NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power...

195
NPTEL Course on Power Quality in Power Distribution Systems Dr. Mahesh Kumar Professor, Department of Electrical Engineering Indian Institute of Technology Madras Chennai - 600 036

Transcript of NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power...

Page 1: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

NPTEL Courseon

Power Quality in Power Distribution Systems

Dr. Mahesh KumarProfessor, Department of Electrical Engineering

Indian Institute of Technology MadrasChennai - 600 036

Page 2: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution
Page 3: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Contents

1 SINGLE PHASE CIRCUITS: POWER DEFINITIONS AND ITS COMPONENTS

(Lectures 1-8) 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Power Terms in a Single Phase System . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Sinusoidal Voltage Source Supplying Non-linear Load Current . . . . . . . . . . . 8

1.4 Non-sinusoidal Voltage Source Supplying Non-linear Loads . . . . . . . . . . . . 12

1.4.1 Active Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4.2 Reactive Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.4.3 Apparent Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.4.4 Non Active Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.4.5 Distortion Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.4.6 Fundamental Power Factor . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.4.7 Power Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 THREE PHASE CIRCUITS: POWER DEFINITIONS AND VARIOUS COMPO-

NENTS

(Lectures 9-18) 27

2.1 Three-phase Sinusoidal Balanced System . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Balanced Three-phase Circuits . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.2 Three Phase Instantaneous Active Power . . . . . . . . . . . . . . . . . . 28

i

Page 4: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2.1.3 Three Phase Instantatneous Reactive Power . . . . . . . . . . . . . . . . . 29

2.1.4 Power Invariance in abc and αβ0 Coordinates . . . . . . . . . . . . . . . . 31

2.2 Instantaneous Active and Reactive Powers for Three-phase Circuits . . . . . . . . 33

2.2.1 Three-Phase Balance System . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.2.2 Three-Phase Unbalance System . . . . . . . . . . . . . . . . . . . . . . . 35

2.3 Symmetrical components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.3.1 Effective Apparent Power . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.3.2 Positive Sequence Powers and Unbalance Power . . . . . . . . . . . . . . 51

2.4 Three-phase Non-sinusoidal Balanced System . . . . . . . . . . . . . . . . . . . . 51

2.4.1 Neutral Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.4.2 Line to Line Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.4.3 Apparent Power with Budeanu Resolution: Balanced Distortion Case . . . 54

2.4.4 Effective Apparent Power for Balanced Non-sinusoidal System . . . . . . 54

2.5 Unbalanced and Non-sinusoidal Three-phase System . . . . . . . . . . . . . . . . 55

2.5.1 Arithmetic and Vector Apparent Power with Budeanu’s Resolution . . . . . 58

2.5.2 Effective Apparent Power . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3 FUNDAMENTAL THEORY OF LOAD COMPENSATION

(Lectures 19-24) 69

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.2 Fundamental Theory of Load Compensation . . . . . . . . . . . . . . . . . . . . . 70

3.2.1 Power Factor and its Correction . . . . . . . . . . . . . . . . . . . . . . . 70

3.2.2 Voltage Regulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.2.3 An Approximation Expression for the Voltage Regulation . . . . . . . . . 75

3.3 Some Practical Aspects of Compensator used as Voltage Regulator . . . . . . . . . 81

3.4 Phase Balancing and Power Factor Correction of Unbalanced Loads . . . . . . . . 85

3.4.1 Three-phase Unbalanced Loads . . . . . . . . . . . . . . . . . . . . . . . 85

ii

Page 5: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

3.4.2 Representation of Three-phase Delta Connected Unbalanced Load . . . . . 88

3.4.3 An Alternate Approach to Determine Phase Currents and Powers . . . . . 90

3.4.4 An Example of Balancing an Unbalanced Delta Connected Load . . . . . . 92

3.5 A Generalized Approach for Load Compensation using Symmetrical Components . 95

3.5.1 Sampling Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

3.5.2 Averaging Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

3.6 Compensator Admittance Represented as Positive and Negative Sequence Admit-

tance Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

4 CONTROL THEORIES FOR LOAD COMPENSATION

(Lectures 25-35) 119

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.1.1 State Space Modeling of the Compensator . . . . . . . . . . . . . . . . . . 126

4.1.2 Switching Control of the VSI . . . . . . . . . . . . . . . . . . . . . . . . 128

4.1.3 Generation of Ploss to maintain dc capacitor voltage . . . . . . . . . . . . . 129

4.1.4 Computation of load average power (Plavg) . . . . . . . . . . . . . . . . . 130

4.2 Some Misconception in Reactive Power Theory . . . . . . . . . . . . . . . . . . . 131

4.3 Theory of Instantaneous Symmetrical Components . . . . . . . . . . . . . . . . . 156

4.3.1 Compensating Star Connected Load . . . . . . . . . . . . . . . . . . . . . 156

4.3.2 Compensating Delta Connected Load . . . . . . . . . . . . . . . . . . . . 162

5 SERIES COMPENSATION: VOLTAGE COMPENSATION USING DVR

(Lectures 36-44) 169

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

5.2 Conventional Methods to Regulate Voltage . . . . . . . . . . . . . . . . . . . . . . 169

5.3 Dynamic Voltage Restorer (DVR) . . . . . . . . . . . . . . . . . . . . . . . . . . 170

5.4 Operating Principle of DVR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

5.4.1 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

iii

Page 6: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

5.5 Mathematical Description to Compute DVR Voltage . . . . . . . . . . . . . . . . 177

5.6 Transient Operation of the DVR . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

5.6.1 Operation of the DVR With Unbalance and Harmonics . . . . . . . . . . . 182

5.7 Realization of DVR voltage using Voltage Source Inverter . . . . . . . . . . . . . 182

5.8 Maximum Compensation Capacity of the DVR Without Real Power Support from

the DC Link . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

iv

Page 7: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Chapter 1

SINGLE PHASE CIRCUITS: POWERDEFINITIONS AND ITS COMPONENTS(Lectures 1-8)

1.1 Introduction

The definitions of power and its various components are very important to understand quantitativeand qualitative power quality aspects in power system [1]–[5]. This is not only necessary from thepoint of view of conceptual clarity but also very much required for practical applications such asmetering, quantification of active, reactive power, power factor and other power quality parametersin power system. These aspects become more important when power system is not ideal i.e. itdeals with unbalance, harmonics, faults and fluctuations in frequency. We therefore, in this chapterexplore the concept and fundamentals of single phase system with some practical applications andillustrations.

1.2 Power Terms in a Single Phase System

Let us consider a single-phase system with sinusoidal system voltage supplying a linear load asshown in Fig. 1.1. The voltage and current are expressed as below.

v(t) =√

2V sinωt

i(t) =√

2I sin(ωt− φ) (1.1)

The instantaneous power can be computed as,

p(t) = v(t) i(t) = V I [2 sinωt sin(ωt− φ)]

= V I[cosφ− cos(2ωt− φ)]

= V I cosφ(1− cos 2ωt)− V I sinφ sin 2ωt (1.2)= P (1− cos 2ωt)−Q sin 2ωt

= pactive(t)− preactive(t)

1

Page 8: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

v i

Fig. 1.1 A single phase system

Here, P = 1T

∫ T+t1t=t1

p(t) dt = average value of pactive(t). This is called as average active power.The reactive power Q is defined as,

Q∆= max preactive(t) (1.3)

It should be noted that the way Q is defined is different from P . The Q is defined as maximumvalue of the second term of (1.2) and not an average value of the second term. This differenceshould always kept in mind.Equation (1.2) shows that instantaneous power can be decomposed into two parts. The first termhas an average value of V I cosφ and an alternating component of V I cos 2ωt, oscillating at twicethe line frequency. This part is never negative and therefore is called unidirectional or dc power.The second term has an alternating component V I sinφ sin 2ωt oscillating at twice frequency witha peak vale of V I sinφ. The second term has zero average value. The equation (1.2) can furtherbe written in the following form.

p(t) = V I cosφ− V I cos(2ωt− φ)

= p(t) + p(t)

= paverage + poscillation

= puseful + pnonuseful (1.4)

With the above definitions of P andQ, the instantaneous power p(t) can be re-written as following.

p(t) = P (1− cos 2ωt)−Q sin 2ωt (1.5)

2

Page 9: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Example 1.1 Consider a sinusoidal supply voltage v(t) =√

2× 230 sinωt supplying a linear loadof impedance ZL = 12 + j13 Ω at ω = 2πf radian per second, f = 50 Hz. Express current i(t) asa function of time. Based on v(t) and i(t) determine the following.

1. Instantaneous power p(t), instantaneous active power pactive(t) and instantaneous reactivepower preactive(t)

2. Compute average real power P , reactive power Q, apparent power S, and power factor pf .

3. Repeat the above when load is ZL = 12− j13 Ω, ZL = 12 Ω, and ZL = j13 Ω

4. Comment upon the results.

Solution: A single phase circuit supplying linear load is shown Fig. 1.1. In general, the current inthe circuit is given as,

i(t) =√

2I sin(ωt− φ)

where φ = tan−1(XL/RL), and I = (V/|ZL|)

Case 1: When load is inductive, ZL = 12 + j13 Ω

|ZL| =√R2L +X2

L =√

122 + 132 = 17.692 Ω, and I = 230/17.692 = 13 Aφ = tan−1(X/R) = tan−1(13/12) = 47.29o

Therefore we have,

v(t) =√

2× 230 sinωt

i(t) =√

2× 13 sin(ωt− 47.29o)

The instantaneous power is given as,

p(t) = V I cosφ(1− cos 2ωt)− V I sinφ sin 2ωt

= 230× 13 cos 47.29o(1− cos(2× 314t))− 230× 13 sin 47.29o sin(2× 314t)

= 2028.23(1− cos(2× 314t))− 2196.9 sin(2× 314t)

= pactive(t)− preactive(t)

The above implies that,

pactive(t) = 2028.23(1− cos(2× 314t))

preactive(t) = 2196.9 sin(2× 314t)

3

Page 10: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Average real power (P ) is given as,

P =1

T

∫ T

0

p(t) dt

P = V I cosφ = 230× 13× cos 47.29o = 2028.23 W

Reactive power (Q) is given as maximum value of preactive, and equals to V I sinφ as given below.

Q = V I sinφ = 230× 13× sin 47.2906 = 2196.9 VAr

Apparant power,S = V I =√P 2 +Q2 = 230× 13 = 2990 VA

Power factor =P

S=

2028.23

2990= 0.6783

For this case, the voltage, current and various components of the power are shown in Fig. 1.2. Asseen from the figure the current lags the voltage due to inductive load. The pactive has an offset of2028.23 W, which is also indicated as P in the right bottom graph. The preactive has zero averagevalue and its maximum value is equal to Q, which is 2196.9 VArs.

0 0.01 0.02 0.03-400

-200

0

200

400

sec

0 0.01 0.02 0.03-20

-10

0

10

20

sec

0 0.01 0.02 0.03-4000

-2000

0

2000

4000

6000

sec

VA

, W, V

Ar

0 0.01 0.02 0.032000

2100

2200

2300

2400

sec

W, V

Ar

Voltge (V)

Current (A)Average Power (W)

Reactive Power (VAr)

p(t)

pact

(t)

preact

(t)

Fig. 1.2 Case 1: Voltage, current and various power components

4

Page 11: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Case 2: When load is Capacitive, ZL = 12 − j13 that implies |ZL| =√

122 + 132 = 17.692Ω,and I = 230/17.692 = 13 A, φ = tan−1(−13/12) = −47.2906o.

v(t) =√

2× 230 sinωt

i(t) =√

2× 13 sin(ωt+ 47.2906o)

p(t) = V I cosφ(1− cos 2ωt)− V I sinφ sin 2ωt

= 230× 13 cos(−47.2906o)(1− cos(2× 314t))− 230× 13 sin(−47.2906o) sin(2× 314t)

= 2028.23(1− cos(2× 314t)) + 2196.9 sin(2× 314t)

pactive(t) = 2028.23(1− cos(2× 314t))

preactive(t) = −2196.9 sin(2× 314t)

P = V I cosφ = 230× 13× cos 47.2906o = 2028.23 WattQ = V I sinφ = 230× 13× sin(−47.2906o) = −2196.9 VArS = V I =

√P 2 +Q2 = 230× 13 = 2990 VA

For Case 2, the voltage, current and various components of the power are shown in Fig. 1.3. Theexplanation given earlier also holds true for this case.

0 0.01 0.02 0.03-400

-200

0

200

400

sec

0 0.01 0.02 0.03-20

-10

0

10

20

sec

0 0.01 0.02 0.03-4000

-2000

0

2000

4000

6000

sec

VA

, W, V

Ar

0 0.01 0.02 0.03 0.04-3000

-2000

-1000

0

1000

2000

3000

sec

W, V

Ar

Voltage (V)

Current (A)

Average Power (W)

Reactive Power (VAr)

p(t)

pact

(t)

preact

(t)

Fig. 1.3 Case 2: Voltage, current and various power components

5

Page 12: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Case 3: When load is resistive, ZL = 12 Ω, |ZL| = 12 Ω, I = (230/12) = 19.167 A, and φ = 0o.Therefore, we have

v(t) =√

2× 230 sinωt

i(t) =√

2× 19.167 sinωt

p(t) = 230× 19.167 cos 0o(1− cos(2× 314t))− 230× 19.167 sin 0o sin(2× 314t)

= 4408.33(1− cos(2× 314t))− 0

pactive(t) = 4408.33(1− cos(2× 314t))

preactive(t) = 0

P = V I cosφ = 230× 19.167× cos 0o = 4408.33 WQ = V I sinφ = 230× 19.167× sin 0o = 0 VArS = V I =

√P 2 +Q2 = 230× 19.167 = 4408.33 VA

Power factor =4408.33

4408.33= 1

For Case 3, the voltage, current and various components of the power are shown in Fig. 1.4. Sincethe load is resistive, as seen from the graph preactive is zero and p(t) is equal to pactive. The averagevalue of p(t) is real power (P ), which is equal to 4408.33 W.

0 0.01 0.02 0.03-400

-200

0

200

400

sec

0 0.01 0.02 0.03-30

-20

-10

0

10

20

30

sec

0 0.01 0.02 0.03

0

2000

4000

6000

8000

10000

sec

VA

, W, V

Ar

0 0.01 0.02 0.03-1000

0

1000

2000

3000

4000

5000

sec

W, V

Ar

Average Power (W)

Reactive Power (VAr)

Voltage (V)

Current (A)

p(t)

pact

(t)

preact

(t)

Fig. 1.4 Case 3: Voltage, current and various power components

6

Page 13: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Case 4: When the load is purely reactive, ZL = j13 Ω , |ZL| = 13 Ω, I = 23013

= 17.692 A, andφ = 90o. Therefore, we have

v(t) =√

2× 230 sinωt

i(t) =√

2× 17.692 sin(ωt− 90o)

p(t) = 230× 17.692 cos 90o(1− cos(2× 314t))− 230× 17.692 sin 90o sin(2× 314t)

= 0− 4069 sin(2× 314t)

pactive(t) = 0

preactive(t) = 4069 sin(2× 314t)

P = V I cosφ = 230× 17.692× cos 90o = 0 WQ = V I sinφ = 230× 17.692× sin 90o = 4069 VArS = V I =

√P 2 +Q2 = 230× 17.692 = 4069 VA

Power factor =0

4069= 0

For Case 4, the voltage, current and various components of the power are shown in Fig. 1.5. Theload in this case is purely reactive, hence their is no average component of p(t). The maximumvalue of p(t) is same as preactive(t) or Q, which is equal to 4069 VArs.

0 0.01 0.02 0.03-400

-200

0

200

400

sec

0 0.01 0.02 0.03-30

-20

-10

0

10

20

30

0 0.01 0.02 0.03-5000

0

5000

VA

, W, V

Ar

0 0.01 0.02 0.030

1000

2000

3000

4000

5000

W, V

Ar

Voltage (V)

Current (A)

p(t)

pact

(t)

preact

(t)

Active Power (W)

Reactive Power (VAr)

Fig. 1.5 Case 4: Voltage, current and various power components

7

Page 14: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

1.3 Sinusoidal Voltage Source Supplying Non-linear Load Current

The load current is now considered as nonlinear load such as single-phase rectifier load. Thevoltage and current are expressed as following.

v(t) =√

2V sinωt

i(t) =√

2∞∑n=1

In sin(nωt− φn) (1.6)

The instantaneous power is therefore given by,

p(t) = v(t) i(t) =√

2V sinωt√

2∞∑n=1

In sin(nωt− φn)

= V∞∑n=1

[In 2 sinωt sin(nωt− φn)]

= V [I1 2 sinωt sin(ωt− φ1)]

+ V∞∑n=2

[In 2 sinωt sin(nωt− φn)] (1.7)

Note that 2 sinA sinB = cos(A − B) − cos(A + B), using this, Eqn. (1.7) can be re-written asthe following.

p(t) = V I1 [cosφ1 − cos(2ωt− φ1)]− V I1 sinφ1 sin 2ωt

+ V∞∑n=2

In[ (cosφn − cos(2nωt− φn))− sinφn sin 2nωt]

= V I1 cosφ1(1− cos 2ωt)− V I1 sinφ1 sin 2ωt (1.8)

+∞∑n=2

V In[ cosφn(1− cos 2nωt)− sinφn sin 2nωt]

= A + B

In above equation, average active power P and reactive power Q are given by,

P = P1 = average value of p(t) = V I1 cosφ1

Q = Q1∆= peak value of second term in A = V I1 sinφ1 (1.9)

The apparent power S is given by

S = V I

S = V√

[I21 + I2

2 + I23 + .....] (1.10)

8

Page 15: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Equation (1.10) can be re-arranged as given below.

S2 = V 2 I21 + V 2 [I2

2 + I23 + I2

4 + ...]

= (V I1 cosφ1)2 + (V I1 sinφ1)2 + V 2 [I22 + I2

3 + I24 + .....]

= P 2 +Q2 +H2 (1.11)

In above equation,H is known as harmonic power and represents V As corresponding to harmonicsand is equal to,

H = V√

[I22 + I2

3 + I24 + .....] (1.12)

The following points are observed from description.

1. P and Q are dependent on the fundamental current components

2. H is dependent on the current harmonic components

3. Power components −V I cos 2ωt and V I1 sinφ1 sin 2ωt are oscillating components and canbe eliminated using appropriately chosen capacitors and inductors

4. There are other terms in (1.10), which are functions of multiple integer of fundamental fre-quency are reflected in ’B’ terms of Eqn. (1.8). These terms can be eliminated using tunedLC filters.

This is represented by power tetrahedron instead of power triangle (in case of voltage and currentof sinusidal nature of fundamental frequency). In this context, some important terms are definedhere.

Displacement Factor or Fundamental Power Factor (DPF) is denoted by cosφ1 and is cosineangle between the fundamental voltage and current.

Distortion Factor (DF) is defined as ratio of fundamental apparent power (V1 I1) to the totalapparent power (V I).

Distortion factor (DF ) =

√(P 2

1 +Q21)

S

=

√(V 2 I2

1 cos2 φ1 + V 2 I21 sin2 φ1)

V I

=

√V 2 I2

1

V I

=I1

I= cos γ (1.13)

The Power Factor (pf ) is defined as ratio of average active power to the total apparent power (V I)and is expressed as,

9

Page 16: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Power Factor (pf) =P

S

=V I1 cosφ1

V I

=

(I1

I

)cosφ1

= cos γ cosφ1 (1.14)

The equation (1.14) shows that power factor becomes less by a factor of cos γ. This is due to thepresence of the harmonics in the load current. The nonlinear load current increases the ampererating of the conductor for same amount of active power transfer with increased VA rating. Suchkind of load is not desired in power system.

Example 1.2 Consider following single phase system supplying a rectifier load as given in Fig.1.6. Given a supply voltage, v(t) =

√2× 230 sinωt and source impedance is negligible, draw the

voltage and current waveforms. Express current using Fourier series. Based on that determine thefollowing.

1. Plot instantaneous power p(t).

2. Plot components of p(t) i.e. pactive(t), preactive(t).

3. Compute average real power, reactive power, apparent power, power factor, displacementfactor (or fundamental power factor) and distortion factor.

4. Comment upon the results in terms of VA rating and power output.

Idi ( )t

v ( )t

Fig. 1.6 A single phase system with non-linear load

Solution: The above system has been simulated using MATLAB/SIMULINK. The supply volt-age and current are shown in Fig. 1.7. The current waveform is of the square type and its Fourier

10

Page 17: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

series expansion is given below.

i(t) =∞∑

n=2h+1

4Idcnπ

sin(nωt) where h = 0, 1, 2 . . .

The instantaneous power is therefore given by,

p(t) = v(t) i(t) =√

2V sinωt∞∑

n=2h+1

4Idcnπ

sin(nωt). (1.15)

By expansion of the above equation, the average active power (P ) and reactive power (Q) are givenas below.

P = P1 = average value ofPactive(t) or p(t) = V I1 cosφ1

= V I1 (since,φ1 = 0, cosφ1 = 1, sinφ1 = 0)

Q = Q1∆= peak value ofPreactive(t) = V I1 sinφ1 = 0

1 1.001 1.002 1.003 1.004 1.005 1.006

x 105

-300

-200

-100

0

100

200

300

Time(Sec)

Supply current (A)

Supply voltage (V)

1 1.001 1.002 1.003 1.004 1.005 1.006

x 105

-0.5

0

0.5

1

1.5

2

2.5

3

x 104

Time (Sec)

instantaneous power

Average power (W)Reactive power (Var)

Fig. 1.7 Supply voltage, current and instantaneous power waveforms

The rms value of fundamental and rms value of the total source current are given below.

Irms = Id = 103.5 A

I1 =2√

2

πId = 93.15 A

11

Page 18: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The real power (P ) is given by

P = V I1

= V × 2√

2

πId = 21424.5 W .

The reactive power (Q) is given by

Q = Q1 = 0.

The apparent power (S) is given by

S = V Irms

= V Id = 23805 VA .

The distortion factor (cos γ) is,

DF =

√(P 2

1 +Q21)

S

=I1

Irms= cos γ = 0.9.

The displacement factor (cosφ1) is,

DPF = cosφ1 = 1.

Therefore power factor is given by,

pf =P

S=P1

S1

S1

S= cos γ cosφ1

= DF ×DPF = 0.9 (lag)

1.4 Non-sinusoidal Voltage Source Supplying Non-linear Loads

The voltage source too may have harmonics transmitted from generation or produced due to non-linear loads in presence of feeder impedance. In this case, we shall consider generalized case ofnon-sinusoidal voltage source supplying nonlinear loads including dc components. These voltagesand currents are represented as,

v(t) = Vdc +∞∑n=1

√2Vn sin(nωt− φvn) (1.16)

and

i(t) = Idc +∞∑n=1

√2In sin(nωt− φin) (1.17)

12

Page 19: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Therefore, instantanteous power p(t) is given by,

p(t) = [Vdc +∞∑n=1

√2Vn sin(nωt− φvn)].[Idc +

∞∑n=1

√2In sin(nωt− φin)] (1.18)

p(t) = VdcIdc︸ ︷︷ ︸I

+Vdc

∞∑n=1

√2In sin(nωt− φin)︸ ︷︷ ︸

II

+ Idc

∞∑n=1

√2Vn sin(nωt− φvn)︸ ︷︷ ︸

III

+∞∑n=1

√2Vn sin(nωt− φvn)

∞∑n=1

√2In sin(nωt− φin)︸ ︷︷ ︸

IV

(1.19)

p(t) = pdc−dc + pdc−ac + pac−dc + pac−ac (1.20)

The I term (pdc−dc) contribute to power from dc components of voltage and current. Terms II(pdc−ac) and III (pac−dc) are result of interaction of dc and ac components of voltage and current.In case, there are no dc components all these power components are zero. In practical cases, dccomponents are very less and the first three terms have negligible value compared to IV term. Thus,we shall focus on IV (pac−ac) term which correspond to ac components present in power system.The IV term can be written as,

IVth term = pac−ac =∞∑n=1

√2Vn sin(nωt− φvn)

∞∑h=1

√2Ih sin(hωt− φih) (1.21)

where n = h = 1, 2, 3..., similar frequency terms will interact. When n 6= h, dissimilar

13

Page 20: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

frequency terms will interact. This is expressed below.

pac−ac(t) =√

2V1 sin(ωt− φv1)√

2I1 sin(ωt− φi1)︸ ︷︷ ︸A

+√

2V1 sin(ωt− φv1)∞∑

h=2,h6=1

√2Ih sin(hωt− φih)︸ ︷︷ ︸

B

+√

2V2 sin(2ωt− φv2)√

2I2 sin(2ωt− φi2)︸ ︷︷ ︸A

+√

2V2 sin(2ωt− φv2)∞∑

h=1,h6=2

√2Ih sin(hωt− φih)︸ ︷︷ ︸

B

+ . . .+ . . .

+√

2Vn sin(nωt− φvn)√

2In sin(nωt− φin)︸ ︷︷ ︸A

+√

2Vn sin(nωt− φvn)∞∑

h=1,h6=n

√2Ih sin(hωt− φih)︸ ︷︷ ︸

B

(1.22)

The terms in A of above equation form similar frequency terms and terms in B form dissimilarfrequency terms, we shall denote them by pac−ac−nn and pac−ac−nh. Thus,

pac−ac−nn(t) =∞∑n=1

VnIn2 sin(nωt− φvn) sin(nωt− φin) (1.23)

and

pac−ac−nh(t) =∞∑n=1

√2Vn sin(nωt− φvn)

∞∑h=1,h6=n

√2In sin(hωt− φih) (1.24)

14

Page 21: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Now, let us simplify pac−ac−nn in

pac−ac−nn(t) =∞∑n=1

VnIn[cos(φin − φvn)− cos(2nωt− φin − φvn)]

=∞∑n=1

VnIn[cos(φn)− cos(2nωt− (φin − φvn)− 2φvn)]

=∞∑n=1

VnIn[cos(φn)− cos (2nωt− 2φvn)− φn]

=∞∑n=1

VnIn[cos(φn)− cos(2nωt− 2φvn) cosφn − sin(2nωt− 2φvn) sinφn]

=∞∑n=1

[VnIn cosφn1− cos(2nωt− 2φvn)

−VnIn sinφn sin(2nωt− 2φvn)] (1.25)

where φn = (φin − φvn) = phase angle between nth harmonic current and voltage.

pac−ac−nn(t) =∞∑n=1

[VnIn cosφn1− cos(2nωt− 2φvn)]

−∞∑n=1

[VnIn sinφn sin(2nωt− 2φvn)] (1.26)

Therefore, the instantaneous power is given by,

p(t) = pdc−dc︸ ︷︷ ︸I

+ pdc−ac︸ ︷︷ ︸II

+ pac−dc︸ ︷︷ ︸III

+ pac−ac−nn︸ ︷︷ ︸IVA

+ pac−ac−nh︸ ︷︷ ︸IVB︸ ︷︷ ︸

IV

p(t) = VdcIdc + Vdc

∞∑n=1

√2In sin(nωt− φin) + Idc

∞∑n=1

√2Vn sin(nωt− φvn)

+∞∑n=1

[VnIn cosφn1− cos(2nωt− φvn)]

−∞∑n=1

[VnIn sinφn. sin(2nωt− φin)] (1.27)

1.4.1 Active Power

Instantaneous active power, pactive(t) is expressed as,

pactive(t) = VdcIdc +∞∑n=1

[VnIn cosφn1− cos(2nωt− φvn)] (1.28)

15

Page 22: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

It has non-negative value with some average component, giving average active power. Therefore,

P =1

T

∫ T

0

p(t) dt

= VdcIdc +∞∑n=1

VnIn cosφn. (1.29)

Reactive power can be defined as

q(t) = preactive(t) = −∞∑n=1

[VnIn sinφn sin(2nωt− 2φvn)] (1.30)

resulting in

Q , max of (1.30) magnitude

=∞∑n=1

VnIn sinφn. (1.31)

From (1.29)

P = Pdc +∞∑n=1

VnIn cosφn

= Pdc + V1I1 cosφ1 + V2I2 cosφ2 + V3I3 cosφ3 + . . .

= Pdc + P1 + P2 + P3 + . . .

= Pdc + P1 + PH (1.32)

In above equation,Pdc = Average active power corresponding to the dc componentsP1 = Average fundamental active powerPH = Average Harmonic active power

Average fundamental active power (P1) can also be found from fundamentals of voltage and cur-rent i.e.,

P1 =1

T

∫ T

0

v1(t) i1(t)dt (1.33)

and harmonic active power (PH) can be found

PH =∞∑n=1

VnIn cosφn = P − P1 (1.34)

16

Page 23: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

1.4.2 Reactive Power

The reactive power or Budeanu’s reactive power (Q) can be found by summing maximum value ofeach term in (1.30). This is given below.

Q =∞∑n=1

VnIn sinφn

= V1I1 sinφ1 + V2I2 sinφ2 + V3I3 sinφ3 + . . .

= Q1 +Q2 +Q3 + . . .

= Q1 +QH (1.35)

Usually this reactive power is referred as Budeanu’s reactive power, and sometimes we use sub-script B’ to indicate that i.e.,

QB = Q1B +QHB (1.36)

The remaining dissimilar terms of (1.27) are accounted using prest(t). Therefore, we can write,

p(t) = pdc−dc + pactive(t) + preactive(t)︸ ︷︷ ︸similar frequency terms

+ prest(t)︸ ︷︷ ︸non-similar frequency terms

(1.37)

where,

pdc−dc = VdcIdc

pactive(t) =∞∑n=1

[VnIn cosφn1− cos(2nωt− 2φvn)]

preactive(t) = −∞∑n=1

[VnIn sinφn. sin(2nωt− 2φvn)]

prest(t) = Vdc

∞∑n=1

√2In sin(nωt− φin) + Idc

∞∑n=1

√2Vn sin(nωt− φvn)

+∞∑n=1

√2Vn sin(nωt− φvn)

∞∑m=1,m 6=n

√2Im sin(mωt− φim) (1.38)

1.4.3 Apparent Power

The scalar apparent power which is defined as product of rms value of voltage and current, isexpressed as following.

S = V I

=√V 2dc + V 2

1 + V 22 + · · ·

√I2dc + I2

1 + I22 + · · · (1.39)

=√V 2dc + V 2

1 + V 2H

√I2dc + I2

1 + I2H

17

Page 24: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Where,

V 2H = V 2

2 + V 23 + · · · =

∞∑n=2

V 2n

I2H = I2

2 + I23 + · · · =

∞∑n=2

I2n (1.40)

VH and IH are denoted as harmonic voltage and harmonic current respectively. Expanding (1.39)we can write

S2 = V 2I2

= (V 2dc + V 2

1 + V 2H)(I2

dc + I21 + I2

H)

= V 2dcI

2dc + V 2

dcI21 + V 2

dcI2H + V 2

1 I21 + V 2

1 I2dc + V 2

1 I2H + V 2

HI2dc + V 2

HI21 + V 2

HI2H

= V 2dcI

2dc + V 2

1 I21 + V 2

HI2H + V 2

dc(I21 + I2

H) + I2dc(V

21 + V 2

H) + V 21 I

2H + V 2

HI21

= S2dc + S2

1 + S2H + S2

D

= S21 + S2

dc + S2H + S2

D︸ ︷︷ ︸= S2

1 + S2N (1.41)

In above equation, the term SN is as following.

S2N = V 2

dcI21 + V 2

dcI2H + V 2

1 I2dc + V 2

1 I2H + V 2

HI2dc + V 2

HI21 + V 2

HI2H + I2

dcI2H + I2

dcV2dc (1.42)

Practically in power systems dc components are negligible. Therefore neglecting the contributionof Vdc and Idc associated terms in (1.42), the following is obtained.

S2N = I2

1V2H + V 2

1 I2H + V 2

HI2H

= D2V +D2

I + S2H (1.43)

The terms DI and DV in (1.43) are known as apparent powers due to distortion in current andvoltage respectively. These are given below.

DV = I1 VH

DI = V1 IH (1.44)

These are further expressed in terms of THD components of voltage and current, as given below.

THDV =VHV1

THDI =IHI1

(1.45)

From (1.45), the harmonic components of current and voltage are expressed below.

VH = THDV V1

IH = THDI I1 (1.46)

18

Page 25: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Using (1.44) and (1.46),

DV = V1 I1 THDV = S1 THDV

DI = V1 I1 THDI = S1 THDI

SH = VHIH = S1 THDI THDV (1.47)

Therefore using (1.43) and (1.47), SN could be expressed as following.

S2N = S2

1 (THD2I + THD2

V + THD2I THD

2V ) (1.48)

Normally in power system, THDV << THDI , therefore,

SN ≈ S1DI (1.49)

The above relationship shows that as the THD content in voltage and current increases, the non fun-damental apparent power SN increases for a given useful transmitted power. This means there aremore losses and hence less efficient power network.

1.4.4 Non Active Power

Non active power is denoted by N and is defined as per following equation.

S2 = P 2 +N2 (1.50)

This power includes both fundamental as well as non fundamental components, and is usuallycomputed by knowing active power (P ) and apparent power (S) as given below.

N =√S2 − P 2 (1.51)

1.4.5 Distortion Power

Due to presence of distortion, the total apparent power S can also be written in terms of activepower (P ), reactive power (Q) and distortion power (D)

S2 = P 2 +Q2 +D2. (1.52)

Therefore,

D =√S2 − P 2 −Q2. (1.53)

1.4.6 Fundamental Power Factor

Fundamental power factor is defined as ratio of fundamental real power (P1) to the fundamentalapparent power (S1). This is given below.

pf1 = cosφ1 =P1

S1

(1.54)

The fundamental power factor as defined above is also known as displacement power factor.

19

Page 26: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

1.4.7 Power Factor

Power factor for the single phase system considered above is the ratio of the total real power (P )to the total apparent power (S) as given by the following equation.

pf =P

S

=P1 + PH√S2

1 + S2N

=(1 + PH/P1)√1 + (SN/S1)2

P1

S1

(1.55)

Substituting SN from (1.48), the power factor can further be simplified to the following equation.

pf =(1 + PH/P1)√

1 + THD2I + THD2

V + THD2ITHD

2V

pf1 (1.56)

Thus, we observe that the power factor of a single phase system depends upon fundamental (P1)and harmonic active power (PH), displacement factor(DPF = pf1) and THDs in voltage andcurrent. Further, we note following points.

1. P/S is also called as utilization factor indicator as it indicates the usage of real power.

2. The term SN/S1 is used to decide the overall degree of harmonic content in the system.

3. The flow of fundamental power can be characterized by measurement of S1, P1, pf1, and Q1.

For a practical power system P1 >> PH and THDV << THDI , the above expression of powerfactor is further simplified as given below.

pf =pf1√

1 + THD2I

(1.57)

Example 1.3 Consider the following voltage and current in single phase system.

vs(t) =√

2× 230 sin(ωt) +√

2× 50 sin(3ωt− 30)

i(t) = 2 +√

2× 10 sin(ωt− 30) +√

2× 5 sin(3ωt− 60)

Determine the following.

(a) Active power, (P )

(b) Reactive power, (Q)

(c) Apparent power, (S)

(d) Power factor, (pf)

20

Page 27: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Solution: Here the source is non-sinusoidal and is feeding a non-linear load. The instantaneouspower is given by,

p(t) = v(t) i(t)

p(t) = Vdc +∞∑n=1

√2Vn sin(nωt− φvn) Idc +

∞∑n=1

√2 In sin(nωt− φin)

(a) The active power ’P’ is given by,

P =1

T

∫ T

0

p(t) dt

= Pdc + V1 I1 cosφ1 + V2 I2 cosφ2 + ......+ Vn In cosφn (1.58)= Pdc + P1 + PH

where,

φn = φin − φvnPdc = Vdc Idc

P1 = V1 I1 cosφ1

PH =∞∑n=2

Vn In cosφn

Here, Vdc = 0, V1 = 230 V, φv1 = 0, V3 = 50 V, φv3 = 30, Idc = 2 A, I1 = 10 A, φi1 = 30,I3 = 5 A, φi3 = 60. Therefore, φ1 = φi1 − φv1 = 30 and φ3 = φi3 − φv3 = 30.Substituting these values in (1.58), the above equation gives,

P = 0× 2 + 230× 10× cos 30 + 50× 5× cos 30 = 2208.36 W.

(b). The reactive power (Q) is given by,

Q =∞∑n=1

Vn In sinφn

= V1 I1 sinφ1 + V2 I2 sinφ2 + .....Vn In sinφn

= 230× 10× sin 30 + 50× 5× sin 30 = 1275 VAr.

(c). The Apparent power S is given by,

S = Vrms Irms

=√V 2dc + V 2

1 + V 22 + ....V 2

n

√I2dc + I2

1 + I22 + .....I2

n

=√V 2dc + V 2

1 + V 2H

√I2dc + I2

1 + I2H

where,

VH =√V 2

2 + V 23 + ....V 2

n

IH =√I2

2 + I23 + .....I2

n

21

Page 28: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Substituting the values of voltage and current components, the apparent power S is computed asfollowing.

S =√

0 + 2302 + 502√

22 + 102 + 52

= 235.37× 11.357 = 2673.31 VA

(d). The power factor is given by

pf =P

S=

2208.36

2673.31= 0.8261 lag

Example 1.4 Consider following system with distorted supply voltages,

v(t) = Vdc +∞∑n=1

√2Vnn2

sin(nωt− φvn)

with Vdc = 10V, Vn/n2 = 230

√2/n2 and φvn = 0 for n = 1, 3, 5, 7, . . .

The voltage source supplies a nonlinear current of,

i(t) = Idc +∞∑n=1

√2Inn

sin(nωt− φin).

with Idc = 2A, In = 20/n A and φin = n× 30o for n = 1, 3, 5, 7, . . .

Compute the following.

1. Plot instantaneous power p(t), pactive(t), preactive(t), Pdc, and prest(t).

2. Compute P, P1, PH (= P3 + P5 + P7 + . . .).

3. Compute Q, Q1, QH (= Q3 + Q5 + Q7 + . . .).

4. Compute S, S1, SH , N, D.

5. Comment upon each result.

Solution: Instantaneous power is given as following.

22

Page 29: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

p(t) = v(t) i(t) =

(10 +

∞∑n=1,3,5

230√

2

n2sin(nωt)

)(2 +

∞∑n=1,3,5

20√

2

nsinn(ωt− 300)

)

= 20︸︷︷︸I

+ 10∞∑

n=1,3,5

20√

2

nsinn(ωt− 300)︸ ︷︷ ︸

II

+ 2∞∑

n=1,3,5

230√

2

n2sin(nωt)︸ ︷︷ ︸

III

+

(∞∑

n=1,3,5

230√

2

n2sin(nωt)

)(∞∑

n=1,3,5

20√

2

nsinn(ωt− 300)

)︸ ︷︷ ︸

IV

= 20︸︷︷︸I

+∞∑

n=1,3,5

200√

2

nsinn(ωt− 300)︸ ︷︷ ︸II

+∞∑

n=1,3,5

460√

2

n2sin(nωt)︸ ︷︷ ︸

III

+∞∑

n=1,3,5

4600

n3(cos(30o n)(1− cos 2nωt)− sin (2nωt) sin(30o n))︸ ︷︷ ︸

IV A

+

(∞∑

n=1,3,5

230√

2

n2sinnωt

)(∞∑

h=1,3,5;h6=n

20√

2

hsinh(ωt− 300)

)︸ ︷︷ ︸

IV B

1. Computation of p(t), pactive(t), preactive(t), Pdc., and prest(t)

pdc−dc(t) = 20 W

pactive(t) =∞∑

n=1,3,5

4600

n3cosn300(1− cos 2nωt)

preactive(t) = −∞∑

n=1,3,5

4600

n3sin(n300) sin(2nωt)

prest(t) =∞∑

n=1,3,5

200√

2

nsinn(ωt− 300) +

∞∑n=1,3,5

460√

2

n2sin(nωt)

+

(∞∑

n=1,3,5

230√

2

n2sinnωt

)(∞∑

h=1,3,5;h6=n

20√

2

hsinh(ωt− 300)

)

23

Page 30: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2. Computation of P, P1, PH

P =1

T

∫ T

0

p(t)dt

= 20 +∞∑

n=1,3,5

4600

n3cos(30o n)

= 20 + 4600 cos 300 +∞∑

n=3,5,7...

4600

n3cos(30o n)

= 20 + 3983.71 + (−43.4841)

= Pdc + P1 + PH

Thus,Active power contributed by dc components of voltage and current, Pdc = 20 W.

Active power contributed by fundamental frequency components of voltage and current, P1 =3983.71 W.

Active power contributed by harmonic frequency components of voltage and current, PH = −43.4841W.

3. Computation of Q, Q1, QH

Q =∞∑

n=1,3,5

4600

n3sin(30o n)

= 4600 sin 300 +∞∑

n=3,5,7...

4600

n3sin(30o n)

= 2300 + 175.7548 VArs= Q1 +QH

The above implies that, Q1 = 4600 VArs and QH =∑∞

n=3,5,7...(4600/n3) sin(30o n) = 175.7548VArs.

4. Computation of Apparent Powers and Distortion Powers

24

Page 31: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The apparent power S is expressed as following.

Vrms =√V 2dc + V 2

1 + V 23 + V 2

5 + V 27 + V 2

9 + ....

=√

102 + 2302 + (230/32)2 + (230/52)2 + (230/72)2 + (230/92)2 + ....

= 231.87 V (up to n = 9)

Irms =√I2dc + I2

1 + I23 + I2

5 + I27 + I2

9 + ....

=√

22 + 202 + (20/3)2 + (20/5)2 + (20/7)2 + (20/9)2 + ....

= 21.85 A (up to n = 9)

The apparent power, S = Vrms Irms = 231.87× 21.85 = 5066.36 VA.Fundamental apparent power, S1 = V1 x I1 = 4600 VA.Apparent power contributed by harmonics SH = VH × IH

VH =√V 2

3 + V 25 + V 2

7 + V 29 + ....

=√

(230/32)2 + (230/52)2 + (230/72)2 + (230/92)2 + ....

= 27.7 V (up to n = 9)

IH =√I2

3 + I25 + I2

7 + I29 + ....

=√

(20/3)2 + (20/5)2 + (20/7)2 + (20/9)2 + ....

= 8.57 A (up to n = 9).

Therefore the harmonic apparent power, SH = VH × IH = 237.5 VA.

Non active power, N =√S2 − P 2 =

√50672 − 3960.22 = 3160.8 VArs (up to n=9)

Distortion Power D =√S2 − P 2 − Q2 =

√50672 − 3960.22 − 2475.772 = 1965.163

VArs (up to n=9).

Displacement power factor (cosφ1)

cosφ1 =P1

S1

=3983.7

(230)(20)= 0.866 lagging

Power factor (cosφ)

cosφ =P

S=

3960.217

5067= 0.781 lagging

The voltage, current, various powers and power factor are plotted in the Fig. 1.8, verifyingabove values.

25

Page 32: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

-200

-100

0

100

200

time (sec)

Vol

tage

(v)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

-20

-10

0

10

20

time (sec)

Cur

rent

(A

)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

-2000

0

2000

4000

time (sec)

Inst

. Pow

er

(W)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.040

1000

2000

3000

4000

time (sec)

Inst

. ac

tive

pow

er (

W)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

-1000

-500

0

500

1000

time (sec)

Inst

. rea

ctiv

e p

ow

er

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-2000

0

2000

time (sec)R

est o

f in

st. p

ower

(W

)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.041000

1500

2000

2500

3000

time (sec)0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

1165

1166

1167

1168

time (sec)

No

n ac

tive

pow

er (

VA

)

Avg. active power (W)

Avg. reactive power (VAr)

Total apparent power (VA)

Distortion power (W)

Fig. 1.8 Various powers

References

[1] IEEE Group, “IEEE trial-use standard definitions for the measurement of electric power quan-tities under sinusoidal, nonsinusoidal, balanced, or unbalanced conditions,” 2000.

[2] E. Watanabe, R. Stephan, and M. Aredes, “New concepts of instantaneous active and reactivepowers in electrical systems with generic loads,” IEEE Transactions on Power Delivery, vol. 8,no. 2, pp. 697–703, 1993.

[3] T. Furuhashi, S. Okuma, and Y. Uchikawa, “A study on the theory of instantaneous reactivepower,” IEEE Transactions on Industrial Electronics, vol. 37, no. 1, pp. 86–90, 1990.

[4] A. Ferrero and G. Superti-Furga, “A new approach to the definition of power components inthree-phase systems under nonsinusoidal conditions,” IEEE Transactions on Instrumentationand Measurement, vol. 40, no. 3, pp. 568–577, 1991.

[5] J. Willems, “A new interpretation of the akagi-nabae power components for nonsinusoidalthree-phase situations,” IEEE Transactions on Instrumentation and Measurement, vol. 41,no. 4, pp. 523–527, 1992.

26

Page 33: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Chapter 2

THREE PHASE CIRCUITS: POWERDEFINITIONS AND VARIOUSCOMPONENTS(Lectures 9-18)

2.1 Three-phase Sinusoidal Balanced System

Usage of three-phase voltage supply is very common for generation, transmission and distributionof bulk electrical power. Almost all industrial loads are supplied by three-phase power supply forits advantages over single phase systems such as cost and efficiency for same amount of power us-age. In principle, any number of phases can be used in polyphase electric system, but three-phasesystem is simpler and giving all advantages of polyphase system is preferred. In previous section,we have seen that instantaneous active power has a constant term V Icosφ as well pulsating termV I cos(2ωt− φ). The pulsating term does not contribute to any real power and thus increases theVA rating of the system.

In the following section, we shall study the various three-phase circuits such as balanced, un-balanced, balanced and unbalanced harmonics and discuss their properties in details [1]–[5].

2.1.1 Balanced Three-phase Circuits

A balanced three-phase system is shown in Fig. 2.1 below.

Three-phase balanced system is expressed using following voltages and currents.

va(t) =√

2V sin(ωt)

vb(t) =√

2V sin(ωt− 120) (2.1)

vc(t) =√

2V sin(ωt+ 120)

27

Page 34: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

a

b

c

a

b

c

Fig. 2.1 A three-phase balanced circuit

and

ia(t) =√

2I sin(ωt)

ib(t) =√

2I sin(ωt− 120) (2.2)

ic(t) =√

2I sin(ωt+ 120)

In (2.1) and (2.2) subscripts a, b and c are used to denote three phases which are balanced. Balancedthree-phase means that the magnitude (V ) is same for all three phases and they have a phase shiftof −120o and 120o. The balanced three phase system has certain interesting properties. These willbe discussed in the following section.

2.1.2 Three Phase Instantaneous Active Power

Three phase instantaneous active power in three phase system is given by,

p3φ(t) = p(t) = va(t)ia(t) + vb(t)ib(t) + vc(t)ic(t)

= pa + pb + pc (2.3)

In above equation, pa(t), pb(t) and pc(t) are expressed similar to single phase system done previ-ously. These are given below.

pa(t) = V I cosφ 1− cos 2ωt − V I sinφ sin 2ωt

pb(t) = V I cosφ 1− cos(2ωt− 120o) − V I sinφ sin 2(ωt− 120o) (2.4)pc(t) = V I cosφ 1− cos(2ωt+ 120o) − V I sinφ sin 2(ωt+ 120o)

Adding three phase instantaneous powers given in (2.4), we get the three-phase instantaneouspower as below.

p(t) = 3V I cosφ − V I cosφcos 2ωt+ cos 2(ωt− 120o) + cos 2(ωt+ 120o)− V I sinφsin 2ωt+ sin 2(ωt− 120o) + sin 2(ωt+ 120o) (2.5)

Summation of terms in curly brackets is always equal to zero. Hence,

p3φ(t) = p(t) = 3V I cosφ. (2.6)

28

Page 35: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

This is quite interesting result. It indicates for balanced three-phase system, the total instantaeouspower is equal to the real power or average active power (P ), which is constant. This is the reasonwe use 3-phase system. It does not involve the pulsating or oscillating components of power as incase of single phase systems. Thus it ensures less VA rating for same amount of power transfer.

Here, total three-phase reactive power can be defined as sum of maximum value of preactive(t)terms in (2.4). Thus,

Q = Qa +Qb +Qc = 3V I sinφ. (2.7)

Is there any attempt to define instantaneous reactive power q(t) similar to p(t) such that Q isaverage value of that term q(t)?. H. Akagi et al. published paper [6], in which authors defined terminstantaneous reactive power. The definition was facilitated through αβ0 transformation. Brieflyit is described in the next subsection.

2.1.3 Three Phase Instantatneous Reactive Power

H.Akagi et.al. [6] attempted to define instantaneous reactive power(q(t)) using αβ0 transforma-tion. This transformation is described below.

The abc coordinates and their equivalent αβ0 coordinates are shown in the Fig. 2.2 below.

av

bv

cv

j

60o

- c /2

- b /2

-j c

j b

O

Fig. 2.2 A abc to αβ0 transformation

Resolving a, b, c quantities along the αβ axis we have,

vα =

√2

3(va −

vb2− vc

2) (2.8)

vβ =

√2

3

√3

2(vb − vc) (2.9)

Here,√

23

is a scaling factor, which ensures power invariant transformation. Along with that, wedefine zero sequence voltage as,

29

Page 36: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

v0 =

√2

3

√1

2(va + vb + vc) (2.10)

Based on Eqns.(4.60)-(2.10) we can write the above equations as follows.

v0(t)vα(t)vβ(t)

=

√2

3

1√2

1√2

1√2

1 −12

−12

0√

32

−√

32

va(t)vb(t)vc(t)

(2.11)

v0

vαvβ

= [Aoαβ]

vavbvc

The above is known as Clarke-Concordia transformation. Thus, va, vb and vc can also be expressedin terms of v0, vα and vβ by pre-multiplying (2.11) by matrix [A0αβ]−1, we have

vavbvc

= [A0αβ]−1

v0

vαvβ

It will be interesting to learn that

[A0αβ]−1 = [Aabc] =

√2

3

1√2

1√2

1√2

1 −12

−12

0√

32

−√

32

−1

[A0αβ]−1 =

√2

3

1√2

1 01√2

−12

√3

21√2−12

−√

32

= [A0αβ]T = [Aabc] (2.12)

Similarly, we can write down instantaneous symmetrical transformation for currents, which isgiven below. i0

iαiβ

=

√2

3

1√2

1√2

1√2

1 −12

−12

0√

32

−√

32

iaibic

(2.13)

Now based on ’0αβ’ transformation, the instantaneous active and reactive powers are defined asfollows. The three-phase instantaneous power p(t) is expressed as the dot product of 0αβ compo-nents of voltage and currents such as given below.

30

Page 37: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

p(t) = vα iα + vβ iβ + v0 i0

=2

3

[(va −

vb2− vc

2

)(ia −

ib2− ic

2

)−√

3

2(vb − vc)

√3

2(ib − ic)

+1√2

(va + vb + vc)1√2

(ia + ib + ic)

]= va ia + vb ib + vc ic (2.14)

Now what about instantaneous reactive power? Is there any concept defining instantaneous reactivepower? In 1983-84,authors H.akagi have attempted to define instantaneous reactive power usingstationary αβ0 frame, as illustrated below. In [6], the instantaneous reactive power q(t) is definesas the cross product of two mutual perpendicular quantities, such as given below.

q(t) = vα × iβ + vβ × iαq(t) = vαiβ − vβiα

=2

3

[(va −

vb2− vc

2

) √3

2(ib − ic)−

√3

2(vb − vc)

(ia −

ib2− ic

2

)]

=2

3

√3

2

[(−vb + vc) ia +

(va −

vb2− vc

2+vb2− vc

2

)ib +

(−va +

vb2

+vc2

+vb2− vc

2

)ic

]= − 1√

3[(vb − vc) ia + (vc − va) ib + (va − vb) ic]

= − [vbcia + vcaib + vabic] /√

3 (2.15)

This is also equal to the following.

q(t) =1√3

[(ib − ic) va +

(−ib

2+ic2− ia +

ib2

+ic2

)vb +

(−ib

2+ic2

+ ia −ib2− ic

2

)vc

]=

1√3

[(ib − ic) va + (ic − ia) vb + (ia − ib) vc] (2.16)

2.1.4 Power Invariance in abc and αβ0 Coordinates

As a check for power invariance, we shall compute the energy content of voltage signals in twotransformations. The energy associated with the abc0 system is given by (v2

a + v2b + v2

c ) and theenergy associated with the αβ0 components is given by

(v2

0 + v2α + v2

β

). The two energies must

be equal to ensure power invariance in two transformations. It is proved below. Using, (2.11) and

31

Page 38: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

squares of the respective components, we have the following.

v2α =

[√2

3

(va −

vb2− vc

2

)]2

v2α =

2

3

v2a +

v2b

4+v2c

4− 2vavb

2+

2vbvc4− 2vavc

2

=

2

3v2a +

v2b

6+v2c

6− 2vavb

3+vbvc

3− 2vavc

3(2.17)

Similary we can find out square of vβ term as given below.

v2β =

[√3

2

√2

3(vb − vc)

]2

=1

2

(v2b + v2

c − 2vbvc)

=v2b

2+v2c

2− vbvc (2.18)

Adding (2.17) and (2.18), we find that,

v2α + v2

β =2

3

(v2a + v2

b + v2c − vcvb − vbvc − vcva

)=

(v2a + v2

b + v2c

)−(v2a

3+v2b

3+v2c

3+

2vavb3

+2vbvc

3+

2vavc3

)=

(v2a + v2

b + v2c

)− 1

3(va + vb + vc)

2

=(v2a + v2

b + v2c

)−

1√3

(va + vb + vc)

2

(2.19)

Since v0 = 1√3(va + vb + vc), the above equation, (2.19) can be written as,

v2α + v2

β + v20 = v2

a + v2b + v2

c . (2.20)

From the above it is implies that the energy associated with the two systems remain same instant toinstant basis. In general the instantaneous power p(t) remain same in both transformations. Thisis proved below.

32

Page 39: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Using (2.14), following can be written.

p(t) = vαiα + vβiβ + voio

p(t) =

v0

vαvβ

T i0iαiβ

=

[Aabc]

vavbvc

T [Aabc]

iaibic

=

vavbvc

T [Aabc]T [Aabc]

iaibic

=

vavbvc

T [Aabc]−1 [Aabc]

iaibic

=

[va vb vc

] iaibic

= vaia + vbib + vcic (2.21)

From (2.12), [Aabc] and its inverse are as following.

[Aabc] =

√2

3

1√2

1√2

1√2

1 −1√2

−1√2

0√

32

−√

32

and (2.22)

[Aabc]−1 = [Aabc]

T =

√2

3

1√2

1 01√2−1√

2

√3

21√2−1√

2−√

32

2.2 Instantaneous Active and Reactive Powers for Three-phase Circuits

In the previous section instantaneous active and reactive powers were defined using αβ0 trans-formation. In this section we shall study these powers for various three-phase circuits such asthree-phase balanced, three-phase unbalanced, balanced three-phase with harmonics and unbal-anced three-phase with harmonics. Each case will be considered and analyzed.

33

Page 40: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2.2.1 Three-Phase Balance System

For three-phase balanced system, three-phase voltages have been expressed by equation (2.1). Forthese phase voltages, the line to line voltages are given as below.

vab =√

3√

2v sin (ωt+ 30)

vbc =√

3√

2v sin(ωt− 90)

vca =√

3√

2v sin (ωt+ 150) (2.23)

The above relationship between phase and line to line voltages is also illustrated in Fig. 2.3. For

0oaV V

bV

bV3

30o

abV

V

cV

Fig. 2.3 Relationship between line-to-line and phase voltage

the above three-phase system, the instantaneous power p(t) can be expressed using (2.21) and it isequal to,

p(t) = vaia + vbib + vcic

= vαiα + vβiβ + v0i0

= 3V I cosφ (2.24)

The instantaneous reactive power q(t) is as following.

q(t) = [√

3√

2V sin (ωt− 90o)√

2I sin (ωt− φ)

+√

3√

2V sin (ωt+ 150o)√

2I sin (ωt− 120o − φ)

+√

3√

2V sin (ωt+ 30o)√

2I sin (ωt+ 120 − φ)]/√

3

= −√

3V I[cos (90 − φ)− cos (2ωt− 90o − φ)

+ cos (90o − φ)− cos (2ωt− 30o − φ)

+ cos (90o − φ)− cos (2ωt+ 150o − φ)]/√

3

= −√

3V I[3 sinφ− cos (2ωt− φ+ 30o)− cos (2ωt− φ+ 30o + 120o)

− cos (2ωt− φ+ 30o − 120o)]/√

3

= −V I [3 sinφ− 0]

q(t) = −3V I sinφ (2.25)

34

Page 41: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The above value of instantaneous reactive power is same as defined by Budeanu’s [1] and is givenin equation (2.7). Thus, instantaneous reactive power given in (2.15) matches with the conven-tional definition of reactive power defined in (2.7). However the time varying part of second termsof each phase in (2.4) has no relevance with the definition given in (2.15).

Another interpretation of line to line voltages in (2.15) is that the voltages vab, vbc and vca have90o phase shift with respect to voltages vc, va and vb respectively. These are expressed as below.

vab =√

3vc∠− 90o

vbc =√

3va∠− 90o (2.26)vca =

√3vb∠− 90o

In above equation, vc∠ − 90o implies that vc∠ − 90o lags vc by 90o. Analyzing each term in(2.15) contributes to,

vbcia =√

3va∠− 90. ia

=√

3√

2V sin (ωt− 90) .√

2I sin (ωt− φ)

=√

3V I 2 sin (ωt− 90) . sin (ωt− φ)

=√

3V I [cos (90 − φ)− cos (2ωt− 90 − φ)]

=√

3V I [sinφ− cos 90 + (2ωt− φ)]=√

3V I [sinφ+ sin (2ωt− φ)]

=√

3V I [sinφ+ sin 2ωt cosφ− cos 2ωt sinφ]

vbcia/√

3 = V I [sinφ (1− cos 2ωt) + cosφ sin 2ωt]

Similarly,

vcaib/√

3 = V I

[sinφ

(1− cos

2

(ωt− 2π

3

))]+V I cosφ. sin 2

(ωt− 2π

3

)vabic/

√3 = V I

[sinφ

(1− cos

2

(ωt+

3

))]+V I cosφ. sin 2

(ωt+

3

)(2.27)

Thus, we see that the role of the coefficients of sinφ and cosφ have reversed. Now if we takeaverage value of (2.27), it is not equal to zero but V I sinφ in each phase. Thus three-phase reactivepower will be 3V I sinφ. The maximum value of second term in (2.27) represents active averagepower i.e., V I cosφ. However, this is not normally convention about the notation of the powers.But, important contribution of this definition is that average reactive power could be defined as theaverage value of terms in (2.27).

2.2.2 Three-Phase Unbalance System

Three-phase unbalance system is not uncommon in power system. Three-phase unbalance mayresult from single-phasing, faults, different loads in three phases. To study three-phase system

35

Page 42: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

with fundamental unbalance, the voltages and currents are expressed as following.

va =√

2Va sin (ωt− φva)vb =

√2Vb sin (ωt− 120o − φvb) (2.28)

vc =√

2Vc sin (ωt+ 120o − φvc)and,

ia =√

2Ia sin (ωt− φia)ib =√

2Ib sin (ωt− 120o − φib) (2.29)ic =√

2Ic sin (ωt+ 120o − φic)For the above system, the three-phase instantaneous power is given by,

p3φ(t) = p(t) = vaia + vbib + vcic

=√

2Va sin (ωt− φva) sin (ωt− φia)+√

2Vb sin (ωt− 120o − φvb)√

2Ib sin (ωt− 120o − φib) (2.30)+√

2Vc sin (ωt+ 120o − φvc)√

2Ic sin (ωt+ 120o − φic)Simplifying above expression we get,

p3φ(t) = VaIa cosφa 1− cos (2ωt− 2φva)︸ ︷︷ ︸pa,active

−VaIa sinφa sin (2ωt− 2φva)︸ ︷︷ ︸pa,reactive

+VbIb cosφb [1− cos 2 (ωt− 120)− 2φvb]−VbIb sinφb sin 2 (ωt− 120)− 2φvb+VcIc cosφc [1− cos 2 (ωt+ 120)− 2φvc]−VcIc sinφc sin 2 (ωt+ 120)− 2φvc (2.31)

where φa = (φia − φva)Therefore,

p3φ (t) = pa,active + pb,active + pc,active + pa,reactive + pb,reactive + pc,reactive

= pa + pb + pc + pa + pb + pc (2.32)

where,

pa = Pa = VaIa cosφa

pb = Pb = VbIb cosφb (2.33)pc = Pc = VcIc cosφc

and

pa = −VaIa cos (2ωt− φa − 2φva)

pb = −VbIb cos (2ωt− 240o − φb − 2φvb) (2.34)pc = −VcIc cos (2ωt+ 240− φc − 2φvc)

36

Page 43: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Also it is noted that,

pa + pb + pc = vaia + vbib + vcic = P (2.35)

and,

pa + pb + pc = −VaIacos(2ωt− φva − φib)−VbIbcos 2(ωt− 120)− φvb − φib−VcIccos 2(ωt+ 120)− φvc − φic

6= 0

This implies that, we no longer get advantage of getting constant power, 3V I cosφ from interactionof three-phase voltages and currents. Now, let us analyze three phase instantaneous reactive powerq(t) as per definition given in (2.15).

q(t) = − 1√3

(vb − vc)ia + (vc − va)ib + (va − vb)ic

= − 2√3

[Vbsin(ωt− 120o − φvb)− Vcsin(ωt+ 120o − φvc) Iasin(ωt− φia)

+ Vcsin(ωt+ 120o − φvc)− Vasin(ωt− φva)√

2Ibsin(ωt− 120o − φib) (2.36)

+Vasin(ωt− 120o − φva)− Vb sin(ωt− 120o − φvb)√

2Ic sin(ωt+ 120o − φic)]

From the above,√

3q(t) = −[VbIa cos(φia − 120o − φvb)− cos(2ωt− 120o − φia − φvb)

−VcIa cos(φia + 120o − φvc)− cos(2ωt+ 120o − φia − φvc)+VcIb cos(φib + 240o − φvc)− cos(2ωt− φib − φvc) (2.37)−VaIb cos(φib − 120o − φva)− cos(2ωt− 120o − φva − φib)+VaIc cos(φic − 120o − φva)− cos(2ωt+ 120o − φva − φic)

−VbIc cos(φic − 240o − φvb)− cos(2ωt− φic − φvb)]

Now looking this expression,we can say that

1

T

∫ T

0

q(t)dt = − 1√3

[VbIa cos(φia − φvb − 120o)

−VcIa cos(φia − φvc + 120o)

+VcIb cos(φib + 240o − φvc)−VaIb cos(φib − 120o − φva)+VaIc cos(φic − 120o − φva)

−VbIc cos(φic − 240o − φvb)]

= qa(t) + qb(t) + qc(t)

6= VaIa cosφa + VbIb cosφb + VcIc cosφc (2.38)

37

Page 44: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Hence the definition of instantaneous reactive power does not match to that defined by Budeanue’sreactive power [1] for three-phase unbalanced circuit. If only voltages or currents are distorted, theabove holds true as given below. Let us consider that only currents are unbalanced, then

va(t) =√

2V sin(ωt)

vb(t) =√

2V sin(ωt− 120) (2.39)

vc(t) =√

2V sin(ωt+ 120)

and

ia(t) =√

2Ia sin(ωt− φa)ib(t) =

√2Ib sin(ωt− 120o − φb) (2.40)

ic(t) =√

2Ic sin(ωt+ 120o − φc)And the instantaneous reactive power is given by,

q(t) = − 1√3[vbcia + vcaib + vabic]

= − 1√3[√

3 va∠−π/2 ia +√

3 vb∠− π/2 ib +√

3 vc∠− π/2 ic]= −[

√2V sin(ωt− π/2)

√2Ia sin(ωt− φia)

+√

2V sin(ωt− 120o − π/2)√

2Ib sin(ωt− 120o − φib)+√

2V sin(ωt+ 120o + π/2)√

2Ic sin(ωt+ 120o − φic)]= −[V Iacos(π/2− φia)− cos π/2− (2ωt− φia)

+V Ibcos(π/2− φib)− cos(2ωt− 240o − π/2− φib)+V Iccos(π/2− φic)− cos(2ωt+ 240o − π/2− φic)]

= −[(V Ia sinφia + V Ib sinφib + V Ic sinφic)+V Ia sin(2ωt− φia) + V Ib sin(2ωt− 240o − φib) + V Ic sin(2ωt+ 240o − φic)]

Thus,

Q =1

T

∫ T

0

q(t)dt = −(V Ia sinφia + V Ib sinφib + V Ic sinφic) (2.41)

Which is similar to Budeanu’s reactive power.

The oscillating term of q(t) which is equal to q(t) is given below.

q(t) = V Ia sin(2ωt− φia) + V Ib sin(2ωt− 240o − φib) + V Ic sin(2ωt+ 240o − φic) (2.42)

which is not similar to what is being defined as reactive component of power in (2.4).

2.3 Symmetrical components

In the previous section, the fundamental unbalance in three phase voltage and currents have beenconsidered. Ideal power systems are not designed for unbalance quantities as it makes power sys-tem components over rated and inefficient. Thus, to understand unbalance three-phase systems,

38

Page 45: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

a concept of symmetrical components introduced by C. L. Fortescue, will be discussed. In 1918,C. L Fortescue, wrote a paper [7] presenting that an unbalanced system of n-related phasors canbe resolved into n system of balanced phasors, called the symmetrical components of the originalphasors. The n phasors of each set of components are equal in length and the angles. Although,the method is applicable to any unbalanced polyphase system, we shall discuss about three phasesystems.

For the discussion of symmetrical components, a complex operator denoted as a is defined as,

a = 1∠120o = ej2π/3 = cos 2π/3 + j sin 2π/3

= −1/2 + j√

3/2

a2 = 1∠240o = 1∠− 120o = ej4π/3 = e−j2π/3 = cos 4π/3 + j sin 4π/3

= −1/2− j√

3/2

a3 = 1∠360o = ej2π = 1

Also note an interesting property relating a, a2 and a3,

a+ a2 + a3 = 0. (2.43)

3 1 oa o

2 1 120oa

1 120oa

o

Fig. 2.4 Phasor representation of a, a2 and a3

These quantities i.e., a, a2 and a3 = 1 also represent three phasors which are shifted by 120o

from each other. This is shown in Fig. 2.4.Knowing the above and using Fortescue theorem, three unbalanced phasor of a three phase un-

balanced system can be resolved into three balanced system phasors.

1. Positive sequence components are the composed of three phasors equal in magnitude and dis-placed from each other by 120 degrees in phase and having a phase sequence of original phasors.

39

Page 46: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2. Negative sequence components consist of three phasors equal in magnitude, phase shift of−120o and 120o between phases and with phase sequence opposite to that of the original phasors.

3. Zero sequence components consist of three phasors equal in magnitude with zero phase shiftfrom each other.

Positive sequence components: V a1, V b1, V c1

Negative sequence components: V a2, V b2, V c2

Zero sequence components: V a0, V b0, V c0

Thus we can write,

V a = V a1 + V a2 + V a0

V b = V b1 + V b2 + V b0 (2.44)V c = V c1 + V c2 + V c0

Graphically, these are represented in Fig. 2.5. Thus if we add the sequence components of eachphase vectorially, we shall get V a, V b and V s as per (2.44). This is illustrated in Fig. 2.6.

1Va

1Vc

1Vb

2Vb

2Va

2Vc

0Va

0Vb

0Vc

(a) (b) (c)

Fig. 2.5 Sequence components (a) positive sequence (b) negative sequence (c) zero sequence

Now knowing all these preliminaries, we can proceed as following. Let V a1 be a reference phasor,therefore V b1 and V c1 can be written as,

V b1 = a2V a1 = V a1∠− 120

V c1 = aV a1 = V a1∠120 (2.45)

Similarly V b2 and V c2 can be expressed in terms of V a2 as following.

V c2 = a2V b2 = V a2∠− 120

V b2 = aV a2 = V a2∠120 (2.46)

40

Page 47: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Va

1Va2Va

0Va

1Vc

2Vc

0Vc

Vc

Vb

1Vb

2Vb

0Vb

o

Fig. 2.6 Unbalanced phasors as vector sum of positive, negative and zero sequence phasors

The zero sequence components have same magnitude and phase angle and therefore these areexpressed as,

V b0 = V c0 = V a0 (2.47)

Using (2.45), (2.46) and (2.47) we have,

V a = V a0 + V a1 + V a2 (2.48)

V b = V b0 + V b1 + V b2

= V a0 + a2 V a1 + a V a2 (2.49)

V c = V c0 + V c1 + V c2

= V a0 + a V a1 + a2 V a2 (2.50)

Equations (2.48)-(2.50) can be written in matrix form as given below.V a

V b

V c

=

1 1 11 a2 a1 a a2

V ao

V a1

V a2

(2.51)

Premultipling by inverse of matrix [Aabc] which is equal to

1 1 11 a2 a1 a a2

, the symmetrical com-

41

Page 48: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

ponents are expressed as given below.V ao

V a1

V a2

=1

3

1 1 11 a a2

1 a2 a

V a

V b

V c

(2.52)

= [A012]

V a

V b

V c

The symmetrical transformation matrices Aabc and A012 are related as following.

[A012] = [Aabc]−1 = 3 [Aabc]

∗ (2.53)

From (2.52), the symmetrical components can therefore be expressed as phase voltages as follow-ing.

V a0 =1

3(V a + V b + V c)

V a1 =1

3(V a + aV b + a2V c) (2.54)

V a2 =1

3(V a + a2V b + aV c)

The other component i.e., V bo, V co, V b1, V c1, V b2, V c2 can be found from V ao, V a1, V a2. It shouldbe noted that quantity V ao does not exist if sum of unbalanced phasors is zero. Since sum of lineto line voltage phasors i.e., V ab + V bc + V ca = (V a − V b) + (V b − V c) + (V c − V a) is alwayszero, hence zero sequence voltage components are never present in the line voltage, regardless ofamount of unbalance. The sum of the three phase voltages, i.e.,V a + V b + V c is not necessarilyzero and hence zero sequence voltage exists.

Similarly sequence components can be written for currents. Denoting three phase currents byIa, Ib, and Ic respectively, the sequence components in matrix form are given below.IaoIa1

Ia2

=1

3

1 1 11 a a2

1 a2 a

IaIbIc

(2.55)

Thus,

Iao =1

3(Ia + Ib + Ic)

Ia1 =1

3(Ia + aIb + a2Ic)

Ia2 =1

3(Ia + a2Ib + aIc)

42

Page 49: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

In three-phase, 4-wire system, the sum of line currents is equal to the neutral current (In). thus,

In =1

3(Ia + Ib + Ic)

= 3Ia0 (2.56)

This current flows in the fourth wire called neutral wire. Again if neutral wire is absent, then zerosequence current is always equal to zero irrespective of unbalance in phase currents. This is illus-trated below.

a

b

c

a

b

c

(a)

a

b

c

a

b

c

(b)

Fig. 2.7 Various three phase systems (a) Three-phase three-wire system (b) Three-phase four-wire system

In 2.7(b), in may or may not be zero. However neutral voltage (VNn) between the system andload neutral is always equal to zero. In 2.7(a), there is no neutral current due to the absence of theneutral wire. But in this configuration the neutral voltage, VNn, may or may not be equal to zerodepending upon the unbalance in the system.

Example 2.1 Consider a balanced 3 φ system with following phase voltages.

V a = 100∠0o

V b = 100∠− 120o

V c = 100∠120o

Using (2.54), it can be easily seen that the zero and negative sequence components are equal tozero, indicating that there is no unbalance in voltages. However the converse may not apply.Now consider the following phase voltages. Compute the sequence components and show that theenergy associated with the voltage components in both system remain constant.

V a = 100∠0o

V b = 150∠− 100o

V c = 75∠100o

43

Page 50: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Solution Using (2.54), sequence components are computed. These are:

V a0 =1

3(V a + V b + V c)

= 31.91∠− 50.48o V

V a1 =1

3(V a + aV b + a2V c)

= 104.16∠4.7o V

V a2 =1

3(V a + a2V b + aV c)

= 28.96∠146.33o V

If you find energy content of two frames that is abc and 012 system, it is found to be constant.

Eabc = k [V 2a + V 2

b + V 2c ] = k.(381.25)

E012 = k 3[V 2a0 + V 2

a1 + V 2a2] = k.(381.25)

Thus, Eabc = E012 with k some constant of proportionality.

The invariance of power can be further shown by following proof.

Sv = P + jQ = [ V a V b V c ]

IaIbIc

=

V a

V b

V c

T IaIbIc

=

[Aabc]

V a0

V a1

V a2

T [Aabc]

Ia0

Ia1

Ia2

=

V a0

V a1

V a2

T [Aabc]T [Aabc]

Ia0

Ia1

Ia2

∗ (2.57)

The term Sv is referred as vector or geometric apparent power. The difference between will begiven in the following. The transformation matrix [Aabc] has following properties.

[Aabc]−1 =

1

3[Aabc]

T and (2.58)

[Aabc]∗ = [Aabc]

44

Page 51: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Therefore using (2.58), (2.57) can be written as the following.

Sv = P + jQ =

V a0

V a1

V a2

T3[I]

Ia0

Ia1

Ia2

= 3

V a0

V a1

V a2

TIa0

Ia1

Ia2

∗Sv = P + jQ = V aI

∗a + V bI

∗b + V cI

∗c

= 3 [V a0I∗a0 + V a1I

∗a1 + V a2I

∗a2] (2.59)

Equation (2.59) indicates that power invariance holds true in both abc and 012 components. But,this is true on phasor basis. Would it be true on the time basis? In this context, concept of instanta-neous symmetrical components will be discussed in the latter section. The equation (2.59) furtherimplies that,

Sv = P + jQ = 3 [ (Va0Ia0 cosφa0 + Va1Ia1 cosφa1 + Va2Ia2 cosφa2)

+j(Va0Ia0 sinφa0 + Va1Ia1 sinφa1 + Va2Ia2 sinφa2) ] (2.60)

The power terms in (2.60) accordingly form positive sequence, negative sequence and zero se-quence powers denoted as following. The positive sequence power is given as,

P+ = Va1Ia1 cosφa1 + Vb1Ib1 cosφb1 + Vc1Ic1 cosφc1

= 3Va1Ia1 cosφa1. (2.61)

Negative sequence power is expressed as,

P− = 3Va2Ia2 cosφa2. (2.62)

The zero sequence power is

P 0 = 3Va0Ia0 cosφa0. (2.63)

Similarly, sequence reactive power are denoted by the following expressions.

Q+ = 3Va1Ia1 sinφa1

Q− = 3Va2Ia2 sinφa2

Q0 = 3Va0Ia0 sinφa0 (2.64)

Thus, following holds true for active and reactive powers.

P = Pa + Pb + Pc = P0 + P1 + P2

Q = Qa +Qb +Qc = Q0 +Q1 +Q2 (2.65)

45

Page 52: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Here, positive sequence, negative sequence and zero sequence apparent powers are denoted as thefollowing.

S+ = |S+| =√P+2 +Q+2 = 3Va1Ia1

S− = |S+| =√P−2 +Q−2 = 3Va2Ia2

S0 = |S+| =√P 02 +Q02 = 3Va0Ia0 (2.66)

The scalar value of vector apparent power (Sv) is given as following.

Sv = |Sa + Sb + Sc| = |S0

+ S+

+ S−|

= |(Pa + Pb + Pc) + j(Qa +Qb +Qc)| (2.67)=√P 2 +Q2

Similarly, arithematic apparent power (SA) is defined as the algebraic sum of each phase or se-quence apparent power, i.e.,

SA = |Sa|+ |Sb|+ |Sc|= |Pa + jQa|+ |Pb + jQb|+ |Pc + jQc| (2.68)

=√P 2a +Q2

a +√P 2b +Q2

b +√P 2c +Q2

c

In terms of sequence components apparent power,

SA = |S0|+ |S+|+ |S−|= |P 0 + jQ0|+ |P+ + jQ+|+ |P− + jQ−| (2.69)

=

√P 02 +Q02 +

√P+2 +Q+2 +

√P−2 +Q−2

Based on these two definitions of the apparent powers, the power factors are defined as the follow-ing.

Vector apparent power = pfv =P

Sv(2.70)

Arithematic apparent power = pfA =P

SA(2.71)

46

Page 53: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Example 2.2 Consider a 3-phase 4 wire system supplying resistive load, shown in Fig. 2.8below. Determine power consumed by the load and feeder losses.

'a

'c

'n

a

b

c

n

Va

Vb

Vc

RIa

In

Ib

Ic

r j x

r j x

r j x

r j x

'

Fig. 2.8 A three-phase unbalanced load

Power dissipated by the load =(√

3V )2

R=

3V 2

R

The current flowing in the line =

√3V

R= |V a − V b

R|

and Ib = −Ia

Therefore losses in the feeder =

(√3V

R

)2

× r +

(√3V

R

)2

× r

= 2( rR

)(3V 2

R

)

Now, consider another example of a 3 phase system supplying 3-phase load, consisting of threeresistors (R) in star as shown in the Fig. 2.9. Let us find out above parameters.

Power supplied to load = 3

(V

R

)2

×R =3V 2

R

Losses in the feeder = 3

(V

R

)2

× r =( rR

)(3V 2

R

)

Thus, it is interesting to see that power dissipated in the unbalanced system is twice the power lossin balanced circuit. This leads to conclusion that power factor in phases would become less thanunity, while for balanced circuit, the power factor is unity. Power analysis of unbalanced circuitshown in Fig. 2.8 is given below.

47

Page 54: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

'a

'c

'n

a

b

c

nVa

R

Ia

r j x

r j x

r j x

r j x

R

R

'b

Vb

Vc

Ib

Ic

In

Fig. 2.9 A three-phase balanced load

The current in phase-a, Ia =V a − V b

R=V ab

R=

√3VaR

∠30

The current in phase-b, Ib = −Ia =

√3V

R∠(30− 180)o

=

√3V

R∠− 150o

The current in phase-c and neutral are zero, Ic = In = 0

The phase voltages are: V a = V ∠0o, V b = V ∠− 120o, V c = V ∠120o.

The phase active and reactive and apparent powers are as following.

Pa = VaIa cosφa = V I cos 30 =

√3

2V I

Qa = VaIa sinφa = V I sin 30 =1

2V I

Sa = VaIa = V I

Pb = VbIb cosφb = V I cos(−30) =

√3

2V I

Qb = VbIb sinφb = V I sin(−30) = −1

2V I

Sb = VbIb = V I

Pc = Qc = Sc = 0

Thus total active power P = Pa + Pb + Pc = 2×√

3

2V I =

√3V I

=√

3V

√3V

R

P =3V 2

RTotal reactive power Q = Qa +Qb +Qc = 0

48

Page 55: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The vector apparent power, Sv =√P 2 +Q2 = 3V 2/R = P

The arithmetic apparent power, SA = Sa + Sb + Sc = 2V I = (2/√

3)P

From the values of Sv and SA, it implies that,

pfv =P

Sv=P

P= 1

pfA =P

SA=

P

(2/√

3)P=

√3

2= 0.866

This difference between the arthmetic and vector power factors will be more due to the unbalancesin the load.

For balance load SA = SV , therefore, pfA = pfV = 1.0. Thus for three-phase electrical cir-cuits, the following holds true.

pfA ≤ pfV (2.72)

2.3.1 Effective Apparent Power

For unbalanced three-phase circuits, their is one more definition of apparent power, which is knownas effective apparent power. The concept assumes that a virtual balanced circuit that has the samepower output and losses as the actual unbalanced circuit. This equivalence leads to the definitionof effective line current Ie and effective line to neutral voltage Ve.

The equivalent three-phase unbalanced and balanced circuits with same power output and lossesare shown in Fig. 2.10. From these figures, to maintain same losses,

'a

'c

'n

n

Va aR

Ia

r j x

r j x

r j x

r j x

'bVb

Vc

Ib

Ic

In

Vn

bR

cR

'a

'c

'n

Vea

Iea

r j x

r j x

r j x

r j x

'b

eR

eR

eRIeb

Iec

I 0n

n

Veb

Vec

(a) (b)

Fig. 2.10 (a) Three-phase with unbalanced voltage and currents (b) Effective equivalent three-phase system

rI2a + rI2

b + rI2c + rI2

n = 3rI2e

The above equation implies the effective rms current in each phase is given as following.

Ie =

√(I2a + I2

b + I2c + I2

n)

3(2.73)

49

Page 56: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

For the original circuit shown in Fig. 2.8, the effective current Ie is computed using above equationand is given below.

Ie =

√(I2a + I2

b )

3since, Ic = 0and In = 0

=

√2 I2

a

3=

√2 (√

3V/R)2

3

=

√2V

R

To account same power output in circuits shown above, the following identity is used with Re = Rin Fig. 2.10.

V 2a

R+V 2b

R+V 2c

R+V 2a + V 2

b + V 2c

3R=

3V 2e

R+

9V 2e

3R(2.74)

From (2.74), the effective rms value of voltage is expressed as,

Ve =

√1

183 (V 2

a + V 2b + V 2

c ) + V 2ab + V 2

bc + V 2ca (2.75)

Assuming, 3 (V 2a + V 2

b + V 2c ) ≈ V 2

ab + V 2bc + V 2

ca, equation (2.75) can be written as,

Ve =

√V 2a + V 2

b + V 2c

3= V (2.76)

Therefore, the effective apparent power (Se), using the values of Ve and Ie, is given by,

Se = 3Ve Ie =3√

2V 2

R

Thus the effective power factor based on the definition of effective apparent power (Se), for thecircuit shown in Fig. 2.8 is given by,

pfe =P

S e=

3V 2/R

3√

2V 2/R=

1√2

= 0.707

Thus, we observe that,

SV ≤ SA ≤ Se,

pfe (0.707) ≤ pfA (0.866) ≤ pfV (1.0).

When the system is balanced,

Va = Vb = Vc = Ven = Ve,

Ia = Ib = Ic = Ie,

In = 0,

and SV = SA = Se.

50

Page 57: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2.3.2 Positive Sequence Powers and Unbalance Power

The unbalance power Su can be expressed in terms of fundamental positive sequence powers P+,Q+ and S+ as given below.

Su =

√S2e − S+2 (2.77)

where S+ = 3V +I+ and S+2= P+2

+Q+2.

2.4 Three-phase Non-sinusoidal Balanced System

A three-phase nonsinusoidal system is represented by following set of equaitons.

va(t) =√

2V1 sin(wt− α1) +√

2∞∑n=2

Vn sin(nwt− αn)

vb(t) =√

2V1 sin(wt− 120 − α1) +√

2∞∑n=2

Vn sin(n(wt− 120)− αn) (2.78)

vc(t) =√

2V1 sin(wt+ 120 − α1) +√

2∞∑n=2

Vn sin(n(wt+ 120)− αn)

Similarly, the line currents can be expressed as,

ia(t) =√

2I1 sin(wt− β1) +√

2∞∑n=2

In sin(nwt− βn)

ib(t) =√

2I1 sin(wt− 120 − β1) +√

2∞∑n=2

In sin(n(wt− 120 )− βn) (2.79)

ic(t) =√

2I1 sin(wt+ 120 − β1) +√

2∞∑n=2

In sin(n(wt+ 120 )− βn)

In this case,

Sa = Sb = Sc,

Pa = Pb = Pc, (2.80)Qa = Qb = Qc,

Da = Db = Dc.

The above equation suggests that such a system has potential to produce significant additionalpower loss in neutral wire and ground path.

51

Page 58: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2.4.1 Neutral Current

The neutral current for three-phase balanced system with harmonics can be given by the followingequation.

in = ia + ib + ic

=√

2 [ Ia1 sin (wt− β1) + Ia2 sin (2wt− β2) + Ia3 sin (3wt− β3)

+Ia1 sin (wt− 120o − β1) + Ia2 sin (2wt− 240o − β2) + Ia3 sin (3wt− 360o − β3)

+Ia1 sin (wt+ 120o − β1) + Ia2 sin (2wt+ 240o − β2) + Ia3 sin (3wt+ 360o − β3)

+Ia4 sin (4wt− β4) + Ia5 sin (5wt− β5) + Ia6 sin (6wt− β6)

+Ia4 sin (wt− 4× 120o − β4) + Ia5 sin (5wt− 5× 120o − β5) + Ia6 sin (6wt− 6× 120o − β6)

+Ia4 sin (wt+ 4× 120o − β4) + Ia5 sin (5wt+ 5× 120o − β5) + Ia6 sin (6wt+ 6× 120o − β6)

(2.81)+Ia7 sin (7wt− β7) + Ia8 sin (8wt− β8) + Ia9 sin (9wt− β9)

+Ia7 sin (7wt− 7× 120o − β7) + Ia8 sin (8wt− 8× 120o − β8) + Ia9 sin (9wt− 9× 120o − β9)

+Ia7 sin (7wt+ 7× 120o − β7) + Ia8 sin (8wt+ 8× 120o − β8) + Ia9 sin (9wt+ 9× 120o − β9) ]

From the above equation, we observe that, the triplen harmonics are added up in the neutral current.All other harmonics except triplen harmonics do not contribute to the neutral current, due to theirbalanced nature. Therefore the neutral current is given by,

in = ia + ib + ic =∞∑

n=3,6,..

3√

2In sin(nwt− βn). (2.82)

The RMS value of the current in neutral wire is therefore given by,

In = 3

[∞∑

n=3,6,..

I2n

]1/2

. (2.83)

Due to dominant triplen harmonics in electrical loads such as UPS, rectifiers and other powerelectronic based loads, the current rating of the neutral wire may be comparable to the phase wires.

It is worth to mention here that all harmonics in three-phase balanced systems can be catego-rized in three groups i.e., (3n + 1), (3n + 2) and 3n (for n = 1, 2, 3, ...) called positive, nega-tive and zero sequence harmonics respectively. This means that balanced fundamental, 4th, 7th10th,... form positive sequence only. Balanced 2nd, 5th, 8th, 11th,... form negative sequence onlyand the balanced triplen harmonics i.e. 3rd, 6th, 9th,... form zero sequence only. But in case ofunbalanced three-phase systems with harmonics, (3n + 1) harmonics may start forming negativeand zero sequence components. Similarly, (3n + 2) may start forming positive and zero sequencecomponents and 3n may start forming positive and negative sequence components.

2.4.2 Line to Line Voltage

For the three-phase balanced system with harmonics, the line-to-line voltages are denoted as vab,vbc and vca. Let us consider, line-to-line voltage between phases a and b. It is given as following.

52

Page 59: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

vab(t) = va(t)− vb(t)

=∞∑n=1

√2Vn sin(nωt− αn)−

∞∑n=1

√2Vn sin(n (ωt− 120o)− αn)

=∞∑n=1

√2Vn sin(nωt− αn)−

∞∑n=1

√2Vn sin((nωt− αn)− n× 120o)

=∞∑n=1

√2Vn [sin(nωt− αn)− sin(nωt− αn) cos(n× 120o)

+ cos(nωt− αn) sin(n× 120o)]

=∞∑

n6=3,6,9...

√2Vn [sin(nωt− αn)− sin(nωt− αn) (−1/2)

+ cos(nωt− αn) (±√

3/2)]

=√

2∞∑

n6=3,6,9...

Vn

[(3/2) sin(nωt− αn) + (±

√3/2) cos(nωt− αn)

]=√

3√

2∞∑

n6=3,6,9...

Vn

[(√

3/2) sin(nωt− αn) + (±1/2) cos(nωt− αn)]

(2.84)

Let√

3/2 = rn cosφn and ±1/2 = rn sinφn. This impliles rn = 1 and φn = ±30o. Using this,equation (2.84) can be written as follows.

vab(t) =√

3√

2∞∑

n6=3,6,9...

Vn [sin(nωt− αn ± 30o)] . (2.85)

In equations (2.84) and (2.85), vab = 0 for n = 3, 6, 9, . . . and for n = 1, 2, 4, 5, 7, . . ., the ± signof 1/2 or sign of 300 changes alternatively. Thus it is observed that triplen harmonics are missingin the line to line voltages, inspite of their presence in phase voltages for balanced three-phasesystem with harmonics. Thus the following identity hold true for this system,

VLL ≤√

3VLn (2.86)

Above equation further implies that,

√3VLL I ≤ 3VLn I. (2.87)

In above equation, I refers the rms value of the phase current. For above case, Ia = Ib = Ic = Iand In = 3

∑∞n=3,6,9... In

2. Therefore, effective rms current, Ie is given by the following.

53

Page 60: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Ie =

√3 I2 + 3

∑∞n=3,6,9... In

2

3

=

√√√√I2 +∞∑

n=3,6,9...

In2 (2.88)

≥ I

2.4.3 Apparent Power with Budeanu Resolution: Balanced Distortion Case

The apparent power is given as,

S = 3VlnI =√P 2 +Q2

B +D2B

=√P 2 +Q2 +D2 (2.89)

where,

P = P1 + PH = P1 + P2 + P3 + ....

= 3V1I1 cosφ1 + 3∞∑n=1

VnIn cosφn

(2.90)

where, φn = βn − αn. Similarly,

Q = QB = QB1 +QBH

= Q1 +QH (2.91)

Where Q in (2.89) is called as Budeanu’s reactive power (VAr) or simply reactive power which isdetailed below.

Q = Q1 +QH = Q1 +Q2 +Q3 + ....

= 3V1I1 sinφ1 + 3∞∑n=1

VnIn sinφn (2.92)

2.4.4 Effective Apparent Power for Balanced Non-sinusoidal System

The effective apparent power Se for the above system is given by,

Se = 3VeIe (2.93)

For a three-phase, three-wire balanced system, the effective apparent power is found after cal-culating effective voltage and current as given below.

Ve =√

(V 2ab + V 2

bc + V 2ca)/9

= Vll/√

3 (2.94)

54

Page 61: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Ie =√

(I2a + I2

b + I2c )/3

= I (2.95)

ThereforeSe = S =

√3VllI (2.96)

For a four-wire system, Ve is same is given (2.94) and Ie is given by (2.88). Therefore, theeffective apparent power is given below.

√3VllI ≤ 3VlnIe (2.97)

The above implies that,

Se ≥ SA. (2.98)

Therefore, it can be further concluded that,

pfe (= P/Se) ≥ pfA (= P/SA). (2.99)

2.5 Unbalanced and Non-sinusoidal Three-phase System

In this system, we shall consider most general case i.e., three-phase system with voltage and currentquantities which are unbalanced and non-sinusoidal. These voltages and currents are expressed asfollowing.

va(t) =∞∑n=1

√2Van sin(nωt− αan)

vb(t) =∞∑n=1

√2Vbn sin n (ωt− 120o)− αbn (2.100)

vc(t) =∞∑n=1

√2Vcn sin n (ωt+ 120o)− αcn

Similarly, currents can be expressed as,

ia(t) =∞∑n=1

√2Ian sin(nωt− βan)

ib(t) =∞∑n=1

√2Ibn sin n (ωt− 120o)− βbn (2.101)

ic(t) =∞∑n=1

√2Icn sin n (ωt+ 120o)− βcn

55

Page 62: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

For the above voltages and currents in three-phase system, instantaneous power is given as follow-ing.

p(t) = va(t)ia(t) + vb(t)ib(t) + vc(t)ic(t)

= pa(t) + pb(t) + pc(t)

=

(∞∑n=1

√2Van sin(nωt− αan)

)(∞∑n=1

√2Ian sin(nωt− βan)

)(2.102)

+

(∞∑n=1

√2Vbn sin n(ωt− 120o)− αbn

)(∞∑n=1

√2Ibn sin n(ωt− 120o)− βbn

)

+

(∞∑n=1

√2Vcn sin n(ωt+ 120o)− αcn

)(∞∑n=1

√2Icn sin n(ωt+ 120o)− βcn

)

In (2.102), each phase power can be found using expressions derived in Section 1.4 of Unit 1. Thedirect result is written as following.

pa(t) =∞∑n=1

VanIan cosφan 1− cos(2nωt− 2αan) −∞∑n=1

VanIan sinφan cos(2nωt− 2αan)

+

(∞∑n=1

√2Van sin(nωt− αan)

)(∞∑

m=1,m 6=n

√2Iam sin(mωt− βam)

)

=∞∑n=1

Pan 1− cos(2nωt− 2αan) −∞∑n=1

Qan cos(2nωt− 2αan)

+

(∞∑n=1

√2Van sin(nωt− αan)

)(∞∑

m=1,m 6=n

√2Iam sin(mωt− βam)

)(2.103)

In the above equation, φan = (βan − αan). Similarly, for phases b and c, the instantaneouspower is expressed as below.

pb(t) =∞∑n=1

Pbn [1− cos 2n(ωt− 120o)− 2αbn]−∞∑n=1

Qbn cos 2n(ωt− 120o)− 2αbn

+

(∞∑n=1

√2Vbn sin n(ωt− 120o)− αbn

)(∞∑

m=1,m 6=n

√2Ibm sin m(ωt− 120o)− βbm

)(2.104)

56

Page 63: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

and

pc(t) =∞∑n=1

Pcn [1− cos 2n(ωt+ 120o)− 2αcn]−∞∑n=1

Qcn cos 2n(ωt+ 120o)− 2αcn

+

(∞∑n=1

√2Vcn sin n(ωt+ 120o)− αcn

)(∞∑

m=1,m 6=n

√2Icm sin m(ωt+ 120o)− βcm

)(2.105)

From equations (2.103), (2.104) and (2.105), the real powers in three phases are given as follows.

Pa =∞∑n=1

VanIan cosφan

Pb =∞∑n=1

VbnIbn cosφbn (2.106)

Pc =∞∑n=1

VcnIcn cosφcn

Similarly, the reactive powers in three phases are given as following.

Qa =∞∑n=1

VanIan sinφan

Qb =∞∑n=1

VbnIbn sinφbn (2.107)

Qc =∞∑n=1

VcnIcn sinφcn

Therefore, the total active and reactive powers are computed by summing the phase powers usingequations (2.106) and (2.107), which are given below.

P = Pa + Pb + Pc =∞∑n=1

(VanIan cosφan + VbnIbn cosφbn + VcnIcn cosφcn)

= Va1Ia1 cosφa1 + Vb1Ib1 cosφb1 + Vc1Ic1 cosφc1

+∞∑n=2

(VanIan cosφan + VbnIbn cosφbn + VcnIcn cosφcn)

= Pa1 + Pb1 + Pc1 +∞∑n=2

(Pan + Pbn + Pcn)

= P1 + PH (2.108)

57

Page 64: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

and,

Q = Qa +Qb +Qc =∞∑n=1

(VanIan sinφan + VbnIbn sinφbn + VcnIcn sinφcn)

= Va1Ia1 sinφa1 + Vb1Ib1 sinφb1 + Vc1Ic1 sinφc1

+∞∑n=2

(VanIan sinφan + VbnIbn sinφbn + VcnIcn sinφcn)

= Qa1 +Qb1 +Qc1 +∞∑n=2

(Qan +Qbn +Qcn)

= Q1 +QH (2.109)

2.5.1 Arithmetic and Vector Apparent Power with Budeanu’s Resolution

Using Budeanu’s resolution, the arithmetic apparent power for phase-a, b and c are expressed asfollowing.

Sa =√P 2a +Q2

a +D2a

Sb =√P 2b +Q2

b +D2b (2.110)

Sc =√P 2c +Q2

c +D2c

The three-phase arithmetic apparent power is arithmetic sum of Sa, Sb and Sc in the above equation.This is given below.

SA = Sa + Sb + Sc (2.111)

The three-phase vector apparent power is given as following.

Sv =√P 2 +Q2 +D2 (2.112)

Where P andQ are given in (2.108) and (2.109) respectively. The total distortion powerD is givenas following.

D = Da +Db +Dc (2.113)

Based on above definitions of the apparent powers, the arithmetic and vector power factors aregiven below.

pfA =P

SA

pfv =P

Sv(2.114)

From equations (2.111), (2.112) and (2.114), it can be inferred that

SA ≥ Sv

pfA ≤ pfv (2.115)

58

Page 65: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2.5.2 Effective Apparent Power

Effective apparent power (Se=3VeIe) for the three-phase unbalanced systems with harmonics canbe found by computing Ve and Ie as following. The effective rms current (Ie) can be resolved intotwo parts i.e., effective fundamental and effective harmonic components as given below.

Ie =√I2e1 + I2

eH (2.116)

Similarly,

Ve =√V 2e1 + V 2

eH (2.117)

For three-phase four-wire system,

Ie =

√I2a + I2

b + I2c + I2

n

3(2.118)

=

√I2a1 + I2

a2 + ...+ I2b1 + I2

b2 + ...+ I2c1 + I2

c2 + ...+ I2n1 + I2

n2 + ...

3

=

√I2a1 + I2

b1 + I2c1 + I2

n1 + ...+ I2a2 + I2

b2 + I2c2 + I2

n2 + ...

3

=

√I2a1 + I2

b1 + I2c1 + I2

n1

3+I2a2 + I2

a3 + ...+ I2b2 + I2

b3 + ...+ I2c2 + I2

c3 + ...+ I2n2 + I2

n3...

3

Ie =√I2e1 + I2

eH

In the above equation,

Ie1 =

√I2a1 + I2

b1 + I2c1 + I2

n1

3

IeH =

√I2aH + I2

bH + I2cH + I2

nH

3(2.119)

Similarly, the effective rms voltage Ve is given as following.

Ve =

√1

18[3(V 2

a + V 2b + V 2

c ) + (V 2ab + V 2

bc + V 2ca)]

=√V 2e1 + V 2

eH (2.120)

Where

Ve1 =

√1

18[3(V 2

a1 + V 2b1 + V 2

c1) + (V 2ab1 + V 2

bc1 + V 2ca1)]

VeH =

√1

18[3(V 2

aH + V 2bH + V 2

cH) + (V 2abH + V 2

bcH + V 2caH)] (2.121)

For three-phase three-wire system, In = 0 = In1 = InH .

59

Page 66: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Ie1 =

√I2a1 + I2

b1 + I2c1

3

IeH =

√I2aH + I2

bH + I2cH

3(2.122)

Similarly

Ve1 =

√V 2ab1 + V 2

bc1 + V 2ca1

9

VeH =

√V 2abH + V 2

bcH + V 2caH

9(2.123)

The expression for effective apparent power Se is given as following.

Se = 3VeIe

= 3√V 2e1 + V 2

eH

√I2e1 + I2

eH

=√

9V 2e1I

2e1 + (9V 2

e1I2eH + 9V 2

eHI2e1 + 9V 2

eHI2eH)

=√S2e1 + S2

eN (2.124)

In the above equation,

Se1 = 3Ve1Ie1 (2.125)

SeN =√S2e − S2

e1

=√D2eV +D2

eI + S2eH

=√

3(I2e1V

2eH) + 3(V 2

e1I2eH) + 3(V 2

eHI2eH) (2.126)

In equation (2.126), distortion powers DeI , DeV and harmonic apparent power SeH are given asfollowing.

DeI = 3Ve1IeH

DeV = 3VeHIe1 (2.127)SeH = 3VeHIeH

By defining above effective voltage and current quantities, the effective total harmonic distortion(THDe) are expressed below.

THDeV =VeHVe1

THDeI =IeHIe1

(2.128)

Substituting VeH and IeH in (2.126),

SeN = Se1

√THD2

e1 + THD2eV + THD2

eITHD2eV . (2.129)

60

Page 67: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

In above equation,

DeI = Se1THDI

DeV = Se1THDV (2.130)SeH = Se1(THDI)(THDV ).

Using (2.124) and (2.129), the effective apparent power is given as below.

Se =√S2e1 + S2

eN = Se1

√1 + THD2

eV + THD2eI + THD2

eV THD2eI (2.131)

Based on above equation, the effective power factor is therefore given as,

pfe =P

Se=

P1 + PH

Se1√

1 + THD2eV + THD2

eI + THD2eV THD

2eI

=(1 + PH/P1)√

1 + THD2eV + THD2

eI + THD2eV THD

2eI

P1

Se1

=(1 + PH/P1)√

1 + THD2eV + THD2

eI + THD2eV THD

2eI

pfe1 (2.132)

Practically, the THDs in voltage are far less than those of currents THDs, therefore THDeV <<THDeI . Using this practical constraint and assuming PH << P1, the above equation can besimplified to,

pfe ≈pfe1√

1 + THD2eI

(2.133)

In the above context, their is another useful term to denote unbalance of the system. This isdefined as fundamental unbalanced power and is given below.

SU1 =√S2e1 − (S+

1 )2 (2.134)

Where, S+1 is fundamental positive sequence apparent power, which is given below.

S+1 =

√(P+

1 )2 + (Q+1 )2 (2.135)

In above, P+1 = 3V +

1 I+1 cosφ+

1 and Q+1 = 3V +

1 I+1 sinφ+

1 . Fundamental positive sequence powerfactor can thus be expressed as a ratio of P+

1 and S+1 as given below.

P+f1 =

P+1

S+1

(2.136)

Example 2.3 Consider the following three-phase system. It is given that voltages V a, V b and V c

are balanced sinusoids with rms value of 220 V. The feeder impedance is rf +jxf = 0.02+j0.1 Ω.The unbalanced load parameters are: RL = 12 Ω and XL = 13 Ω. Compute the following.

a. The currents in each phase, i.e., Ia, Ib and Ic and neutral current, In.

61

Page 68: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

f fr jxav

bv

cv

nv

aV

bV

cV

nV

LX

LR

aI

bI

cI

LOA

D

nI

Fig. 2.11 An unbalanced three-phase circuit

b. Losses in the system.

c. The active and reactive powers in each phase and total three-phase active and reactive powers.

d. Arithmetic, vector and effective apparent powers and power factors based on them.

Solution:

a. Computation of currents

va (t) = 220√

2 sin (ωt)

vb (t) = 220√

2 sin (ωt− 120)

vc (t) = 220√

2 sin (ωt+ 120)

vab (t) = 220√

6 sin (ωt+ 30)

Therefore,

Ia =220√

3∠30

13∠90= 29.31∠−60A

Ib = −Ia = −29.311∠−60 = 29.31∠120A

Ic =220∠120

12= 18.33∠120A.

Thus, the instantaneous expressions of phase currents can be given as following.

ia(t) = 41.45 sin (ωt− 60)

ib(t) = −ia(t) = −41.45 sin (ωt− 60) = 41.45 sin (ωt+ 120)

ic(t) = 25.93 sin (ωt+ 120)

b. Computation of losses

62

Page 69: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The losses occur due to resistance of the feeder impedance. These are computed as below.

Losses = rf (I2a + I2

b + I2c + I2

n)

= 0.02 (29.312 + 29.312 + 18.332 + 18.332) = 47.80 W

c. Computation of various powers

Phase-a active and reactive power:

Sa = V a I∗a = 220∠0 × 29.31∠60 = 3224.21 + j5584.49

implies that, Pa = 3224.1 W, Qa = 5584.30 VAr

Similarly,

Sb = V b I∗b = 220∠−120 × 29.31∠60 = −3224.21 + j5584.49

implies that, Pb = −3224.1 W, Qb = 5584.30 VAr

For phase-c,

Sc = V c I∗c = 220∠120 × 18.33∠−120 = 4032.6 + j0

implies that, Pc = 4032.6 W, Qc = 0 VAr

Total three-phase active and reactive powers are given by,

P3−phase = Pa + Pb + Pc = 3224.1− 3224.1 + 4032.6 = 4032.6 WQ3−phase = Qa +Qb +Qc = 5584.30 + 5584.30 + 0 = 11168.60 VAr.

d. Various apparent powers and power factors

The arithmetic, vector and effective apparent powers are computed as below.

SA = |Sa|+ |Sb|+ |Sc|= 6448.12 + 6448.12 + 4032.6 = 16928.84 VA

Sv = |Sa + Sb + Sc|= |4032.6 + j11168.6| = |11874.32∠70.14| = 11874.32 VA

Se = 3VeIe = 3× 220×√I2a + I2

b + I2c + I2

n

3

= 3× 220×√

29.312 + 29.312 + 18.332 + 18.332

3= 3× 220× 28.22

= 18629.19 VA

63

Page 70: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Based on the above apparent powers, the arithmetic, vector and effective apparent power factorsare computed as below.

pfA =P3−phase

SA=

4032.6

16928.84= 0.2382

pfv =P3−phase

Sv=

4032.6

11874.32= 0.3396

pfe =P3−phase

Se=

4032.6

18629.19= 0.2165

In the above computation, the effective voltage and current are found as given in the following.

Ve =

√V 2a + V 2

b + V 2c

3= 220 V

Ie =

√I2a + I2

b + I2c + I2

n

3= 28.226 A

Example 2.4 A 3-phase, 3-wire system is shown in Fig. 2.12. The 3-phase voltages are balancedsinusoids with RMS value of 230 V. The 3-phase loads connected in star are given as following.Za = 5 + j12 Ω, Zb = 6 + j8 Ω and Zc = 12− j5 Ω.

Compute the following.

a. Line currents, i.e., I la, I lb and I lc and their instantaneous expressions.

b. Load active and reactive powers and power factor of each phase.

c. Compute various apparent powers and power factors based on them.

Na

c bZ

laI

lbI

lcI

saV

scV

sbV

Fig. 2.12 A star connected three-phase unbalanced load

Solution:

a. Computation of currents

64

Page 71: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Given that Za = 5 + j 12 Ω, Zb = 6 + j 8 Ω, Zc = 12− j 5 Ω.

V sa = 230∠0VV sb = 230∠−120VV sc = 230∠120V

V nN =1

1Za

+ 1Zb

+ 1Zc

(V sa

Za+V sb

Zb+V sc

Zc

)=

11

5+j12+ 1

6+j8+ 1

12−j5

(230∠0

5 + j12+

230∠−120

6 + 8j+

230∠120

12− j5

)=

1

0.2013∠−37.0931.23∠−164.50

= −94.22− j123.18 = 155.09∠−127.41V

Now the line currents are computed as below.

Ial =V sa − V nN

Za=

230∠0 − 155.09∠−127.41

5 + j12= 26.67∠−46.56A

Ibl =V sb − V nN

Zb=

230∠−120 − 155.09∠−127.41

6 + j8= 7.88∠−158.43A

Icl =V sc − V nN

Zc=

230∠120 − 155.09∠−127.41

12− j5= 24.85∠116.3A

Thus, the instantaneous expressions of line currents can be given as following.

ial (t) = 37.72 sin (ωt− 46.56)

ibl (t) = 11.14 sin (ωt− 158.43)

icl (t) = 35.14 sin (ωt+ 116.3)

b. Computation of load active and reactive powers

Sa = V aI∗a = 230∠0 × 26.67∠46.56 = 4218.03 + j4456.8

Sb = V bI∗b = 230∠−120 × 7.88∠158.43 = 1419.82 + j1126.06

Sc = V cI∗c = 230∠120 × 24.85∠−116.3 = 5703.43 + j368.11

implies that,Pa = 4218.03 W, Qa = 4456.8 VArPb = 1419.82 W, Qb = 1126.06 VArPc = 5703.43 W, Qc = 368.11 VAr

65

Page 72: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Total three-phase active and reactive powers are given by,

P3−phase = Pa + Pb + Pc = 4218.03 + 1419.82 + 5703.43 = 11341.29 WQ3−phase = Qa +Qb +Qc = 4456.8 + 1126.06 + 368.11 = 5950.99 VAr.

The power factors for phases a, b and c are given as follows.

pfa =Pa

|Sa|=

4218.03√4218.032 + 4456.82

=4218.03

6136.3= 0.6873 (lag)

pfb =Pb

|Sb|=

1419.82

1419.822 + 1126.062=

1419.82

1812.16= 0.7835 (lag)

pfc =Pc

|Sc|=

5703.43

5703.432 + 368.112=

5703.43

5715.30= 0.9979 (lag)

c. Computation of various apparent powers and power factors

The arithmetic, vector and effective apparent powers are computed as below.

SA = |Sa|+ |Sb|+ |Sc|= 6136.3 + 1812.16 + 5715.30 = 13663.82 VA

Sv = |Sa + Sb + Sc|= |11341.29 + j5909.92| = 12807.78 VA

Se = 3VeIe = 3× 230×√I2la + I2

lb + I2lc + I2

ln

3

= 3× 220×√

26.672 + 7.882 + 24.852 + 02

3= 3× 230× 21.53

= 14859.7 VA

The arithmetic, vector and effective apparent power factors are computed as below.

pfA =P3−phase

SA=

11341.29

13663.82= 0.8300

pfv =P3−phase

Sv=

11341.29

12807.78= 0.8855

pfe =P3−phase

Se=

11341.29

14859.7= 0.7632

References

[1] IEEE Group, “IEEE trial-use standard definitions for the measurement of electric power quan-tities under sinusoidal, nonsinusoidal, balanced, or unbalanced conditions,” 2000.

66

Page 73: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

[2] E. Watanabe, R. Stephan, and M. Aredes, “New concepts of instantaneous active and reactivepowers in electrical systems with generic loads,” IEEE Transactions on Power Delivery, vol. 8,no. 2, pp. 697–703, 1993.

[3] T. Furuhashi, S. Okuma, and Y. Uchikawa, “A study on the theory of instantaneous reactivepower,” IEEE Transactions on Industrial Electronics, vol. 37, no. 1, pp. 86–90, 1990.

[4] A. Ferrero and G. Superti-Furga, “A new approach to the definition of power components inthree-phase systems under nonsinusoidal conditions,” IEEE Transactions on Instrumentationand Measurement, vol. 40, no. 3, pp. 568–577, 1991.

[5] J. Willems, “A new interpretation of the akagi-nabae power components for nonsinusoidalthree-phase situations,” IEEE Transactions on Instrumentation and Measurement, vol. 41,no. 4, pp. 523–527, 1992.

[6] H. Akagi, Y. Kanazawa, and A. Nabae, “Instantaneous reactive power compensators compris-ing switching devices without energy storage components,” IEEE Transactions on IndustryApplications, no. 3, pp. 625–630, 1984.

[7] C. L. Fortesque, “Method of symmetrical co-ordinates applied to the solution of polyphasenetworks,” AIEE, 1918.

67

Page 74: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

68

Page 75: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Chapter 3

FUNDAMENTAL THEORY OF LOADCOMPENSATION(Lectures 19-24)

3.1 Introduction

In general, the loads which cause fluctuations in the supply voltage due to poor power factor,unbalanced and harmonics, d.c components require compensation. Typical loads requiring com-pensation are arc furnaces, induction furnaces, arc welders, steel rolling mills, winders, very largemotors, which start and stop frequently, high energy physics experiments,which require pulse highpower supplies. All these loads can be classified into three basic categories.

1. Unbalanced ac load2. Unbalanced ac + non linear load3. Unbalanced ac + nonlinear ac + dc component load.

The dc component is generally caused by the usage of have wave rectifiers. These loads, par-ticularly nonlinear loads generate harmonics as well as fundamental frequency voltage variations.For example arc furnaces generate significant amount of harmonics at the load bus.Other serious loads which degrade power quality are adjustable speed drives which include powerelectronic circuitry, all power electronics based converters such as thyristor controlled drives, rec-tifiers, cyclo converters etc.. In general, following aspects are important, while we do provide theload compensation in order to improve the power quality [1].

1. Type of Load (unbalance , harmonics and dc component)2. Real and Reactive power requirements (maximum, minimum and concurrence of maximum realand reactive power requirements in multiple loads)3. Rate of change of real and reactive power etc.

In this unit, we however, discuss fundamental load compensation techniques for unbalanced linearloads such as combination of resistance, inductance and capacitance and their combinations. The

69

Page 76: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

objective here will be to maintain currents balanced and unity factor with their voltages.

3.2 Fundamental Theory of Load Compensation

We shall find some fundamental relation ship between supply system the load and the compensator.We shall start with the principle of power factor correction, which in its simplest form, and can bestudied without reference to supply system [2]–[5].

The supply system, the load and the compensator can be modeled in various ways. The supplysystem can be modeled as a Thevenin’s equivalent circuit with an open circuit voltage and a se-ries impedance, (its current or power and reactive power) requirements. The compensator can bemodeled as variable impedance or as a variable source (or sink) of reactive current. The choice ofmodel varied according to the requirements. The modeling and analysis done here is on the basisof steady state and phasor quantities are used to note the various parameters in system.

3.2.1 Power Factor and its Correction

Consider a single phase system shown in 3.1(a) shown below. The load admittance is represented

VsI

lI

VRI

lIXI

l

Fig. 3.1 (a) Single line diagram of electrical system (b) Phasor diagram

by Yl = Gl + jBl supplied from a load bus at voltage V = V ∠0. The load current is I l is given as,

I l = V (Gl + jBl) = V Gl + jV Bl (3.1)= IR + jIX

According to the above equation, the load current has a two components, i.e. the resistive or inphase component and reactive component or phase quadrature component and are represented byIR and IX respectively. The current, IX will lag 90o for inductive load and it will lead 90o fromthe reference voltage. This is shown in 3.1(b). The load apparent power can be expressed in terms

70

Page 77: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

of bus voltage V and load current Il as given below.

Sl = V (Il)∗

= V (IR + j IX)∗

= V (IR − j IX)

= V (Il cosφl − j Il sinφl)= V Il cosφl − j V Il sinφl= Sl cosφl − j Sl sinφl (3.2)

From (3.1), I l = V (Gl + jBl) = V Gl + jV Bl, equation (3.2) can also be written as following.

Sl = V (Il)∗

= V (V Gl + jV Bl)∗

= V (V Gl − jV Bl)

= V 2Gl − j V 2Bl

= Pl + jQl (3.3)

From equation (3.129), load active (Pl) and reactive power (Ql) are given as,

Pl = V 2Gl

Ql = −V 2Bl (3.4)

Now suppose a compensator is connected across the load such that the compensator current, Iγ isequal to −IX , thus,

Iγ = V Yγ = V (Gγ + jBγ) = −IX= −j V Bl (3.5)

The above condition implies that Gγ = 0 and Bγ = −Bl. The source current Is, can thereforegiven by,

Is = I l + Iγ = IR (3.6)

Therefore due to compensator action, the source supplies only in phase component of the loadcurrent. The source power factor is unity. This reduces the rating of the power conductor andlosses due to the feeder impedance. The rating of the compensator is given by the followingexpression.

Sγ = Pγ + j Qγ = V (Iγ)∗

= V (−j V Bl)∗

= jV 2Bl (3.7)

Using (3.4), the above equation indicates the Pγ = 0 and Qγ = −Ql. This is an interestinginference that the compensator generates the reactive power which is equal and opposite to theload reactive and it has no effect on active power of the load. This is shown in Fig. 3.2.

71

Page 78: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

XI

I

s RI IVsI

lIIlI

l

XI

Fig. 3.2 (a) Single line diagram of compensated system (b) Phasor diagram

Using (3.2) and (3.7), the compensator rating can further be expressed as,

Qγ = −Ql = −Sl sinφl = −Sl√

1− cos2 φl VArs (3.8)

From (3.8),

|Qγ| =√

1− cos2 φl (3.9)

If |Qγ| < |Ql| or |Bγ| < |Bl|, then load is partially compensated. The compensator of fixedadmittance is incapable of following variations in the reactive power requirement of the load. Inpractical however a compensator such as a bank of capacitors can be divided into parallel sections,each of switched separately, so that discrete changes in the reactive power compensation can bemade according to the load. Some sophisticated compensators can be used to provide smooth anddynamic control of reactive power.

Here voltage of supply is being assumed to be constant. In general if supply voltage varies, theQγ will not vary separately with the load and compensator error will be there. In the followingdiscussion, voltage variations are examined and some additional features of the ideal compensatorwill be studied.

3.2.2 Voltage Regulation

Voltage regulation can be defined as the proportional change in voltage magnitude at the load busdue to change in load current (say from no load to full load). The voltage drop is caused due tofeeder impedance carrying the load current as illustrated in Fig. 3.3(a). If the supply voltage isrepresented by Thevenin’s equivalent, then the voltage regulation (VR) is given by,

V R =

∣∣E∣∣− ∣∣V ∣∣∣∣V ∣∣ =

∣∣E∣∣− |V ||V |

(3.10)

for V being a reference phasor.In absence of compensator, the source and load currents are same and the voltage drop due to the

72

Page 79: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

feeder is given by,

∆V = E − V = ZsIl (3.11)

The feeder impedance, Zs = Rs + jXs. The relationship between the load powers and its voltageand current is expressed below.

Sl = V (I l)∗ = Pl + jQl (3.12)

Since V = V , the load current is expressed as following.

I l =Pl − jQl

V(3.13)

Substituting, Il from above equation into (3.11), we get

∆V = E − V = (Rs + jXs)

(Pl − jQl

V

)=

RsPl +XsQl

V+ j

XsPl −RsQl

V= ∆VR + j∆VX (3.14)

Thus, the voltage drop across the feeder has two components, one in phase (∆VR) and another isin phase quadrature (∆VX). This is illustrated in Fig. 3.3(b).

E

V RV

Xj V

lIlR I l ljX I

l

lI l l lS P jQ

l l lY G jB

Load

sI

V

E

Fee

der

s s sZ R jX

Fig. 3.3 (a) Single phase system with feeder impedance (b) Phasor diagram

From the above it is evident that load bus voltage (V ) is dependent on the value of the feederimpedance, magnitude and phase angle of the load current. In other words, voltage change (∆V )depends upon the real and reactive power flow of the load and the value of the feeder impedance.

Now let us add compensator in parallel with the load as shown in Fig. 3.4(a). The question is:whether it is possible to make

∣∣E∣∣ =∣∣V ∣∣, in order to achieve zero voltage regulation irrespective of

change in the load. The answer is yes, if the compensator consisting of purely reactive components,has enough capacity to supply to required amount of the reactive power. This situation is shownusing phasor diagram in Fig. 3.4(b).The net reactive at the load bus is now Qs = Qγ + Ql. The compensator reactive power (Qγ) hasto be adjusted in such a way as to rotate the phasor ∆V until

∣∣E∣∣ =∣∣V ∣∣.

73

Page 80: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

E

V

lI

l

lI l l lS P jQ

l l lY G jB

Load

sI V

E

Fee

der

s s sZ R jX

Co

mp.

I

Q

VI sI

s sR I

s sjX I

Fig. 3.4 (a) Voltage with compensator (b) Phasor diagram

From (3.14) and Fig. 3.3(b),

E∠δ =

(V +

Rs Pl +XsQs

V

)+ j

(Xs Pl −RsQs

V

)(3.15)

The above equation implies that,

E2 =

(V +

Rs Pl +XsQs

V

)2

+

(Xs Pl −RsQs

V

)2

(3.16)

The above equation can be simplified to,

E2V 2 = (V 2 +Rs Pl)2 +X2

s Q2s + 2(V 2 +Rs Pl)XsQs

+X2s P

2l +R2

s Q2s −(((((

((2Xs PlRsQs (3.17)

Above equation, rearranged in the powers of Qs, is written as following.

(R2s +X2

s )Q2s + 2V 2XsQs + (V 2 +Rs Pl)

2 + (Xs Pl)2 − E2 V 2 = 0 (3.18)

Thus the above equation is quadratic inQs and can be represented using coefficients ofQs as givenbelow.

aQ2s + bQs + c = 0 (3.19)

Where a = R2s +X2

s , b = 2V 2Xs and c = (V 2 +Rs Pl)2 +X2

s P2l − E2 V 2.

Thus the solution of above equation is as following.

Qs =−b±

√(b2 − 4ac)

2a(3.20)

In the actual compensator, this value would be determined automatically by control loop. Theequation also indicates that, we can find the value of Qs by subjecting a condition such as E = V

74

Page 81: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

irrespective of the requirement of the load powers (Pl, Ql). This leads to the following conclusionthat a purely reactive compensator can eliminate supply voltage variation caused by changes inboth the real and reactive power of the load, provided that there is sufficient range and rate of Qs

both in lagging and leading pf. This compensator therefore acts as an ideal voltage regulator. Itis mentioned here that we are regulating magnitude of voltage and not its phase angle. In fact itsphase angle is continuously varying depending upon the load current.

It is instructive to consider this principle from different point of view. We have seen that com-pensator can be made to supply all load reactive power and it acts as power factor correction device.If the compensator is designed to compensate power factor, then Qs = Ql + Qγ = 0. This im-plies that Qγ = −Ql. Substituting Qs = 0 for Ql in (3.14) to achieve this condition, we get thefollowing.

∆V =(Rs + jXs)

VPl (3.21)

From above equation, it is observed that ∆V is independent of Ql. Thus we conclude that a purelyreactive compensator cannot maintain both constant voltage and unity power factor simultaneously.Of course the exception to this rule is a trivial case when Pl = 0.

3.2.3 An Approximation Expression for the Voltage Regulation

Consider a supply system with short circuit capacity (Ssc) at the load bus. This short circuit capac-ity can be expressed in terms of short circuit active and reactive powers as given below.

Ssc = Psc + jQsc = E I∗sc = E

(E

Zsc

)∗=E2

Z∗sc(3.22)

Where Zsc = Rs + jXs and Isc is the short circuit current. From the above equation

|Zsc| =E2

Ssc

Therefore, Rs =E2

Ssccosφsc

Xs =E2

Sscsinφsc

tanφsc =Xs

Rs

(3.23)

Substituting above values of Rs and Xs, (3.14) can be written in the following form.

∆V

V=

(Pl cosφsc +Ql sinφsc

V 2+ j

Pl sinφsc −Ql cosφscV 2

)E2

Ssc

∆V

V=

∆VRV

+ j∆VXV

(3.24)

75

Page 82: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Using an approximation that E ≈ V , the above equation reduces to the following.

∆V

V=

(Pl cosφsc +Ql sinφsc

Ssc+ j

Pl sinφsc −Ql cosφscSsc

)(3.25)

The above implies that,

∆VRV≈ Pl cosφsc +Ql sinφsc

Ssc∆VXV≈ Pl sinφsc −Ql cosφsc

Ssc

Often (∆VX/V ) is ignored on the ground that the phase quadrature component contributes negli-gible to the magnitude of overall phasor. It mainly contributes to the phase angle. Therefore theequation (3.25) is simplified to the following.

∆V

V=

∆VRV

=Pl cosφsc +Ql sinφsc

Ssc(3.26)

Implying that the major change in voltage regulation occurs due to in phase component, ∆VR.Although approximate, the above expression is quite useful in terms of short circuit level (Ssc),(Xs/Rs, active and reactive power of the load.

On the basis of incremental changes in active and reactive powers of the load, i.e., 0 → Pl and0→ Ql, the above equation can further be written as,

∆V

V=

∆VRV

=∆Pl cosφsc + ∆Ql sinφsc

Ssc. (3.27)

Further, feeder reactance (Xs) is far greater than feeder resistance (Rs), i.e., Xs >> Rs. Thisimplies that, φsc → 90o, sinφsc → 1 and cosφsc → 0. Using this approximation the voltageregulation is given as following.

∆V

V≈ ∆VR

V≈ ∆Ql

Sscsinφsc ≈

∆Ql

Ssc. (3.28)

That is, per unit voltage change is equal to the ratio of the reactive power swing to the short circuitlevel of the supply system. Representing ∆V approximately by E − V , the equation (3.28) can bewrittwn as,

E − VV

≈ Ql

Ssc. (3.29)

The above leads to the following expression,

V ' E

(1 + ∆QlSsc

)' E(1− Ql

Ssc) (3.30)

76

Page 83: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

with the assumption that, Ql/Ssc << 1. Although above relationship is obtained with approxi-mations, however it is very useful in visualizing the action of compensator on the voltage. Theabove equation is graphically represented as Fig. 3.5. The nature of voltage variation is droopingwith increase in inductive reactive power of the load. This is shown by negative slope −E/Ssc asindicated in the figure.

The above characteristics also explain that when load is capacitive, Ql is negative. This makesV > E. This is similar to Ferranti effect due to lightly loaded electric lines.

0lQ

V

sc

E

S

E

Fig. 3.5 Voltage variation with reactive power of the load

Example 3.1 Consider a supply at 10 kV line to neutral voltage with short circuit level of 250MVA and Xs/Rs ratio of 5, supplying a star connected load inductive load whose mean power is25 MW and whose reactive power varies from 0 to 50 MVAr, all quantities per phase.

(a) Find the load bus voltage (V ) and the voltage drop (∆V ) in the supply feeder. Thus determineload current (I l), power factor and system voltage (E ).

(b) It is required to maintain the load bus voltage to be same as supply bus voltage i.e. V =10 kV.Calculate reactive power supplies by the compensator.

(c) What should be the load bus voltage and compensator current if it is required to maintain theunity power factor at the supply?

Solution: The feeder resistance and reactance are computed as following.Zs = E2

s/Ssc = (10 kV)2/250 = 0.4 Ω/phaseIt is given that, Xs/Rs = tanφsc = 5, therefore φsc = tan−1 5 = 78.69o. From this,

Rs = Zs cosφsc = 0.4 cos(78.69o) = 0.0784 Ω

Xs = Zs sinφsc = 0.4 sin(78.69o) = 0.3922 Ω

(a) Without compensation Qs = Ql, Qγ = 0To know ∆V , first the voltage at the load bus has to be computed. This is done by rearranging

77

Page 84: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

(3.18) in powers of voltage V . This is given below.

(R2s +X2

s )Q2l + 2V 2XsQl + (V 2 +Rs Pl)

2 +X2s P

2l − E2 V 2 = 0

(R2s +X2

s )Q2l︸ ︷︷ ︸

III

+ 2V 2XsQl︸ ︷︷ ︸II

+ V 4︸︷︷︸I

+R2s P

2l︸ ︷︷ ︸

III

+ 2V 2Rs Pl︸ ︷︷ ︸II

+X2s P

2l︸ ︷︷ ︸

III

−E2 V 2︸ ︷︷ ︸II

= 0

Combining the I, II and III terms in the above equation, we get the following.

V 4 +

2(RsPl +XsQl)− E2V 2 + (R2

s +X2s )(Q2

l + P 2l ) = 0 (3.31)

Now substituting values of Rs, Xs, Pl, Ql and E in above equation, we get,

V 4 +

2 [0.0784× 25 + 0.3922× 50]− 102V 2 + (0.07842 + 0.39222)(252 + 502) = 0

After simplifying the above, we have the following equation.

V 4 − 56.86V 2 + 500 = 0

Therefore

V 2 =56.86±

√56.86− 4× 500

2= 45.985, 10.875

and V = ±6.78 kV, ±3.297 kV

Since rms value cannot be negative and maximum rms value must be a feasible solution, thereforeV = 6.78 kV.Now we can compute ∆V using (3.14), as it is given below.

∆V =Rs Pl +XsQl

V+ j

Xs Pl −RsQl

V

=0.0784 25 + 0.392 50

6.78+ j

0.3922 25− 0.0784 50

6.78= 3.1814 + j0.8677 kV = 3.2976∠15.25o kV

Now the line current can be found out as following.

I l =Pl −Ql

V=

25− j50

6.782= 3.86− j7.3746 kA= 8.242∠− 63.44o kA

The power factor of load is cos (tan−1(Ql/Pl)) = 0.4472 lagging. The phasor diagram for thiscase is similar to what is shown in Fig. 3.3(b).

(b) Compensator as a voltage regulator

Now it is required to maintain V = E = 10.0 kV at the load bus. For this let their be reac-tive power Qγ supplied by the compensator at the load bus. Therefore the net reactive power at theload bus is equal to Qs, which is given below.

Qs = Ql +Qγ

78

Page 85: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Thus from (3.18), we get,

(R2s +X2

s )Q2s + 2V 2XsQs + (V 2 +RsPl)

2 +X2sP

2l − E2V 2 = 0

(0.7842 + 0.39222)2Q2s + 2× 102 × 0.3922×Qs +

(102 + 0.784× 25)2 + 0.39222 × 252 − 104

= 0

From the above we have,

0.16Q2s + 78.44Qs + 491.98 = 0.

Solving the above equation we get,

Qs =−78.44±

√78.442 − 4× 0.16× 491.98

2× 0.16= −6.35 or− 484 MVAr.

The feasible solution is Qs = −6.35 MVAr because it requires less rating of the compensator.Therefore the reactive power of the compensator (Qγ) is,

Qγ = Qs −Ql = −6.35− 50 = −56.35 MVAr.

With Qs = −6.35 MVAr, the ∆V is computed by replacing Qs for Ql in (3.14) as given below.

∆V =RsPl +XsQs

V+ j

XsPl −RsQs

V

=0.0784× 25 + 0.39225×−6.35

10+ j

0.39225× 25− 0.0784× (−6.35)

10

=1.96− 2.4

10+ j

9.805 + 0.4978

10= −0.0532 + j1.030 kV = 1.03137∠92.95o kV

Now, we can find supply voltage E as given below.

E = ∆V + V

= 10− 0.0532 + j1.030

= 9.9468 + j1.030 = 10∠5.91o kV

The supply current is,

Is =Pl − jQs

V=

25− j(−6.35)

10= 2.5 + j0.635 kA = 2.579∠14.25o kA.

This indicates that power factor is not unity for perfect voltage regulation i.e., E = V . For thiscase the compensator current is given below.

Iγ =−jQγ

V=−j(−56.35)

10Iγ = j5.635 kA

79

Page 86: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The load current is computed as below.

I l =Pl − jQl

V=

25− j50

10= 2.5− j5.0 = IlR + jIlX = 5.59∠63.44o kA

The phasor diagram is similar to the one shown in Fig. 3.4(b). The phasor diagram shown hasinteresting features. The voltage at the load bus is maintained to 1.0 pu. It is observed that thereactive power of the compensator Qγ is not equal to load reactive power (Ql). It exceeds by 6.35MVAr. As a result of this compensation, the voltage regulation is perfect, however power factor isnot unity. The phase angle between V and Is is cos−1 0.969 = 14.25o as computed above. There-fore the angle between E and Is is (14.25o − 5.91o = 8.34o). Thus, source power factor (φs) iscos(8.340) = 0.9956 leading.

(c) Compensation for unity power factor

To achieve unity power factor at the load bus, the condition Qγ = −Ql must be satisfied, whichfurther implies that the net reactive power at the load bus is zero. Therefore substituting Ql = 0 in(3.31), we get the following.

V 4 +

2(RsPl − E2)V 2 + (R2

s +X2s )(P 2

l +Q2l ) = 0

V 4 + (2× 0.0784× 25− 102)V 2 + (0.07842 + 0.39222) 252 = 0

From the above,

V 4 + 96.08V 2 + 99.79 = 0

The solution of the above equation is,

V 2 =96.08± 93.97

2= 95.02, 1.052

V = ±9.747 kV,±1.0256 kV.

Since rms value cannot be negative and maximum rms value must be a feasible solution, thereforeV = 9.747 kV. Thus it is seen that for obtaining unity power factor at the load bus does not ensuredesired voltage regulation. Now the other quantities are computed as given below.

I l =Pl − jQ

V=

25− j50

9.747= 2.5648− j5.129 = 5.7345∠−63.43o kA

SinceQγ = −Ql, this implies that Iγ = −jQγ/V = jQl/V = j5.129 kA. The voltage drop acrossthe feeder is given as following.

∆V =Rs Pl +XsQl

V+ j

Xs Pl −RsQl

V

=(0.784× 25 + j0.3922× 25)

9.747= 0.201 + j1.005 = 1.0249∠5.01o kV

80

Page 87: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The phasor diagram for the above case is shown in Fig. 3.6.

5.1

3kA

Ij

10 kVE

V =9.75 kV

5.1

3kA

lXIj

= 2.56 kAlRI RV

XV

5.73 63.43 kAlI

5.77

sI

Fig. 3.6 Phasor diagram for system with compensator in voltage regulation mode

The percentage voltage change = (10 − 9.748)/10 × 100 = 2.5. Thus we see that power factorimproves voltage regulation enormously compared with uncompensated case. In many cases, de-gree of improvement is adequate and the compensator can be designed to provide reactive powerrequirement of load rather than as a ideal voltage regulator.

3.3 Some Practical Aspects of Compensator used as Voltage Regulator

In this section, some practical aspects of the compensator in voltage regulation mode will be dis-cussed. The important parameters of the compensator which play significant role in obtainingdesired voltage regulation are: Knee point (V k), maximum or rated reactive power Qγmax and thecompensator gain Kγ .

The compensator gain Kγ is defined as the rate of change of compensator reactive power Qγ

with change in the voltage (V ), as given below.

Kγ =dQr

dV(3.32)

For linear relationship between Qγ and V with incremental change, the above equation be writtenas the following.

∆Qγ = ∆V Kγ (3.33)

81

Page 88: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Assuming compensator characteristics to be linear with Qγ ≤ Qγmax limit, the voltage can berepresented as,

V = Vk +Qγ

(3.34)

This is re-written as,

Qγ = Kγ(V − Vk) (3.35)

Flat V-Q characteristics imply that Kγ → ∞. That means the compensator which can absorb orgenerate exactly right amount of reactive power to maintain supply voltage constant as the loadvaries without any constraint. We shall now see the regulating properties of the compensator,when compensator has finite gain Kr operating on supply system with a finite short circuit level,Ssc. The further which are made in the following study are: high Xs/Rs ratio and negligible loadpower fluctuations. The net reactive power at the load bus is sum of the load and the compensatorreactive power as given below.

Ql +Qγ = Qs (3.36)

Using earlier voltage and reactive power relationship from equation (3.30), it can be written as thefollowing.

V ' E(1− Qs

Ssc) (3.37)

The compensator voltage represented by (3.34) and system voltage represented by (3.37) are shownin Fig. 3.7(a) and (b) respectively.

kV

Q

k

QV V

K

V

E

sQ

1- s

SC

QV E

S

(a) (b)

Fig. 3.7 (a) Voltage characterstics of compensator (b) System voltage characteristics

Differentiating V with respect to Qs, we get, intrinsic sensitivity of the supply voltage withvariation in Qs as given below.

dV

dQs

= − E

Ssc(3.38)

82

Page 89: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

It is seen from the above equation that high value of short circuit level Ssc reduces the voltagesensitivity, making voltage variation flat irrespective of Ql. With compensator replacing Qs =Qγ +Ql in (3.37), we have the following.

V ' E

(1− Ql +Qγ

Ssc

)(3.39)

Substituting Qγ from (3.35), we get the following equation.

V ' E

[1 +KγVk/Ssc1 + EKγ/Ssc

− Ql/Ssc1 + EKγ/Ssc

](3.40)

Although approximate, above equation gives the effects of all the major parameters such as loadreactive power, the compensator characteristics Vγ and Kγ and the system characteristics E andSsc. As we discussed, V-Q characteristics is flat for high or infinite value Kγ . However the highervalue of the gain Kγ means large rating and quick rate of change of the reactive power with varia-tion in the system voltage. This makes cost of the compensator high.

The compensator has two effects as seen from (3.40), i.e., it alters the no load supply voltage(E) and it modifies the sensitivity of supply point voltage to the variation in the load reactive power.Differentiating (3.40) with respect to Ql, we get,

dV

dQl

= − E/Ssc1 +KrE/Ssc

(3.41)

which is voltage sensitivity of supply point voltage to the load reactive power. It can be seen thatthe voltage sensitivity is reduced as compared to the voltage sensitivity without compensator asindicated in (3.38).

It is useful to express the slope (−E/Ssc) by a term in a form similar to Kγ = dQγ/dV , as givenbelow.

Ks = −SscE

Thus,1

Ks

= − E

Ssc(3.42)

Substituting V from (3.39) into (3.35), the following is obtained.

Qγ = Kγ

[E

(1− Ql +Qγ

Ssc

)− Vk

](3.43)

Collecting the coefficients of Qγ from both sides of the above equation, we get

Qγ =Kγ

1 +Kγ (E/Ssc)

[E

(1− Ql

Ssc

)− Vk

](3.44)

83

Page 90: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Setting knee voltage Vk of the compensator equal to system voltage E i.e., Vk = E, the aboveequation is simplified to,

Qγ = − Kγ (E/Ssc)

1 +Kγ (E/Ssc)Ql

= −(

Kγ/Ks

1 +Kγ/Ks

)Ql. (3.45)

From the above equation it is observed that, when compensator gain Kγ → ∞, Qγ → −Ql. Thisindicates perfect compensation of the load reactive power in order to regulate the load bus voltage.

Example 3.2 Consider a three-phase system with line-line voltage 11 kV and short circuit capacityof 480 MVA. With compensator gain of 100 pu determine voltage sensitivity with and withoutcompensator. For each case, if a load reactive power changes by 10 MVArs, find out the change inload bus voltage assuming linear relationship between V-Q characteristics. Also find relationshipbetween compensator and load reactive powers.

Solution: The voltage sensitivity can be computed using the following equation.

dV

dQl

= − E/Ssc1 +KγE/Ssc

Without compensator Kγ = 0, E = (11/√

3) = 6.35 kV and Ssc = 480/3 = 160 MVA.Substituting these values in the above equation, the voltage sensitivity is given below.

dV

dQl

= − 6.35/160

1 + 0× 6.35/160= −0.039

The change in voltage due to variation of reactive power by 10 MVArs, ∆V = −0.039 × 10 =−0.39 kV.With compensator, Kγ = 100

dV

dQl

= − 6.35/160

1 + 100× 6.35/160= −0.0078

The change in voltage due to variation of reactive power by 10 MVArs, ∆V = −0.0078 × 10 =−0.078 kV.Thus it is seen that, with finite compensator gain their is quite reduction in the voltage sensitivity,which means that the load bus is fairly constant for considerable change in the load reactive power.The compensator reactive power Qγ and load reactive power Ql are related by equation (3.45) andis given below.

Qγ = − Kγ (E/Ssc)

1 +Kγ (E/Ssc)Ql = − 100× (6.35/160)

1 + 100× (6.35/160)Ql

= −0.79Ql

It can be observed that when compensator gain, (Qγ) is quite large, then compensator reactivepowerQγ is equal and opposite to that of load reactive power i.e.,Qγ = −Ql. It is further observed

84

Page 91: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

that due to finite compensator gain i.e., Kγ = 100, reactive power is partially compensated Thecompensator reactive power varies from 0 to 7.9 MVAr for 0 to 10 MVAr change in the loadreactive power.

3.4 Phase Balancing and Power Factor Correction of Unbalanced Loads

So far we have discussed voltage regulation and power factor correction for single phase systems.In this section we will focus on balancing of three-phase unbalanced loads. In considering unbal-anced loads, both load and compensator are modeled in terms of their admittances and impedances.

3.4.1 Three-phase Unbalanced Loads

Consider a three-phase three-wire system suppling unbalanced load as shown in Fig. 3.8.

anV

aZ

bZ

cZ

bnV

cnV

N n

aI

1I

2I

bI

cI

Fig. 3.8 Three-phase unbalanced load

Applying Kirchoff’s voltage law for the two loops shown in the figure, we have the followingequations.

−V an + Za I1 + Zb (I1 − I2) + V bn = 0

−V bn + Zb I2 + Zb (I2 − I1) + V cn = 0 (3.46)

Rearranging above, we get the following.

V an − V bn = (Za + Zb) I1 − Zb I2

V bn − V cn = (Zb + Zc) I2 − Zb I1 (3.47)

The above can be represented in matrix form as given below.[V an − V bn

V bn − V cn

]=

[(Za + Zb) −Zb−Zb (Zb + Zc)

] [I1

I2

](3.48)

85

Page 92: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Therefore the currents are given as below.[I1

I2

]=

[(Za + Zb) −Zb−Zb (Zb + Zc)

]−1 [V an − V bn

V bn − V cn

]=

1

∆Z

[(Zb + Zc) Zb

Zb (Za + Zb)

] [V an − V bn

V bn − V cn

]Therefore,

[I1

I2

]=

1

∆Z

[(Zb + Zc) Zb

Zb (Za + Zb)

] [V an − V bn

V bn − V cn

]. (3.49)

Where, ∆Z = (Zb + Zc)(Za + Zb)− Z2b = Za Zb + Zb Zc + Zc Za. The current I1 is given below.

I1 =1

∆Z

[(Zb + Zc)(V an − V bn) + Zb(V bn − V cn)

]=

1

∆Z

[(Zb + Zc)V an − ZcV bn − ZbV cn

](3.50)

Similarly,

I2 =1

∆Z

[Zb(V an − V bn) + (Za + Zb)(V bn − V cn)

]=

1

∆Z

[ZbV an + ZaV bn − (Za + Zb)V cn

](3.51)

Now,

Ia = I1 =1

∆Z

[(Zb + Zc)V an − ZcV bn − ZbV cn

]Ib = I2 − I1

=1

∆Z

[

ZbV an + ZaV bn − (Za + Zb)V cn − (Zb + Zc)V an + ZcV bn + ZbV cn

]=

[(Zc + Za)V bn − ZaV cn − ZcVan

]∆Z

=(Zc + Za)V bn − ZcV an − ZaV cn

∆Z

(3.52)

and

Ic = −I2 = −Ib − Ia =(Za + Zb)V cn − ZbV an − ZaV bn

∆Z

(3.53)

Alternatively phase currents can be expressed as following.

Ia =V an − V Nn

Za

Ib =V bn − V Nn

Zb(3.54)

Ic =V cn − V Nn

Zc

86

Page 93: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Applying Kirchoff’s current law at node N , we get Ia + Ib + Ic = 0. Therefore from the aboveequation,

V an − V Nn

Za+V bn − V Nn

Zb+V cn − V Nn

Zc= 0. (3.55)

Which implies that,

V an

Za+V bn

Zb+V cn

Zc=

[1

Za+

1

Zb+

1

Zc

]V Nn =

ZaZb + ZbZc + ZcZaZaZbZc

V Nn (3.56)

From the above equation the voltage between the load and system neutral can be found. It is givenbelow.

V Nn =ZaZbZc

∆Z

[V an

Za+V bn

Zb+V cn

Zc

]=

11Za

+ 1Zb

+ 1Zc

[V an

Za+V bn

Zb+V cn

Zc

](3.57)

Some interesting points are observed from the above formulation.

1. If both source voltage and load impedances are balanced i.e., Za = Zb = Zc = Z, thenV Nn = 1

3

(V an + V bn + V cn

)= 0. Thus their will not be any voltage between two neutrals.

2. If supply voltage are balanced and load impedances are unbalanced, then V Nn 6=0 and isgiven by the above equation.

3. If supply voltages are not balanced but load impedances are identical, then V Nn = 13

(V an + V bn + V cn

).

This equivalent to zero sequence voltage V 0.

It is interesting to note that if the two neutrals are connected together i.e., V Nn = 0, then eachphase become independent through neutral. Such configuration is called three-phase four-wiresystem. In general, three-phase four-wire system has following properties.

V Nn = 0

Ia + Ib + Ic = INn 6= 0

(3.58)

The current INn is equivalent to zero sequence current (I0) and it will flow in the neutral wire.For three-phase three-wire system, the zero sequence current is always zero and therefore followingproperties are satisfied.

V Nn 6= 0

Ia + Ib + Ic = 0

(3.59)

Thus, it is interesting to observe that three-phase three-wire and three-phase four-wire system havedual properties in regard to neutral voltage and current.

87

Page 94: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

cI

bI

aIab

lY

bclY

ca

lY

c

ab

cI

bI

aI

clZ

c

ab

alZ

blZ n

(a) (b)

Fig. 3.9 (a) An unbalanced delta connected load (b) Its equivalent star connected load

3.4.2 Representation of Three-phase Delta Connected Unbalanced Load

A three-phase delta connected unbalanced and its equivalent star connected load are shown in Fig.3.9(a) and (b) respectively. The three-phase load is represented by line-line admittances as givenbelow.

Y abl = Gab

l + jBabl

Y bcl = Gbc

l + jBbcl

Y cal = Gca

l + jBcal

(3.60)

The delta connected load can be equivalently converted to star connected load using followingexpressions.

Zal =

Zabl Zca

l

Zabl + Zbc

l + Zcal

Zbl =

Zbcl Z

abl

Zabl + Zbc

l + Zcal

Zcl =

Zcal Zbc

l

Zabl + Zbc

l + Zcal

(3.61)

Where Zabl = 1/Y ab

l , Zbcl = 1/Y bc

l and Zcal = 1/Y ca

l . The above equation can also be written inadmittance form

Y al =

Y abl Y bc

l + Y bcl Y ca

l + Y cal Y ab

l

Y bcl

Y bl =

Y abl Y bc

l + Y bcl Y ca

l + Y cal Y ab

l

Y cal

Y cl =

Y abl Y bc

l + Y bcl Y ca

l + Y cal Y ab

l

Y abl

(3.62)

Example 3.3 Consider three-phase system supply a delta connected unbalanced load with Z la =

Ra = 10 Ω, Z lb = Rb = 15 Ω and Z l

c = Rc = 30 Ω as shown in Fig. 3.8. Determine the voltage

88

Page 95: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

between neutrals and find the phase currents. Assume a balance supply voltage with rms value of230 V. Find out the vector and arithmetic power factor. Comment upon the results.Solution: The voltage between neutrals V Nn is given as following.

VNn =RaRbRc

RaRb +RbRc +RcRa

[V an

Ra

+V bn

Rb

+V cn

Rc

]=

10× 15× 30

10× 15 + 15× 30 + 30× 10

[V ∠0o

10+V ∠−120o

15+V ∠120o

30

]=

4500

900

[3V ∠0o + 2V ∠−120o + V ∠120o

30

]=

4500

900

1

30V

[3 + 2

(−1

2− j√

3

2

)+

(−1

2+ j

√3

2

)]

=V

6

[3− 1− 1

2− j2×

√3

2+ j

√3

2

]

=V

6

[3

2− j√

3

2

]= V

[1

4− j 1

4√

3

]= V [0.25− j0.1443]

VNn =V

2√

3∠−30o = 66.39∠−30o Volts

Knowing this voltage, we can find phase currents as following.

Ia =V an − V Nn

Ra

=V ∠0o − V/(2

√3)∠−30o

10

=V [1− 0.25 + j0.1443]

10= 230× [0.075 + j0.01443]

= 17.56∠10.89o Amps

Similarly,

Ib =V bn − V Nn

Rb

=V ∠− 120o − V/(2

√3)∠−30o

15= 230× [−0.05− j0.04811]

= 15.94∠−136.1o Amps

and

Ic =V cn − V Nn

Zc=V ∠120o − V/(2

√3)∠−30o

30= 230× [−0.025 + j0.03367]

= 9.64∠126.58o Amps

89

Page 96: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

It can been seen that Ia + Ib + Ic = 0. The phase powers are computed as below.

Sa = V a (Ia)∗ = Pa + jQa = 230× 17.56∠− 10.81o = 3976.12− j757.48 VA

Sb = V b (Ib)∗ = Pb + jQb = 230× 15.94∠(−120o + 136.1o = 3522.4 + j1016.69 VA

Sc = V c (Ic)∗ = Pc + jQc = 230× 9.64∠(120o − 126.58o) = 2202.59− j254.06 VA

From the above the total apparent power SV = Sa + Sb + Sc = 9692.11 + j0 VA. Therefore,SV =

∣∣Sa + Sb + Sc∣∣ = 9692.11 VA.

The total arithmetic apparent power SA =∣∣Sa∣∣ +

∣∣Sb∣∣ +∣∣Sc∣∣ == 9922.2 VA. Therefore, the

arithmetic and vector apparent power factors are given by,

pfA =P

SA=

9692.11

9922.2= 0.9768

pfV =P

SV=

9622.11

9622.11= 1.00.

It is interesting to note that although the load in each phase is resistive but each phase has somereactive power. This is due to unbalance of the load currents. This apparently increases the ratingof power conductors for given amount of power transfer. It is also to be noted that the net reactivepower Q = Qa + Qb + Qc = 0 leading to the unity vector apparent power factor . However thearithmetic apparent power factor is less than unity showing the effect of the unbalance loads on thepower factor.

3.4.3 An Alternate Approach to Determine Phase Currents and Powers

In this section, an alternate approach will be discussed to solve phase currents and powers directlywithout computing the neutral voltage for the system shown in Fig.3.9(a). First we express three-phase voltage in the following form.

V a = V ∠0o

V b = V ∠− 120o = α2V (3.63)V c = V ∠1200 = αV

Where, in above equation, α is known as complex operator and value of α and α2 are given below.

α = ej2π/3 = 1∠120o = −1/2 + j√

3/2

α2 = ej4π/3 = 1∠240 = 1∠− 120 = −1/2− j√

3/2 (3.64)

Also note the following property,

1 + α + α2 = 0. (3.65)

Using the above, the line to line voltages can be expressed as following.

90

Page 97: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

V ab = V a − V b = (1− α2)V

V bc = V b − V c = (α2 − α)V (3.66)V ca = V c − V a = (α− 1)V

Therefore, currents in line ab, bc and ca are given as,

Iabl = Y abl V ab = Y ab

l (1− α2)V

Ibcl = Y bcl V bc = Y bc

l (α2 − α)V (3.67)Ical = Y ca

l V ca = Y cal (α− 1)V

Hence line currents are given as,

Ial = Iabl − Ical = [Y abl (1− α2)− Y ca

l (α− 1)]V

Ibl = Ibcl − Iabl = [Y bcl (α2 − α)− Y ab

l (1− α2)]V (3.68)Icl = Ical − Ibcl = [Y ca

l (α− 1)− Y bcl (α2 − α)]V

Example 3.4 Compute line currents by using above expressions directly for the problem in Exam-ple 3.3.Solution: To compute line currents directly from the above expressions, we need to compute Y ab

l .These are given below

Y abl =

1

Zabl

=Zcl

Z la Z

lb + Z l

b Zlc + Z l

c Zla

Y bcl =

1

Zbcl

=Zal

Z la Z

lb + Z l

b Zlc + Z l

c Zla

(3.69)

Y cal =

1

Zcal

=Zbl

Z la Z

lb + Z l

b Zlc + Z l

c Zla

Substituting, Z la = Ra = 10 Ω, Z l

b = Rb = 15 Ω and Z lc = Rc = 30 Ω into above equation, we get

the following.

Y abl = Gab =

1

30Ω

Y bcl = Gbc =

1

90Ω

Y cal = Gca =

1

60Ω

Substituting above values of the admittances in (3.68) , line currents are computed as below.

91

Page 98: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Ia =

[1

30

1− (−1

2− j√

3

2)

− 1

60

(−1

2+ j

√3

2)− 1

]V

= V (0.075 + j0.0144)

= 0.07637V ∠10.89o

= 17.56∠10.89o Amps, for V=230 V

Similarly for Phase-b current,

Ib =

[1

90

(−1

2− j√

3

2)− (−1

2+ j

√3

2)

− 1

30

1− (−1

2− j√

3

2)

]V

= V (−0.05− j0.0481)

= 0.06933V ∠− 136.1o

= 15.94∠− 136.91o Amps, for V=230 V

Similarly for Phase-c current,

Ic =

[1

60

(−1

2+ j

√3

2)− 1)

− 1

90

(−1

2− j√

3

2)− (−1

2+ j

√3

2)

]V

= V (−0.025 + j0.0336)

= 0.04194V ∠126.58o

= 9.64∠126.58o Amps, for V=230 V

Thus it is found that the above values are similar to what have been found in previous Example3.3. The other quantities such as powers and power factors are same.

3.4.4 An Example of Balancing an Unbalanced Delta Connected Load

An unbalanced delta connected load is shown in Fig. 3.10(a). As can be seen from the figure thatbetween phase-a and b there is admittance Y ab

l = Gabl and other two branches are open. This is

an example of extreme unbalanced load. Obviously for this load, line currents will be extremelyunbalanced. Now we aim to make these line currents to be balanced and in phase with their phasevoltages. So, let us assume that we add admittances Y ab

γ , Y bcγ and Y ca

γ between phases ab, bc andca respectively as shown in Fig. 3.10(b) and (c). Let values of compensator susceptances are givenby,

Y abγ = 0

Y bcγ = jGab

l /√

3

Y caγ = −jGab

l /√

3

92

Page 99: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

cI

bI

aIab ab

l lY G

c

b

0cal

Y

0bc

lY

a

cI

bI

aIab ab

l lY G

c

ab

/3

ca

abl

Y

jG

/

3

bc

abl

YjG

0abY

0calY0

bclY

cI

bI

aI

c

ab

/3

ca

abl

Y

jG/

3

bc

abl

YjG

ab ablY G

(a) (b) (c)

Fig. 3.10 (a) An unbalanced three-phase load (b) With compensator (c) Compensated system

Thus total admittances between lines are given by,

Y ab = Y abl + Y ab

γ = Gabl + 0 = Gab

l

Y bc = Y bcl + Y bc

γ = 0 + jGabl /√

3 = jGabl /√

3

Y ca = Y cal + Y ca

γ = 0− jGabl /√

3 = −jGabl /√

3.

Therefore the impedances between load lines are given by,

Zab =1

Y ab=

1

Gabl

Zbc =1

Y bc=−j√

3

Gabl

Zca =1

Y ca=j√

3

Gabl

Note that Zab + Zbc + Zca = 1/Gabl − j

√3/Gab

l + j√

3/Gabl = 1/Gab

l .The impedances, Za, Zb and Zc of equivalent star connected load are given as follows.

Za =Zab × Zca

Zab + Zbc + Zca

= (1

Gabl

× j√

3

Gabl

)/(1

Gabl

)

=j√

3

Gabl

93

Page 100: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Zb =Zbc × Zab

Zab + Zbc + Zca

= (1

Gabl

× −j√

3

Gabl

)/(1

Gabl

)

=−j√

3

Gabl

Zc =Zca × Zbc

Zab + Zbc + Zca

= (−j√

3

Gabl

× j√

3

Gabl

)/(1

Gabl

)

=3

Gabl

The above impedances seen from the load side are shown in Fig. 3.11(a) below. Using (3.57),

cI

bI

aI

3cl ab

l

ZG

c

ab

3al ab

l

Z jG

n

3bl ab

l

Z jG

cI

bI

aI

c

ab

N1ablG

1ablG

1ablG

Source

Load

Fig. 3.11 Compensated system (a) Load side (b) Source side

the voltage between load and system neutral of delta equivalent star load as shown in Fig. 3.11, iscomputed as below.

VNn =1

1Za

+ 1Zb

+ 1Zc

[V an

Za+V bn

Zb+V cn

Zc

]=

11

(j√

3/Gabl )+ 1

(−j√

3/Gabl )+ 3

Gabl

[V ∠0o

j√

3/Gabl

+V ∠− 120o

−j√

3/Gabl

+V ∠120o

3/Gabl

]=

3V

Gabl

(Gabl

3− jG

abl√3

)= 2V

(1

2− j√

3

2

)= 2V ∠− 60o

94

Page 101: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Using above value of neutral voltage the line currents are computed as following.

Ia =V an − V Nn

Za

=[V ∠0o − 2V ∠− 60o]

j√

3Gabl

= Gabl V = Gab

l V a

Ib =V bn − V Nn

Zb

=[V ∠− 120o − 2V ∠− 60o]

−j√

3Gabl

= Gabl V ∠240o

= Gabl V b

Ic =V cn − V Nn

Zc

=[V ∠120o − 2V ∠− 60o]

3Gabl

= Gabl V ∠120o

= Gabl V c

From the above example, it is seen that the currents in each phase are balanced and in phase withtheir respective voltages. This is equivalently shown in Fig. 3.11(b). It is to be mentioned here thatthe two neutrals in Fig. 3.11 are not same. In Fig. 3.11(b), the neutral N is same as the systemneutral as shown in Fig. 3.8, whereas in Fig. 3.11(a), V Nn = 2V ∠−60o. However the readermay be curious to know why Y ab

γ = 0, Y bcγ = jGab

l /√

3 and γca = −jGabl /√

3 have been chosenas compensator admittance values. The answer of the question can be found by going followingsections.

3.5 A Generalized Approach for Load Compensation using SymmetricalComponents

In the previous section, we have expressed line currents Ia, Ib and Ic, in terms load admittancesand the voltage V for a delta connected unbalanced load as shown in Fig 3.12(a). For the sake ofcompleteness, theses are reproduced below.

95

Page 102: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Ial = Iabl − Ical = [Y abl (1− α2)− Y ca

l (α− 1)]V

Ibl = Ibcl − Iabl = [Y bcl (α2 − α)− Y ab

l (1− α2)]V (3.70)Icl = Ical − Ibcl = [Y ca

l (α− 1)− Y bcl (α2 − α)]V

cI

bI

aI

ca

Y

ablYabY

bcY bclY

ca

lY

c

ab

clI

blI

alI

ablY

bclY

ca

lY

c

ab

(a) (b)

Fig. 3.12 (a) An unbalanced delta connested load (b) Compensated system

Since loads are currents are unbalanced, these will have positive and negative currents. The zerosequence current will be zero as it is three-phase and three-wire system. These symmetrical com-ponents of the load currents are expressed as following.

I0l

I1l

I2l

=1√3

1 1 11 a a2

1 a2 a

IalIblIcl

(3.71)

From the above equation, zero sequence current is given below.

I0l =(Ial + Ial + Ial

)/√

3

The positive sequence current is as follows.

I1l =1√3

[Ial + αIbl + α2Icl

]=

1√3

[Y abl (1− α2)− Y ca

l (α− 1) + αY bcl (α2 − α)− Y ab

l (1− α2)

+α2Y cal (α− 1)− Y bc

l (α− α2)

]V

=1√3

[Y abl − α2Y ab

l + Y cal − αY ca

l + α3Y bcl − α2Y bc

l − αY abl − α3Y ab

l + α3Y cal − α2Y ca

l

−α4Y bcl + α3Y bc

l ]V

=(Y abl + Y bc

l + Y cal

)V√

3

96

Page 103: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Similarly negative sequence component of the current is,

I2l =1√3

[Ial + α2Ibl + αIcl

]=

1√3

[Y abl (1− α2)− Y ca

l (α− 1) + α2Y bcl (α2 − α)− Y ab

l (1− α2)

+αY cal (α− 1)− Y bc

l (α2 − α)

]V

=1√3

[Y abl − α2Y ab

l − αY cal + Y ca

l + α4Y bcl − α3Y bc

l − α2Y abl

+α4Y abl + α2Y ca

l − αY cal − α3Y bc

l + α2Y bcl ]V

=1√3

[−3α2Y abl − 3Y bc

l − 3αY cal ]V

= −[α2Y abl + Y bc

l + αY cal ]√

3V

From the above, it can be written that,

I0l = 0

I1l =(Y abl + Y bc

l + Y cal

)√3V (3.72)

I2l = −(α2Y ab

l + Y bcl + αY ca

l

)√3V

When compensator is used, three delta branches Y abγ , Y bc

γ and Y caγ are added as shown in Fig.

3.12(b). Using above analysis, the sequence components of the compensator currents can be givenas below.

I0γ = 0

I1γ =(Y abγ + Y bc

γ + Y caγ

) √3V (3.73)

I2γ = −(α2Y ab

γ + Y bcγ + αY ca

γ

) √3V

Since, compensator currents are purely reactive, i.e., Gabγ = Gbc

γ = Gcaγ = 0,

Y abγ = Gab

γ + jBabγ = jBab

γ

Y bcγ = Gbc

γ + jBbcγ = jBbc

γ (3.74)Y caγ = Gca

γ + jBcaγ = jBca

γ .

Using above, the compensated sequence currents can be written as,

I0γ = 0

I1γ = j(Babγ +Bbc

γ +Bcaγ

) √3V (3.75)

I2γ = −j(α2Babγ +Bbc

γ + αBcaγ )√

3V

Knowing nature of compensator and load currents, we can set compensation objectives as fol-lowing.

1. All negative sequence component of the load current must be supplied from the compensatornegative current, i.e.,

I2l = −I2γ (3.76)

97

Page 104: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The above further implies that,

Re(I2l

)+ j Im

(I2l

)= −Re

(I2γ

)− j Im

(I2γ

)(3.77)

2. The total positive sequence current, which is source current should have desired power factorfrom the source, i.e.,

Im (I1l + I1γ)

Re (I1l + I1γ)= tanφ = β (3.78)

Where, φ is the desired phase angle between the line currents and the supply voltages. The aboveequation thus implies that,

Im (I1l + I1γ) = β Re (I1l + I1γ) (3.79)

Since Re (I1γ) = 0, the above equation is rewritten as following.

Im (I1l)− β Re (I1l) = −Im (I1γ) (3.80)

The equation (3.77) gives two conditions and equation (3.79) gives one condition. There arethree unknown variables, i.e., Bab

γ , Bbcγ and Bca

γ and three conditions. Therefore the unknownvariables can be solved. This is described in the following section. Using (3.75), the current I2γ isexpressed as following.

I2γ = −j[α2Babγ +Bbc

γ + αBcaγ ]√

3V

= −j

[(−1

2− j√

3

2

)Babγ +Bbc

γ +

(−1

2+ j

√3

2

)Bcaγ

]√

3V

=

[(−√

3

2Babγ +

√3

2Bcaγ

)− j

(−1

2Babγ +Bbc

γ −1

2Bcaγ

)] √3V (3.81)

= Re(I2γ

)− j Im

(I2γ

)Thus the above equation implies that(√

3

2Babγ −

√3

2Bcaγ

)= − 1√

3VRe(I2γ

)=

1√3V

Re(I2l

)(3.82)

and, (−1

2Babγ +Bbc

γ −1

2Bcaγ

)= − 1√

3VIm(I2γ

)=

1√3V

Im(I2l

)(3.83)

Or (−Bab

γ + 2Bbcγ −Bca

γ

)=

1√3V

2 Im(I2l

)(3.84)

98

Page 105: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From (3.75), Im (I1γ) can be written as,

Im (I1γ) =(Babγ +Bbc

γ +Bcaγ

) √3V. (3.85)

Substituting Im (I1γ) from above equation into (3.85), we get the following.

−(Babγ +Bbc

γ +Bcaγ ) = − 1√

3VIm (I1γ) =

1√3V

Im (I1l)− β Re (I1l)

(3.86)

Subtracting (3.86) from (3.84), the following is obtained,

Bbcγ =

−1

3√

3V[Im (I1l)− 2 Im (I2l)− β Re (I1l)]. (3.87)

Now, from (3.82) we have

−1

2Babγ −

1

2Bcaγ =

1√3V

Im (I2l)−Bbcγ

=1√3V

Im (I2l)−[−1

3√

3V

Im (I1l)− β Re (I1l)− 2 Im (I2l)

]=

1

3√

3V

Im (I2l) + Im (I1l)− β Re (I1l)

(3.88)

Reconsidering (3.88) and (3.82), we have

−1

2Babγ −

1

2Bcaγ =

1

3√

3V

Im (I1l) + Im (I2l)− β Re (I1l)

1

2Babγ −

1

2Bcaγ =

1

3√

3V[√

3Re (I2l)]

Adding above equations, we get

Bcaγ =

−1

3√

3V[Im (I1l) + Im (I2l) +

√3Re (I2l)− βRe (I1l)]. (3.89)

Therefore,

Babγ = Bca

γ +2

3√

3V[√

3Re (I2l)] (3.90)

=−1

3√

3V[Im (I1l) + Im (I2l) +

√3Re (I2l)− βRe (I1l)− 2

√3Re (I2l)]

=−1

3√

3V[Im (I1l) + Im (I2l)−

√3Re (I2l)− βRe (I1l)] (3.91)

From the above, the compensator susceptances in terms of real and imaginary parts of the load

99

Page 106: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

current can be written as following.

Babγ =

−1

3√

3V[Im (I1l) + Im (I2l)−

√3 Re (I2l)− β Re (I1l)]

Bbcγ =

−1

3√

3V[Im (I1l)− β Re (I1l)− 2Im (I2l − β Re (I1l)] (3.92)

Bcaγ =

−1

3√

3V[Im (I1l) + Im (I2l) +

√3 Re (I2l)− β Re (I1l)]

In the above equation, the susceptances of the compensator are expressed in terms of real andimaginary parts of symmetrical components of load currents. It is however advantageous to expressthese susceptances in terms of instantaneous values of voltages and currents from implementationpoint of view. The first step to achieve this is to express these susceptances in terms of loadcurrents, i.e., Ial, Ibl and Icl, which is described below. Using equation (3.71), the sequencecomponents of the load currents are expressed as,

I0l =1√3

[Ial + Ibl + Icl

]I1l =

1√3

[Ial + αIbl + α2Icl

](3.93)

I2l =1√3

[Ial + α2I1l + αIcl

].

Substituting these values of sequence components of load currents, in (3.92), we can obtain com-pensator susceptances in terms of real and inaginary components of the load currents. Let us startfrom the Bbc

γ , as obtained following.

Bbcγ = − 1

3√

3V

[Im(I1l)− 2Im(I2l)− β Re(I1l)

]= − 1

3√

3V

[Im(Ial + αIbl + α2Icl√

3

)− 2 Im

(Ial + α2Ibl + αIcl√

3

)− β Re

(Ial + αIbl + α2Icl√

3

)]= − 1

9V

[Im

(−Ial + (2 + 3α)Ibl + (2 + 3α2)Icl− β Re (Ial + αIbl + α2Icl)

]= − 1

9V

[Im−Ial + 2Ibl + 2Icl + 3αIbl + 3α2Icl

− β Re (Ial + αIbl + α2Icl)

]By adding and subracting Ibl and Icl in the above equation we get,

Bbcγ = − 1

9V

[Im

(−Ial − Ibl − Icl) + 3Ibl + 3Icl + 3α Ibl + α2 Icl− β Re(Ial + αIbl + α2Icl)

]We know that Ial+Ibl+Icl=0, therefore Ial + Ibl = −Icl.

Bbcγ = − 1

3V

[−Im

(Ial)

+ Im(αIbl

)+ Im

(α2Icl

)− β

3Re(Ial + αIbl + α2Icl

])(3.94)

100

Page 107: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Similarly, it can be proved that,

Bcaγ = − 1

3V

[Im(Ial)− Im

(αIbl

)+ Im

(α2Icl

)− β

3Re(Ial + αIbl + α2Icl

])(3.95)

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)− β

3Re(Ial + αIbl + α2Icl

])(3.96)

The above expressions forBcaγ andBab

γ are proved below. For convenience, the last term associatedwith β is not considered. For the sake simplicity in equations (3.95) and (3.96) are proved to thosegiven in equations (3.92).

Bcaγ =

−1

3V

[Im(Ial)− Im

(αIbl

)+ Im

(α2Icl

)]= − 1

3V

[Im

(I0l + I1l + I2l

)√

3− α

(I0l + α2I1l + αI2l

)√

3+ α2

(I0l + αI1l + α2I2l

)√

3

]

Since I0l=0

Bcaγ = − 1

3√

3VIm[I1l − 2α2I2l

]= − 1

3√

3VIm

[I1l − 2

(−1

2− j√

3

2

)I2l

]= − 1

3√

3VIm[I1l + I2l + j

√3I2l

]= − 1

3√

3V

[Im(I1l

)+ Im

(I2l

)+√

3Re(I2l

)]Note that Im(jI2l) = Re(I2l). Adding β term, we get the following.

Bcaγ = − 1

3√

3V

[Im(I1l) + Im(I2l) +

√3Re

(I2l

)− βRe(I1l)

]Similarly,

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)]= − 1

3V

[Im

(I0l + I1l + I2l

)√

3+ α

(I0l + α2I1l + αI2l

)√

3− α2

(I0l + αI1l + α2I2l

)√

3

]101

Page 108: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Babγ = − 1

3√

3VIm[I1l − 2αI2l

]= − 1

3√

3VIm

[I1l − 2

(−1

2+ j

√3

2

)I2l

]= − 1

3√

3VIm[I1l + I2l − j

√3I2l

]= − 1

3√

3V

[Im(I1l

)+ Im

(I2l

)−√

3Re(I2l

)]Thus, Compensator susceptances are expressed as following.

Babγ = − 1

3V

[Im(Ial) + Im(αIbl)− Im(α2Icl)−

β

3Re(Ial + αIbl + α2Icl)

]Bbcγ = − 1

3V

[−Im(Ial) + Im(αIbl) + Im(α2Icl)−

β

3Re(Ial + αIbl + α2Icl)

](3.97)

Bcaγ = − 1

3V

[Im(Ial)− Im(αIbl) + Im(α2Icl)−

β

3Re(Ial + αIbl + α2Icl)

]An unity power factor is desired from the source. For this cosφl = 1, implying tanφl = 0 henceβ = 0. Thus we have,

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)]Bbcγ = − 1

3V

[−Im

(Ial)

+ Im(αIbl

)+ Im

(α2Icl

)](3.98)

Bcaγ = − 1

3V

[Im(Ial)− Im

(αIbl

)+ Im

(α2Icl

)]The above equations are easy to realize in order to find compensator susceptances. As mentionedabove, sampling and averaging techniques will be used to convert above equation into their timeequivalents. These are described below.

3.5.1 Sampling Method

Each current phasor in above equation can be expressed as,

Ial = Re(Ial) + jIm(Ial)

= Ial,R + jIal,X (3.99)

102

Page 109: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

An instantaneous phase current is written as follows.

ial(t) =√

2 Ial Im (ejωt)

=√

2 Im (Ial ejωt)

=√

2 Im[(Ial,R + jIal,X) ejωt

]=√

2 Im [(Ial,R + jIal,X)(cosωt+ j sinωt)]

=√

2 Im [(Ial,R cosωt− Ial,X sinωt) + j(Ial,R sinωt+ Ial,X cosωt)]

=√

2 [(Ial,R sinωt+ Ial,X cosωt)] (3.100)

Im (Ial) = Ial,X =ial(t)√

2at sinωt = 0, cosωt = 1 (3.101)

From equation (3.63), the phase voltages can be expressed as below.

va(t) =√

2V sinωt

vb(t) =√

2V sin(ωt− 120o) (3.102)vc(t) =

√2V sin(ωt+ 120o)

From above voltage expressions, it is to be noted that, sinωt = 0, cosωt = 1 implies that thephase-a voltage, va(t) is going through a positive zero crossing, hence, va(t) = 0 and d

dtva(t) = 0.

Therefore, equation (3.101), can be expressed as following.

Ia,l =ial(t)√

2when, va(t) = 0, dva/dt > 0 (3.103)

Similarly,

Ib,l = Ibl,R + jIbl,X (3.104)

Therefore,

α (Ib,l) = α(Ibl,R + jIbl,X)

=

(−1

2+ j

√3

2

)(Ibl,R + jIbl,X)

=

(−1

2Ibl,R −

√3

2Ibl,X

)+ j

(√3

2Ibl,R −

1

2Ibl,X

)(3.105)

From the above,

Im α (Ibl) =

√3

2Ibl,R −

1

2Ibl,X (3.106)

103

Page 110: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Similar to equation, (3.100), we can express phase-b current in terms Im (α Ibl), as given below.

ibl(t) =√

2 Im (Ibl ejωt)

=√

2 Im (α Ibl ejwtα−1)

=√

2 Im (αIbl ej(wt−1200))

=√

2 Im

[(−1

2+ j

√3

2

)(Ibl,R + jIbl,X) ej(wt−1200)

]

=√

2 Im [

(−1

2Ibl,R −

√3

2Ibl,X

)+ j

(√3

2Ibl,R −

1

2Ibl,X

)

(cos(wt− 1200) + j sin(wt− 1200))

] (3.107)

=√

2

[(−1

2Ibl,R −

√3

2Ibl,X) sin(wt− 1200) + (

√3

2Ibl,R −

1

2Ibl,X) cos(wt− 1200)

]From the above equation, we get the following.

Im (αIbl) =ibl(t)√

2when, vb(t) = 0, dvb/dt > 0 (3.108)

Similarly for Phase-c, it can proved that,

Im (α2Icl) =icl(t)√

2when, vc(t) = 0, dvc/dt > 0 (3.109)

Substituting Im (Ial) , Im (αIbl) and Im (α2Icl) from (3.103), (3.108) and (3.109) respectively, in(3.98), we get the following.

Babr = − 1

3√

2V

[ia|(va=0, dva

dt>0) + ib|(vb=0,

dvbdt>0)− ic|(vc=0, dvc

dt>0)

]Bbcr = − 1

3√

2V

[−ia|(va=0, dva

dt>0) + ib|(vb=0,

dvbdt>0)

+ ic|(vc=0, dvcdt>0)

](3.110)

Bcar = − 1

3√

2V

[ia|(va=0, dva

dt>0) − ib|(vb=0,

dvbdt>0)

+ ic|(vc=0, dvcdt>0)

]Thus the desired compensating susceptances are expressed in terms of the three line currents sam-pled at instants defined by positive-going zero crossings of the line-neutral voltages va, vb, vc. Anartificial neutral at ground potential may be created measuring voltages va, vb and vc to implementabove algorithm.

3.5.2 Averaging Method

In this method, we express the compensator susceptances in terms of real and reactive power termsand finally expressed them in time domain through averaging process. The method is described

104

Page 111: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

below.From equation (3.98), susceptance, Bab

γ , can be re-written as following.

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)]= − 1

3V 2

[V Im

(Ial)

+ V Im(αIbl

)− V Im

(α2Icl

)]= − 1

3V 2

[Im(V Ial

)+ Im

(V αIbl

)− Im

(V α2Icl

)](3.111)

Note the following property of phasors and applying it for the simplification of the above expres-sion.

Im(V I)

= −Im(V∗I∗)

(3.112)

Using above equation (3.111) can be written as,

Babγ =

1

3V 2

[Im (V Ial)

∗ + Im (αV a Ibl)∗ − Im (α2V Icl)

∗]=

1

3V 2

[Im (V

∗a I∗al) + Im (α∗V

∗a I∗bl)− Im ((α2)∗V

∗a I∗cl)]

=1

3V 2

[Im (V

∗a I∗al) + Im (α2V

∗a I∗bl)− Im (αV

∗a I∗cl)]

Since V a = V ∠0o is a reference phasor, therefore V a = V∗a = V , α2 V a = V

∗b and αV a = V

∗c .

Using this, the above equation can be written as following.

Babγ =

1

3V 2

[Im (V a I

∗al) + Im (V b I

∗bl)− Im (V c I

∗cl)]

Similarly, Bbcγ =

1

3V 2

[−Im (V a I

∗al) + Im (V b I

∗bl) + Im (V c I

∗cl)]

(3.113)

Bcaγ =

1

3V 2

[Im (V a I

∗al)− Im (V b I

∗bl) + Im (V c I

∗cl)]

It can be further proved that,

Im (V a I∗a) =

1

T

∫ T

0

va(t)∠(−π/2) ial(t) dt

Im (V b I∗b) =

1

T

∫ T

0

vb(t)∠(−π/2) ibl(t) dt (3.114)

Im (V c I∗c) =

1

T

∫ T

0

vc(t)∠(−π/2) icl(t) dt

105

Page 112: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Since,

va(t)∠(−π/2) = vbc(t)/√

3

vb(t)∠(−π/2) = vca(t)/√

3

vc(t)∠(−π/2) = vca(t)/√

3

Equation (3.115) can be written as,

Im (V a I∗a) =

1√3T

∫ T

0

vbc(t) ial(t) dt

Im (V b I∗b) =

1√3T

∫ T

0

vca(t) ibl(t) dt (3.115)

Im (V c I∗c) =

1√3T

∫ T

0

vab(t) icl(t) dt

Substituting above values of Im (V a I∗a), Im (V b I

∗b) and Im (V c I

∗c) into (3.113), we get the fol-

lowing.

Babγ =

1

(3√

3V 2)

1

T

∫ T

0

(vbc ial + vca ibl − vab icl) dt

Bbcγ =

1

(3√

3V 2)

1

T

∫ T

0

(−vbc ial + vca ibl + vab icl) dt (3.116)

Bcaγ =

1

(3√

3V 2)

1

T

∫ T

0

(vbc ial − vca ibl + vab icl) dt

The above equations can directly be used to know the the compensator susceptances by performingthe averaging on line the product of the line to line voltages and phase load currents. The term∫ T

0=∫ t1+T

t1can be implemented using moving average of one cycle. This improves transient

response by computing average value at each instant. But in this case the controller responsewhich changes the susceptance value, should match to that of the above computing algorithm.

3.6 Compensator Admittance Represented as Positive and Negative SequenceAdmittance Network

Recalling the following relations from equation (3.92) for unity power factor operation i.e. β = 0,we get the following.

Babγ =

−1

3√

3V[Im (I1l) + Im (I2l)−

√3 Re (I2l)]

Bbcγ =

−1

3√

3V[Im (I1l)− β Re (I1l)− 2Im (I2l)] (3.117)

Bcaγ =

−1

3√

3V[Im (I1l) + Im (I2l) +

√3 Re (I2l)]

106

Page 113: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From these equations, it is evident the the first terms form the positive sequence suceptance as theyinvolve I1l terms. Similarly, the second and third terms in above equation form negative sequencesusceptance of the compensator, as these involve I2l terms. Thus, we can write,

Babγ = Bab

γ1 +Babγ2

Babγ = Bab

γ1 +Babγ2 (3.118)

Babγ = Bab

γ1 +Babγ2

Therefore,

Babγ1 = Bbc

γ1 = Bcaγ1 = − 1

3√

3V

[Im(I1l)

](3.119)

And,

Babγ2 = − 1

3√

3V

(Im(I2l)−

√3Re(I2l)

)Bbcγ2 = − 1

3√

3V

(−2 Im(I2l)

)(3.120)

Bcaγ2 = − 1

3√

3V

(Im(I2l) +

√3Re(I2l)

)Earlier in equation, (3.121), it was established that,

I0l = 0

I1l =(Y abl + Y bc

l + Y cal

)√3V

I2l = −(α2Y ab

l + Y bcl + αY ca

l

)√3V

Noting that,

Y abl = Gab

l + jBabl

Y bcl = Gbc

l + jBbcl

Y cal = Gca

l + jBcal

Therefore,

Im (I1l) = Im ((Y abl + Y bc

l + Y cal )√

3V )

= (Babl +Bbc

l +Bcal )√

3V (3.121)

Thus equation (3.119) is re-written as following.

Babγ1 = Bbc

γ1 = Bcaγ1 = −1

3

(Babl +Bab

l +Bcal

)(3.122)

107

Page 114: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Now we shall compute Babγ2, Bbc

γ2 and Bcaγ2 using equations (3.120) as following. Knowing that,

I2l = −(α2Y ab

l + Y bcl + αY ca

l

)√3V

= −

[(−1

2− j√

3

2

)(Gabl + jBab

l

)+(Gbcl + jBbc

l

)+

(−1

2+j√

3

2

)(Gca

l + jBcal )

]√

3V

= −

[−G

abl

2+

√3

2Babl +Gbc

l −Gcal

2−√

3

2Bcal − j

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)]√

3V

=

[Gabl

2−√

3

2Babl −Gbc

l +Gcal

2+

√3

2Bcal + j

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)]√

3V

(3.123)

The above implies that,

Im (I2l) =

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)√

3V

−√

3 Re(I2l) =

(−√

3

2Gabl +

3

2Babl +√

3Gbcl −√

3

2Gcal −

3

2Bcal

)√

3V

Thus, Babγ2 can be given as,

Babγ2 = − 1

3√

3V

[2Bab

l −Bbcl −Bca

l +√

3Gbcl −√

3Gcal

]√3V

= −1

3

[2Bab

l −Bbcl −Bca

l +√

3(Gbcl −Gca

l

)]=

1√3

(Gcal −Gbc

l

)+

1

3

(Bbcl +Bca

l − 2Babl

)(3.124)

Similarly,

Bcaγ2 = − 1

3√

3V

[Im(I2l) +

√3 Re(I2l)

]= − 1

3√

3V

[√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2+

√3

2Gabl −

3

2Babl −√

3Gbcl +

√3

2Gcal +

3

2Bcal

]√

3V

= −1

3

[√3Gab

l −√

3Gbcl −Bab

l −Bbcl + 2Bca

l

]=

1√3

(Gbcl −Gab

l

)+

1

3

(Babl +Bbc

l − 2Bcal

)(3.125)

And, Bbcγ2 is computed as below.

108

Page 115: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Bbcγ2 = − 1

3√

3V

[−2Im(I2l)

]=

2

3√

3V

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)V√

3

=1√3

(Gabl −Gca

l

)+

1

3

(Babl +Bca

l − 2Bbcl

)(3.126)

Using (3.119), (3.124)-(3.126), We therefore can find the overall compensator susceptances asfollowing.

Babγ = Bab

γ1 +Babγ2

= −1

3

(Babγ +Bbc

γ +Bcaγ

)+

1√3

(Gcal −Gbc

l

)+

1

3

(Bcal +Bbc

l − 2Babl

)= −Bab

l +1√3

(Gcal −Gbc

l

)Similarly,

Bbcγ = Bbc

γ1 +Bbcγ2

= −1

3

(Babγ +Bbc

γ +Bcaγ

)+

1√3

(Gabl −Gca

l

)+

1

3

(Babl +Bca

l − 2Bbcl

)= −Bbc

l +1√3

(Gabl −Gca

l

)And,

Bcaγ = Bca

γ1 +Bcaγ2

= −1

3

(Babγ +Bbc

γ +Bcaγ

)+

1√3

(Gbcl −Gab

l

)+

1

3

(Bbcl +Bab

l − 2Bcal

)= −Bca

l +1√3

(Gbcl −Gab

l

)Thus, the compensator susceptances in terms of load parameters are given as in the following.

Babγ = −Bab

l +1√3

(Gcal −Gbc

l

)Bbcγ = −Bbc

l +1√3

(Gabl −Gca

l

)(3.127)

Bcaγ = −Bca

l +1√3

(Gbcl −Gab

l

)109

Page 116: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

It is interesting to observe above equations. The first parts of the equation nullifies the effect of theload susceptances and the second parts of the equations correspond to the unbalance in resistiveload. The two terms together make source current balanced and in phase with the supply voltages.The compensator’s positive and negative sequence networks are shown in Fig. 3.13 .What happens if we just use the following values of the compensator susceptances as given below?

Babγ = −Bab

l

Bbcγ = −Bbc

l (3.128)Bcaγ = −Bca

l

In above case, load suceptance parts of the admittance are fully compensated. However the sourcecurrents after compensation remain unbalanced due to unbalance conductance parts of the load.

Load

1,abcI

sai

sci

sbi

lai

lci

lbi

2,abcI

1abB

1caB

1bcB

2abB

2caB

2bcB

Sou

rce

, ,ab bc caB B B

Fig. 3.13 Sequence networks of the compensator

Example 3.5 For a delta connected load shown in 3.14 , the load admittances are given as follow-ing,

Y abl = Gab

l + jBabl

Y bcl = Gbc

l + jBbcl

Y cal = Gca

l + jBcal

Given the load parameters:

Zabl = 1/Y ab

l = 5 + j12 Ω

Zbcl = 1/Y bc

l = 3 + j4 Ω

Zcal = 1/Y ca

l = 9− j12 Ω

Determine compensator susceptances (Babγ , B

bcγ , B

caγ ) so that the supply sees the load as balanced

and unity power factor. Also find the line currents and source active and reactive powers beforeand after compensation.

110

Page 117: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

caY

abY

a

bc

bcY

abB

bcB

caB

N

av

cv

bv

ci

ai

bi

Fig. 3.14 A delta connected load

Solution:

Zabl = 5 + j12 Ω ⇒ Y ab

l = 0.03− j0.0710 fZbcl = 3 + j4 Ω ⇒ Y bc

l = 0.12− j0.16 fZcal = 9− j13 Ω ⇒ Y ca

l = 0.04 + j0.0533 f

Once we know the admittances we know,

Gabl = 0.03, Bab

l = −0.0710

Gbcl = 0.12, Bbc

l = −0.16

Gcal = 0.04, Bca

l = 0.0533

Babγ = −Bab

l +1√3

(Gcal −Gbc

l ) = 0.0248 f

Bbcγ = −Bbc

l +1√3

(Gabl −Gca

l ) = 0.1540 f

Bcaγ = −Bca

l +1√3

(Gbcl −Gab

l ) = −0.0011 f

Total admittances are:

Y ab = Y abl + Y ab

γ = 0.03− j0.0462 fY bc = Y bc

l + Y bcγ = 0.12− j0.006 f

Y ca = Y cal + Y ca

γ = 0.04 + j0.0522 f

Knowing these total admittances, we can find line currents using following expressions.Current Before Compensation

111

Page 118: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Ia = Iabl − Ical =[(1− α2)Y ab

l − (α− 1)Y cal

]V = 0.2150V ∠−9.51oA

Ib = Ibcl − Iabl =[(α2 − α)Y bc

l − (1− α2)Y abl

]V = 0.4035V ∠−161.66oA

Ic = Ical − Ibcl =[(α− 1)Y ca

l − (α2 − α)Y bcl

]V = 0.2358V ∠43.54oA

Powers Before Compensation

Sa = V a (Ial)∗ = Pa + jQa = V (0.2121 + j0.0355)

Sb = V b (Ibl)∗ = Pb + jQb = V (0.3014 + j0.2682)

Sc = V c (Icl)∗ = Pc + jQc = V (0.0552 + j0.2293)

Total real power, P = Pa + Pb + Pc = V 0.5688W

Total reactive power, Q = Qa +Qb +Qc = V 0.5330 VArPower factor in phase-a, pfa = cosφa = cos(9.51o) = 0.9863 lagPower factor in phase-b, pfb = cosφb = cos(41.63o) = 0.7471 lagPower factor in phase-c, pfc = cosφc = cos(76.45o) = 0.2334 lag

Thus we observe that the phases draw reactive power from the lines and currents are unbalancedin magnitude and phase angles.

After Compensation

Ia = Iab − Ica =[(1− α2)Y ab − (α− 1)Y ca

]V = 0.1896V ∠0oA

Ib = Ibc − Iab =[(α2 − α)Y bc − (1− α2)Y ab

]V = 0.1896V ∠−120oA

Ia = Ica − Ibc =[(α− 1)Y ca

l − (α2 − α)Y bc]V = 0.1896V ∠120oA

Powers After Compensation

Sa = V a (Ial)∗ = Pa + jQa = V (0.1986 + j0.0)

Sb = V b (Ibl)∗ = Pb + jQb = V (0.1986 + j0.0)

Sc = V c (Icl)∗ = Pc + jQc = V (0.1986 + j0.0)

Total real power, P = Pa + Pb + Pc = (V × 0.5688 )W

Total reactive power, Q = Qa +Qb +Qc = 0 VArPower factor in phase-a, pfa = cosφa = cos(0o) = 1.0

Power factor in phase-b, pfb = cosφb = cos(0o) = 1.0

Power factor in phase-c, pfc = cosφc = cos(0o) = 1.0

From above results we observe that after placing compensator of suitable values as calculatedabove, the line currents become balanced and have unity power factor relationship with their volt-ages.

112

Page 119: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Example 3.6 Consider the following 3-phase, 3-wire system. The 3-phase voltages are balancedsinusoids with RMS value of 230 V at 50 Hz. The load impedances are Za = 3 + j 4 Ω, Zb =5 + j 12 Ω, Zc = 12− j 5 Ω. Compute the following.

1. The line currents I la, I lb, I lc.

2. The active (P ) and reactive (Q) powers of each phase.

3. The compensator susceptance ( Babγ , B

bcγ , B

caγ ), so that the supply sees the load balanced and

unity power factor.

4. For case (3), compute the source, load, compensator active and reactive powers (after com-pensation).

Na

c bZ

laI

lbI

lcI

saV

scV

sbV

Fig. 3.15 An unbalanced three-phase three-wire star connected load

Solution:

Given that Za = 3 + j 4 Ω, Zb = 5 + j 12 Ω, Zc = 12− j 5 Ω.

1. Line currents I la, I lb, andI lc are found by first computing neutral voltage as given below.

V nN =1

( 1Za

+ 1Zb

+ 1Zc

)(V a

Za+V b

Zb+V c

Zc)

= (ZaZbZc

ZaZb + ZbZc + ZcZa)(V a

Za+V b

Zb+V c

Zc)

=Zabc∆Z

(V a

Za+V b

Zb+V c

Zc)

=50∠97.82o

252.41∠55.5o(24.12∠−99.55o)

= 43.79− j 67.04V

= 80.75∠−57.15o V

Now the line currents are computed as below.

113

Page 120: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

I la =V a − V nN

Za

=230∠0o − 80.75∠−57.15o

3 + j 4= 33.2− j 21.65

= 39.63∠−33.11oA

I lb =V b − V nN

Zb

=230∠−120o − 80.75∠−57.15o

5 + j 12= 15.85∠152.21oA

I lc =V c − V nN

Zc

=230∠120o − 80.75∠−57.15o

12− j 5= 23.89∠143.35oA

2. Active and Reactive Powers

For phase a

Pa = VaIa cosφa = 230× 39.63× cos(33.11o) = 7635.9W

Qa = VaIa sinφa = 230× 39.63× sin(33.11o) = 4980 VAr

For phase b

Pb = VbIb cosφb = 230× 15.85× cos(−152.21o − 120o) = 140.92W

Qb = VbIb sinφb = 230× 15.85× sin(−152.21o − 120o) = 3643.2 VAr

For phase c

Pc = VcIc cosφc = 230× 23.89× cos(−143.35o + 120o) = 5046.7W

114

Page 121: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Qc = VcIc sinφc = 230× 23.89× sin(−143.35o + 120o) = −2179.3 VAr

Total three phase powers

P = Pa + Pb + Pc = 12823W

Q = Qa +Qb +Qc = 6443.8 VAr

3. Compensator Susceptance

First we convert star connected load to a delta load as given below.

Zabl = ZaZb+ZbZc+ZcZa

Zc= 4 + j 19 = 19.42∠77.11o Ω

Zbcl = ∆Z

Za= 50.44 + j 2.08 = 50.42∠2.36o Ω

Zcal = ∆Z

Zb= 19.0− j 4.0 = 19.42∠− 11.89o Ω

The above implies that,

Y abl = 1/Zab

l = Gabl + j Bab

l = 0.0106− j0.050f

Y bcl = 1/Zbc

l = Gbcl + j Bbc

l = 0.0198− j0.0008f

Y cal = 1/Zca

l = Gcal + j Bca

l = 0.0504 + j0.0106f

From the above, the compensator susceptances are computed as following.

Babγ = −Bab

l +(Gcal −G

bcl )√

3= 0.0681f

Bbcγ = −Bbc

l +(Gabl −G

cal )√

3= −0.0222f

Bcaγ = −Bca

l +(Gbcl −G

abl )√

3= −0.0053f

Zabγ = −j 14.69 Ω (capacitance)

Zbcγ = j 45.13 Ω (inductance)

Zcaγ = j 188.36 Ω (inductance)

115

Page 122: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

4. After Compensation

Zab′

l = Zabl ||Zab

γ = 24.97− j 41.59 = 48.52∠− 59.01o Ω

Zbc′

l = Zbcl ||Zbc

γ = 21.52− j 24.98 = 32.97∠49.25o Ω

Zca′

l = Zcal ||Zca

γ = 24.97− j 41.59 = 19.7332∠− 6.0o Ω

Let us convert delta connected impedances to star connected.

Za =Zab′ × Zca′

Zab′ + Zbc′ + Zca′= 9.0947− j 10.55

= 13.93∠− 49.25o Ω

Zb =Zbc′ × Zab′

Zab′ + Zbc′ + Zca′= 23.15 + j 2.43

= 23.28∠6.06o Ω

Zc =Zca′ × Zbc′

Zab′ + Zbc′ + Zca′= 4.8755 + j 8.12

= 9.47∠59.01o Ω

The new voltage between the load and system neutral after compensation is given by,

V′

nN =1

( 1Za

+ 1Zb

+ 1Zc

)(V a

Za+V b

Zb+V c

Zc)

= 205.42∠−57.15o V

Based on the above, the line currents are computed as following.

Ia =V a − V

nN

Za= 18.58∠0o A

Ib =V b − V

nN

Zb= 18.58∠− 120o A

Ic =V c − V

nN

Zc= 18.58∠120o A

116

Page 123: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Thus, it is seen that after compensation, the source currents are balanced and have unity powerfactor with respective supply voltages.

Source powers after compensation

Pa = Pb = Pc = 230× 18.58 = 4272.14 W

P = 3Pa = 12820.2 W

Qa = Qb = Qc = 0

Q = 0 VAr

Compensator powers

Sab

γ = V ab(Iab

γ

∗)

= V ab(V ab∗.Y abγ

∗)

= V2

abYabγ

= (230×√

3)2 × (−j 0.0068)

= −j 10802 VA

Sbc

γ = V 2bcY

bcγ

= (230×√

3)2 × (j 0.0222)

= j 3516 VASca

γ = V 2caY

caγ∗

= (230×√

3)2 × (j 0.0053)

= j 842 VA

References

[1] A. Ghosh and G. Ledwich, Power quality enhancement using custom power devices. KluwerAcademic Pub, 2002, vol. 701.

[2] T. J. E. Miller, Reactive power control in electric systems. Wiley, 1982.

[3] L. Gyugyi, “Reactive power generation and control by thyristor circuits,” IEEE Transactionson Industry Applications, no. 5, pp. 521–532, 1979.

117

Page 124: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

[4] R. Otto, T. Putman, and L. Gyugyi, “Principles and applications of static, thyristor-controlledshunt compensators,” IEEE Transactions on Power Apparatus and Systems, no. 5, pp. 1935–1945, 1978.

[5] N. Hingorani and L. Gyugyi, Understanding FACTS: Concepts and Technology of Flexible ACTransmission Systems, 2000. Wiley-IEEE Press, 1999.

118

Page 125: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Chapter 4

CONTROL THEORIES FOR LOADCOMPENSATION(Lectures 25-35)

4.1 Introduction

In the previous chapter, we studied the methods of load compensation. These methods can elimi-nate only the fundamental reactive power and unbalance in the steady state. These kinds of com-pensators can be realized using passive LC filters and thyristor controlled devices. However, whenharmonics are present in the system, these methods fails to provide correct compensation. To cor-rect load with unbalance and harmonics, instantaneous load compensation methods are used. Thetwo important theories in this context are Instantaneous Theory of load compensation often knownas pq theory [1] and Instantaneous Symmetrical Component Theory for load compensation [2].These theories will be discussed in this chapter. Their merits and demerits and applications will beexplored in detail.

To begin with pq theory, we shall first recall the αβ0 transformation, which was discussed inChapter 2. For three-phase system shown in Fig. 4.1, the αβ0 transformations for voltages andcurrents are given below.

a

b

c

n

a

b

c

Fig. 4.1 A three phase system

119

Page 126: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

v0

vαvβ

=

√2

3

1√2

1√2

1√2

1 −12−1

2

0√

32−√

32

vavbvc

(4.1)

i0iαiβ

=

√2

3

1√2

1√2

1√2

1 −12−1

2

0√

32−√

32

iaibic

(4.2)

The instantaneous active, p(t) and reactive, q(t) powers were defined in Chapter 2 through equa-tions (2.14)-(2.15) respectively. For the sake of completeness these are given below.

p3φ(t) = vaia + vbib + vcic

= vαiα + vβvβ + v0i0

= pα + pβ + p0

= pαβ + p0 (4.3)

Where, pαβ = pα + pβ = vαiα + vβiβ and p0 = v0i0.In the instantaneous reactive power theory, as discussed in Chapter 2, the instantaneous reactive

power, q(t) was defined as,

q(t) = qαβ = vα × iβ + vβ × iα= vαiβ − vβiα

= − 1√3

[vbcia + vcaib + vabic] (4.4)

Therefore, powers po, pαβ and qαβ can be expressed in matrix form as given below. p0

pαβqαβ

=

v0 0 00 vα vβ0 −vβ vα

i0iαiβ

(4.5)

From the above equation, the currents, i0, iα and iβ are computed as given below.

i0iαiβ

=

v0 0 00 vα vβ0 −vβ vα

−1 p0

pαβqαβ

=1

v0(v2α + v2

β)

v2α + v2

β 0 00 v0vα −v0vβ0 v0vβ v0vα

p0

pαβqαβ

(4.6)

In the above equation,

i0 =p0(v2

α + v2β)

v0(vα2 + vβ2)=p0

v0

=v0i0v0

= i0 (4.7)

iα =vα

v2α + v2

β

pαβ +−vβ

vα2 + vβ2(qαβ)

= iαp + iαq (4.8)

120

Page 127: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

iβ =vβ

v2α + v2

β

pαβ +vα

vα2 + vβ2qαβ

= iβp + iβq (4.9)

wherei0 = zero sequence instantaneous currentiαp = α-phase instantaneous active current = vα

v2α+v2βp

iβp = β-phase instantaneous active current =vβ

v2α+v2βp

iαq = α-phase instantaneous reactive current = − vβv2α+v2β

q

iβq = β-phase instantaneous reactive current = vαv2α+v2β

q

Using above definitions of various components of currents, the three phase instantaneous powercan be expressed as,

p3φ = v0i0 + vαiα + vβiβ

= v0i0 + vα(iαp + iαq) + vβ(iβp + iβq)

= v0i0 + vα

[vα

v2α + v2

β

pαβ +−vβ

v2α + v2

β

qαβ

]+ vβ

[vβ

v2α + v2

β

pαβ +vα

v2α + v2

β

qαβ

]= v0i0 + vαiαp + vαiαq + vβiβp + vβiβq

= v0i0 + (pαp + pαq) + (pβp + pβq)

= v0i0 + (pαp + pβp) (4.10)

In the above equation,

pαq + pβq = vαiαq + vβiβq = 0 (4.11)

If referred to compensator (or filter), the equation (4.6) can be written as,

if0

ifαifβ

=1

v0(v2α + v2

β)

v2α + v2

β 0 00 v0vα −v0vβ0 v0vβ v0vα

pf0

pfαβqfαβ

(4.12)

Since the compensator does not supply any instantaneous real power, therefore,

pf3φ = pf0 + pfαβ = 0 (4.13)

The instantaneous zero sequence power exchanges between the load and the compensator andcompensator reactive power must be equal to load reactive power. Therefore we have,

pfo = plo = voilo (4.14)pfαβ = −plo = −voilo (4.15)qfαβ = ql = vαilβ − vβilα (4.16)

121

Page 128: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Since over all real power from the compensator is equal to zero, therefore the following should besatisfied.

pfo + pfαβ = 0 (4.17)

The power flow description is shown in Fig. 4.2.

l lop p

C

lop

o

lp

P

lq

Fig. 4.2 Power flow description of three-phase 4-wire compensated system

Also the zero sequence current should be circulated through the compensator, therefore,

ifo = ilo (4.18)

Using the conditions of compensator powers as given in the above, the α and β components ofcompensator currents can be given as following.

ifα =vα

v2α + v2

β

pfαβ +−vβ

v2α + v2

β

qfαβ

=1

v2α + v2

β

[vα(−voilo)− vβ(vαilβ − vβilα)]

=1

v2α + v2

β

[−vαvoilo − vβvαilβ + v2βilα)] (4.19)

Similarly,

ifβ =1

v2α + v2

β

[vβ( pfαβ) + vα( qfαβ)]

=1

v2α + v2

β

[vβ(−voilo) + vα(vαilβ − vβilα)]

=1

v2α + v2

β

[−vovβilo − vαvβilα + v2αilβ] (4.20)

122

Page 129: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The above equations are derived based on assumption that in general vlo 6= 0 If vo = 0 ,then

if0 = ilo

ifα =1

v2α + v2

β

[v2β ilα − vαvβ ilβ] (4.21)

ifβ =1

v2α + v2

β

[−vαvβilα + v2α ilβ]

Once the compensator currents, ifo, ifα and ifβ are known, they are transformed back to the abcframe in order to implement in real time. This transformation is given below.

i∗fai∗fbi∗fc

=

√2

3

1√2

1 01√2−1

2

√3

21√2−1

2−√

32

ifoifαifβ

(4.22)

These reference currents are shown in Fig. 4.3. Once reference compensator currents are known,these are tracked using voltage source inverter (VSI). The other details of the scheme can be asfollowing.Thus, compensator powers can be expressed in terms of load powers as following.

sbi

sci

sbv

scv

lbi

lci

LOAD

LOAD

LOAD

N

'n

n

*fci*

fai *fbi

sav sai lai

Fig. 4.3 A three-phase four-wire compensated system with ideal compensator

pfo = plo

pfαβ = (pl − plavg)− plo = pl

pf = pfo + pfαβ = pl

123

Page 130: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The power components and various components of currents are related as following.[pαpβ

]=

[vαiαvβiβ

]=

[vα(iαp + iαq)vβ(iβp + iβq)

]=

[vαiαpvβiβp

]+

[vαiαqvβiβq

]=

[pαppβp

]+

[pαqpβq

](4.23)

The following quantities are defined.α - axis instantaneous active power = pαp = vα iαpα - axis instantaneous reactive power = pαq= vα iαqβ - axis instantaneous active power = pβp = vβ iβpβ - axis instantaneous reactive power = pβq = vβ iβq

It is seen that,

pαp + pβp = vαiαp + vβiβp =(vα)(vα)

v2α + v2

β

p+(vβ)(vβ)

v2α + v2

β

p = (v2α + v2

β

v2α + v2

β

)p = p (4.24)

and

pαq + pβq = vαiαq + vβiβq =(−vβ)(vα)

vα2 + vβ2

p+(vβ)(vα)

vα2 + vβ2

p = 0 (4.25)

Thus, it can be observed that the sum of pαp and pβp is equal to total instantaneous real power p(t)and the sum of pαq and pβq is equal to zero. Therefore,

p3φ = p+ po

= pα + pβ + po

= pαp + pβp + po (4.26)

For an ideal compensator,

pfo = plo = voilo = po

pfαβ = −plo (4.27)qfαβ = ql

For practical compensator, the switching and ohmic losses should be considered. These lossesshould be met from the source in order to maintain the dc link voltage constant. Let these lossesare denoted by Ploss, then the following formulation is used to include this term. Let the average

124

Page 131: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

power that must be supplied to the compensator be ∆p, then ∆p is given as following.

∆p = po + Ploss (4.28)

Thus, the compensator powers can be expressed as,

pf0 = pl0

pfαβ = pl −∆p (4.29)qfαβ = ql

Once these compensator powers are obtained, the compensator currents if0, ifα and ifβ are com-puted using (4.18), (4.19) and (4.20). Knowing these currents, we can obtain compensator currentsin abc frame using equation (4.22). These currents are realized using voltage source inverter (VSI).One of the common VSI topology is illustrated in Fig. 4.4. This VSI topology is known as neutralclamped inverter.

3S 5S

a

c

dc1C

fai fbi fci

sai

sbi

sci

sav

sbv

scv

lai

lbi

lci

Nn

dc2C

4S 6S 2S

1i

LOAD

LOAD

LOAD

1S

2i

b

fR

n

fL

fR

fL

PCC

dc2V

dc1V oi

Fig. 4.4 Voltage Source Inverter

While realizing compensator using voltage source inverter, there are switching and other lossesin the inverter circuit. Therefore, a fraction of total power is required to maintain dc capacitorvoltage to a reference value by generating Ploss term. Once reference filter currents (if0, ifα, ifβ)are obtained, the filter currents in abc system are obtained as below.

i∗fai∗fbi∗fc

=

√2

3

1√2

1 01√2−1

2

√3

21√2−1

2−√

32

ifoifαifβ

As discussed above, the compensator powers are substituted in equation (4.29), the compensator

currents are expressed as below.

125

Page 132: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

ifoifαifβ

=1

v0(v2α + v2

β)

v2α + v2

β 0 00 vovα −v0vβ0 v0vβ v0vα

plopl + p0 − ploss

qlαβ

ifoifαifβ

=1

v0(v2α + v2

β)

v2α + v2

β 0 00 vovα −v0vβ0 v0vβ v0vα

plopl + p0 − ploss

qlαβ

Now once we know i∗fa, i

∗fb, i

∗fc signals, these have to be synthesized using voltage source

inverter. A typical voltage source inverter (VSI) along with a three-phase compensated system asshown in Fig. 4.5.

Voltage regulator

Active filtercontroller

lai

lbi

lci

sav sbv scv

fai

fci

lossP

Voltage sourceinverter

Dymanic hysteresis current

control

fai

fci

dcVdc refV

LOAD

Source

sai lai

faifci

scisbisav

sbv

0sni

nli

scv

6S1S

Fig. 4.5 Control algorithm for three-phase compensated system

4.1.1 State Space Modeling of the Compensator

There are different VSI topologies to realize [3]. The most commonly used is neutral clampedinverter topology as shown in Fig. 4.4. Since this is a three phase four-wire system each phase canbe considered independently. Therefore, to analyze above circuit, only one phase is considered,which is shown in Fig. 4.6. The other phases work similarly. In Fig. 4.6(a), for switch S1 is closedand switch S4 is open, the KVL can be written as below.

Lfdifadt

+Rf ifa + vsa − Vdc1 = 0 (4.30)

From the above equation,

difadt

= −Rf

Lfifa −

vsaLf

+Vdc1Lf

(4.31)

126

Page 133: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

2dcC

1dcC

fR fL savfai

-

+dc1V

-

+dc2V

2dcC

1dcC

4 (ON)S

fR fL savfai

-

+dc1V

-

+dc2V

1 (OFF)S1 (ON)S

4 (OFF)S

(a) (b)

Fig. 4.6 Equivalent circuit (a) S1 (ON), S4 (OFF) (b) S1 (OFF), S4 (ON)

Similarly, when S1 is open and S4 is closed as shown in Fig. 4.6 (b),

difadt

= −Rf

Lfifa −

vsaLf− Vdc2

Lf. (4.32)

The above two equations can be combined into one by using switching signals Sa, Sa, as givenbelow.

difadt

= −Rf

Lfifa + Sa

Vdc1Lf− Sa

Vdc2Lf− vsaLf

(4.33)

Similarly, for phases b and c, the first order derivative of filter currents can be written as following.

difbdt

= −Rf

Lfifb + Sb

Vdc1Lf− Sb

Vdc2Lf− vsbLf

(4.34)

difcdt

= −Rf

Lfifc + Sc

Vdc1Lf− Sc

Vdc2Lf− vscLf

(4.35)

where, Sa = 0 and Sa = 1 implies that the top switch is open and bottom switch is closed and,Sa = 1 and Sa = 0 implies that the top switch is closed and bottom switch is open. The two logicsignals Sa and Sa are complementary to each other. This logic also holds for the other two phases.

The inverter currents i1 and i2 as shown in Fig. 4.4, can be expressed in terms of filter currentsand switching signals. These are given below.

i1 = Sa ifa + Sb ifb + Sc ifc

i2 = Sa ifa + Sb ifb + Sc ifc (4.36)

The relationship between DC capacitor voltages Vdc1, Vdc2 and inverter currents i1 and i2 is givenas below.

Cdc1dVdc1dt

= −i1

Cdc2dVdc2dt

= i2 (4.37)

127

Page 134: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Considering Cdc1 = Cdc2 = Cdc and substituting i1 and i2 from (4.36), the above equations can bewritten as,

dVdc1dt

= − SaCdc

ifa −SbCdc

ifb −ScCdc

ifc (4.38)

dVdc2dt

=SaCdc

ifa +SbCdc

ifb +ScCdc

ifc (4.39)

The equations (4.33), (4.34), (4.35), (4.38) and (4.39) can be represented in state space form asgiven below.

d

dt

ifaifbifcVdc1Vdc2

=

−RfLf

0 0 SaLf−SaLf

0 −RfLf

0 SbLf− SbLf

0 0 −RfLf

ScLf− ScLf

−SaC−Sb

C−Sc

C0 0

SaC

SbC

ScC

0 0

ifaifbifcVdc1Vdc2

+

− 1Lf

0 0

0 − 1Lf

0

0 0 − 1Lf

0 0 00 0 0

vsavsbvsc

(4.40)

The above equation is in the form,

x = Ax+B u. (4.41)

Where, x is a state vector, A is system matrix, B is input matrix and u is input vector. This statespace equation can be solved using MATLAB to implement the compensator for simulation study.

4.1.2 Switching Control of the VSI

In the equation (4.40), the switching signals Sa, Sa, Sb, Sb, Sc, and Sc are generated using a hys-teresis band current control. This is described as following. The upper and lower bands of thereference filter current (say phase-a) are formed using hysteresis h i.e., i∗fa + h and i∗fa − h. Then,following logic is used to generate switching signals.

If ifa ≥ (i∗fa + h)

Sa = 0 and Sa = 1else if ifa ≤ (i∗fa − h)

Sa = 1 and Sa = 0else if (i∗fa − h) < ifa < (i∗fa + h)Retain the current status of the switches.end

The first order derivative of state variables can be easily solved using ‘c2d’ (continuous to dis-crete) command in MATLAB. It is given below.

[Ad Bd] = c2d(A, B, td) (4.42)

128

Page 135: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The value of the state vector is updated using the following equation.

x[(k + 1)T ] = Ad x[kT ] +Bd u[kT ] (4.43)

Where x(k + 1) refers the value of the state vector at (k + 1)th sample. The Ad and Bd computedby ‘c2d′ function as described above can be expressed as below.The solution of state equation given by (4.41) is given as following [Nagrath].

x(t) = eA(t−t0) x(t0) +

t∫to

eA(t−τ)B u(τ)dτ (4.44)

Where to represents initial time and t represents final time. The above equation can be re-writtenas following.

x(t) = φ(t− to)x(to) +

t∫to

φ(t− τ)B u(τ) dτ (4.45)

Writing above equation for small time interval, kT ≤ t ≤ (k+1)T with to = kT and t = (k+1)T ,

x[(k + 1)T ] = eAT x[kT ] +

(k+1)T∫kT

eA (k+1)T−τB u[kT ] dτ

= eAT x[kT ] +

(k+1)T∫kT

eA (k+1)T−τB dτ

u[kT ] (4.46)

Comparing (4.43) and (4.46), the discrete matrices Ad and Bd computed by ‘c2d′ MATLAB func-tion can be written as following.

Ad = eAT

Bd =

(k+1)T∫kT

eA (k+1)T−τB dτ (4.47)

4.1.3 Generation of Ploss to maintain dc capacitor voltage

The next step is to determine Ploss in order to maintain the dc link voltage close to its referencevalue. In compensation, what could be an indication of Ploss to account losses in the inverter. Theaverage voltage variation of dc link may be an indicator of Ploss in the inverter. If losses are morethan supplied by the inverter, the dc link voltage, i.e., Vdc = Vdc1 + Vdc2, will decline towards zeroand vice versa. For proper operation of compensator, we need to maintain dc capacitor voltageto two times of the reference value of each capacitor voltage i.e., Vdc1 + Vdc2 = Vdc = 2Vdcref .Thus, we have to replenish losses in inverter and sustain dc capacitor voltage to 2Vdcref with each

129

Page 136: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

capacitor voltage to Vdcref . This is achieved with the help of proportional integral (PI) controllerdescribed below [4]. Lets define an error signal as following.

eV dc = 2Vdcref − (Vdc1 + Vdc2) = 2Vdcref − Vdc

Then, the term Ploss is computed as following.

Ploss = Kp eV dc +Ki

∫ Td

0

eV dc dt

This control loop need not be very fast and may be updated once in a voltage cycle, preferably atthe positive of phase-a voltage and generate Ploss term at these points. The above controller can beimplemented using digital domain as following.

Ploss (k) = Kp eV dc (k) +Ki

k∑j=0

eV dc (j)Td. (4.48)

In the above equation, k represents the kth sample of error, eV dc. For k = 1, the above equation canbe written as,

Ploss (1) = Kp eV dc (1) +Ki

1∑j=0

eV dc (j)Td.

= Kp eV dc (1) +Ki[eV dc(0) + eV dc(0)]Td. (4.49)

Similarly for k = 2, we can write,

Ploss (2) = Kp eV dc (1) +Ki [eV dc (0) + eV dc (1) + eV dc (2)]Td. (4.50)

Replacing Ki[eV dc(0) + eV dc(0)] from (4.49), we get,

Ki [eV dc (0) + eV dc (1)]Td = Ploss (1)−Kp eV dc(1) (4.51)

Substituting above value in (4.50),we obtain the following.

Ploss (2) = Kp eV dc (2) + Ploss (1)−KpeV dc (1) +KieV dc (2)Td

= Ploss (1) +Kp [eV dc (2)− eV dc (1)] +Ki eV dc (2)Td (4.52)

In general, for kth sample of Ploss,

Ploss (k) = Ploss (k − 1) +Kp [eV dc (k)− eV dc (k − 1)] +Ki eV dc (k)Td. (4.53)

The algorithm can be used to implement PI controler to generate Ploss. The control action can beupdated at every positive zero crossing of phase-a voltage for example.

4.1.4 Computation of load average power (Plavg)

In reference current expressions, the average load power (Plavg) is required to be computed. Al-though low pass filter can be used to find this, however the dynamic response is quite slow. The

130

Page 137: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

dynamic performance of computation of Plavg plays significant role in compensation. For thisreason, a moving average algorithm can be used, which is described below.

〈pl〉 = Plavg = 〈va ila + vb ilb + vc ilc〉

=1

T

∫ T

0

(va ila + vb ilb + vc ilc) dt (4.54)

The above equation can be written with integration from t1 to t1 + T as given in the following.

Plavg =1

T

∫ t1+T

t1

(vaila + vbilb + vcilc) dt (4.55)

This is known as moving average filter (MAF). Any change in variables instantly reflected withsettling time of one cycle.

4.2 Some Misconception in Reactive Power Theory

The instantaneous reactive power theory, that has evolved from Fortesque, Park and Clarke Trans-formations of voltage and current specified in phases-a, b and c coordinates [5]. In general, for3-phase, 4-wire system, vovα

=

√2

3

1√2

1√2

1√2

1 −12−1

2

0√

32−√

32

vavbvc

(4.56)

Similarly, for currents, the αβ0 components are given as following.ioiαiβ

=

√2

3

1√2

1√2

1√2

1 −12−1

2

0√

32−√

32

iaibic

(4.57)

For balanced system v0 = (va+vb+vc)/√

3 = 0. For three-phase, three-wire system, ia+ib+ic =0, which implies that i0 = 0. Using these details, the above transformations in equations (4.56)and (4.57) result to the following.

[vαvβ

]=

√2

3

[1 −1

2−1

2

0√

32−√

32

]vavbvc

(4.58)

[iαiβ

]=

√2

3

[1 −1

2−1

2

0√

32−√

32

]iaibic

(4.59)

131

Page 138: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From equation (4.58), and using va + vb + vc = 0, we get the following.

vα =

√2

3

[va −

vb + vc2

]=

√2

3

[va −

vb2− vc

2

]=

√2

3

[va −

vb2−(−va

2− vb

2

)]=

√3

2va (4.60)

vβ =

√2

3

[√3

2vb −

√3

2vc

]

=

√2

3

[√3

2vb −

√3

2(−va − vb)

]

=

√2

3

[√3

2vb +

√3

2va +

√3

2vb

]=

1√2va +

√2vb (4.61)

Writing equations (4.60) and (4.61) in matrix form we get,[vαvβ

]=

[√3

20

1√2

√2

] [vavb

](4.62)

Similary, using ia + ib + ic = 0 the following can be written.[iαiβ

]=

[√3

20

1√2

√2

] [iaib

](4.63)

According to the pq theory, the abc components of voltages and currents are transformed to the αand β coordinates and the instantaneous powers p and q of the load can be expressed as following.

p = vαiα + vβiβ (4.64)q = vαiβ − vβiα (4.65)

The above equations representing instantaneous active and reactive powers can be expressed inmatrix form as following [1]. [

pq

]=

[vα vβ−vβ vα

] [iαiβ

](4.66)

132

Page 139: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Therefore from above equation (4.66), the α β components of currents can be expressed as follow-ing. [

iαiβ

]=

[vα vβ−vβ vα

]−1 [pq

]

The matrix,[vα vβ−vβ vα

]−1

is given as following.

[vα vβ−vβ vα

]−1

=1

v2α + v2

β

[vα −vβvβ vα

](4.67)

From the above equation,

iα =vα

v2α + v2

β

p− vβv2α + v2

β

q (4.68)

iβ =vβ

v2α + v2

β

p+vα

v2α + v2

β

q (4.69)

Which can further be written as,

iα = iαp + iαq (4.70)iβ = iβp + iβq (4.71)

In the above equation,

iαp =vα

v2α + v2

β

p (4.72)

iαq = − vβv2α + v2

β

q (4.73)

iβp =vβ

v2α + v2

β

p (4.74)

iβp =vα

v2α + v2

β

q (4.75)

The instantaneous active and reactive components of currents in supplying line can be calculatedfrom the α and β components of the current as given in the following.[

iaib

]=

[√3

20

1√2

√2

]−1 [iαiβ

]=

[√23

0

− 1√6

1√2

][iαp + iαqiβp + iβq

]

=

[√23

0

− 1√6

1√2

][iαpiβp

]+

[√23

0

− 1√6

1√2

] [iαqiβq

]

[iapibp

]=

[√23

0

− 1√6

1√2

][iαpiβp

](4.76)

133

Page 140: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

and,

[iaqibq

]=

[√23

0

− 1√6

1√2

][iαqiβq

](4.77)

The active and reactive components of the line currents must be consistent to the basic defini-tions. However, these components of currents have little in common with the reactive power of theload as defined in [5]. This is shown in the following illustrations.

Example 4.1 Assume a resistive load connected as shown in the 4.7 below. It is supplied froma symmetrical source of a sinusoidal balanced voltage with va =

√2V sinωt, with V = 230 Volts.

Express the voltage and currents for primary and secondary side of the transformer. Express theactive and reactive component of the currents, powers and discuss them.

230 0AV V

n

AI

BI

CI

BV

bI

cI

CV

aI aV

bV

cV

Fig. 4.7 An unbalanced resistive load supplied by three-phase delta-star connected transformer

Solution: With he above given values, the primary side phase voltages with respect to virtualground could be expressed as the following.

vA =√

2V sinωt = 230√

2 sinωt

vB =√

2V sin(ωt− 120o) = 230√

2 sin(ωt− 120o) (4.78)

vC =√

2V sin(ωt+ 120o) = 230√

2 sin(ωt+ 120o)

In phasor form,

VA = 230∠0VVB = 230∠−120VVC = 230∠120V

Therefor the primary side line-to-line voltage are expressed as following.

vAB =√

2√

3V sin(ωt+ 30o)

vBC =√

2√

3V sin(ωt− 90o) (4.79)

vCA =√

2√

3V sin(ωt+ 150o)

134

Page 141: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

In phasor form,

V AB = 398.37∠30VV BC = 398.37∠−90VV CA = 398.37∠150V

These voltages are transformed to the secondaries and are expressed below.

va =√

2√

3V sinωt = 398.37√

2 sin(ωt+ 30)

vb =√

2√

3V sin(ωt− 120o) = 398.37√

2 sin(ωt− 90) (4.80)

vc =√

2√

3V sin(ωt+ 120o) = 398.37√

2 sin(ωt+ 150)

In phasor form,

Va = 398.37∠30VVb = 398.37∠−90VVc = 398.37∠150V

Therefore the currents on the secondary side are given below.

ia =

√2√

3V

Rcosωt

ib = 0 (4.81)ic = 0

Taking V = 230 V and R = 4 Ω, the currents on the secondary side of the transformer are given asfollowing.

ia =vaR

=

√2√

3V

4sinωt = 99.56

√2 sin(ωt+ 30) =

√2I sin(ωt+ 30) (4.82)

ib = 0

ic = 0

In phasor form, the above can be expressed as,

Ia = 99.59∠30A (4.83)Ib = 0 A (4.84)Ic = 0 A (4.85)

This phase-a current ia in the secondary side of transformer is transformed to the primary of thedelta connected winding, therefore the currents on the secondary side of transformer are given asfollowing.

iA =√

2I sin(ωt+ 30) (4.86)

iB = −iA = −√

2I sin(ωt+ 30)

iC = 0

135

Page 142: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The above can be written in phasor form as given below.

IA = 99.59∠30o = I∠30o AIB = −IA = 1∠−180o × 99.59∠30o

= 99.59∠− 150o = I∠− 150o A (4.87)IC = 0 A

After, knowing the voltages and currents of the primary side of the transformer, their α and βcomponents are expressed as below.[

vαvβ

]=

[√32

01√2

√2

][vAvB

](4.88)

Substituting vA and vB from (4.102) in the above equation, we get the following.[vαvβ

]=

[√32

01√2

√2

] [ √2V sinωt√

2V sin(ωt− 120)

]=

[ √3V sinωt

−√

3V cos(ωt)

](4.89)

And, [iαiβ

]=

[√32

01√2

√2

] [iAiB

]=

[√32

01√2

√2

][ √2 I sin(ωt+ 30)

−√

2 I sin(ωt+ 30)

]=

[√3 I sin(ωt+ 30)−I sin(ωt+ 30)

](4.90)

Based on the above the active and reactive powers are computed as below.

p(t) = vαiα + vβiβ

=√

3V sin(ωt)√

3I sin(ωt+ 30)−√

3V cos(ωt)(−I) sin(ωt+ 30))

= 2√

3V I sin(ωt+ 30)

[√3

2sin(ωt) +

1

2cos(ωt)

]=√

3V I [2 sin(ωt+ 30) sin(ωt+ 30)]

=√

3V I[2 sin2(ωt+ 30)

]=√

3V I [1− cos 2(ωt+ 30)] (4.91)

q(t) = vαiβ − vβiα=√

3V sin(ωt) −I sin(ωt+ 30) − (−√

3V cos(ωt))√

3I sin(ωt+ 30)

= −√

3V I 2 sin(ωt+ 30)

[1

2sinωt−

√3

2cosωt

]= −

√3V I 2 sin(ωt+ 30)(− cos(ωt+ 30))

=√

3V I sin 2(ωt+ 30) (4.92)

136

Page 143: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Based on above values of p and q powers, the α and β components of active and reactive compo-nents are given below. [

iαiβ

]=

[iαp + iαqiβp + iβq

]Where,

iαp =vα

v2α + v2

β

p

=

√3V sin(ωt)

(√

3V sin(ωt))2 + (−√

3V cos(ωt))2p

=1√3V

sin(ωt)√

3V I(1− cos 2(ωt+ 30))

= I sinωt 1− cos 2(ωt+ 30) (4.93)

Similarly,

iβp =vβ

v2α + v2

β

p

=−√

3V cos(ωt)

(√

3V I sin(ωt))2 + (−√

3V cos(ωt))2p

=−√

3V cos(ωt)

3V 2

√3V I(1− cos 2(ωt+ 30))

= −I cosωt 1− cos 2(ωt+ 30) (4.94)

iαq =−vβ

v2α + v2

β

q

=−(−√

3V cosωt)

3V 2

√3V I sin 2(ωt+ 30)

= I cosωt sin 2(ωt+ 30) (4.95)

iβq =vα

v2α + v2

β

q

=

√3V sinωt

3V 2

√3V I sin 2(ωt+ 30)

= I sinωt sin 2(ωt+ 30) (4.96)

Thus knowing iαp, iαq, iβp and iβq we can determine active and reactive components of currents onthe source side as giben below. [

iapibp

]=[C]−1[iαpiβp

]137

Page 144: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Where,

[C]−1 =

[√32

01√2

√2

]−1

=

[√23

0−1√

61√2

]

Using above equation, we can find out the active and reactive components of the current, as givenbelow.

[iapibp

]=

[√23

0−1√

61√2

] [I sinωt 1− cos 2(ωt+ 30)−I cosωt 1− cos 2(ωt+ 30)

]From the above,

iap =

√2

3I sinωt 1− cos 2(ωt+ 30)

=

√2

3

I

22 sinωt− 2 sinωt cos 2(ωt+ 30)

=I√62 sinωt− sin(3ωt+ 60)− sin(−ωt− 60)

=I√62 sinωt+ sin(ωt+ 60)− sin(3ωt+ 60)

ibp = − 1√6iαp +

1√2iβp

= − 1√6I sinωt 1− cos 2(ωt+ 30)+

1√2

(−I cosωt) 1− cos 2(ωt+ 30)

= −I 1− cos 2(ωt+ 30)[

sinωt√6

+cosωt√

2

]= − 2I√

61− cos 2(ωt+ 30)

[1

2sinωt+

√3

2cosωt

]= − 2I√

61− cos 2(ωt+ 30) sin(ωt+ 60)

= − 2I√6sin(ωt+ 60)− sin(ωt+ 60) cos 2(ωt+ 30)

= − 2I√6

sin(ωt)

1

2+ cos(ωt)

√3

2+

1

2sin(ωt)− 1

2sin(3ωt+ 120)

= − I√

6sin(ωt) + 2 sin(ωt+ 60)− sin(3ωt+ 120)

138

Page 145: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

iaq =

√2

3iαq

=

√2

3

I

22 sin 2(ωt+ 30) cosωt

=I√6sin(ωt+ 60) + sin(3ωt+ 60)

ibq = − 1√6iαq +

1√2iβq

= − 1√6I cosωt sin 2(ωt+ 30)+

1√2

(I sinωt) sin 2(ωt+ 30)

= − I√6

2 sin 2(ωt+ 30)

−1

2cosωt+

√3

2sinωt)

=

I√6

2 sin 2(ωt+ 30) sin(ωt− 30)

=I√6cos(ωt+ 90)− cos(3ωt+ 30)

=I√6− sin(ωt)− cos(3ωt+ 30)

Thus we have,

iap =I√62 sinωt+ sin(ωt+ 60)− sin(3ωt+ 60) (4.97)

ibp = − I√6sin(ωt) + 2 sin(ωt+ 60)− sin(3ωt+ 120) (4.98)

iaq =I√6sin(ωt+ 60) + sin(3ωt+ 60) (4.99)

ibq =I√6− sin(ωt)− cos(3ωt+ 30) . (4.100)

From the above equations, the names instantaneous active current and instantaneous reactive cur-rent given in the pq theory do not have commonality with the notion of active and reactive currentsused in electrical engineering. Also, the reactive current iq occurs in supply lines of load in spiteof the absence of the reactive element of the in the load. Furthermore, the nature of load is linearand harmonics are absent, still resolutions of active and reactive components of the current basedon pq theory gives harmonics. For example, in the above discussion,

iap =I√62 sinωt+ sin(ωt+ 60)− sin(3ωt+ 60) (4.101)

is the active current component in the line ‘a’ and it contains the third order harmonic. Thiscontradicts the basic notion of the active current that was introduced to electrical engineering by

139

Page 146: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Fryze [5]. Thus, it seems a major misconception of electrical phenomenon in three phase circuitswith balanced sinusoidal voltages for linear load that do not have harmonics. Moreover, the activecurrent ip that results from pq theory is not the current that should remain in the supply lines afterthe load is compensated to unity power factor as defined by Fryze. Therefore it can not be a com-pensation goal.

Also, it is evident that the instantaneous reactive power q(t) as defined by pq theory does notreally identify the power properties of load instantaneously. For example for the above discussion,the active and reactive power are given as following.

p(t) =√

3V I 1− cos 2(ωt+ 30)q(t) =

√3V I sin 2(ωt+ 30)

The following points are noted.

1. The active components of currents (iap, ibp, icp) and reactive components of currents, (iaq, ibq, icq)contain third harmonic, which is not possible for a linear load as discussed above.

2. The sum of reactive components of currents, iaq and ibq is not equal to zero, i.e., iaq+ibq 6= 0),even though no overall reactive power is required from the load.

3. The instantaneous reactive power q(t) defined by pq theory does not identify the power prop-erties of the load instantaneously. Both powers p(t) and q(t) are time varying quantities, sothat a pair of their values at any single point of time does not identify the power propertiesof load. The possibility of instantaneous identification of active and reactive power p(t) andq(t) does not mean that power properties of load are identified instantaneously. For example,

At (ωτ + 30) = 90,

p(t) = 2

√3V I

q(t) = 0

The above implies that as if it is resistive load.

Similarly at (ωτ + 30) = 0,

p(t) = 0q(t) = 0

Which implies as there is no load.

And when (ωτ + 30) = 105,

p(t) =

√3V I (1 +

√32)

q(t) = −√

3V I(12)

implies as it is capacitive load.

Similarly when (ωτ + 30) = 75,

p(t) =

√3V I (1 +

√32)

q(t) =√

3V I(12)

implies as if the load is inductive.

140

Page 147: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

We therefore conclude that power properties cannot be identified without monitoring of the ofp(t) and q(t) powers over the entire cycle period. For example in above case, the instantaneousreactive power q(t) has occurred is not because of load reactive power Q but because of voltageunbalance. This unbalance nature of load can not be identified by instantaneous reactive powerq(t) values. Therefore, pq theory gives no advantage with respect to the time interval needed toidentify the nature of load and its property over the the over power theories based on time domainor frequency domain approach that required the system to be monitored over one period.

Thus, we have seen that each phase has some reactive power. But there is no reactive element.This reactive power appear because of unbalance in the system and not because of reactive compo-nent. So this is an additional information what is required. From this illustration, it is evident thatthe instantaneous reactive power current has commonality with the load reactive power Q. It alsoappears that the instantaneous active current in pq theory (iap, ibp, icp) have no commonality withthe load active power P .

Powers computation

The secondary side powers are given as following.

Sa = Pa + jQa = V a I∗a = 398.37∠30 99.59∠− 30 = 39675 VA

Thus,Pa = 39675 W, Qa = 0 VAr

Sb = Pb + jQb = V b I∗b = 398.37∠− 90 × 0 = 0 VA

Pb = 0 W, Qb = 0 VAr

Sc = Pc + jQc = V c I∗c = 398.37∠150 × 0 = 0 VA

Pc = 0 W, Qc = 0 VAr

The total active and reactive powers on the secondary side are given as following.

P = Pa + Pb + Pc = 39675 WQ = Qa +Qb +Qc = 0 VAr

Svect = Sarith = P = 39675 VApfvect = pfarith = P/S = 1.0

The primary side powers are given as following.

SA = PA + jQA = V a I∗a = 230∠0 99.59∠− 30 = 19837.50− j11453.16 VA

Thus,PA = 19837.50 W, QA = −11453.160 VAr

SB = PB + jQB = V b I∗b = 230∠− 120 (−99.59∠30)∗ =

PB = 19837.50 W, QB = −11453.160 VAr

SC = PC + jQC = V C I∗C = 230∠120 × 0

PC = 0 W, QC = 0 VAr

141

Page 148: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The total active and reactive powers on the primary side are given as following.

P = Pa + Pb + Pc = 39675 WQ = Qa +Qb +Qc = 0 VAr

Svect =∣∣SA + SB + SC

∣∣ = P = 39675 VA

Sarith =∣∣SA∣∣+

∣∣SB∣∣+∣∣SC∣∣ = 22906 + 22906 + 0 = 45813 VA

pfvect = P/Svect = 1.0

pfarith = P/S = 39675/45813 = 0.866

Example 4.2: Assume an Inductive load connected as shown in the 4.7 below. It is supplied froma symmetrical source of a sinusoidal balanced voltage with va =

√2V sinωt, with V = 230 Volts.

Express the voltage and currents for primary and secondary side of the transformer. Express theactive and reactive component of the currents, powers and discuss them.

X

230 0AV V

n

AI

BI

CI

BV

CV

aI aV

bV

cVbI

cI

Fig. 4.8 An unbalanced reactive load supplied by three-phase delta-star connected transformer

Solution: With the above given values, the primary side phase voltages with respect to virtualground could be expressed as the following.

vA =√

2V sinωt = 230√

2 sinωt

vB =√

2V sin(ωt− 120o) = 230√

2 sin(ωt− 120o)

vC =√

2V sin(ωt+ 120o) = 230√

2 sin(ωt+ 120o)

Therefore, the primary side line-to-line voltage are expressed as following.

vAB =√

2√

3V sin(ωt+ 30)

vBC =√

2√

3V sin(ωt− 90)

vCA =√

2√

3V sin(ωt+ 150)

These voltages are transformed to the secondaries and are expressed below.

va =√

2√

3V sinωt = 398.37√

2 sin(ωt+ 30)

vb =√

2√

3V sin(ωt− 120o) = 398.37√

2 sin(ωt− 90)

vc =√

2√

3V sin(ωt+ 120o) = 398.37√

2 sin(ωt+ 150)

142

Page 149: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

In phasor form, the above voltages are expressed as below.

V a =√

3V ∠30

V b =√

3V ∠− 90

V c =√

3V ∠150

Therefore, the currents on the secondary side are given below.

ia =

√2√

3V

Xsin(ωt− 60) = 99.59

√2 sin(ωt− 60)

ib = 0

ic = 0

In phasor form, the above can be expressed as,

Ia = 99.59∠− 60A.

The above phase-a current (ia) is transformed to the primary of the delta connected winding. Sincethe currents should have 90 phase shift with respect to the voltages across the windings as givenby (4.102), therefore the currents on the secondary side of transformer are given as following.

iA = = iAB = ia =√

2I sin(ωt− 60o)

iB = −iA = −√

2I sin(ωt− 60)

iC = 0

In phasor form, the above can be expressed as,

IA = I∠− 60 = 99.59∠− 60AIB = −I∠− 60 = 99.59∠− 60AIC = 0 A.

After, knowing the voltages and currents of the primary side of the transformer, their α and βcomponents are expressed as below.[

vαvβ

]=

[√32

01√2

√2

][vAvB

]Substituting vA and vB from (4.102) in the above equation, we get the following.[

vαvβ

]=

[√32

01√2

√2

] [ √2V sinωt√

2V sin(ωt− 120)

]=

[ √3V sinωt

−√

3V cos(ωt)

](4.102)

And, [iαiβ

]=

[√32

01√2

√2

] [iAiB

]=

[√32

01√2

√2

][ √2 I sin(ωt− 60)

−√

2 I sin(ωt− 60)

]=

[√3 I sin(ωt− 60)−I sin(ωt− 60)

](4.103)

143

Page 150: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Based on the above, the active and reactive powers are computed as below.

p(t) = pαβ = vαiα + vβiβ

=√

3V sin(ωt)√

3I sin(ωt− 60) + (−√

3V cosωt)(−I sin(ωt− 60))

= 2√

3V I sin(ωt− 60)

[√3

2sinωt+

1

2cosωt

]=√

3V I [2 sin(ωt− 60) cos(ωt− 60)]

=√

3V I sin 2(ωt− 60) (4.104)

q(t) = vαiβ − vβiα=√

3V sinωt −I sin(ωt− 60) − (−√

3V cosωt)√

3I sin(ωt− 60)

= −√

3V I 2 sin(ωt− 60)

[1

2sinωt−

√3

2cosωt

]= −

√3V I 2 sin(ωt− 60)(sin(ωt− 60))

= −√

3V I 2 sin2(ωt− 60)

= −√

3V I 1− cos 2(ωt− 60) (4.105)

Based on above values of p and q powers, the α and β components of active and reactive compo-nents are given below.

[iαiβ

]=

[iαp + iαqiβp + iβq

]

Where,

iαp =vα

v2α + v2

β

p

=

√3V sinωt

(√

3V sinωt)2 + (−√

3V cosωt)2p

=1√3V

sinωt√

3V I(sin 2(ωt− 60))

= I sinωt sin 2(ωt− 60) (4.106)

Similarly,

144

Page 151: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

iβp =vβ

v2α + v2

β

p

=−√

3V cosωt

(√

3V I sinωt)2 + (−√

3V cosωt)2p

=−√

3V cosωt

3V 2

√3V I sin 2(ωt− 60)

= −I cosωt sin 2(ωt− 60) (4.107)

iαq =−vβ

v2α + v2

β

q

=−(−√

3V cosωt)

3V 2

[−√

3V I 1− cos 2(ωt− 60)]

= −I cosωt(1− cos 2(ωt− 60)) (4.108)

iβq =vα

v2α + v2

β

q

=

√3V sinωt

3V 2

[−√

3V I 1− cos 2(ωt− 60)]

= −I sinωt 1− cos 2(ωt− 60) (4.109)

Thus, knowing iαp, iαq, iβp and iβq, we can determine active and reactive components of currentson the source side as given below. [

iapibp

]=[C]−1[iαpiβp

]Where,

[C]−1 =

[√32

01√2

√2

]−1

=

[√23

0−1√

61√2

](4.110)

Using above equation, we can find out the active and reactive components of the current, as givenbelow.

[iapibp

]=

[√23

0−1√

61√2

][I sinωt sin 2(ωt− 60)−I cosωt sin2(ωt− 60)

]From the above,

145

Page 152: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

iap =

√2

3I sinωt sin 2(ωt− 60)

=

√2

3

I

22 sinωt sin 2(ωt− 60)

=I√6cos(ωt− 2ωt+ 120)− cos(3ωt− 120)

=I√6cos(ωt− 120)− cos(3ωt− 120) (4.111)

ibp = − 1√6iαp +

1√2iβp

= − 1√6I sinωt sin 2(ωt− 60)+

1√2

(−I cosωt) sin 2(ωt− 60)

= − I√62 sin 2(ωt− 60)

[1

2sinωt+

√3

2cosωt

]= − I√

62 sin 2(ωt− 60) sin(ωt+ 60)

= − I√6cos(ωt− 180)− cos(3ωt− 60)

=I√6cosωt+ cos(3ωt− 60)

iaq =

√2

3iαq

= −√

2

3I 1− cos 2(ωt− 60) cosωt

= − I√62 cosωt− 2 cosωt cos 2(ωt− 60)

= − I√62 cosωt− cos(3ωt− 120)− cos(ωt− 120)

=I√6−2 cosωt+ cos(ωt− 120) + cos(3ωt− 120) (4.112)

146

Page 153: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

ibq = − 1√6iαq +

1√2iβq

= − 1√6

[−I cosωt 1− cos 2(ωt− 60)] +1√2

[−I sinωt 1− cos 2(ωt− 60)]

= − I√6

2 1− cos 2(ωt− 60)

−1

2cosωt−

√3

2sinωt)

=I√6

2 1− cos 2(ωt− 60)

1

2cosωt+

√3

2sinωt)

=

I√6

2 1− cos 2(ωt− 60 cos(ωt− 60)

=I√62 cos(ωt− 60)− 2 cos(ωt− 60) cos 2(ωt− 60)

=I√62 cos(ωt− 60) + cos 3ωt− cos(ωt− 60) (4.113)

Thus we have,

iap =I√6cos(ωt− 120)− cos(3ωt− 120)

ibp =I√6cos(ωt) + cos(3ωt− 60)

iaq =I√6−2 cosωt+ cos(ωt− 120) + cos(3ωt− 120)

ibq =I√62 cos(ωt− 60) + cos 3ωt− cos(ωt− 60) .

From above equations, it is clear that there exist active components of current which also have thirdharmonic component. Also, the reactive components too have third harmonics. This again doesnot match with definitions of active and reactive components of currents proposed by Fryze [5].

Powers computationThe secondary side powers are given as following.

Sa = Pa + jQa = V a I∗a = 398.37∠30 99.59∠60 = 39673.8361∠90VA

Thus,Pa = 0 W, Qa = 39673.8361 VAr

Sb = Pb + jQb = V b I∗b = 398.37∠− 90 × 0 = 0 VA

Pb = 0 W, Qb = 0 VAr

Sc = Pc + jQc = V c I∗c = 398.37∠150 × 0 = 0 VA

Pc = 0 W, Qc = 0 VAr

147

Page 154: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The total active and reactive powers on the secondary side are given as following.

P = Pa + Pb + Pc = 0 WQ = Qa +Qb +Qc = 39673.83 VAr

Svect = Sarith = Q = 39673.83 VApfvect = pfarith = P/S = 0

The primary side powers are given as following.

SA = PA + jQA = V a I∗a = 230∠0 99.59∠60 = 11452.85 + j19836.91 VA

Thus,PA = 11452.85 W, QA = 19836.91 VAr

SB = PB + jQB = V b I∗b = 230∠− 120 (−99.59∠− 120)∗ =

PB = −11452.85 W, QB = 19836.91 VAr

SC = PC + jQC = V C I∗C = 230∠120 × 0

PC = 0 W, QC = 0 VAr

The total active and reactive powers on the primary side are given as following.

P = Pa + Pb + Pc = 0 WQ = Qa +Qb +Qc = 39673.82 VAr

Svect =∣∣SA + SB + SC

∣∣ = 39673.82 VA

Sarith =∣∣SA∣∣+

∣∣SB∣∣+∣∣SC∣∣ = 22906 + 22906 + 0 = 45811.4 VA

pfvect = P/Svect = 0

pfarith = P/S = 0

Example 4.3: Consider the star-delta connected ideal transformer with 1:1 turn ratio as shown inFig. 2. The secondary side of transformer, a load of 2 ohms is connected between the phase-a andb. Compute the following.

(a) Time domain expressions of currents in each phase on both primary and secondary side.

(b) Does the load require reactive power from the source? If any, find its value. Also computethe reactive power on each phase of either side of transformer.

(c) Also determine active powers on each phase and overall active power on either side of thetransformer.

(d) If you have similar arrangement with balanced load and same output power, comment uponthe rating of line conductors and transformers.

Solution In this example, we have three phase star-delta connected transformer of turns ratio 1:1with star side connected to three phase balanced voltage source and neutral connected to ground.

148

Page 155: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Thus, in this three phase star delta connected transformer, delta side phase voltages equal the starside line voltages. The primary side instantaneous phase voltages are given by,

vA = 230√

2 sin(ωt) = 325.27 sin(ωt)

vB = 230√

2 sin(ωt− 120) = 325.27 sin(ωt− 120).

vC = 230√

2 sin(ωt+ 120) = 325.27 sin(ωt+ 120)

Therefore, instantaneous line to line voltages at delta side (secondary) are given by,

vab = 230√

2 sin(ωt) = 325.27 sin(ωt)

vbc = 230√

2 sin(ωt− 120) = 325.27 sin(ωt− 120)

vca = 230√

2 sin(ωt+ 120) = 325.27 sin(ωt+ 120).

Therefore, instantaneous phase voltages with respect to ground are given as follows.

va = 230√

2√3

sin(ωt− 30) = 132.79√

2 sin(ωt− 30)

vb = 230√

2√3

sin(ωt− 150) = 132.79√

2 sin(ωt− 150)

vc = 230√

2√3

sin(ωt+ 90) = 132.79√

2 sin(ωt+ 90)

(a) On delta side, we have a resistive load of R = 3 Ω connected between terminals a and b. Thus,expression for instantaneous currents flowing out of terminals a, b and c of the transformer aregiven by,

ia =vabR

=230√

2

3sinωt = 76.67

√2 sinωt

ib = −ia = −vabR

= −230√

2

3sinωt = −76.67

√2 sinωt

ic = 0

Therefore, the winding currents on the secondary side are given as below.

iab = 76.67√

2 sinωt = 108.41 sinωt

ibc = 0

ica = 0

These currents are transformed to the primary windings. Thus, the time domain expressions ofthese currents are given as below.

149

Page 156: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

iA = 76.67√

2 sinωt

iB = 0

iC = 0

(b) Load does not require any reactive power from the source because it is a purely resistive load.This fact can also be verified by the looking at the expressions for instantaneous phase voltagesand currents on star side.

(c) Similarly, no current (and hence power) is drawn by the load from phases-B and C. Thus,various powers on the primary side are as follows.

For phase-A,SA = PA + jQA = V A I

∗A = 230∠0 × 76.67∠0 = 17634.1 VA

PA = 17634.1 W, QA = 0 VArFor phase-B,

SB = PB + jQB = V B I∗B = 230∠− 120 × 0 = 0 VA

PB = 0 W, QB = 0 VArFor phase-C,

SC = PC + jQC = V C I∗C = 230∠120 × 0 = 0 VA

PC = 0 W, QC = 0 VAr

Thus, various powers on the secondary side are as follows.

For phase-a,Sa = Pa + jQa = V a I

∗a = 132.79∠− 30 × 76.67∠0 = 8817.05− j5090.53 VA

Pa = 8817.05 W, Qa = −5090.53 VArFor phase-b,

Sb = Pb + jQb = V b I∗b = 132.79∠− 150 × (−76.67∠0) = 8817.05 + j5090.53 VA

Pb = 8817.05 W, Qb = 5090.53 VArFor phase-c,

Sc = Pc + jQc = V c I∗c = 132.79∠90 × 0 = 0 VA

Pc = 0 W, Qc = 0 VAr

For the above analysis, it is observed that the total active power, P = PA + PB + PC = Pa +Pb + Pc = 17634.1 W and total reactive power Q = QA + QB + QC = Qa + Qb + Qc = 0 VAr.However, due to unbalanced load, on delta side of the transformer phase-b and phase-c experiencereactive power as calculated above. This creates some power factor in each phase.

(d) Because, the load which is currently getting power from one phase on delta side will be shared

150

Page 157: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

by all the three phases phase voltages being the same. Thus, required current rating of the lineconductors and transformers will be reduced.

Example 4.4Consider the a three-phase balanced system shown in Fig. 4.9. The supply voltages are: V a =220∠0, V b = 220∠ − 120, V c = 220∠120 with balance load resistance of 1.32 Ω. The outputpower Po is 100 kW and the losses in the feeder are 5% of te output power. The reactance of feederis not considered in the study.

(a) Determine voltages at the load points, phase currents and feeder resistance.

(b) Now let us say that the load is unbalanced by connecting a resistive load between phases-a andb with same power output. With this configuration, compute active, reactive, various apparentpowers and power factor based on them. Compute losses in the system and comment upon theresult.

(c) Now, with same losses and power output, find an equivalent three-phase balanced circuit andrepeat (b). Comment on the result.

N

aI

cI

bI

' 230 0oaV

'bV

r

r

r

aR

bR

cR'cV

bV

cV

aV

Fig. 4.9 Study on three-phase unbalanced system

Solution:(a) For the given system the phase voltages are V a = 220∠0, V b = 220∠−120, V c = 220∠120.The load active power is 100 kW with 5% losses in the feeder, i.e., ∆Ps = 5 kW. Using aboveparameters the RMS value of the phase voltage is computed as following.

Po = 100× 1000 = 3× V 2

1.32From the above equation, value of V is given as following.

V =

√100000× 1.32

3= 209.76 V

Thus, the rms value of phase currents is given by,

I =V

R=

209.76

1.32= 159.9 A

151

Page 158: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The voltage drop across the feeder is found using following equation.

∆V = 230− 209.76 = 20.23 V

For 5% losses in the feeder, the value of feeder resistance can be computed as below.

3× (158.9)2 × r = 0.05× 100× 1000

which implies, r = 0.066 Ω

Since the load is balanced, the various apparent powers (arithmetic, vector and effective) are same.These are computed below.

SA = Sa + Sb + Sc

= Va Ia + Vb Ib + Vc Ic

= 3Va Ia

= 3× 209.5× 158.9

= 100 kVA

Now we compute the vector apparent power (Sv) as given below.

Sv =√P 2 +Q2

=√

(Va Ia cosφa + Vb Ib cosφb + Vc Ic cosφc)2 + (Va Ia sinφa + Vb Ib sinφb + Vc Ic sinφc)2

=√

(3Va Ia cosφa)2 + (3Va Ia sinφa)2

= 3Va Ia

= 100 kVA

Similarly, the effective apparent power can be computed as below.

Se = 3Ve Ie = 3Va Ia = 100 kVA

The power factors based on above apparent powers are given below.

pfA = pfv = pfe = 1.0

Now an unbalanced circuit with same power output is considered. This is achieved by placing aresistance of 1.17 Ω between any two phases (say between phases a and b). This is shown in Fig.4.10.

For this circuit, the line currents are computed as below. Let I be the RMS value of the linecurrent, then following equation must be satisfied.

I2 ×R = 100× 1000

(4.114)

From the above equation, I is given as below.

I =

√100000

1.17= 292.35 A

152

Page 159: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

'a

'c

'n

a

b

c

n

Va

Vb

Vc

RIa

In

Ib

Ic

'

r

r

r

r

Fig. 4.10 Three-phase unbalanved circuit

Thus, phase currents are as following.

Ia = 292.35∠30AIb = −Ia = −292.35∠30AIc = 0 A

Knowing the currents in the circuit we can compute the voltages V a, V b and V c as following.

V a = 220∠0 − 292.25∠30 × 0.066

= 220− (16.67 + j9.9296)

= 223.33− j9.6296

= 203.55∠− 2.70 V

Similarly, voltages V b and V c can be computed which are given below.

V b = 220∠− 120 − (−292.25∠30)× 0.066

= 203.55∠− 117.29 V

V c = 220∠120 − 0× 0.066 = 220∠120

Thus, Va = Vb = 203.55 V and Vc = 0 V. Knowing three-phase voltages and currents, the activeand reactive powers are computed as following.

Sa = V a I∗a = 203.55∠− 2.70 × 292.35∠− 30

= 50088.79− j 32156.42

= Pa + jQa

Thus, Sa =√P 2a +Q2

a = 59522.46 VA

Sb = V b I∗b = 203.55∠− 117.3 × 292.35∠150

= 50088.79 + j 32156.42

= Pb + jQb

Thus, Sb =√P 2b +Q2

b = 59522.46 VA

153

Page 160: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Sc = V c I∗c = 220∠120 × 0

= 0 + j 0

= Pc + jQc

Thus, Sc =√P 2c +Q2

c = 0 VA

Based on the computations of active and reactive powers in the above, we shall compute arithmetic,vector and effective apparent powers and corresponding power factors.

SA = Sa + Sb + Sc = 59522.46 + 59522.46 + 0 = 119094.92 VASv = |Sv| = |Sa + Sb + Sc| = 100177.58 VA

The effective apparent power is computed as following.

Se = 3× Ve × Ie

= 3×√V 2a + V 2

b + V 2c

3×√I2a + I2

b + I2c

3

=√V 2a + V 2

b + V 2c ×

√I2a + I2

b + I2c

=√

203.62 + 203.62 + 2202 ×√

292.352 + 292.352 + 02

= 149816.05 VA

Based on above apparent powers, the power factor are as following.

pfA = 100000/119044.92 = 0.84

pfv = 1000000/100000 = 1.0

pfe = 100000/149816.05 = 0.667

The power loss in the feeders is computed as below.

Ploss = I2a r + I2

b r + I2c r

= 292.352 × 0.066 + 292.352 × 0.066 + 0

= 11260 = 11.26 kW

Thus it is observed that when the energy is delivered to the unbalanced load, the power loss inthe supply increases from 5 to 11.26 kW. It means that currents on the lines have increased and thisimplies that an unbalanced purely resistive load cannot be considered as unity power factor load.The calculation of the power factor using the vector apparent power leads to a value which is equalto unity. This disqualifies, the vector apparent power Sv as an acceptable definition of the apparentpower in the presence of the load unbalance.

Since the value of the apparent power of balanced load is independent of this power definition,we could ask the question: what is apparent power of a balanced load with the active power P =100 kW , that causes same power loss of 11.26 kW, as the unbalanced load discussed earlier. For

154

Page 161: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

power loss of 11.26 kW the following equation holds true.

3× I2 × r = 11260

Therefore, I =

√11260

3× 0.066= 283.01 A

Now we use the condition of output power as below.

P0 = 100000 = 3× I2 ×R

Implying that, R =

√100000

3× 238.015= 0.59 Ω

The total input power must be equal to output power + losses. Therefore,

Pi = 11260 = 3× 220× I × cosφ

Implying cosφ =112600

220× 238.05= 0.7078 =⇒ φ = 44.94

From the above equation,

tanφ = tan 44.94 = 0.9978 =X

R + rTherefore, X = 0.9978× (0.59 + 0.0659) = 0.65 Ω.

Therefore, load impedance (Z) is given as below.

Z = 0.59 + j0.65 Ω

(4.115)

Knowing above parameters the voltages at the load points can be computed which are given below.

V a = 220∠0 − 238.015∠44.94 × 0.0659

= 220− 11.1− j11.07 = 208.89− j 11.07

= 209.18∠− 3.03VSimilarly, V b = 209.18∠− 123.03V

V c = 209.18∠116.97V

The phase currents are as following.

Ia = 238.015∠− 44.94AIb = 238.015∠− 164.94AIc = 238.015∠− 75.06A

For this equivalent balanced circuit (with same output power and power loss), the three apparentpowers i.e., arithmetic, vector and effective are same and these are as following.

SA = Sv = Se = 3× Va × Ia = 3× 209.18× 238.015 = 149.363 kVAand P = Pa + Pb + Pc = 100 kW

Thus the power factor based on the above apparent powers will also be same. Therefore,

pfa = pfv = pfe = 100/149.363 = 0.67

155

Page 162: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

4.3 Theory of Instantaneous Symmetrical Components

The theory of instantaneous symmetrical components can be used for the purpose of load balanc-ing, harmonic suppression, and power factor correction [2], [4]. The control algorithms based oninstantaneous symmetrical component theory can practically compensate any kind of unbalanceand harmonics in the load, provided we have a high band width current source to track the filterreference currents. These algorithms have been derived in this section. For any set of three-phaseinstantaneous currents or voltages, the instantaneous symmetrical components are defined by,

ia0

ia+

ia−

=1

3

1 1 11 a a2

1 a2 a

iaibic

(4.116)

Similarly for three-phase instantaneous voltages, we have,va0

va+

va−

=1

3

1 1 11 a a2

1 a2 a

vavbvc

(4.117)

In the above equations, a is a complex operator and it is given by a = ej 2π/3 and a2 = ej 4π/3.It is to be noted that the instantaneous components of currents, ia+ and ia− are complex timevarying quantities also they are complex conjugate of each other. This same is true for va+ andva− quantities. The terms ia0 and va0 are real quantities, however (-) has been used as upper scriptfor the sake of uniformity of notation. These instantaneous symmetrical components are used toformulate equations for load compensation. First a three-phase, four-wire system supplying starconnected load is considered.

4.3.1 Compensating Star Connected Load

A three-phase four wire compensated system is shown in Fig. 4.11. In the figure, three-phase loadcurrents (ila, ilb and ilc), can be unbalanced and nonlinear load. The objective in either three orfour-wire system compensation is to provide balanced supply current such that its zero sequencecomponent is zero. We therefore have,

sbi

sci

sbv

scv

lbi

lci

LOAD

LOAD

N

n

n

*fci*

fai *fbi

savsai lai

Ideal compensator

LOAD

Fig. 4.11 A three-phase four-wire compensated system

156

Page 163: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

isa + isb + isc = 0 (4.118)

Using equations (4.116)-(4.117), instantaneous positive sequence voltage (va+) and current (ia+)are computed from instantaneous values of vsa, vsb, vsc and isa, isb, isc respectively. To have apredefined power factor from the source, the relationship between the angle of va+ and ia+ is givenas following.

∠va+ = ∠ia+ + φ+ (4.119)

Where φ+ is desired phase angle between va+ and ia+. The above equation is rewritten asfollows.

(1

3[vsa + a vsb + a2 vsc]

)= ∠

(1

3[isa + a isb + a2 isc]

)+ φ+

L.H.S = R.H.S

L.H.S of the above equation is expressed as below

L.H.S = ∠

[1

3

vsa +

(−1

2+ j

√3

2

)vsb +

(−1

2− j√

3

2

)vsc

]

= ∠

[1

3

(vsa −

vsb2− vsc

2

)+ j

√3

2(vsb − vsc)

]

= tan−1 (√

3/2) (vsb − vsc)(vsa − vsb/2− vsc/2)

= tan−1 K1

K2

(4.120)

Where, K1 = (√

3/2) (vsb − vsc) and K2 = (vsa − vsb/2− vsc/2). Similarly R.H.S of the equa-tion is expanded as below.

R.H.S = ∠

[1

3

isa +

(−1

2+ j

√3

2

)isb +

(−1

2− j√

3

2

)isc

]+ φ+

= ∠

[1

3

(isa −

isb2− isc

2

)+ j

√3

2(isb − isc)

]+ φ+

= tan−1 (√

3/2) (isb − isc)(isa − isb/2− isc/2)

+ φ+

= tan−1 K3

K4

+ φ+ (4.121)

157

Page 164: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Where, K3 = (√

3/2) (isb − isc) and K4 = (isa − isb/2− isc/2). Equating (4.120) and (4.121),we get the following.

tan−1 K1

K2

= tan−1 K3

K4

+ φ+

Taking tangent on both sides, the following is obtained.

tan

(tan−1 K1

K2

)= tan

(tan−1 K3

K4

+ φ+

)Therefore,

K1

K2

=(K3/K4) + tanφ+

1− (K3/K4)× tanφ+

K1

K2

=K3 +K4 tanφ+

K4 −K3 × tanφ+

The above equation implies that,

K1K4 −K1K3 tanφ+ −K2K3 −K2K4 tanφ+ = 0

Substituting the values of K1, K2, K3, K4 in the above equation, the following expression is ob-tained.

√3

2(vsb − vsc)

(isa −

isb2− isc

2

)− 3

4(vsb − vsc) (isb − isc) tanφ+

−√

3

2

(vsa −

vsb2− vsc

2

)(isb − isc)−

(vsa −

vsb2− vsc

2

) (isa −

isb2− isc

2

)tanφ+ = 0

Above equation can be arranged with terms associated with isa, isb and isc. This is given below.√3

2(vsb − vsc) +

tanφ+

2(vsb + vsc − 2 vsa)

isa

+

√3

2(vsc − vsa) +

tanφ+

2(vsc + vsa − 2 vsb)

isb

+

√3

2(vsa − vsb) +

tanφ+

2(vsa + vsb − 2 vsc)

isc = 0

Dividing above equation by√

32

, it can be written as follows.(vsb − vsc) +

tanφ+√3

(vsb + vsc − 2 vsa)

isa

+

(vsc − vsa) +

tanφ+√3

(vsc + vsa − 2 vsb)

isb

+

(vsa − vsb) +

tanφ+√3

(vsa + vsb − 2 vsc)

isc = 0

158

Page 165: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Assume β = tanφ+/√

3, the above equation is further simplified to,

(vsb − vsc) + β (vsb + vsc − 2 vsa) isa+ (vsc − vsa) + β (vsc + vsa − 2 vsb) isb+ (vsa − vsb) + β (vsa + vsb − 2 vsc) isc = 0. (4.122)

Adding and subtracting vsa, vsb and vsc in β terms and expressing vs0 = 3(vsa + vsb + vsc) in aboveequation, we get the following.

(vsb − vsc)− 3β (vsa − vs0) isa+ (vsc − vsa)− 3β (vsb − vs0) isb+ (vsa − vsb)− 3β (vsc + vs0) isc = 0. (4.123)

The third objective of compensation is that the power supplied from the source (ps) must be equalto average load power (Plavg). Therefore the following holds true.

ps = vsa isa + vsb isb + vsc isc = Plavg (4.124)

The above equation has important implications. For example when supply voltage are balanced,the above equation is satisfied for balanced source currents. However if supply voltage are unbal-anced and distorted, above equation gives set of currents which are also not balanced and sinusoidalin order to supply constant power.

Equations (4.118), (4.124) and (4.123), can be written in matrix form as given below. 1 1 1(vsb − vsc) + β(vsb + vsc − 2vsa) (vsc − vsa) + β(vsc + vsa − 2vsb) (vsa − vsb) + β(vsa + vsb − 2vsc)

vsa vsb vsc

isaisbisc

=

00

Plavg

Which can be further written as, [

A] [isabc

]=[Plavg

](4.125)

Therefore,

[isabc

]=[A−1

] [Plavg

]=

1

∆A

ac11 ac12 ac13

ac21 ac22 ac23

ac31 ac32 ac33

T 00

Plavg

=

1

∆A

ac11 ac21 ac31

ac12 ac22 ac32

ac13 ac23 ac33

00

Plavg

Where acij is the cofactor of ith row and jth column of A matrix in (4.125) and ∆A is the deter-minant of matrix A. Due to the presence of zero elements in first two rows of column vector with

159

Page 166: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

power elements, the cofactors in first two columns need not to be computed. These are indicatedby dots in the following matrix.

[isabc

]=

isaisbisc

=1

∆A

· · ac31

· · ac32

· · ac33

00

Plavg

=1

∆A

ac31

ac32

ac33

PlavgThe determinant of matrix A is computed as below.

∆A = [(vsc − vsa) + β(vsc + vsa − 2vsb)]vsc − [(vsa − vsb) + β(vsa + vsb − 2vsc)]vsb

−[(vsb − vsc) + β(vsb + vsc − 2vsa)]vsc + [(vsa − vsb) + β(vsa + vsb − 2vsc)]vsa

+[(vsb − vsc) + β(vsb + vsc − 2vsa)]vsb − [(vsc − vsa) + β(vsc + vsa − 2vsb)]vsa

= β[v2sc + vsavsc − 2vsbvsc − vsavsb − v2

sb + 2vsbvsc − vsbvsc − v2sc + 2vsavsc

+v2sa + vsbvsa − 2vscvsa + v2

sb + vscvsb − 2vsavsb − vscvsa − v2sa + 2vsbvsa]

+(vsc − vsa)vsc − (vsa − vsb)vsb − (vsb − vsc)vsc+(vsa − vsb)vsa + (vsb − vsc)vsb − (vsc − vsa)vsa

The above equation can be further simplified to,

∆A = β · 0 + v2sc − vsavsc − vsbvsa + v2

sb − vsbvsc + v2sc + v2

sa

−vsavsb + v2sb − vscvsb − vsavsc + v2

sa

= 2v2sa + 2v2

sb + 2v2sc − 2vsavsb − 2vsbvsc − 2vscvsa

= v2sa + v2

sb − 2vsavsb + v2sb + v2

sc − 2vsbvsc + v2sc + v2

sa − 2vscvsa

= (vsa − vsb)2 + (vsb − vsc)2 + (vsc − vsa)2

= (v2sab + v2

sbc + v2sca)

Further adding and subtracting, v2sa + v2

sb + v2sc in above equation, we get the following.

∆A = 3(v2sa + v2

sb + v2sc)− (v2

sa + v2sb + v2

sc + 2vsavsb + 2vsbvsc + 2vscvsa) (4.126)

Further,

(vsa + vsb + vsc)2 = v2

sa + v2sb + v2

sc + 2vsavsb + 2vsbvsc + 2vscvsa (4.127)

Using equations (4.126) and (4.127), we obtain the following.

∆A = 3(v2sa + v2

sb + v2sc)− (vsa + vsb + vsc)

2

= 3∑j=a,b,c

v2sj − 9v2

so

= 3

[ ∑j=a,b,c

v2sj − 3v2

so

](4.128)

160

Page 167: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

In above equation, the term vso is the instantaneous zero sequence component of the source voltageand it is given as following.

vso =(vsa + vsb + vsc)

3(4.129)

The determinant of matrix A, using (4.128), can also be expressed as,

∆A = 3

[v2sa + v2

sb + v2sc − 3 vso

(vsa + vsb + vsc)

3

]= 3

[v2sa + v2

sb + v2sc + 3v2

so − 2× 3v2so

]= 3[v2

sa + v2sb + v2

sc + v2so + v2

so + v2so − 2vso(vsa + vsb + vsc)]

= 3[(v2sa + v2

so − 2vsavso) + (v2sb + v2

so − 2vsbvso) + (v2sc + v2

so − 2vscvso)]

= 3[(v2sa − v2

so) + (v2sb − v2

so) + (v2sc − v2

so)]

The cofactors ac31, ac32, and ac33 are computed as below.

ac31 = [−(vsc − vsa)− β(vsc + vsa − 2vsb) + (vsa − vsb) + β(vsa + vsb − 2vsc)]

= vsa − vsb − vsc + vsa + β(−vsc − vsa + 2vsb + vsa + vsb − 2vsc)

= (2vsa − vsb − vsc) + 3β(vsb − vsc)= (2vsa + vsa − vsa − vsb − vsc) + 3β(vsb − vsc)= 3 (vsa − vs0) + 3 β(vsb − vsc)

Similarly,

ac32 = [−(vsa − vsb)− β(vsa + vsb − 2vsc) + (vsb − vsc) + β(vsb + vsc − 2vsa)]

= (−vsa + vsb + vsb − vsc) + β(−vsa − vsb + 2vsc + vsb + vsc − 2vsa)

= (2vsb − vsc − vsa) + 3β(vsc − vsa)= 3(vsb − vso) + 3β(vsc − vsa)

And,

ac33 = [(vsc − vsa) + β(vsc + vsa − 2vsb)− (vsb − vsc)− β(vsb + vsc − 2vsa)]

= (2vsc − vsa − vsb) + 3β(vsa − vsb)= 3(vsc − vso) + 3β(vsa − vsb)

Knowing the value of cofactors, we now have,isaisbisc

=1

3[∑

j=a,b,c v2sj − 3v2

so

]3(vsa − vs0) + 3β(vsb − vsc)

3(vsb − vs0) + 3β(vsc − vsa)3(vsc − vs0) + 3β(vsa − vsb)

[Plavg]

=1[∑

j=a,b,c v2sj − 3v2

so

](vsa − vso) + β(vsb − vsc)

(vsb − vso) + β(vsc − vsa)(vsc − vso) + β(vsa − vsb)

[Plavg] (4.130)

161

Page 168: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From the above equation, the desired source currents can be written as following.

isa =(vsa − vso) + β(vsb − vsc)∑

j=a,b,c v2sj − 3 v2

so

Plavg (4.131)

isb =(vsb − vso) + β(vsc − vsa)∑

j=a,b,c v2sj − 3 v2

so

Plavg (4.132)

isc =(vsc − vso) + β(vsa − vsb)∑

j=a,b,c v2sj − 3 v2

so

Plavg (4.133)

Applying Kirchoff’s current law at the point of common coupling (PCC), we have,

i∗fa = ila − isa (4.134)i∗fb = ilb − isb (4.135)i∗fc = ilc − isc (4.136)

Replacing isa, isb and isc from equations (4.131)-(4.133), we obtain the reference filter currents asgiven in the following.

i∗fa = ila − isa = ila −(vsa − vso) + β(vsb − vsc)∑

j=a,b,c v2sj − 3 v2

so

Plavg (4.137)

i∗fb = ilb − isb = ilb −(vsb − vso) + β(vsc − vsa)∑

j=a,b,c v2sj − 3 v2

so

Plavg (4.138)

i∗fc = ilc − isc = ilc −(vsc − vso) + β(vsa − vsb)∑

j=a,b,c v2sj − 3 v2

so

Plavg (4.139)

4.3.2 Compensating Delta Connected Load

The balancing of an unbalanced ∆-connected load is a generic problem and the theory of instan-taneous symmetrical components can be used to balance the load. The schematic diagram of thiscompensated scheme is shown in Fig. 4.12. This compensator is connected between the phasors of

Load

lbci

*fbci

labi

*fabi

Loa d

Loa d

*fcai

lcai

sai

sci

sbi

sav

sbv

scv

Fig. 4.12 A compensation for delta connected load

this scheme. The aim is to generate the three reference current waveforms denoted by i∗fab, i∗fbc, i

∗fca

162

Page 169: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

respectively based on the measurement of system voltages and load currents such that the supplysees balanced load. As was in previous case, the requirements for compensating source currentsare same. Therefore the following equations are valid.

isa + isb + isc = 0 (4.140)

Applying Kirchoff’s current law at nodes, we can express isa, isb and isc respectively as following.

isa = (ilab − i∗fab)− (ilca − i∗fca)isb = (ilbc − i∗fbc)− (ilab − i∗fcb) (4.141)isc = (ilca − i∗fca)− (ilbc − i∗fbc)

As can be seen from above equations, equation (4.140) is satisfied. Since in ∆ connected load,zero sequence current cannot flow, therefore

(ilab − i∗fab) + (ilbc − i∗fbc) + (ilca − i∗fca) = 0 (4.142)

The source supplies the average load power, Plavg and following equation is satisfied.

vsaisa + vsbisb + vscisc = Plavg (4.143)

From equation (4.123), the power factor between the source voltages and currents should be met.Thus we have,

(vsb − vsc)− 3β (vsa − vs0) isa+ (vsc − vsa)− 3β (vsb − vs0) isb+ (vsa − vsb)− 3β (vsc + vs0) isc = 0. (4.144)

Replacing isa, isb and isc from (4.141), above equation can be simplified to the following.

(vsb − vsc)− 3β(vsa − vs0) [(ilab − i∗fab)− (ilca − i∗fca)]+(vsc − vsa)− 3β(vsb − vsa) [(ilbc − i∗fbc)− (ilab − i∗fab)]+(vsa − vsb)− 3β(vsb − vsa) [(ilca − i∗fca)− (ilbc − i∗fbc)] = 0 (4.145)

Simplifying above expression we get

[(vsb − vsc) + β(vsb + vsc − 2vsa) − (vsc − vsa) + β(vsc + vsa − 2vsb)] (ilab − i∗fab)︸ ︷︷ ︸I

+ [(vsc − vsa) + β(vsc + vsa − 2vsb) − (vsa − vsb) + β(vsa − vsb − 2vsc)] (ilbc − i∗fbc)︸ ︷︷ ︸II

+ [(vsa − vsb) + β(vsa − vsb − 2vsc) − (vsb − vsc) + β(vsb + vsc − 2vsa)] (ilca − i∗fca)︸ ︷︷ ︸III

(4.146)

The first term I is as follows:

I = (vsb − vsc − vsc + vsa) + β(vsb + vsc − 2vsa − vsc − vsa + 2vsb) (ilab − i∗fab)= (vsa + vsb − 2vsc)− 3β(vsa − vsb) (ilab − i∗fab)= −3 (vsc − vs0) + β (vsa − vsb) (ilab − i∗fab) (4.147)

163

Page 170: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Similarly, the second and third terms are given as below.

II = (vsc − vsa − vsa + vsb) + β(vsc + vsa − 2vsb − vsa − vsb + 2vsc) (ilbc − i∗fbc)= (vsb + vsc − 2vsa)− 3β(vsb − vsc) (ilbc − i∗fbc)= −3 (vsa − vs0) + β (vsb − vsc) (ilbc − i∗fbc) (4.148)

III = (vsa − vsb − vsb + vsc) + β(vsa + vsb − 2vsc − vsb − vsc + 2vsa) (ilca − i∗fca)= (vsc + vsa − 2vsb)− 3β(vsc − vsa) (ilca − i∗fca)= −3 (vsb − vs0) + β (vsc − vsa) (ilca − i∗fca) (4.149)

Summing above three terms and simplifying we get,

(vsc − vs0) + β(vsa − vsb) (ilab − i∗fab)+ (vsa − vs0) + β(vsb − vsc) (ilbc − i∗fbc)+ (vsb − vs0) + β(vsc − vsa) (ilca − i∗fca) = 0 (4.150)

The third condition for load compensation ensures that the average load power should be suppliedfrom the sources. Therefore,

vsaisa + +vsbisb + vscisc = Plavg (4.151)

The terms isa, isb and isc are substituted from (4.141) in the above equation and the modifiedequation is given below.

vsa

(ilab − i∗fab)− (ilca − i∗fca)

+ vsb

(ilbc − i∗fbc)− (ilab − i∗fab)

+vsc

(ilca − i∗fca)− (ilbc − i∗fbc)

= 0 (4.152)(4.153)

The above is simplified to,

(vsa − vsb)(ilab − i∗fab) + (vsb − vsc)(ilbc − i∗fbc) + (vsc − vsa)(ilca − i∗fca) = 0 (4.154)

Equations (4.142), (4.150), (4.154) can be written in the matrix form as given below.

1 1 1(vsc − vs0) + β(vsa − vsb) (vsa − vs0) + β(vsb − vsc) (vsb − vs0) + β(vsc − vsa)

vsa − vsb vsb − vsc vsc − vsa

ilab − i∗fabilbc − i∗fbcilca − i∗fca

=

00

Plavg

(4.155)

The above equation can be written in the following form.

[A∆]

ilab − i∗fabilbc − i∗fbcilca − i∗fca

=

00

Plavg

(4.156)

164

Page 171: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Therefore, ilab − i∗fabilbc − i∗fbcilca − i∗fca

= [A∆]−1

00

Plavg

(4.157)

The above equation is solved by finding the determinate ofA∆ and the cofactors transpose as givenbelow. ilab − i∗fabilbc − i∗fbc

ilca − i∗fca

=1

|A∆|

ac11 ac12 ac13

ac21 ac22 ac23

ac31 ac32 ac33

T 00

Plavg

=

1

|A∆|

ac11 ac21 ac31

ac12 ac22 ac32

ac13 ac23 ac33

00

Plavg

(4.158)

The determinant |A∆| and cofactors in above equation are calculated below.

|A∆| = [(vsa − vso) + β(vsb − vsc)](vsc − vsa)− [(vsb − vso) + β(vsc − vsa)](vsb − vsc)−[(vsc − vso) + β(vsa − vsb)](vsc − vsa) + [(vsb − vso) + β(vsc − vsa)](vsa − vsb)+[(vsc − vso) + β(vsa − vsb)](vsb − vsc)− [(vsa − vso) + β(vsb − vsc)](vsa − vsb)

Separating all the terms containing β and rearranging the above equation, we get,

|A∆| = (vsa − vso)(vsc − vsa)− (vsb − vso)(vsb − vsc)− (vsc − vso)(vsc − vsa)+(vsb − vso)(vsa − vsb) + (vsc − vso)(vsb − vsc)− (vsa − vso)(vsa − vsb)+β [(vsb − vsc)(vsc − vsa)− (vsc − vsa)(vsb − vsc)− (vsa − vsb)(vsc − vsa)+(vsc − vsa)(vsa − vsb) + (vsa − vsb)(vsb − vsc)− (vsb − vsc)(vsa − vsb)]

It is seen that the terms containing β cancel each other and give zero. Thus, above equationbecomes,

|A∆| = (vsa − vso)(vsc − vsa − vsa + vsb) + (vsb − vso)(−vsb + vsc + vsa − vsb)(4.159)+(vsc − vso)(−vsc + vsa + vsb − vsc) + β × 0

In the above, using (vsa + vsb + vsc) = 3vs0,

vsc − vsa − vsa + vsb = vsa + vsb + vsc − 3vsa = −3(vsa − vso)−vsb + vsc + vsa − vsb = vsa + vsb + vsc − 3vsb = −3(vsb − vso)−vsc + vsa + vsb − vsc = vsa + vsb + vsc − 3vsc = −3(vsc − vso).

Replacing above term in (4.159), we get the following.

A∆ = −3 [(vsa − vs0)(vsa − vs0) + (vsb − vs0)(vsb − vs0) + (vsc − vs0)(vsc − vs0)]

= −3[(vsa − vs0)2 + (vsb − vs0)2 + (vsc − vs0)2]

= −3∑j=a,b,c

(vsj − vs0)2 (4.160)

165

Page 172: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The above equation can also be written as,

|A∆| = −3 [v2sa + v2

sb + v2sc + 3 v2

s0 − 2 vs0 (vsa + vsb + vsc)]

= −3 [v2sa + v2

sb + v2sc + 3 v2

s0 − 6 v2s0]

= −3 [v2sa + v2

sb + v2sc − 3v2

s0]

= −3

( ∑j=a,b,c

v2sj − 3v2

s0

)(4.161)

Calculation of cofactors of A∆

We need to calculate ac31, ac32 and ac33. These are computed as following.

ac31 = [(vsb − vso) + β(vsc − vsa)− (vsa − vso) + β(vsb − vsc)]= [(vsb − vsa)− β(vsa + vsb − 2 vsc)]

= [(vsb − vsa)− β(vsa + vsb + vsc − 3 vsc)]

= [(vsb − vsa)− β(3 vs0 − 3 vsc)]

= − [vsab − 3 β(vsc − vs0)]

Similarly,

ac32 = −[(vsb − vso) + β(vsc − vsa)− (vsc − vso) + β(vsa − vsb)]= −[(vsb − vsc) + β(vsc + vsb − 2 vsa)]

= −[(vsb − vsc) + β(vsa + vsb + vsc − 3 vsa)]

= −[(vsb − vsc) + β(3 vs0 − 3 vsa)]

= − [vsbc − 3β(vsa − vs0)]

and

ac33 = [(vsa − vso) + β(vsb − vsc)− (vsc − vso) + β(vsa − vsb)]= [(vsa − vsc)− β(vsa + vsc − 2 vsb)]

= [(vsa − vsc)− β(vsa + vsb + vsc − 3 vsb)]

= [(vsa − vsc)− β(3 vs0 − 3 vsb)]

= − [vsca − 3β(vsb − vs0)]

Therefore, the solution of the equation is given by,ilab − i∗fabilbc − i∗fbcilca − i∗fca

=1

A∆

ac11 ac12 ac13

ac21 ac22 ac23

ac31 ac32 ac33

T 00

Plavg

=

1

A∆

ac11 ac21 ac31

ac12 ac22 ac32

ac13 ac23 ac33

00

Plavg

(4.162)

166

Page 173: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From the above equation and substituting the values of cofactors obtained above, we get the fol-lowing.

ilab − i∗fab =ac31

|A∆|Plavg

=− [vsab − 3β(vsc − vs0)]

−3[∑

j=a,b,c v2sj − 3 v2

s0

] Plavg=

[vsab/3− β(vsc − vs0)]∑j=a,b,c v

2sj − 3 v2

s0

Plavg

From the above equation, the reference compensator current (i∗fab) can be given as follows.

i∗fab = ilab −vsab/3− β(vsc − vs0)∑

j=a,b,c v2sj − 3 v2

s0

Plavg (4.163)

Similarly,

i∗fbc = ilbc −vsbc/3− β(vsa − vs0)∑

j=a,b,c v2sj − 3 v2

s0

Plavg (4.164)

and

i∗fca = ilca −vsca/3− β(vsb − vs0)∑

j=a,b,c v2sj − 3 v2

s0

Plavg (4.165)

When the source power factor is unity, β = 0, for balanced source voltages (fundamental) vs0 = 0.Substituting these values in above equations, we get,

i∗fab = ilab −vsab

3∑

j=a,b,c v2sj

Plavg

i∗fbc = ilbc −vsbc

3∑

j=a,b,c v2sj

Plavg (4.166)

i∗fca = ilca −vsca

3∑

j=a,b,c v2sj

Plavg

Further it can be seen that,

v2sab + v2

sbc + v2sca = (vsa − vsb)2 + (vsb − vsc)2 + (vsc − vsa)2

= v2sa + v2

sb − 2vsavsb + v2sb + v2

sc − 2vsbvsc + v2sc + v2

sa − 2vscvsa

= 2(v2sa + v2

sb + v2sc)− (2vsavsb + 2vsbvsc + 2vscvsa)

= 3(v2sa + v2

sb + v2sc)− (v2

sa + v2sb + v2

sc + 2vsavsb + 2vsbvsc + 2vscvsa)

= 3(v2sa + v2

sb + v2sc)− (vsa + vsb + vsc)

2

= 3[(v2sa + v2

sb + v2sc)− 3 v2

s0]

= 3

[ ∑j=a,b,c

v2sj − 3 v2

s0

](4.167)

167

Page 174: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Replacing 3[∑

j=a,b,c v2sj − 3 v2

s0

]in equations (4.163), (4.164) and (4.165), we get the following.

i∗fab = ilab −vsab − 3β (vsc − vs0)

3(v2sab + v2

sbc + v2sca)

Plavg

i∗fbc = ilbc −vsbc − 3β (vsa − vs0)

3(v2sab + v2

sbc + v2sca)

Plavg (4.168)

i∗fca = ilca −vsca − 3β (vsb − vs0)

3(v2sab + v2

sbc + v2sca)

Plavg

For unity power factor and balanced source voltages (fundamental), the reference compensatorcurrents are given as following.

i∗fab = ilab −vsab

27V 2Plavg

i∗fbc = ilbc −vsbc

27V 2Plavg (4.169)

i∗fca = ilca −vsca

27V 2Plavg

Where V is the rms value of phase voltages.

References

[1] H. Akagi, Y. Kanazawa, and A. Nabae, “Instantaneous reactive power compensators compris-ing switching devices without energy storage components,” IEEE Transactions on IndustryApplications, no. 3, pp. 625–630, 1984.

[2] A. Ghosh and G. Ledwich, “Load compensating DSTATCOM in weak ac systems,” IEEETransactions on Power Delivery, vol. 18, no. 4, pp. 1302–1309, Oct. 2003.

[3] V. George and Mahesh K. Mishra, “DSTATCOM topologies for three-phase high power appli-cations,” International Journal of Power Electronics, vol. 2, no. 2, pp. 107–124, 2010.

[4] A. Ghosh and A. Joshi, “A new approach to load balancing and power factor correction inpower distribution system,” IEEE Transactions on Power Delivery, vol. 15, no. 1, pp. 417–422, 2000.

[5] L. S. Czarnecki, “Budeanu and Fryze: Two frameworks for interpreting power properties ofcircuits with nonsinusoidal voltages and currents,” Electrical Engineering (Archiv fur Elek-trotechnik), vol. 80, no. 6, pp. 359–367, 1997.

168

Page 175: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

Chapter 5

SERIES COMPENSATION: VOLTAGECOMPENSATION USING DVR(Lectures 36-44)

5.1 Introduction

Power system should ensure good quality of electric power supply, which means voltage and cur-rent waveforms should be balanced and sinusoidal. Furthermore, the voltage levels on the systemshould be within reasonable limits, generally within 100±5% of their rated value. If the voltage ismore or less than this pre-specified value, performance of equipments is sacrificed. In case of lowvoltages, picture on television starts rolling, the torque of induction motor reduces to the square ofvoltage and therefore there is need for voltage compensation.

5.2 Conventional Methods to Regulate Voltage

In order to keep load bus voltage constant, many conventional compensating devices such as listedbelow can be used. In general, these can be referred as VAR compensator.

1. Shunt Capacitors

2. Series Capacitors

3. Synchronous Capacitor

4. Tap Changing Transformer

5. Booster Transformer

6. Static Synchronous Series Capacitor

7. Dynamic Voltage Restorer

169

Page 176: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The first six methods are employed at transmission level while the last method is by employingDynamic Voltage Restorer (DVR), is mostly employed in power distribution network to protect anyvoltage variation at the load bus connected to the sensitive and critical electrical units. The DVRis a series connected custom power device used to mitigate the voltage unbalance, sags, swells,harmonics and any abrupt changes due to abnormal conditions in the system. In the followingsection, dynamic voltage restorer will be described in detail.

5.3 Dynamic Voltage Restorer (DVR)

A dynamic voltage restorer (DVR) is a solid state inverter based on injection of voltage in serieswith a power distribution system [1], [2]. The DC side of DVR is connected to an energy source oran energy storage device, while its ac side is connected to the distribution feeder by a three-phaseinterfacing transformer. A single line diagram of a DVR connected power distribution system isshown in the figure below. In this figure, vs(t) represents supply voltage, vt(t) represents terminalvoltage and vl(t) represents the load voltage. Since DVR is a series connected device, the sourcecurrent, is(t) is same as load current, il(t). Also note in the figure, vf (t) is DVR injected voltagein series with line such that the load voltage is maintained at sinusoidal nominal value.

( )sv t s sR jX ( )lv t

s li iLOAD

( )tv t

( )fv t

Fig. 5.1 A single-line diagram of DVR compensated system

The three-phase DVR compensated system is shown in Fig. 5.2 below. It is assumed that thetransmission line has same impedance in all three phases. A DVR unit which is represented in Fig.5.1, have following components [3]- [5].

1. Voltage Source Inverter

2. Filter capacitors and inductors

3. Injection transformer

4. DC storage system

These components are shown in Fig. 5.3. Some other important issues i.e., how much voltageshould be injected in series using appropriate algorithm, choice of suitable power converter topol-ogy to synthesize voltage and design of filter capacitor and inductor components have to be ad-dressed while designing the DVR unit.

170

Page 177: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

savs sR jX fav

lav

A

sbvs sR jX fbv lbv

B

s sR jXC

sai

sbi

sci

scv fcv

N n

lcv

Fig. 5.2 A single-line diagram of DVR compensated system

Load

vs

vt vl

LsRs Injecion transformer

vf

VSI

Cdc

Vdc

+-

Energystorage

PCC

Fig. 5.3 Schematic diagram of a DVR based compensation in a distribution system

5.4 Operating Principle of DVR

Consider a DVR compensated single phase system as shown in Fig. 5.4. Let us assume that sourcevoltage is 1.0 pu and we want to regulate the load voltage to 1.0 pu. Let us denote the phase anglebetween V s and V l as δ. In this analysis, harmonics are not considered. Further we assume thatduring DVR operation, real power is not required except some losses in the inverter and the nonideal filter components. These losses for the time being are considered to be zero. This conditionimplies that the phase difference between V f and Is should be 90o. Let us first consider a generalcase to understand the concept.

For the circuit shown in Fig. 5.4, applying Kirchoff’s voltage law,

V s + V f = Is (Rs + jXs) + V l

= Is Zs + V l. (5.1)

171

Page 178: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

sV s sR jX lV

sILOAD

tV

fV

Fig. 5.4 Schematic diagram of a DVR based compensation in a distribution system

Note that in above curcuit Is = I l = I . The load voltage V l can be written in terms of load currentand load impedance as given below.

V s + V f = I (Zs + Zl) (5.2)

Therefore equation (5.1) can be written as following.

V s + V f = I (Rs + jXs) + V l

= I Zs + V l (5.3)

Above equation can be re-written as following.

V s = V l + I Rs − (V f − jI Xs) (5.4)

With the help of above equation, the relationship between load voltage and the source and DVRvoltages can be expressed as below.

V l =

(V s + V f

Zs + Zl

)Zl (5.5)

Example 5.1 Let us apply condition to maintain load voltage same as source voltage i.e., V l = V s

or Vl = Vs. Discuss the feasibility of injected voltage in series with the line as shown in Fig. 5.4,to obtain load voltage same as source voltage. Consider the following cases.

a. Line resistance is negligible with Zs = j0.25 pu and Zl = 0.5 + j0.25 pu.

b. When the load is purely resistive with Zs = 0.45 + j0.25 pu and Zl = 0.5 pu.

Solution:

(a) When line resistance is negligible

172

Page 179: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The above condition implies that, Xs = 0. Without DVR, the load terminal voltage V l can begiven as following.

I l =V s

Zl + jZs=

1.0∠0o

0.5 + j0.5= 1.0− j1.0 = 1.4142∠− 45o pu

Therefore the load voltage is given as following. V l = Zl I l = 0.5590∠26.56o × 1.4142∠45o =0.7906∠− 18.43o pu. This is illustrated in Fig. 5.5(a). Thus the load voltage has reduced by 21%.Now it is desired to maintain load voltage same as supply voltage in magnitude and phase angle.Thus, substituting V s = V l in equation (5.3), we get,

V s + V f = I (Rs + jXs) +V l

⇒ V f = I (Rs + jXs)

V f =jXs

ZlV l, since Rs = 0 and I = V l/Zl

Neglecting resistance part of the feeder impedance, Zs = j0.25, the DVR voltage can be computedas above.

V f =j0.25

0.5 + j0.251.0∠0o for V l = 1.0∠0o

= 0.4472∠63.4349o pu.

From the above the, line current is computed as following.

Is =V l

Zl=

1.0∠0o

0.5590∠26.56o= 1.7889∠− 26.56o pu.

It is to be noted that, although V s = V l = 1.0∠0o pu, it does not imply that no power flows fromsource to load. In fact the total effective source voltage is V

s = V s + V f = 1.2649∠18.8 pu.Therefore it implies that the effective source voltage is leading the load voltage by an angle of18.43o. This ensures the power flow from the source to load. This is illustrated by drawing phasordiagram in the Fig. 5.5(b). below.

(b) When load is purely resistive

For this case Xl = 0, therefore Zl = Rl = 0.5 pu. Substituting V s = V l in (5.1), we get thefollowing.

V f = (Rs + jXs)I

Substituting I = V l/Rl, we get,

V f =Rs + jXs

Rl

V l = Rs

(V l

Rl

)+ jXs

(V l

Rl

)= Rs Is + jXsIs.

173

Page 180: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

=1.0 0olV

I

sjX I

f

s

VjX

IsV

26.56o

=1.0 0osV

I

sjX I45o

18.8o

(a) (b)

lV

Fig. 5.5 Terminal voltage (a) Without DVR (b) With DVR

From this above equation, it is indicated that DVR voltage has two components, the one is in phasewith Is and the other is in phase quadrature with Is. This implies that for purely resistive load, itis not possible to maintain V l = V s without active power supplied from the DVR to the load. Thisis due to the presence of in phase component of the DVR voltage in the above equation. This isillustrated in the phasor diagram given in Fig. 5.6 below.

lVI sI R

sjX IsV

Fig. 5.6 When load is purely resistive

Although it is not possible to maintain V l = V s without injection of active power from theDVR, however it is possible to maintain the magnitudes of load voltage and source voltages to thesame value i.e., Vs = Vl. However, this may be true for a limited range of the load resistance.

5.4.1 General Case

In general, it is desired to maintain the magnitude of the load voltage equal to the source voltagei.e., 1.0 pu. The voltage equation in general relating the source, load and DVR has been expressedin (5.4) and is given below.

V s = V l + I Rs − (V f − jI Xs)

The above equation is illustrated using phasor diagram description in Fig. 5.7 given below. Threecases of voltage compensation are discussed below.

Case 1: When Rs I < CD

For this case, it is always possible to maintain load voltage same as source voltage i.e., Vl = Vs.The DVR is expected to supply enough range reactive power to meet this condition. When Rs Is

174

Page 181: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

I

sV

lV

sI R

sjX I

1

1

1

f sV jX I

l

sV

Fig. 5.7 Compensation using DVR: General case

is quite smaller than CD, the above condition can be met by suppling less reactive power from theDVR. For this condition there are two solutions. Graphically, these solutions are represented bypoints A an B in the Fig. 5.7.

Case 2: When Rs I > CD

For this condition, it is not possible to meet Vl = Vs. This is shown by lines passing throughpoints D and D1. This may take place due to the higher feeder resistance or high current, thusmaking product of I Rs relatively large.

Case 3: When Rs I = CD

This is limiting case of compensation to obtain Vs = Vl. This condition is now satisfied at onlyone point when CD=Rs I . This is indicated by point D in the Fig. 5.7.

Now let us set the following objective for the load compensation.

Vl = Vs = V = 1.0 pu (5.6)

From Fig. 5.7, OA = V cosφl = cosφl.Therefore, CD = OD − OC = V (1 − cosφl) = (1 − cosφl) pu. In order to meet the condition

175

Page 182: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

given by (5.6), the following must be satisfied.

Rs I ≤ V (1− cosφl) (5.7)

The above implies that

Rs ≤V (1− cosφl)

I(5.8)

or I ≤ V (1− cosφl)

Rs

(5.9)

Thus it is observed that for a given power factor, the DVR characteristics can be obtained byvarying Rs and keeping I constant or vice versa. This is described below. Let us consider threeconditions Rs = 0.04 pu, Rs = 0.1 pu and Rs = 0.4 pu. For these values of feeder resistance, theline currents are expressed as following using (5.9).

I = 25 (1− cosφl) pu forRs = 0.04 puI = 10 (1− cosφl) pu forRs = 0.1 puI = 2.5 (1− cosφl) pu forRs = 0.4 pu.

The above currents are plotted as function of load power factor and are shown in Fig. 5.8. SinceRs I = Vl(1− cosφl), when Rs increases, I has to decrease to make Vl(1− cosφl) to be a constantfor a given power factor. Thus if the load requires more current than the permissible value, theDVR will not be able to regulate the load voltage at the nominal value, i.e., 1.0 pu. Howeverwe can regulate bus voltage less than 1.0 pu. For regulating the load voltage less than 1.0 pu thecurrent drawing capacity of the load increases.

-80 -60 -40 -20 0 20 40 600

2

4

6

8

10

12

14

16

18

-300-600 00300 600

Fig. 5.8 DVR characteristics for different load power factor and feeder resistance

176

Page 183: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

5.5 Mathematical Description to Compute DVR Voltage

The previous section explains DVR characteristics and describes the feasibility of realizing DVRvoltage graphically under different operating conditions. In this section, a feasible solution forthe DVR voltage is presented with a mathematical description. This plays significant role whileimplementing DVR on real time basis. Reproducing equation (5.3) for sake of completeness,

V s + V f = V l + (Rs + jXs) I. (5.10)

Denoting, (Rs + jXs) I = a2 + jb2 and V f = Vf∠V f = Vf (a1 + jb1), the above equation can bewritten as following.

V s = V l + (a2 + jb2)− V f

= V l + (a2 + jb2)− Vf (a1 + jb1)

= 1.0∠0o + (a2 + jb2)− Vf (a1 + jb1) (5.11)

Since, source voltage and load voltage have to be maintained at nominal value i.e., 1.0 pu, thereforeV s = Vs∠δ = 1.0∠δ. Substituting this value of V s in above equation, we get,

V s = 1.0∠δ = cos δ + j sin δ = (1 + a2)− Vf a1+ j(b2 − Vf b1) (5.12)

Squaring and adding the real and imaginary parts from both the sides of the above equation, weget,

(1 + a2)2 + V 2f a

21 − 2(1 + a2) a1 Vf + b2

2 + V 2f b

21 − 2 b1 b2 Vf − 1.0 = 0 (5.13)

Since a21 + b2

1 = 1, therefore summation of underlines terms, V 2f (a2

1 + b21). = V 2

f . Using this andrearranging above equation in the power of Vf , we get the following.

V 2f − 2 (1 + a2) a1 + b1 b2 Vf + (1 + a2)2 + b2

2 − 1.0 = 0 (5.14)

The above equation gives two solutions for Vf . These are equivalent to two points A and B shownin the Fig. 5.7. However, the feasible value of the voltage is chosen on the basis of the rating ofthe DVR.

Example 5.2 Consider a system with supply voltage 230 V = 1.0 pu, 50 Hz as shown in the Fig.5.9. Consider feeder impedance as Zs = 0.05 + j0.3 pu and load impedance Zl = 0.5 + j0.3 pu.

1. Compute the load voltage without DVR.

2. Compute the current and DVR voltage such that Vl = Vs.

3. Compute the effective source voltage including DVR. Explain the power flow in the circuit.

4. Compute the terminal voltage with DVR compensation.

177

Page 184: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

( )sv t s sR jX ( )lv t

s li i

( )tv t

( )fv t

ll

RjX

Fig. 5.9 A DVR compensated system

Solution: 1. When DVR is not connected.

The system parameters are given as following. The supply voltage Vs = 1.0∠0o pu, Zs = Rs +j Xs = 0.05 + j0.3 and Zl = Rl + j Xl = 0.5 + j0.3 pu. The current in the circuit is given by,

Is = I = I l =V s

Zs + Zl=

1.0∠0

0.55 + j0.6= 0.83− j0.91 = 1.2286∠− 47.49o pu

The load voltage is therefore given by,

V l = Zs Is = 1.2286∠− 47.490 × 0.5 + j0.3

= 0.7164∠− 16.53o pu

Thus we observe that the load voltage is 71% of the rated value. Due to reduction in the loadvoltage, the load may not perform to the expected level.

2. When DVR is connected

It is desired to maintain Vl = Vs by connecting the DVR. Taking V l as reference phasor i.e.,V l = 1.0∠0o, The line current is computed as below.

I =1.0∠0

0.5 + j0.3= 1.47− j0.88 = 1.715∠− 30.96o pu

Writing KVL for the circuit shown in (5.9),

V s + V f = V l + (Rs + jXs) I.

The DVR voltage V f can be expressed as following.

V f = Vf∠(∠Is + 90o)

178

Page 185: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The angle of V f is taken as (∠Is + 90o) so that DVR do not exchange any active power with thesystem.

V f = Vf∠(∠Is + 90o)

= Vf∠(−30.96o + 90o)

= Vf∠59.04o = Vf (0.51 + j0.86) = Vf (a1 + jb1) pu

The above equation implies that a1 = 0.51, b1 = 0.86.Let us now compute (Rs + jXs) I .

(Rs + jXs) I = (0.05 + j0.3) (1.46− j0.86)

= 0.3041∠80.54o × 1.715∠− 30.96o

= 0.5199∠49.58o

= 0.3370 + j0.3958 pu

The above implies that a2 = 0.337 and b2 = 0.3958. As discussed in previous section, the equationV s + V f = V l + (Rs + jXs) I can be written in following form.

V 2f − 2 (1 + a2) a1 + b1 b2 Vf + (1 + a2)2 + b2

2 − 1.0 = 0.

Substituting a1, b1, a2, b2 in the above equation, we get the following quadratic equation for theDVR.

V 2f − 2.0463Vf + 0.9442 = 0

Solving the above equation, we get Vf = 0.7028, 1.3434 pu as two values of the DVR voltage.These two values correspond to the points A and B respectively in Fig. 5.7. However, the feasiblesolution is Vf = 0.7028 pu, as it ensures less rating of the DVR.Therefore,

V f = 0.7028∠59.04o

= 0.3614 + j0.6028 pu.

The source voltage can be computed using the following equation.

V s = V l + (Rs + jXs) I − V f .

= 1.0∠0o + (0.05 + j0.3) 1.715∠− 30.96o − 0.7028∠59.04o

= 0.9767− j0.2056 = 1.0∠− 11.89o pu

3. Effective source voltage

It is seen that the magnitude of V s is 1.0 pu which is satisfying the condition Vs = Vl. How-ever the angle of V s is ∠ − 11.89o which implies that power is flowing from load to the source.This is not true because the effective source voltage is now V

s = V s+V f . This is computed below.

V s + V f = V‘

s = 0.9767− j0.2056 + 0.3614 + j0.6028

= 1.3382 + j0.397

= 1.3958∠16.52o pu

179

Page 186: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From above it is evident that the effective source voltage has magnitude of 1.3958 pu and anangle of ∠16.52 which ensures that power flows from source to the load. For this the equivalentcircuit is shown in the Fig. 5.10 below.

's s fV V V

s sR jX 1.0 0olV

s lI I

ll

RjX

1.39 16.52o

Fig. 5.10 A DVR compensated system

4. Terminal voltage with DVR compensation

The terminal voltage can also be computed as following.

V t = V s − Zs I = V l − V f

= 1.0∠0o − 0.6983∠59.06o

= 0.8796∠− 43.28o pu

This indicates that for rated current flowing in the load, the terminal voltage is less than the 1.0 puand needs compensation. After compensation the load voltage is 1.0 pu as shown in the Fig. 5.10.

The details of voltages are depicted in the following figure.

5.6 Transient Operation of the DVR

In the previous section the operation of the DVR in the steady state was discussed with assumptionthat full system information is available. While implementing the DVR compensation scheme, theabove discussed method should be implemented on the real time basis. For the single phase DVRoperation, following steps are required.

1. Define a reference quantity such as the terminal voltage Vl(t) and other quantities are syn-chronized to it.

2. To compute phase angle of the DVR voltage, a fundamental of line current is extracted withrespect to reference quantity.

3. Then DVR voltage is computed using Equation (5.14), which is reproduced below.

V 2f − 2 (1 + a2) a1 + b1 b2 Vf + (1 + a2)2 + b2

2 − 1.0 = 0

180

Page 187: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

4. DVR voltage Vf is then synthesized using magnitude Vf from the above equation and phaseangle that leads the fundamental of the line current by 90o.

The above method can be refereed as Type 1 control [1]. The method assumes that all circuitparameters are known along with the information of the source impedance. This however may notbe feasible in all circumstances. To solve this problem Type 2 control is suggested. In Type 2control only local quantities are required to compute the DVR voltage. The method is describedbelow.The terminal voltage, which is local quantity to the DVR as shown in Fig. 5.1 can be expressed asfollowing.

V t = V l − V f

= Vl∠0o − Vf (a1 + jb1)

= (Vl − a1 Vf )− jb1 Vf (5.15)

Since, V t = Vt∠δt = Vt cos δt + jVt sin δt, the above equation is written as following.

Vt cos δt + jVt sin δt = (Vl − a1 Vf )− jb1 Vf (5.16)

Squaring adding both sides we get,

V 2t = (Vl − a1 Vf )

2 + b21 V

2f

= V 2l + a2

1 V2f + b2

1 V2f − 2a1 Vl Vf

= V 2l + V 2

f − 2a1 Vl Vf (since a21 + b2

1 = 1). (5.17)

The above equation can be arranged in the powers of the DVR voltage as given below.

V 2f − 2a1 Vl Vf + V 2

l − V 2t = 0 (5.18)

To implement DVR for unbalanced three-phase system without harmonics, the positive sequencecurrents (I

+

a , I+

b and I+

c ) of line currents are extracted using Fourier transform. Based on thesevalues of currents the angles of DVR voltages are found by shifting current angles by 90o i.e.,

∠V fa = ∠I+

a + 90o

∠V fb = ∠I+

b + 90o (5.19)

∠V fc = ∠I+

c + 90o

The magnitude of DVR voltage can be found using equations (5.14) and (5.18) for Type 1 andType 2 control respectively.

Based on above the DVR voltages vfa, vfb, vfc can be expressed in time domain as given below.

vfa =√

2Vfa sin(ω t+ ∠V fa)

vfb =√

2Vfb sin(ω t− 120o + ∠V fb) (5.20)

vfc =√

2Vfc sin(ω t+ 120o + ∠V fc)

181

Page 188: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

5.6.1 Operation of the DVR With Unbalance and Harmonics

In the previous analysis, it was assumed that the supply voltages are unbalanced without harmonics.In this section the operation of the DVR with harmonics will be discussed. The terminal voltages(vta, vtb and vtc) are resolved into their fundamental positive sequence voltages and the rest part, asgiven below.

vta = v+ta1 + vta rest

vtb = v+tb1 + vtb rest (5.21)

vtc = v+tc1 + vtc rest

The angles of fundamental DVR voltages (∠V fa1,∠V fb1 and∠V fc1) can be extracted as explainedabove. The magnitudes of the fundamental DVR voltages (Vfa1, Vfb1 andVfc1) can be computedusing equations (5.14) and (5.18) for Type 1 and Type 2 control respectively. For example, usingType 2 control the fundamental phase-a DVR voltage is computed as per following equation.

V 2fa1 − 2 aa1 Vl Vfa1 + V 2

l − V +ta1

2= 0 (5.22)

In above equation aa1 + jba1 = ∠V fa1 and V +ta1 is fundamental positive sequence phase-a terminal

voltage as given above in (5.21). Similar expression can be written for phase-b and phase-c. Thisequation gives solution only for fundamental component of the DVR voltage. The rest of the DVRvoltages which consist of harmonics and unbalance must be equal and opposite to that of the restpart of the terminal voltages i.e., vta rest, vtb rest and vtc rest. Therefore these can be given usingfollowing equations.

vfa rest = −vta restvfb rest = −vtb rest (5.23)vfc rest = −vtc rest

Thus, the total DVR voltage to be injected can be given as following.

vfa = vfa1 + vfa rest

vfa = vfb1 + vfb rest (5.24)vfa = vfc1 + vfc rest

In above equation, vfa1, vfb1, vfc1 are constructed using equation (5.20). Once vfa, vfb and vfc areknown, these voltages are synthesized using suitable power electronic circuit. It will be discussedin the following section.

5.7 Realization of DVR voltage using Voltage Source Inverter

In the previous section, a reference voltage of DVR was extracted using discussed control algo-rithms. This DVR voltage however should be realized in practice. This is achieved with the helpof power electronic converter which is also known as voltage source inverter. Various componentsof the DVR were listed in the beginning of chapter. They are shown in detail in the Fig. 5.11.

182

Page 189: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

The transformer injects the required voltage in series with the line to maintain the load bus voltageat the nominal value. The transformer not only reduces the voltage requirement but also providesisolation between the inverters. The filter components of the DVR such as external inductance (Lt)which also includes the leakage of the transformer on the primary side and ac filter capacitor onthe secondary side play significant role in the performance of the DVR [Sasitharan Thesis].

Load bus

+

+

fv

sv

litvLsRs

pv

lv

invitL

D1

D4

dcV

dcCinvv

S4

S3

S2

S1

D2

D3

Loa

d

To otherphases

tR

Cf

Fig. 5.11 The DVR circuit details

The same DC link can be extended to other phases as shown in Fig. 5.11. The single phaseequivalent of the DVR is shown in the Fig. 5.12.

tvlv

Cf

invv tX tR

faci

Fig. 5.12 Equivalent circuit of the DVR

In Figs. 5.11 and 5.12, vinv denotes the switched voltage generated at the inverter output termi-nals, the inductance, Lt represents the total inductance and resistance including leakage inductanceand resistance of transformer. The resistance, Rt models the switching losses of the inverter andthe copper loss of the connected transformer. The voltage source inverter (VSI) is operated in aswitching band voltage control mode to track the reference voltages generated using control logicas discussed below.Let V ∗f be the reference voltage of a phase that DVR needs to inject in series with the line with help

183

Page 190: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

of the VSI explained above. We form a voltage hysteresis band of ±h over this reference value.Thus, the upper and lower limits within which the DVR has to track the voltage can be given asfollowing.

vf up = v∗f + h

vf dn = v∗f − h (5.25)

The following switching logic is used to synthesize the reference DVR voltage.

If vf ≥ vfupS1 − S2 OFF and S3 − S4 ON (‘-1’ state)

else if vf ≤ vfdnS1 − S2 ON and S3 − S4 OFF (‘+1’ state)

else if vfdn ≥ vf ≤ vfupretain the current switching status of switches

end.

It is to be noted that switches status S1 − S2 ON and S3 − S4 OFF is denoted by ‘+1’ stateand it gives vinv = +Vdc. The switches status S1 − S2 OFF and S3 − S4 ON corresponds to ‘-1’state providing vinv = −Vdc as shown in Fig. 5.11. The above switching logic is very basic andhas scope to be refined. For example ‘0’ state of the switches of the VSI as shown in Fig. 5.11, canalso be used to have smooth switching and to minimize switching losses. In the zero state, vinv = 0and refers switches status as S3D1 or S4D2 for positive inverter current (iinv > 0). Similarly, fornegative inverter current (iinv < 0), ‘0’ state is obtained through S1D3 or S2D4. With the additionof ‘0’ state, the switching logic becomes as follows.

If v∗f > 0if vf ≥ vfup

‘0’ stateelse if vf ≤ vfdn

‘+1’ stateend

else if v∗f < 0if vf ≥ vfup

‘-1’ stateelse if vf ≤ vfdn

‘0’ stateend

end.

In order to improve the switching performance one more term is added in the above equation

184

Page 191: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

based on the feedback of filter capacitor current.

vf up = v∗f + h+ α ifac

vf dn = v∗f − h+ α ifac (5.26)

Where α is a proportional gain given to smoothen and stabilize the switching performance of theVSI [2]. The dimension of α is Ω and is thus is equivalent to virtual resistance, whose effect todamp out and smoothen the DVR voltage trajectory resulted from the switching of the inverter[4]. The value of hysteresis band (h) should be chosen in such a way that it limits switchingfrequency within the prescribed maximum value. This kind of voltage control using VSI is called asswitching band control. The actual DVR voltage is compared with these upper and lower bands ofthe voltage (Vf up, Vf dn) and accordingly switching commands to the power switch are generated.The switching control logic is described in the Table 5.1. To minimize switching frequency ofthe VSI, three level logic has been used. For this an additional check of polarity of the referencevoltage has been taken into consideration. Based on this switching status, the inverter supplies+Vdc, 0 and −Vdc levels of voltage corresponding to the 1, 0 and -1 given in the table, in order tosynthesis the reference DVR voltage.

Table 5.1 Three level switching logic for the VSI

Conditions Switching value

V ∗f ≥ 0 Vf > Vup 0

V ∗f ≥ 0 Vf < Vdn 1

V ∗f < 0 Vf > Vup -1

V ∗f < 0 Vf < Vdn 0

In addition to switching band control, an additional loop is required to correct the voltage inthe dc storage capacitor against losses in the inverter and transformer. During transients, the dccapacitor voltage may rise or fall from the reference value due to real power flow for a shortduration. To correct this voltage deviation, a small amount of real power must be drawn from thesource to replenish the losses. To accomplish this, a simple proportional-plus-integral controller(PI) is used. The signal uc is generated from this PI controller as given below.

uc = Kp eV dc +Ki

∫eV dcdt (5.27)

Where, eV dc = Vdc ref − Vdc. This control loop need not to be too fast. It may be updated oncein a cycle preferably synchronized to positive zero crossing of phase-a voltage. Based on thisinformation the variable uc will be included in generation of the fundamental of DVR voltage asgiven below.

V f1 = Vf1∠(∠Is + 90o − uc) = Vf1 (a1 + jb1) (5.28)

Then the equation (5.18), is modified to the following.

V 2f1 − 2 a1 Vl Vf1 + V 2

l − V 2t1 = 0 (5.29)

The above equation is used to find the DVR voltage. It can be found that the phase differencebetween line current and DVR voltage differs slightly from 90o in order to account the losses inthe inverter.

185

Page 192: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

5.8 Maximum Compensation Capacity of the DVR Without Real PowerSupport from the DC Link

There is direct relationship between the terminal voltage, power factor of the load and the maxi-mum possible achievable load voltage, with assumption that no real power is required from the dcbus. Referring to quadratic equation in (5.29), for given value of Vt1 and a target load bus voltageVl, The equation gives two real values of Vf1 for feasible solution. In case solution is not feasible,the equation gives two complex conjugate roots. This concludes that the maximum voltage thatDVR can compensate corresponds to the single solution of the above equation, which is givenbelow. This solution corresponds to point ‘D’ in Fig. 5.7.

Vf1 =2 a1 Vl ±

√(2 a1 Vl)2 − 4 (V 2

l − V 2t1)

2(5.30)

Since, voltage should not be complex number, the value of the terms within square root must notbe negative. Therefore

(2 a1 Vl)2 ≥ 4 (V 2

l − V 2t1) (5.31)

The above equation implies that

Vl =Vt1√1− a2

1

. (5.32)

And therefore, the DVR voltage is given by the following equation.

Vf1 = a1 Vl (5.33)

With no losses in the VSI, uc = 0,

Vl =Vt1√1− a2

1

=Vt1√1− a2

1

(5.34)

Since, a1 + jb1 = 1∠(90o + φl) = cos(90o + φl) + j sin(90o + φl) = − sinφl + j cosφl. Thisimplies a1 = − sinφl, therefore

√1− a2

1 =√

1− (− sinφl)2 = cosφl. using this relation, theabove equation can be written as following.

Vl =Vt1

cosφl(5.35)

Example 5.3 A DVR is shown in Fig. 5.13. The feeder impedance of the line 0.1 + j0.5 pu.Assume ih to be load current represented by square waveform approximated by the followingexpression.

ih = 1. sin(ω t− 30o) + 0.3 sin(3ω t− 90o) pu

1. Find the load voltage v(t) without DVR compensation i.e., vf = 0.

2. Is it possible to maintain load voltage, Vl to be 1.0 pu sinusoidal waveform? If yes what is theDVR voltage, vf (t)?

3. If no, how much maximum voltage can be maintained at load terminal with the DVR withouttaking any real power from the dc bus?

186

Page 193: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

TT

s

f lt

h

Fig. 5.13 A DVR compensated system

Solution:

1. When Vf = 0

vt = vs −∑h=1,3

Zsh ih

The impedance at the fundamental frequency, Zs1 = 0.1 + j0.5 = 0.51∠78.7o pu.The impedance at third harmonic, Zs3 = 0.1 + j1.5 = 1.50∠86.18o pu.Therefore the voltage drop due to fundamental component of the current,

Vzs1 = (0.1 + j0.5)× 0.707∠−30o

= 0.51∠78.69o × 0.707∠−30o

= 0.36∠48.69 pu.

The voltage drop due to third harmonic component of the current,

Vzs3 = (0.1 + j1.5)× 0.21∠−90o

= 0.31∠−3.820 pu.

The load voltage thus can be given by

vt = vs − (is1 Zs1 + is3 Zs3)

= 1.0 sinωt− 0.51 sin(ωt+ 48.69o)︸ ︷︷ ︸−0.45 sin(3ωt− 3.82o)

= 0.7947 sin(ωt− 28.81o)− 0.45 sin(3ωt− 3.82o) pu= vt1(t) + vth(t).

Implying that,

V t1 =0.7947√

2∠−28.81o = 0.5619∠−28.81o pu

2. With DVR

187

Page 194: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

From the above equation, Vt1 = 0.5619. With load voltage vl = 1.0 sin (ωt− φl), the DVRvoltage Vf1 can be solved using quadratic equation as mentioned in Type 2 control. Further,

V f1 = Vf1∠(∠Is1 + 90o) = Vf1∠(−30o + 90o) = Vf1∠60o

= Vf1 (cos 60o + j sin 60o) = Vf1 (0.5 + j0.8666) = Vf1 (a1 + jb1) pu.

The above implies a1 = 0.5, b1 = 0.866.. Knowing this, we can solve Vf1 using followingquadratic equation.

V 2f1 − 2 a1 Vl Vf1 + V 2

l − V 2t1 = 0

From the above,

Vf1 = a1 Vl ±√a2

1 V2l − (V 2

l − V 2t1)

= 0.5× 1√2±

√√√√(0.5)2 ×(

1√2

)2

(1√2

)2

− (0.5619)2

= 0.35±

√0.125− 0.1842

The above solution is complex quantity, which implies that it is not possible to maintain load volt-age at 1.0 sin (ωt− φl).

3. Maximum possible load voltage

The maximum load voltage that can be obtained with the DVR, without any real power fromthe dc bus can be given as following.

Vl =Vt1√1− a2

1

=0.5619√1− 0.52

= 0.6488 pu.

In the time domain the load voltage vl = vl1 =√

2× 0.6488 sin(ωt− φl) = 0.9175 sin(ωt− φl).For this load voltage the DVR voltage is given as following.

Vf1 = a1 Vl = 0.5× 0.6488 = 0.3244 pu.

This implies

V f1 = 0.3244∠60o pu.

The time domain repression for the fundamental DVR voltage is given as,

vf1(t) =√

2× 0.3244 sin(ωt+ 60o) = 0.4587 sin(ωt+ 60o) pu.

The harmonic voltage that DVR compensates is as following.

vfh(t) = −vth = 0.45 sin (3ωt− 3.82o) pu.

The total DVR voltage is given as below.

vf (t) = vf1(t) + vth(t)

= 0.4587 sin(ωt+ 60o) + 0.45 sin (3ωt− 3.82o) pu

188

Page 195: NPTEL Course on Power Quality in Power Distribution Systems …nptel.ac.in/courses/108106025/Power quality_in_power_distribution... · NPTEL Course on Power Quality in Power Distribution

References

[1] A. Ghosh and G. Ledwich, “Compensation of distribution system voltage using dvr,” IEEETransactions on Power Delivery, vol. 17, no. 4, pp. 1030–1036, Oct. 2002.

[2] A. Ghosh and G. Ledwich, “Structures and control of a dynamic voltage regulator (dvr),” inIEEE Power Engineering Society Winter Meeting, vol. 3. IEEE, 2001, pp. 1027–1032.

[3] A. Ghosh, A. Jindal, and A. Joshi, “Design of a capacitor-supported dynamic voltage restorer(dvr) for unbalanced and distorted loads,” IEEE Transactions on Power Delivery, vol. 19, no. 1,pp. 405–413, Jan. 2004.

[4] S. Sasitharan and Mahesh K. Mishra, “Constant switching frequency band controller for dy-namic voltage restorer,” IET Power Electronics, vol. 3, no. 5, pp. 657–667, Sept. 2010.

[5] S. Sasitharan, Mahesh K. Mishra, B. Kalyan Kumar, and V. Jayashankar, “Rating and designissues of dvr injection transformer,” International Journal of Power Electronics, vol. 2, no. 2,pp. 143–163, 2010.

189