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 powerusage. In principle, any number of phases can be used in polyphase electric system, howeverthree-phase system is simpler and giving all advantages of polyphase system. 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− φ)

ib(t) =√

2I sin(ωt− 120 − φ) (2.2)

ic(t) =√

2I sin(ωt+ 120 − φ)

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 voltage or current magnitude (V or I) are same for all three phases andthey have a phase shift of −120o and 120o. The currents are assumed to have φ degree lag withtheir respective phase voltages. The balanced three phase system has certain interesting properties.These will be 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 2(ωt− 120o) − V I sinφ sin 2(ωt− 120o) (2.4)pc(t) = V I cosφ 1− cos 2(ωt+ 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,

28

p3φ(t) = p(t) = 3V I cosφ. (2.6)

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)

29

Here,√

23

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

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√3

(va + vb + vc)1√3

(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)

In the above, the following property of matrices of from (2.12), is used.

[Aabc]T [Aabc] = [Aabc]

−1 [Aabc] = I (2.22)

In above, I is identity matrix.

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.

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)

33

0oaV V

bV

bV3

30o

abV

V

cV

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

The above relationship between phase and line to line voltages is also illustrated in Fig. 2.3. Forthe 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) = − 1√3

[√

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)]

= −V I [cos (90 − φ)− cos (2ωt− 90o − φ)

+ cos (90o − φ)− cos (2ωt− 30o − φ)

+ cos (90o − φ)− cos (2ωt+ 150o − φ)]

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

− cos (2ωt− φ+ 30o − 120o)]

= −V I [3 sinφ− 0]

q(t) = −3V I sinφ (2.25)

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 have

34

90o 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 systemwith 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)

35

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)

Also it is noted that,

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

36

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,√

3 q(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 sinφa + VbIb sinφb + VcIc sinφc (2.38)

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, the

37

above 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 composed of three phasors, equal in magnitude, phase shift

39

of −120o and 120o between phases with phase sequence same to that of the original phasors.

2. Negative sequence components consist of three phasors equal in magnitude, phase shift of120o and −120o between phases 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.

These are denoted as following.

Positive sequence components: V a+, V b+, V c+

Negative sequence components: V a−, V b−, V c−

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

Thus, we can write,

V a = V a+ + V a− + V a0

V b = V b+ + V b− + V b0 (2.44)V c = V c+ + V c− + 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.

Va

Vc

Vb

Vb

Va

Vc

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 a+ be a reference phasor,therefore V b+ and V c+ can be written as,

V b+ = a2V a+ = V a+∠− 120

V c+ = aV a+ = V a+∠120 (2.45)

Similarly V b− and V c− can be expressed in terms of V a− as following.

40

Va

VaVa

0Va

Vc

Vc

0Vc

Vc

Vb

Vb

Vb

0Vb

o

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

V b− = aV a− = V a−∠120

V c− = a2V a− = V a−∠− 120 (2.46)

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 a+ + V a− (2.48)

V b = V b0 + V b+ + V b−

= V a0 + a2 V a+ + a V a− (2.49)

V c = V c0 + V c+ + V c−

= V a0 + a V a+ + a2 V a− (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 a0

V a+

V a−

(2.51)

41

Premultipling by inverse of matrix [Asabc] =

1 1 11 a2 a1 a a2

, the symmetrical components are

expressed as given below. V a0

V a+

V a−

=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 A012 and Asabc are related by the following expression.

[A012] = [Asabc]−1 = [Asabc]

∗ (2.53)

From (2.52), the symmetrical components can therefore be expressed as the following.

V a0 =1

3(V a + V b + V c)

V a+ =1

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

V a− =1

3(V a + a2V b + aV c)

The other component i.e., V b0, V c0, V b+, V c+, V b−, V c− can be found from V a0, V a+, V a+. Itshould be noted that quantity V a0 does not exist if sum of unbalanced phasors is zero. Since sumof line to 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.Ia0

Ia+

Ia−

=1

3

1 1 11 a a2

1 a2 a

IaIbIc

(2.55)

Thus,

Ia0 =1

3(Ia + Ib + Ic)

Ia+ =1

3(Ia + aIb + a2Ic)

42

Ia− =1

3(Ia + a2Ib + aIc)

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

In = 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 a+ =1

3(V a + aV b + a2V c)

= 104.16∠4.7o V

V a− =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 ] = 381.25 k

E0+− = 3 k [V 2a0 + V 2

a+ + V 2a−] = 381.25 k

Thus, Eabc = E0+−with k as 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

∗

=

[Asabc]

V a0

V a+

V a−

T [Asabc]

Ia0

Ia+

Ia−

∗

=

V a0

V a+

V a−

T [Asabc]T [Asabc]

∗

Ia0

Ia+

Ia−

∗ (2.57)

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

[Asabc]T [Asabc]

∗ = 3 [I] (2.58)

44

The matrix, [I], is identity matrix. Using (2.58), (2.57) can be written as the following.

Sv = P + jQ =

V a0

V a+

V a−

T3[I]

Ia0

Ia+

Ia−

∗

= 3

V a0

V a+

V a−

TIa0

Ia+

Ia−

∗Sv = P + jQ = V aI

∗a + V bI

∗b + V cI

∗c

= 3 [V a0I∗a0 + V a+I

∗a+ + V a−I

∗a−] (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 + Va+Ia+ cosφa+ + Va−Ia− cosφa−)

+j(Va0Ia0 sinφa0 + Va+Ia+ sinφa+ + Va−Ia− sinφa−) ] (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+ = Va+Ia+ cosφa+ + Vb+Ib+ cosφb+ + Vc+Ic+ cosφc+

= 3Va+Ia+ cosφa+. (2.61)

Negative sequence power is expressed as,

P− = 3Va−Ia− cosφa−. (2.62)

The zero sequence power is

P 0 = 3Va0Ia0 cosφa0. (2.63)

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

Q+ = 3Va+Ia+ sinφa+

Q− = 3Va−Ia− sinφa−

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 = 3Va+Ia+

S− = |S+| =√P−2 +Q−2 = 3Va−Ia−

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 2ab + V 2

bc + V 2ca

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.

In above the terms Da, Db and Dc are known as distortion powers in phase-a, b, c respectively. Thedefinition of The distortion power, D, is given in Section 1.4.5. The above equation suggests thatsuch a system has potential to produce significant additional power loss in neutral wire and groundpath.

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 + V 2

e1I2eH + 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.

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