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101

IntrIntr

Intr

oductionoduction

oduction

he 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 sys-

tem from power station generator down to the

utilisation devices. In order to ensure most

favourable conditions for a supply system from

engineering and economical standpoint, it is im-

portant to have power factor as close to unity as

possible. In this chapter, we shall discuss the

various methods of power factor improvement.

6.16.1

6.1

Power FactorPower Factor

Power Factor

The cosine of angle between voltage and current

in an a.c. circuit is known as power factor.

In an a.c. circuit, there is generally a phase

difference φ between voltage and current. The

term cos φ is called the power factor of the cir-

cuit. If the circuit is inductive, the current lags

behind the voltage and the power factor is referred

CHAPTERCHAPTER

CHAPTER

Power Factor Improvement

6.1 Power Factor

6.2 Power Triangle

6.3 Disadvantages of Low Power Factor

6.4 Causes of Low Power Factor

6.5 Power Factor Improvement

6.6 Power Factor Improvement Equip-

ment

6.7 Calculations of Power Factor Correc-

tion

6.8 Importance of Power Factor Improve-

ment

6.9 Most Economical Power Factor

6.10 Meeting the Increased kW Demand

on Power Stations

CONTENTS

102 Principles of Power System

to as lagging. However, in a capacitive circuit, current leads the volt-

age and power factor is said to be leading.

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. 6.1. The circuit current I can be resolved into

two perpendicular components, namely ;

(a) I cos φ in phase with V

(b) I sin φ 90

out of phase with V

The component I cos φ is known as active or wattful component,

whereas component I sin φ 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 cos φ will be high. Therefore, a circuit having small reactive current (i.e., I sin φ) will

have high power factor and vice-versa. It may be noted that value of power factor can never be more

than unity.

(i) 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.

(ii) Sometimes power factor is expressed as a percentage. Thus 0·8 lagging power factor may

be expressed as 80% lagging.

6.26.2

6.2

P P

er er

iangleiangle

iangle

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. 6.1 is multiplied by voltage V, then we get the power triangle OAB

shown in Fig. 6.2 where

OA = VI cos φ and represents the active power in watts or kW

AB = VI sin φ and represents the reactive power in VAR or kVAR

OB = VI and represents the apparent power in VA or kVA

The following points may be noted form the power triangle :

(i) The apparent power in an a.c. circuit has two components viz.,

active and reactive power at right angles to each other.

= OA

+ AB

or (apparent power)

= (active power)

+ (reactive power)

or (kVA)

= (kW)

+ (kVAR)

(ii) Power factor, cos φ =

active

power

apparent power

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, what-

ever be the waveform.

(iii) 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 =

sin φ

∴ kVAR = kW tan φ

* If the current lags behind the voltage, the reactive power drawn is known as lagging reactive power. How-

ever, if the circuit current leads the voltage, the reactive power is known as leading reactive power.

Power Factor Improvement 103

(iv) 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.

(v) 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 =

Resistance

Impedanc

cos

Active pow

Apparent Powe

(vi) 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.

Illustration. Let us illustrate the power relations in an a.c. 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 × 10 = 2000 VA

Active power = VI cos φ = 200 × 10 × 0·8 = 1600 W

Reactive power = VI sin φ = 200 × 10 × 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 sin φ).

6.36.3

6.3

Disadvantages of Low Power Factor Disadvantages of Low Power Factor

Disadvantages of Low Power Factor

The power factor plays an importance role in a.c. circuits since power consumed depends upon this

factor.

P = V

cos φ (For single phase supply)

∴ I

cos

...(i)

P =

cos φ (For 3 phase supply)

∴ I

3cos

...(ii)

It is clear from above that for fixed power and voltage, the load current is inversely proportional

to the power factor. Lower the power factor, higher is the load current and vice-versa. A power factor

less than unity results in the following disadvantages :

(i) Large kVA rating of equipment. The electrical machinery (e.g., alternators, transformers,

switchgear) is always rated in *kVA.

Now, kVA =

It is clear that kVA rating of the equipment is inversely proportional to power factor. The smaller

the power factor, the larger is the kVA rating. Therefore, at low power factor, the kVA rating of the

equipment has to be made more, making the equipment larger and expensive.

(ii) Greater conductor size. To transmit or distribute a fixed amount of power at constant

voltage, the conductor will have to carry more current at low power factor. This necessitates

* The electrical machinery is rated in kVA because the power factor of the load is not known when the

machinery is manufactured in the factory.

104 Principles of Power System

large conductor size. For example, take the case of a single phase a.c. motor having an input

of 10 kW on full load, the terminal voltage being 250 V. At unity p.f., the input full load

current would be 10,000/250 = 40 A. At 0·8 p.f; the kVA input would be 10/0·8 = 12·5 and

the current input 12,500/250 = 50 A. If the motor is worked at a low power factor of 0·8, the

cross-sectional area of the supply cables and motor conductors would have to be based upon

a current of 50 A instead of 40 A which would be required at unity power factor.

(iii) Large copper losses. The large current at low power factor causes more I

R losses in all the

elements of the supply system. This results in poor efficiency.

(iv) Poor voltage regulation. The large current at low lagging power factor causes greater

voltage drops in alternators, transformers, transmission lines and distributors. This results

in the decreased voltage available at the supply end, thus impairing the performance of

utilisation devices. In order to keep the receiving end voltage within permissible limits,

extra equipment (i.e., voltage regulators) is required.

(v) Reduced handling capacity of system. The lagging power factor reduces the handling

capacity of all the elements of the system. It is because the reactive component of current

prevents the full utilisation of installed capacity.

The above discussion leads to the conclusion that low power factor is an objectionable feature in

the supply system

6.46.4

6.4

Causes of Low Power Factor Causes of Low Power Factor

Causes of Low Power Factor

Low power factor is undesirable from economic point of view. Normally, the power factor of the

whole load on the supply system in lower than 0·8. The following are the causes of low power factor:

(i) Most of the a.c. motors are of induction type (1φ and 3φ induction motors) which have low

lagging power factor. These motors work at a power factor which is extremely small on

light load (0·2 to 0·3) and rises to 0·8 or 0·9 at full load.

(ii) Arc lamps, electric discharge lamps and industrial heating furnaces operate at low lagging

power factor.

(iii) The load on the power system is varying ; being high during morning and evening and low at

other times. During low load period, supply voltage is increased which increases the

magnetisation current. This results in the decreased power factor.

6.56.5

6.5

P P

er Fer F

er F

actor Impractor Impr

actor Impr

ementement

ement

The low power factor is mainly due to the fact that most of the power loads are inductive and, there-

fore, 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.

Power Factor Improvement 105

Illustration. To illustrate the power factor improvement by a capacitor, consider a single *phase

load taking lagging current I at a power factor cos φ

as shown in Fig. 6.3.

The capacitor C is connected in parallel with the load. The capacitor draws current I

which

leads the supply voltage by 90

. The resulting line current I′ is the phasor sum of I and I

and its angle

of lag is φ

as shown in the phasor diagram of Fig. 6.3. (iii). It is clear that φ

is less than φ

, so that

cos φ

is greater than cos φ

. Hence, the power factor of the load is improved. The following points

are worth noting :

(i) The circuit current

′

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.

∴ I cos φ

′

cos φ

(iii) The lagging reactive component is reduced after p.f. improvement and is equal to the differ-

ence between lagging reactive component of load (I sin φ

) and capacitor current (I

) i.e.,

′

sin φ

= I sin φ

− I

(iv) As I cos φ

′

cos φ

∴ VI cos φ

= V

′

cos φ

[Multiplying by V]

Therefore, active power (kW) remains unchanged due to power factor improvement.

(v)

′

sin φ

= I sin φ

− I

∴ V

′

sin φ

= VI sin φ

− VI

[Multiplying by V]

i.e., Net kVAR after p.f. correction = Lagging kVAR before p.f. correction − leading kVAR of

equipment

6.66.6

6.6

P P

er Fer F

er F

actor Impractor Impr

actor Impr

ement Equipmentement Equipment

ement 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 take special steps to

improve the power factor. This can be achieved by the following equipment :

1. Static capacitors. 2. Synchronous condenser. 3. Phase advancers.

1. 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**

* The treatment can be used for 3-phase balanced loads e.g., 3-φ induction motor. In a balanced 3-φ load,

analysis of one phase leads to the desired results.

** To distinguish from the so called synchronous condenser which is a synchrnous motor running at no load

and taking leading current.

106 Principles of Power System

capacitor) draws a leading current and partly or completely neutralises the lagging reactive compo-

nent of load current. This raises the power factor of the load. For three-phase loads, the capacitors

can be connected in delta or star as shown in Fig. 6.4. Static capacitors are invariably used for power

factor improvement in factories.

Advantages

(i) They have low losses.

(ii) They require little maintenance as there are no rotating parts.

(iii) They can be easily installed as they are light and require no foundation.

(iv) They can work under ordinary atmospheric conditions.

Disadvantages

(i) They have short service life ranging from 8 to 10 years.

(ii) They are easily damaged if the voltage exceeds the rated value.

(iii) Once the capacitors are damaged, their repair is uneconomical.

2. Synchronous condenser. A synchronous motor takes a leading current when over-excited

and, therefore, behaves as a capacitor. An over-excited 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 6.5 shows the power factor improvement by synchronous condenser method. The 3φ load takes

current I

at low lagging power factor cos φ

. The synchronous condenser takes a current I

which

leads the voltage by an angle φ

*. The resultant current I is the phasor sum of I

and I

and lags

behind the voltage by an angle φ. It is clear that φ is less than φ

so that cos φ is greater than cos φ

Thus the power factor is increased from cos φ

to cos φ. Synchronous condensers are generally used

at major bulk supply substations for power factor improvement.

Advantages

(i) By varying the field excitation, the magnitude of current drawn by the motor can be changed

by any amount. This helps in achieving stepless † control of power factor.

* If the motor is ideal i.e., there are no losses, then φ

= 90

. However, in actual practice, losses do occur in

the motor even at no load. Therefore, the currents I

leads the voltage by an angle less than 90

† The p.f. improvement with capacitors can only be done in steps by switching on the capacitors in various

groupings. However, with synchronous motor, any amount of capacitive reactance can be provided by

changing the field excitation.

Power Factor Improvement 107

(ii) The motor windings have high thermal stability to short circuit currents.

(iii) The faults can be removed easily.

Disadvantages

(i) There are considerable losses in the motor.

(ii) The maintenance cost is high.

(iii) It produces noise.

(iv) Except in sizes above 500 kVA, the cost is greater than that of static capacitors of the same

rating.

(v) As a synchronous motor has no self-starting torque, therefore, an auxiliary equipment has to

be provided for this purpose.

Note. The reactive power taken by a synchronous motor depends upon two factors, the d.c. field excitation

and the mechanical load delivered by the motor. Maximum leading power is taken by a synchronous motor with

maximum excitation and zero load.

3. Phase advancers. 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

its stator winding draws exciting current which lags be-

hind the supply voltage by 90

. If the exciting ampere

turns can be provided from some other a.c. source, then

the stator winding will be relieved of exciting current and

the power factor of the motor can be improved. This job

is accomplished by the phase advancer which is simply an a.c. 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 over-excited

synchronous motor.

Phase advancers have two principal advantages. Firstly, as the exciting ampere turns are sup-

plied at slip frequency, therefore, lagging kVAR drawn by the motor are considerably reduced. Sec-

ondly, 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

below 200 H.P.

Synchronous Condenser

Static Capacitor

108 Principles of Power System

6.76.7

6.7

CalculaCalcula

Calcula

tions of Ptions of P

tions of P

er Fer F

er F

actor Corractor Corr

actor Corr

ectionection

ection

Consider an inductive load taking a lagging current I at a power factor cos φ

. In order to improve the

power factor of this circuit, the remedy is to connect such an equipment in parallel with the load

which takes a leading reactive component and partly cancels the lagging reactive component of the

load. Fig. 6.6 (i) shows a capacitor connected across the load. The capacitor takes a current I

which

leads the supply voltage V by 90

. The current I

partly cancels the lagging reactive component of

the load current as shown in the phasor diagram in Fig. 6.6 (ii). The resultant circuit current becomes

I′ and its angle of lag is φ

. It is clear that φ

is less than φ

so that new p.f. cos φ

is more than the

previous p.f. cos φ

From the phasor diagram, it is clear that after p.f. correction, the lagging reactive component of

the load is reduced to I′sin φ

Obviously, I′ sin φ

= I sin φ

− I

or I

= I sin φ

− I′ sin φ

∴ Capacitance of capacitor to improve p.f. from cos φ

to cos φ

∵ X

Power triangle. The power factor correction can also be illustrated from power triangle. Thus

referring to Fig. 6.7, the power triangle OAB is for the power factor cos φ

, whereas power triangle

OAC is for the improved power factor cos φ

. It may be seen that

active power (OA) does not change with power factor improve-

ment. However, the lagging kVAR of the load is reduced by the

p.f. correction equipment, thus improving the p.f. to cos φ

Leading kVAR supplied by p.f. correction equipment

= BC = AB − AC

=kVAR

− kVAR

= OA (tan φ

− tan φ

)

= kW (tan φ

− tan φ

)

Knowing the leading kVAR supplied by the p.f. correction equipment, the desired results can be

obtained.

Example 6.1 An alternator is supplying a load of 300 kW at a p.f. of 0·6 lagging. If the power

factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading ?

Power Factor Improvement 109

Solution :

kVA =

⋅

300

= 500 kVA

kW at 0·6 p.f. = 300 kW

kW at 1 p.f. = 500 × 1 = 500 kW

∴ Increased power supplied by the alternator

= 500 − 300 = 200 kW

Note the importance of power factor improvement. When the p.f. of the alternator is unity, the

500 kVA are also 500 kW and the engine driving the alternator has to be capable of developing this

power together with the losses in the alternator. But when the power factor of the load is 0·6, the

power is only 300 kW. Therefore, the engine is developing only 300 kW, though the alternator is

supplying its rated output of 500 kVA.

Example 6.2 A single phase motor connected to 400 V, 50 Hz supply takes 31·7A at a power

factor of 0·7 lagging. Calculate the capacitance required in parallel with the motor to raise the

power factor to 0·9 lagging.

Solution : The circuit and phasor diagrams are shown in Figs. 6.8 and 6.9 respectively. Here

motor M is taking a current I

of 31·7A. The current I

taken by the capacitor must be such that when

combined with I

, the resultant current I lags the voltage by an angle φ where cos φ = 0·9.

Referring to the phasor diagram in Fig. 6.9,

Active component of I

= I

cos φ

= 31·7 × 0·7 = 22·19A

Active component of I = I cos φ = I × 0·9

These components are represented by OA in Fig. 6.9.

∴ I =

22 19

⋅

= 24·65A

Reactive component of I

= I

sin φ

= 31·7 × 0·714* = 22·6A

Reactive component of I = I sin φ = 24·65

109

−⋅

= 24·65 × 0·436 = 10·75 A

It is clear from Fig. 6.9 that :

= Reactive component of I

− Reactive component of I

= 22·6 − 10·75 = 11·85A

But I

= V × 2π f C

or 11·85 = 400 × 2π × 50 × C

∴ C = 94·3 × 10

−6

F = 94·3

µµ

µF

* sin φ

()

1 cos 1 0 7

−φ=−⋅

= 0·714

110 Principles of Power System

Note the effect of connecting a 94·3 µF capacitor in parallel with the motor. The current taken

from the supply is reduced from 31·7 A to 24·65 A without altering the current or power taken by the

motor. This enables an economy to be affected in the size of generating plant and in the cross-

sectional area of the conductors.

Example 6.3 A single phase a.c. generator supplies the following loads :

(i) Lighting load of 20 kW at unity power factor.

(ii) Induction motor load of 100 kW at p.f. 0·707 lagging.

(iii) Synchronous motor load of 50 kW at p.f. 0·9 leading.

Calculate the total kW and kVA delivered by the generator and the power factor at which it works.

Solution : Using the suffixes 1, 2 and 3 to indicate the different loads, we have,

kVA

= 20 kVA

kVA

100

070

⋅

= 141·4 kVA

kVA

⋅

= 55·6 kVA

These loads are represented in Fig. 6.10. The three kVAs’ are not in phase. In order to find the

total kVA, we resolve each kVA into rectangular components – kW and kVAR as shown in Fig. 6.10.

The total kW and kVAR may then be combined to obtain total kVA.

kVAR

=kVA

sin φ

= 20 × 0 = 0

kVAR

=kVA

sin φ

= −141·4 × 0·707 = − 100 kVAR

kVAR

=kVA

sin φ

= + 55·6 × 0·436 = + 24·3 kVAR

Note that kVAR

and kVAR

are in opposite directions ; kVAR

being a lagging while kVAR

being a leading kVAR.

Total kW = 20 + 100 + 50 = 170 kW

Total kVAR = 0 − 100 + 24·3 = −75·7 kVAR

Total kVA =

kW kVAR

afa f afa f

222

170 75 7

+=+⋅

= 186 kVA

Power factor =

Total kW

Total kVA

170

186

= 0·914 lagging

The power factor must be lagging since the resultant kVAR is lagging.

Power Factor Improvement 111

Example 6.4 A 3-phase, 5 kW induction motor has a p.f. of 0·75 lagging. A bank of capacitors

is connected in delta across the supply terminals and p.f. raised to 0·9 lagging. Determine the kVAR

rating of the capacitors connected in each phase.

Solution :

Original p.f., cos φ

= 0·75 lag ; Motor input, P = 5 kW

Final p.f., cos φ

= 0·9 lag ; Efficiency, η = 100 % (assumed)

= cos

−1

(0·75) = 41·41

; tan φ

= tan 41·41º = 0·8819

= cos

−1

(0·9) = 25·84

; tan φ

= tan 25·84º = 0·4843

Leading kVAR taken by the condenser bank

= P (tan φ

− tan φ

)

= 5 (0·8819 − 0·4843) = 1·99 kVAR

∴ Rating of capacitors connected in each phase

= 1·99/3 = 0·663 kVAR

Example 6.5 A 3-phase, 50 Hz, 400 V motor develops 100 H.P. (74·6 kW), the power factor

being 0·75 lagging and efficiency 93%. A bank of capacitors is connected in delta across the supply

terminals and power factor raised to 0·95 lagging. Each of the capacitance units is built of 4 similar

100 V capacitors. Determine the capacitance of each capacitor.

Solution :

Original p.f., cos φ

= 0·75 lag ; Final p.f., cos φ

= 0·95 lag

Motor input, P = output/η = 74·6/0·93 = 80 kW

= cos

−1

(0·75) = 41·41

tan φ

= tan 41·41º = 0·8819

= cos

−1

(0·95) = 18·19

tan φ

= tan 18·19

= 0·3288

Leading kVAR taken by the condenser bank

= P (tan φ

− tan φ

)

= 80 (0·8819 − 0·3288) = 44·25 kVAR

Leading kVAR taken by each of three sets

= 44·25/3 = 14·75 kVAR ... (i)

Fig. 6.11 shows the delta* connected condenser bank. Let C farad be the capacitance of 4

capacitors in each phase.

Phase current of capacitor is

= V

= 2π f C V

=2π × 50 × C × 400

= 1,25,600 C amperes

kVAR/phase =

ph CP

1000

400

600

1000

= 50240 C ... (ii)

* In practice, capacitors are always connected in delta since the capacitance of the capacitor required is one-

third of that required for star connection.

112 Principles of Power System

Equating exps. (i) and (ii), we get,

50240 C = 14·75

∴ C = 14·75/50,240 = 293·4 × 10

−6

F = 293·4 µ F

Since it is the combined capacitance of four equal capacitors joined in series,

∴ Capacitance of each capacitor = 4 × 293·4 = 1173·6

µµ

µF

Example 6.6. The load on an installation is 800 kW, 0·8 lagging p.f. which works for 3000

hours per annum. The tariff is Rs 100 per kVA plus 20 paise per kWh. If the power factor is improved

to 0·9 lagging by means of loss-free capacitors costing Rs 60 per kVAR, calculate the annual saving

effected. Allow 10% per annum for interest and depreciation on capacitors.

Solution.

Load, P = 800 kW

cos φ

= 0·8 ; tan φ

= tan (cos

−1

0·8) = 0·75

cos φ

= 0·9 ; tan φ

= tan (cos

−1

0·9) = 0·4843

Leading kVAR taken by the capacitors

= P (tan φ

− tan φ

) = 800 (0·75 − 0·4843) = 212·56

Annual cost before p.f. correction

Max. kVA demand = 800/0·8 = 1000

kVA demand charges = Rs 100 × 1000 = Rs 1,00,000

Units consumed/year = 800 × 3000 = 24,00,000 kWh

Energy charges/year = Rs 0·2 × 24,00,000 = Rs 4,80,000

Total annual cost = Rs (1,00,000 + 4,80,000) = Rs 5,80,000

Annual cost after p.f. correction

Max. kVA demand = 800/0·9 = 888·89

kVA demand charges = Rs 100 × 888·89 = Rs 88,889

Energy charges = Same as before i.e., Rs 4,80,000

Capital cost of capacitors = Rs 60 × 212·56 = Rs 12,750

Annual interest and depreciation = Rs 0·1 × 12750 = Rs 1275

Total annual cost = Rs (88,889 + 4,80,000 + 1275) = Rs 5,70,164

∴ Annual saving = Rs (5,80,000 − 5,70,164) = Rs 9836

Example 6.7. A factory takes a load of 200 kW at 0·85 p.f. lagging for 2500 hours per annum.

The traiff is Rs 150 per kVA plus 5 paise per kWh consumed. If the p.f. is improved to 0·9 lagging by

means of capacitors costing Rs 420 per kVAR and having a power loss of 100 W per kVA, calculate

the annual saving effected by their use. Allow 10% per annum for interest and depreciation.

Solution :

Factory load, P

= 200 kW

cos φ

= 0·85 ; tan φ

= 0·62

cos φ

= 0·9 ; tan φ

= 0·4843

Suppose the leading kVAR taken by the capacitors is x.

∴ Capacitor loss =

100

1000

= 0·1 x kW

Total power, P

= (200 + 0·1x) kW

Leading kVAR taken by the capacitors is

x = P

tan φ

− P

tan φ

= 200 × 0·62 − (200 + 0·1x) × 0·4843

Power Factor Improvement 113

or x = 124 − 96·86 − 0·04843 x

∴ x = 27·14/1·04843 = 25·89 kVAR

Annual cost before p.f. improvement

Max. kVA demand = 200/0.85 = 235.3

kVA demand charges = Rs 150 × 235·3 = Rs 35,295

Units consumed/year = 200 × 2500 = 5,00,000 kWh

Energy charges = Rs 0·05 × 5,00,000 = Rs 25,000

Total annual cost = Rs (35,295 + 25,000) = Rs 60,295

Annual cost after p.f. improvement

Max. kVA demand = 200/0·9 = 222·2

kVA demand charges = Rs 150 × 222·2 = Rs 33,330

Energy charges = same as before i.e., Rs 25,000

Annual interest and depreciation = Rs 420 × 25·89 × 0·1 = Rs 1087

Annual energy loss in capacitors = 0·1 x × 2500 = 0·1 × 25·89 × 2500 = 6472 kWh

Annual cost of losses occurring in capacitors

= Rs 0·05 × 6472 = Rs 323

∴ Total annual cost = Rs (33,330 + 25,000 + 1087 + 323) = Rs 59,740

Annual saving = Rs (60,295 − 59,740) = Rs 555

Example 6.8. A factory operates at 0·8 p.f. lagging and has a monthly demand of 750 kVA. The

monthly power rate is Rs 8·50 per kVA. To improve the power factor, 250 kVA capacitors are in-

stalled in which there is negligible power loss. The installed cost of equipment is Rs 20,000 and fixed

charges are estimated at 10% per year. Calculate the annual saving effected by the use of capaci-

tors.

Solution.

Monthly demand is 750 kVA.

cos φ = 0·8 ; sin φ = sin (cos

−1

0·8) = 0·6

kW component of demand = kVA × cos φ = 750 × 0·8 = 600

kVAR component of demand = kVA × sin φ = 750 × 0·6 = 450

Leading kVAR taken by the capacitors is 250 kVAR. Therefore, net kVAR after p.f. improve-

ment is 450 − 250 = 200.

∴ kVA after p.f. improvement =

600 200

afaf

= 632·45

Reduction in kVA = 750 − 632·45 = 117·5

Monthly saving on kVA charges = Rs 8·5 × 117·5 = Rs 998·75

Yearly saving on kVA charges = Rs 998·75 × 12 = Rs 11,985

Fixed charges/year = Rs 0·1 × 20,000 = Rs 2000

Net annual saving = Rs (11,985 − 2000) = Rs 9,985

Example 6.9. A synchronous motor improves the power factor of a load of 200 kW from 0.8

lagging to 0.9 lagging. Simultaneously the motor carries a load of 80 kW. Find (i) the leading kVAR

taken by the motor (ii) kVA rating of the motor and (iii) power factor at which the motor operates.

Solution.

Load, P

= 200 kW ; Motor load, P

= 80 kW

p.f. of load, cos φ

= 0·8 lag

p.f. of combined load, cos φ

= 0·9 lag

114 Principles of Power System

Combined load, P = P

+ P

= 200 + 80 = 280 kW

In Fig. 6.12, ∆ OAB is the power triangle for load,

∆ ODC for combined load and ∆ BEC for the motor.

(i) Leading kVAR taken by the motor

= CE = DE − DC = AB − DC

[ AB = DE]

= P

tan φ

− P* tan φ

= 200 tan (cos

−1

0·8) − 280 tan (cos

−1

0·9)

= 200 × 0·75 − 280 × 0·4843

= 14·4 kVAR

(ii) kVA rating of the motor = BC =

BE EC

af af af a f

+= +⋅

= 81·28 kVA

(iii) p.f. of motor, cos φ

Motor kW

Motor kVA

⋅

= 0·984 leading

Example 6.10. A factory load consists of the following :

(i) an induction motor of 50 H.P. (37·3 kW) with 0·8 p.f. and efficiency 0·85.

(ii) a synchronous motor of 25 H.P. (18·65 kW) with 0·9 p.f. leading and efficiency 0·9.

(iii) lighting load of 10 kW at unity p.f.

Find the annual electrical charges if the tariff is Rs 60 per kVA of maximum demand per annum

plus 5 paise per kWh ; assuming the load to be steady for 2000 hours in a year.

Solution.

Input power to induction motor = 37·3/0·85 = 43·88 kW

Lagging kVAR taken by induction motor = 43·88 tan (cos

−1

0·8) = 32·91

Input power to synchronous motor

= 18·65/0·9 = 20·72 kW

Leading kVAR taken by synchronous motor

= 20·72 tan (cos

−1

0·9) = 10

Since lighting load works at unity p.f., its lagging kVAR = 0.

Net lagging kVAR = 32·91 − 10 = 22·91

Total active power = 43·88 + 20·72 + 10 = 74·6 kW

Total kVA =

74 6 22 91

⋅+⋅

afa f

= 78

Annual kVA demand charges = Rs 60 × 78 = Rs 4,680

Energy consumed/year = 74·6 × 2000 = 1,49,200 kWh

Annual Energy charges = Rs 0·05 × 1,49,200 = Rs 7,460

Total annual bill = kVA demand charges + Energy charges

= Rs (4680 + 7460) = Rs 12,140

Example 6.11. A supply system feeds the following loads (i) a lighting load of 500 kW (ii) a load

of 400 kW at a p.f. of 0·707 lagging (iii) a load of 800 kW at a p.f. of 0·8 leading (iv) a load of 500 kW

at a p.f. 0·6 lagging (v) a synchronous motor driving a 540 kW d.c. generator and having an overall

efficiency of 90%. Calculate the power factor of synchronous motor so that the station power factor

may become unity.

* In right angled triangle OAB, AB = P

tan φ

In right angled triangle ODC, DC = OD tan φ

= (P

+ P

) tan φ

= P tan φ

Power Factor Improvement 115

Solution. The lighting load works at unity p.f. and, therefore, its lagging kVAR is zero. The

lagging kVAR are taken by the loads (ii) and (iv), whereas loads (iii) and (v) take the leading kVAR.

For station power factor to be unity, the total lagging kVAR must be neutralised by the total leading

kVAR. We know that kVAR = kW tan φ.

∴ Total lagging kVAR taken by loads (ii) and (iv)

= 400 tan (cos

−1

0·707) + 500 tan (cos

−1

0·6)

= 400 × 1 + 500 × 1·33 = 1065

Leading kVAR taken by the load (iii)

= 800 tan (cos

−1

0·8) = 800 × 0·75 = 600

∴ Leading kVAR to be taken by synchronous motor

= 1065 − 600 = 465 kVAR

Motor input = output/efficiency = 540/0·9 = 600 kW

If φ is the phase angle of synchronous motor, then,

tan φ = kVAR/kW = 465/600 = 0·775

∴φ= tan

−1

0·775 = 37·77

∴ p.f. of synchronous motor = cos φ = cos 37·77

= 0·79 leading

Therefore, in order that the station power factor may become unity, the synchronous motor should

be operated at a p.f. of 0·79 leading.

Example 6.12. An industrial load consists of (i) a synchronous motor of 100 metric h.p. (ii)

induction motors aggregating 200 metric h.p., 0·707 power factor lagging and 82% efficiency and

(iii) lighting load aggregating 30 kW.

The tariff is Rs 100 per annum per kVA maximum demand plus 6 paise per kWh. Find the annual

saving in cost if the synchronous motor operates at 0·8 p.f. leading, 93% efficiency instead of 0·8 p.f.

lagging at 93% efficiency.

Solution. The annual power bill will be calculated under two conditions viz., (a) when synchro-

nous motor runs with lagging p.f. and (b) when synchronous motor runs with a leading p.f.

(a) When synchronous motor runs at p.f. 0·8 lagging. We shall find the combined kW and

then calculate total kVA maximum demand using the relation :

kVA =

kW kVAR

afa f

Input to synchronous motor =

100

735

93 100

×⋅

⋅×

= 79 kW

*Lagging kVAR taken by the synchronous motor

= 79 tan (cos

−1

0·8) = 79 × 0·75 = 59·25 kVAR

Input to induction motors =

200

735

0 82 100

×⋅

⋅×

= 179.4 kW

Lagging kVAR taken by induction motors

= 179·4 tan (cos

−1

0·707) = 179·4 × 1 = 179·4 kVAR

Since lighting load works at unity p.f., its lagging kVAR is zero.

∴ Total lagging kVAR = 59·25 + 179·4 = 238·65 kVAR

Total active power = 79 + 179·4 + 30 = 288·4 kW

Total kVA =

23865 288 4

afaf

+⋅

= 374·4 kVA

Annual kVA demand charges = Rs 100 × 374·4 = Rs 37,440

* Since the synchronous motor in this case runs at lagging p.f., it takes lagging kVAR.

116 Principles of Power System

Energy consumed/year = 288·4 × 8760 = 25,26384 kWh

Annual energy charges = Rs 0·06 × 25,26,384 = Rs 1,51,583

Total annual bill = Rs (37,440 + 1,51,583) = Rs 1,89,023

(b) When synchronous motor runs at p.f. 0·8 leading. As the synchronous motor runs at

leading p.f. of 0·8 (instead of 0·8 p.f. lagging), therefore, it takes now 59·25 leading kVAR. The

lagging kVAR taken by induction motors are the same as before i.e., 179·4.

∴ Net lagging kVAR = 179·4 − 59·25 = 120·15

Total active power = Same as before i.e., 288·4 kW

∴ Total kVA =

12015 288 4

afaf

+⋅

= 312·4

Annual kVA demand charges = Rs 100 × 312·4 = Rs 31,240

Annual energy charges = Same as before i.e., Rs 1,51,583

Total annual bill = Rs (31,240 + 1,51,583) = Rs 1,82,823

∴ Annual saving = Rs (1,89,023 − 1,82,823) = Rs 6200

TUTORIAL PROBLEMSTUTORIAL PROBLEMS

TUTORIAL PROBLEMS

1. What should be the kVA rating of a capacitor which would raise the power factor of load of 100 kW from

0·5 lagging to 0·9 lagging ? [125 kVA]

2. A 3-phase, 50 Hz, 3300 V star connected induction motor develops 250 H.P. (186·5 kW), the power

factor being 0·707 lagging and the efficiency 0·86. Three capacitors in delta are connected across the

supply terminals and power factor raised to 0·9 lagging. Calculate :

(i) the kVAR rating of the capacitor bank.

(ii) the capacitance of each unit. [(i) 111·8 kVAR (ii) 10·9

µµ

µF]

3. A 3-phase, 50 Hz, 3000 V motor develops 600 H.P. (447·6 kW), the power factor being 0·75 lagging and

the efficiency 0·93. A bank of capacitors is connected in delta across the supply terminals and power

factor raised to 0·95 lagging. Each of the capacitance units is built of five similar 600-V capacitors.

Determine the capacitance of each capacitor. [156

µµ

µF]

4. A factory takes a load of 800 kW at 0·8 p.f. (lagging) for 3000 hours per annum and buys energy on tariff

of Rs 100 per kVA plus 10 paise per kWh. If the power factor is improved to 0·9 lagging by means of

capacitors costing Rs 60 per kVAR and having a power loss of 100 W per kVA, calculate the annual

saving effected by their use. Allow 10% per annum for interest and depreciation on the capacitors.

[Rs 3972]

5. A station supplies 250 kVA at a lagging power factor of 0·8. A synchronous motor is connected in

parallel with the load. If the combined load is 250 kW with a lagging p.f. of 0.9, determine :

(i) the leading kVAR taken by the motor.

(ii) kVA rating of the motor.

(iii) p.f. at which the motor operates. [(i) 28·9 kVAR (ii) 57·75 kVA (iii) 0·866 lead]

6. A generating station supplies power to the following :

(i) a lighting load of 100 kW;

(ii) an induction motor 800 h.p. (596·8 kW) p.f. 0·8 lagging, efficiency 92%;

(iii) a rotary converter giving 150 A at 400 V at an efficiency of 0·95.

What must be the power factor of the rotary convertor in order that power factor of the supply station may

become unity ? [0·128 leading]

7. A 3-phase, 400 V synchronous motor having a power consumption of 50 kW is connected in parallel

with an induction motor which takes 200 kW at a power factor of 0·8 lagging.

(i) Calculate the current drawn from the mains when the power factor of the synchronous motor is

unity.

Power Factor Improvement 117

(ii) At what power factor should the synchronous motor operate so that the current drawn from the

mains is minimum. ? [(i) 421 A (ii) 0·316 leading]

8. A factory load consists of the following :

(i) an induction motor of 150 h.p. (111·9 kW) with 0·7 p.f. lagging and 80% efficiency ;

(ii) a synchronous motor of 100 h.p. (74·6 kW) with 0·85 p.f. leading at 90% efficiency ;

(iii) a lighting load of 50 kW.

Find the annual electric charges if the tariff is Rs 100 per annum per kVA maximum demand plus 7 paise

per kWh ; assuming the load to be steady throughout the year. [Rs 1,96,070]

9. A 3-phase synchronous motor having a mechanical load (including losses) of 122 kW is connected in

parallel with a load of 510 kW at 0·8 p.f. lagging. The excitation of the motor is adjusted so that the kVA

input to the motor becomes 140 kVA. Determine the new power factor of the whole system.

[0·8956 lagging]

10. A 3-phase synchronous motor is connected in parallel with a load of 700 kW at 0·7 power factor lagging

and its excitation is adjusted till it raises the total p.f. to 0.9 lagging. Mechanical load on the motor

including losses is 150 kW. Find the power factor of the synchronous motor. [0·444 leading]

6.86.8

6.8

Impor Impor

Impor

tance of Ptance of P

tance of P

er Fer F

er F

actor Impractor Impr

actor Impr

ementement

ement

The improvement of power factor is very important for both consumers and generating stations as

discussed below :

(i) For consumers. A consumer* has to pay electricity charges for his maximum demand in

kVA plus the units consumed. If the consumer imporves 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.

(ii) For generating stations. A generating station is as much concerned with power factor im-

provement 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 × cos φ, 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.

6.96.9

6.9

Most Economical Power Factor Most Economical Power Factor

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.

Consider a consumer taking a peak load of P kW at a power factor of cos φ

and charged at a rate

of Rs x per kVA of maximum demand per annum. Suppose the consumer improves the power factor

* This is not applicable to domestic consumers because the domestic load (e.g., lighting load) has a p.f. very

close to unity. Here, consumer means industrial and other big consumers.

† Max. demand in kVA =

Peak kW

cos

If cos φ is more, maximum kVA demand will be less and vice-versa.

118 Principles of Power System

to cos φ

by installing p.f. correction equipment. Let expenditure

incurred on the p.f. correction equipment be Rs y per kVAR per

annum. The power triangle at the original p.f. cos φ

is OAB and

for the improved p.f. cos φ

, it is OAC [See Fig. 6.13].

kVA max. demand at cos φ

, kVA

= P/cos φ

= P sec φ

kVA max. demand at cos φ

, kVA

= P/cos φ

= P sec φ

Annual saving in maximum demand charges

= Rs x (kVA

− kVA

)

= Rs x (P sec φ

− P sec φ

)

= Rs x P (sec φ

− sec φ

) ...(i)

Reactive power at cos φ

, kVAR

= P tan φ

Reactive power at cos φ

, kVAR

= P tan φ

Leading kVAR taken by p.f. correction equipment

= P (tan φ

− tan φ

)

Annual cost of p.f. correction equipment

= Rs Py (tan φ

− tan φ

) ...(ii)

Net annual saving, S = exp. (i) − exp. (ii)

= xP (sec φ

− sec φ

) − yP (tan φ

− tan φ

)

In this expression, only φ

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

is zero i.e.

(S)=0

[xP (sec φ

− sec φ

) − yP (tan φ

− tan φ

)] = 0

(xP sec φ

) −

(xP sec φ

) −

(yP tan φ

) + yP

(tan φ

) = 0

or 0 − xP sec φ

tan φ

− 0 + yP sec

or −x tan φ

+ y sec φ

or tan φ

sec φ

or sin φ

= y/x

∴ Most economical power factor, cos φ

−=−

sin ( /

)

It may be noted that the most economical power factor (cos φ

) depends upon the relative costs

of supply and p.f. correction equipment but is independent of the original p.f. cos φ

Example 6.13 A factory which has a maximum demand of 175 kW at a power factor of 0·75

lagging is charged at Rs 72 per kVA per annum. If the phase advancing equipment costs Rs 120 per

kVAR, find the most economical power factor at which the factory should operate. Interest and

depreciation total 10% of the capital investment on the phase advancing equipment.

Solution :

Power factor of the factory, cos φ

= 0·75 lagging

Max. demand charges, x = Rs 72 per kVA per annum

Expenditure on phase advancing equipment, y = Rs 120 × 0·1 = Rs 12* /kVAR/annum

* The total investment for producing 1 kVAR is Rs 120. The annual interest and depreciation is 10%. It

means that an expenditure of Rs 120 × 10/100 = Rs 12 is incurred on 1 kVAR per annum.

Power Factor Improvement 119

∴ Most economical p.f. at which factory should operate is

cos φ

1 1 12 72

−=−

yx/(/

)

= 0·986 lagging

Example 6.14 A consumer has an average demand of 400 kW at a p.f. of 0·8 lagging and

annual load factor of 50%. The tariff is Rs 50 per kVA of maximum demand per annum plus 5 paise

per kWh. If the power factor is improved to 0·95 lagging by installing phase advancing equipment,

calculate :

(i) the capacity of the phase advancing equipment

(ii) the annual saving effected

The phase advancing equipment costs Rs 100 per kVAR and the annual interest and deprecia-

tion together amount to 10%.

Solution :

Max. kW demand, P = 400/0·5 = 800 kW

Original p.f., cos φ

= 0·8 lag ; Final p.f., cos φ

= 0·95 lag

= cos

−1

(0·8) = 36·9

; tan φ

= tan 36·9

= 0·75

= cos

−1

(0·95) = 18·2

; tan φ

= tan 18·2

= 0·328

(i) Leading kVAR taken by phase advancing equipment

= P (tan φ

− tan φ

) = 800 (0·75 − 0·328) = 337 kVAR

∴ Capacity of phase advancing equipment should be 337 kVAR.

(ii) Max. demand charges, x = Rs 50/kVA/annum

Expenditure on phase advancing equipment

y = Rs 0·1 × 100 = Rs 10/kVAR/annum

Max. kVA demand at 0·8 p.f. = 800/0·8 = 1000 kVA

Max. kVA demand at 0·95 p.f. = 800/0·95 = 842 kVA

Annual saving in maximum demand charges

= Rs 50 (1000 − 842) = Rs 7900

Annual expenditure on phase advancing equipment

= Rs (y × capacity of equipment)

= Rs 10 × 337 = 3370

∴ Net annual saving = Rs (7900 − 3370) = Rs 4530

Example 6.15 A factory has an average demand of 50 kW and an annual load factor of 0·5.

The power factor is 0·75 lagging. The tariff is Rs 100 per kVA of maximum demand per annum plus

5 paise per kWh. If loss free capacitors costing Rs 600 per kVAR are to be utilised, find the value of

power factor at which maximum saving will result. The interest and depreciation together amount to

10%. Also determine the annual saving effected by improving the p.f. to this value.

Solution :

Max. demand charge, x = Rs 100/kVA/annum

Expenditure on capacitors, y = Rs 0·1 × 600 = Rs 60/kVAR/annum

Most economical p.f., cos φ

1 1 60 100

−=−

yx//

af a f

= 0·8 lag

Max. kW demand = 50/0·5 = 100 kW

The maximum kVA demand at 0·75 p.f. is = 100/0·75 = 133·34 kVA, whereas it is = 100/0·8 =

125 kVA at 0·8 p.f.

∴ Annual saving = Rs 100 (133·34 − 125) = Rs 834

120 Principles of Power System

Example 6.16 A factory takes a steady load of 200 kW at a lagging power factor of 0·8. The

tariff is Rs 100 per kVA of maximum demand per annum plus 5 paise per kWh. The phase advancing

plant costs Rs 500 per kVAR and the annual interest and depreciation together amount to 10%.

Find:

(i) the value to which the power factor be improved so that annual expenditure is minimum

(ii) the capacity of the phase advancing plant

(iii) the new bill for energy, assuming that the factory works for 5000 hours per annum.

Solution :

Peak load of factory, P = 200 kW

Original power factor, cos φ

= 0·8 lagging

Max. demand charges, x = Rs 100/kVA/annum

Charges on phase advancing plant, y = Rs 500 × 0·1

= Rs 50/kVAR/annum

(i) Most economical power factor, cos φ

1 1 50 100

−=−

yx//

af a f

= 0·866 lagging

(ii) Capacity of phase advancing plant = P [tan φ

− tan φ

]

= 200 [tan (cos

−1

0·8) −tan (cos

−1

0·866)]

= 200 [0·75 − 0·5774] = 34·52 kVAR

(iii) Units consumed/year = 200 × 5000 = 10

kWh

Annual energy charges = Rs 0·05 × 10

= Rs 50,000

Annual cost of phase advancing plant = Rs y × Capacity of plant

= Rs 50 × 34·52 = Rs 1726

Max. demand charge = Rs x × P/cos φ

= Rs 100 × 200/0·866 = Rs 23,094

Annual bill for energy = Rs (50,000 + 1726 + 23,094) = Rs 74,820

Example 6.17 An industrial load takes 80,000 units in a year, the average power factor being

0·707 lagging. The recorded maximum demand is 500 kVA. The tariff is Rs 120 per kVA of maximum

demand plus 2·5 paise per kWh. Calculate the annual cost of supply and find out the annual saving

in cost by installing phase advancing plant costing Rs 50 per kVAR which raises the p.f. from 0·707

to 0·9 lagging. Allow 10% per year on the cost of phase advancing plant to cover all additional

costs.

Solution.

Energy consumed/year = 80,000 kWh

Maximum kVA demand = 500

Annual cost of supply = M.D. Charges + Energy charges

= Rs (120 × 500 + 0·025 × 80,000)

= Rs (60,000 + 2000) = Rs 62,000

cos φ

= 0·707 lag ; cos φ

= 0·9 lag

Max. kW demand at 0·707 p.f.,P = 500 × 0·707 = 353·3 kW

Leading kVAR taken by phase advancing equipment

= P [tan φ

− tan φ

]

= 353·3 [tan (cos

−1

0·707) − tan (cos

−1

0·9)]

= 353·3 [1 − 0·484] = 182·3 kVAR

Annual cost of phase advancing equipment

= Rs 182·3 × 50 × 0·1 = Rs 912

Power Factor Improvement 121

When p.f. is raised from 0·707 lag to 0·9 lag, new maximum kVA demand is = 353·3/0·9 = 392·6

kVA.

Reduction in kVA demand = 500 − 392·6 = 107·4

Annual saving in kVA charges = Rs 120 × 107·4 = Rs 12,888

As the units consumed remain the same, therefore, saving will be equal to saving in M.D. charges

minus annual cost of phase advancing plant.

∴ Annual saving = Rs (12,882 − 912) = Rs 11,976

TUTORIAL PORBLEMSTUTORIAL PORBLEMS

TUTORIAL PORBLEMS

1. A factory which has a maximum demand of 175 kW at a power factor of 0·75 lagging is charged at Rs 72

per kVA per annum. If the phase advancing equipment costs Rs 120 per kVAR, find the most economical

power factor at which the factory should operate. Interest and depreciation total 10% of the capital

investment on the phase advancing equipment. [0·986 leading]

2. A consumer has a steady load of 500 kW at a power factor of 0·8 lagging. The tariff in force is Rs 60 per

kVA of maximum demand plus 5 paise per kWh. If the power factor is improved to 0·95 lagging by

installing phase advancing equipment, calculate :

(i) The capacity of the phase advancing equipment.

(ii) The annual saving effected.

The phase advancing equipment costs Rs 100 per kVAR and the annual interest and depreciation together

amount to 10%. [(i) 210·6 kVAR (ii) Rs. 3,815]

3. A factory has an average demand of 320 kW and an annual load factor of 50%. The power factor is 0·8

lagging. The traiff is Rs 80 per annum per kVA of maximum demand plus 5 paise per kWh. If the loss

free capacitors costing Rs 100 per kVAR are to be utilised, find the value of power factor at which

maximum saving will result. The interest and depreciation together amount to 12%. Also determine the

annual saving effected by improving the power factor to this value. [0·988 lagging ; Rs 3040]

4. What will be the kVA rating of a phase advancing plant if it improves p.f. from 0·8 lagging to 0·891

lagging ? The consumer load is 1000 kW and the current taken by the phase advancer leads the supply

voltage at a p.f. of 0·1. [230 kVA]

5. A consumer takes a steady load of 300 kW at a lagging power factor of 0·7 for 3000 hours a year. The

tariff is Rs 130 per kVA of maximum demand annually and 4 paise per kWh. The annual cost of phase

advancing plant is Rs 13 per kVAR. Determine the annual saving if the power factor of the load is

improved ? [Rs 12929·8]

6.106.10

6.10

Meeting the Incr Meeting the Incr

Meeting the Incr

eased eased

eased

W Demand on PW Demand on P

W Demand on P

er Staer Sta

er Sta

tionstions

tions

The useful output of a power station is the kW output delivered by it to the supply system.

Sometimes, a power station is required to deliver more kW to meet the increase in power demand.

This can be achieved by either of the following two methods :

(i) By increasing the kVA capacity of the power station at the same power factor (say cos φ

Obviously, extra cost will be incurred to increase the kVA capacity of the station.

(ii) By improving the power factor of the station from cos φ

to cos φ

without increasing the

kVA capacity of the station. This will also involve extra cost on account of power factor

correction equipment.

Economical comparison of two methods. It is clear that each method of increasing kW capac-

ity of the station involves extra cost. It is, therefore, desirable to make economical comparison of the

two methods. Suppose a power station of rating P kVA is supplying load at p.f. of cos φ

. Let us

suppose that the new power demand can be met either by increasing the p.f. to cos φ

at P kVA or by

122 Principles of Power System

increasing the kVA rating of the station at the original p.f. cos φ

. The power* triangles for the whole

situation are shown in Fig. 6.14.

(i) Cost of increasing kVA capacity of station. Referring to Fig. 6.14, the increase in kVA

capacity of the station at cos φ

to meet the new demand is given by :

Increase in kVA capacity

= BD =

φφ

cos

( BF = AC)

−

cos

φφ

−

P cos cos

cos

φφ

−

[  OE = OB = P]

If Rs x is the annual cost per kVA of the station, then,

Annual cost due to increase in kVA capacity

= Rs

xP cos cos

cos

φφ

−

...(i)

(ii) Cost of p.f. correction equipment. Referring to Fig. 6.14, the new demand in kW can be

met by increasing the p.f. from cos φ

to cos φ

at the original kVA of the station. The leading kVAR

to be taken by the p.f. correction equipment is given by ED i.e.

Leading kVAR taken by p.f. correction equipment

= ED = CD − CE

= OD sin φ

− OE sin φ

sin φ

− OE sin φ

OE cos

cos

sin φ

− OE sin φ

= OE (tan φ

cos φ

− sin φ

)

= P (tan φ

cos φ

− sin φ

)

If Rs. y is the annual cost per kVAR of the p.f. correction equipment, then,

Annual cost on p.f. correction equipment

= Rs y P (tan φ

cos φ

− sin φ

) ...(ii)

Different cases

(a) The p.f. correction equipment will be cheaper if

exp. (ii) < exp. (i)

or yP (tan φ

cos φ

− sin φ

xP cos cos

cos

φφ

−

* Note the construction. Here ∆ OAB is the power triangle for the station supplying P kVA at cos φ

. The

demand on the station is OA kW. The new demand is OC kW. This can be met :

(i) either by increasing the kVA demand of the station to OD at the same p.f. cos φ

. Obviously, ∆ OCD

is the power triangle when the station is supplying OC kW at cos φ

(ii) or by increasing the p.f. from cos φ

to cos φ

at same kVA i.e., P kVA. Obviously, OB = OE.

Therefore, ∆ OCE is the power triangle when the station is supplying OC kW at improved p.f. cos φ

Power Factor Improvement 123

or y (tan φ

cos φ

− sin φ

)<x

cos cos

cos

φφ

−

(b) The maximum annual cost per kVAR (i.e., y) of p.f. correction equipment that would justify

its installation is when

exp. (i) = exp. (ii)

or yP (tan φ

cos φ

− sin φ

xP (cos

cos

)

cos

φφ

−

sin

cos

cos sin

φφ

−

x cos cos

cos

φφ

−

sin cos sin cos

cos

φφ φφ

12 21

−

x cos cos

cos

φφ

−

or y sin (φ

− φ

)=x (cos φ

− cos φ

)

∴ y =

cos cos

sin

φφ

−

Example 6.18 A power plant is working at its maximum kVA capacity with a lagging p.f. of 0·7.

It is now required to increase its kW capacity to meet the demand of additional load. This can be

done :

(i) by increasing the p.f. to 0·85 lagging by p.f. correction equipment

(ii) by installing additional generation plant costing Rs 800 per kVA.

What is the maximum cost per kVA of p.f. correction equipment to

make its use more economical than the additional plant ?

Soloution. Let the initial capacity of the plant be OB kVA at a p.f.

cos φ

. Referring to Fig. 6.15, the new kW demand (OC) can be met by

increasing the p.f. from 0·7 (cos φ

) to 0·85 lagging (cos φ

) at OB kVA or

by increasing the capacity of the station to OD kVA at cos φ

Cost of increasing plant capacity. Referring to Fig. 6.15, the in-

crease in kVA capacity is BD.

Now OE cos φ

= OD cos φ

or OB cos φ

= OD cos φ

( OE = OB)

∴ OD = OB × cos φ

/cos φ

= OB × 0·85/0·7 = 1·2143 OB

Increase in the kVA capacity of the plant is

BD = OD − OB = 1·2143 × OB − OB = 0·2143 OB

∴ Total cost of increasing the plant capacity

= Rs 800 × 0·2143 × OB

= Rs 171·44 × OB ...(i)

Cost of p.f. correction equipment.

cos φ

= 0·7 ∴ sin φ

= 0·714

cos φ

= 0·85 ∴ sin φ

= 0·527

Leading kVAR taken by p.f. correction equipment is

ED = CD − CE = OD sin φ

− OE sin φ

= 1·2143 × OB sin φ

− OB sin φ

= OB (1·2143 × 0·714 − 0·527) = 0·34 × OB

124 Principles of Power System

Let the cost per kVAR of the equipment be Rs y.

∴ Total cost of p.f. correction equipment

= Rs 0·34 × OB × y ...(ii)

The cost per kVAR of the equipment that would justify its installation is when exp. (i) = exp. (ii)

i.e.,

171·44 × OB = 0·34 × OB × y

∴ y = Rs 171·44/0·34 = Rs 504·2 per kVAR

If the losses in p.f. correction equipment are neglected, then its kVAR = kVA. Therefore, the

maximum cost per kVA of p.f. correction equipment that can be paid is Rs 504·2.

Example 6.19. A system is working at its maximum kVA capacity with a lagging power factor

0·7. An anticipated increase of load can be met by one of the following two methods :

(i) By raising the p.f. of the system to 0·866 by installing phase advancing equipment.

(ii) By installing extra generating plant.

If the total cost of generating plant is Rs 100 per kVA, estimate the limiting cost per kVA of phase

advancing equipment to make its use more economical than the additional generating plant. Interest

and depreciation charges may be assumed 10% in each case.

Solution. The original demand is OA and the increased demand is OC. Fig. 6.16 shows the two

methods of meeting the increased kW demand (OC).

Cost of increasing plant capacity

BD = OD − OB

= OB ×

0 866

070

⋅

− OB

= OB (1·237 − 1)

= 0·237 × OB

∴ Annual cost of increasing the plant capacity

= Rs 10 × 0·237 × OB

= Rs. 2·37 × OB ...(i)

Cost of phase advancing equipment. Leading kVAR taken by phase advancing equipment,

ED = CD − CE

= OD* sin φ

− OE sin φ

= 1·237 × OB × sin φ

− OB sin φ

= OB (1·237 × 0·174 − 0·5) = 0·383 × OB

Let the cost per kVAR of the equipment be Rs y.

Annual cost of phase advancing equipment

= Rs 0·1 × y × 0·383 × OB ...(ii)

For economy, the two costs should be equal i.e., exp. (i) = exp. (ii).

∴ 0·1 × y × 0·383 × OB = 2·37 × OB

or y = Rs

237

1038

⋅

⋅×⋅

= Rs 61·88

If the losses in the phase advancing equipment are neglected, then its kVAR = kVA. Hence, the

maximum cost per kVA of phase advancing equipment that can be paid is Rs 61·88.

* OD = OB + BD = OB + 0·237 × OB = 1·237 × OB

Power Factor Improvement 125

TUTORIAL PROBLEMSTUTORIAL PROBLEMS

TUTORIAL PROBLEMS

1. A system is working at its maximum capacity with a lagging power factor of 0·707. An anticipated

increase in load can be met by (i) raising the power factor of the system to 0·87 lagging by the installation

of phase advancers and (ii) by installing extra generating cables etc. to meet the increased power demand.

The total cost of the latter method is Rs 110 per kVA. Estimate the limiting cost per kVA of the phase

advancing plant which would justify the installation. [Rs 76·26 per kVAR]

2. For increasing the kW capacity of a power station working at 0·7 lagging power factor, the necessary

increase in power can be obtained by raising power factor to 0·9 lagging or by installing additional plant.

What is the maximum cost per kVA of power factor correction apparatus to make its use more economi-

cal than the additional plant at Rs 800 per kVA ? [Rs 474 per kVA]

3. An electrical system is working at its maximum kVA capacity with a lagging p.f. of 0·8. An anticipated

increase of load can be met either by raising the p.f. of the system to 0·95 lagging by the installation of

phase advancing plant or by erecting an extra generating plant and the required accessories. The total

cost of the latter method is Rs 80 per kVA. Determine the economic limit cost per kVA of the phase

advancing plant. Interest and depreciation may be assumed 12% in either case. [Rs 37.50 per kVA]

SELF-TESTSELF-TEST

SELF-TEST

1. Fill in the blanks by inserting appropriate words/figures.

(i) The power factor of an a.c. circuit is given by ............... power divided by ............... power.

(ii) The lagging power factor is due to ............... power drawn by the circuit.

(iii) Power factor can be improved by installing such a device in parallel with load which takes ........... .

(iv) The major reason for low lagging power factor of supply system is due to the use of ............... motors.

(v) An over-excited synchronous motor on no load is known as ...............

2. Pick up the correct words/figures from the brackets and fill in the blanks.

(i) The smaller the lagging reactive power drawn by a circuit. the ............... is its power factor.

(smaller, greater)

(ii) The maximum value of power factor can be ............... (1, 0·5, 0·9)

(iii) KVAR = ............... tan φ (kW, KVA)

(iv) By improving the power factor of the system, the kilowatts delivered by the generating station are

............... (decreased, increased, not changed)

(v) The most economical power factor for a consumer is generally ...............

(0·95 lagging, unity, 0·6 lagging)

ANSWER TO SELF-TESTANSWER TO SELF-TEST

ANSWER TO SELF-TEST

1. (i) active, apparent, (ii) lagging reactive (iii) leading reactive power, (iv) induction (v) synchronous

condenser.

2. (i) greater, (ii) 1, (iii) kW, (iv) increased, (v) 0·95 lagging.

CHAPTER REVIEW TOPICSCHAPTER REVIEW TOPICS

CHAPTER REVIEW TOPICS

1. Why is there phase difference between voltage and current in an a.c. circuit ? Explain the concept of

power factor.

2. Discuss the disadvantages of a low power factor.

3. Explain the causes of low power factor of the supply system.

4. Discuss the various methods for power factor improvement.

5. Derive an expression for the most economical value of power factor which may be attained by a

consumer.

126 Principles of Power System

6. Show that the economical limit to which the power factor of a load can be raised is independent of the

original value of power factor when the tariff consists of a fixed charge per kVA of maximum demand

plus a flat rate per kWh.

7. Write short notes on the following :

(i) Power factor improvement by synchronous condenser

(ii) Importance of p.f. improvement

(iii) Economics of p.f. improvement

DISCUSSION QUESTIONSDISCUSSION QUESTIONS

DISCUSSION QUESTIONS

1. What is the importance of power factor in the supply system ?

2. Why is the power factor not more than unity ?

3. What is the effect of low power factor on the generating stations ?

4. Why is unity power factor not the most economical p.f. ?

5. Why a consumer having low power factor is charged at higher rates ?

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