INTRODUCTION
The electrical energy is almost exclusively generated, transmitted and distributed in the form of alternating current. Therefore, the question of power factor immediately comes into picture. Most of the loads (e.g. induction motors, arc lamps) are inductive in nature and hence have low lagging power factor. The low power factor is highly undesirable as it causes an increase in current, resulting in additional losses of active power in all the elements of power system from power station generator down to the utilization devices. In order to ensure most favorable conditions for a supply system from engineering and economical standpoint, it is important to have power factor as close to unity as possible.
1.1 POWER FACTOR
The cosine of angle between voltage and current in an a.c. circuit is known as
Power factor.

Fig: 1.1
Consider an inductive circuit taking a lagging current I from sup ply voltage V; the angle of lag being φ. The phasor diagram of the circuit is shown in Fig. 1.1
The circuit current I can be resolved into two perpendicular components, namely;
a) Icos φ in phase with V
b)I sin φ 90° out of phase with V
The component I cosφ is known as active or wattful component, Fig. 1.1 whereas component I sinφ is called the reactive or wattless component. The reactive component is a measure of the power factor. If the reactive component is small, the phase angle φ is small and hence power factor cosφ will be high. Therefore, a circuit having small reactive current (i.e., I sinφ) will have high power factor and viceversa. It may be noted that value of power factor can never be more than unity.
1)It is a usual practice to attach the word ‘lagging’ or ‘leading’ with the numerical value of power factor to signify whether the current lags or leads the voltage. Thus if the circuit has a p.f. of 0.5 and the current lags the voltage, we generally write p.f. as 0.5 lagging.
2)Sometimes power factor is expressed as a percentage. Thus 0.8 lagging power factor may be expressed as 80% lagging.
1.2 IMPORTANCE OF POWER FACTOR IMPROVEMENT
The improvement of power factor is very important for both consumers and generating stations as discussed below:
1)For Consumers: A consumer has to pay electricity charges for his maximum demand in kVA plus the units consumed. If the consumer improves the power factor, then there is a reduction in his maximum kVA demand and consequently there will be annual saving due to maximum demand charges. Although power factor improvement involves extra annual expenditure on account of p.f. correction equipment, yet improvement of p.f. to a proper value results in the net annual saving for the consumer.
2)For generating stations: A generating station is as much concerned with power factor improvement as the consumer. The generators in a power station are rated in kVA but the useful output depends upon kW output. As station output is kW= kVA x cosφ , therefore, number of units supplied by it depends upon the power factor. The greater the power factor of the generating station, the higher is the kWh it delivers to the system. This leads to the conclusion that improved power factor increases the earning capacity of the power station.
1.3 MOST ECONOMICAL POWER FACTOR
If a consumer improves the power factor, there is reduction in his maximum kVA demand and hence there will be annual saving over the maximum demand charges. However, when power factor is improved, it involves capital investment on the power factor correction equipment. The consumer will incur expenditure every year in the shape of annual interest and depreciation on the investment made over the p.f. correction equipment. Therefore, the net annual saving will be equal to the annual saving in maximum demand charges minus annual expenditure incurred on p.f. correction equipment.
The value to which the power factor should be improved so as to have maximum net annual saving is known as the most Economical Power Factor.

Fig: 1.2
kVA max. demand at cos φ_{1} , kVA_{1} = P/cos φ_{1} = P sec φ_{1}
kVA max. demand at cos φ_{2} , kVA_{2} = P/cos φ_{2} = P sec φ_{2}
Annual saving in maximum demand charges
= Rs x (kVA_{1} kVA_{2})
= Rs x (P sec φ_{1} P sec φ_{2})
= Rs x P(sec φ_{1} sec φ_{2)}
Reactive power at cos φ_{1} , kVAR_{1} = P tan φ_{1}
Reactive power at cos φ_{2}’ kVAR_{2} = P tan φ_{2}
Leading kVAR taken by p.f. correction equipment
= P (tan φ_{1} tan φ_{2})
Annual cost of p.f. correction equipment
= Rs Pv (tan φ_{1} tan φ_{2})
Net annual saving, S = exp. (i) exp. (ii)
= xP (sec φ_{1_} sec φ_{2}) yP (tan φ_{1} tan φ_{1})
In this expression, only is variable while all other quantities are fixed. Therefore, the net annual saving will be maximum if differentiation of above expression w.r:t. φ_{1} is zero i.e.
d/dφ_{2 (}S) = 0
or d/dφ_{2 }[xP (sec φ_{1} sec φ_{2 }) yP (tan φ_{1} tan φ_{2}) = 0
or d/dφ_{2 }(xP sec φ_{1}) d/dφ_{2 }(xP sec φ_{2}) d/dφ_{2 }(yP tan φ_{1}) + yP (tan φ_{2}) = 0
or 0 xP sec φ_{2} tan φ_{2} 0 + yP sec^{2} φ_{2 }) = 0
or x tan φ_{2 }+ y sec φ_{2} = 0
or tan φ_{2} = y/x sec φ_{2}
or sin φ_{2} = y/x
\Most economical power factor, Cos φ_{2} = Öi sin^{2} φ_{2} = Öi (y/ x)^{2}
It may be noted that the most economical power factor (cos φ_{2 }) depends upon the relative costs of supply and p.f. correction equipment but is independent of the original p.f. cos φ_{1.}
1.4 POWER TRIANGLE
The analysis of power factor can also be made in terms of power drawn by the a.c. circuit. If each side of the current triangle oab of Fig. 1.3 is multiplied by voltage V, then we get the power triangle OAB shown in Fig. 1.3 where

1) VI Sinφ The apparent power in an ac. circuit has two components viz,
active and reactive power at right angles to each other.
OB^{2 }= OA^{2}+AB^{2}
or (apparent power)^{2} = (active power)^{2 }+ (reactive power)^{2 }
or (kVA)^{2 }= (kW)^{2 }+ (kVAR)^{2}
2) Power factor, cos φ = OA⁄OB= active power ⁄ apparent power =KW ⁄ KVA
Thus the power factor of a circuit may also be defined as the ratio of active power to the apparent power. This is a perfectly general definition and can be applied to all cases, whatever be the waveform.
3) The lagging reactive power is responsible for the low power factor. It is clear from the power triangle that smaller the reactive power component, the higher is the power factor of the circuit.
kVAR = kVA sin φ = cos φ sin φ
kVAR = kWtanφ
4) For leading currents, the power triangle becomes reversed. This fact provides a key to the power factor improvement. If a device taking leading reactive power (e.g. capacitor) is connected in parallel with the load, then the lagging reactive power of the load will be partly neutralised, thus improving the power factor of the load.
5) The power factor of a circuit can be defined in one of the following three ways:
a)Power factor = cos φ = cosine of angle between V and I
b)Power factor = R/Z= Resistance ⁄ Impedance
c)Power factor = VI cos φ ⁄ VI= Active power ∕ Apparent Power
6) The reactive power is neither consumed in the circuit nor it does any useful work. It merely flows back and forth in both directions in the circuit. A wattmeter does not measure reactive power.
Let us illustrate the power relations in an ac. circuit with an example. Suppose a
circuit draws a current of 10 A at a voltage of 200 V and its p.f. is 0.8 lagging. Then,
Apparent power = VI = 200 x 10 = 2000 VA
Active power = VI cos φ = 200 x 10 x 0.8 = 1600 W
Reactive power = VI sin φ = 200 x 10 x 0.6 = 1200 VAR
The circuit receives an apparent power of 2000 VA and is able to convert only 1600 watts into active power. The reactive power is 1200 VAR and does no useful work. It merely flows into and out of the circuit periodically. In fact, reactive power is a liability on the source because the source has to supply the additional current (i.e., I sinφ).
1.5 POWER FACTOR LINEAR CIRCUIT
The power factor of an AC electric power system is defined as the ratio of the real power to the apparent power, and is a number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf). Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a nonlinear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power.
In an electric power system, a load with low power factor draws more current than a load with a high power factor, for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor.
Linear loads with low power factor (such as induction motors) can be corrected with a passive network of capacitors or inductors. Nonlinear loads, such as rectifiers, distort the current drawn from the system. In such cases, active power factor correction is used to counteract the distortion and raise power factor. The devices for correction of power factor may be at a central substation, or spread out over a distribution system, or built into power using equipment.
Instantaneous and average power calculated from AC voltage and current with a unity power factor (φ=0, cosφ=1)
Instantaneous and average power calculated from AC voltage and current with a zero power factor (φ=90, cosφ=0)
Instantaneous and average power calculated from AC voltage and current with a lagging power factor (φ=45, cosφ=0.71)
In a purely resistive AC circuit, voltage and current waveforms are in step (or in phase), changing polarity at the same instant in each cycle. Where reactive loads are present, such as with capacitors or inductors, energy storage in the loads result in a time difference between the current and voltage waveforms. This stored energy returns to the source and is not available to do work at the load. Thus, a circuit with a low power factor will have higher currents to transfer a given quantity of real power than a circuit with a high power factor. A linear load does not change the shape of the waveform of the current, but may change the relative timing (phase) between voltage and current.
Circuits containing purely resistive heating elements (filament lamps, strip heaters, cooking stoves, etc.) have a power factor of 1.0. Circuits containing inductive or capacitive elements (lamp ballasts, motors, etc.) often have a power factor below 1.0.
DEFINITION AND CALCULATION
The power factor is defined as:
In the case of a perfectly sinusoidal waveform, P, Q and S can be expressed as vectors that form a vector triangle such that:
If φ is the phase angle between the current and voltage, then the power factor is equal to, and
Since the units are consistent, the power factor is by definition a dimensionless number between 0 and 1. When power factor is equal to 0, the energy flow is entirely reactive, and stored energy in the load returns to the source on each cycle. When the power factor is 1, all the energy supplied by the source is consumed by the load. Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle, where leading indicates a negative sign.
If a purely resistive load is connected to a power supply, current and voltage will change polarity in step, the power factor will be unity (1), and the electrical energy flows in a single direction across the network in each cycle. Inductive loads such as transformers and motors (any type of wound coil) consume reactive power with current waveform lagging the voltage. Capacitive loads such as capacitor banks or buried cable generate reactive power with current phase leading the voltage. Both types of loads will absorb energy during part of the AC cycle, which is stored in the device's magnetic or electric field, only to return this energy back to the source during the rest of the cycle.
For example, to get 1 kW of real power, if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.2 power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷ 0.2 = 5 kVA). This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes.
1.5.1 LINEAR LOADS
The reactive power increases the current flowing between the power source and the load, which increases the power losses through transmission and distribution lines. This results in additional costs for power companies. Reactive power can require the use of wiring, switches, circuit breakers, transformers and transmission lines with higher current capacities. Therefore, power companies require their customers, especially those with large loads, to maintain their power factors above a specified amount (usually 0.90 or higher) or be subject to additional charges. Electricity utilities measure reactive power used by high demand customers and charge higher rates accordingly. Some consumers install power factor correction schemes at their factories to cut down on these higher costs.
1.5.2 POWER FACTOR CORRECTION OF LINEAR LOADS
It is often desirable to adjust the power factor of a system to near 1.0. This power factor correction is achieved by switching in or out banks of inductors or capacitors. For example the inductive effect of motor loads may be offset by locally connected capacitors. When reactive elements supply or absorb reactive power near the load, the apparent power is reduced.
Power factor correction may be applied by an electrical power transmission utility to improve the stability and efficiency of the transmission network. Correction equipment may be installed by individual electrical customers to reduce the costs charged to them by their electricity supplier. A high power factor is generally desirable in a transmission system to reduce transmission losses and improve voltage regulation at the load.
Power factor correction brings the power factor of an AC power circuit closer to 1 by supplying reactive power of opposite sign, adding capacitors or inductors which act to cancel the inductive or capacitive effects of the load, respectively. For example, the inductive effect of motor loads may be offset by locally connected capacitors. If a load had a capacitive value, inductors (also known as reactors in this context) are connected to correct the power factor. In the electricity industry, inductors are said to consume reactive power and capacitors are said to supply it, even though the reactive power is actually just moving back and forth on each AC cycle.
The reactive elements can create voltage fluctuations and harmonic noise when switched on or off. They will supply or sink reactive power regardless of whether there is a corresponding load operating nearby, increasing the system's noload losses. In a worst case, reactive elements can interact with the system and with each other to create resonant conditions, resulting in system instability and severe over voltage fluctuations. As such, reactive elements cannot simply be applied at will, and power factor correction is normally subject to engineering analysis.
An automatic power factor correction unit is used to improve power factor. A power factor correction unit usually consists of a number of capacitors that are switched by means of contactors. These contactors are controlled by a regulator that measures power factor in an electrical network. To be able to measure 'power factor', the regulator uses a CT (Current transformer) to measure the current in one phase.
Depending on the load and power factor of the network, the power factor controller will switch the necessary blocks of capacitors in steps to make sure the power factor stays above 0.9 or other selected values (usually demanded by the energy supplier).
Legend:
 Reactive Power Control Relay
 Network connection points
 Slowblow Fuses
 Inrush Limiting Contactors
 Capacitors (singlephase or threephase units, deltaconnection)
 Transformer Suitable voltage transformation to suit control power (contactors, ventilation)
POWER FACTOR OF VARIOUS PERPOSE
2.1 Power in resistive Load
Fig: 2.1
Ac Source drives a purely resistive load.
In this example, the current to the load would be 2 amps, RMS. The power dissipated at the load would be 240 watts. Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations look similar to that in an equivalent DC circuit. If we were to plot the voltage, current, and power waveforms for this circuit, it would look like Figure below.
Fig: 2.2 Current is in phase with voltage in a resistive circuit.
Note that the waveform for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads. If the source were a mechanical generator, it would take 240 watts worth of mechanical energy (about 1/3 horsepower) to turn the shaft.
Also note that the waveform for power is not at the same frequency as the voltage or current! Rather, its frequency is double that of either the voltage or current waveforms. This different frequency prohibits our expression of power in an AC circuit using the same complex (rectangular or polar) notation as used for voltage, current, and impedance, because this form of mathematical symbolism implies unchanging phase relationships. When frequencies are not the same, phase relationships constantly change.
As strange as it may seem, the best way to proceed with AC power calculations is to use scalar notation, and to handle any relevant phase relationships with trigonometry.
2.2 Power in reactive Load
Consider a circuit for a singlephase AC power system, where a 120 volt, 60 Hz AC voltage source is delivering power to a reactive AC circuits: (Figure below)
For comparison, let's consider a simple AC circuit with a purely reactive load in Figure below.
Fig: 2.3
AC circuit with a purely reactive (inductive) load.
Fig: 2.4
Power is not dissipated in a purely reactive load. Though it is alternately absorbed from and returned to the source.
Note that the power alternates equally between cycles of positive and negative. (Figure above) This means that power is being alternately absorbed from and returned to the source. If the source were a mechanical generator, it would take (practically) no net mechanical energy to turn the shaft, because no power would be used by the load. The generator shaft would be easy to spin, and the inductor would not become warm as a resistor would.
Now, let's consider an AC circuit with a load consisting of both inductance and resistance in Figure below.
Fig: 2.5
AC circuit with both reactance and resistance.
At a frequency of 60 Hz, the 160 mill henry of inductance gives us 60.319 Ω of inductive reactance. This reactance combines with the 60 Ω of resistance to form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152^{o}. If we're not concerned with phase angles (which we're not at this point), we may calculate current in the circuit by taking the polar magnitude of the voltage source (120 volts) and dividing it by the polar magnitude of the impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load current is 1.410 amps. This is the figure an RMS ammeter would indicate if connected in series with the resistor and inductor.
We already know that reactive components dissipate zero power, as they equally absorb power from, and return power to, the rest of the circuit. Therefore, any inductive reactance in this load will likewise dissipate zero power. The only thing left to dissipate power here is the resistive portion of the load impedance. If we look at the waveform plot of voltage, current, and total power for this circuit, we see how this combination works in Figure below.
Fig: 2.6
A combined resistive/reactive circuit dissipates more power than it returns to the source. The reactance dissipates no power; though, the resistor does.
As with any reactive circuit, the power alternates between positive and negative instantaneous values over time. In a purely reactive circuit that alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source.
Looking at the waveform plot for power, it should be evident that the wave spends more time on the positive side of the center line than on the negative, indicating that there is more power absorbed by the load than there is returned to the circuit. What little returning of power that occurs is due to the reactance; the imbalance of positive versus negative power is due to the resistance as it dissipates energy outside of the circuit (usually in the form of heat). If the source were a mechanical generator, the amount of mechanical energy needed to turn the shaft would be the amount of power averaged between the positive and negative power cycles.
Mathematically representing power in an AC circuit is a challenge, because the power wave isn't at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance that we've worked with so far.
 REVIEW:
 In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
 In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90^{o} out of phase with each other.
 In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0^{o} and 90^{o}.
2.3 True, Reactive, and Apparent power
As a rule, true power is a function of a circuit's dissipative elements, usually resistances (R). Reactive power is a function of a circuit's reactance (X). Apparent power is a function of a circuit's total impedance (Z). Since we're dealing with scalar quantities for power calculation, any complex starting quantities such as voltage, current, and impedance must be represented by their polar magnitudes, not by real or imaginary rectangular components. For instance, if I'm calculating true power from current and resistance, I must use the polar magnitude for current, and not merely the “real” or “imaginary” portion of the current. If I'm calculating apparent power from voltage and impedance, both of these formerly complex quantities must be reduced to their polar magnitudes for the scalar arithmetic.
There are several power equations relating the three types of power to resistance, reactance, and impedance (all using scalar quantities)
P= true power, P= I^{2}R, P= E^{2}/R
Measured in units of watts
Q= Reactive power, Q= I^{2}X , Q= E^{2}/X
Measured in units of volt ampsReactive (VAR)
S= Apparent power, S= I^{2}Z , S= E^{2}/Z, S=IE
Measured in units of volt amps (VA)
Please note that there are two equations each for the calculation of true and reactive power. There are three equations available for the calculation of apparent power, P=IE being useful only for that purpose. Examine the following circuits and see how these three types of power interrelate for: a purely resistive load in Figure below, a purely reactive load in Figure below, and a resistive/reactive load in Figure below.
Resistive load only:
Fig: 2.7
True power, reactive power, and apparent power for a purely resistive load.
Reactive load only:
Fig: 2.8
True power, reactive power, and apparent power for a purely reactive load.
Resistive/reactive load:
Fig: 2.9
True power, reactive power, and apparent power for a resistive/reactive load.
These three types of power — true, reactive, and apparent — relate to one another in trigonometric form. We call this the power triangle
Fig: 2.10
Power triangle relating apparent power to true power and reactive power.
Using the laws of trigonometry, we can solve for the length of any side (amount of any type of power), given the lengths of the other two sides, or the length of one side and an angle.
POWER FACTOR IMPROVEMENT
3.1 POWER FACTOR IMPROVEMENT
The low power factor is mainly due to the fact that most of the power loads are inductive and, therefore, take lagging currents. In order to improve the power factor, some device taking leading power should be connected in parallel with the load. One of such devices can be a capacitor. The capacitor draws a leading current and partly or completely neutralises the lagging reactive component of load current. This raises the power factor of the load.
(i) (ii) (iii)
Fig: 3.1
To illustrate the power factor improvement by a capacitor, consider a single phase load taking lagging current I at a power factor cos φ_{1 }as shown in Fig. 3.1
The capacitor C is connected in parallel with the load. The capacitor draws current I_{C} which leads the supply voltage by 90°. The resulting line current I΄ is the phasor sum of I and I_{C} and its angle lag is φ_{2} as shown in the phasor diagram of Fig. 3.1 (iii). It is clear that φ_{2 }is less than φ_{1}, so that cos φ_{2} is greater than cos φ_{1}. Hence, the power factor of the load is improved. The following points worth noting:
I.The circuit current I΄after p.f. correction is less than the original circuit current I.
II. The active or wattful component remains the same before and after p.f. correction because only the lagging reactive component is reduced by the capacitor.
Icos φ_{1 = }I΄cos φ_{2}
III.The lagging reactive component is reduced after p.f. improvement and is equal to the difference between lagging reactive component of load (I sin φ_{1}) and capacitor current (I_{C}) i.e.,
I΄sin φ_{2 }= I sin φ_{1} I_{C}
IV.As I cos φ_{1} = I΄ cos φ_{2}
VI cos φ_{1} = V I΄ cos φ_{2 } [Multiplying by V]
Therefore, active power (kW) remains unchanged due to power factor improvement.
V. I΄ sin φ_{2} = I sin φ_{1 } I
V I΄ sin φ_{2} = VI sin φ_{1 } VI_{C} [Multiplying by V]
Net kVAR after p.f. correction = Lagging kVAR before p.f. correction leading kVAR of equipment.
3.1.1 POWER FACTOR IMPROVEMENT EQUIPMENT
Normally, the power factor of the whole load on a large generating station is in the region of 0.8 to 0.9. However, sometimes it is lower and in such cases it is generally desirable to ke special steps to improve the power factor. This can be achieved by the following equipment:
 Static capacitors.
 Synchronous condenser.
 phase advancers.
3.1.2 STATIC CAPACITOR
The power factor can be improved by connecting capacitors in parallel with the equipment operating at lagging power factor. The capacitor (generally known as static capacitor) draws a leading current and partly or completely neutralises the lagging reactive component of load current. This raises the power factor of the load. For threephase loads, the capacitors can be connected in delta or star as shown in Fig. Static capacitors are invariably used for power factor improvement in factories.
Fig: 3.2
Advantages
 They have low losses.
 They require little maintenance as there are no rotating parts.
 They can be easily installed as they are light and require no foundation.
 They can work under ordinary atmospheric conditions.
Disadvantages
 They have short service life ranging from 8 to 10 years.
 They are easily damaged if the voltage exceeds the rated value.
 Once the capacitors are damaged, their repair is uneconomical.
STATIC CAPACITOR BANKS
Static capacitor banks are intended to increase bandwidth and stability of power transmission lines, to stabilize voltage in load carrying components, to decrease power loss, to improve quality of distribution and reactive power compensation.
Capacitor banks can be operated separate units and as part of static thyristor compensators, filtering/compensating devices and in connection with shunting reactors, and with shunt reactors, controlled reactors and vacuumreactor groups.
Technical Data:
 Nominal Capacity, MVAr: 5 – 200
 Nominal Voltage, kV: 6, 10, 35, 110
 Neutral Terminal: isolated or deadgrounded
 Enclosure: in accordance to GOST 1515069 У1, УХЛ1
Fig: 3.3
3.1.3 SYNCHRONOUS CONDENSER
A synchronous motor takes a leading current when overexcited and, therefore, behaves as a capacitor. An overexcited synchronous motor running on no load is known as Synchronous condenser. When such a machine is connected in parallel with the supply, it takes a leading current which partly neutralises the lagging reactive component of the load. Thus the power factor is improved.
Fig shows the power factor improvement by synchronous condenser method. The 3φ load takes current I_{L }at low lagging power factor cos φ_{L.} The synchronous condenser takes a current I_{m} which leads the voltage by an angle φ_{m.}. The resultant current I is the phasor sum of I_{m} and I_{L} and lags behind the voltage by an angle φ. It is clear that φ is less than φ_{L} so that cos φ is greater than cos φ_{L. }Thus the power factor is increased from cos φ_{L} to cos φ . Synchronous condensers are generally used at major bulk supply substations for power factor improvement.
Fig: 3.4
Advantage
 By varying the field excitation, the magnitude of current drawn by the motor can be changed by any amount. This helps in achieving step less control of power factor.
 The motor windings have high thermal stability to circuit current .
 The fauls can be removed easily.
Disadvantage
 There are considerable losses in the motor.
 The maintenance cost is high.
 It produces noise.
 Except in sizes above 500 kVA, the cost is greater than that of static capacitors of the same rating.
 As a synchronous motor has no selfstarting torque, therefore, auxiliary equipment has to be provided for this purpose.
3.1.4 PHASE ADVANCERS
Fig:3.5
Phase advancers are used to improve the power factor of induction motors. The low power factor of an induction motor is due to the fact that stator winding draws exciting current which lags behind the supply voltage by 90º . If the exciting ampere turns can be provided from some other a.c. source, then a stator winding will be relieved of exciting current and a power factor of the motor can be improved. This job accomplished by the phase advancer which is simply an ac. exciter. The phase advancer is mounted on the same shaft as the main motor and is connected in the rotor circuit of the motor. It provides exciting ampere turns to the rotor circuit at slip frequency. By providing more ampere turns than required. the induction motor can be made to operate on leading power factor like an overexcited synchronous motor.
Phase advancers have two principal advantages. Firstly. As the exciting ampere turns are supplied at slip frequency, therefore, lagging kVAR drawn by the motor are considerably reduced. Secondly, phase advancer can be conveniently used where the use of synchronous motors is unadmissible. However, the major disadvantage of phase advancers is that they are not economical for motors beIow 200 H.P.
SYNCHRONOUS CONDENSER
INTRODUCTION
In electrical engineering, a synchronous condenser (sometimes synchronous compensator) is a synchronous motor that is not attached to any driven equipment. Its field is controlled by a voltage regulator to either generate or absorb reactive power as needed to support a system’s voltage or to maintain the system power factor at a specified level. The condenser’s installation and operation are identical to large electric motors.
Increasing the device's field excitation results in its furnishing magnetizing power (kilovars) to the system. Its principal advantage is the ease with which the amount of correction can be adjusted. The energy stored in the rotor of the machine can also help stabilize a power system during short circuits or rapidly fluctuating loads such as electric arc furnaces. Large installations of synchronous condensers are sometimes used in association with highvoltage direct current converter stations to supply reactive power.
Synchronous motors are used principally in large power applications because of their high operating efficiency, reliability, controllable power factor, and relatively low sensitivity to voltage dips. They are constantspeed machines with applications in mills, refineries, power plants, and the like, to drive pumps, compressors, fans, pulverizers, and other large loads, and to assist in powerfactor correction. One very interesting .application is the 44MW, bky, 60Hz, 50pole, 144 r/min synchronous motors used to drive the Queen Elizabeth 11 passenger ship. A solidstate volts/hertz drive circuit provides speed control through frequency adjustment.
Synchronous machines designed specifically for powerfactor control have no external shafts and are called synchronous condensers. They “float” on the bus, supplying reactive power to the system. The direction of the reactive power, and hence the power factor of the system, is adjusted by changing the field excitation of the machine.
Unless otherwise specified, when discussing the behavior of individual motors and/or generators, it will be assumed that the machine is connected to a power source of unlimited capacity and zero impedance, called an infinite bus. The terminal voltage and frequency of the infinite bus remain constant and are unaffected by any power drawn from or supplied to the infinite bus. Large power systems in the United States and other highly industrialized countries may be considered to approximate those of the infinite bus
4.1 SYNCHRONOUS MOTOR STARTING
When starting, the rotor is allowed to accelerate to its maximum speed as an induction motor using its builtin damper windings.2 At this speed the slip is very small and the rotating flux of the stator moves very slowly relative to the revolving rotor. Direct current is then applied to the magnet windings, forming alternate north and south poles that “lock” in rotational synchronism with the corresponding opposite poles of the rotating flux, the slip is zero, all inductionmotor action ceases, and the machine operates at synchronous speed. The synchronous speed, as developed in Section 4.5 in Chapter 4, is expressed in terms of the stator frequency and the number of poles:
n_{s}= r/min (1)
If the rotor magnets are energized before the machine reaches its maximum speed as an induction motor, the rotor may not synchronize and severe vibration will occur; every time a pole of the rotating stator flux passes a rotor pole, alternate attraction and repulsion occurs. Such outofstep operation, called pole slipping, also causes cyclic current surges and torque pulses at slip frequency in the armature windings.
The adverse effects of frequent starting, or operating with unbalanced threephase line voltages, are the same as for induction motors.
A simplified circuit diagram for rotor and stator connections is shown in Figure 4.1 A varistor, or resistor and switch, connected across the field leads during locked rotor and acceleration, prevents a high induced emf in the field windings, and the induced current in the circuit formed by the field windings and external resistor provides additional inductionmotor torque.
The circuit for a brushless excitation system is shown in Figure 8.6. A frequency sensitive solidstate control circuit monitors the frequency of the emf induced in the motor field winding by the rotating flux of the stator. The frequency of the emf is the same as that in the squirrelcage winding and is a function of the frequency of the applied stator voltage and the slip. That is,
f_{s =} sf_{s } (2)
At locked rotor s = 1 .0, causing the rotor frequency to equal the applied stator frequency. As the rotor accelerates, the slip becomes smaller, and the frequency decreases.
When the frequencysensitive circuit observes the “high frequency” generated in the rotor coils at locked rotor, and during highslip acceleration, it closes SCR2 and opens SCR I; opening SCR 1 blocks the exciter current, and closing SCR2 connects the discharge resistor across the field windings.
Fig: 4.1
Simplified circuit diagrams for rotor and stator connections of a synchronous motor.
At nearsynchronous speed, the rotor frequency is very low (approaching zero), and the control circuit opens closes SCRl, opening SCR2 disconnects the discharge resistor, and admits current to the field windings. The solidstate switching is programmed to close SCRl at an instant that will ensure that the rotor poles will be facing opposite polarity, thus preventing pole slipping.
Fig: 4.2
Circuit diagram of a brushless excitation system for a synchronous machine. (Courtesy Dresser Rand, Electric Machinery)
Synchronous motors without builtin starting components are not energized until they have been accelerated to approximately synchronous speed by an auxiliary motor or turbine. No load is connected to a synchronous motor during the acceleration period. When near synchronous speed, the magnet windings are energized, and the field current adjusted to obtain an induced stator voltage approximately equal to the supply voltage. The circuit breaker connecting the stator to the threephase supply is dosed when the induced stator voltage is in phase with the supply voltage. This procedure, called paralleling.
Reversing a Synchronous Motor
The direction of rotation of a synchronous motor is determined by its starting direction, as initiated by inductionmotor action. Thus, to reverse the direction of a threephase synchronous motor, it is necessary to first stop the motor and then reverse the phase sequence of the threephase connections at the stator. Reversing the current to the field windings will not affect the direction of rotation.
4.2 SHAFT LOAD, POWER ANGLE, AND DEVELOPED TORQUE
Although the rotor of a synchronous motor rotates in synchronism with the rotating flux of the stator. Increases in shaft load cause the rotor magnets to change their angular position with respect to the rotating flux. This displacement angle can be seen viewing the rotor with a strobe light synchronized with the stator frequency.
Figure 4.3 shows a “strobe view” of a salientpole motor operating at synchronous speed with no load on the shaft. The direction of rotation is counterclockwise (CCW). As the machine is loaded, the rotor changes its relative position with respect to the rotating flux of the stator. Lagging behind it by angle & Angle & expressed in electrical degrees, is called the power angle, load angle, or torque angle. The broken line on the South Pole in Figure 8.7 shows the position of the rotating magnets relative to the rotating flux of the stator for a representative load.
A synchronous motor operates at the same average speed for all values of load from no load to peak load. When the load on a synchronous motor is increased, the motor slows down just enough to allow the rotor to change its angular position in relation to the rotating flux of the stator, and then goes back to synchronous speed. Similarly, when the load is removed, it accelerates Just enough to cause the rotor to decrease its angle of lag in relation to the rotating flux, and then goes back to synchronous speed. When the peak load that the machine can handle is exceeded, the rotor pulls out of synchronism.
The torque developed by all synchronous motors has two components: a reluctancetorque component and a magnettorque component. The reluctancetorque component is due to the normal characteristic of magnetic materials in a magnetic field to align themselves so that the reluctance of the magnetic circuit is a minimum
Fig: 4.3
Strobe view of a salientpole motor operating at synchronous speed Shaft. (Courtesy Dresser Rand, Electric Machinery) The magnettorque component is due to the magnetic attraction between the field poles (magnets) on the rotor and the corresponding opposite poles of the rotaring stator flux.
The analysis of synchronousmotor behavior in this chapter is based on magnet torque alone. This is justified for cylindricalrotor motors because surface provides an insignificant reluctance torque. It is also justified motors operating from 50 percent rated load to above 100 percent rated load with power factors ranging from unity to leading; the reluctance torque significantly smaller than the magnet torque. The performance salientpole motors.
4.3 COUNTEREMF AND ARMATUREREACTON VOLTAGE
The airgap flux in a synchronous motor includes a rotating field flux φ_{f }due to DC current (I_{f}) in the rotating magnets, and a rotating armature flux. Called armature reaction flux (φ_{ar}), caused by the threephase armature currents in the tator windings. The magnitude and phase angle of the armaturereaction flux is a function of the magnitude and phase angle of the armature current. Both the magnet flux and the armature reaction flux rotate in the same direction, and at synchronous speed with respect to the armature windings.
The rotating fluxes φ_{f and} φ_{ar} generates speed voltages in the stator conductor, as visualized in Fig 4.4 for one armature phase of a threephase synchronous motor.
(a)
(b)
Fig: 4.4
Separate circuits showing emfs generated by component magnetic fields for one phase of a synchronous motor: (a) due to rotating magnets: (b) due to rotating armature reaction flux motor. Although shown as separate diagrams for ease of visualization, both φ_{f }and φ_{ar} rotate within the same threephase stator winding.
4.3.1 COUNTEREMF
The rotor magnets sweeping the stator conductors, as shown in Figure 8.8(a), generate a speedvoltage, called a counteremf (cemf) or excitation voltage, that acts in opposition to the applied voltage. The speedvoltage is proportional to the field flux and the speed of rotation. Expressed in equation form:
E_{f }= n_{s} φ_{f }E_{ f} ————————– (3)
Expressing the flux in terms of magnetomotive force (mmf) and the reluctance of the magnetic circuit,
E_{f =} n_{s} K_{f } (4)
Where: E, = excitation voltage/phase (V)
n_{s} = synchronous speed (r/min)
φ_{f }= pole flux (Wb)
N_{f}= number of turns of conductor in field coil
I_{f }= DC field current (A)
= reluctance of magnetic circuit (At/Wb)
K_{f}= constant
The speed of a synchronous motor is equal to the speed of the rotating therefore constant for a given frequency. Thus, as indicated in Eq. (i), the excitation voltage is a function of the field current alone. Note, however, that because of magnetic saturation effects, the reluctance of the magnetic circuit is not constant. Hence φ_{f }and E_{f} are not proportional to I_{f}.
The cemf plays a significant role in synchronousmotor operation. And as will be explained later, may be less than, equal to, or greater than the applied stator voltage. Its adjustment, through changes in field current, is also used to correct system power factor.
4.3.2 ARMATUREREACTION VOLTAGE The rotating armaturereaction flux sweeping the stator conductors, as visualized in Fig 4.4(b), generates a speed voltage called the armaturereaction voltage. The armaturereaction speed voltage, expressed in terms of armaturereaction flux is
E_{ar} = n_{s }φ_{ar} K_{a } ————————– (5)
Where: E_{ar }= armaturereaction voltage (V)
φ_{ar }= armaturereaction flux (Wb)
n_{s }= synchronous speed (r/min)
K_{a }= constant
Neglecting the effects of magnetic saturation, the armaturereaction flux is proportional to the armature current. Hence, the armaturereaction voltage may be conveniently expressed in terms of the armature current and a fictitious armature reaction. Thus,
E_{ar} = I_{a . }J X_{ar} ————————– (6)
where: X_{ar} = armaturereaction reactance (Ω/phase).
4.4 EQUIVALENTCIRCUIT MODEL AND PHASOR DIAGRAM A SYNCHRONOUSMOTOR ARMATURE
The equivalentcircuit model for one armature phase of a cylindrical rotor synchronous motor is shown in Figure 4.5(a). All values are per phase. Applying Kirchhoff’s voltage law to Figure 4.5(a),
Into Eq.
(a)
(b)
Fig: 4.5
(a) Equivalentcircuit model for one phase of a synchronousmotor armature; (b) hasor diagram corresponding to the equivalentcircuit model in (a).
where: R_{a} = armature resistance (Ω /phase)
X_{t} = armature leakage reactance (Ω /phase)
X_{s} = synchronous reactance (Ω l/phase’)
Z_{s} = synchronous impedance (Ω /phase)
V_{T} = applied voltage/phase (V)
A phasor diagram showing the component phasors and the tiptotai1 determination of V_{t }is shown in Figure 4.4(b). The phase angle δ of the excitation voltage Fig 4.4(b) is equal to the torque angle in Figure 4.3. The torque angle is also called the load angle or power angle.
4.5 SYNCHRONOUSMOTOR POWER EQUATION (MAGNET POWER)
Except for very small machines, the armature resistance of a synchronous motor is
relatively insignificant compared to its synchronous reactance, enabling Eq(10) be approximated as
VT= E_{f }+ I_{a }JX_{s ————————–} (11)
The equivalentcircuit and phasor diagram corresponding to Eq.(11) Figure 4.6 and are normally used for the analysis of synchronousmotor behavior as the motor responds to changes in load and/or changes in field excitation.
From the geometry of the phasor diagram,
I_{a} X_{s} cos θ_{i} = E_{f} sin δ (12)
Multiplying through by V_{T }and rearranging terms,
V_{T}I_{a} cos θ_{i} = sin δ (13)
Since the left side of Eq. (13) is an expression for active power input, the magnet power/phase developed by the synchronous motor can be expressed as
P_{in.Iφ}= V_{T}I_{a} cos θ_{i} (14)
or
P_{in.Iφ}= sin δ (15)
Thus, for a threephase synchronous motor,
P_{in} = 3 * V_{T}I_{a} cos θ_{i} (16)
or
P_{in} = sin δ (17)
(a)
(b)
Fig: 4.6
(a) equivalentcircuit (b) phasor diagram for a synchronous motor, assuming armature resistance is negligible.
Equation (15), called the synchronousmachine power equation, expresses the magnet power/phase developed by a cylindricalrotor motor, in terms of its excitation voltage and power angle. Assuming a constant source voltage and constant frequency, Eqs. (14) and (15) may be expressed as proportionalities that are very useful in synchronousmachine analysis:
P α I_{a} cos θ (18)
P α E_{f} sin δ (19)
4.6 EFFECT OF CHANGES IN SHAFT LOAD ON ARMATURE CURRENT, POWER ANGLE, AND POWER FACTOR
The effect that changes in shaft load have on armature current, power angle and power factor are illustrated in Figure 4.7 the applied stator voltage, frequency, and field excitation are assumed constant. The heavy lines indicate the initial load conditions, and the light lines indicate the new steadystate conditions that correspond to doubling the shaft load. Note: In accordance with Eqs. (18) and (19) doubling the shaft load doubles both I_{a} cos θ_{i} and E_{f} sin δ. When adjusting phasor diagrams to show new steadystate conditions, the line of action of the new I_{a}JX_{s} phasor must be perpendicular to the new I_{a} phasor. Furthermore, as shown in Figure 4.7. if the excitation is not changed, increasing the shaft load causes the locus of the E_{f} phasor to describe a circular arc, increasing its phase angle with increasing shaft load. Note also that an increase in shaft load is also accompanied by a decrease in θ_{i} resulting in an increase in power factor.
Fig: 4.7
Phasor diagram showing effect of changes in shaft load on armature current, power angle, and power factor of a synchronous motor.
As additional load is placed on the machine, the rotor continues to increase its lag relative to the rotating flux, thereby increasing both the angle of lag of the cemf phasor and the magnitude of the stator current. During all this loading, however, except for the transient conditions whereby the rotor assumes a new position in relation to the rotating flux, the average speed of the machine does not change. Finally, with increased loading, a point is reached at which a further increase in fails to cause a corresponding increase in motor torque, and the rotor pulls out of synchronism. The point of maximum torque occurs at a power angle of approximately 90° for a cylindricalrotor machine, as is evidenced in Eq. (17). The critical value of torque that causes a synchronous motor to pull out of synchronism is called the pullout torque.
4.7 EFFECT OF CHANGES IN FIELD EXCITATION ON SYNCHRONOUSMOTOR PERFORMANCE
Intuitively we can expect that increasing the strength of the magnets will increase the magnetic attraction, and thereby cause the rotor magnets to have a closer alignment with the corresponding opposite poles of the rotating stator flux; the result is a smaller power angle. Proof of this behavior can be seen in Eq. (17). Assuming a constant shaft load, the steadystate value of E_{f} sin δ must be constant. A step increase in E_{f} will cause a transient increase in E_{f} sin δ. and the rotor will accelerate. As the rotor changes its angular position, δ decreases until E_{f} sin δ has the same steadystate value as be fore, at which time the rotor is again operating at synchronous speed angular position of the rotor magnets relative to the rotating flux of the a fraction of a second.
The effect of changes in field excitation on armature current, pc power factor of a synchronous motor operating with a constant shaft load, from a constant voltage, constant frequency supply, is illustrated in Figure – (19), for a constant shaft load,
Fig: 4.8
Phasor diagram showing the effect of changes in field excitation on armature current power angle, and power factor of a synchronous motor
E_{f1} sin δ_{1} = E_{f2} sin δ_{2 }= E_{f3} sin δ_{3 }= E_{f} sin δ (20)
This is shown in Figure 4.8 where the locus of the tip of the E1 phasor is a straight line parallel to the V_{t} phasor. Similarly, from Eq. (18), for a constant shaft load.
I_{a1 }cos θ_{i1} = I_{a2 }cos θ_{i2 }= I_{a3 }cos θ_{i3 }= I_{a }cos θ_{i} (21)
This is shown in Figure 4.8, where the locus of the tip of the I_{a} phasor is a line perpendicular to the V_{t} phasor.
Note that increasing the excitation from E_{f1} to E_{f3} in Figure 4.8 caused the angle of the current phasor (and hence the power factor) to go from lagging to leading. The value of field excitation that results in unity power factor is called normal excitation. Excitation greater than normal is called over excitation, and excitation
less than normal is called under excitation. Furthermore, as indicated in Figure 4.8. when operating in the overexcited mode, E_{f} > V_{t} .
4.8 V CURVES
Curves of armature current vs. field current or armature current vs. excitation voltage are called V curves, and are shown in Figure 4.9. for representative synchronous motor loads. The curves are related to the phasor diagram in Figure 4.8. and illustrate the effect of different values of field excitation on armature current and power factor for specific shaft loads. Note that increases in shaft load require increases in field excitation in order to maintain unity power factor.
The locus of the left endpoints of the V curves in Figure 4.9 represents the stability limit (δ = 90°). Any reduction in excitation below the stability limit for a particular load will cause the rotor to pull out of synchronism.
The constantload V curves shown in Figure 4.9 may be plotted from laboratory data (I_{a} vs I_{f}) as I_{f} is varied, or graphically by plotting I_{a} vs E_{f} from a family of phasor diagrams such as that shown in Figure 8.12, or from the following mathematical expression for the V curves6:
I_{a}= . [ E^{2}_{f }+ V^{2}_{T} – 2 ] ^{0.5} (22)
Equation () is based on the geometry of the phasor diagram and assumes Ra S negligible. Note: If the developed torque is less than the shaft load plus wind age and friction, instability will occur, and the expression under the square root sign will be negative.
The family of V curves shown in Figure 4.9 represents computer plots of Eq. (22), assuming a threephase, 40hp, 220V, 60Hz synchronous motor with a synchronous reactance of 1.27 (Ω /phase).
Fig: 4.9
Family of representative V curve for a Synchronous motor
EXPERIMENT SETUP
5.1 EXPERIMENT SET UP
Fig: 5.1
5.2 NECESSARY EQUIPMENTS
 Variac
 Ammeter (DC)
 Ammeter (AC)
 Variable register AC supply voltage
 DC supply voltage
 Tacometer
 Connector
5.3 COLLECT DATA WITHOUT SYNCHRONIZING
When supply voltage= 315V then r.p.m = 1120 rpm
When supply voltage= 345V then r.p.m = 1230 rpm
When supply voltage= 430V then r.p.m = 1500 rpm
Thus
N= 120f⁄P here
= 120×50⁄4 f= 50 Hz
=1500 r.p.m P= 4
5.5 V CURVES FOR A SYNCHRONOUS MOTOR
Fig: 5.2
COLLECT DATA FROM CURVE
 When excitation voltage vf = 0 then the armature current= 0.78 A
 When excitation voltage vf = 50 then the armature current= 0.52 A
 When excitation voltage vf =100 then the armature current= 0.2 A
 When excitation voltage vf =125 then the armature current= Unity
 When excitation voltage vf = 150 then the armature current= 0.15A
 When excitation voltage vf = 200 then the armature current= 0.8 A
 When excitation voltage vf = 220 then the armature current= 1.2A
The magnitude of armature current varies with excitation. The current has large value both for low and high values of excitation. In between, it has minimum value corresponding to a certain excitation. The variations of I with excitation are known as V curves because of their shape.
For the same output load, the armature current varies over a wide range and so causes the power factor also to vary accordingly. When overexcited, the motor runs with leading power factor and with lagging power factor when underexcited. In between, the power factor is unity. The minimum armature current corresponds to unity power factor.
APPLICATION
A synchronous condenser provides stepless automatic power factor correction with the ability to produce up to 150% additional MVARs. The system produces no switching transients and is not affected by system electrical harmonics (some harmonics can even be absorbed by synchronous condensers). They will not produce excessive voltage levels and are not susceptible to electrical resonances. Because of the rotating inertia of the condenser, it can provide limited voltage support during short power outages.
The use of rotating synchronous condensers was common through the 1950s. They remain an alternative (or a supplement) to capacitors for power factor correction because of problems that have been experienced with harmonics causing capacitor overheating and catastrophic failures. Synchronous condensers are also very good for supporting voltage. The reactive power produced by a capacitor bank is in direct proportion to the square of its terminal voltage, where a synchronous condenser's reactive power declines less rapidly, and can be adjusted to compensate for falling terminal voltage. This reactive power improves voltage regulation in situations such as starting large motors, or where power must travel long distances from where it is generated to where it is used, as is the case with power wheeling (distribution of electric power from one geographical location to another within an electric power distribution system).
Synchronous Condensers may also be referred to as Dynamic Power Factor Correction systems. These machines can prove very effective when advanced controls are utilized. A PLC based controller with pf controller and regulator will allow the system to be set to meet a given power factor or can be set to produce a specified amount of KVAR. In both cases, the output of the machine obviously cannot exceed its capability.
CALCULATION POWER FACTOR
CALCULATING POWER FACTOR
It should be noted that power factor, like all ratio measurements, is a unitless quantity.
For the purely resistive circuit, the power factor is 1 (perfect), because the reactive power equals zero. Here, the power triangle would look like a horizontal line, because the opposite (reactive power) side would have zero length.
For the purely inductive circuit, the power factor is zero, because true power equals zero. Here, the power triangle would look like a vertical line, because the adjacent (true power) side would have zero length.
The same could be said for a purely capacitive circuit. If there are no dissipative (resistive) components in the circuit, then the true power must be equal to zero, making any power in the circuit purely reactive. The power triangle for a purely capacitive circuit would again be a vertical line (pointing down instead of up as it was for the purely inductive circuit).
Power factor can be an important aspect to consider in an AC circuit, because any power factor less than 1 means that the circuit's wiring has to carry more current than what would be necessary with zero reactance in the circuit to deliver the same amount of (true) power to the resistive load. If our last example circuit had been purely resistive, we would have been able to deliver a full 169.256 watts to the load with the same 1.410 amps of current, rather than the mere 119.365 watts that it is presently dissipating with that same current quantity. The poor power factor makes for an inefficient power delivery system.
Poor power factor can be corrected, paradoxically, by adding another load to the circuit drawing an equal and opposite amount of reactive power, to cancel out the effects of the load's inductive reactance. Inductive reactance can only be canceled by capacitive reactance, so we have to add a capacitor in parallel to our example circuit as the additional load. The effect of these two opposing reactances in parallel is to bring the circuit's total impedance equal to its total resistance (to make the impedance phase angle equal, or at least closer, to zero).
Since we know that the (uncorrected) reactive power is 119.998 VAR (inductive), we need to calculate the correct capacitor size to produce the same quantity of (capacitive) reactive power. Since this capacitor will be directly in parallel with the source (of known voltage), we'll use the power formula which starts from voltage and reactance:
Let's use a rounded capacitor value of 22 µF and see what happens to our circuit: (Figure below)
Fig:6.1
Parallel capacitor corrects lagging power factor of inductive load. V2 and node numbers: 0, 1, 2, and 3 are SPICE related, and may be ignored for the moment.
The power factor for the circuit, overall, has been substantially improved. The main current has been decreased from 1.41 amps to 994.7 milliamps, while the power dissipated at the load resistor remains unchanged at 119.365 watts. The power factor is much closer to being 1:
Since the impedance angle is still a positive number, we know that the circuit, overall, is still more inductive than it is capacitive. If our power factor correction efforts had been perfectly ontarget, we would have arrived at an impedance angle of exactly zero, or purely resistive. If we had added too large of a capacitor in parallel, we would have ended up with an impedance angle that was negative, indicating that the circuit was more capacitive than inductive.
A SPICE simulation of the circuit of (Figure above) shows total voltage and total current are nearly in phase. The SPICE circuit file has a zero volt voltagesource (V2) in series with the capacitor so that the capacitor current may be measured. The start time of 200 msec ( instead of 0) in the transient analysis statement allows the DC conditions to stabilize before collecting data. See SPICE listing “pf.cir power factor”.
pf.cir power factor
V1 1 0 sin(0 170 60)
C1 1 3 22uF
v2 3 0 0
L1 1 2 160mH
R1 2 0 60
# resolution stop start
tran 1m 200m 160m
end
Fig:6.2
Zero phase angle due to inphase V_{total} and I_{total} . The lagging I_{L} with respect to V_{total} is corrected by a leading I_{C} .
Note that the total current (I_{total}) is in phase with the applied voltage (V_{total}), indicating a phase angle of near zero. This is no coincidence. Note that the lagging current, I_{L} of the inductor would have caused the total current to have a lagging phase somewhere between (I_{total}) and I_{L}. However, the leading capacitor current, I_{C}, compensates for the lagging inductor current. The result is a total current phaseangle somewhere between the inductor and capacitor currents. Moreover, that total current (I_{total}) was forced to be inphase with the total applied voltage (V_{total}), by the calculation of an appropriate capacitor value.
Since the total voltage and current are in phase, the product of these two waveforms, power, will always be positive throughout a 60 Hz cycle, real power as in Figure above. Had the phaseangle not been corrected to zero (PF=1), the product would have been negative where positive portions of one waveform overlapped negative portions of the other as in Figure above. Negative power is fed back to the generator. It cannot be sold; though, it does waste power in the resistance of electric lines between load and generator. The parallel capacitor corrects this problem.
Note that reduction of line losses applies to the lines from the generator to the point where the power factor correction capacitor is applied. In other words, there is still circulating current between the capacitor and the inductive load. This is not normally a problem because the power factor correction is applied close to the offending load, like an induction motor.
It should be noted that too much capacitance in an AC circuit will result in a low power factor just as well as too much inductance. You must be careful not to overcorrect when adding capacitance to an AC circuit. You must also be very careful to use the proper capacitors for the job (rated adequately for power system voltages and the occasional voltage spike from lightning strikes, for continuous AC service, and capable of handling the expected levels of current).
If a circuit is predominantly inductive, we say that its power factor is lagging (because the current wave for the circuit lags behind the applied voltage wave). Conversely, if a circuit is predominantly capacitive, we say that its power factor is leading. Thus, our example circuit started out with a power factor of 0.705 lagging, and was corrected to a power factor of 0.999 lagging.
 REVIEW:
 Poor power factor in an AC circuit may be “corrected”, or reestablished at a value close to 1, by adding a parallel reactance opposite the effect of the load's reactance. If the load's reactance is inductive in nature (which is almost always will be), parallel capacitance is what is needed to correct poor power factor.
6.1 PRACTICAL POWER FACTOR CORRECTION
Most likely, you will have access to a wattmeter for measuring true power, whose reading you can compare against a calculation of apparent power (from multiplying total voltage and total current measurements). From the values of true and apparent power, you can determine reactive power and power factor. Let's do an example problem to see how this works: (Figure below)
Fig: 6.3
Wattmeter reads true power; product of voltmeter and ammeter readings yields appearant power.
First, we need to calculate the apparent power in kVA. We can do this by multiplying load voltage by load current:
As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which tells us that the power factor in this circuit is rather poor (substantially less than 1). Now, we figure the power factor of this load by dividing the true power by the apparent power:
Using this value for power factor, we can draw a power triangle, and from that determine the reactive power of this load: (Figure below)
Fig: 6.4
Reactive power may be calculated from true power and appearant power.
To determine the unknown (reactive power) triangle quantity, we use the Pythagorean Theorem “backwards,” given the length of the hypotenuse (apparent power) and the length of the adjacent side (true power):
If this load is an electric motor, or most any other industrial AC load, it will have a lagging (inductive) power factor, which means that we'll have to correct for it with a capacitor of appropriate size, wired in parallel. Now that we know the amount of reactive power (1.754 kVAR), we can calculate the size of capacitor needed to counteract its effects:
Rounding this answer off to 80 µF, we can place that size of capacitor in the circuit and calculate the results: (Figure below)
Fig: 6.5
Parallel capacitor corrects lagging (inductive) load.
An 80 µF capacitor will have a capacitive reactance of 33.157 Ω, giving a current of 7.238 amps, and a corresponding reactive power of 1.737 kVAR (for the capacitor only). Since the capacitor's current is 180^{o} out of phase from the the load's inductive contribution to current draw, the capacitor's reactive power will directly subtract from the load's reactive power, resulting in:
This correction, of course, will not change the amount of true power consumed by the load, but it will result in a substantial reduction of apparent power, and of the total current drawn from the 240 Volt source: (Figure below)
Fig: 6.6
Power triangle before and after capacitor correction.
The new apparent power can be found from the true and new reactive power values, using the standard form of the Pythagorean Theorem:
This gives a corrected power factor of (1.5kW / 1.5009 kVA), or 0.99994, and a new total current of (1.50009 kVA / 240 Volts), or 6.25 amps, a substantial improvement over the uncorrected value of 9.615 amps! This lower total current will translate to less heat losses in the circuit wiring, meaning greater system efficiency (less power wasted).
6.2 POWER FACTOR CORRECTION CONCEPT
With respect to Fluorescent and High intensity Discharge Lamps.
POWER FACTOR:
All Discharge Lamps, such as Fluorescent Lamps, High Pressure Mercury Vapour Lamps, Sodium Lamps, Metal Halide Lamps, etc., require ballasts (chokes) or transformers for their operation. These devices are inductive in nature. When a Discharge Lamp is switched on, it draws Apparent Power from the mains. This Apparent Power (VA) has two components; one is the Active Power (W) actually being consumed by the lamp for illuminating it, and the other is the Reactive Power (VAr) feeding the electromagnetic circuit of the control gear.
Power factor is the ratio of Active Power (W) to the Apparent Power (VA) (Figure 1).
Power factor is also denoted by p.f. or cos Ø . Typical p.f. of ballasts is from 0.35 to 0.5.
POWER FACTOR CORRECTION
As mentioned above, the inductive components, such as ballasts, draw Reactive Power (VAr) from the mains. It lags behind the Active Power (W) by 90^{o} (Figure 1). A capacitor, if connected across the mains, will also draw Reactive Power [VAr_{(c)}], but it leads the Active Power (W) by 90^{o}. The direction of the capacitive Reactive Power [VAr_{(c)}] is opposite to the direction of the inductive Reactive Power (VAr) (Figure 6.7).
Fig: 6.7
If a capacitor is connected in parallel with an inductive load, it will draw capacitive Reactive Power [VAr_{(c)}]. The effective Reactive Power drawn by the circuit will reduce to the extent of the capacitive Reactive Power [VAr_{(c)}], resulting in reduction of Apparent Power from VA to VA_{1}. The phase angle between the Active Power and the new Apparent Power VA_{1} will also reduce from Ø to Ø_{1} (Fig: 6.8). Thus the power factor will increase from cos Ø to cos Ø_{1}.New
p.f. = cos Ø_{1} =
By selecting a capacitor of an appropriate value, the power factor can be corrected to 1. However, in practice, the power factor is improved to fall between 0.9 and 0.95.
Fig: 6.8
6.3 POWER FACTOR CORRECTION
If you are paying power factor charges, it will probably be worthwhile investing in power factor correction equipment. Often your investment in such equipment can be recovered in a 23 years and from then on will provide ongoing savings to you business.
The electrical power used by motors and fluorescent tubes, has two components; active power and reactive power. The active power is converted into useful work – turning motors or producing light but the reactive component is only used to “energize” the magnetic or electrostatic properties of the equipment. Reactive power still has to be produced by the generator and transmitted to your business over the electricity network. Producing and transmitting the reactive power creates costs for the electricity companies which are passed on to you as power factor penalties or surcharges.
6.4 POWER FLOW WITHOUT POWER FACTOR CORRECTION
Fig:6.9
The Power Factor is, roughly speaking, a measure of what proportion of the total power supplied to your business is actually used by the equipment. The power factor value can vary between just above 0 and 1. The aim is keep the power factor as close as possible to 1, as power factor penalties generally cut in if it drops below 0.85.
6.5 POWER FLOW WITH POWER FACTOR CORRECTION
Fig: 6.10
To effectively reduce your power factor, you will need to obtain a technical assessment and design parameters for the power factor correcting capacitors to ensure they match the requirements of your equipment. Your energy supplier may be able to assist you in supplying this technical assistance, but there are also technical consultants available who will have the expertise.
Why not consider having a complete energy audit done for you business, which also addresses your power correction requirements? Generally the cost savings from energy efficiency actions identified in the audit will more than pay for the cost of the audit.
6.6 POWER FACTOR FUNDAMENTALS
 If you use equipment containing electric motors, or florescent lights extensively, you may have a problem with your power factor.
 Check your bill for power factor penalties or peak load measured in kVA.
 A poor power factor will result in additional electricity costs to your business.
 Obtain expert technical advice to determine how your power factor can be corrected and install correction equipment if required.
 Consider also conducting a full energy audit, as these generally identify cost effective energy savings opportunities.
The main advantages of the Power Factor Correction are:
1. The electrical load on the Utility is reduced, thereby allowing the Utility to supply the surplus power to other consumers, without increasing its generation capacity.
2. Most of the Utilities impose low power factor penalties. By correcting the power factor, this penalty can be avoided.
3. High power factor reduces the load currents. Therefore, a considerable saving is made in the hardware cost, such as cables, switchgear, substation transformers, etc.
CONCLUSION
Synchronous motors load the power line with a leading power factor. This is often usefull in cancelling out the more commonly encountered lagging power factor caused by induction motors and other inductive loads. Originally, large industrial synchronous motors came into wide use because of this ability to correct the lagging power factor of induction motors.
This leading power factor can be exaggerated by removing the mechanical load and over exciting the field of the synchronous motor. Such a device is known as a synchronous condenser. Furthermore, the leading power factor can be adjusted by varying the field excitation. This makes it possible to nearly cancel an arbitrary lagging power factor to unity by paralleling the lagging load with a synchronous motor. A synchronous condenser is operated in a borderline condition between a motor and a generator with no mechanical load to fulfill this function. It can compensate either a leading or lagging power factor, by absorbing or supplying reactive power to the line. This enhances power line voltage regulation.
In industrial application where both synchronous motors and induction motors are used. The synchronous motor is usually operated at a leading power factor to compensate for the lagging power factor of the induction motors. In those cases where a synchronous motor is operated whithout load, merely for the perpose of improving the power factor of a system, the machine is called a synchronous condenser.machines specifically designed for this application are built whithout external shafts.
The electrical power used by motors and fluorescent tubes, has two components; active power and reactive power. The active power is converted into useful work – turning motors or producing light but the reactive component is only used to “energize” the magnetic or electrostatic properties of the equipment. Reactive power still has to be produced by the generator and transmitted to your business over the electricity network. Producing and transmitting the reactive power creates costs for the electricity companies which are passed on to you as power factor penalties or surcharges.
REFERENCE
Books
 Charles I Hubert . ELECTRICAL MACHINE theory ,operation,
 V.K Mehata . Rohit Mehata. PRINCIPAL OF POWER SYSTEM
 Mc Pherson, G. An INTRODUCTION TO ELECTRICAL MACHINES AND TRANSFORMES Wiley New York 1981
 Ratcliffe R. the changspeed PAM MOTOR AND ITS APPLICATION IN THE RUBBER AND PLASTICS INDUSTRIES IEEE TRANS.INDUSTRY GENERAL APPLICATION, Vol IGA6 No mar/Apr.1970
 Jhon j.Gainger, William D. Stevenson.JR POWER SYSTEM ANALYSES
 Tony R Kuphaldp. Under the terms & condition of the Design science License POWER ELECTRIC CIRCUIT (Copy Right)
Internet
http://www. Google.com
 http://www. Synchronous condenser.com
 http://www. Static Capacitor.com
 http://www. Power Factor.com
 http://www. Phase advancers.com