a result of the design of the
equipment for example:

1.B Operating Conditions
The following operating conditions will
affect the value of PF:

1.B.1 Loading
The power factor of an electrical motor reaches its maximum
value under full load. The power factor decreases rapidly when the load
decreases. Fig.1 below illustrates the effect of the loading on a motor’s
power factor and Table1 gives values of PF for different motor loading
factors.
Fig.1: effect of the loading on a
motor’s power factor
Table1: values of PF for different motor
loading factors.

1.B.2 Line voltage
Increasing the line
voltage on motors and transformers above the rated voltage will increase the
consumption of reactive energy. The result will be reduction of power factor.
For example, an increase of 10% on the rated voltage can result in 20%
reduction of the power factor.

2 Bad Impacts
of Low Power Factor

There are many bad impacts of low power factor as follows:

2.1 Loss in efficiency of the
equipment
When an installation operates with a low power factor, the
amount of useful power available inside the installation at the distribution transformers
is considerably reduced due to the amount of reactive energy that the
transformers have to carry. In this case, we say that there is a loss in the
efficiency of the transformer.
Fig.2 below indicates the available actual power of distribution
equipment designed to supply power to 1000 KW installation.
Fig.2
So, if you have distribution equipment designed to supply a 1000
KW installation at PF=1, the same equipment will supply only 80% of the power
(800 KW) for the same installation at PF=0.8. The loss in the efficiency of
the distribution equipment in this case will be 20%.

2.2 Loss in distribution
capacity
In general terms, as the power factor of a three phase system
decreases the current rises. The heat dissipation in the system rises
proportionately by a factor equivalent to the square of the current rise.
Fig.3 below graphically displays the variation of the I^{2}R
losses in feeders and branches. Losses are expressed in percent as a function
of power factor.
Fig.3

2.3 Larger Investment for
future Expansion
In case of expansion, a larger investment is required in the
equipment needed to increase distribution capability of the installation,
such as oversized transformers, cables and switchgears.
2.3.A Transformers
For an installation which requires 800KW, the transformer should
be approximately:
Low power factor means you’re not
fully utilizing the electrical power you’re paying for. Another way to look
at it is that at 0.5 power factor, it
takes 100% more KVA to do the same work when power factor = 1.
This means that with a high power factor, the effective capacity
of your local electricity network potentially deferring future investment in
electrical infrastructure and allowing you to connect more machinery to the
same utility connection.
2.3.B Large size conductors
Fig.4 below shows a variation of a cross section of a conductor
as a function of the power factor for a given useful power. This illustrates
that when the power factor of an installation is low, the surcharge on the
electricity bill is only part of the problem.
Fig.4
So, for an installation with a power factor = 0.7, the
crosssection of the conductor must be twice as large as it would be if the
power factor =1.

2.4 Increase the running costs of
the installation
The
running costs of an installation with low power factor will be increased due
to the following reasons:

3 Benefits
of Power Factor correction

There are many benefits of power factor
correction like:

3.1 Increase the efficiency of
distribution equipment
If
the efficiency of a distribution equipment = Active power used by equipment /equipment’s
total working power  (equation1)
And
Equipment’s
total working power = Active power used by equipment + Reactive power
produced in equipment  (equation2)
So,
in equation2, if we decrease/ compensate for the reactive power by increasing
the PF, the equipment’s total working power will be decreased when Active
power used by equipment is fixed. Hence, from equation1, the efficiency of
the equipment will be increased.
Decreasing
the total power will make good savings in the utility bill.
Example#1:
Table2
shows 600 KW working power vs KVA required for different power factors
Table2

3.2
Release more system capacity to connect additional loads
When
power factor capacitors are located at the terminals of an inductive load,
they will deliver all or most of the reactive power required by the load.
This means a reduction in system current will occur, permitting additional
load to be connected to the system without increasing the size of the
transformers, switchboards and other distribution equipment.
Table3 shows the increased available power at the
transformer output by improvement of Power Factor from 0.7 to 1.
Table3
Often,
this release in the system capacity is reason enough to warrant an
improvement in power factor. Especially when conductors or panels are
overheating or where overcurrent devices frequently opened.
The
percent released capacity resulting from an improvement in PF is:
% released system capacity = 100 x (1 – PF_{O} / PF_{f})
Where:
PF_{O}
= original power factor before improvement,
PF_{f}
= final power factor after improvement.
Example#2:
Determine
the system capacity released by improving power factor from 0.5 to 0.8.
Solution:
%
released system capacity = 100 x (1 – PF_{O} / PF_{f})
%
released system capacity = 100 x (1 – 0.5 / 0.8) = 37.5 %
This
means that the KVA load or line current is reduced by 37.5% of what it was
before PF correction which means that we add 37.5% KVA capacity to the system
without exceeding the total power used by the system before PF correction.
Also,
we can get % released system capacity by using Laws as follows:
KVA _{Released} = kW (1/ PF_{O}  1/ PF_{f})
% KVA _{Released} = KVA _{Released}
/ KVA _{Original}
Example#3:
Calculate
the system capacity released for above example using Laws noting that the
load is 200 KW.
Solution:
From
laws:
KVA
_{Released} = kW (1/ PF_{O}  1/ PF_{f})
= 200 (1/0.5 – 1/0.8) = 200 x 0.75 = 150 KVA
Initial
KVA = 200 KW/0.5 = 400 KVA
%
released system capacity = 150/400 = 37.5% which is the same answer in above
example.
Also,
from the following fig.5 you can find the KVA_{ Released} factor for
any power factor improvement.
Fig.5
Example#4:
From
example#3, determine the %KVA _{Released}
Solution:
KVA _{Released}
factor = 0.75
KVA _{Released
}= 200 KW x 0.75 = 150 KVA
Initial
KVA = 200 KW/0.5 = 400 KVA
%
released system capacity = 150/400 = 37.5% which is the same answer in above
example.
Another
curve (Fig.6) can be used to find the % reduction in current for any power
factor improvement.
Fig.6
Example#5:
Determine
the system capacity released by improving power factor from 0.7 to 0.95.
Solution:
% released
system capacity = 100 x (1 – PF_{O} / PF_{f})
%
released system capacity = 100 x (1 – 0.7 / 0.95) = 27 %
From
the curve you can get the same answer.

3.3
Reduced power losses
For
the older power system networks, the kilowatt losses (I^{2}R) can
account for 2 to 5 % of the total load. Since the power losses are
proportional to I^{2} and the current is inversely proportional to
PF, so an improvement in power factor will cause a reduction in system
current, a reduction in system losses and reduced power bills.
This
reduction can be approximated as:
% loss reduction in power = 100 x (1 – PF_{O}^{2} / PF_{f}^{2})
Where:
PF_{O}
= original power factor before improvement,
PF_{f}
= final power factor after improvement.
From
the following curve (Fig.7) you can find the % reduction in losses for any
power factor improvement.
Fig.7
Example#6:
Improve
power factor from 70 percent to 95 percent, calculate the reduction in power
losses.
Solution:
%
loss reduction in power = 100 x (1 – PF_{O}^{2}
/ PF_{f}^{2})
%
loss reduction in power = 100 x (1 –0.7^{2} / 0.95^{2})
= 46%
From
the above curve, you can get the same solution.
Also,
you can calculate the reduction in power losses at transformers from the
following curve (Fig.8):
Fig.8
Where:
3=
transformer with normal losses
2=
transformer with reduced losses
1=
transformer with low losses
In
any transformer, there are two types of losses:
Example#7:
500
kVA Transformer with Actual load 300 kW, if we correct PF from 0.7 to 0.95,
calculate the % reduction in power losses.
Solution:
Situation
before power factor correction:
Iron
losses: 1150 kW (independent of PF) from above curve
Copper
losses = 6000* (actual load/rated load)² = 6000* (300/0.7 / 500)² = 4410 W
Situation
after power factor correction to 0.9:
Iron
losses = 1150 kW (independent of PF) from above curve
Copper
losses = 6000* (300/0.95 / 500)² = 2393.35 W
Reduction
in Copper losses = 4410 – 2393.35 = 2016.65 W
%Reduction
in power losses = 2016.65 / 4410 =45.7%
The
same result we got it from above example#6.

3.4
Voltage improvement Due to Reduced Line Voltage Drop
When
capacitors are added to the power system, the voltage level will increase.
The percent voltage rise associated with an improvement in PF can be
approximated as:
% voltage rise = (capacitor KVAR x Transformer % IZ) / Transformer
KVA
Where:
Transformer
% IZ = Transformer impedance % from nameplate.
From
the following curve (fig.9) you can find the % voltage drop for any power factor
improvement.
Fig.9
Example#8:
Improve
power factor from 60 percent to 90 percent, calculate the reduction in
voltage drop.
Solution:
From
the above curve, at PF = 0.6 the, Voltage drop = 5.1%
And
at PF = 0.9, the voltage drop = 3.6%
So,
the reduction in voltage drop = 5.13.6 = 1.5%
Notes:

3.5
Reduction in Size of Transformers, Cables and Switchgear in New Installations
and Less Investment
Installing power factor correction equipment allows the
conductors crosssection to be reduced, since less current is absorbed by the
compensated installation for the same active power.
Fig10 below represents the increasing size of conductors required to carry the same
100 KW at various power factors.
Fig10
So, for an installation with a 100 KW load, at a power
factor = 1, the crosssection of the conductor is half size as it would be if
the power factor =0.7.
This will lead to less Investment
for new installations.

In the next article, we will continue
explaining the technical part for Power Factor Correction Capacitors Sizing Calculations. Please,
keep following.
The
previous and related articles are listed in below table:
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