Symmetrical Components in Electrical Engineering


Introduction

Symmetrical components method was discovered by C. L. Fortescue. The knowledge of symmetrical components is very useful for the study of unsymmetrical faults in three phase power networks. The concept  is also useful for studying the three phase machine behavior under unbalanced condition.

This article requires a little bit  knowledge about phasor representation that we already discussed.

Symmetrical Components

Why symmetrical components? Symmetrical component technique is used for analyzing unbalanced three phase systems. When the system is balanced, analysis is very simple. We do not analyze for all the three phases instead we analyze it as single phase system. So the three phase system is reduced to simpler single phase system. Symmetrical component method helps to apply the single phase analysis tools also to the unbalanced three phase system. How?

In Symmetrical component method, any unbalanced three phase system can be resolved into three sets of symmetrical components. These three sets are positive sequence, negative sequence and zero sequence.

Considering counterclockwise abc sequence as positive sequence, then acb will be negative sequence (See Fig-A). Both the positive and negative sequence components are balanced. It means that the three phasors have the same magnitude and the phase angle between any two phasors is 120 degrees. The three phasors of zero sequence are of same magnitude and aligned in the same direction. So, in case of zero sequence the angle between any two phasors is zero. All these phasors rotate counterclockwise with frequency of the system. So the relative position between the phasors remains the same. .


For identification purposes we have used the symbols +, - and 0 for positive, negative and zero sequence components respectively.

Every three phase unbalanced system can be decomposed to three balanced systems as in Fig-A.

A three phase unbalanced system is shown in Fig-B.
The unbalnced system in Fig-B can be resolved to symmetrical components like Fig-A. In Figure-C just see  how each unbalanced component is made up of +ve, -ve and 0 sequence components.

From the diagram above it is easy to verify the below equations.
                          

Now is the time to apply the phasor operator  a  that we learned previously.
(Phasor operator when applied to a phasor rotates the phasor anticlockwise by 120 degrees).

So the above equation can be written as below
      

In the above equation we have eliminated both b and c phase positive, negative and zero sequence components.

When the unbalanced system is known. we know Va, Vb and Vc.
Of course we also know the value of phasor operator a which is constant.

So the above three equations has three unknown Va+, Va- and Va0.
We can solve the equations and find the three unknowns by using school maths.

Now from Va+, Va- and Va0 that we calculated we can construct the full symmetrical components as in Fig-A. It will simplify for per phase analysis.

Operators j and a in Electrical Engineering




We have already discussed about phasors and its simple properties. Perhaps now it is the time that we want to explore a little more. Every effort is made to keep it as simple as our previous article. 

Before proceeding further I want to clarify that here we are mainly concerned about phasor multiplication and 'j' and 'a' operators.This article will also help us better appreciate the use of symmetrical components ( for analysis of unbalanced 3-phase systems)  and subsequently other phenomena  in transformer and AC circuits.


We know that phasor in the form x+j y is drawn as an arrow from origin to (x, y) point.


Till now I represented the phasor in x+j y form also called rectangular form. A phasor can also be represented in polar form. In the polar form we also need two parameters, these are length of phasor (r) and angle(phi) it makes with the +ve horizontal axis . See the Figure-A.





Phasor Multiplication

I have already discussed the use of j in phasor representation.
We know that j is equal to square root of -1.

or    j = sqrt(-1) so j.j = -1


Now consider two phasors A = 2 + j 3 and B = -1 + j 2

Let us multiply A and B

 A.B = (2+j 3) . (-1 + j2) = -2 + j 4 -j 3 + j.j (3.2) = -2 + j 1 - 6 = -8 + j 1 

Directly multiply each of real and imaginary parts from A with each of B. It is simple!

It is even easier to multiply in polar form. See the example below. As illustrated in figure-A,  we represent below the phasors A and B in the polar form. For phasor A, 4 is its length and it makes 20 degrees with x-axis. Similarly B is of length 3 units and it makes 40 degrees with +ve horizontal axis









( 20, 40 and 60 are angles in degree)

Representing in polar form, the multiplication has become extremely easy.
Just multiply the lengths and add the angles to get the new phasor. 
You can convert it back to the rectangular form.

                                      A.B = 12(cos 60 + sin 60)   
j and a Operators

What we will get, if a phasor is multiplied with j?

for example
      if    A = 3 + j 4
Then    j A =j(3 + j 4) = j 3 + j.j 4 = -4 + j 3   ( As j.j =-1)

Now draw the phasor -4 + j 3. It will be observed that the angle between 3 + j 4 and -4 + j 3 is 90 degrees.



Any phasor when multiplied by j  will rotate the original phasor by 90 degrees in anticlockwise direction. Now if the resultant phasor is again multiplied by j then the phasor is again rotated by 90 degrees in anticlockwise direction, so on.
In our example j(-4 + j 3) = -j4 -3 = -(3 + j 4), which is in opposite direction to (p + j q). So clearly the phasor has again undergone 90 degrees anticlockwise rotation. See the figure. Every time we apply j, we rotate the phasor counterclockwise by 90 degrees.

Now let us consider about another operator ' a ' (standard symbol). It has the capacity to rotate a phasor counterclockwise by 120 degrees. applying ' a ' twice the phasor is rotated by 240 degrees, by applying thrice the original phasor is rotated 360 degrees or one complete rotation, so the original phasor.

It is clear that as the phasor is rotated 120 degrees (magnitude remains the same) then in polar form

                              a = 1/120deg  

in rectangular form   a = 1.cos 120 + j 1.sin 120
                          or  a = -0.5 + j 0.866


see the Fig-C how a phasor A is rotated by 120 degrees when applied with operator a.



I colored them red green and blue to recall our balanced three phase system.



Clearly we are able to get the phasors B and C  by applying the operator a repeatedly on phasor A.
Otherwise we can say that, the balanced system of A-B-C sequence can be equally represented in terms of 'a' and A only. The operator a will be used more in our article symmetrical components.