Download Low power factor

Contents
Introduction
2
Low power factor
3
Effects of low power factor
3
Power factor measurement
4
Low power factor circuits
4
Power factor improvement
6
Power factor improvement capacitors
10
Power factor correction methods
14
Summary
16
Answers
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EGG202A: 8 Improve the power factor of ac circuits
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1
Introduction
In this section, we will look at power factor, which is very important in
electrical theory. In all domestic, industrial and commercial applications,
power is the prime consideration of the installation.
Power must be supplied as efficiently as possible, and this is where the
power factor comes in. Power factor control is an effective way of
conserving power, and it also determines the cable size of the conductors in
both the final sub-circuit and the main supply line.
We will consider power factor, looking at the relevant terms and the
methods used to accomplish power factor control.
At the end of this section you should be able to:
2

describe the effects of low power factor.

describe the requirements for power factor improvement.

list the methods used to improve low power factor of a installation.

state local supply authority and AS/NZS 3000 requirements
regarding the power factor of an installation and power factor
improvement equipment.

describe the methods used to measure single phase power factor.

using manufacturers catalogues, select power factor equipment for a
particular installation.
EGG202A: 8 Improve the power factor of ac circuits
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Low power factor
Consider a parallel circuit that begins as a resistive circuit, but has more and
more inductance added to it. The phase angle of this circuit will start at 0
degrees, but will increase as we add more inductance.
Figure 1: The power triangle
As a circuit becomes more reactive, the phase angle increases. Looking at
the power triangle in figure 1, you can see the effect this has on the power
relations. As φ grows, the magnitude of the reactive power Q grows relative
to P. This means that a smaller and smaller component of the apparent
power S is devoted to supplying the true power of the circuit.
This situation is known as ‘low power factor’, because as φ grows from 0
toward 90° the power factor (cos φ) decreases from 1 towards 0.
Thus if the voltage is constant, and the power factor decreases, the current
must increase to maintain the same real power in the circuit.
Effects of low power factor
There are definite drawbacks to a low power factor. It results in the
following disadvantages:

There is an increased voltage drop in the supply conductors. This can
only be compensated for by increasing the cable size, which will lead
to higher costs for the larger cable and its accessories.

Alternator efficiency is reduced. This is due to the internal power
losses of the alternator I2R having a higher value. Real power
EGG202A: 8 Improve the power factor of ac circuits
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delivered to the load is also decreased in proportion to the power
factor where apparent power is constant, since P = S cos Again
this is due to the fact that the load current is high on low power
factor loads. To overcome this drop in efficiency and decreased
output of the alternator, a larger unit must be installed; again, the
cost of the machine and its accessories increases.

To handle the increased current, larger control devices are required.
These control devices comprise the switch gear and transformers and
their associated protection devices. Again, the cost of the installation
escalates.

There are larger power losses in the supply cables, due to the
increased current. This is minimised by increasing the cable size
(higher cost), but as P = I 2 Rcable, the cable size would have to vary
at an exponential rate to keep the loss in the cable constant.
Power factor measurement
Consumers try to ensure that the installation is not creating excess cost
through having low power factor.
To this end they need to be aware of the actual power factor at all times. A
simple solution is to compare energy meter readings against voltmeter and
ammeter readings to determine the power factor.
An installation may also justify the extra expense of installing specifically
designed meters that measure actual power factor.
Low power factor circuits
The types of circuit which cause or produce a low circuit power factor are
given below.
4

Large fluorescent lighting loads. Inside the lamp circuit of each
fitting, a very inductive ballast is used to reduce the lamp current so
as to avoid wasting true power by using resistors. This is one of the
most common forms of artificial lighting.

Lightly loaded motors. The motor inductive losses are relatively
constant, but the output from the shaft, in mechanical watts, varies
with the load. Normally, in the fully loaded motor, the mechanical
power far exceeds the inductive losses within the motor, and the
power factor is as high as 0.85, but on light loads this ratio
decreases, and so the power factor falls. In some installations this
condition occurs as a consequence of the duty cycle of the motor.

Lightly loaded transformers. Since a transformer operates on the
same basis as a motor, the considerations outlined above apply.
EGG202A: 8 Improve the power factor of ac circuits
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
Capacitance in long transmission lines. Remember how a
capacitor is constructed: two conductors separated by an insulator.
This is what a transmission line is. There are three separate phases
(conductors) separated by air (an insulator) from each other. Each
phase is also separated from the ground (also a conductor) by the
same insulator. Thus a three-phase system can be thought of as
having six capacitors, one between each phase and the ground and
one between each of the phase combinations. This is illustrated in
Figure 2.
Figure 2: Transmission line capacitance
EGG202A: 8 Improve the power factor of ac circuits
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Power factor improvement
The local supply authority often requires consumers to maintain the power
factor of loads at a nominal value. In any case, the power factor of the load
should generally be around 0.8 (unless there is a particular reason for it), as
this is the most cost effective value. A lower power factor will lead to higher
losses, which will reduce the efficiency of the circuit. But if the power factor
was forced higher, the rise in efficiency may not outweigh the increased cost
of the installation.
AS/NZS 3000:2000 does not quote a minimum value for power factor but
does however give guidelines for the installation of power factor correction
equipment and this is found in clause 4.6 ‘Capacitors’. To find what the
minimum value of power factor required for your area is, consult your local
Supply Authority. By way of example, the NSW Service and Installation
Rules require a power factor of 0.9 lagging.
To overcome the disadvantages of a lower power factor being generated, a
technique is used whereby the reactive components of the circuit cancel each
other out.
You will remember that a practical circuit generally contains either a
resistive or a resistive/inductive load. As power factor improvement is not
required for resistive circuits, the examples dealing with power factor
improvement refer to resistive/inductive circuits only. This is illustrated in
Figure 3.
Figure 3: Single phase circuit with PF improvement
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So, as we are looking at a circuit containing inductance, the power factor of
the circuit will be lagging.
The power triangle for the circuit in Figure 3 is given in Figure 4, showing
apparent power, true power, reactive power and power factor angle.
Figure 4: Power triangle for PF improvement
To improve the power factor of the circuit, the circuit phase angle must be
altered. It can be seen from the power triangle that a change in either the true
power or the reactive power would have this effect. The true power cannot
be altered, however; it is the normal output or consumption of the load. This
means that the only way is to alter the reactive power of the load.
As the two types of reactive power, inductive and capacitive, are 180° out of
phase, they may be directly added to each other. This simplifies the
calculation considerably.
Remember that the total reactive power is always less than the largest type
of reactive power if power factor improvement is used.
Example
A 240 V, 50 Hz single phase installation draws a current of 40 A from a
supply at a power factor of 0.4 lagging. Determine the VAR rating of a
capacitor bank to be connected in parallel with the load to achieve an
installation power factor of:
(a) 0.866 lagging
(b) unity
Solution
Draw the circuit (Figure 5).
EGG202A: 8 Improve the power factor of ac circuits
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Figure 5
cos   0.4
  66.4
S  VI
 240  40
 9.6 kVA
P  S cos 
 9600  0.4
 3.84 kW
Q  S sin 
 9600  sin 66.4
 8.8 kVAr
(a) New cos φ is 0.866:
 φ = 30°
Now, as P is constant in the load
Q  P tan 30
 3840 tan 30
 2.22 kVAr
Q new  Qold  Qcapacitor
Qcapacitor  Qold  Q new
 8.88  2.22
 6.58 kVAr
(b)
New cos φ is 1:
 φ = 0°
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EGG202A: 8 Improve the power factor of ac circuits
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Now, as P is constant in the load:
Q
= P tan 0
= 3.84 tan 0
= 0 VAr
Q capacitor
= Qold – Q new
= 8.8 – 0
= 8.8 kVAr
Student exercise 1
1
A 240 V, 50 Hz single phase installation draws a current of 16 A from a supply at a
power factor of 0.52 lagging. Determine the VAR rating of a capacitor bank to be
connected in parallel with the load to achieve an installation power factor of:
(a) 0.8 lagging
___________________________________________________________________
___________________________________________________________________
(b) unity.
___________________________________________________________________
___________________________________________________________________
2
A 240 V, 50 Hz single phase installation comprises of an inductive load which draws a
current of 32 A from a supply at a power factor of 0.6 lagging. Determine the VAR
rating of a capacitor bank to be connected in parallel with the load to achieve an
installation power factor of:
(a) 0.75 lagging
___________________________________________________________________
___________________________________________________________________
(b) 0.9 lagging.
___________________________________________________________________
___________________________________________________________________
Check your answers with those given at the end of the section.
EGG202A: 8 Improve the power factor of ac circuits
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Power factor improvement capacitors
To determine the actual value of the capacitance used to correct the power
factor of the circuit, Ohm’s law and the equation for capacitance are
required. Both of these were covered in previous modules, but we will
revise them here.
To determine the capacitive reactance, first use the Ohm’s law equation for
reactive power to determine the capacitor current.
I capacitor 
capacitor VAr
supply voltage
This can be written as:
IC 
QC
VC
Next, calculate the value of capacitive reactance:
XC 
VC
IC
Finally, determine the capacitance of the capacitor:
C=
1
2fXC
Example
A 240 V, 50 Hz single phase motor draws a current of 10 A from a supply at
a power factor of 0.6 lagging. Determine the capacitance required to be
connected in parallel with the load in order to achieve an installation power
factor of:
(a) 0.8 lagging
(b) 0.9 lagging
10
EGG202A: 8 Improve the power factor of ac circuits
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Solution
Draw a circuit diagram (Figure 6).
Figure 6
cos   0.6
  53.13
S  VI
 240 10
 2.4 kVA
P  S cos 
 2400  cos 53.15
 1.44 kW
Q  S sin 
 2400  sin 53.15
 1.92 kVAr
(a)
New cos  is 0.8
 = 36.86°
Now, as P is constant in the load
Q  P tan
= 1440 tan 36.86°
= 1.08 kVAR
Qcapacitor = Qold  Qnew
 1.92 kVAr -1.08 kVAr
= 840 VAr
EGG202A: 8 Improve the power factor of ac circuits
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IC 
QC
VC
840
240
 3.5 A

XC 
VC
IC
240
3.5
 68.6 

1
2 fX C
C
1
2    50  68.6
 46.4 μF

(b)
New cos  is 0.9
  
Now, as P is constant in the load
Q  P tan 
 1.44 tan 25.84
 697 VAr
Q capacitor  Qold  Qnew
 1920  697
 1.22 kVAr
IC 
QC
VC
1220
240
 5.08 A

XC 
VC
IC
240
5.08
 47 

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EGG202A: 8 Improve the power factor of ac circuits
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C
1
2 fX C
1
2    50  68.6
 46.4 μF

Student exercise 2
1
A 240 V, 50 Hz single phase motor draws a current of 8 A from a supply at a power
factor of 0.7 lagging. Determine the capacitance required to be connected in parallel
with the load to achieve an installation power factor of:
(a) 0.8 lagging
___________________________________________________________________
___________________________________________________________________
(b) 0.85 lagging.
___________________________________________________________________
___________________________________________________________________
2
A 240 V, 50 Hz single supply is connected to fluorescent lighting circuit which draws
a current of 13.3 A at a power factor of 0.55 lagging. Determine the capacitance
required to be connected in parallel with the load to achieve an installation power
factor of:
(a) 0.75 lagging
___________________________________________________________________
___________________________________________________________________
(b) 0.85 lagging.
___________________________________________________________________
___________________________________________________________________
Check your answers with those given at the end of the section.
EGG202A: 8 Improve the power factor of ac circuits
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13
Power factor correction methods
The power factor correction device used is dependent upon the size of the
installation.
Once you have determined what device is most appropriate and the size
required, you need to refer to manufacturer’s catalogues for the most cost
effective solution.
Capacitors
For most discharge lighting applications, even though the total load is large
it is comprised of small individual loads, and a small value capacitor
connected in every second (or in every fitting) is the typical circuit
arrangement. The total capacitance for capacitors in parallel is the numerical
sum of the capacitance of each capacitor.
C t  C1  C 2  C3  .....Cn
Example
A single phase fluorescent lighting circuit comprises 14 twin 36 W fittings.
If a 30 F capacitor is connected in every second fitting, as illustrated in
Figure 7, determine the total capacitance of the circuit.
Figure 7
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EGG202A: 8 Improve the power factor of ac circuits
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Solution
The total capacitance of the circuit:
C t  C1 + C 2 + C3 + C 4 + C 5 + C6 + C7
 30 F +F +F + 30 F +F +F+F
= 210 F
We can see that instead of using seven 30 F capacitors, we could have used
a single 210 F capacitor. In some applications this is what is actually done,
but in most cases it is too expensive, both in materials and installation costs.
For large single function loads either a large capacitor or a capacitor bank is
installed.
Synchronous motors
Another way of improving the power factor of a load involves the use of a
synchronous motor. Basically the synchronous motor consists of two
windings, a stator winding (stationary winding) supplied with alternating
current and a rotor winding (rotating winding) supplied with direct current.
If the dc supply to the rotor winding is increased beyond the normal full load
dc value, over-excitation occurs and the motor supply current changes its
phase angle from a lagging angle to a leading angle. Thus the power factor
of the motor goes from a lagging value to a leading value. The reactive
VARs produced by the motor are now capacitive and may be used to correct
or cancel the inductive VAR in the installation. In this application, the motor
is called a synchronous capacitor or synchronous condenser. It will be
covered in greater detail in dealing with ac machines and synchronous
motors.
The single drawback to this method is the cost, both very high initial costs
and high ongoing maintenance costs. Thus this method only finds
application in the largest types of loads—typically in the MVA range—in
large factories and at power stations.
An advantage of course is that this method may incorporate a synchronous
motor driving a load such as a blower which results in a machine serving
more than one purpose.
EGG202A: 8 Improve the power factor of ac circuits
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Summary

The power factor is the cosine of the phase angle (φ) between the
current and the voltage in a circuit.

The power factor is described in terms of:
–
a number ranging from 0 to 1, plus
a statement that the current leads or lags the voltage (unless the
power factor is 1, where the two are in phase.

The disadvantages of a low power factor are:
–
–
–
–
–
–
reduced overall efficiency
increased power losses
larger alternators and transformers
larger transmission conductors
higher rated control and protection equipment
increased voltage drop
–
overall increased costs

To improve the power factor of a load, the inductive VARs are cancelled
by the application of capacitive VARs.

The equations used to determine the value of the capacitor required are:
XC 

The capacitor VARs are implemented by using:
–
–
–
16
VC
1
and C =
IC
2fX C
capacitors individually in a small load
capacitors arranged in a bank (parallel and parallel/series) for
larger single loads
synchronous capacitors (motors) for very large installations.
EGG202A: 8 Improve the power factor of ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3817
Check your progress
In questions 1–5, place the letter matching your answer in the brackets provided.
Part A
1
The power factor of an ac circuit cannot exceed:
(a) 0.5
(b) 0.707
(c) 1.0
(d) 2.0 .
2
( )
Reducing the value of inductance in parallel with a resistor connected to an ac supply
will cause the power factor to:
(a) remain the same
(b) reach unity
(c) increase
(d) decrease.
3
( )
A main disadvantage of a low power factor in a supply system is that:
(a) larger currents are necessary for the same power
(b) the waveshape of the EMF changes from a sine waveform
(c) the true power increases
(d) smaller conductors must be used.
4
( )
Connecting a capacitor in parallel to a fluorescent lamp circuit will increase the:
(a) circuit power factor.
(b) apparent power
(c) true power
(d) reactive power.
EGG202A: 8 Improve the power factor of ac circuits
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( )
17
5
When capacitors improve the power factor of an inductive load the total power will:
(a) remain the same
(b) fluctuate
(c) increase
(d) decrease.
( )
Part B
1
The fluorescent lighting load in a shopping centre is measured at 6000 W at
0.26 power factor. Calculate the rating of a capacitor bank to improve the power factor
to 0.8.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
2 A 1.5 kW single phase load is to be connected to a portable alternator. Determine the
rating of the alternator and the current carrying capacity of the connecting leads if the
load is supplied at 240 V and has a power factor of:
(a) unity
___________________________________________________________________
___________________________________________________________________
(b) 0.8 lag
___________________________________________________________________
___________________________________________________________________
(c) 0.6 lag.
___________________________________________________________________
___________________________________________________________________
Answers to Check your progress are at the end of the module.
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EGG202A: 8 Improve the power factor of ac circuits
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Answers
Student exercise 1
1 (a)
1783 VAR (1.78 kVAR)
(b)
3280 VAR (3.28 kVAR)
2 (a)
2.08 kVAR
(b)
3.91 kVAR
Student exercise 2
1 (a)
20.1 µF
(b)
35.2 µF
2 (a)
(b)
61 µF
87.2 µF
Check your progress
Part A
1 (c)
4 (a)
2 (d)
5 (a)
3 (a)
Part B
1 17.8 kVAR
2 (a)
(b)
1 5 kVA, 62.5 A
18.75 kVA, 78.125 A
EGG202A: 8 Improve the power factor of ac circuits
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(c)
2
25 kVA, 104.17 A
EGG202A: 8 Improve the power factor of ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3817