EDS Unit 5.pptx

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY
Jonnada, Andhra Pradesh- 535005
UNIT–V: Compensation for Power
Factor Improvement
Department of Electrical and Electronics Engineering

Syllabus
 Capacitive compensation for power–factor control
 Different types of power capacitors
 shunt and series capacitors
 Effect of shunt capacitors (Fixed and switched)
 Power factor correction
 Capacitor allocation
 Economic justification
 Procedure to determine the best capacitor location.

Capacitive compensation for power–factor control
Power Factor Correction (Power Factor Compensation) uses parallel connected capacitors to oppose
the effects of inductive elements and reduce the phase shift between the voltage and current
Power Factor Correction is a technique which uses capacitors to reduce the reactive power
component of an AC circuit in order to improve its efficiency and reduce current.
When dealing with direct current (DC) circuits, the power dissipated by the connected load is
simply calculated as the product of the DC voltage times the DC current, that is V*I, given in watts
(W).
For a fixed resistive load, current is proportional to the applied voltage so the electrical power
dissipated by the resistive load will be linear.
For an AC circuit, the power dissipated in watts at any instant in time is equal to the product of the
volts and amperes at that exact same instant.

In a DC circuit the average power is simply V*I, but the average power of an AC circuit is not the same
value as many AC loads have inductive elements, such as coils, windings, transformers, etc. where the
current is out of phase with the voltage by some degrees resulting in the actual power dissipated in
watts being less than the product of the voltage and current.
For an AC circuit containing an inductor, coil, or solenoid or some other form of inductive load, its
inductive reactance (XL) creates a phase angle with the current lagging behind the voltage by 90o.

RL Series Circuit R and L Vector Diagrams

Phasor Diagram for the Series RL Circuit

Power Factor Correction Example No1
An RL series circuit consists of a resistance of 15Ω and an inductor which has an inductive reactance
of 26Ω. If a current of 5 amperes flows around the circuit, calculate:
1) the supply voltage.
2) the phase angle between the supply voltage and circuit current.
3) Draw the resulting phasor diagram.
Solution: 1). The supply voltage VS

We can double check this answer of 150Vrms using the impedances of the circuit as follows:
2). The phase angle (phi) using Trigonometry functions is:

3). The resulting phasor diagram showing VS

Power in a RL Series Circuit
The power consumed by the resistor in watts is called the “real power” so is given the symbol “P”
(or W). Watts can also be calculated as I2R, where R is the total resistance of the circuit.
Real Power, P = V*I cos(Θ)
Inductive Power Triangle

Capacitive Power Triangle

Power Triangle Equations

Power Factor Correction Example No. 2
A coil has a resistance of 10Ω and an inductance of 46mH. If it draws a current of 5 Amperes when
connected to a 100Vrms, 60Hz supply, calculate:
1) the voltages across the components.
2) the phase angle of the circuit.
3) the different powers consumed by the series RL circuit.

First find the impedances
1). The voltages across the resistor, VR and inductor, VL

2). The phase angle of the circuit
3). The circuit power

Power Factor Correction
• Power Factor Correction (Power Factor Compensation) improves the phase angle between the supply
voltage and current while the real power consumption in watts remains the same.
• Adding an impedance in the form of capacitive reactance in parallel with the coil above will
decrease Θ and thus increases the power factor which in turn reduces the circuits rms current
drawn from the supply.
• The power factor of an AC circuit can vary from between 0 and 1 depending on the strength of the
inductive load but in reality it can never be less than about 0.2 for the heaviest of inductive loads.
• Adding a capacitor in parallel with the coil will not only reduce this unwanted reactive power, but
will also reduce the total amount of current taken from the source supply.

A power factor of 0.95 is equal to a phase angle of: cos(0.95) = 18.2o thus the amount of VAR required
is:
If the original uncorrected VAR value was 433VAR and the new calculated value is 82.2VAR, we need
a reduction of 433 – 82.2 = 350.8 VAR(capacitive).
Therefore:

Therefore the capacitance of the capacitor is calculated as:
So to improve the power factor of the coil in example no2 from 0.5 to 0.95 requires a parallel
connected capacitor of 93uF.
New Volt-Amperes Value

Power Triangle
If the circuits apparent power has been reduced from 500VA to just 263VA, we can calculate the rms
current supplied as:
S = V*I, therefore: I = S/V = 263/100 = 2.63 Amperes

Final Power Factor Correction Circuit
This has the effect of reducing the circuits power factor, that is the ratio of active power to the
apparent power, as well as improving the power quality of the circuit and reduces the amount of
source current required. This technique is called “Power Factor Correction”.

ADVANTAGES AND BENEFITS OF POWER FACTOR IMPROVEMENT
(i) The cost and bill for the energy is reduced since there is penalty for poor power factor (<0.85)
(ii) Voltage improvement and better regulation
(iii) Overall system losses becomes reduced and hence is important from conservation of energy
Table 1: Typical power factor values of industrial loads and plants

Different types of power capacitors
• A power capacitor is a highly technical and complex device in that very thin dielectric materials
and high electric stresses are involved, coupled with highly sophisticated processing techniques.
• In the past, most power capacitors were constructed with two sheets of pure aluminum foil
separated by three or more layers of chemically impregnated kraft paper.
• Capacitor sizes have increased from the 15–25 kvar range to the 200–300 kvar range (capacitor
banks are usually supplied in sizes ranging from 300 to 1800 kvar).
• Capacitors are getting more attention today than ever before, partly due to a new dimension
added in the analysis: changeout economics.
• Under certain circumstances, even replacement of older capacitors can be justified on the basis of
lower-loss evaluations of the modern capacitor design.

Power capacitors are passive electronic components that provide
a static source of reactive power in electrical distribution systems.
Capacitance, a measure of energy storage ability, is typically
expressed as
C = K A/D,
where A is the area of the electrodes, D is their separation,
and K is a function of the dielectric between the electrodes.
Power capacitors are used in:
•the aerospace and automotive industries
•power factor correction and lighting circuits
•power supplies
•motor starters

1. Aluminum Electrolytic Capacitors: Aluminum electrolytic capacitors use an electrolytic process
to form the dielectric. Wet electrolytic capacitors have a moist electrolyte. Dry or solid electrolytic
capacitors do not.
2. Ceramic Capacitors: Ceramic capacitors have a dielectric made of ceramic materials.
3. Chip Capacitors: Chip capacitors or surface mount capacitors do not have leads.
4. Film Capacitors: Film capacitors are insulated with polyester, polycarbonate, polypropylene,
polystyrene, or other dielectric materials.
5. High Voltage Capacitors: High voltage capacitors are used for storing charge and energy in high
voltage applications.
6. Supercapacitors: Ultracapacitors store charges (energy) by physically separating positive and
negative charges (unlike batteries which do so chemically). Very high power densities can be
achieved by this method.

7. Tantalum Capacitors: Tantalum capacitors are used in smaller electronic devices
including portable telephones, pagers, personal computers, and automotive electronics.

Shunt and Series capacitors
1. Series Capacitors
Series capacitors, that is, capacitors connected in
series with lines, have been used to a very limited
extent on distribution circuits due to being a
more specialized type of apparatus with a limited
range of application.
Fig. 1: Voltage phasor diagrams for a feeder circuit of
lagging power factor: (a) and (c) without and (b) and (d)
with series capacitors.
• A series capacitor is a negative (capacitive)
reactance in series with the circuit’s positive
(inductive) reactance with the effect of
compensating for part or all of it.

• A series capacitor can even be considered as a voltage regulator that provides for a voltage boost
that is proportional to the magnitude and power factor of the through current.
• A series capacitor produces more net voltage rise than a shunt capacitor at lower power factors,
which creates more voltage drop.

Fig. 1: Voltage phasor diagrams for a feeder circuit of lagging power
factor: (a) and (c) without and (b) and (d) with series capacitors.
Consider the feeder circuit and its
voltage phasor diagram as shown in
Figure 1 (a) and (c).
The voltage drop through the feeder can
be expressed approximately as
cos sin
L
VD IR IX
 
  (1)
where
R is the resistance of the feeder circuit
XL is the inductive reactance of the feeder
circuit
cos θ is the receiving-end power factor
sin θ is the sine of the receiving-end power
factor angle

However, when a series capacitor is applied, as shown in Figure 1 (b) and (d), the resultant lower
voltage drop can be calculated as
cos ( )sin
L C
VD IR I X X
 
   (2)
where Xc is the capacitive reactance of the series capacitor.
1.1 Overcompensation
The series-capacitor size is selected for a distribution feeder application in such a way that the
resultant capacitive reactance is smaller than the inductive reactance of the feeder circuit.
However, in certain applications (where the resistance of the feeder circuit is larger than its inductive
reactance), the reverse might be preferred so that the resultant voltage drop is
cos ( )sin
C L
VD IR I X X
 
   (3)
The resultant condition is known as overcompensation.

Fig. 2 Overcompensation of the receiving-end voltage: (a) at
normal load and (b) at the start of a large motor.

1.2 Leading Power Factor
To decrease the voltage drop considerably between the sending and receiving ends by the application
of a series capacitor, the load current must have a lagging power factor.
Fig. 3 Voltage phasor diagram with leading power factor: (a)
without series capacitors and (b) with series capacitors.
As can be seen from the figure, the receiving
end voltage is reduced as a result of having
series capacitors.
When cos θ = 1.0, sin θ ≅ 0, and therefore,
( )sin 0
L C
I X X 
 
hence, Equation (2) becomes
VD IR
 (4)
Thus, in such applications, series capacitors practically have no value.

2. Shunt Capacitors
• Shunt capacitors, that is,
capacitors connected in
parallel with lines, are used
extensively in distribution
systems.
• Shunt capacitors supply the
type of reactive power or
current to counteract the out
of- phase component of
current required by an
inductive load. Fig. 4 Voltage phasor diagrams for a feeder circuit of lagging power
factor: (a) and (c) without and (b) and (d) with shunt capacitors.

Voltage drop in feeders, or in short transmission lines, with lagging power factor can be
approximated as
R X L
VD I R I X
  (5)
where
R is the total resistance of the feeder circuit, Ω
XL is the total inductive reactance of the feeder circuit, Ω
IR is the real power (or in-phase) component of the current, A
IX is the reactive (or out-of-phase) component of the current lagging the voltage by 90°, A

Example 1
Consider the right-angle triangle shown in Figure 5 (b). Determine the power factor of the load on a
460 V three-phase system, if the ammeter reads 100 A and the wattmeter reads 70 kW.
Fig. 5 (a) Phasor diagram and (b) power triangle for a typical
distribution load.
Solution:
3( )( )
1000
V I
S 
3(460 )(100 )
1000
V A

79.67 kVA


Thus,
cos
P
pf
S

 
70
79.67
kW
kVAR

0.88 88 %
or


Effect of shunt capacitors (Fixed and switched)
Fixed Capacitor Bank
• There are certain loads mainly certain industrial loads which need fixed reactive power to meet
power factor.
• In this type feeder fixed capacitor bank is used.
• These banks do not have separate control system to switch ON or OFF.
• These banks run with feeders.
• The banks are connected to the feeders as long as the feeders are live.

Effect of shunt capacitors (Fixed and switched)
Switched Capacitor Banks
• In high voltage power system, compensation of reactive power is mainly required during peak load
condition of system. There may be reverse effect if the bank is connected to the system at mean load
condition.
• At low load condition, the capacitive effect of bank may increase the reactive power of the system
instead of decreasing it.
• In this situation capacitors bank must be switched ON during peak load poor power factor
condition and must also be switched OFF during low load and high power factor condition. Here
switched capacitor banks are used.

The cosine of the angle between current and sending voltage is known as the power factor of the
circuit.
Assume that a load is supplied with a real power P, lagging reactive power Q1, and apparent power S1
at a lagging power factor of
(1)
or
When a shunt capacitor of Qc kVA is installed at the load, the power factor can be improved from cos θ1
to cos θ2, where

or
(2)
2.1 Concept of Leading and Lagging Power Factors
In general, for a given load, the power factor
is lagging if the load withdraws reactive
power; on the other hand, it is leading if the
load supplies reactive power.
Fig. 6 Examples of some of the sources of leading
and lagging reactive power at the load.

2.2 Economic Power Factor
• As can be observed from Figure 7 (b), the apparent power and the reactive power are decreased from S1
to S2 kVA and from Q1 to Q2 kVAR (by providing a reactive power of Q), respectively.
• The reduction of reactive current results in a reduced total current, which in turn causes less power
losses.
Fig. 7 Illustration of power factor correction.
• The economic power factor is the
point at which the economic benefits
of adding shunt capacitors just equal
the cost of the capacitors.
• In the past, this economic power factor was around 95%.
• Today’s high plant and fuel costs have pushed the economic power factor toward unity.

Example 2
Assume that a substation supplies three different kinds of loads, mainly, incandescent lights, induction motors,
and synchronous motors, as shown in Figure 8. The substation power factor is found from the total reactive and
real powers of the various loads that are connected. Based on the given data in Figure 8, determine the following:
a. The apparent, real, and kVARs of each load
b. The total apparent, real, and reactive powers of the power that should be supplied by the substation
c. The total power factor of the substation
Fig. 8: For Example 8.2: (a)
connection diagram, (b) phasor
diagrams of individual loads,
and (c) phasor diagram of
combined loads.

Solution
a.
1. For a 100 kVA lighting load Since incandescent lights are basically a unity power factor load, it is
assumed that all the current is kilowatt current. Hence,
1 1
100 100
S P
kVA kW


2. For 400 hp of connected induction motor loads
Assume that for the motor loads,

3. 200 hp motor with a 0.8 leading power factor
At full load, assume kVA = motor-hp rating = 200 kVA:

b. At the substation, the total real power is

c. The kVA of the substation is
The power factor of the substation is

Example 3
Assume that a load withdraws 80 kW and 60 kvar at a 0.8 power factor. It is required that its power
factor is to be improved from 80% to 90% by using capacitors. Determine the amount of the reactive
power to be provided by using capacitors.
Without capacitors at PF = 0.8
Solutions
kW = 80
kvar = 60
Thus, the kVA requirement of the load is
kVA = (802 + 602)1/2 = 100 kVA

With capacitors at PF = 0.9
kW = 80
kVA ≅ 88.9
Hence, the line supplies 38.7 kvar and the load needs 60 kvar, and the capacitor supplies the
difference, or

Example 4
Assume that a certain load withdraws a kilowatt current of 2 A and kilovar current of 2 A. Determine
the amount of total current that it withdraws.
Solution:
(kvar current)^2+ (kW current)^2= (total current)^2
(2 A)^2+ (2 A)^2= (total current)^2
or
4 + 4 = (total current)^2
Hence, Total current = 8^1/2= 2.83 A

Example 5
Assume that a 460 V cable circuit is rated at 240 A but is carrying a load of 320 A at 0.65 power factor.
Determine the kvar of capacitor that is needed to reduce the current to 240 A.
Solution
The kVA corresponding to 240 A is

Thus, the operating power factor corresponding to the new load is
The capacitor kvar required is

Example 6
Assume that a 1000 kW turbine unit (turbogenerator set) has a turbine capability of 1250 kW. It is
operating at a rated load of 1250 kVA at 0.85 power factor. An additional load of 150 kW at 0.85 power
factor is to be added. Determine the value of capacitors needed in order not to overload the turbine nor
the generator.
Solution:
Original load

Additional load
Total load

The minimum operating power factor for a load of 1150 kW and not exceeding the kVA rating of the
generator is
The maximum load kvar for this situation is

where 0.426 is the tangent corresponding to the maximum power factor of 0.935.
Thus, the capacitors must provide the difference between the total load kvar and the permissible
generator kvar, or

OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS
For a given power Pk the peak power supplied, Qc required capacitor bank kVAR, is obtained by
Where tan is the original p.f and tan the optimum p.f. tan is obtained with the following
procedure.
 1
 
(i) The best desired p.f (a value chosen between 0.90 to 0.98) is set.
(ii) Additional savings in ‘kW’ losses at the set p.f value, when all capacitors are connected to the
substation or distribution bus is determined.
(iii) Reduction in losses due to capacitors applied to substation buses and additional losses due to
capacitors are computed.
(iv) Reduction in KVA demand and hence the additional capacitors needed at buses and feeders is
computed.

v. Total annual savings in demand reduction due to additional capacitors applied, is computed
(Rs/year).
vi. Total annual saving due to additional released capacity per year and hence the annual savings due
to energy losses (Rs/year).
vii. Total annual losses due to cost of capacitors, energy losses in capacitors and other operating
expenses are computed.
viii.Net savings (annual) is arrived at (Rs/year)
ix. Is the power factor set economical? For this, the procedure is repeated by setting a new value.
x. Best p.f is one which gives maximum value for step viii.

1. Best Capacitor Location and Optimum Capacitor Size
The capacitor size and location is usually determined by minimizing (i) power loss (ii) voltage
regulation, and (iii) additional cost and other expenses for illustration.
The procedure usually followed is
(i) to determine the desired p.f for the given load or circuit from the data, viz., kW or KVA and p.f
(ii) the feeder or line voltage and improvement needed
(iii) to determine correction factor needed to correct either load or feeder circuit p.f from the existing
one
(iv) to check the existing capacitors or other p.f correcting devices, their capacity, location, etc., and
improvement given

Optimal Location of Capacitor
• To obtain the best location for the capacitor banks, the feeder line is decided into a number of
sections with a combination of both uniformly distributed loads and concentrated loads.
• This component is counterbalanced by a capacitor bank in each section or segment such that power
loss due to inductive current is reduced by the capacitor allocated in that section.
• The ratio of the reactive (capacitive) ‘kVAR’ of the capacitor to the total reactive load is determined
along with the ratio reactive current at the end of the line segment to the beginning of the line
segment (a).
• The distance of capacitor bank from the feeding point of the line segment is obtained by considering
(a) the ratio, (b) original losses due to reactive current, (c) capacitance, and (d) compensation ratio.
• The optimum number of capacitor banks is obtained by making the cost function of the above
variables a minimum.

2 Benefits with Capacitor Installation in Distribution Systems
In general, the benefits from installation of capacitors are
(i) The generation plant capacity, together with transmission and distribution substations and network
capacity is released and hence increased to that extent
(ii) Reduced energy losses (power losses)
(iii) Reduced voltage drop and improved voltage regulation
(iv) Increase in revenue due to voltage improvement, reduced losses and hence supply of more energy
Quantitative estimate of the above benefit and hence the revenue increase is a more involved procedure and
complex calculations are needed.
The total benefits due to capacitors installation can be put as total financial (benefit) obtained due to (i)
Demand reduction + (ii) Energy savings (reduction) + Revenue increase due to release of additional capacity.

Review Questions
1. Explain the need for p.f improvement in distribution systems.
2. How in the capacitor bank ratings obtained when the load p.f is to be improved from cos theta1 to cos
theta 2?
3. Explain how reduction in line current and hence power losses are obtained with p.f improvement.
4. What are the different locations for p.f improvement capacitors? Discuss their relative advantages and
disadvantages.
5. How does p.f improvement help in reduction in % VD and hence voltage regulation of distribution
transformers? With transformer of 6% reactance and 2% magnetizing current, what will be the probable
ratings of capacitor bank to compensate for magnetizing current and improve regulation by 2%?
6. What is series capacitor compensation in feeder lines ? How does it improve the regulation of the lines?
Discuss with suitable examples.

8. Discuss how voltage profi le of a long feeder can be improved by connecting shunt capacitor
banks at the end of the feeders.
9. How is economical power factor arrived at for a given distribution system with different loads ?
10. What is the justification for p.f improve and what are the benefits ?
11. Compare and explain role of shunt and series capacitors in p.f correction.

EDS Unit 5.pptx

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EDS Unit 5.pptx