Improvement of Load Power Factor Using FACTS …shodhganga.inflibnet.ac.in/bitstream/10603/12223/11/8-chapter 3.pdfChapter 3 Improvement of Load Power Factor Using FACTS Controllers

Chapter 3 Improvement of Load Power Factor Using FACTS Controllers

Ph.D Thesis submitted to Jawaharlal Nehru Technological University Anantapur

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CHAPTER 3

IMPROVEMENT OF LOAD POWER FACTOR USING FACTS

CONTROLLERS

3.1 INTRODUCTION

The low power factor effects on transmission line, switchgear, transformers etc. It is observed

that if the power plant works on low power factor, the capital cost of plant in generation,

transmission and distribution systems is increased. Higher the capital charges means, higher

the annual fixed charges, which will increase the cost per unit or the effect of all headed over

the consumer who has to pay more. Thus it is always an advantage for both the consumers and

the suppliers to work at higher power factors. Usually the suppliers encourage the people to

work at improved power factors by adopting a two part tariff, charging the consumer on his

maximum demand in kVA and the number of units consumed by him. The maximum demand

of the consumer is measured with a maximum demand meter, installed at consumer‟s premises;

the reading of meter is taken annually. If the consumers will try to work at low power factor,

for the same power from the mains, he will draw more current or his kVA demand is increased

for which he has to pay extra. This is how the consumer is discouraged to have low power

factors. The power factor is the ratio of active power component (kW) & apparent power

component (kVA) of any A.C. system.

Fig 3.1 Power Triangle

B

kVAr kVA

kW o



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Referring Fig 3.1 Power triangle OAB, OA-represents active or real component of power

(kW).

OB- represents reactive or wattless components of power (kVAr) & AB-represents the

apparent power component (kVA). Basically power factor is defined as the cosine of power

triangle which must be right angled triangle. Therefore: from Fig 3.1

kW = kVA cos Φ (3.1)

kVAr= kVA.sin Φ (3.2)

kVAr = kW tan Φ (3.3)

kVA= (kW) (3.4)

cosΦ

The power factor will be of leading nature, if the current is leading the voltage, it will be of

lagging nature, if current is lagging the voltage and unity if current and voltage are in phase

with each other.[28]

3.2 FACTORS AFFECTING LOW POWER FACTOR

a) As most of the electrical applications are with induction motors, they work, on lagging

power factor of the power supplied.

b) The transformers at power stations, sub-station etc draw the magnetizing current which

causes the total current of the line to be lagging the line voltage

c) The industrial heating furnaces particularly induction heating will have very low

lagging power factor consuming system.

d) Arc lamps, fluorescent lamps, mercury vapour lamps etc. operate at lagging power

factor.

e) Transmission and distribution lines & feeders will also have more inductive effect,

hence main power flow through both the systems will be at low power factor.



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3.3 EFFECT OF LOW POWER FACTOR ON POWER FLOW & POWER

CONSUMPTION

Let us consider an alternator delivering 1000 A at 500 V. (single phase).

Then its Rating = VI/1000 = 1000x500/ 100 = 500kVA (3.5)

If the alternator is loaded at unity power factor, then the load supplied:

kVA x p.f = 500 x 1= 500 kW (3.6)

If the load p.f is 0.6 lagging, then the power supplied by alternator is:

Load supplied = kVA x p.f = 500 x 0.6 = 300 kW At 0.6 p.f lagging (3.7)

By the above computation, it is observed that alternator is developing its maximum current and

voltage even at 0.6 p.f lagging and supplying only 60% of its total capacity therefore in order

to supply its actual power say (500 kW) the alternator is to be over leaded and the conductors

connected between alternator and load must be provided with maximum cross-sectional area to

withstand the overload current. Hence for the given amount of power generation or

transmission, the size of alternator is bigger and large conductors are to be used for the power

transmission. In other words, greater will be the cost of generation and transmission. That is

the reason that the suppliers always stress the consumers to increase the power factor with their

utilities[28].

3.4 EFFECTS OF LOW POWER FACTOR ON TRANSMISSION LINES

Effect on generators:

The generated kVA and kW capacities will have low power factor. Because of this, the power

supplied by the exciters is increased, copper losses in the generator winding are increased, and

so, the efficiency of generator is decreased.



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Effect on transmission lines :

For the transmission of power, more current has to be sent at low power factor. As the line has

to carry more current, its cross sectional area should have to be increased, which increases the

capital cost of the transmission line. Also increased current increases the line losses and

decreases the line efficiency. The line drop is also increased.

Effect on transformers:

The transformers which are connected with transmission lines and with distribution feeders

will have the effect of decrease in kW capacity with the decrease in power factor also increase

in line voltage

Effect on switchgear and bus bar

The cross-sectional area of bus bars, and the contact bars enlarged for the same amount of

power to be delivered at low power factors.

Effects on prime movers

The generator will develop more reactive (kVAr) or wattles power with low power factor, but

certain amount of energy is needed to develop this power, which is being supplied by the prime

mover. This energy supplied by the prime mover is idle and represents as dead investment.

Working on low power factor decreases the efficiency of prime mover. [29]

3.5 BENEFITS OF POWER FACTOR IMPROVEMENT

The following are the benefits, listed with improved power factor.

i) The kW capacity of transformers and the lines is increased

ii) The efficiency of generating plant is increased

iii) The overall consuming cost per unit is decreased

iv) The regulation of transmission lines and distribution feeder is improved



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v) The kW capacity of alternator is increased.

vi) The energy of the prime mover is better utilized.

3.6 METHOD OF IMPROVING POWER FACTOR

By the use of static capacitors

The static capacitors are connected in parallel with the supply mains, which draw the current,

leading the voltage by 900

and neutralize the reactive lagging current component of the load

current, hence to improve the power factor nearest to unity.

By the help of synchronous condenser

The synchronous condenser is also called as synchronous motor. This motor draws the current

from the mains at leading power factor, there by neutralizing lagging reactive component of

the load current. It also develops the mechanical power

Phase advancer

These are the special commutator machines, which improve the power factor of induction

motors.

FACTS controller

These are the powerful power electronics devices which are used to improve the power factor

of bulk or small quantity of power transmission and distribution lines [30]

Reactive Current for the improvement of p.f

Considering an a.c. circuit with inductive load as shown in Fig. 3.2 and its vector diagram as in

Fig 3.3

I

Ic c

Ir (Iµ)

r

L230V150 HZ

a.c. supply

K

Load

Fig 3.2 Circuit diagrams with inductive load



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Fig 3.3 vector diagram to represent current components

Let the current supplied to the circuit is „I‟ Amp. and let this current is lagging the voltage by

an angle of Φ0.

The current supplied to this circuit can be resolved into two components one

along the voltage vector and the other is in quadrature. The component along the voltage vector

is known as in phase vector or active component of current which is “I.cos Φ” and the other is

known as reactive component of current which is “-I sin Φ”. They are shown in Fig 3.3

In order to improve the power factor, angle „Φ‟should be decreased to zero for unity power

factor as cos Φ= cos.0 =1

In order to decrease the angle „Φ‟ the reactive component of current, I sin Φ is to be decreased.

This is obtained by introducing leading current, Ic through capacitor of magnitude equal to the

reactive component (I) in the circuit as in Fig 3.2 observed in Fig 3.3. This leading current, Ic

will lead the voltage exactly by 900 and will be in phase opposition to I the reactive inductive

component of current. Now Ic= +I sin Φ & I = -I sin Φ will neutralize each other leaving

supply current (I) in phase with V. Theory gives the power factor is unity. Thus the required

leading reactive current to compensate the existing lagging reactive current is given as

Ic = I [1-(p.f)2] [ 31] (3.8)

Active components

Reactive components

A

V

I

IsinΦ

IcosΦ



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3.7 POWER FACTOR IMPROVEMENT

By the method of introducing leading components of current in common component of current

is commonly employed. It is obtained by connecting a number of capacitors in parallel across

the main supply at the end. The value of the total capacitance required for improving the

power factor nearest to unity for a given power P, in the network at a frequency f and voltage

V is determined as follows

P= VI (p.f) (3.9)

I = P (3.10)

V(p.f)

C = P .tan Φ (3.11)

2v2

It is observed from equation (3.11) that the capacitance required for improving the power

factor is inversely proportional to frequency „f‟ this shows that the static capacitors are best

suited for high frequencies also it is seen that the capacitance required is inversely proportional

to the square of the operating voltage. Thus the total value of the capacitance required per

phase in three phase system depends upon the nature of connections, whether star connected or

delta connected. In practice it is observed that the delta connection is preferable. A calculation

of capacitance of a capacitor to improve the power factor from 0.73 lagging to 0.93 lagging is

done.

Fig 3.4 Delta connected static capacitors bank

Capacitor bank delta connected

3-ph supply

3-ph I.M

Star connected



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Motor & elements parameters

Out put of I.M =400 hp

Voltage (L-L) = 2000V

Frequency of supply = 50 Hz

Working p.f = 0.73 lagging

Improved p.f = 0.93 lagging

Capacitance of each capacitor = 57.47 µF

Total number of capacitor in series = 4

Each capacitor voltage = 500 V

Frequency of supply = 50 Hz.

Efficiency of motor = 85%

Motor line current with working p.f = IL1 = 142.8A (3.12)

Ip1 = 99.96-J 101.98 (3.13)

Motor line current with required improved p.f = 0.93

IL2 = 107.43 A (3.14)

Phase current (Ip2) =Ip2 = 99.91- J39.48 (3.15)

Reactive current to be neutralized = Iµ = Iµ1- Iµ2 = 62.56A (3.16)

The bank of capacitors used to improve the power factor is connected in delta, therefore phase

voltage (Vph) = 2000V, but each unit of capacitors connected in series is as shown in Fig

3.4.The reactive current in each phase of the bank is

Iµp = 62.56 = 36.12 A (3.17)

3

Let Xc be the capacitive reactance of each capacitor

Xc = 500 = 13.846 (3.18)

36.12



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C = 229.95µF (3.19)

Thus the capacitance of each capacitor of the bank is

[C =229.95 = 57.48µF] (3.20)

4

Hence the capacitor bank of each capacitor must have 57.48µF capacitance to improve the p.f

from 0.73 lagging to 0.93 lagging. [32]

3.7.1 Improvement of p.f by FACTS Controller approach

Fig 3.5 FACTS Controller Circuit Diagram

Supply current (Is) = kVA = 98.1 x 1000 = 84.93A (3.21)

Vp 1155

Thus Rating of rectifier = Vp Ip/1000=S = 98.1 kVA. (3.22)

Diode or Inductor current (Id) =Rectifier Rating = Id = 62.89A (3.23)

1.35x Vp

Peak current (Ipk) = Form factor x Id = Ipk = 69.81A (3.24)

Voltage across diode & inductor (Vd) = Vph = Vd = 816.71V (3.25)

2

Inductance of inductor (Ld) = Vd = 58.5mH (3.26)

4fIpk

Switching rating = rectifier rating = 98.1 kVA. (3.27)

Capacitor rating Data In2

= (38.11) 2

, ωn = 2π50 = 314.16 rad/sec (3.28)

Where In = 1.55x kVAr/Vd = 38.11A

Single phase diode

bridge rectifier

Booster converter

98.1 kVA 20.10 kVAr

62.89

Vph=115v

D

C S

A

B N

ph 98.1 kVA

58.5mH

229.95 F

Ld

Id Is



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C = 229.95 µF (3.29)

Therefore capacitor rating = In2 = Q = 20.10 kVAr (3.30)

2fc

Fig 3.6 Power Triangle for FACTS

sin Ф = kVar = 20.10 = 0.20 (3.31)

kVr 98.10

Ф = sin-1

(0.20) = 11.530

(3.32)

p.f = cos Ф = cos (11.53

0) = 0.98 [33] (3.33)

Q=20.1 kVAr S=98.1 kVA

B

P A o



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3.8 SIMULATION RESULTS

Fig 3.7 Simulation diagram for improvement of Power Factor

Fig 3.8 Circuit Diagram with Inductive load

current 1

model Voltage

measure-

ment

current

current 2

Active &

Reactive power

PQ

V

I

continuous

power

magnitude

signal

angle

V_fundamental

I_fundamental

magnitude

signal

angle

k- cos

sin

1

2

X

X

X

Deg->Rad

1

PQ Gain

I/2

Display

0.6283

V

I

X



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Fig 3.9 Input current waveform for Inductive Load

(Time on x-axis, Current on y-axis)

Fig 3.10 Real and Reactive power waveforms

(Yellow colour line = Q, Purple colour line=P)

(Time on x-axis, P-Q on y-axis)

Magnitude

Signal angle

V_fundamental

V 1

2 I

I_fundamental

Signal

Magnitude

angle

X

+

- K- cos

sin

X

X

Deg->Rad

Display 1

Display

0.7333

Gain

1

PQ

I/2

-0.6799



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Fig 3.11 Simulated diagram of Delta connected static capacitors bank

Fig 3.12 Waveform of Improved active & reactive power quality response



Fig 3.13 Simulated diagram of FACTS (UPFC)-Single phase rectifier

100mH

D1

D2

D3

D4

D5

D6

120V

60Hz

Vo

200µF

Scope1

Scope2 Active &

Reactive

power

Vd1 Vd2

PQ V

I

D2&D4;

Circuit

T

100VA

120V/24V

Max

Current

measurement



53

Fig 3.14 Simulated diagram of FACTS Controller Approach (RP/AP= tan Ф)

X

1/sqrt(3)

Display1

PQ 1

[1.1 4e+007]

X

1

2

Vabc

Iabc

+

-

+

-

+

-

P

Display

1.16 4e+007

The Fig 3.14 computes the 3-ph instantaneous real power and reactive power using

the following equations:

1) P=Va1la+Vb

1lb+Vc

1lc

2) Q=1/sqrt(3) (Vbc1la+Vca

1lb+Vab

1lc)

Note: Equation-2 is valid only for a balanced & harmonics-free system.



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Fig 3.15 Real & Reactive power waveform of UPFC with Max. p.f



Fig 3.16 Output waveform of diode rectifier voltage with FACTS controllers

(Time on x-axis, Vdc on y-axis)

3.9 CONCLUSION

In this work the power factor is computed by static capacitors approach and FACTS controllers

approach theoretically & verified the results by simulation. The simulation results are noted. In

comparison, it is found that the FACTS controller approach is better than the static capacitor

approach for power factor correction. Hence the FACTS controllers methods can be

implemented for the power factor improvement in both transmission & distribution lines

effectively. The comparative evaluated results of power factor are tabulated in Table 3.1

Table 3.1 Evaluation of Power factor



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S.No Name of the circuit Power factor

1 Inductive load 0.73

2 Induction motor with C- bank 0.93

3 Using FACTS (UPFC controller) 0.98

Improvement of Load Power Factor Using FACTS …shodhganga.inflibnet.ac.in/bitstream/10603/12223/11/8-chapter 3.pdfChapter 3 Improvement of Load Power Factor Using FACTS Controllers

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