NPTEL Courseon
Power Quality in Power Distribution Systems
Dr. Mahesh KumarProfessor, Department of Electrical Engineering
Indian Institute of Technology MadrasChennai - 600 036
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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+
2π
3
))]+V I cosφ. sin 2
(ωt+
2π
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
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
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
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
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
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
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
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
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
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
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
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
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
'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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
[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
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
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
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
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
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
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
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
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
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
(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
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
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
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
Assuming compensator characteristics to be linear with Qγ ≤ Qγmax limit, the voltage can berepresented as,
V = Vk +Qγ
Kγ
(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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
[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
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
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
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
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
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
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
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
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
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
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
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
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
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
vβ
=
√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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
'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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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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
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
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)
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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
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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.
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=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
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
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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
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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.
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( )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)
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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
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
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)
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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
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
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
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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.
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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?
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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
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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
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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.
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