Symmetrical Components implementation to PowerX,
an Educational Software Tool
Thomas R. Walsh and Zejin Li
Eastern Washington University
ABSTRACT
This paper describes a student research project that contributes functionality to an existing
software application, named PowerX, which is used for educational instruction in the courses
EENG 350: Energy Systems and EENG 450: Power Systems Analysis at Eastern Washington
University. PowerX is a software application that was originally developed under an internal
grant from EWU to do research in the area of Synchrophasors for the Power Grid. Over time,
more and more functionality was added to PowerX and eventually it started being used in classes
like EENG 350 and EENG 450 to give the students a better physical feel for realistic numbers
when doing numerical computations as well as displaying graphical representation of typical
power system problems. Students were very excited to use PowerX and some were interested in
not only using the software but understanding how the software was written. This resulted in a
student doing a research project, EENG 491: Research Project, where the student would first
learn C#, then learn about software development and design and ultimately add some
functionality to PowerX related to the Power Grid. This paper describes a research project where
the student added a graphical component that investigates faults on the power grid and computes
and displays symmetrical components for a variety of these faults.
Introduction
The purpose of this student research project is to develop a software component using C# that
will give the user the ability to visualize symmetrical components which are used to analyze
faults in a 3-phase power system. This component will be integrated into an existing application
called PowerX. Faults such as single line-to-ground, line-to-line, double line-to-ground and 3phase faults are considered.
The software allows the user to generate three-phase voltages, select a particular voltage and
change its magnitude or phase. The user may also simulate a fault on the line. Symmetrical
Components, for both voltage and current will be calculated and displayed graphically.
Subsequent fault currents will then be computed.
Overview of PowerX
PowerX is a software tool that is used in the classes EENG 350: Energy Systems and EENG 450:
Power Systems Analysis that helps the students understand power system concepts. PowerX
provides functionality that is aligned with the topics discussed in these courses.
Some of the functionality that PowerX provides is a GUI to build one-line diagrams, a unit’s
converter, a complex calculator, computation of R, L and C for a transmission line as well as
computing power flow.
For example, PowerX provides icons which allow the user to build a one-line diagram. When
building a one-line diagram, the user can add generators, bus bars, transformers, loads, phasor
measurement units (PMU) and phasor data concentrators (PDC) which are all available under the
“One-Line Diagram” menu item. A simple one-line diagram is shown in the following figure.
Figure 1: One Line Diagram
Another feature in PowerX is the ability to compute both the AC and DC resistance of a given
conductor as a function of resistivity and temperature. This is shown in the following figure:
Figure 2: AC and DC Resistance of a Conductor
Symmetrical Components
Fortescue [1] showed that unbalanced phasors can be represented as a sum of balanced phasors.
Symmetrical Components is analogous to Fourier series in that a set of unbalanced phasor can be
expressed as a sum of balanced phasors. Symmetrical Components use this result to analyze
faults in a 3-phase power system.
The balanced set of phasors, equal magnitudes and separated by equal angles, includes the zero
sequence, positive sequence and negative sequence components. These components are typically
referenced with respect to the ‘a’ phase. The ‘a’ phase is considered the principal phase. The
zero, positive and negative sequence components are denoted by Va0, Va1, and Va2 , respectively.
The line-to-neutral voltages, Va, Vb and Vc , ,of a typical power system, can be expressed in
terms of the sequence components as given by:
𝑉𝑎
1 1
[𝑉𝑏 ] = [1 𝑎2
𝑉𝑐
1 𝑎
1 𝑉𝑎0
𝑎 ] [𝑉𝑎1 ] , where a = 1 at 120 degrees.
𝑎2 𝑉𝑎2
Given the line-to neutral voltages Va, Vb and Vc , you can compute the sequence components,
Va0, Va1, and Va2 as shown below:
𝑉𝑎0
1 1
𝑉
[ 𝑎1 ] = [1
3
𝑉𝑎2
1
1
𝑎
𝑎2
1 𝑉𝑎
𝑎2 ] [𝑉𝑏 ]
𝑎 𝑉𝑐
Similarly, the phase currents can also be expressed in terms of their sequence components as
follows:
𝐼𝑎
1
[𝐼𝑏 ] = [1
𝐼𝑐
1
1
𝑎2
𝑎
1 𝐼𝑎0
𝑎 ] [𝐼𝑎1 ]
𝑎2 𝐼𝑎2
And the sequence components of the current are related to the original phase currents by:
𝐼𝑎0
1 1
[𝐼𝑎1 ] = [1
3
𝐼𝑎2
1
1
𝑎2
𝑎
1 𝐼𝑎
𝑎 ] [𝐼𝑏 ]
𝑎2 𝐼𝑐
For a four-line 3-phase system, phase conductors have line impedances denoted by Zaa, Zbb , Zcc
and Znn, as well as the mutual impedance between phase conductors denoted as Zab, Zbc and Zca.
In addition, Zan, Zbn and Zcn, are the mutual impedances between phase conductors and the
neutral line. For a balanced system, it is assumed that the phase conductors are all equal, and the
mutual impedances are equal as well. The zero, positive and negative sequence impedances Z0,
Z1 and Z2 can be shown [2] to be given by:
Z0 = Zaa +2Zab - 6Zan+3Znn,
Z1 = Zaa - Zab,
Z2 = Zaa - Zab
Given the sequence impedances and a source voltage, one has everything that they need to
determine the fault currents, given a specific type of fault, which are subsequently used in
determining the rating on protective devices employed in the power system.
Single Line-to-Ground Fault
The first case that was considered is the single line-to-ground fault, SLG. This occurs when one
of the phases from a balanced 3-phase system shorts to ground. For instance, when phase ‘a’
shorts to ground, we get the following
Va = 0,
Ib = Ic=0
This is illustrated in the following figure:
Va=0
Vb
Vc
Figure 3: Single Line-to-Ground Fault
The sequence components associated with the phase currents are found by using
𝐼𝑎0
1 1 1
[𝐼𝑎1 ] = [1 𝑎2
3
𝐼𝑎2
1 𝑎
1 𝐼𝑎
1 1 1
𝑎 ] [𝐼𝑏 ] = [1 𝑎2
3
𝑎2 𝐼𝑐
1 𝑎
1 𝐼𝑎
1 𝐼𝑎
𝑎 ] [ 0 ] = [𝐼𝑎 ]
3 𝐼
𝑎2 0
𝑎
Therefore, it is seen that the sequence components of the current are all equal to each other and
equal to one-third of the fault current, Ia.
1
Ia0 = Ia1 = Ia2 = 3 𝐼𝑎
The voltage and current of the principle phase ‘a’ is given as
𝑉𝑎 = 𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2 = 0
𝐼𝑎 = 𝐼𝑎0 + 𝐼𝑎1 + 𝐼𝑎2
It can be shown that the fault current, Ia , is found by connecting the equivalent sequence
networks [2] in series and solving the following equation:
𝐼𝑎 = 3
𝐸𝑎1
𝑍1 + 𝑍2 + 𝑍0
Line-to-Line Fault
A line-to-line fault involves a short circuit between two phase conductors that are assumed to be
phases 'b' and 'c'. Assuming prefault currents are negligible, the voltage and current relationships
describing the line-to-line fault are given by:
Vb = Vc
Ib = -Ic, and Ia=0
This is illustrated in the following figure:
+
Va
Vb=Vc
-
Figure 4: Line-to-Line Fault
The sequence components for this case are:
𝐼𝑎0
1 1 1
[𝐼𝑎1 ] = [1 𝑎
3
𝐼𝑎2
1 𝑎2
1 𝐼𝑎
1 1
2 ] [𝐼 ] = [
𝑎
1
𝑏
3
𝑎 𝐼𝑐
1
1
𝑎
𝑎2
0
0
0
1
1
1
2 ] [ 𝐼 ] = [𝐼 (𝑎 − 𝑎 2 )] = [ 𝑗𝐼 √3 ]
𝑏
𝑎
𝑏
𝑏
3
3
𝐼𝑏 (𝑎2 − 𝑎)
𝑎 −𝐼𝑏
−𝑗𝐼𝑏 √3
The negative-sequence component of the current is equal to the negative of the positive sequence
component and there is no zero sequence component:
Ia2 = -Ia1
Ia0=0
The voltage of the shorted phase 'b' and 'c', may be expressed in terms of the sequence
components:
Vb = Va0 + a2 Va1 + aVa2
Vc = Va0 + a2 Va2 + aVa1
Solving these equations, one can show that Va1 = Va2. Using the sequence networks [2], and
computing Ia0, Ia1 and Ia2, the fault currents may be shown to be equal to:
𝐼𝑏 = −𝐼𝑐 = −𝑗𝐼𝑎1 √3
Double Line-to-Ground Fault
A double line-to-ground (2LG) fault involves a short circuit between two phase conductors 'b'
and 'c' and ground. This is illustrated in the following figure:
Va
Vb
Vc
Figure 5: Double Line-to-Ground Fault
For a double line-to-ground fault with prefault current assumed negligible, the voltage and
current relationships are:
Vb = Vc= 0
Ia=0
The sequence components for the phase voltages are given as:
𝑉𝑎0
1 1 1
[𝑉𝑎1 ] = [1 𝑎
3
𝑉𝑎2
1 𝑎2
1 𝑉𝑎
1 𝑉𝑎
2 ] [ ] = [𝑉 ]
𝑎
0
𝑎
3 𝑉
𝑎 0
𝑎
The sequence components of the voltage at the fault are all equal:
Va1 = Va2 = Va0 =
1
V
3 a
Neglecting prefault load currents, the current in phase 'a' is:
Ia1 + Ia2 + Ia0 =
1
I =0
3 a
The sequence networks are to be interconnected in parallel [2] for a double line-to-line ground
fault. This results in a current divider network where Ia1 is divided up into the equivalent positive
and negative sequence networks. The first step is to compute Ia1.
𝐼𝑎1 =
𝐸𝑎1
𝑍0 +𝑍1 ||𝑍2
And then using current division, we have
𝐼𝑎2 = −𝐼𝑎1
𝑍2
𝑍1
𝑎𝑛𝑑 𝐼𝑎0 = −𝐼𝑎1
𝑍1 +𝑍2
𝑍1 +𝑍2
Given the current sequence components, it is straight forward to compute the fault currents, Ib
and Ic using the matrix equations given earlier in this paper.
Three-Phase Fault
The last case considered is a three-phase fault which is a symmetrical fault. We may still use the
method of symmetrical components to analyze this situation. This is illustrated in the following
figure:
+
Va
+
Vb
+
Vc
Figure 6: Three-Phase Fault
Assuming the prefault currents are negligible, the voltage and current relationships describing
the fault are:
Va = Vb = Vc
Ia + Ib + Ic = 3Ia0 = 0
There is no zero-sequence current for a 3-phase fault. The positive and negative sequence
components of voltage at the fault location are:
1
1
Va1 = (Va + aVb + a2 Vc ) = (1 + a + a2 )Va = 0
3
3
1
1
Va2 = (Va + aVc + a2 Vb ) = (1 + a + a2 )Va = 0
3
3
The fault current is simply the positive sequence component, Ia = Ia1.
Software
This section describes the software that was developed during the course of this research project.
The software allows the user to generate a 3-phase voltage. The user can then click a button to
simulate a particular type of fault. Symmetrical Components are then computed from which a
resulting fault current is computed.
The figure below shows the form that was developed to calculate fault currents for a 3-phase
system using Symmetrical Components. The “Create 3-Phase” button allows users to generate a
balanced 3-phase voltage. As soon as the user clicks the “Create 3-Phase” button, the magnitude
and phase of both voltage and current can be found in the lower left corner even though they are
not being displayed. The balanced 3-phases are labeled as A, B and C, respectively. There are
four check boxes that allow the user to select which set of phasors are rendered on the plot. The
following figure shows a balanced three phase system operating under normal conditions.
Phase B
Create 3-phase
Phase A
Faults
Phase C
Line-to-Neutral Voltages
Line Currents
Figure 7: Symmetrical Components Form
The general procedure for computing a fault current is as follows:
1. Compute the current sequence components, Ia0, Ia1, and Ia2 from the sequence networks.
2. Compute the voltage sequence components, Va0, Va1 and Va2 from the sequence networks
and the values computed in 1.
3. Compute the phase currents, Ia, Ib, and Ic , from the matrix equations given earlier in this
paper.
4. Compute the phase voltages, Va, Vb, and Vc, from the matrix equations given earlier in
this paper.
5. Determine the fault currents from 3.
Single Line-to-Ground Fault
The following figure shows the line-to-neutral voltage and phase currents when the user clicks
on the “SLG Fault” button.
Single Line-to-Ground Fault
Fault Current
Figure 8: Single Line-to-Ground Fault
The fault current in this case, is Ia, which is 3.529 p.u. at -90 degrees. This means that if the
nominal current of the line is 100 Amps, then the fault current will be 3.529 times this value or
approximately 352.9 Amps. It is this value that is used to design protection into the power
system using protective relays and it is this value that will determine the trip settings on these
relays and breakers. When comparing the phase voltages with figure 7, before the fault, we had a
balanced 3-phase system but after the fault, Va = 0.0 and the “b’ and “c” phases have reduced
values and are at +128 and -128 degrees, respectively.
To clear the fault and reset the system so that we have a balanced three-0system, we simply click
on the “Clear 3-Phase” button and that will return the system to what is displayed in figure 7.
Double Line-to-Line Fault
The following figure shows the line-to-neutral voltage and phase currents when the user clicks
on the “LL Fault” button.
Double Line-to-Line Fault
Fault Current
Figure 9: Double Line-to-Line Fault
The fault current here is Ib and Ic which are equal but negative of each other. The line-to-neutral
voltage is behind the plot of the phase current but can be read from the text boxes in the lower
left corner, Van=08, Vbn=0.4, and Vcn=0.4. Under normal conditions, these voltages should all be
at 1.0 p.u. and 120 degrees out of phase. Here we see that all phases have been reduced and “b”
and “c” reduced more than phase “a” and the phase relationship is no longer 120 degrees. Phase
“b” and “c” are now in phase which makes sense since it is a line-to-line fault.
Double Line-To-Ground Fault
The following figure shows the line-to-neutral voltage and phase currents when the user clicks
on the “2LineGND” button.
Double Line-to-Ground Fault
Fault Current
Figure 10: Double Line-to-Ground Fault
Similar to the previous case, the fault current here is Ib and Ic which are equal in magnitude but
have different phases. The line-to-neutral voltage for phase “b” and “c” are both zero which is
expected since both of these phases have been shorted to ground. Phase “a” has been reduced
from 1.0 p.u to 0.894 p.u. Under normal conditions, these voltages should all be at 1.0 p.u. and
120 degrees out of phase.
Three-Phase Fault
The following figure shows the line-to-neutral voltage and phase currents when the user clicks
on the “3-Phase Fault” button.
Three-Phase Fault
Fault Current
Figure 11: Three-Phase Fault
Here we have a three-phase fault, which is a symmetrical fault. The fact that it is a symmetrical
fault means that we do not need to use Symmetrical Components to analyze the circuit but rather
we can view the 3-phase system as a single-phase system and use conventional circuit theory
techniques for a single phase system. None the less, we should get the same results if we choose
to use the method of Symmetrical Components. Here we observe voltage collapse as all the lineto neutral voltages are zero and all the phase currents are equal still maintaining a 120 phase
relationship which is expected since the fault is symmetrical. The fault current on all three lines
is equal to 3.333 p.u.
Future Work and Integration into Courses
This work assumed a 3-phase generator operating at no load and that the fault occurred on an
open circuit transmission line. The next step would be to allow user to connect a 3-phase load to
the generator as well as various elements such as step-up and step-down transformers.
Also, the course, EENG 450: Power Systems Analysis, is the course that teaches symmetrical
components. At present, we are currently inserting screens shots from this program into the class
lecture notes. We are also designing weekly homework assignments that require the student to
use various features in PowerX to complete the assignment. Next quarter will be the first time
that the students will have a symmetrical component tool available to them. Hopefully, this will
further their understanding of this topic which is typically difficult to master during the students
first exposure.
References
[1] Fortescue, C.L., ”Method of Symmetrical Co-ordinates Applied to the solution of Polyphase Networks ”. AIEE
Transactions, Vol.37, Part ІІ, Page 1027-1140 (1918).
[2] Yamayee and Bala, “Electromechanical Energy Devices and Power Systems”, Wiley, 1993.