Overview
An unbalanced three-phase system can be resolved into three balanced systems in the sinusoidal steady state. This method of resolving an unbalanced system into three balanced phasor system has been proposed by C. L. Fortescue. This method is called resolving symmetrical components of the original phasors or simply symmetrical components.
In this chapter we shall discuss symmetrical components transformation and then will present how unbalanced components like Y- or Δ -connected loads, transformers, generators and transmission lines can be resolved into symmetrical components. We can then combine all these components together to form what are called sequence networks .
Section I: Symmetrical Components Symmetrical Components A system of three unbalanced phasors can be resolved in the following three symmetrical components:
Fig. 7.1 depicts a set of three unbalanced phasors that are resolved into the three sequence components mentioned above. In this the original set of three phasors are denoted by Va , Vb and Vc , while their positive, negative and zero sequence components are denoted by the subscripts 1, 2 and 0 respectively. This implies that the positive, negative and zero sequence components of phase-a are denoted by Va1 , Va2 and Va0 respectively. Note that just like the voltage phasors given in Fig. 7.1 we can also resolve three unbalanced current phasors into three symmetrical components.
Fig. 7.1 Representation of (a) an unbalanced network, its (b) positive sequence, (c) negative sequence and (d) zero sequence. | ||
Symmetrical Component Transformation
Before we discuss the symmetrical component transformation, let us first define the α -operator. This has been given in (1.34) and is reproduced below.
 | (7.1) |
Note that for the above operator the following relations hold
 | (7.2) |
Also note that we have
 | (7.3) |
Using the a -operator we can write from Fig. 7.1 (b)
 | (7.4) |
Similarly from Fig. 7.1 (c) we get
 | (7.5) |
Finally from Fig. 7.1 (d) we get
 | (7.6) |
 | (7.7) |
 | (7.8) |
 | (7.9) |
Therefore,
 | (7.10) |
 | (7.11) |
The symmetrical component transformation matrix is then given by
 | (7.12) |
Defining the vectors V a012 and V abc as
|
we can write (7.4) as
 |
where C is the symmetrical component transformation matrix and is given by
 | (7.13) |
The original phasor components can be obtained from the inverse symmetrical component transformation, i.e.,
 | (7.14) |
Finally, if we define a set of unbalanced current phasors as Iabc and their symmetrical components as Ia012 , we can then define
 | (7.15) |
Real and Reactive Power
The three-phase power in the original unbalanced system is given by
| (7.16) |
where I* is the complex conjugate of the vector I . Now from (7.10) and (7.15) we get
| (7.17) |
From (7.11) we get
|
Therefore from (7.17) we get
| (7.18) |
We then find that the complex power is three times the summation of the complex power of the three phase sequences.
Orthogonal Transformation
Instead of the transformation matrix given in (7.13), let us instead use the transformation matrix
| (7.19) |
We then have
| (7.20) |
Note from (7.19) and (7.20) that C -1 = ( CT )* . We can therefore state C( CT )* = I3 , where I3 is (3 × 3) identity matrix. Therefore the transformation matrices given in (7.19) and (7.20) are orthogonal. Now since
|
we can write from (7.17)
| (7.21) |
We shall now discuss how different elements of a power system are represented in terms of their sequence components. In fact we shall show that each element is represented by three equivalent circuits, one for each symmetrical component sequence.
Sequence Circuits for Loads
In this section we shall construct sequence circuits for both Y and Δ -connected loads separately.
Sequence Circuit for a Y-Connected Load
Consider the balanced Y-connected load that is shown in Fig. 7.2. The neutral point (n) of the windings are grounded through an impedance Zn . The load in each phase is denoted by ZY . Let us consider phase-a of the load. The voltage between line and ground is denoted by Va , the line-to-neutral voltage is denoted by Van and voltage between the neutral and ground is denoted by Vn . The neutral current is then
| (7.22) |
Therefore there will not be any positive or negative sequence current flowing out of the neutral point.
|
Fig. 7.2 Schematic diagram of a balanced Y-connected load.
The voltage drop between the neutral and ground is
| (7.23) |
Now
| (7.24) |
We can write similar expression for the other two phases. We can therefore write
| (7.25) |
Pre-multiplying both sides of the above equation by the matrix C and using (7.8) we get
| (7.26) |
Now since
|
We get from (7.26)
| (7.27) |
We then find that the zero, positive and negative sequence voltages only depend on their respective sequence component currents. The sequence component equivalent circuits are shown in Fig. 7.3. While the positive and negative sequence impedances are both equal to ZY , the zero sequence impedance is equal to
| (7.28) |
If the neutral is grounded directly (i.e., Zn = 0), then Z0 = ZY . On the other hand, if the neutral is kept floating (i.e., Zn = ∞ ), then there will not be any zero sequence current flowing in the circuit at all.
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Fig. 7.3 Sequence circuits of Y-connected load: (a) positive, (b) negative and (c) zero sequence.
Sequence Circuit for a Δ -Connected Load
Consider the balanced Δ -connected load shown in Fig. 7.4 in which the load in each phase is denoted by ZΔ . The line-to-line voltages are given by
| (7.29) |
Adding these three voltages we get
|
Fig. 7.4 Schematic diagram of a balanced Δ -connected load.
| (7.30) |
Denoting the zero sequence component Vab , Vbc and Vca as Vab0 and that of Iab , Ibc and Ica as Iab0 we can rewrite (7.30) as
| (7.31) |
Again since
|
We find from (7.31) Vab0 = Iab0 = 0. Hence a Δ -connected load with no mutual coupling has not any zero sequence circulating current. Note that the positive and negative sequence impedance for this load will be equal to ZΔ .
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