Power Quality in Power Distribution Systems

7/29/2019 Power Quality in Power Distribution Systems

1/195

NPTEL Course

onPower Quality in Power Distribution Systems

Dr. Mahesh KumarProfessor, Department of Electrical Engineering

Indian Institute of Technology MadrasChennai - 600 036


2/195


3/195

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


4/195

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

2.1.4 Power Invariance in abc and 0 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 Budeanus 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


5/195

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 ofPloss 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


6/195

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


7/195

Chapter 1

SINGLE PHASE CIRCUITS: POWER

DEFINITIONS AND ITS COMPONENTS

(Lectures 1-8)

1.1 Introduction

The definitions of power and its various components are very important to understand quantitative

and qualitative power quality aspects in power system [1][5]. This is not only necessary from the

point of view of conceptual clarity but also very much required for practical applications such as

metering, quantification of active, reactive power, power factor and other power quality parameters

in power system. These aspects become more important when power system is not ideal i.e. it

deals with unbalance, harmonics, faults and fluctuations in frequency. We therefore, in this chapter

explore the concept and fundamentals of single phase system with some practical applications and

illustrations.

1.2 Power Terms in a Single Phase System

Let us consider a single-phase system with sinusoidal system voltage supplying a linear load as

shown in Fig. 1.1. The voltage and current are expressed as below.

v(t) =

2V sin t

i(t) =

2Isin(t ) (1.1)The instantaneous power can be computed as,

p(t) = v(t) i(t) = V I[2sin t sin(t )]= V I[cos cos(2t )]= V Icos (1 cos2t) V Isin sin2t (1.2)= P(1 cos2t) Q sin2t= pactive(t) preactive(t)

1


8/195

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 difference

should always kept in mind.Equation (1.2) shows that instantaneous power can be decomposed into two parts. The first term

has an average value ofV Icos and an alternating component of V Icos2t, 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 Isin sin2t oscillating at twice frequency witha peak vale of V Isin . The second term has zero average value. The equation (1.2) can furtherbe written in the following form.

p(t) = V Icos V Icos(2t )= p(t) + p(t)

= paverage +poscillation

= puseful +pnonuseful (1.4)

With the above definitions ofP and Q, the instantaneous powerp(t) can be re-written as following.

p(t) = P(1 cos2t) Q sin2t (1.5)

2


9/195

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

2 230 sin t supplying a linear loadof impedance ZL = 12 +j13 at = 2f 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 , ZL = 12 , and ZL = j13

4. Comment upon the results.

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

the circuit is given as,

i(t) =

2Isin(t )where = tan1(XL/RL), and I = (V /|ZL|)

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

|ZL| =

R2L + X2L =

122 + 132 = 17.692 , and I = 230/17.692 = 13 A

= tan1(X/R) = tan1(13/12) = 47.29o

Therefore we have,

v(t) =

2 230 sin ti(t) =

2 13 sin(t 47.29o)

The instantaneous power is given as,

p(t) = V Icos (1 cos2t) V Isin sin2t= 230 13cos47.29o(1 cos(2 314t)) 230 13sin47.29o 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


10/195

Average real power (P) is given as,

P =1

T

T0

p(t) dt

P = V Icos = 230 13 cos47.29o = 2028.23 WReactive power (Q) is given as maximum value ofpreactive, and equals to V Isin as given below.

Q = V Isin = 230 13 sin47.2906 = 2196.9 VArApparant power, S = V I =

P2 + 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. As

seen 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,

VAr

0 0.01 0.02 0.032000

2100

2200

2300

2400

sec

W,

VAr

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


11/195

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

122 + 132 = 17.692,and I = 230/17.692 = 13 A, = tan1(13/12) = 47.2906o.

v(t) =

2 230 sin ti(t) =

2 13 sin(t + 47.2906o)

p(t) = V Icos (1 cos2t) V Isin sin2t= 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) = 2196.9sin(2 314t)

P = V Icos = 230 13 cos 47.2906o = 2028.23 WattQ = V Isin = 230 13 sin(47.2906o) = 2196.9 VArS = V I = P2 + Q2 = 230 13 = 2990 VA

For Case 2, the voltage, current and various components of the power are shown in Fig. 1.3. The

explanation 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,

VAr

0 0.01 0.02 0.03 0.04-3000

-2000

-1000

0

1000

2000

3000

sec

W,

VAr

Voltage (V)

Current (A)

Average Power (W)


p(t)

pact

(t)

preact

(t)


5


12/195

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

v(t) =

2 230 sin ti(t) =

2 19.167 sin t

p(t) = 230 19.167cos0o

(1 cos(2 314t)) 230 19.167sin0o

sin(2 314t)= 4408.33(1 cos(2 314t)) 0

pactive(t) = 4408.33(1 cos(2 314t))preactive(t) = 0

P = V Icos = 230 19.167 cos0o = 4408.33 WQ = V Isin = 230 19.167 sin0o = 0 VArS = V I =

P2 + Q2 = 230 19.167 = 4408.33 VA

Power factor = 4408.334408.33

= 1

For Case 3, the voltage, current and various components of the power are shown in Fig. 1.4. Since

the load is resistive, as seen from the graph preactive is zero and p(t) is equal to pactive. The averagevalue ofp(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,

VAr

0 0.01 0.02 0.03-1000

0

1000

2000

3000

4000

5000

sec

W,

VAr

Average Power (W)


Voltage (V)

Current (A)

p(t)

pact

(t)

preact

(t)


6


13/195

Case 4: When the load is purely reactive, ZL = j13 , |ZL| = 13 , I = 23013 = 17.692 A, and = 90o. Therefore, we have

v(t) =

2 230sin ti(t) =

2 17.692 sin(t 90o)

p(t) = 230 17.692cos90o

(1 cos(2 314t)) 230 17.692sin90o

sin(2 314t)= 0 4069 sin(2 314t)

pactive(t) = 0

preactive(t) = 4069 sin(2 314t)

P = V Icos = 230 17.692 cos 90o = 0 WQ = V Isin = 230 17.692 sin90o = 4069 VArS = V I =

P2 + 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. The

load in this case is purely reactive, hence their is no average component of p(t). The maximumvalue ofp(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,

VAr

0 0.01 0.02 0.030

1000

2000

3000

4000

5000

W,

VAr

Voltage (V)

Current (A)

p(t)

pact

(t)

preact

(t)

Active Power (W)



7


14/195

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. The

voltage and current are expressed as following.

v(t) = 2V sin ti(t) =

2

n=1

In sin(nt n) (1.6)

The instantaneous power is therefore given by,

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

2V sin t

2

n=1

In sin(nt n)

= V

n=1[In 2sin t sin(nt n)]= V [I1 2sin t sin(t 1)]

+ V

n=2

[In 2sin t sin(nt n)] (1.7)

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

p(t) = V I1 [cos 1 cos(2t 1)] V I1 sin 1 sin2t

+ V n=2

In[(cos n cos(2nt n)) sin n sin2nt]

= V I1 cos 1(1 cos2t) V I1 sin 1 sin2t (1.8)

+

n=2

V In[cos n(1 cos2nt) sin n sin2nt]

= A + B

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

P = P1

= average value ofp(t) = V I1

cos 1

Q = Q1= peak value of second term in A = V I1 sin 1 (1.9)

The apparent power S is given by

S = V I

S = V

[I21 + I22 + I

23 + .....] (1.10)

8


15/195

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

S2 = V2 I21 + V2 [I22 + I

23 + I

24 + ...]

= (V I1 cos 1)2 + (V I1 sin 1)

2 + V2 [I22 + I23 + I

24 + .....]

= P2

+ Q2

+ H2

(1.11)

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

H = V

[I22 + I23 + I

24 + .....] (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 cos2t and V I1 sin 1 sin2t 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 tuned

LC filters.

This is represented by power tetrahedron instead of power triangle (in case of voltage and current

of sinusidal nature of fundamental frequency). In this context, some important terms are defined

here.

Displacement Factor or Fundamental Power Factor (DPF) is denoted by cos 1 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) =

(P21 + Q

21)

S

=

(V2 I21 cos

2 1 + V2 I21 sin2 1)

V I

= V2 I21V I

=I1I

= cos (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


16/195

Power Factor (pf) =P

S

=V I1 cos 1

V I

=I1

I

cos 1

= cos cos 1 (1.14)

The equation (1.14) shows that power factor becomes less by a factor of cos . This is due to thepresence of the harmonics in the load current. The nonlinear load current increases the ampere

rating of the conductor for same amount of active power transfer with increased VA rating. Such

kind 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 sin t and source impedance is negligible, draw thevoltage and current waveforms. Express current using Fourier series. Based on that determine the

following.

1. Plot instantaneous power p(t).

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

3. Compute average real power, reactive power, apparent power, power factor, displacement

factor (or fundamental power factor) and distortion factor.

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

Id

i ( )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


17/195

series expansion is given below.

i(t) =

n=2h+1

4Idcn

sin(nt) where h = 0, 1, 2 . . .

The instantaneous power is therefore given by,

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

2V sin t

n=2h+1

4Idcn

sin(nt). (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 cos 1

= V I1 (since, 1 = 0, cos 1 = 1, sin 1 = 0)

Q = Q1= peak value ofPreactive(t) = V I1 sin 1 = 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

Id = 93.15 A

11


18/195

The real power (P) is given by

P = V I1

= V 2

2

Id = 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 ) is,

DF = (P21 + Q21)S=

I1Irms

= cos = 0.9.

The displacement factor (cos 1) is,

DP F = cos 1 = 1.

Therefore power factor is given by,

pf =P

S=

P1S1

S1S

= cos cos 1

= DF DP F = 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 of

non-sinusoidal voltage source supplying nonlinear loads including dc components. These voltages

and currents are represented as,

v(t) = Vdc + n=1

2Vn sin(nt vn) (1.16)

and

i(t) = Idc +

n=1

2In sin(nt in) (1.17)

12


19/195

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

p(t) = [Vdc +

n=1

2Vn sin(nt

vn)].[Idc +

n=1

2In sin(nt

in)] (1.18)

p(t) = VdcIdc I

+ Vdc

n=1

2In sin(nt in)

II

+ Idc

n=1

2Vn sin(nt vn)

III

+

n=1

2Vn sin(nt vn)

n=1

2In sin(nt in) IV

(1.19)

p(t) = pdcdc +pdcac +pacdc +pacac (1.20)

The I term (pdcdc) contribute to power from dc components of voltage and current. Terms II(pdcac) and III (pacdc) 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, dc

components are very less and the first three terms have negligible value compared to IV term. Thus,

we shall focus on IV (pacac) term which correspond to ac components present in power system.The IV term can be written as,

IVth term = pacac = n=1

2Vn sin(nt vn) h=1

2Ih sin(ht ih) (1.21)

where n = h = 1, 2, 3..., similar frequency terms will interact. When n = h, dissimilar13


20/195

frequency terms will interact. This is expressed below.

pac

ac(t) =

2V1 sin(t

v1)

2I1 sin(t

i1) A

+

2V1 sin(t v1)

h=2,h=1

2Ih sin(ht ih)

B

+

2V2 sin(2t v2)

2I2 sin(2t i2) A

+

2V2 sin(2t v2)

h=1,h=2

2Ih sin(ht ih)

B+ . . . + . . .

+

2Vn sin(nt vn)

2In sin(nt in) A

+

2Vn sin(nt vn)

h=1,h=n

2Ih sin(ht ih)

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 pacacnn and pacacnh. Thus,

pacacnn(t) =

n=1

VnIn2sin(nt vn)sin(nt in) (1.23)

and

pacacnh(t) =

n=1

2Vn sin(nt vn)

h=1,h=n

2In sin(ht ih) (1.24)

14


21/195

Now, let us simplify pacacnn in

pacacnn(t) =

n=1

VnIn[cos(in vn) cos(2nt in vn)]

=

n=1 VnIn[cos(n) cos(2nt (in vn) 2vn)]=

n=1

VnIn[cos(n) cos(2nt 2vn) n]

=

n=1

VnIn[cos(n) cos(2nt 2vn) cos n sin(2nt 2vn) sin n]

=

n=1

[VnIn cos n{1 cos(2nt 2vn)}

VnIn sin n sin(2nt

2vn)] (1.25)

where n = (in vn) = phase angle between nth harmonic current and voltage.

pacacnn(t) =

n=1

[VnIn cos n{1 cos(2nt 2vn)}]

n=1

[VnIn sin n sin(2nt 2vn)] (1.26)

Therefore, the instantaneous power is given by,

p(t) = pdcdc

I

+pdcac

II

+pacdc

III

+pacacnn

IVA

+pacacnh

IVB IV

p(t) = VdcIdc + Vdc

n=1

2In sin(nt in) + Idc

n=1

2Vn sin(nt vn)

+

n=1

[VnIn cos n{1 cos(2nt vn)}]

n=1[VnIn sin n. sin(2nt in)] (1.27)

1.4.1 Active Power

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

pactive(t) = VdcIdc +

n=1

[VnIn cos n{1 cos(2nt vn)}] (1.28)

15


22/195

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

P =1

T

T0

p(t) dt

= VdcIdc +

n=1 VnIn cos n. (1.29)Reactive power can be defined as

q(t) = preactive(t) =

n=1

[VnIn sin n sin(2nt 2vn)] (1.30)

resulting in

Q max of (1.30) magnitude

= n=1

VnIn sin n. (1.31)

From (1.29)

P = Pdc +

n=1

VnIn cos n

= Pdc + V1I1 cos 1 + V2I2 cos 2 + V3I3 cos 3 + . . .

= 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 cos n = P P1 (1.34)

16


23/195

1.4.2 Reactive Power

The reactive power or Budeanus reactive power (Q) can be found by summing maximum value of

each term in (1.30). This is given below.

Q = n=1

VnIn sin n

= V1I1 sin 1 + V2I2 sin 2 + V3I3 sin 3 + . . .

= Q1 + Q2 + Q3 + . . .

= Q1 + QH (1.35)

Usually this reactive power is referred as Budeanus 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) = pdcdc +pactive(t) +preactive(t)

similar frequency terms

+ prest(t) non-similar frequency terms

(1.37)

where,

pdcdc = VdcIdc

pactive(t) =

n=1

[VnIn cos n{1 cos(2nt 2 vn)}]

preactive(t) =

n=1[VnIn sin n. sin(2nt 2 vn)]prest(t) = Vdc

n=1

2In sin(nt in) + Idc

n=1

2Vn sin(nt vn)

+

n=1

2Vn sin(nt vn)

m=1,m=n

2Im sin(mt im) (1.38)

1.4.3 Apparent Power

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

expressed as following.

S = V I

=

V2dc + V21 + V

22 +

I2dc + I

21 + I

22 + (1.39)

=

V2dc + V21 + V

2H

I2dc + I

21 + I

2H

17


24/195

Where,

V2H = V22 + V

23 + =

n=2

V2n

I

2

H = I

2

2 + I

2

3 + =

n=2 I2n (1.40)VH and IH are denoted as harmonic voltage and harmonic current respectively. Expanding (1.39)we can write

S2 = V2I2

= (V2dc + V21 + V

2H)(I

2dc + I

21 + I

2H)

= V2dcI2dc + V

2dcI

21 + V

2dcI

2H + V

21 I

21 + V

21 I

2dc + V

21 I

2H + V

2HI

2dc + V

2HI

21 + V

2HI

2H

= V2dcI2dc + V

21 I

21 + V

2HI

2H + V

2dc(I

21 + I

2H) + I

2dc(V

21 + V

2H) + V

21 I

2H + V

2HI

21

= S2dc + S21 + S

2H + S

2D

= S21 + S2dc + S

2H + S

2D

= S21 + S2N (1.41)

In above equation, the term SN is as following.

S2N = V2

dcI21 + V

2dcI

2H + V

21 I

2dc + V

21 I

2H + V

2HI

2dc + V

2HI

21 + V

2HI

2H + I

2dcI

2H + I

2dcV

2dc (1.42)

Practically in power systems dc components are negligible. Therefore neglecting the contribution

ofVdc and Idc associated terms in (1.42), the following is obtained.

S2N = I21V

2H + V

21 I

2H + V

2HI

2H

= D2V + D2I + 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.

T HDV =VHV1

T HDI =IHI1

(1.45)

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

VH = T HDV V1

IH = T HDI I1 (1.46)

18


25/195

Using (1.44) and (1.46),

DV = V1 I1 T HDV = S1 T HDV

DI = V1 I1 T HDI = S1 T HDI

SH = VHIH = S1 T HDI T HDV (1.47)

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

S2N = S21 (T HD

2I + T HD

2V + T HD

2I T HD

2V) (1.48)

Normally in power system, T HDV


26/195

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 + PHS21 + S

2N

=(1 + PH/P1)1 + (SN/S1)2

P1S1

(1.55)

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

pf =(1 + PH/P1)

1 + T HD2I + T HD

2V + T HD

2IT HD

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(DP F = 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 T HDV


27/195

Solution: Here the source is non-sinusoidal and is feeding a non-linear load. The instantaneous

power is given by,

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

p(t) = {Vdc +

n=1

2 Vn sin(nt vn)} {Idc +

n=1

2 In sin(nt in)}

(a) The active power P is given by,

P =1

T

T0

p(t) dt

= Pdc + V1 I1 cos 1 + V2 I2 cos 2 + ...... + Vn In cos n (1.58)

= Pdc + P1 + PH

where,

n = in vnPdc = Vdc Idc

P1 = V1 I1 cos 1

PH =

n=2

Vn In cos n

Here, Vdc = 0, V1 = 230 V, v1 = 0, V3 = 50 V, v3 = 30, Idc = 2 A, I1 = 10 A, i1 = 30,

I3 = 5 A, i3 = 60. Therefore, 1 = i1 v1 = 30 and 3 = i3 v3 = 30.

Substituting these values in (1.58), the above equation gives,

P = 0 2 + 230 10 cos30 + 50 5 cos 30 = 2208.36 W.(b). The reactive power (Q) is given by,

Q =

n=1

Vn In sin n

= V1 I1 sin 1 + V2 I2 sin 2 + .....Vn In sin n

= 230 10 sin30 + 50 5 sin30 = 1275 VAr.(c). The Apparent power S is given by,

S = Vrms Irms

=

V2dc + V21 + V

22 + ....V

2n

I2dc + I

21 + I

22 + .....I

2n

= V2dc + V21 + V2H I2dc + I21 + I2Hwhere,

VH =

V22 + V23 + ....V

2n

IH =

I22 + I23 + .....I

2n

21


28/195

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

2Vn

n2 sin(nt vn)

with Vdc = 10 V, Vn/n2 = 230

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

The voltage source supplies a nonlinear current of,

i(t) = Idc +

n=1

2Inn

sin(nt in).

with Idc = 2 A, In = 20/n A and in = 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


29/195

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

10 +

n=1,3,5230

2

n2sin(nt)

2 +

n=1,3,520

2

nsin n(t 300)

= 20I

+ 10

n=1,3,5

20 2n

sin n(t 300) II

+ 2

n=1,3,5

230 2n2

sin(nt) III

+

n=1,3,5

230

2

n2sin(nt)

n=1,3,5

20

2

nsin n(t 300)

IV

= 20

I+

n=1,3,5200

2

nsin n(t 300)

II+

n=1,3,5460

2

n2sin(nt)

III+

n=1,3,5

4600

n3(cos(30o n)(1 cos 2nt) sin (2nt) sin(30o n))

IV A

+

n=1,3,5

230

2

n2sin nt

h=1,3,5;h=n

20

2

hsin h(t 300)

IV B

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

pdcdc(t) = 20 W

pactive(t) =

n=1,3,5

4600

n3cos n300(1 cos 2nt)

preactive(t) =

n=1,3,54600

n3sin(n300) sin(2nt)

prest(t) =

n=1,3,5

200 2n

sin n(t 300) +

n=1,3,5

460 2n2

sin(nt)

+

n=1,3,5

230

2

n2sin nt

h=1,3,5;h=n

20

2

hsin h(t 300)

23


30/195

2. Computation ofP, P1, PH

P =1

T

T0

p(t)dt

= 20 +

n=1,3,5

4600n3

cos(30o n)

= 20 + 4600cos 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 ofQ, 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) = 175.7548

VArs.

4. Computation of Apparent Powers and Distortion Powers

24


31/195

The apparent power S is expressed as following.

Vrms =

V2dc + V21 + V

23 + V

25 + V

27 + V

29 + ....

=

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

= 231.87 V (up to n = 9)

Irms =

I2dc + I21 + I

23 + I

25 + I

27 + I

29 + ....

=

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 =

V23 + V25 + V

27 + V

29 + ....

=

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

= 27.7 V (up to n = 9)

IH =

I23 + I25 + I

27 + I

29 + ....

=

(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 P2 = 50672 3960.22 = 3160.8 VArs (up to n=9)Distortion Power D =

S2 P2 Q2 = 50672 3960.22 2475.772 = 1965.163

VArs (up to n=9).

Displacement power factor (cos 1)

cos 1 =P1

S1=

3983.7

(230)(20)= 0.866 lagging

Power factor (cos )

cos =P

S=

3960.217

5067= 0.781 lagging

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

above values.

25


32/195

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

-200

-100

0

100

200

time (sec)

Voltage(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)

Current(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.Power(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.activepower(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.reactivepower

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-2000

0

2000

time (sec)

Re

stofinst.power(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)

Nonactivepower(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 reactive

powers in electrical systems with generic loads, IEEE Transactions on Power Delivery, vol. 8,

no. 2, pp. 697703, 1993.

[3] T. Furuhashi, S. Okuma, and Y. Uchikawa, A study on the theory of instantaneous reactive

power, IEEE Transactions on Industrial Electronics, vol. 37, no. 1, pp. 8690, 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 Instrumentation

and Measurement, vol. 40, no. 3, pp. 568577, 1991.

[5] J. Willems, A new interpretation of the akagi-nabae power components for nonsinusoidal

three-phase situations, IEEE Transactions on Instrumentation and Measurement, vol. 41,

no. 4, pp. 523527, 1992.

26


33/195

Chapter 2

THREE PHASE CIRCUITS: POWER

DEFINITIONS AND VARIOUS

COMPONENTS

(Lectures 9-18)

2.1 Three-phase Sinusoidal Balanced System

Usage of three-phase voltage supply is very common for generation, transmission and distribution

of bulk electrical power. Almost all industrial loads are supplied by three-phase power supply for

its 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-phase

system 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 Icos as well pulsating termV Icos(2t ). 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(t)

vb(t) =

2V sin(t 120) (2.1)vc(t) =

2V sin(t + 120)

27


34/195

a

b

c

a

b

c

Fig. 2.1 A three-phase balanced circuit

and

ia(t) =

2Isin(t)

ib(t) =

2Isin(t 120) (2.2)ic(t) =

2Isin(t + 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 shiftof120o 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,

p3(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 Icos {1 cos2t} V Isin sin2tpb(t) = V Icos {1 cos(2t 120o)} V Isin sin2(t 120o) (2.4)pc(t) = V Icos {1 cos(2t + 120o)} V Isin sin2(t + 120o)

Adding three phase instantaneous powers given in (2.4), we get the three-phase instantaneous

power as below.

p(t) = 3 V Icos V Icos {cos2t + cos 2(t 120o) + cos 2(t + 120o)} V Isin {sin2t + sin 2(t 120o) + sin 2(t + 120o)} (2.5)

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

p3(t) = p(t) = 3V Icos . (2.6)

28


35/195

This is quite interesting result. It indicates for balanced three-phase system, the total instantaeous

power 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 in

case 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 Isin . (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 0 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 0 transforma-tion. This transformation is described below.

The abc coordinates and their equivalent 0 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 0 transformation

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

v = 23 (va vb2 vc2 ) (2.8)v =

2

3

3

2(vb vc) (2.9)

Here,

23

is a scaling factor, which ensures power invariant transformation. Along with that, we

define zero sequence voltage as,

29


36/195

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)v(t)v(t)

= 23

12 12 121 12

12

032

32

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

(2.11) v0v

v

= [Ao] vavb

vc

The above is known as Clarke-Concordia transformation. Thus, va, vb and vc can also be expressedin terms ofv0, v and v by pre-multiplying (2.11) by matrix [A0]1, we have

vavbvc

= [A0]1 v0v

v

It will be interesting to learn that

[A0]1 = [Aabc] = 2

3 12 12 121 12 12

032

32

1

[A0]1 =

23

12

1 012

12

32

12

12

32

= [A0]T = [Aabc] (2.12)

Similarly, we can write down instantaneous symmetrical transformation for currents, which is

given below.

i0ii

= 23 12 12 121 12 12

032

32

iaibic

(2.13)Now based on 0 transformation, the instantaneous active and reactive powers are defined asfollows. The three-phase instantaneous power p(t) is expressed as the dot product of 0 compo-nents of voltage and currents such as given below.

30


37/195

p(t) = v i + v i + v0 i0

=2

3 va vb

2

vc

2 ia ib

2

ic

2

3

2

(vb

vc)

3

2

(ib

ic)

+1

2(va + vb + vc)

12

(ia + ib + ic)

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

Now what about instantaneous reactive power? Is there any concept defining instantaneous reactive

power? In 1983-84,authors H.akagi have attempted to define instantaneous reactive power using

stationary 0 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) = v i + v iq(t) = vi vi

=2

3

va vb

2 vc

2

32

(ib ic)

3

2(vb vc)

ia ib

2 ic

2

=2

3

3

2

(vb + vc) ia +

va vb

2 vc

2+

vb2

vc2

ib +

va + vb

2+

vc2

+vb2

vc2

ic

= 13

[(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

ic2

vc

=

13

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

2.1.4 Power Invariance in abc and 0 Coordinates

As a check for power invariance, we shall compute the energy content of voltage signals in two

transformations. The energy associated with the abc0 system is given by (v2a + v2b + v

2c ) and the

energy associated with the 0 components is given by

v20 + v2 + v

2

. The two energies must

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

31


38/195

squares of the respective components, we have the following.

v2 =

2

3

va vb

2 vc

2

2v2 =

23v2a + v2b4 + v2c4 2vavb2 + 2vbvc4 2vavc2

=2

3v2a +

v2b6

+v2c6

2vavb3

+vbvc

3 2vavc

3(2.17)

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

v2 =

3

2

2

3(vb vc)

2=

12

v2b + v2c 2vbvc

=v2b2

+v2c2

vbvc (2.18)

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

v2 + v2 =

2

3 v2a + v2b + v2c vcvb vbvc vcva=

v2a + v

2b + v

2c

v2a3

+v2b3

+v2c3

+2vavb

3+

2vbvc3

+2vavc

3

=

v2a + v

2b + v

2c

13

(va + vb + vc)2

=

v2a + v2b + v

2c

13

(va + vb + vc)

2(2.19)

Since v0 =13

(va + vb + vc), the above equation, (2.19) can be written as,

v2 + v2 + v

20 = v

2a + v

2b + v

2c . (2.20)

From the above it is implies that the energy associated with the two systems remain same instant to

instant basis. In general the instantaneous power p(t) remain same in both transformations. Thisis proved below.

32


39/195

Using (2.14), following can be written.

p(t) = vi + vi + voio

p(t) = v0vv

T

i0ii

=

[Aabc] vavb

vc

T[Aabc] iaib

ic

=

vavbvc

T [Aabc]T [Aabc] iaib

ic

=

vavb

vc T

[Aabc]1 [Aabc]

iaib

ic =

va vb vc

iaibic

= vaia + vbib + vcic (2.21)

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

[Aabc] = 23

1

21

21

21 12

12

032

32

and (2.22)

[Aabc]1 = [Aabc]T =

2

3

12

1 012

12

32

12

12

32

2.2 Instantaneous Active and Reactive Powers for Three-phase Circuits

In the previous section instantaneous active and reactive powers were defined using 0 trans-formation. In this section we shall study these powers for various three-phase circuits such as

three-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


40/195

2.2.1 Three-Phase Balance System

For three-phase balanced system, three-phase voltages have been expressed by equation (2.1). For

these phase voltages, the line to line voltages are given as below.

vab =

3

2v sin(t + 30)

vbc = 32v sin(t 90)vca =

3

2v sin(t + 150) (2.23)

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

0o

aV 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

= vi + vi + v0i0

= 3 V I cos (2.24)

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

q(t) = [

3

2V sin(t 90o)

2Isin(t )+

3

2V sin(t + 150o)

2Isin(t 120o )+

3

2V sin(t + 30o)

2Isin(t + 120 )]/

3

=

3V I[cos (90 ) cos (2t 90o )+cos(90o

)

cos(2t

30o

)

+cos(90o ) cos(2t + 150o )]/3=

3V I[3sin cos(2t + 30o) cos (2t + 30o + 120o)

cos (2t + 30o 120o)]/

3

= V I[3sin 0]q(t) = 3V Isin (2.25)

34


41/195

The above value of instantaneous reactive power is same as defined by Budeanus [1] and is given

in 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 terms

of 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 90ovbc =

3va 90o (2.26)

vca =

3vb 90oIn 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(t 90) .

2Isin(t )= 3V I2 sin(t 90) . sin(t )=

3V I [cos (90 ) cos (2t 90 )]

=

3V I [sin cos {90 + (2t )}]=

3V I[sin + sin (2t )]

=

3V I [sin + sin 2t cos cos2t sin ]vbcia/

3 = V I [sin (1 cos2t) + cos sin2t]

Similarly,

vcaib/

3 = V I

sin

1 cos

2

t 2

3

+V I cos . sin2t 2

3

vabic/

3 = V I

sin

1 cos

2

t +

2

3

+V I cos . sin2

t +

2

3

(2.27)

Thus, we see that the role of the coefficients of sin and cos have reversed. Now if we takeaverage value of (2.27), it is not equal to zero but V Isin in each phase. Thus three-phase reactivepower will be 3V Isin . The maximum value of second term in (2.27) represents active averagepower i.e., V Icos . 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 may

result from single-phasing, faults, different loads in three phases. To study three-phase system

35


42/195

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

va =

2Va sin(t va)vb =

2Vb sin(t 120o vb) (2.28)

vc =

2Vc sin(t + 120

o

vc)and,

ia =

2Ia sin(t ia)ib =

2Ib sin(t 120o ib) (2.29)

ic =

2Ic sin(t + 120o ic)

For the above system, the three-phase instantaneous power is given by,

p3(t) = p(t) = vaia + vbib + vcic

=

2Va sin(t va)sin(t ia)+

2Vb sin(t

120o

vb)

2Ib sin(t

120o

ib) (2.30)

+2Vc sin(t + 120o vc) 2Ic sin(t + 120o ic)Simplifying above expression we get,

p3(t) = VaIa cos a {1 cos(2t 2va)} pa,active

VaIa sin a sin (2t 2va) pa,reactive

+VbIb cos b [1 cos {2 (t 120) 2vb}]VbIb sin b sin {2 (t 120) 2vb}

+VcIc cos c [1 cos {2 (t + 120) 2vc}]VcIc sin c sin {2 (t + 120) 2vc} (2.31)where a = (ia va)

Therefore,

p3 (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 cos a

pb = Pb = VbIb cos b (2.33)

pc = Pc = VcIc cos c

and pa = VaIa cos (2 t a 2 va)pb = VbIb cos (2 t 240o b 2 vb) (2.34)pc = VcIc cos(2t + 240 c 2 vc)36


43/195

Also it is noted that,

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

and,

pa + pb + pc = VaIacos(2t va ib)VbIbcos {2(t 120) vb ib}VcIccos {2(t + 120) vc ic}

= 0This implies that, we no longer get advantage of getting constant power, 3V Icos from interactionof three-phase voltages and currents. Now, let us analyze three phase instantaneous reactive power

q(t) as per definition given in (2.15).

q(t) = 13

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

= 2

3{Vbsin(t 120o vb) Vcsin(t + 120o vc)} Iasin(t ia)+ {Vcsin(t + 120o vc) Vasin(t va)}

2Ibsin(t 120o ib) (2.36)

+{Vasin(t 120o va) Vb sin(t 120o vb)}

2Ic sin(t + 120o ic)

From the above,

3q(t) =

VbIa {cos(ia 120o vb) cos(2t 120o ia vb)}

VcIa {cos(ia + 120o vc) cos(2t + 120o ia vc)}+VcIb {cos(ib + 240o vc) cos(2t ib vc)} (2.37)

VaIb {cos(ib 120o

va) cos(2t 120o

va ib)}+VaIc {cos(ic 120o va) cos(2t + 120o va ic)}VbIc {cos(ic 240o vb) cos(2t ic vb)}

Now looking this expression,we can say that

1

T

T0

q(t)dt = 13

VbIa cos(ia vb 120o)

VcIa cos(ia vc + 120o)+VcIb cos(ib + 240

o

vc)

VaIb cos(ib 120o va)+VaIc cos(ic 120o va)VbIc cos(ic 240o vb)

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

= VaIa cos a + VbIb cos b + VcIc cos c (2.38)37


44/195

Hence the definition of instantaneous reactive power does not match to that defined by Budeanues

reactive power [1] for three-phase unbalanced circuit. If only voltages or currents are distorted, the

above holds true as given below. Let us consider that only currents are unbalanced, then

va(t) =

2V sin(t)

vb(t) = 2V sin(t 120) (2.39)vc(t) =

2V sin(t + 120)

and

ia(t) =

2Ia sin(t a)ib(t) =

2Ib sin(t 120o b) (2.40)

ic(t) =

2Ic sin(t + 120o c)

And the instantaneous reactive power is given by,

q(t) = 13

[vbcia + vcaib + vabic]

= 13 [3 va/2 ia + 3 vb /2 ib + 3 vc /2 ic]= [2V sin(t /2)2Ia sin(t ia)

+

2V sin(t 120o /2)2Ib sin(t 120o ib)+

2V sin(t + 120o + /2)

2Ic sin(t + 120

o ic)]= [V Iacos(/2 ia) cos {/2 (2t ia)}

+V Ibcos(/2 ib) cos(2t 240o /2 ib)+V Iccos(/2 ic) cos(2t + 240o /2 ic)]

= [(V Ia sin ia + V Ib sin ib + V Ic sin ic)+V Ia sin(2t ia) + V Ib sin(2t 240o ib) + V Ic sin(2t + 240o ic)]

Thus,

Q =1

T

T0

q(t)dt = (V Ia sin ia + V Ib sin ib + V Ic sin ic) (2.41)Which is similar to Budeanus reactive power.

The oscillating term ofq(t) which is equal to q(t) is given below.q(t) = V Ia sin(2t ia) + V Ib sin(2t 240o ib) + V Ic sin(2t + 240o ic) (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 been

considered. 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


45/195

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 can

be 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 phase

systems.

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

a = 1120o = ej2/3 = cos 2/3 +j sin 2/3

= 1/2 +j

3/2

a2 = 1240o = 1 120o = ej4/3 = ej2/3 = cos 4/3 +j sin 4/3= 1/2 j

3/2

a3 = 1360o = ej2 = 1

Also note an interesting property relating a, a2

and a3

,

a + a2 + a3 = 0. (2.43)

31

o

a o

21 120

o

a

1 120o

a

o

Fig. 2.4 Phasor representation ofa, 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


46/195

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 shift

from each other.

Positive sequence components: Va1, Vb1, Vc1Negative sequence components: Va2, Vb2, Vc2Zero sequence components: Va0, Vb0, Vc0

Thus we can write,

Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0 (2.44)

Vc = Vc1 + Vc2 + Vc0

Graphically, these are represented in Fig. 2.5. Thus if we add the sequence components of each

phase vectorially, we shall get Va, Vb and Vs as per (2.44). This is illustrated in Fig. 2.6.

1Va

1Vc

1Vb

2Vb

2Va

2Vc

0Va0Vb

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 Va1 be a reference phasor,therefore Vb1 and Vc1 can be written as,

Vb1 = a2Va1 = Va1 120

Vc1 = aVa1 = Va1120 (2.45)

Similarly Vb2 and Vc2 can be expressed in terms ofVa2 as following.

Vc2 = a2Vb2 = Va2 120

Vb2 = aVa2 = Va2120 (2.46)

40


47/195

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,

Vb0 = Vc0 = Va0 (2.47)

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

Va = Va0 + Va1 + Va2 (2.48)

Vb = Vb0 + Vb1 + Vb2

= Va0 + a2 Va1 + a Va2 (2.49)

Vc = Vc0 + Vc1 + Vc2

= Va0 + a Va1 + a2 Va2 (2.50)

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

VaVbVc

= 1 1 11 a2 a1 a a2

VaoVa1Va2

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

1 1 11 a2 a1 a a2

, the symmetrical com-41


48/195

ponents are expressed as given below.

VaoVa1Va2

=

1

3

1 1 11 a a2

1 a2 a

VaVbVc

(2.52)

= [A012]VaVb

Vc

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.

Va0 =

1

3(Va + Vb + Vc)

Va1 =1

3(Va + aVb + a

2Vc) (2.54)

Va2 =1

3(Va + a

2Vb + aVc)

The other component i.e., Vbo, Vco, Vb1, Vc1, Vb2, Vc2 can be found from Vao, Va1, Va2. It shouldbe noted that quantity Vao does not exist if sum of unbalanced phasors is zero. Since sum of lineto line voltage phasors i.e., Vab + Vbc + Vca = (Va Vb) + (Vb Vc) + (Vc Va) is alwayszero, hence zero sequence voltage components are never present in the line voltage, regardless of

amount of unbalance. The sum of the three phase voltages, i.e., Va + Vb + Vc is not necessarily

zero and hence zero sequence voltage exists.

Similarly sequence components can be written for currents. Denoting three phase currents by

Ia, Ib, and Ic respectively, the sequence components in matrix form are given below.IaoIa1Ia2

= 13

1 1 11 a a21 a2 a

IaIbIc

(2.55)Thus,

Iao =1

3(Ia + Ib + Ic)

Ia1 =1

3(Ia + aIb + a

2Ic)

Ia2 =1

3(Ia + a

2Ib + aIc)

42


49/195

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 zero

sequence 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 the

neutral wire. But in this configuration the neutral voltage, VN n, may or may not be equal to zerodepending upon the unbalance in the system.

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

Va = 1000o

Vb = 100 120oVc = 100120

o

Using (2.54), it can be easily seen that the zero and negative sequence components are equal to

zero, 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 the

energy associated with the voltage components in both system remain constant.

Va = 1000o

Vb = 150 100oVc = 75100

o

43


50/195

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

Va0 =1

3(Va + Vb + Vc)

= 31.91 50.48o VVa1 = 1

3(Va + aVb + a2Vc)

= 104.164.7o V

Va2 =1

3(Va + a

2Vb + aVc)

= 28.96146.33o V

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

Eabc = k [V2

a + V2

b + V2

c ] = k.(381.25)

E012 = k 3[V2

a0 + V2

a1 + V2

a2] = 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 = [ Va Vb Vc ]

IaIbIc

=

VaVb

Vc

T

IaIb

Ic

=

[Aabc]Va0Va1

Va2

T[Aabc] Ia0Ia1

Ia2

=

Va0Va1Va2

T[Aabc]T[Aabc] Ia0Ia1

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


51/195

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

Sv = P +jQ =

Va0Va1Va2

T3[I]Ia0Ia1

Ia2

= 3Va0Va1

Va2

TIa0Ia1Ia2

Sv = P +jQ = VaI

a + VbI

b + VcI

c

= 3 [Va0Ia0 + Va1I

a1 + Va2I

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) further

implies that,

Sv = P +jQ = 3 [ (Va0Ia0 cos a0 + Va1Ia1 cos a1 + Va2Ia2 cos a2)

+j(Va0Ia0 sin a0 + Va1Ia1 sin a1 + Va2Ia2 sin a2) ] (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 cos a1 + Vb1Ib1 cos b1 + Vc1Ic1 cos c1

= 3Va1Ia1 cos a1. (2.61)

Negative sequence power is expressed as,

P = 3Va2Ia2 cos a2. (2.62)

The zero sequence power is

P0 = 3Va0Ia0 cos a0. (2.63)

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

Q+ = 3Va1Ia1 sin a1

Q = 3Va2Ia2 sin a2Q0 = 3Va0Ia0 sin a0 (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


52/195

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

following.

S+ = |S+| =

P+2 + Q+2 = 3Va1Ia1

S = |S+| =

P2 + Q2 = 3Va2Ia2

S0 = |S+

| = P02 + 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)=

P2 + 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)=

P2a + Q2a +

P2b + Q

2b +

P2c + Q

2c

In terms of sequence components apparent power,

SA = |S0| + |S+| + |S|= |P0 +jQ0| + |P+ +jQ+| + |P +jQ| (2.69)= P

02 + Q02 +P+2 + Q+2 +P

2 + Q2

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


53/195

Example 2.2 Consider a 3-phase 4 wire system supplying resistive load, shown in Fig. 2.8

below. 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=

3V2

R

The current flowing in the line =

3V

R= |Va Vb

R|

and Ib = Ia

Therefore losses in the feeder =

3V

R

2 r +

3V

R

2 r

= 2 rR3 V2R Now, consider another example of a 3 phase system supplying 3-phase load, consisting of three

resistors (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

R2

r =

r

R3 V2

R Thus, it is interesting to see that power dissipated in the unbalanced system is twice the power loss

in balanced circuit. This leads to conclusion that power factor in phases would become less than

unity, while for balanced circuit, the power factor is unity. Power analysis of unbalanced circuit

shown in Fig. 2.8 is given below.

47


54/195

'a

'c

'n

a

b

c

nV

a

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 =Va Vb

R=

VabR

=

3 VaR

30

The current in phase-b, Ib = Ia =

3 V

R (30 180)o

=

3 V

R 150o

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

The phase voltages are: Va = V0o, Vb = V 120o, Vc = V120o.

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

Pa = VaIa cos a = V Icos30 =

3

2

V I

Qa = VaIa sin a = V Isin30 =

1

2V I

Sa = VaIa = V I

Pb = VbIb cos b = V Icos(30) =

3

2V I

Qb = VbIb sin b = V Isin(30) = 12

V I

Sb = VbIb = V I

Pc = Qc = Sc = 0

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

3

2 V I = 3 V I=

3 V

3 V

R

P =3 V2

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

48


55/195

The vector apparent power, Sv =

P2 + Q2 = 3 V2/R = P

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

3) P

From the values ofSv 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 unbalances

in 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 known

as effective apparent power. The concept assumes that a virtual balanced circuit that has the same

power output and losses as the actual unbalanced circuit. This equivalence leads to the definition

of effective line current Ie and effective line to neutral voltage Ve.

The equivalent three-phase unbalanced and balanced circuits with same power output and losses

are shown in Fig. 2.10. From these figures, to maintain same losses,

'a

'c

'n

n

Va aRIa

rj 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 + rI2b + rI

2c + rI

2n = 3rI

2e

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

Ie =

(I2a + I

2b + I

2c + I

2n)

3(2.73)

49


56/195

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

Ie =

(I2a + I

2b )

3since, Ic = 0and In = 0

=2 I2a

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.

V2aR

+V2bR

+V2cR

+V2a + V

2b + V

2c

3R=

3V2eR

+9V2e3R

(2.74)

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

Ve =

1

18{3 (V2a + V2b + V2c ) + V2ab + V2bc + V2ca} (2.75)

Assuming, 3 (V2a + V2

b + V2

c ) V2ab + V2bc + V2ca, equation (2.75) can be written as,

Ve =

V2a + V

2b + V

2c

3= V (2.76)

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

Se = 3 Ve Ie =3

2 V2

RThus 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

Se=

3 V2/R

3

2 V2/R=

12

= 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


57/195

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+ = 3 V+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 1) + 2

n=2

Vn sin(nwt n)

vb(t) =

2V1 sin(wt 120 1) +

2

n=2

Vn sin(n(wt 120) n) (2.78)

vc(t) =

2V1 sin(wt + 120 1) +

2

n=2

Vn sin(n(wt + 120) n)

Similarly, the line currents can be expressed as,

ia(t) = 2I1 sin(wt 1) + 2

n=2

In sin(nwt n)

ib(t) =

2I1 sin(wt 120 1) +

2

n=2

In sin(n(wt 120 ) n) (2.79)

ic(t) =

2I1 sin(wt + 120 1) +

2

n=2

In sin(n(wt + 120 ) n)

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 additional

power loss in neutral wire and ground path.

51


58/195

2.4.1 Neutral Current

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

equation.

in = ia + ib + ic

= 2 [ Ia1 sin(wt 1) + Ia2 sin(2wt 2) + Ia3 sin (3wt 3)+Ia1 sin(wt 120o 1) + Ia2 sin (2wt 240o 2) + Ia3 sin (3wt 360o 3)+Ia1 sin(wt + 120

o 1) + Ia2 sin(2wt + 240o 2) + Ia3 sin (3wt + 360o 3)

+Ia4 sin(4wt 4) + Ia5 sin (5wt 5) + Ia6 sin(6wt 6)+Ia4 sin(wt 4 120o 4) + Ia5 sin(5wt 5 120o 5) + Ia6 sin(6wt 6 120o +Ia4 sin(wt + 4 120o 4) + Ia5 sin (5wt + 5 120o 5) + Ia6 sin(6wt + 6 120o

(2

+Ia7 sin(7wt 7) + Ia8 sin (8wt 8) + Ia9 sin(9wt 9)+Ia7 sin(7wt 7 120

o

7) + Ia8 sin(8wt 8 120o

8) + Ia9 sin(9wt 9 120o

+Ia7 sin(7wt + 7 120o 7) + Ia8 sin(8wt + 8 120o 8) + Ia9 sin (9wt + 9 120o

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 their

balanced nature. Therefore the neutral current is given by,

in = ia + ib + ic =

n=3,6,..

3

2In sin(nwt n). (2.82)

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

In = 3 n=3,6,..

I2n1/2 . (2.83)Due to dominant triplen harmonics in electrical loads such as UPS, rectifiers and other power

electronic 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, 7th

10th,... form positive sequence only. Balanced 2nd, 5th, 8th, 11th,... form negative sequence only

and the balanced triplen harmonics i.e. 3rd, 6th, 9th,... form zero sequence only. But in case of

unbalanced three-phase systems with harmonics, (3n + 1) harmonics may start forming negative

and 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


59/195

vab(t) = va(t) vb(t)

=

n=1

2Vn sin(n t n)

n=1

2Vn sin(n (t 120o) n)

=

n=1

2Vn sin(n t n)

n=1

2Vn sin((n t n) n 120o)

=

n=1

2Vn [sin(n t n) sin(n t n) cos(n 120o)

+ cos(n t n) sin(n 120o)]

=

n=3,6,9...

2Vn [sin(n t n) sin(n t n) (1/2)

+ cos(n t

n) (

3/2)

=

2

n=3,6,9...Vn

(3/2) sin(n t n) + (

3/2) cos(n t n)

=

3

2

n=3,6,9...Vn

(

3/2) sin(n t n) + (1/2) cos(n t n)

(2.84)

Let

3/2 = rn cos n and 1/2 = rn sin n. This impliles rn = 1 and n = 30o. Using this,equation (2.84) can be written as follows.

vab(t) = 32

n=3,6,9...Vn [sin(n t n 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 signof1/2 or sign of300 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-phase

system with harmonics. Thus the following identity hold true for this system,

VLL

3 VLn (2.86)

Above equation further implies that,

3 VLL I 3 VLn 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


60/195

Ie =

3 I2 + 3

n=3,6,9... In

2

3

= I2 +

n=3,6,9...In

2

(2.88)

I

2.4.3 Apparent Power with Budeanu Resolution: Balanced Distortion Case

The apparent power is given as,

S = 3VlnI =

P2 + Q2B + D2B

= P2 + Q2 + D2 (2.89)where,P = P1 + PH = P1 + P2 + P3 + ....

= 3V1I1 cos 1 + 3

n=1

VnIn cos n

(2.90)

where, n = n n. Similarly,Q = QB = QB1 + QBH

= Q1 + QH (2.91)

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

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

= 3V1I1 sin 1 + 3

n=1

VnIn sin n (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 =

(V2ab + V2

bc + V2

ca)/9

= Vll/

3 (2.94)

54


61/195

Ie =

(I2a + I2b + I

2c )/3

= I (2.95)

Therefore

Se = 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 3 VlnIe (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 current

quantities which are unbalanced and non-sinusoidal. These voltages and currents are expressed as

following.

va(t) =

n=1

2Van sin(n t an)

vb(t) =

n=1

2Vbn sin {n (t 120o) bn} (2.100)

vc(t) =

n=1

2Vcn sin {n (t + 120o) cn}

Similarly, currents can be expressed as,

ia

(t) =

n=1

2Ian

sin(n t

an

)

ib(t) =

n=1

2Ibn sin {n (t 120o) bn} (2.101)

ic(t) =

n=1

2Icn sin {n (t + 120o) cn}

55


62/195

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(nt an)

n=1

2Ian sin(nt an)

(2.102)

+

n=1

2Vbn sin {n(t 120o) bn}

n=1

2Ibn sin {n(t 120o) bn}

+

n=1

2Vcn sin {n(t + 120o) cn}

n=1

2Icn sin {n(t + 120o) cn}

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 cos an {1 cos(2nt 2an)}

n=1

VanIan sin an cos(2nt 2an)

+

n=1

2Van sin(nt an)

m=1, m=n

2Iam sin(mt am)

=

n=1

Pan {1 cos(2nt 2an)}

n=1

Qan cos(2nt 2an)

+

n=1

2Van sin(nt an)

m=1, m=n

2Iam sin(mt am)

(2.103)

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

pb(t) =

n=1

Pbn [1 cos {2n(t 120o) 2bn}]

n=1

Qbn cos {2n(t 120o) 2bn}

+

n=1

2Vbn sin {n(t 120o) bn}

m=1, m=n

2Ibm sin {m(t 120o) bm}

(2.104)

56


63/195

and

pc(t) =

n=1

Pcn [1 cos {2n(

Documents

Power Quality in Power Distribution Systems