Power System Analysis Prof. Debapriya Das Department of Electrical Engineering Indian Institute of Technology, Kharagpur Lecture–49 Symmetrical components So next in that previous topic we have seen that your that 3 phase fault, but in that case our objective was mainly for building algorithm and few numericals we have solved right, but question is that in the that we have to have matrix, soin anyway. So, today we will start the other thing that is your symmetrical component because symmetrical component is more important for your our fault studies,because most of the fault actually unsymmetrical or all unbalanced type of faults right therefore, symmetricalyour what you call that symmetric unsymmetrical asymmetrical component will be that your is easier, I mean will be applicable for such kind of fault analysis. (Refer Slide Time: 01:22) So, let us come to that your symmetrical component. So, generally in a balanced system analysis can be done on a single phase basis that we have seen also for three phase fault. The knowledge of voltage and current in one phase is sufficient to determine the voltage and current in other 2 phases because it is a balanced system.Real and reactive powers are three times the corresponding per phase values. So, for that is for balanced system, but when the system is unbalanced, the voltages, currents and the phase impedances are
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Power System Analysis Prof. Debapriya Das
Department of Electrical Engineering Indian Institute of Technology, Kharagpur
Lecture–49
Symmetrical components
So next in that previous topic we have seen that your that 3 phase fault, but in that case
our objective was mainly for 𝑍𝑍𝐵𝐵𝐵𝐵𝐵𝐵building algorithm and few numericals we have solved
right, but question is that in the 𝑍𝑍𝐵𝐵𝐵𝐵𝐵𝐵 that we have to have 𝑍𝑍𝐵𝐵𝐵𝐵𝐵𝐵matrix, soin anyway. So,
today we will start the other thing that is your symmetrical component because
symmetrical component is more important for your our fault studies,because most of the
fault actually unsymmetrical or all unbalanced type of faults right therefore,
symmetricalyour what you call that symmetric unsymmetrical asymmetrical component
will be that your is easier, I mean will be applicable for such kind of fault analysis.
(Refer Slide Time: 01:22)
So, let us come to that your symmetrical component. So, generally in a balanced system
analysis can be done on a single phase basis that we have seen also for three phase fault.
The knowledge of voltage and current in one phase is sufficient to determine the voltage
and current in other 2 phases because it is a balanced system.Real and reactive powers
are three times the corresponding per phase values. So, for that is for balanced system,
but when the system is unbalanced, the voltages, currents and the phase impedances are
in general unequal. So, unbalanced operation can result whenloads are unbalanced, this is
one thing. So, loads are unbalanced means that I mean all the three phase loads may be
different, and this is more your what you call more applicable to the your distribution
lowvoltage distribution system for example, in our country say 11 kV distribution system
or in other parts of this world.
So, sometimes what happened that in three phase distribution loads are not balanced, this
is one thing second thing in many places that from the same substation same feeder right
single phase 2 phase or 3 phase loads are supplied. So, in that case yourthat things are
become system become unbalanced and another aspect is that unbalanced system
operation can result that is due to unsymmetrical fault.
(Refer Slide Time: 02:53)
For example, that your line to line fault that is called𝐿𝐿 − 𝐿𝐿 − 𝐺𝐺 right then double line to
ground fault or single line to ground fault right. Such an unbalanced operation can be
analyzed through symmetrical components,where the unbalanced threephase voltages
and currents are transformed in to three sets of balanced voltages and currents called
symmetrical components. So, another type of fault happens, but very rare that is your
conductor opening.Suppose one phase conductor your opening means cut right. So, these
are very rare, but question is that is also called open conductor your fault, but that we
will see when we will go for unbalanced or unsymmetrical fault analysis. So,
symmetrical components of anyour unbalanced threephase system right this one. So, the
unbalanced phasors of a threephase system can be resolved in to the following three
component sets of balancedphases, which possess certain symmetry; that means, for
unbalanced phasors, voltage or current right can we can dissolve in to following three
component sets of balanced phasor and it has some symmetry.
(Refer Slide Time: 04:18)
That means, first thing is that is set of your three phasors equal in magnitude, displaced
from each other by 120 degree in phase and having the same phase sequence as the
original unbalanced phasors. And this set of balanced phasorthat is called positive
sequence sequence component. So, a set of three phasors equal in magnitude, but
displaced from each other by 120 degree. The way we have solved the balanced system
right and your and the; or the original unbalanced phasor this set of balanced phasor is
called positive sequence component. Similarly a set of three phasors again equal in
magnitudedisplaced again from each other by 180 degree or sorry 120 degree in phase,
and having the phase sequence opposite to that of the original phasor I mean if for this
one for positive sequence if it is abc say for this one opposite it will be ac b sequence
right and this set of balanced phasor is called negative sequence component.
And last one that is a set of 3 phases they are equal in magnitude with 0 phase
displacement. There is no phase displacement between them, but they have equal
magnitude from each and this set is calledzerosequence component. So, the component
of this set are all identical, I mean they all are identical because they have equal
magnitude.
(Refer Slide Time: 05:50)
So, therefore, these three sets of balanced phasors are called symmetrical components of
the original unbalanced phasor. So, assume that thethree phasors are represented by
𝑎𝑎 𝑏𝑏 𝑐𝑐 and such that phase sequence 𝑎𝑎 𝑏𝑏 𝑐𝑐 is called positive sequence. So, this is this you
know, this you know already we have done it right, this that three phase system. So, this
is abc. So, if 𝑎𝑎 𝑏𝑏 𝑐𝑐 sequence this is actually positive sequence.
Now, say𝑉𝑉𝑎𝑎 ,𝑉𝑉𝑏𝑏 ,𝑉𝑉𝑐𝑐are balanced voltages that is phasors, characterized by equal magnitude
and interphase difference of 120 degree this also you know. Then the set is said to have
phase sequence 𝑎𝑎 𝑏𝑏 𝑐𝑐 or positive sequence right, that is whatever we have studied little
bit for solving say 3 phase transmission line or this that almost all are abc all the
𝑎𝑎 𝑏𝑏 𝑐𝑐yourwhat you call phase sequence. So, if𝑉𝑉𝑏𝑏 lags𝑉𝑉𝑎𝑎by 120∘, this you know
and𝑉𝑉𝑐𝑐lags𝑉𝑉𝑏𝑏by120∘, that alsoyou know, then we can make this kind of your definition.
(Refer Slide Time: 07:11)
That is that assume your𝑉𝑉𝑎𝑎 is a reference phasor. Suppose if we put like this say𝑉𝑉𝑎𝑎 is
reference phasor. So,𝑉𝑉𝑏𝑏 lags from𝑉𝑉𝑎𝑎here by120∘, and𝑉𝑉𝑐𝑐 lags from𝑉𝑉𝑎𝑎by240∘, or
otherwise𝑉𝑉𝑐𝑐leads𝑉𝑉𝑎𝑎by120∘. So, if we if I define like this say V a is equal to V a,a
reference phasor V b is equal to beta square V a I am coming to that and V c is equal to
beta V a right. Where the complex power a complex operated beta is defined as we have
taken 𝛽𝛽 = 𝑒𝑒𝑗𝑗120∘ ,; that means, this 𝛽𝛽 has the following properties. 𝛽𝛽2 = 𝑒𝑒𝑗𝑗240∘ =
𝑒𝑒−𝑗𝑗120∘ = 𝛽𝛽∗. Similarly (𝛽𝛽2)∗ = 𝛽𝛽and 𝛽𝛽3 = 1. So, it is one right and1 + 𝛽𝛽 + 𝛽𝛽2 = 0this
conditions will be applicable now right. Now in this case that𝑉𝑉𝑏𝑏 = 𝑉𝑉𝑎𝑎∠ − 120∘ here this
figure.
That means𝑉𝑉𝑏𝑏 = 𝑉𝑉𝑎𝑎𝑒𝑒−𝑗𝑗120∘ = 𝛽𝛽2𝑉𝑉𝑎𝑎because 𝛽𝛽2 = 𝑒𝑒𝑗𝑗240∘ = 𝑒𝑒−𝑗𝑗120∘. So, we can write𝑉𝑉𝑏𝑏 =
𝛽𝛽2𝑉𝑉𝑎𝑎 that is why we writing here𝑉𝑉𝑏𝑏 = 𝛽𝛽2𝑉𝑉𝑎𝑎 .Similarly𝑉𝑉𝑐𝑐 = 𝑉𝑉𝑎𝑎∠ − 240∘ because
these𝑉𝑉𝑐𝑐lags from𝑉𝑉𝑎𝑎 that is 240 degree. So,𝑉𝑉𝑐𝑐 = 𝑉𝑉𝑎𝑎∠ − 240∘ that is we have given that
your this𝑉𝑉𝑐𝑐 = 𝑉𝑉𝑎𝑎∠120∘that is𝑉𝑉𝑎𝑎𝑒𝑒𝑗𝑗120 ∘, that is𝑉𝑉𝑐𝑐 = 𝛽𝛽𝑉𝑉𝑎𝑎 ; that means, we are writing
here𝑉𝑉𝑐𝑐 = 𝛽𝛽𝑉𝑉𝑎𝑎 .So; that means, that means this𝑉𝑉𝑎𝑎 ,𝑉𝑉𝑏𝑏 ,𝑉𝑉𝑐𝑐 this𝑉𝑉𝑏𝑏 ,𝑉𝑉𝑐𝑐we are putting in terms of
a complex operator 𝛽𝛽,where𝛽𝛽 = 𝑒𝑒𝑗𝑗120 ∘.
(Refer Slide Time: 10:03)
Next is if the sequence is 𝑎𝑎 𝑐𝑐 𝑏𝑏 I mean it is if it is a negative sequence, right then𝑉𝑉𝑎𝑎 =
𝑉𝑉𝑎𝑎now𝑉𝑉𝑏𝑏 = 𝛽𝛽𝑉𝑉𝑎𝑎and𝑉𝑉𝑐𝑐 = 𝛽𝛽2𝑉𝑉𝑎𝑎 .Now sequence is 𝑎𝑎 𝑐𝑐 𝑏𝑏; that means,𝑉𝑉𝑐𝑐 lags
from𝑉𝑉𝑎𝑎by120∘this is your 120∘ and this one also your120∘. So, and𝑉𝑉𝑏𝑏 lags
from𝑉𝑉𝑎𝑎by240∘. Just whatever we have done previously it was 𝛽𝛽2 now it will 𝛽𝛽 it was 𝛽𝛽
now it will be 𝛽𝛽2therefore,𝑉𝑉𝑏𝑏 = 𝑉𝑉𝑎𝑎∠ − 240∘ = 𝑉𝑉𝑎𝑎𝑒𝑒𝑗𝑗120∘ = 𝛽𝛽𝑉𝑉𝑎𝑎 ,