INTERNATIONAL BURCH UNIVERSITY
FACULTY OF ENGINEERING AND NATURAL SCIENCES
ELECTRICAL AND ELECTRONICS ENGINEERING
POWER SYSTEM ANALYSIS
FINAL PROJECT
Name and Surname: Adis Dedić
Student ID: 20002395
Date:16.01.2023
Score:_________
Table of Contents
1. INTRODUCTION............................................................................................................................3
1.1. SYMETRICAL COMPONENTS.............................................................................................4
1.2. THREE-PHASE LINE-TO-GROUND FAULT.......................................................................6
1.3. POWER FLOW........................................................................................................................9
1.3.1. NEWTON RAPSON METHOD........................................................................................9
1.4. TASK AND SYSTEM DESCRIPTION.................................................................................10
2. PROCEDURES..............................................................................................................................13
2.1. PSAT MODEL........................................................................................................................13
2.1.1. COMPONENTS PARAMETERS....................................................................................14
2.2. SIMULINK MODEL..............................................................................................................18
3. RESULTS.......................................................................................................................................23
3.1. POWER FLOW RESULTS....................................................................................................23
3.1.1. POWER FLOW REPORT................................................................................................26
3.2. FAULT EFFECTS...................................................................................................................29
3.2.1. FAULT VOLTAGES AND CURRENTS.........................................................................30
4. CONCLUSION..............................................................................................................................33
5. REFERENCES...............................................................................................................................34
2
1. INTRODUCTION
When excessively high currents pass through an abnormal path as a result of partial or complete
insulation failure at one or more locations in the system, a system that is otherwise running
normally may be disturbed or disrupted. Short circuit or fault are terms used to describe total
insulation failure.
When energized conductors come into contact with one another or the ground, short circuits can
happen. A fault causes currents to flow through the system to the point of the fault at an abnormally
high magnitude. The overhead transmission lines in power systems are one of the weakest spots
since they are primarily made of bare conductors. These overhead transmission lines experience a
high number of faults. Other possible causes of faults include wind, sleet, conductor clashes brought
on by conductor galloping, small animals (like snakes, birds), trees, cranes, airplanes, vandalism,
collisions of vehicles with poles or towers, line breaks brought on by excessive ice loading, or other
harm to supporting structures.
Typically, there are two types of faults: permanent and transient. The equipment is harmed by
persistent flaws. Because enough energy goes into the short circuit in a brief period of time to
generate an explosion, the destruction to the equipment is typically violent. Conductors can be
burned or melted as a result of such a continuous fault. Fortunately, the majority of failures on
overhead transmission lines are transient or ephemeral in nature. As a result, the service can be
quickly restored by quickly isolating and then re-closing the problematic line part.
The goal of the fault analysis, also known as the short-circuit study or analysis, is to determine the
maximum and minimum fault currents and voltages at various points throughout the power system
for various fault types in order to choose the proper protective schemes, relays, and circuit breakers
in order to quickly restore the system to normal operation.
3
1.1. SYMETRICAL COMPONENTS
Figure 1 illustrates the solution of any imbalanced three-phase system of phasors into three
balanced systems of phasors: (1) positive-sequence system, (2) negative-sequence system, and (3)
zero-sequence system.
Figure 1: Analysis and synthesis of set of three unbalanced voltage phasors: (a) original system
of unbalanced phasors; (b) positive-sequence components; (c) negative-sequence components; (d)
zero-sequence components; (e) graphical addition of phasors to get original unbalanced phasors.
4
A balanced system of phasors with the same phase sequence as the original unbalanced system
serves as a representation for the positive-sequence system. The positive-sequence system's
phasors have equal magnitudes and are 120 degrees apart.
A balanced system of phasors with the original system's phase sequence reversed serves as the
representation for the negative-sequence system. The negative-sequence system's phasors are also
120 degrees apart and equal in magnitude.
Three single phasors that are equal in magnitude and angular displacements are used to represent
the zero-sequence system.
It is necessary to have a unit phasor (or operator) that will rotate another phasor by 120 degrees
counterclockwise while maintaining its magnitude when multiplied by the phasor in order to
apply the symmetrical components theory to three-phase systems. Such an operator is represented
by a complex number of unit magnitude and an angle of 120 degrees, and is defined by:
a=1∠120°
5
1.2. THREE-PHASE LINE-TO-GROUND FAULT
Generally speaking, the three-phase fault is symmetrical and not imbalanced. The three-phase fault,
on the other hand, is a balanced (i.e., symmetrical) fault that might also be studied using symmetric
components. A balanced three-phase fault at fault location F with impedances Zf and Zg is depicted
generally in Figure 2a.
Figure 2: Three-phase fault: (a) general representation; (b) interconnection of sequence
networks
Figure 2b demonstrates how the generated sequence networks are not interconnected. Instead, the
sequence networks are separated from one another because they are short-circuited over their own
fault impedances. The positive-, negative-, and zero-sequence currents can be described as follows
since only the positive-sequence network is thought to have an internal voltage source.
If the fault impedance Zf is zero, then
Substituting Equations into equation,
6
and we get
Sequence networks short-circuit over their own fault impedances as a result.
Substituting into equation
and,
Hence, the line-to-line voltages become
7
You should take note that if Zf = 0,
and
8
1.3. POWER FLOW
The answer to an electric power system's typical balanced three-phase steady-state operating
circumstances is power flow. Power flow calculations are typically done in connection with system
operation and control, power system planning, and operational planning. In order to study the
normal operating mode, contingency analysis, outage security assessment, and optimal dispatching
and stability, data from power flow studies is used. Remember that typically a "contingency" is
believed to be the loss of a significant transmission component or a sizable producing unit.
Before 1929, all calculations involving power flow (or load flow, as it was known at the time) were
done manually. Power flow estimates were made in 1929 using network calculators or network
analyzers made by Westinghouse. Most of the early iterative methods were based on the Y-matrix of
the Gauss-Seidel method when the first publication presenting the first digital method to solve the
power flow problem was published in 1954. It needed the least number of iterations for small
networks and the least amount of computer storage.
The Newton-Raphson approach was created as a result of new power flow studies and the declining
reliability of the old methods as network size increased.
1.3.1. NEWTON RAPSON METHOD
The Newton-Raphson approach uses the Newton-Raphson algorithm to resolve the power network's
simultaneous quadratic equations. It requires more time each iteration than the Gauss-Seidel
algorithm, but only a few, and is significantly network size independent.
It is now the approach for solving non-linear algebraic equations that is most frequently utilized.
Based on an approximate initial state and the application of the Taylor series expression, the
Newton-Raphson approach. With each consecutive iteration, assuming the original answer is some
constant, the solution will become more accurate and closer to the true value.
It is customary when solving manually to go through as many iterations as necessary until there is
little to no difference between the results of the current and preceding iteration. Nowadays,
MATLAB is used for the majority of the calculations, while manual solutions are rarely used. Since
MATLAB can complete many iterations quickly, it can produce very accurate solutions.
The fundamental idea behind the Newton-Raphson approach is that it starts by identifying the
function's tangent line at the initial guess point. The next step, known as iteration, involves
observing the point where the tangent line intersects the x-axis and using that value as our new
guess values. For each iteration, the same process is carried out until a certain condition is reached.
9
1.4. TASK AND SYSTEM DESCRIPTION
Prepare a formal, final report for the power system analysis project. All figures should have figure
numbers and captions and tables should have table numbers and captions. The report should be
organized as follows:
Introduction:
Describe the system being studied and the studies that were to be performed using PSAT and
Simulink.
Procedures:
Describe what you did for the power flow study and for the fault analysis study. Describe how you
prepared your data for entry into PSAT, and Simulink and include your data tables.
Results:
Provide results from the power flow study of the base system with figures and tables. Provide
results of the fault studies.
Summary and Conclusions:
Summarize and conlcude your report. Describe your confidence in your simulation of the power
system.
Project 10 – Bus 8 - Three-phase to ground fault
Figure 3
10
The slack is on bus number 3, base power 100 MVA, Base voltage 230 kV.
Nominal voltage of generators (SM1, SM2, SM3) and primary voltage of transformers
(T1,T2,T3) are 24 kV, 18 kV, 13.8 kV, not like in the table.
System has 3 generators, 3 transformers, 9 buses, 6 transmission lines and 3 loads. It has base
power of 100 MVA and base voltage of 230 kV.
Generator 1 has 512 MVA nominal power and 16.5 kV nominal voltage. Generator 2 has 270 MVA
nominal power and 18 kV nominal voltage. Generator 3 has 125 MVA nominal power and 13.8 kV
nominal voltage.
Transformer 1 has 24 kV, Transformer 2 has 18 kV and Transformer 3 has 13.8 kV nominal primary
voltage. Transformers have 230 kV nominal secondary voltage.
Load at bus 5 has 125 MW nominal active power and 50 MVA r nominal reactive power. Load
connected to bus 6 has 90 MW nominal active power and 30 MVA r nominal reactive power. Load at
bus 8 has 100 MW nominal active power and 35 MVAr nominal reactive power.
Line parameters for modeling system are shown in given Table 4 CP line parameters:
Bus 1 and bus 2 are generator buses in the system, whereas bus 3 is a slack bus. The load buses are
bus numbers 5, 6, and 8. To balance the system's active and reactive power, a slack bus is used.
Slack bus can also be referred to as reference bus or swing bus. All other buses in the system will
utilize the slack bus as an angle reference, which is set to 0°. (degrees).
11
Main components in system:
1. Generator is device which converts motive power (mechanical energy) into electrical power for
use in external circuit.
2. Transformer is static electrical machine which transfers C electrical power from one circuit to
other circuit at constant frequency, but voltage level depends on altering, i.e. increasing or
decreasing according needs.
3. Bus is any node of single-line diagram at which voltage, current, power flow or other quantities
are to be evaluated. This may correspond to physical busbars in substation.
4. Transmission line is specialized cable or other structure designed to conduct electromagnetic
waves in contained manner. Term applies when conductors are long enough that wave nature of
transmission must be taken into consideration.
5. Electrical load is electrical component or portion of circuit that consumes (active) electric
power. This is opposed to power source, such as battery or generator, which produces power. In
electric power, circuit examples of load are appliances and lights.
12
2. PROCEDURES
2.1. PSAT MODEL
To create model of network in PSAT, we will use following components from PSAT’s Simulink
Library:
1)
2)
3)
4)
1) PV bus for load flow studies
2) Generator (defining 4th order synchronous machine)
3) Transformer
4) Slack bus (V-theta bus)
13
5)
6)
7)
5) Bus
6) Constant power load
7) π -model for 3-phase line
In end, when we connect all components, model looks like as it is shown in Figure 4 below:
Figure 4
2.1.1. COMPONENTS PARAMETERS
And input all the parameters from the table in the schematic components. Firs the generator
Figure 5 generator 1
14
generator 2
generator 3
Then the slack and two PV elements connected with the generators as well.
Figure 6 Slack I PV
15
After the generator parameters were set, next we needed to adjust the transformer parameters.
Figure 7 Transformers
16
To get the parameter for the line resistance we need to convert them into p.u values of resistance,
reactance and susceptance for each line. With the formula
Z base=
V base ² 230 kV ²
=
=529 [ohm ]
S base 100 MVA
The per-unit values for the line parameters are now calculated to the following expressions:
R pu=
X pu =
R
Z base
∗L
X
∗L
Z base
Y pu=Y ∗Z base∗L
I am just showing one line from 4 to 5, because it is the same calculation for each of them
Figure 8 Line
17
2.2. SIMULINK MODEL
The main elements that were used to model the system in MATLAB Simulink are shown in Figure
9. The model consists of three generators, three transformers, six pi-section lines, nine busses and
three loads.
Figure 9: Main elements used for the Simulink model: (a) Generator; (b) Transformer; (c)
Pi-Section Line; (d) Tree- Phase Parallel RLC Load; (e) Bus; (f) Load Flow Bus;
Upon connecting all the elements according to the schematic diagram, the final Simulink model
looked like shown in Figure 10 below:
18
Figure 10: Simulink model of the given system
In order to model the fault that occurs at BUS 8, a three-phase breaker was used along with a
resistive element that represents the fault resistance. Also, we are doing current and voltage
measurements as shown in the figure bellow:
Figure 11: Fault breaker and Fault Resistance on BUS 5
19
The breaker connecting all three phases with ground is a three-phase type since the fault is
programmed to affect all three phases with ground and it is a three-phase line-to-ground fault. Rf is
initially set at 0.001Ω , which is quite near to 0Ω , but other values and their consequences must
also be studied.
The next step was to enter the parameters for each element.
The Phase-to-Phase rms voltage, the frequency (set to 50 Hz for the entire system), and the
generator type were the most crucial settings for the generators. In contrast to Generators 1 and 2,
which are PV generators, Generator 3 is a swing type generator because it is connected to the slack
bus. The block window configurations for each of the three generators are shown in Figure 21.
The source resistance and inductance values were left at their default settings. It was crucial to put
the power numbers for Generators 1 and 2, which are PV generators, under the Load Flow Tab as
well. 512 MW for the first generator and 270 MW for the second generator were the values
specified for the active power. Both generators' minimum and maximum reactive powers were
programmed to be -inf and inf, respectively.
Figure 12: Generator configurations
Next, we had to configure the transformer elements in a similar manner. All three transformers have
a Yg connection at both windings. The other parameters are set to values as shown in the following
figures:
20
Figure 13: Transformer configurations
The line parameters were all obtained from values given in Table 4. The block parameters that
were used in the Simulink model are frequency, the positive- and zero-sequence values for
resistance, inductance and capacitance and the length of each line.
Figure 14: Example of block parameters for the 7-8 pi-section line
21
The loads at BUS 5, BUS 6, and BUS 8 were our final stops. The nominal phase-to-phase voltages,
the nominal frequency, and the active and reactive power values of the tree loads were entered. The
loads were specified to be constant PQ loads. The values entered are depicted in the following
figures:
Figure 15: Load parameters
22
3. RESULTS
3.1. POWER FLOW RESULTS
We'll utilize the PSAT model to calculate the system's power flow. PSAT toolbox is opened by
running >>psat in the MATLAB Command Window. We import our model as a ".mdl" file when
PSAT Window appears, as demonstrated in Figure 16 below:
Figure 16 PSAT setup
after this we click the power flow button an analyse the results.
23
a)
b)
Figure 17: Static report a) in p.u b) in units
All the voltages are around one 0.9-1.1 in P.U that means that the model is correct
24
Now I will also post the real power, reactive power , voltages, angles
Voltage Magnitude Profile
1.2
Voltage Magnitude Profile
250
1
200
0.8
V [kV]
V [p.u.]
150
0.6
100
0.4
50
0.2
0
1
2
3
4
5
6
7
8
0
9
1
2
3
4
5
6
7
8
9
Bus #
Bus #
Figure 18 Voltages
Real Power Profile
5
Real Power Profile
500
4
400
3
300
200
1
PG - PL [MW]
PG - PL [p.u.]
2
0
100
0
-1
-100
-2
-200
-3
1
2
3
4
5
6
7
8
9
Bus #
-300
1
2
3
4
5
6
7
8
9
Bus #
Figure 19 Real power
Reactive Power Profile
3
Reactive Power Profile
300
2
200
1
Q G - Q L [MVar]
QG - QL [p.u.]
100
0
0
-1
-100
-2
-3
-200
1
2
3
4
5
Bus #
6
7
8
9
-300
1
3
4
5
Bus #
Figure 20 Reactive power
25
2
6
7
8
9
3.1.1. POWER FLOW REPORT
P S A T 2.1.11
Author: Federico Milano, (c) 2002-2019
e-mail: federico.milano@ucd.ie
website: faraday1.ucd.ie/psat.html
File: /home/adis/Documents/Fax/Semestar 5/PSA/LABS/LAB4/Adis/LAB4_adis_V3_mdl
Date: 16-Jan-2023 20:08:02
NETWORK STATISTICS
Buses:
Lines:
Transformers:
Generators:
Loads:
9
6
3
3
3
SOLUTION STATISTICS
Number of Iterations:
4
Maximum P mismatch [p.u.] 0
Maximum Q mismatch [p.u.] 0
Power rate [MVA]
100
POWER FLOW RESULTS
Bus
Bus1
Bus2
Bus3
Bus4
Bus5
Bus6
Bus7
Bus8
Bus9
V
[p.u.]
phase
P gen
Q gen
P load
Q load
[rad]
[p.u.]
[p.u.]
[p.u.]
[p.u.]
1
0.18417 4.096
-2.7694
1
0.1064
2.16
1.7648
1
0
-2.5227
2.7238
1.0158
0.16148 0
0
0.99719 0.14635 0
0
0.9804
0.13676 0
0
0.97986 0.08085 0
0
0.95752 0.07257 0
0
0.93802 0.06308 0
0
STATE VARIABLES
delta_Syn_1
omega_Syn_1
e1q_Syn_1
e1d_Syn_1
e2q_Syn_1
e2d_Syn_1
26
1.6714
1
0.29103
0.71436
0.23629
0.87837
0
0
0
0
1.25
0.9
0
1
0
0
0
0
0
0.5
0.3
0
0.35
0
delta_Syn_2
omega_Syn_2
e1q_Syn_2
e1d_Syn_2
e2q_Syn_2
e2d_Syn_2
delta_Syn_3
omega_Syn_3
e1q_Syn_3
e1d_Syn_3
e2q_Syn_3
e2d_Syn_3
0.66757
1
1.0999
0.45254
1.0283
0.47279
-0.58866
1
1.3488
-0.39438
1.2228
-0.48073
OTHER ALGEBRAIC VARIABLES
vf_Syn_1
pm_Syn_1
p_Syn_1
q_Syn_1
vf_Syn_2
pm_Syn_2
p_Syn_2
q_Syn_2
vf_Syn_3
pm_Syn_3
p_Syn_3
q_Syn_3
1.3644
4.1151
4.096
-2.7694
2.2155
2.1646
2.16
1.7648
3.3769
-2.4786
-2.5227
2.7238
LINE FLOWS
From Bus To Bus
Bus5
Bus6
Bus7
Bus9
Bus7
Bus8
Bus1
Bus2
Bus3
Bus4
Bus4
Bus5
Bus6
Bus8
Bus9
Bus4
Bus7
Bus9
Line
[p.u.]
1
2
3
4
5
6
7
8
9
P Flow
Q Flow P Loss
[p.u.]
[p.u.]
[p.u.]
-1.8821
-2.0463
-0.49179
-0.97275
2.6518
1.5887
4.096
2.16
-2.5227
Q Loss
1.4903
0.05794 0.00084
1.4185
0.10967 0.00116
1.9914
0.14032 0.00108
1.7197
0.17353 0.00113
-0.31685 0.06311 0.01386
-0.68071 0.03875 0.00038
-2.7694
0
0.13751
1.7648
0
0.09019
2.7238
0
0.32308
LINE FLOWS
From Bus To Bus
Bus4
Bus4
27
Bus5
Bus6
Line
[p.u.]
1
2
P Flow
Q Flow P Loss
[p.u.]
[p.u.]
[p.u.]
1.9401
2.156
-1.4895
-1.4174
0.05794
0.10967
Q Loss
0.00084
0.00116
Bus5
Bus6
Bus8
Bus9
Bus4
Bus7
Bus9
Bus7
Bus9
Bus7
Bus8
Bus1
Bus2
Bus3
3
4
5
6
7
8
9
0.63211
1.1463
-2.5887
-1.5499
-4.096
-2.16
2.5227
-1.9903
0.14032 0.00108
-1.7185
0.17353 0.00113
0.33071 0.06311 0.01386
0.68109 0.03875 0.00038
2.9069
0
0.13751
-1.6746
0
0.09019
-2.4008
0
0.32308
GLOBAL SUMMARY REPORT
TOTAL GENERATION
REAL POWER [p.u.]
3.7333
REACTIVE POWER [p.u.]
1.7192
TOTAL LOAD
REAL POWER [p.u.]
3.15
REACTIVE POWER [p.u.]
1.15
TOTAL LOSSES
REAL POWER [p.u.]
0.58332
REACTIVE POWER [p.u.]
0.56923
28
3.2. FAULT EFFECTS
We will now look into how the three-phase line-to-ground fault has affected the system.
We will examine several sites in the system and examine graphical values of voltages and currents
before and after the fault occurs in order to see the effects such a malfunction has on the system.
Inserted voltage measurement blocks between all phases and current measurement blocks in each of
the three phases to monitor the voltages and currents. The following diagram displays those blocks:
Figure 21: Voltage and current measurements
The fault current and fault voltage are the most significant numbers when examining a three-phase
line-to-ground fault. As previously noted, a three-phase breaker with a fault resistance Rf in series
will be used to simulate the failure. The ground is then linked to those two components. Since it is
typically ignored when investigating the 3LG fault, the initial value of Rf that will be analyzed will
be very small, almost nil. In the future, greater values will also be examined to observe how the
fault current functions. The simulation was run in three different scenarios, with the following
outcomes:
29
3.2.1. FAULT VOLTAGES AND CURRENTS
Figure 22: Fault voltage for Rf = 0.001Ω
Figure 23: Fault current for Rf = 0.001Ω
30
Figure 24: Fault voltage for Rf = 0.1Ω
Figure 25: Fault current for Rf = 0.1Ω
31
Figure 26: Fault voltage for Rf = 1Ω
Figure 27: Fault current for Rf = 1Ω
32
4. CONCLUSION
Our study must conclude with an analysis of the power flow and fault currents/voltages data. If
the bus voltages of the system for power flow analysis are between 0,9 p.u. and 1,1 p.u., it will
function successfully. According to the static data, the bus voltages in our situation range from
0.94 p.u. to 1.02 p.u. It is clear from the power flow calculation findings that there is no issue and
that the system is operating as it should.
The fault currents in all three phases that were abnormally high after we generated the fault were
then examined in relation to how the work would typically operate. Additionally, we can see how
Rf affects the fault current from the plots. Higher (larger than 0) Rf resistances result in reduced
ripple and fault current values in moments that are always measured from the time the fault first
occurred. Additionally, we can see that currents spike to an excessively high amount at the exact
moment when a failure occurs.
After inducing the fault, we looked at the fault currents in all three phases that were abnormally
high compared to how things would normally operate. Additionally, we can see from the graphs
how Rf affects the fault current. Greater (larger than 0) values of Rf result in smaller ripple and a
higher value for the fault current in moments that are always from the time the fault first
occurred. We may also see that currents go "sky-high" to an excessively high number at the exact
moment a fault occurs.
Finally, we noticed variations in the current and voltage levels at different buses. We can see that,
even for smaller Rf values, the system exhibits very slight variances. Additionally, we can see
that the voltages in all three phases are rippled and that the current is slightly increasing. After
the fault and right as it happens, the ripple is largest.
33
5. REFERENCES
PSA Project Report Lejla Imširović Sarajevo, 2019
34