101
IntrIntr
IntrIntr
Intr
oductionoduction
oductionoduction
oduction
T
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
6.16.1
6.1
Power FactorPower Factor
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
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
CONTENTS
CONTENTS
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
o
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
6.26.2
6.2
P P
P P
P
oo
oo
o
ww
ww
w
er er
er er
er
TT
TT
T
rr
rr
r
iangleiangle
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.
OB
2
= OA
2
+ AB
2
or (apparent power)
2
= (active power)
2
+ (reactive power)
2
or (kVA)
2
= (kW)
2
+ (kVAR)
2
(ii) Power factor, cos φ =
O
A
O
B
==
active
power
apparent power
kW
kV
A
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 =
kW
c
os
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 =
R
Z
=
Resistance
Impedanc
e
(c) Power factor =
VI
Active pow
er
Apparent Powe
r
φ
=
(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
6.36.3
6.3
Disadvantages of Low Power Factor Disadvantages of Low Power Factor
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
L
I
L
cos φ (For single phase supply)
I
L
=
P
V
L
cos
φ
...(i)
P =
3
V
L
I
L
cos φ (For 3 phase supply)
I
L
=
P
V
L
3cos
φ
...(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 =
kW
c
os
φ
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
2
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
6.46.4
6.4
Causes of Low Power Factor Causes of Low Power Factor
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
6.56.5
6.5
P P
P P
P
oo
oo
o
ww
ww
w
er Fer F
er Fer F
er F
actor Impractor Impr
actor Impractor Impr
actor Impr
oo
oo
o
vv
vv
v
ementement
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
1
as shown in Fig. 6.3.
The capacitor C is connected in parallel with the load. The capacitor draws current I
C
which
leads the supply voltage by 90
o
. The resulting line current I is the phasor sum of I and I
C
and its angle
of lag is φ
2
as shown in the phasor diagram of Fig. 6.3. (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
are 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.
I cos φ
1
=
I
cos φ
2
(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
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
C
V
I
sin φ
2
= VI sin φ
1
VI
C
[Multiplying by V]
i.e., Net kVAR after p.f. correction = Lagging kVAR before p.f. correction leading kVAR of
equipment
6.66.6
6.66.6
6.6
P P
P P
P
oo
oo
o
ww
ww
w
er Fer F
er Fer F
er F
actor Impractor Impr
actor Impractor Impr
actor Impr
oo
oo
o
vv
vv
v
ement Equipmentement Equipment
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
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.
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
m
= 90
o
. However, in actual practice, losses do occur in
the motor even at no load. Therefore, the currents I
m
leads the voltage by an angle less than 90
o
.
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
o
. 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
6.76.7
6.7
CalculaCalcula
CalculaCalcula
Calcula
tions of Ptions of P
tions of Ptions of P
tions of P
oo
oo
o
ww
ww
w
er Fer F
er Fer F
er F
actor Corractor Corr
actor Corractor Corr
actor Corr
ectionection
ectionection
ection
Consider an inductive load taking a lagging current I at a power factor cos φ
1
. 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
C
which
leads the supply voltage V by 90
o
. The current I
C
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
2
. It is clear that φ
2
is less than φ
1
so that new p.f. cos φ
2
is more than the
previous p.f. cos φ
1
.
From the phasor diagram, it is clear that after p.f. correction, the lagging reactive component of
the load is reduced to Isin φ
2
.
Obviously, I sin φ
2
= I sin φ
1
I
C
or I
C
= I sin φ
1
I sin φ
2
Capacitance of capacitor to improve p.f. from cos φ
1
to cos φ
2
=
I
V
C
ω
X
V
IC
C
C
==
F
H
G
I
K
J
1
ω
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
1
, whereas power triangle
OAC is for the improved power factor cos φ
2
. 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
2
.
Leading kVAR supplied by p.f. correction equipment
= BC = AB AC
=kVAR
1
kVAR
2
= OA (tan φ
1
tan φ
2
)
= kW (tan φ
1
tan φ
2
)
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 =
kW
c
os
φ
=
300
0
6
= 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
M
of 31·7A. The current I
C
taken by the capacitor must be such that when
combined with I
M
, 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
M
= I
M
cos φ
M
= 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
09
= 24·65A
Reactive component of I
M
= I
M
sin φ
M
= 31·7 × 0·714* = 22·6A
Reactive component of I = I sin φ = 24·65
109
2
−⋅
af
= 24·65 × 0·436 = 10·75 A
It is clear from Fig. 6.9 that :
I
C
= Reactive component of I
M
Reactive component of I
= 22·6 10·75 = 11·85A
But I
C
=
V
X
C
= V × 2π f C
or 11·85 = 400 × 2π × 50 × C
C = 94·3 × 10
6
F = 94·3
µµ
µµ
µF
* sin φ
M
=
()
2
2
M
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
1
=
kW
1
1
2
0
1
c
os
φ
=
= 20 kVA
kVA
2
=
kW
2
2
100
070
7
c
os
φ
=
= 141·4 kVA
kVA
3
=
kW
3
3
5
0
0
9
c
os
φ
=
= 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
1
=kVA
1
sin φ
1
= 20 × 0 = 0
kVAR
2
=kVA
2
sin φ
2
= 141·4 × 0·707 = 100 kVAR
kVAR
3
=kVA
3
sin φ
3
= + 55·6 × 0·436 = + 24·3 kVAR
Note that kVAR
2
and kVAR
3
are in opposite directions ; kVAR
2
being a lagging while kVAR
3
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
2
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
1
= 0·75 lag ; Motor input, P = 5 kW
Final p.f., cos φ
2
= 0·9 lag ; Efficiency, η = 100 % (assumed)
φ
1
= cos
1
(0·75) = 41·41
o
; tan φ
1
= tan 41·41º = 0·8819
φ
2
= cos
1
(0·9) = 25·84
o
; tan φ
2
= tan 25·84º = 0·4843
Leading kVAR taken by the condenser bank
= P (tan φ
1
tan φ
2
)
= 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
1
= 0·75 lag ; Final p.f., cos φ
2
= 0·95 lag
Motor input, P = output/η = 74·6/0·93 = 80 kW
φ
1
= cos
1
(0·75) = 41·41
o
tan φ
1
= tan 41·41º = 0·8819
φ
2
= cos
1
(0·95) = 18·19
o
tan φ
2
= tan 18·19
o
= 0·3288
Leading kVAR taken by the condenser bank
= P (tan φ
1
tan φ
2
)
= 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
I
CP
= V
ph
/X
C
= 2π f C V
ph
=2π × 50 × C × 400
= 1,25,600 C amperes
kVAR/phase =
V
I
ph CP
1000
=
400
1
25
600
1000
×
,
,
C
= 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
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
1
= 0·8 ; tan φ
1
= tan (cos
1
0·8) = 0·75
cos φ
2
= 0·9 ; tan φ
2
= tan (cos
1
0·9) = 0·4843
Leading kVAR taken by the capacitors
= P (tan φ
1
tan φ
2
) = 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
1
= 200 kW
cos φ
1
= 0·85 ; tan φ
1
= 0·62
cos φ
2
= 0·9 ; tan φ
2
= 0·4843
Suppose the leading kVAR taken by the capacitors is x.
Capacitor loss =
100
1000
×
x
= 0·1 x kW
Total power, P
2
= (200 + 0·1x) kW
Leading kVAR taken by the capacitors is
x = P
1
tan φ
1
P
2
tan φ
2
= 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
2
2
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
1
= 200 kW ; Motor load, P
2
= 80 kW
p.f. of load, cos φ
1
= 0·8 lag
p.f. of combined load, cos φ
2
= 0·9 lag
114 Principles of Power System
Combined load, P = P
1
+ P
2
= 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
1
tan φ
1
P* tan φ
2
= 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
2
2
2
2
80
14
4
+= +
= 81·28 kVA
(iii) p.f. of motor, cos φ
m
=
Motor kW
Motor kVA
=
80
81
28
= 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
22
⋅+
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
1
tan φ
1
In right angled triangle ODC, DC = OD tan φ
2
= (P
1
+ P
2
) tan φ
2
= P tan φ
2
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
o
p.f. of synchronous motor = cos φ = cos 37·77
o
= 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
2
2
afa f
+
Input to synchronous motor =
100
735
5
0
93 100
0
×⋅
⋅×
= 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
5
0 82 100
0
×⋅
⋅×
= 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
2
2
.
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
2
2
.
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 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
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
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
6.86.8
6.8
Impor Impor
Impor Impor
Impor
tance of Ptance of P
tance of Ptance of P
tance of P
oo
oo
o
ww
ww
w
er Fer F
er Fer F
er F
actor Impractor Impr
actor Impractor Impr
actor Impr
oo
oo
o
vv
vv
v
ementement
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
6.96.9
6.9
Most Economical Power Factor Most Economical Power Factor
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
1
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
φ
If cos φ is more, maximum kVA demand will be less and vice-versa.
118 Principles of Power System
to cos φ
2
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
1
is OAB and
for the improved p.f. cos φ
2
, it is OAC [See Fig. 6.13].
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
) ...(i)
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 Py (tan φ
1
tan φ
2
) ...(ii)
Net annual saving, S = exp. (i) exp. (ii)
= xP (sec φ
1
sec φ
2
) yP (tan φ
1
tan φ
2
)
In this expression, only φ
2
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
2
is zero i.e.
d
d
φ
2
(S)=0
or
d
d
φ
2
[xP (sec φ
1
sec φ
2
) yP (tan φ
1
tan φ
2
)] = 0
or
d
d
φ
2
(xP sec φ
1
)
d
d
φ
2
(xP sec φ
2
)
d
d
φ
2
(yP tan φ
1
) + yP
d
d
φ
2
(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
=
11
2
2
2
−=
sin ( /
)
φ
yx
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
.
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
1
= 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
2
=
1 1 12 72
2
2
−=
yx/(/
)
af
= 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
1
= 0·8 lag ; Final p.f., cos φ
2
= 0·95 lag
φ
1
= cos
1
(0·8) = 36·9
o
; tan φ
1
= tan 36·9
o
= 0·75
φ
2
= cos
1
(0·95) = 18·2
o
; tan φ
2
= tan 18·2
o
= 0·328
(i) Leading kVAR taken by phase advancing equipment
= P (tan φ
1
tan φ
2
) = 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
2
=
1 1 60 100
2
2
−=
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
1
= 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
2
=
1 1 50 100
2
2
−=
yx//
af a f
= 0·866 lagging
(ii) Capacity of phase advancing plant = P [tan φ
1
tan φ
2
]
= 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
6
kWh
Annual energy charges = Rs 0·05 × 10
6
= 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
2
= 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
1
= 0·707 lag ; cos φ
2
= 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
1
tan φ
2
]
= 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 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
6.106.10
6.10
Meeting the Incr Meeting the Incr
Meeting the Incr Meeting the Incr
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eased
kk
kk
k
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W Demand on PW Demand on P
W Demand on P
oo
oo
o
ww
ww
w
er Staer Sta
er Staer Sta
er Sta
tionstions
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
1
).
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
1
to cos φ
2
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
1
. Let us
suppose that the new power demand can be met either by increasing the p.f. to cos φ
2
at P kVA or by
122 Principles of Power System
increasing the kVA rating of the station at the original p.f. cos φ
1
. 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
1
to meet the new demand is given by :
Increase in kVA capacity
= BD =
BF
AC
c
os
φφ
1
=
cos
1
( BF = AC)
=
OC
OA
cos
φ
1
=
OE
OB
cos
cos
cos
φφ
φ
2
1
1
=
P cos cos
cos
φφ
φ
21
1
ch
[ 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
φ
21
1
ch
...(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
1
to cos φ
2
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
1
OE sin φ
2
=
OC
c
os
φ
1
sin φ
1
OE sin φ
2
=
OE cos
cos
φ
φ
2
1
sin φ
1
OE sin φ
2
= OE (tan φ
1
cos φ
2
sin φ
2
)
= P (tan φ
1
cos φ
2
sin φ
2
)
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
1
cos φ
2
sin φ
2
) ...(ii)
Different cases
(a) The p.f. correction equipment will be cheaper if
exp. (ii) < exp. (i)
or yP (tan φ
1
cos φ
2
sin φ
2
)<
xP cos cos
cos
φφ
φ
21
1
ch
* Note the construction. Here OAB is the power triangle for the station supplying P kVA at cos φ
1
. 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
1
. Obviously, OCD
is the power triangle when the station is supplying OC kW at cos φ
1
.
(ii) or by increasing the p.f. from cos φ
1
to cos φ
2
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
2
.
Power Factor Improvement 123
or y (tan φ
1
cos φ
2
sin φ
2
)<x
cos cos
cos
φφ
φ
21
1
ch
(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
1
cos φ
2
sin φ
2
)=
xP (cos
cos
)
cos
φφ
φ
21
1
or
y
sin
cos
cos sin
φ
φ
φφ
1
1
22
F
H
G
I
K
J
=
x cos cos
cos
φφ
φ
21
1
ch
or
y
sin cos sin cos
cos
φφ φφ
φ
12 21
1
F
H
G
I
K
J
=
x cos cos
cos
φφ
φ
21
1
ch
or y sin (φ
1
φ
2
)=x (cos φ
2
cos φ
1
)
y =
x
cos cos
sin
φφ
φφ
21
12
ch
ch
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
or
(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
1
. Referring to Fig. 6.15, the new kW demand (OC) can be met by
increasing the p.f. from 0·7 (cos φ
1
) to 0·85 lagging (cos φ
2
) at OB kVA or
by increasing the capacity of the station to OD kVA at cos φ
1
.
Cost of increasing plant capacity. Referring to Fig. 6.15, the in-
crease in kVA capacity is BD.
Now OE cos φ
2
= OD cos φ
1
or OB cos φ
2
= OD cos φ
1
( OE = OB)
OD = OB × cos φ
2
/cos φ
1
= 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
1
= 0·7 sin φ
1
= 0·714
cos φ
2
= 0·85 sin φ
2
= 0·527
Leading kVAR taken by p.f. correction equipment is
ED = CD CE = OD sin φ
1
OE sin φ
2
= 1·2143 × OB sin φ
1
OB sin φ
2
= 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
1
OE sin φ
2
= 1·237 × OB × sin φ
1
OB sin φ
2
= 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
0
1038
3
⋅×
= 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 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-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-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 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 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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