Chapter 2
THREE PHASE CIRCUITS: POWER
DEFINITIONS AND VARIOUS
COMPONENTS
(Lectures 9-18)
2.1
Three-phase Sinusoidal Balanced System
Usage of three-phase voltage supply is very common for generation, transmission and distribution
of bulk electrical power. Almost all industrial loads are supplied by three-phase power supply for
its advantages over single phase systems such as cost and efficiency for same amount of power
usage. In principle, any number of phases can be used in polyphase electric system, however
three-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 term
V I cos(2ωt − φ). The pulsating term does not contribute to any real power and thus increases the
VA rating of the system.
In the following section, we shall study the various three-phase circuits such as balanced, unbalanced, 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◦ )
√
vc (t) =
2V sin(ωt + 120◦ )
27
(2.1)
a
a
b
b
c
c
Fig. 2.1 A three-phase balanced circuit
and
√
2I sin(ωt − φ)
√
ib (t) =
2I sin(ωt − 120◦ − φ)
√
ic (t) =
2I sin(ωt + 120◦ − φ)
ia (t) =
(2.2)
In (2.1) and (2.2) subscripts a, b and c are used to denote three phases which are balanced. Balanced
three-phase means that the voltage or current magnitude (V or I) are same for all three phases and
they have a phase shift of −120o and 120o . The currents are assumed to have φ degree lag with
their 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 previously. These are given below.
pa (t) = V I cos φ {1 − cos 2ωt} − V I sin φ sin 2ωt
pb (t) = V I cos φ {1 − cos 2(ωt − 120o )} − V I sin φ sin 2(ωt − 120o )
pc (t) = V I cos φ {1 − cos 2(ωt + 120o )} − V I sin φ sin 2(ωt + 120o )
(2.4)
Adding three phase instantaneous powers given in (2.4), we get the three-phase instantaneous
power as below.
p(t) = 3 V I cos φ − V I cos φ{cos 2ωt + cos 2(ωt − 120o ) + cos 2(ωt + 120o )}
− V I sin φ{sin 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 cos φ.
(2.6)
This is quite interesting result. It indicates for balanced three-phase system, the total instantaeous
power is equal to the real power or average active power (P ), which is constant. This is the reason
we use 3-phase system. It does not involve the pulsating or oscillating components of power as in
case of single phase systems. Thus it ensures less VA rating for same amount of power transfer.
Here, total three-phase reactive power can be defined as sum of maximum value of preactive (t)
terms in (2.4). Thus,
Q = Qa + Qb + Qc = 3V I sin φ.
(2.7)
Is there any attempt to define instantaneous reactive power q(t) similar to p(t) such that Q is
average value of that term q(t)?. H. Akagi et al. published paper [6], in which authors defined term
instantaneous reactive power. The definition was facilitated through αβ0 transformation. Briefly
it 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 transformation. This transformation is described below.
The abc coordinates and their equivalent αβ0 coordinates are shown in the Fig. 2.2 below.

vc
-j
c
- c /2
- b /2
vb
60
o
va 
O
j
b
j
Fig. 2.2 A abc to αβ0 transformation
Resolving a, b, c quantities along the αβ axis we have,
r
2
vb vc
vα =
(va − − )
3
2
2
r √
2 3
vβ =
(vb − vc )
3 2
29
(2.8)
(2.9)
q
Here, 23 is a scaling factor, which ensures power invariant transformation. Along with that, we
define zero sequence voltage as,
r r
2 1
v0 =
(va + vb + vc )
(2.10)
3 2
Based on Eqns.(4.60)-(2.10) we can write the above equations as follows.

r  √1
v0 (t)
2 2
 vα (t)  =
1
3
vβ (t)
0

√1
2
−1
√2
3
2
√1
2
−1
2
√
− 3
2


va (t)
  vb (t) 
vc (t)
(2.11)




v0
va
 vα  = [Aoαβ ]  vb 
vβ
vc
The above is known as Clarke-Concordia transformation. Thus, va , vb and vc can also be expressed
in terms of v0 , vα and vβ by pre-multiplying (2.11) by matrix [A0αβ ]−1 , we have



v0
va
 vb  = [A0αβ ]−1  vα 
vc
vβ

It will be interesting to learn that
[A0αβ ]−1
[A0αβ ]−1
−1
r  1
√
√1
√1
2
2
2
2
−1
−1 
1 √
= [Aabc ] = 
2
2
√
3
3
− 3
0
2

 1
 2
√
1
0
r
2
 2  √1 −1 √3 
T
= 
 2 2
 = [A0αβ ] = [Aabc ]
2
√
3
− 3
1
−1
√
2
2
(2.12)
2
Similarly, we can write down instantaneous symmetrical transformation for currents, which is
given below.

 r  1


√
√1
√1
i0
i
a
2
2
2
 iα  = 2  1 −1 −1   ib 
(2.13)
2
2
√
√
3
3
−
3
iβ
ic
0
2
2
Now based on ’0αβ’ transformation, the instantaneous active and reactive powers are defined as
follows. The three-phase instantaneous power p(t) is expressed as the dot product of 0αβ components of voltage and currents such as given below.
30
p(t) = vα iα + vβ iβ + v0 i0
"
√
√
2 vb vc ib ic
3
3
va − −
ia − −
+
(vb − vc )
(ib − ic )
=
3
2
2
2
2
2
2
1
1
+ √ (va + vb + vc ) √ (ia + ib + ic )
3
3
= va ia + vb ib + vc ic
(2.14)
Now what about instantaneous reactive power? Is there any concept defining instantaneous reactive
power? In 1983-84,authors H.akagi have attempted to define instantaneous reactive power using
stationary αβ0 frame, as illustrated below. In [6], the instantaneous reactive power q(t) is defines
as the cross product of two mutual perpendicular quantities, such as given below.
q(t) = vα × iβ + vβ × iα
q(t) = vα iβ − vβ iα
"
√
√
#
2 vb vc 3
3
ib ic
=
va − −
(ib − ic ) −
(vb − vc ) ia − −
3
2
2
2
2
2
2
√ h
2 3
vb vc vb vc vb vc vb vc i
=
(−vb + vc ) ia + va − − + −
ib + −va + + + −
ic
3 2
2
2
2
2
2
2
2
2
1
= − √ [(vb − vc ) ia + (vc − va ) ib + (va − vb ) ic ]
3
√
(2.15)
= − [vbc ia + vca ib + vab ic ] / 3
This is also equal to the following.
ib ic
ib ic
ib ic
1
ib ic
q(t) = √ (ib − ic ) va + − + − ia + +
vb + − + + ia − −
vc
2
2
2
2
2
2
2
2
3
1
= √ [(ib − ic ) va + (ic − ia ) vb + (ia − ib ) vc ]
(2.16)
3
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 two
transformations. The energy associated with the abc0 system is given by (va2 + vb2 + vc2 ) and the
energy associated with the αβ0 components is given by v02 + vα2 + vβ2 . 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.
"r
vα2 =
vα2
vb vc 2
va − −
3
2
2
#2
vb2 vc2 2va vb 2vb vc 2va vc
2
2
=
v +
+
−
+
−
3 a
4
4
2
4
2
2
2
v
2va vb vb vc 2va vc
2 2 vb
va +
+ c −
+
−
=
3
6
6
3
3
3
(2.17)
Similary we can find out square of vβ term as given below.
vβ2
"√ r
#2
3 2
=
(vb − vc )
2
3
1 2
=
vb + vc2 − 2vb vc
2
v2 v2
= b + c − vb vc
2
2
(2.18)
Adding (2.17) and (2.18), we find that,
2 2
va + vb2 + vc2 − vc vb − vb vc − vc va
3
2
va vb2 vc2 2va vb 2vb vc 2va vc
2
2
2
= va + vb + vc −
+
+
+
+
+
3
3
3
3
3
3
1
= va2 + vb2 + vc2 − (va + vb + vc )2
3
2
1
2
2
2
= va + vb + vc − √ (va + vb + vc )
3
vα2 + vβ2 =
Since v0 =
√1 (va
3
(2.19)
+ vb + vc ), the above equation, (2.19) can be written as,
vα2 + vβ2 + v02 = va2 + vb2 + vc2 .
(2.20)
From the above it is implies that the energy associated with the two systems remain same instant to
instant basis. In general the instantaneous power p(t) remain same in both transformations. This
is proved below.
32
Using (2.14), following can be written.
p(t) = vα iα + vβ iβ + vo io

T 

v0
i0
p(t) =  vα   iα 
vβ
iβ


T 


va
ia
= [Aabc ]  vb  [Aabc ]  ib 
vc
ic

T


va
ia
=  vb  [Aabc ]T [Aabc ]  ib 
vc
ic

T


va
ia
=  vb  [Aabc ]−1 [Aabc ]  ib 
vc
ic


ia
va vb vc  ib 
=
ic
= va ia + vb ib + vc ic
(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 transformation. In this section we shall study these powers for various three-phase circuits such as
three-phase balanced, three-phase unbalanced, balanced three-phase with harmonics and unbalanced 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). For
these 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
Vc
o
Vb
V
ab

3
30
V
Va  V0o
Vb
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. For
the above three-phase system, the instantaneous power p(t) can be expressed using (2.21) and it is
equal to,
p(t) = va ia + vb ib + vc ic
= vα iα + vβ iβ + v0 i0
= 3 V I cosφ
(2.24)
The instantaneous reactive power q(t) is as following.
√
1 √ √
q(t) = − √ [ 3 2V sin (ωt − 90o ) 2I sin (ωt − φ)
3
√ √
√
+ 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 sin φ − cos (2ωt − φ + 30o ) − cos (2ωt − φ + 30o + 120o )
− cos (2ωt − φ + 30o − 120o )]
= −V I [3 sin φ − 0]
q(t) = −3V I sin φ
(2.25)
The above value of instantaneous reactive power is same as defined by Budeanu’s [1] and is given
in equation (2.7). Thus, instantaneous reactive power given in (2.15) matches with the conventional definition of reactive power defined in (2.7). However the time varying part of second terms
of each phase in (2.4) has no relevance with the definition given in (2.15).
Another interpretation of line to line voltages in (2.15) is that the voltages vab , vbc and vca 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)
√
o
vca = 3vb ∠ − 90
In above equation, vc ∠ − 90o implies that vc ∠ − 90o lags vc by 90o . Analyzing each term in
(2.15) contributes to,
√
vbc ia = 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 [sin φ − cos {90◦ + (2ωt − φ)}]
√
= 3V I [sin φ + sin (2ωt − φ)]
√
= 3V I [sin φ + sin 2ωt cos φ − cos 2ωt sin φ]
√
vbc ia / 3 = V I [sin φ (1 − cos 2ωt) + cos φ sin 2ωt]
Similarly,
√
2π
vca ib / 3 = V I sin φ 1 − cos 2 ωt −
3
2π
+V I cos φ. sin 2 ωt −
3
√
2π
vab ic / 3 = V I sin φ 1 − cos 2 ωt +
3
2π
+V I cos φ. sin 2 ωt +
(2.27)
3
Thus, we see that the role of the coefficients of sin φ and cos φ have reversed. Now if we take
average value of (2.27), it is not equal to zero but V I sin φ in each phase. Thus three-phase reactive
power will be 3V I sin φ. The maximum value of second term in (2.27) represents active average
power i.e., V I cos φ. 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 the
average value of terms in (2.27).
2.2.2
Three-Phase Unbalance System
Three-phase unbalance system is not uncommon in power system. Three-phase unbalance may
result from single-phasing, faults, different loads in three phases. To study three-phase system
with fundamental unbalance, the voltages and currents are expressed as following.
√
va = 2Va sin (ωt − φva )
√
vb = 2Vb sin (ωt − 120o − φvb )
√
vc = 2Vc sin (ωt + 120o − φvc )
35
(2.28)
and,
√
ia = 2Ia sin (ωt − φia )
√
ib = 2Ib sin (ωt − 120o − φib )
√
ic = 2Ic sin (ωt + 120o − φic )
(2.29)
For the above system, the three-phase instantaneous power is given by,
p3φ (t) = p(t) = va ia + vb ib + vc ic
√
= 2Va sin (ωt − φva ) sin (ωt − φia )
√
√
+ 2Vb sin (ωt − 120o − φvb ) 2Ib sin (ωt − 120o − φib )
√
√
+ 2Vc sin (ωt + 120o − φvc ) 2Ic sin (ωt + 120o − φic )
(2.30)
Simplifying above expression we get,
p3φ (t) = Va Ia cos φa {1 − cos (2ωt − 2φva )}
{z
}
|
pa,active
−V I sin φa sin (2ωt − 2φva )
| a a
{z
}
pa,reactive
+Vb Ib cos φb [1 − cos {2 (ωt − 120◦ ) − 2φvb }]
−Vb Ib sin φb sin {2 (ωt − 120◦ ) − 2φvb }
+Vc Ic cos φc [1 − cos {2 (ωt + 120◦ ) − 2φvc }]
−Vc Ic sin φc sin {2 (ωt + 120◦ ) − 2φvc }
where φa = (φia − φva )
(2.31)
Therefore,
p3φ (t) = pa,active + pb,active + pc,active + pa,reactive + pb,reactive + pc,reactive
= pa + pb + pc + pea + peb + pec
(2.32)
where,
pa = Pa = Va Ia cos φa
pb = Pb = Vb Ib cos φb
pc = Pc = Vc Ic cos φc
(2.33)
pea = −Va Ia cos (2 ωt − φa − 2 φva )
peb = −Vb Ib cos (2 ωt − 240o − φb − 2 φvb )
pec = −Vc Ic cos (2ωt + 240 − φc − 2 φvc )
(2.34)
and
Also it is noted that,
pa + pb + p c
= va ia + vb ib + vc ic = P
36
(2.35)
and,
pea + peb + pec
= −Va Ia cos(2ωt − φva − φib )
−Vb Ib cos {2(ωt − 120) − φvb − φib }
−Vc Ic cos {2(ωt + 120) − φvc − φic }
6= 0
This implies that, we no longer get advantage of getting constant power, 3V I cos φ from interaction
of three-phase voltages and currents. Now, let us analyze three phase instantaneous reactive power
q(t) as per definition given in (2.15).
1
q(t) = − √ (vb − vc )ia + (vc − va )ib + (va − vb )ic
3
2 h
= − √ {Vb sin(ωt − 120o − φvb ) − Vc sin(ωt + 120o − φvc )} Ia sin(ωt − φia )
3
√
+ {Vc sin(ωt + 120o − φvc ) − Va sin(ωt − φva )} 2Ib sin(ωt − 120o − φib )
(2.36)
i
√
o
o
o
+{Va sin(ωt − 120 − φva ) − Vb sin(ωt − 120 − φvb )} 2Ic sin(ωt + 120 − φic )
From the above,
h
√
3 q(t) = − Vb Ia {cos(φia − 120o − φvb ) − cos(2ωt − 120o − φia − φvb )}
−Vc Ia {cos(φia + 120o − φvc ) − cos(2ωt + 120o − φia − φvc )}
+Vc Ib {cos(φib + 240o − φvc ) − cos(2ωt − φib − φvc )}
−Va Ib {cos(φib − 120o − φva ) − cos(2ωt − 120o − φva − φib )}
+Va Ic {cos(φic − 120o − φva ) − cos(2ωt + 120o − φva − φic )}
i
o
−Vb Ic {cos(φic − 240 − φvb ) − cos(2ωt − φic − φvb )}
(2.37)
Now looking this expression,we can say that
1
T
Z
0
T
1 h
q(t)dt = − √ Vb Ia cos(φia − φvb − 120o )
3
−Vc Ia cos(φia − φvc + 120o )
+Vc Ib cos(φib + 240o − φvc )
−Va Ib cos(φib − 120o − φva )
+Va Ic cos(φic − 120o − φva )
i
o
−Vb Ic cos(φic − 240 − φvb )
= q a (t) + q b (t) + q c (t)
6
=
Va Ia sin φa + Vb Ib sin φb + Vc Ic sin φc
(2.38)
Hence the definition of instantaneous reactive power does not match to that defined by Budeanue’s
reactive 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) =
vb (t) =
vc (t) =
√
√
√
2V sin(ωt)
2V sin(ωt − 120◦ )
(2.39)
◦
2V sin(ωt + 120 )
and
ia (t) =
ib (t) =
ic (t) =
√
√
√
2Ia sin(ωt − φa )
2Ib sin(ωt − 120o − φb )
(2.40)
o
2Ic sin(ωt + 120 − φc )
And the instantaneous reactive power is given by,
q(t)
= − √13 [vbc ia + vca ib + vab ic ]
√
√
√
= − √13 [ 3 va ∠−π/2 ia + 3 vb ∠ − π/2 ib + 3 vc ∠ − π/2 ic ]
√
√
= −[√ 2V sin(ωt − π/2) 2Ia sin(ωt
− φia )
√
o
+√2V sin(ωt − 120 − π/2)√ 2Ib sin(ωt − 120o − φib )
+ 2V sin(ωt + 120o + π/2) 2Ic sin(ωt + 120o − φic )]
= −[V Ia cos(π/2 − φia ) − cos {π/2 − (2ωt − φia )}
+V Ib cos(π/2 − φib ) − cos(2ωt − 240o − π/2 − φib )
+V Ic cos(π/2 − φic ) − cos(2ωt + 240o − π/2 − φic )]
= −[(V Ia sin φia + V Ib sin φib + V Ic sin φic )
+V Ia sin(2ωt − φia ) + V Ib sin(2ωt − 240o − φib ) + V Ic sin(2ωt + 240o − φic )]
Thus,
1
Q=
T
Z
T
q(t)dt = −(V Ia sin φia + V Ib sin φib + V Ic sin φic )
(2.41)
0
Which is similar to Budeanu’s reactive power.
The oscillating term of q(t) which is equal to qe(t) is given below.
qe(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 been
considered. Ideal power systems are not designed for unbalance quantities as it makes power system 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 can
be resolved into n system of balanced phasors, called the symmetrical components of the original
phasors. The n phasors of each set of components are equal in length and the angles. Although,
the method is applicable to any unbalanced polyphase system, we shall discuss about three phase
systems.
For the discussion of symmetrical components, a complex operator denoted as a is defined as,
a
a2
a3
= 1∠120o = ej2π/3 = cos 2π/3 + j sin 2π/3
√
= −1/2 + j 3/2
= 1∠240o = 1∠ − 120o = ej4π/3 = e−j2π/3 = cos 4π/3 + j sin 4π/3
√
= −1/2 − j 3/2
= 1∠360o = ej2π = 1
Also note an interesting property relating a, a2 and a3 ,
a + a2 + a3 = 0.
a  1120
(2.43)
o
o
a  1o
3
o
a 2  1  120o
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 unbalanced 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 of
120o 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 shift
from 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
V c = V c+ + V c− + V c0
(2.44)
Graphically, these are represented in Fig. 2.5. Thus if we add the sequence components of each
phase vectorially, we shall get V a , V b and V s as per (2.44). This is illustrated in Fig. 2.6.
V c
V b
Va0
V a
V b
(a)
Va
V c
(b)
Vb 0
Vc0
(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+ = a2 V a+ = V a+ ∠ − 120◦
V c+ = aV a+ = V a+ ∠120◦
Similarly V b− and V c− can be expressed in terms of V a− as following.
40
(2.45)
V c
Vc0
V c
Vc
V a
V a
o
Va0
Va
Vb
Vb
Vb
Vb 0
Fig. 2.6 Unbalanced phasors as vector sum of positive, negative and zero sequence phasors
V b− = aV a− = V a− ∠120◦
V c− = a2 V a− = V a− ∠ − 120◦
(2.46)
The zero sequence components have same magnitude and phase angle and therefore these are
expressed 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.
  


Va
V a0
1 1 1
 V b  = 1 a2 a  V a+ 
1 a a2
Vc
V a−
41
(2.51)


1 1 1
Premultipling by inverse of matrix [Asabc ] = 1 a2 a , the symmetrical components are
1 a a2
expressed as given below.



 
V a0
Va
1 1 1
V a+  = 1 1 a a2   V b 
(2.52)
3 1 a2 a
V a−
Vc
 
Va
= [A012 ]  V b 
Vc
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.
1
V a0 = (V a + V b + V c )
3
1
V a+ = (V a + aV b + a2 V c )
3
1
V a− = (V a + a2 V b + aV c )
3
(2.54)
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+ . It
should be noted that quantity V a0 does not exist if sum of unbalanced phasors is zero. Since sum
of 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 always
zero, hence zero sequence voltage components are never present in the line voltage, regardless of
amount of unbalance. The sum of the three phase voltages, i.e., V a + V b + V c is not necessarily
zero and hence zero sequence voltage exists.
Similarly sequence components can be written for currents. Denoting three phase currents by
I a , I b , and I c respectively, the sequence components in matrix form are given below.
 

 
I a0
Ia
1 1 1
I a+  = 1 1 a a2   I b 
(2.55)
3 1 a2 a
I a−
Ic
Thus,
1
I a0 = (I a + I b + I c )
3
1
I a+ = (I a + aI b + a2 I c )
3
42
1
I a− = (I a + a2 I b + aI c )
3
In three-phase, 4-wire system, the sum of line currents is equal to the neutral current (I n ). thus,
In = Ia + Ib + Ic
= 3I a0
(2.56)
This current flows in the fourth wire called neutral wire. Again if neutral wire is absent, then zero
sequence current is always equal to zero irrespective of unbalance in phase currents. This is illustrated below.
a
a
b
b
c
a
a
b
b
c
c
c
(a)
(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 (VN n ) between the system and
load neutral is always equal to zero. In 2.7(a), there is no neutral current due to the absence of the
neutral wire. But in this configuration the neutral voltage, VN n , may or may not be equal to zero
depending 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 to
zero, indicating that there is no unbalance in voltages. However the converse may not apply.
Now consider the following phase voltages. Compute the sequence components and show that the
energy associated with the voltage components in both system remain constant.
V a = 100∠0o
V b = 150∠ − 100o
V c = 75∠100o
43
Solution Using (2.54), sequence components are computed. These are:
1
V a0 = (V a + V b + V c )
3
= 31.91∠ − 50.48o V
1
V a+ = (V a + aV b + a2 V c )
3
= 104.16∠4.7o V
1
V a− = (V a + a2 V b + aV c )
3
= 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 [Va2 + Vb2 + Vc2 ] = 381.25 k
2
2
2
E0+− = 3 k [Va0
+ Va+
+ Va−
] = 381.25 k
Thus, Eabc = E0+− with k as some constant of proportionality.
The invariance of power can be further shown by following proof.
∗
Ia
S v = P + jQ = [ V a V b V c ] I b 
Ic
 T  ∗
Va
Ia



= Vb
Ib
Vc
Ic

T 
∗


V a0
I a0
= [Asabc ]  V a+  [Asabc ]  I a+ 
V a−
I a−

T

∗
V a0
I a0
=  V a+  [Asabc ]T [Asabc ]∗  I a+ 
V a−
I a−

(2.57)
The term S v is referred as vector or geometric apparent power. The difference between will be
given in the following. The transformation matrix [Asabc ] has following properties.
[Asabc ]T [Asabc ]∗ = 3 [I]
44
(2.58)
The matrix, [I], is identity matrix. Using (2.58), (2.57) can be written as the following.

T
 ∗
V a0
I a0
S v = P + jQ = V a+  3[I]I a+ 
V a−
I a−

T  ∗
V a0
I a0



= 3 V a+
I a+ 
V a−
I a−
∗
∗
∗
S v = P + jQ = V a I a + V b I b + V c I c
∗
∗
∗
= 3 [V a0 I 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 instantaneous symmetrical components will be discussed in the latter section. The equation (2.59) further
implies that,
S v = P + jQ = 3 [ (Va0 Ia0 cos φa0 + Va+ Ia+ cos φa+ + Va− Ia− cos φa− )
+j(Va0 Ia0 sin φa0 + Va+ Ia+ sin φa+ + Va− Ia− sin φa− ) ]
(2.60)
The power terms in (2.60) accordingly form positive sequence, negative sequence and zero sequence powers denoted as following. The positive sequence power is given as,
P + = Va+ Ia+ cos φa+ + Vb+ Ib+ cos φb+ + Vc+ Ic+ cos φc+
= 3Va+ Ia+ cos φa+ .
(2.61)
Negative sequence power is expressed as,
P − = 3Va− Ia− cos φa− .
(2.62)
P 0 = 3Va0 Ia0 cos φa0 .
(2.63)
The zero sequence power is
Similarly, sequence reactive power are denoted by the following expressions.
Q+ = 3Va+ Ia+ sin φa+
Q− = 3Va− Ia− sin φa−
Q0 = 3Va0 Ia0 sin φa0
(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
45
(2.65)
Here, positive sequence, negative sequence and zero sequence apparent powers are denoted as the
following.
p
+
S + = |S | = P +2 + Q+2 = 3Va+ Ia+
p
+
S − = |S | = P −2 + Q−2 = 3Va− Ia−
p
+
S 0 = |S | = P 02 + Q02 = 3Va0 Ia0
(2.66)
The scalar value of vector apparent power (S v ) is given as following.
0
+
−
Sv = |S a + S b + S c | = |S + S + S |
= |(Pa + Pb + Pc ) + j(Qa + Qb + Qc )|
p
= P 2 + Q2
(2.67)
Similarly, arithematic apparent power (S A ) is defined as the algebraic sum of each phase or sequence apparent power, i.e.,
SA = |S a | + |S b | + |S c |
= |Pa + jQa | + |Pb + jQb | + |Pc + jQc |
q
p
p
= Pa2 + Q2a + Pb2 + Q2b + Pc2 + Q2c
(2.68)
In terms of sequence components apparent power,
0
+
−
SA = |S | + |S | + |S |
= |P 0 + jQ0 | + |P + + jQ+ | + |P − + jQ− |
q
q
q
2
2
2
2
0
0
+
+
= P + Q + P + Q + P − 2 + Q− 2
(2.69)
Based on these two definitions of the apparent powers, the power factors are defined as the following.
P
Sv
P
Arithematic apparent power = pfA =
SA
Vector apparent power = pfv =
46
(2.70)
(2.71)
Example 2.2 Consider a 3-phase 4 wire system supplying resistive load, shown in Fig. 2.8
below. Determine power consumed by the load and feeder losses.
a'
'
c'
n
'
r
jx
a
r
jx
Va
r
jx
Vb c
r
jx
Vc
Ia
b
n
R
Ib
Ic
In
Fig. 2.8 A three-phase unbalanced load
√
( 3V )2
3V 2
Power dissipated by the load =
=
R
√ R
3V
Va−Vb
The current flowing in the line =
=|
|
R
R
and I b = −I a
√ !2
√ !2
3V
3V
×r+
×r
Therefore losses in the feeder =
R
R
r 3 V 2 =2
R
R
Now, consider another example of a 3 phase system supplying 3-phase load, consisting of three
resistors (R) in star as shown in the Fig. 2.9. Let us find out above parameters.
2
V
3V 2
Power supplied to load = 3
×R=
R
R
2
V
r 3V 2
Losses in the feeder = 3
×r =
R
R
R
Thus, it is interesting to see that power dissipated in the unbalanced system is twice the power loss
in balanced circuit. This leads to conclusion that power factor in phases would become less than
unity, while for balanced circuit, the power factor is unity. Power analysis of unbalanced circuit
shown in Fig. 2.8 is given below.
47
a'
b'
c'
n'
r
jx
r
jx
R
a
Ia
Va b
R
r
jx
Vb c
r
jx
Vc
n
Ib
R
Ic
In
Fig. 2.9 A three-phase balanced load
√
Va−Vb
V ab
3 Va
The current in phase-a, I a =
=
=
∠30◦
R √
R
R
3V
The current in phase-b, I b = −I a =
∠(30 − 180)o
R
√
3V
=
∠ − 150o
R
The current in phase-c and neutral are zero, I c = I n = 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 = Va Ia cos φa = V I cos 30 =
Qa = Va Ia sin φa = V I sin 30◦ =
3
VI
2
1
VI
2
Sa = Va Ia = V I
√
3
VI
2
1
Vb Ib sin φb = V I sin(−30)◦ = − V I
2
Vb Ib = V I
Qc = Sc = 0
√
√
3
Pa + Pb + Pc = 2 ×
V I = 3V I
2
√
√
3V
3V
R
3V 2
R
Qa + Qb + Qc = 0
◦
Pb = Vb Ib cos φb = V I cos(−30) =
Qb =
Sb =
Pc =
Thus total active power P =
=
P =
Total reactive power Q =
48
p
The vector apparent power, Sv = P 2 + Q2 = 3 V 2 /R = P
√
The arithmetic apparent power, SA = Sa + Sb + Sc = 2 V I = (2/ 3) P
From the values of Sv and SA , it implies that,
P
P
=
=1
Sv
P
√
P
P
3
√
=
=
= 0.866
=
SA
2
(2/ 3) P
pfv =
pfA
This difference between the arthmetic and vector power factors will be more due to the unbalances
in the load.
For balance load SA = SV , therefore, pfA = pfV = 1.0. Thus for three-phase electrical circuits, the following holds true.
pfA ≤ pfV
2.3.1
(2.72)
Effective Apparent Power
For unbalanced three-phase circuits, their is one more definition of apparent power, which is known
as effective apparent power. The concept assumes that a virtual balanced circuit that has the same
power output and losses as the actual unbalanced circuit. This equivalence leads to the definition
of effective line current Ie and effective line to neutral voltage Ve .
The equivalent three-phase unbalanced and balanced circuits with same power output and losses
are shown in Fig. 2.10. From these figures, to maintain same losses,
a'
b
'
c'
n
'
r
jx
Va
r
jx
Vb
r
r
jx
jx
Ra
Ia
Ib
Vc
Vn
(a)
Rb
a'
b'
n
Rc
c'
Ic
n'
r
jx
r
jx
r
jx
r
jx
In
Re
Iea
Vea
Veb
Vec
(b)
Ieb
Re
n
Re
Iec
In  0
Fig. 2.10 (a) Three-phase with unbalanced voltage and currents (b) Effective equivalent three-phase system
rIa2 + rIb2 + rIc2 + rIn2 = 3rIe2
The above equation implies the effective rms current in each phase is given as following.
r
(Ia2 + Ib2 + Ic2 + In2 )
Ie =
3
49
(2.73)
For the original circuit shown in Fig. 2.8, the effective current Ie is computed using above equation
and is given below.
r
(Ia2 + Ib2 )
since, Ic = 0and In = 0
Ie =
3 s
r
√
2
2 Ia2
2 ( 3V /R)
=
=
3
3
√
2V
=
R
To account same power output in circuits shown above, the following identity is used with Re = R
in Fig. 2.10.
Va2 Vb2 Vc2 Vab2 + Vbc2 + Vca2
3Ve2 9Ve2
+
+
+
=
+
R
R
R
3R
R
3R
From (2.74), the effective rms value of voltage is expressed as,
r
1
Ve =
{3 (Va2 + Vb2 + Vc2 ) + Vab2 + Vbc2 + Vca2 }
18
Assuming, 3 (Va2 + Vb2 + Vc2 ) ≈ Vab2 + Vbc2 + Vca2 , equation (2.75) can be written as,
r
Va2 + Vb2 + Vc2
Ve =
=V
3
(2.74)
(2.75)
(2.76)
Therefore, the effective apparent power (Se ), using the values of Ve and Ie , is given by,
√
3 2V 2
Se = 3 Ve Ie =
R
Thus the effective power factor based on the definition of effective apparent power (Se ), for the
circuit shown in Fig. 2.8 is given by,
pfe =
3 V 2 /R
1
P
= √
= √ = 0.707
2
S e 3 2 V /R
2
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.
q
Su = Se2 − S + 2
(2.77)
2
2
2
where S + = 3 V + I + and S + = P + + Q+ .
2.4
Three-phase Non-sinusoidal Balanced System
A three-phase nonsinusoidal system is represented by following set of equaitons.
va (t) =
√
∞
√ X
2V1 sin(wt − α1 ) + 2
Vn sin(nwt − αn )
n=2
vb (t) =
√
2V1 sin(wt − 120◦ − α1 ) +
∞
√ X
2
Vn sin(n(wt − 120◦ ) − αn )
(2.78)
n=2
vc (t) =
√
∞
√ X
◦
Vn sin(n(wt + 120◦ ) − αn )
2V1 sin(wt + 120 − α1 ) + 2
n=2
Similarly, the line currents can be expressed as,
ia (t) =
√
∞
√ X
In sin(nwt − βn )
2I1 sin(wt − β1 ) + 2
n=2
ib (t) =
√
2I1 sin(wt − 120◦ − β1 ) +
∞
√ X
In sin(n(wt − 120◦ ) − βn )
2
(2.79)
n=2
ic (t) =
√
∞
√ X
◦
2I1 sin(wt + 120 − β1 ) + 2
In sin(n(wt + 120◦ ) − βn )
n=2
In this case,
Sa
Pa
Qa
Da
= Sb = Sc ,
= P b = Pc ,
= Qb = Qc ,
= Db = Dc .
(2.80)
In above the terms Da , Db and Dc are known as distortion powers in phase-a, b, c respectively. The
definition of The distortion power, D, is given in Section 1.4.5. The above equation suggests that
such a system has potential to produce significant additional power 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 following
equation.
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 their
balanced nature. Therefore the neutral current is given by,
in = ia + ib + ic =
∞
X
√
3 2In sin(nwt − βn ).
(2.82)
n=3,6,..
The RMS value of the current in neutral wire is therefore given by,
" ∞
#1/2
X
In = 3
In2
.
(2.83)
n=3,6,..
Due to dominant triplen harmonics in electrical loads such as UPS, rectifiers and other power
electronic based loads, the current rating of the neutral wire may be comparable to the phase wires.
It is worth to mention here that all harmonics in three-phase balanced systems can be categorized in three groups i.e., (3n + 1), (3n + 2) and 3n (for n = 1, 2, 3, ...) called positive, negative and zero sequence harmonics respectively. This means that balanced fundamental, 4th, 7th
10th,... form positive sequence only. Balanced 2nd, 5th, 8th, 11th,... form negative sequence only
and the balanced triplen harmonics i.e. 3rd, 6th, 9th,... form zero sequence only. But in case of
unbalanced three-phase systems with harmonics, (3n + 1) harmonics may start forming negative
and zero sequence components. Similarly, (3n + 2) may start forming positive and zero sequence
components 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)
∞
∞
X
X
√
√
=
2Vn sin(n ωt − αn ) −
2Vn sin(n (ωt − 120o ) − αn )
=
=
n=1
n=1
∞
X
√
∞
X
√
n=1
∞
X
2Vn sin(n ωt − αn ) −
2Vn sin((n ωt − αn ) − n × 120o )
n=1
√
2Vn [sin(n ωt − αn ) − sin(n ωt − αn ) cos(n × 120o )
n=1
+ cos(n ωt − αn ) sin(n × 120o )]
=
∞
X
√
2Vn [sin(n ωt − αn ) − sin(n ωt − αn ) (−1/2)
n6=3,6,9...
=
√
2
∞
X
i
√
+ cos(n ωt − αn ) (± 3/2)
h
i
√
Vn (3/2) sin(n ωt − αn ) + (± 3/2) cos(n ωt − αn )
n6=3,6,9...
∞
X
√ √
=
3 2
h√
i
Vn ( 3/2) sin(n ωt − αn ) + (±1/2) cos(n ωt − αn ) (2.84)
n6=3,6,9...
√
Let 3/2 = rn cos φn and ±1/2 = rn sin φn . This impliles rn = 1 and φn = ±30o . Using this,
equation (2.84) can be written as follows.
vab (t) =
∞
√ √ X
3 2
Vn [sin(n ωt − αn ± 30o )] .
(2.85)
n6=3,6,9...
In equations (2.84) and (2.85), vab = 0 for n = 3, 6, 9, . . . and for n = 1, 2, 4, 5, 7, . . ., the ± sign
of 1/2 or sign of 300 changes alternatively. Thus it is observed that triplen harmonics are missing
in the line to line voltages, inspite of their presence in phase voltages for balanced three-phase
system with harmonics. Thus the following identity hold true for this system,
VLL ≤
√
3 VLn
(2.86)
Above equation further implies that,
√
3 VLL I ≤ 3 VLn I.
(2.87)
In above equation,
I refers the rms value of the phase current. For above case, Ia = Ib = Ic = I
P
2
and In = 3 ∞
n=3,6,9... In . Therefore, effective rms current, Ie is given by the following.
53
s
Ie =
3 I2 + 3
P∞
n=3,6,9... In
2
3
v
u
u
= tI 2 +
∞
X
In 2
(2.88)
n=3,6,9...
≥I
2.4.3
Apparent Power with Budeanu Resolution: Balanced Distortion Case
The apparent power is given as,
q
2
S = 3Vln I = P 2 + Q2B + DB
p
=
P 2 + Q2 + D2
(2.89)
where,
P = P1 + PH = P1 + P2 + P3 + ....
∞
X
Vn In cos φn
= 3V1 I1 cos φ1 + 3
n=1
(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 is
detailed below.
Q = Q1 + QH = Q1 + Q2 + Q3 + ....
∞
X
= 3V1 I1 sin φ1 + 3
Vn In sin φn
(2.92)
n=1
2.4.4
Effective Apparent Power for Balanced Non-sinusoidal System
The effective apparent power Se for the above system is given by,
Se = 3Ve Ie
(2.93)
For a three-phase, three-wire balanced system, the effective apparent power is found after calculating effective voltage and current as given below.
q
Ve =
(Vab2 + Vbc2 + Vca2 )/9
√
= Vll / 3
(2.94)
54
q
(Ia2 + Ib2 + Ic2 )/3
Ie =
= I
(2.95)
Therefore
Se = S =
√
3Vll I
(2.96)
For a four-wire system, Ve is same is given (2.94) and Ie is given by (2.88). Therefore, the
effective apparent power is given below.
√
3Vll I ≤ 3 Vln Ie
(2.97)
The above implies that,
Se ≥ SA .
(2.98)
Therefore, it can be further concluded that,
pfe (= P/Se ) ≤ pfA (= P/SA ).
2.5
(2.99)
Unbalanced and Non-sinusoidal Three-phase System
In this system, we shall consider most general case i.e., three-phase system with voltage and current
quantities which are unbalanced and non-sinusoidal. These voltages and currents are expressed as
following.
∞
X
√
va (t) =
2Van sin(n ωt − αan )
n=1
∞
X
√
vb (t) =
2Vbn sin {n (ωt − 120o ) − αbn }
(2.100)
n=1
∞
X
√
2Vcn sin {n (ωt + 120o ) − αcn }
vc (t) =
n=1
Similarly, currents can be expressed as,
ia (t) =
∞
X
√
2Ian sin(n ωt − βan )
n=1
ib (t) =
∞
X
√
2Ibn sin {n (ωt − 120o ) − βbn }
n=1
ic (t) =
∞
X
√
2Icn sin {n (ωt + 120o ) − βcn }
n=1
55
(2.101)
For the above voltages and currents in three-phase system, instantaneous power is given as following.
p(t) = va (t)ia (t) + vb (t)ib (t) + vc (t)ic (t)
= pa (t) + pb (t) + pc (t)
! ∞
!
∞
X
X√
√
=
2Van sin(nωt − αan )
2Ian sin(nωt − βan )
n=1
+
∞
X
√
n=1
!
∞
X
√
2Vbn sin {n(ωt − 120o ) − αbn }
n=1
+
∞
X
√
(2.102)
!
2Ibn sin {n(ωt − 120o ) − βbn }
n=1
∞
X
√
2Icn sin {n(ωt + 120o ) − βcn }
!
2Vcn sin {n(ωt + 120o ) − αcn }
!
n=1
n=1
In (2.102), each phase power can be found using expressions derived in Section 1.4 of Unit 1. The
direct result is written as following.
pa (t) =
∞
X
Van Ian cos φan {1 − cos(2nωt − 2αan )} −
∞
X
√
!
=
!
2Iam sin(mωt − βam )
Qan cos(2nωt − 2αan )
Pan {1 − cos(2nωt − 2αan )} −
∞
X
√
√
m=1, m6=n
∞
X
n=1
+
∞
X
2Van sin(nωt − αan )
n=1
∞
X
Van Ian sin φan cos(2nωt − 2αan )
n=1
n=1
+
∞
X
!
2Van sin(nωt − αan )
n=1
n=1
∞
X
√
!
2Iam sin(mωt − βam )
(2.103)
m=1, m6=n
In the above equation, φan = (βan − αan ). Similarly, for phases b and c, the instantaneous
power is expressed as below.
pb (t) =
∞
X
o
Pbn [1 − cos {2n(ωt − 120 ) − 2αbn }] −
n=1
+
∞
X
Qbn cos {2n(ωt − 120o ) − 2αbn }
n=1
∞
X
√
2Vbn sin {n(ωt − 120o ) − αbn }
n=1
!
∞
X
√
!
2Ibm sin {m(ωt − 120o ) − βbm }
m=1, m6=n
(2.104)
56
and
pc (t) =
∞
X
o
Pcn [1 − cos {2n(ωt + 120 ) − 2αcn }] −
n=1
+
∞
X
Qcn cos {2n(ωt + 120o ) − 2αcn }
n=1
∞
X
!
∞
X
√
2Vcn sin {n(ωt + 120o ) − αcn }
n=1
!
√
2Icm sin {m(ωt + 120o ) − βcm }
m=1, m6=n
(2.105)
From equations (2.103), (2.104) and (2.105), the real powers in three phases are given as follows.
Pa =
∞
X
Van Ian cos φan
n=1
Pb =
Pc =
∞
X
Vbn Ibn cos φbn
n=1
∞
X
(2.106)
Vcn Icn cos φcn
n=1
Similarly, the reactive powers in three phases are given as following.
Qa =
∞
X
Van Ian sin φan
n=1
Qb =
Qc =
∞
X
n=1
∞
X
Vbn Ibn sin φbn
(2.107)
Vcn Icn sin φcn
n=1
Therefore, the total active and reactive powers are computed by summing the phase powers using
equations (2.106) and (2.107), which are given below.
P = P a + Pb + Pc =
∞
X
(Van Ian cos φan + Vbn Ibn cos φbn + Vcn Icn cos φcn )
n=1
= Va1 Ia1 cos φa1 + Vb1 Ib1 cos φb1 + Vc1 Ic1 cos φc1
∞
X
+
(Van Ian cos φan + Vbn Ibn cos φbn + Vcn Icn cos φcn )
n=2
= Pa1 + Pb1 + Pc1 +
∞
X
(Pan + Pbn + Pcn )
n=2
= P1 + PH
(2.108)
57
and,
Q = Qa + Qb + Qc =
∞
X
(Van Ian sin φan + Vbn Ibn sin φbn + Vcn Icn sin φcn )
n=1
= Va1 Ia1 sin φa1 + Vb1 Ib1 sin φb1 + Vc1 Ic1 sin φc1
∞
X
+
(Van Ian sin φan + Vbn Ibn sin φbn + Vcn Icn sin φcn )
n=2
= Qa1 + Qb1 + Qc1 +
∞
X
(Qan + Qbn + Qcn )
n=2
= Q1 + QH
2.5.1
(2.109)
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 as
following.
Sa =
Sb =
Sc =
p
Pa2 + Q2a + Da2
q
Pb2 + Q2b + Db2
p
Pc2 + Q2c + Dc2
(2.110)
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
The three-phase vector apparent power is given as following.
p
Sv = P 2 + Q2 + D2
(2.111)
(2.112)
Where P and Q are given in (2.108) and (2.109) respectively. The total distortion power D is given
as following.
D = Da + Db + Dc
(2.113)
Based on above definitions of the apparent powers, the arithmetic and vector power factors are
given below.
P
SA
P
=
Sv
pfA =
pfv
(2.114)
From equations (2.111), (2.112) and (2.114), it can be inferred that
SA ≥ Sv
pfA ≤ pfv
58
(2.115)
2.5.2
Effective Apparent Power
Effective apparent power (Se =3Ve Ie ) for the three-phase unbalanced systems with harmonics can
be found by computing Ve and Ie as following. The effective rms current (Ie ) can be resolved into
two parts i.e., effective fundamental and effective harmonic components as given below.
q
2
2
+ IeH
(2.116)
Ie =
Ie1
Similarly,
Ve
q
2
=
Ve12 + VeH
(2.117)
For three-phase four-wire system,
r
Ia2 + Ib2 + Ic2 + In2
Ie =
(2.118)
3
r
2
2
2
2
2
2
2
2
Ia1
+ Ia2
+ ... + Ib1
+ Ib2
+ ... + Ic1
+ Ic2
+ ... + In1
+ In2
+ ...
=
3
r
2
2
2
2
2
2
2
2
Ia1 + Ib1 + Ic1 + In1 + ... + Ia2 + Ib2
+ Ic2
+ In2
+ ...
=
3
r
2
2
2
2
2
2
2
2
2
2
2
2
Ia1 + Ib1 + Ic1 + In1 Ia2 + Ia3
+ ... + Ib2
+ Ib3
+ ... + Ic2
+ Ic3
+ ... + In2
+ In3
...
=
+
3
3
q
Ie =
2
2
Ie1
+ IeH
In the above equation,
r
2
2
2
2
Ia1
+ Ib1
+ Ic1
+ In1
3
r
2
2
2
2
IaH + IbH + IcH
+ InH
=
3
Ie1 =
IeH
Similarly, the effective rms voltage Ve is given as following.
r
1
Ve =
[3(Va2 + Vb2 + Vc2 ) + (Vab2 + Vbc2 + Vca2 )]
18
q
2
=
Ve12 + VeH
(2.119)
(2.120)
Where
r
1
2
2
2
2
[3(Va1
+ Vb12 + Vc12 ) + (Vab1
+ Vbc1
+ Vca1
)]
18
r
1
2
2
2
2
2
2
=
[3(VaH
+ VbH
+ VcH
) + (VabH
+ VbcH
+ VcaH
)]
18
Ve1 =
VeH
For three-phase three-wire system, In = 0 = In1 = InH .
59
(2.121)
r
2
2
2
+ Ic1
+ Ib1
Ia1
3
r
2
2
2
+ IcH
IaH + IbH
=
3
Ie1 =
IeH
(2.122)
Similarly
r
2
2
2
+ Vca1
+ Vbc1
Vab1
9
r
2
2
2
+ VcaH
VabH + VbcH
VeH =
9
The expression for effective apparent power Se is given as following.
Ve1 =
Se = 3 Ve Ie
q
q
2
2
2
+ IeH
Ie1
= 3 Ve12 + VeH
q
2
2
2 2
2 2
=
9 Ve12 Ie1
+ (9Ve12 IeH
+ 9VeH
Ie1 + 9VeH
IeH )
q
2
2
Se1
+ SeN
=
(2.123)
(2.124)
In the above equation,
Se1 = 3 Ve1 Ie1
q
2
SeN =
Se2 − Se1
q
2
2
2
=
DeV
+ DeI
+ SeH
q
2
2
2
2 2
= 3 Ie1
VeH
+ Ve12 IeH
+ VeH
IeH
(2.125)
(2.126)
In equation (2.126), distortion powers DeI , DeV and harmonic apparent power SeH are given as
following.
DeI = 3Ve1 IeH
DeV = 3VeH Ie1
SeH = 3VeH IeH
(2.127)
By defining above effective voltage and current quantities, the effective total harmonic distortion
(T HDe ) are expressed below.
T HDeV
T HDeI
VeH
Ve1
IeH
=
Ie1
=
Substituting VeH and IeH in (2.126),
q
2
2
2
2
SeN = Se1 T HDe1
+ T HDeV
+ T HDeI
T HDeV
.
60
(2.128)
(2.129)
In above equation,
DeI = Se1 T HDI
DeV = Se1 T HDV
SeH = Se1 (T HDI )(T HDV ).
Using (2.124) and (2.129), the effective apparent power is given as below.
q
q
2
2
2
2
2
2
Se = Se1 + SeN = Se1 1 + T HDeV
+ T HDeI
+ T HDeV
T HDeI
(2.130)
(2.131)
Based on above equation, the effective power factor is therefore given as,
P1 + P H
P
p
=
2
2
2
2
Se
Se1 1 + T HDeV
+ T HDeI
+ T HDeV
T HDeI
P1
(1 + PH /P1 )
= p
2 S
2
2
2
1 + T HDeV + T HDeI + T HDeV T HDeI e1
(1 + PH /P1 )
= p
pfe1
2
2
2
2
1 + T HDeV + T HDeI
+ T HDeV
T HDeI
pfe =
(2.132)
Practically, the THDs in voltage are far less than those of currents THDs, therefore T HDeV <<
T HDeI . Using this practical constraint and assuming PH << P1 , the above equation can be
simplified to,
pfe1
pfe ≈ p
2
1 + T HDeI
(2.133)
In the above context, their is another useful term to denote unbalance of the system. This is
defined as fundamental unbalanced power and is given below.
q
2
SU 1 =
Se1
− (S1+ )2
(2.134)
Where, S1+ is fundamental positive sequence apparent power, which is given below.
q
+
2
(P1+ )2 + (Q+
S1 =
1)
(2.135)
+
+
+ +
In above, P1+ = 3V1+ I1+ cos φ+
1 and Q1 = 3V1 I1 sin φ1 . Fundamental positive sequence power
+
+
factor can thus be expressed as a ratio of P1 and S1 as given below.
Pf+1 =
P1+
S1+
(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., I a , I b and I c and neutral current, I n .
61
r f  jx f
Va
Ia
vb
vc
vn
Vb
Ib
Ic
XL
LOAD
va
Vc
Vn
RL
In
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
Ib
Ic
√
220 3∠30
=
= 29.31∠−60◦ A
◦
13∠90
= −I a = −29.311∠−60◦ = 29.31∠120◦ A
220∠120◦
=
= 18.33∠120◦ A.
12
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 (Ia2 + Ib2 + Ic2 + In2 )
= 0.02 (29.312 + 29.312 + 18.332 + 18.332 ) = 47.80 W
c. Computation of various powers
Phase-a active and reactive power:
∗
implies that,
S a = V a I a = 220∠0◦ × 29.31∠60◦ = 3224.21 + j5584.49
Pa = 3224.1 W, Qa = 5584.30 VAr
Similarly,
∗
implies that,
S b = V b I b = 220∠−120◦ × 29.31∠60◦ = −3224.21 + j5584.49
Pb = −3224.1 W, Qb = 5584.30 VAr
For phase-c,
∗
implies that,
S c = V c I c = 220∠120◦ × 18.33∠−120◦ = 4032.6 + j0
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 W
Q3−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
r
Ia2 + Ib2 + Ic2 + In2
Se = 3Ve Ie = 3 × 220 ×
3
r
2
2
29.31 + 29.31 + 18.332 + 18.332
= 3 × 220 ×
= 3 × 220 × 28.22
3
= 18629.19 VA
63
Based on the above apparent powers, the arithmetic, vector and effective apparent power factors
are computed as below.
4032.6
P3−phase
=
= 0.2382
SA
16928.84
P3−phase
4032.6
=
=
= 0.3396
Sv
11874.32
4032.6
P3−phase
=
= 0.2165
=
Se
18629.19
pfA =
pfv
pfe
In the above computation, the effective voltage and current are found as given in the following.
r
Va2 + Vb2 + Vc2
Ve =
= 220 V
3
r
Ie =
Ia2 + Ib2 + Ic2 + In2
= 28.226 A
3
Example 2.4 A 3-phase, 3-wire system is shown in Fig. 2.12. The 3-phase voltages are balanced
sinusoids 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.
N
Vsa
I la
Vsc
I lc
Vsb
I lb
a
c
Zb
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∠0◦ V
V sb = 230∠−120◦ V
V sc = 230∠120◦ V
1
V sa V sb V sc
V nN = 1
+
+
Zb
Zc
+ Z1b + Z1c Za
Za
230∠0◦ 230∠−120◦ 230∠120◦
1
=
+
+
1
1
1
5 + j12
6 + 8j
12 − j5
+ 6+j8
+ 12−j5
5+j12
1
31.23∠−164.50◦
=
◦
0.2013∠−37.09
= −94.22 − j123.18 = 155.09∠−127.41◦ V
Now the line currents are computed as below.
V sa − V nN
230∠0◦ − 155.09∠−127.41◦
=
= 26.67∠−46.56◦ A
Za
5 + j12
230∠−120◦ − 155.09∠−127.41◦
V sb − V nN
=
= 7.88∠−158.43◦ A
=
Zb
6 + j8
◦
V sc − V nN
230∠120 − 155.09∠−127.41◦
= 24.85∠116.3◦ A
=
=
Zc
12 − j5
I al =
I bl
I cl
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 a I a = 230∠0◦ × 26.67∠46.56◦ = 4218.03 + j4456.8
∗
Sb = V b I b = 230∠−120◦ × 7.88∠158.43◦ = 1419.82 + j1126.06
Sc
implies that,
Pa
Pb
Pc
∗
= V c I c = 230∠120◦ × 24.85∠−116.3◦ = 5703.43 + j368.11
= 4218.03 W,
= 1419.82 W,
= 5703.43 W,
Qa = 4456.8 VAr
Qb = 1126.06 VAr
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 W
Q3−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.
Pa
4218.03
4218.03
= 0.6873 (lag)
=√
=
2
2
6136.3
|Sa |
4218.03 + 4456.8
Pb
1419.82
1419.82
=
= 0.7835 (lag)
=
=
2
2
1419.82 + 1126.06
1812.16
|Sb |
Pc
5703.43
5703.43
=
=
= 0.9979 (lag)
=
2
2
5703.43 + 368.11
5715.30
|Sc |
pfa =
pfb
pfc
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
r
2
2
Ila
+ Ilb2 + Ilc2 + Iln
Se = 3Ve Ie = 3 × 230 ×
3
r
2
2
26.67 + 7.88 + 24.852 + 02
= 3 × 220 ×
= 3 × 230 × 21.53
3
= 14859.7 VA
The arithmetic, vector and effective apparent power factors are computed as below.
P3−phase
11341.29
=
= 0.8300
SA
13663.82
P3−phase
11341.29
=
=
= 0.8855
Sv
12807.78
P3−phase
11341.29
=
=
= 0.7632
Se
14859.7
pfA =
pfv
pfe
References
[1] IEEE Group, “IEEE trial-use standard definitions for the measurement of electric power quantities under sinusoidal, nonsinusoidal, balanced, or unbalanced conditions,” 2000.
66
[2] E. Watanabe, R. Stephan, and M. Aredes, “New concepts of instantaneous active and reactive
powers in electrical systems with generic loads,” IEEE Transactions on Power Delivery, vol. 8,
no. 2, pp. 697–703, 1993.
[3] T. Furuhashi, S. Okuma, and Y. Uchikawa, “A study on the theory of instantaneous reactive
power,” IEEE Transactions on Industrial Electronics, vol. 37, no. 1, pp. 86–90, 1990.
[4] A. Ferrero and G. Superti-Furga, “A new approach to the definition of power components in
three-phase systems under nonsinusoidal conditions,” IEEE Transactions on Instrumentation
and Measurement, vol. 40, no. 3, pp. 568–577, 1991.
[5] J. Willems, “A new interpretation of the akagi-nabae power components for nonsinusoidal
three-phase situations,” IEEE Transactions on Instrumentation and Measurement, vol. 41,
no. 4, pp. 523–527, 1992.
[6] H. Akagi, Y. Kanazawa, and A. Nabae, “Instantaneous reactive power compensators comprising switching devices without energy storage components,” IEEE Transactions on Industry
Applications, no. 3, pp. 625–630, 1984.
[7] C. L. Fortesque, “Method of symmetrical co-ordinates applied to the solution of polyphase
networks,” AIEE, 1918.
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