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LABORATORY MANUAL
EE0405 SIMULATION LAB
PREPARED BY
J.PREETHA ROSELYN
(AP/Sr.G/EEE)
DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING
FACULTY OF ENGINEERING & TECHNOLOGY
SRM UNIVERSITY, Kattankulathur 603 203
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LIST OF EXPERIMENTS
S.No. Name of the Experiments Page No.
1 Single phase half controlled converter using R and RL load
using MATLAB / SIMULINK
2 Single phase fully controlled converter using R and RL load
using MATLAB / SIMULINK
3 Three phase fully controlled converter using R and RL load
using MATLAB / SIMULINK
4 Single phase AC voltage regulator using MATLAB /SIMULINK
5 Formation of Y bus matrix by inspection / analytical method
using MATLAB Software
6 Formation of Z bus using building algorithm using MATLAB
Software
7 Gauss Seidal load flow analysis using MATLAB Software
8 Newton Raphson method of load flow analysis using
MATLAB Software
9 Fast decoupled load flow analysis using MATLAB Software
10 Fault analysis using MATLAB Software
11 Economic dispatch using MATLAB Software
12 Load flow analysis using ETAP Software
13 Fault analysis using MIPOWER Software
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TABLE OF CONTENTS
1. Syllabus
2. Mapping of Program Outcomes with Instructional Objectives
3. Mapping of Program Educational Objectives with Program Outcomes
4. Session plan
5. Laboratory policies & Report format.
6. Evaluation sheet
7. Each experiment should be prefixed with prelab questions
with answer key and suffixed with post lab questions with
answer key.
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Syllabus
EE 0405 SIMULATION LAB L T P C
Prerequisite 0 0 3 2
EE 0302,EE 0308
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PURPOSE
To enable the students gain a fair knowledge on the programming and simulation of PowerElectronics and Power Systems.
INSTRUCTIONAL OBJECTIVESAt the end of course the students will be able to:
1. Acquire skills of using computer packages MATLAB coding and SIMULINK in powerelectronics and power system studies.
2. Acquire skills of using ETAP software for power system studies.
LIST OF EXPERIMENTS
1) Use of MATLAB for the following
1. Single phase half controlled converter with R and RL load.2. Single phase fully controlled converter with R and RL load
3. Three phase fully controlled converter with R and RL load.4. Single phase AC voltage controller with R and RL load.
2) Use of MATLAB coding for solving the following
1. Formation of YBus by inspection method/analytical method.2. Formation of ZBus matrix.
3. Load flow analysis for GS, NR and FDLF methods
3) Use of ETAP software for the following
1. Load flow solution for GS, NR and FDLF
2. Symmetrical and unsymmetrical fault analysis3. Transient stability analysis
TOTAL
REFERENCE
Laboratory Manual
Course designed by Department of Electrical and Electronics Engineering
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Programoutcomes
a B c d e f g h i j k
X X X X X
Category
General
(G)
Basic
Sciences
(B)
Engineering
Sciences
and
TechnicalArts(E)
Professional
Subjects(P)
X
Broad area (for Pcategory)
Electrical
Machines
Circuits
and
Systems
Electronics Power System
X X
Staff responsible for preparing the
syllabus
Mr.K.Vijayakumar
Date of preparation December 2006
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Mapping of Course Outcomes
with Instructional Objectives
Mapping of Program Instructional Objectives Vs Program Outcomes
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Program Outcomes
Program Instructional objectives
Acquire skillsof using
computerpackages
MATLABcoding inPower
System
studies
Acquire skills of using computerpackages MATLAB /SIMULINK
in Power Electronics studies.
Acquire skills ofusing ETAP software
for Power Systemstudies
a)An ability to applyknowledge of
mathematics, science,and engineering.
X X X
b) An ability to design
and conductexperiments, as well as
to analyze and interpret
results.
X X X
c)An ability to design a
system, component, orprocess to meet desired
needs within realistic
constraints such aseconomic,environment
al,social, political,
ethical, health and
safety,manufacturability, and
sustainability.
X X
e)An ability to identify,
formulate, and solve
engineering problems
X X
h)The broad education
necessary to understandthe impact of
engineering solutionsin a global perspective
X X X
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Mapping of Program
Educational Objectives withProgram Outcomes
Mapping of Program Educational Objectives Vs Program Outcomes
PROGRAM EDUCATIONAL OBJECTIVES
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1. Graduates are equipped with the fundamental knowledge of Mathematics, Basic sciences
and Electrical and Electronics Engineering.
2. Graduates learn and adapt themselves to the constantly evolving technology by pursuing
higher studies.
3. Graduates are better employable and achieve success in their chosen areas of Electrical
and Electronics Engineering and related fields.
4. Graduates are good leaders and managers by effectively communicating at both technical
and interpersonal levels.
The student outcomes are linked with the program educational objectives as shown below:
PROGRAM OUTCOMES(ak OUTCOMES)
PROGRAM EDUCATIONALOBJECTIVES
1 2 3 4
(a) an ability to apply knowledge of
mathematics, science, and engineering X
(b) an ability to design and conductexperiments, as well as to analyze and
interpret data
X
(c) an ability to design a system, component,
or process to meet desired needs withinrealistic constraints such as economic,
environmental, social, political, ethical, health
and safety, manufacturability, and
sustainability
X
(d) an ability to function on multidisciplinary
teamsX X
(e) an ability to identify, formulate, and solveengineering problems
X
(f) an understanding of professional and
ethical responsibilityX
(g) an ability to communicate effectively inboth verbal and written form.
X
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(h) the broad education necessary to
understand the impact of engineering solutionsin a global perspective.
X
(i) a recognition of the need for, and an abilityto engage in life-long learning
X
(j) a knowledge of contemporary issues X
(k) an ability to use the techniques, skills, andmodern engineering tools necessary for
engineering practice.
X X
Academic Course Description
SRM University, KattankulathurFaculty of Engineering and Technology
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Department of Electrical and Electronics Engineering
COURSE : EE0405
TITLE : SIMULATION LAB
CREDIT : 02
LOCATION : ESB simulation lab
PREREQUISITES COURSES : EE0302-Power Electronics
EE0308-Power System Analysis
PREREQUISITIES BY TOPIC : Load flow studies, Fault analysis, Transient
stability analysis, Single phase and three
phase converters, AC voltage regulators.
Outcomes
Students who have successfully completed this course
Instructional Objective Program outcome
The students will be able to:
1. Acquire skills of using computer packages
MATLAB coding and SIMULINK in
Power Electronics and Power Systemstudies.
2. Acquire skills of using ETAP software for
Power System Studies.
a)An ability to apply knowledge of
mathematics, science, and engineering
b) An ability to design and conductexperiments, as well as to analyze and
interpret results.
c)An ability to design a system,component, or process to meet desired
needs within realistic constraints such
as economic,environmental,social,political, ethical, health and safety,
manufacturability, and sustainability.
e)An ability to identify, formulate, andsolve engineering problemsh)The broad education necessary to
understand the impact of engineering
solutions in a global perspective
Text book(s) and/or required materials:
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1. P.S.Bimbhra, Power Electronics
2. Nagrath and Kothari, Power System Analysis3. B.R.Gupta, Power System Analysis and Design
Web Resources:
www.power-analysis.com
www.4shared.com/powersystem analysis
www.power-electronics.com
Professional component:
General - 0%
Basic Sciences - 0%Engineering sciences & Technical arts - 0%
Professional subject - 100%
Session Plan:
WEEK NAME OF THE
EXPERIMENT
REFERENCE OBJECTIVE
I Single phase half controlled
converter using R and RL load
using MATLAB/ SIMULINK
Power electronics
P.S.Bimbhra
Acquire skills ofusing computer
packages MATLAB
/SIMULINK inpower electronics.
II Single phase fully controlled
converter using R and RL loadusing MATLAB/ SIMULINK
III Three phase fully controlled
converter using R and RL loadusing MATLAB/ SIMULINK
IV Single phase AC voltage
regulator using MATLAB/SIMULINK
V Formation of Y bus matrix byinspection/analytical method
using MATLAB Software
Power system analysis-
Nagrath and Kothari
Acquire skills of
using computerpackages usingMATLAB in
power systems.VI Formation of Zbus matrix using
building algorithm using
MATLAB Software
VII Gauss Seidal load flow analysisusing MATLAB Software
VIII Fast decoupled load flowanalysis using MATLAB
Software
IX Symmetrical Fault analysis
using MATLAB Software
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X Economic Dispatch usingMATLAB Software
XI Load flow analysis using ETAPSoftware
Acquire skills ofusing ETAP
software for power
system studiesXII Fault analysis using MIPOWER
Software
Acquire skills of
using MIPOWERsoftware for power
system studies.
EVALUATION METHOD:
Prelab Test - 5%
Inlab Performance - 35%
Postlab Test - 5%
Attendance - 5%
Record - 10%
Model Exam - 15%
Final Exam - 25%
Total - 100%
LABORATORY POLICIES AND REPORT FORMAT:
1. Lab reports should be submitted on A4 paper. Your report is a professionalpresentation of your work in the lab. Neatness, organization, and completeness will be
rewarded. Points will be deducted for any part that is not clear.
2. The lab reports will be written individually. Please use the following format for your labreports.
a. Cover Page: Include your name, Subject Code, Subject title, Name of
the university.
b. Evaluation Sheet: Gives your internal mark split up.
c. Index Sheet: Includes the name of all the experiments.d. Experiment documentation: It includes experiment name, date,
objective, flowchart, algorithm, formulae used, Model calculation,problem solution, simulated output and print outs.
e. Prelab and Postlab question should be written before and after
completing the experiments.
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3. Your work must be original and prepared independently. However, if you need any
guidance or have any questions or problems, please do not hesitate to approach your staff
in charge during office hours. The students should follow the dress code in the Labsession.
4. Labs will be graded as per the following grading policy:
Prelab Questions - 5%
Preparation of observation/Record 10%
Model Calculation - 10%
Execution - 15%
Postlab Questions - 5%
Attendance - 5%
Model Exam - 25%
University Exam - 25%
Total - 100%
5. Reports Due Dates: Reports should be submitted immediately after next week of the
experiment. A late lab report will have 20% of the points deducted for being one day late.
If a report is 3 days late, a grade of 0 will be assigned.
6. Systems of Tests: Regular laboratory class work over the full semester will carry a
weightage of 75%. The remaining 25% weightage will be given by conducting an end
semester practical examination for every individual student. Prelab questions will beasked at the beginning of each cycle as a viva-voce and the post lab questions should be
available in the observation and record after the completion of the experiment.
DEPT. OF ELECTRICAL & ELECTRONICS ENGINEERING
SRM UNIVERSITY, Kattankulathur 603203.
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Title of Experiment :
Name of the candidate :
Register Number :
Date of Experiment :
Date of submission :
S.No: Marks split up Maximum Marks
(50)
Marks Obtained
1 Attendance 5
2 Preparation of observation/record 10
3 Pre viva questions 5
4 Model Calculation 105 Execution 15
6 Post viva questions 5
TOTAL 50
Signature of the staff
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S.NO. 1
Single Phase Half Wave Rectifier with R & RL load
Aim:
To simulate the 1 half controlled rectifier circuit with R & RL load and obtain the
corresponding waveforms using MATLAB/SIMULINK.
Formulae used:
Average dc voltage, Vdc=Vm(1+cos) (volts) Rms output voltage,Vrms=Vm ((-)+sin2/2)
1/2(volts)
2
Average output current, Idc=Vdc/R (Amps)RMS output current, Irms=Vrms/R (Amps)Where,
Vm is the maximum input voltage
is the firing angle of the SCR.Operation:The phase controlled rectifiers using SCRs are used to obtain controlled dc output voltages from
the fixed ac mains input voltage. The circuit diagram of a half controlled converter is shown in
Figure 1. The output voltage is varied by controlling the firing angle of SCRs. The single phasehalf controlled converter consists of two SCRs and two diodes. During positive half cycle, SCR1
and Diode 2 are forward biased. Current flows through the load when SCR1 is triggered into
conduction. During negative half cycle, SCR3 and D1 are forward biased. If the load is resistive,the load voltage and load current are similar.
If the load is inductive, the current will continue to flow even when the supply voltage
reverses polarity due to the stored energy in the inductor. At the end of positive half cycle, D2 isreverse biased and D1 is forward biased. As SCR1 is not turned off the freewheeling current due
to the stored energy in the inductor will flow through the diode D1 and SCR1. When SCR3 is
triggered, the current gets transferred from SCR1 to SCR3. Load current now flows from supplyvia SCR3, load and D4. At the end of negative half cycle, the freewheeling current will flow
through the diode D2 and SCR3.
Circuit Diagram:
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Model Graph:
Resistive Load
Inductive load:
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Result:
Thus the Single Phase half controlled Rectifier with R & RL Load circuit is simulated usingMATLAB/SIMULINK and the corresponding waveforms are obtained.
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S.NO.2
Single Phase Full Wave Rectifier with R & RL Load
Aim:
To simulate the 1 fully Controlled rectifier circuit with R & RL load and obtain thecorresponding waveforms using MATLAB/SIMULINK.
Formulae used:
Average dc voltage, Vdc=Vm(1+cos) (volts) Rms output voltage,Vrms=Vm ((-)+sin2/2)
1/2(volts)
2Average output current, Idc=Vdc/R (Amps)
RMS output current, Irms=Vrms/R (Amps)Where,
Vm is the maximum input voltage is the firing angle of the SCR.
Operation:The phase controlled rectifiers using SCRs are used to obtain controlled dc output voltages from
the fixed ac mains input voltage. The circuit diagram of a fully controlled converter is shown in
Figure 2. The output voltage is varied by controlling the firing angle of SCRs. The single phasefully controlled converter consists of four SCRs. During positive half cycle, SCR1 and SCR 2
are forward biased. Current flows through the load when SCR1 and SCR2 is triggered into
conduction. During negative half cycle, SCR3 and SCR4 are forward biased. If the load isresistive, the load voltage and load current are similar.
When the load is inductive, SCR1 and SCR2 conduct from to . The nature of the load
current depends on the values of R and L in the inductive load. Because of the inductance, theload current keeps on increasing and becomes maximum at . At , the supply voltage reversesbut SCRs 1 and 2 does not turn off. This is because the load inductance does not allow the
current to go to zero instantly. Thus the energy stored in the inductance flows against the supply
mains. The output voltage is negative from to + since supply voltage is negative.Circuit Diagram:
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Model Graph:
Resistive load
Inductive load :
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Result:
Thus the Single Phase fully controlled Rectifier with R & RL Load circuit is simulated usingMATLAB/SIMULINK and the corresponding waveforms are obtained.
S.NO.3
Three Phase Fully controlled Rectifier with R & RL Load
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Aim:
To simulate the 3 fully Controlled rectifier circuit with R & RL load and obtain the
corresponding waveforms using MATLAB/SIMULINK
Theory:
The three phase full bridge converter works as three phase AC-DC converter for firing angledelay 0
0
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Model Graph:
Resistive load:
Inductive load:
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Result:
Thus the three phase fully controlled Rectifier with R & RL Load circuit is simulated using
MATLAB/SIMULINK and the corresponding waveforms are obtained.
S.NO.4
SINGLE PHASE AC VOLTAGE REGULATORAim:
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To simulate the 1 AC voltage regulator circuit and obtain the suitable waveforms using
MATLAB/SIMULINK
Theory:AC regulators are used to get variable AC voltage from the fixed mains voltage. Some of the
important applications of AC regulators are: domestic and industrial heating, induction heating inmetallurgical industries, induction motor speed control for fan and pump drives, transformer tap
changers in utility systems, static reactive power compensators, lighting control etc., Earlier, auto
transformers, transformers with taps and magnetic amplifiers were employed in theseapplications because of high efficiency, compact size, flexibility in control etc. Two thyristors in
anti parallel are employed for full wave control. In this case, isolation between control and power
circuit is most essential because of the fact that the cathodes of the two thyristors are connected
to the common point. For low power applications, a triac may be used. In this case isolationbetween control and power circuitry is not necessary.
Formulae Used:
The triggering pulse is generated at the point at which the associated cosine wave becomes
instantaneously equal to the control voltage.In other words,
2V sin (-t) = VRAt this instant t= and hence2V sin (-) = VR
= - sin- (VR/2V)Rmax=22V/CVR
Where, VR- breakdown voltage of the Diac
- firing angle delayV- Supply voltage
Circuit Diagram:
Operation:
MT2
MT1
C
RL
RD
MT1 MT2
G
AC
line
R
Fig 1. Single phase ac regulator
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A triac control circuit for lamp dimmers is shown in Fig.1. A diac is a gateless triac designed to
breakdown at a low voltage. During the positive half cycle, the triac requires a positive gate
pulse for turning it on. This is provided by the capacitor C. When its voltage is above thebreakdown voltage of the diac, the capacitor C discharges through the triac gate. When the triac
turns on, the capacitor Voltage will be reset to zero. A similar operation takes place in the
negative half cycles, and a negative gate pulse will be applied when the diac breaks down in thereverse direction. Adjustment of series resistance, R determines the charging rate of capacitor C
and hence the value of the phase angle delay. The output power and thus light intensity are
varied by controlling the phase of conduction of the triac.
Model
Graph:
Result:
Thus the 1 AC Voltage regulator with R load circuit is executed with the help of MATLAB
software and the graph is plotted.
VS
V0
wt
wt
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S.NO.5
Formation of Bus Admittance Matrix using MATLAB Software
Aim:
To develop a computer program to form the bus admittance matrix, Ybus of a power system.
Theory:
The Ybus /Zbus matrix constitutes the models of the passive portions of the power network. Ybusmatrix is often used in solving load flow problems. It has gained widespread applications owingto its simplicity of data preparation and the ease with which the bus admittance matrix can be
formed and modified for network changes. Of course, sparsity is one of its greatest advantages as
it heavily reduces computer memory and time requirements. In short circuit analysis, the
generator and transformer impedances must also be taken into account. In contingency analysis,the shunt elements are neglected, while forming the Z-bus matrix, which is used to compute the
outage distribution factors.This can be easily obtained by inverting the Y-bus matrix formed by inspection method or by
analytical method. The impedance matrix is a full matrix and is most useful for short circuit
studies. Initially, the Y-bus matrix is formed by inspection method by considering line data only.
After forming the Y-bus matrix, the modified Y-bus matrix is formed by adding the generatorand transformer admittances to the respective diagonal elements and is inverted to form the Z-
bus matrix.
The performance equation for a n-bus system in terms of admittance matrix can bewritten as,
nnnnn
n
In
nV
VV
YYY
YYYYYY
I
II
.
.
....
..
..
........
.
.
2
1
21
22221
1211
2
1
(or)
I = Ybus.VThe admittances Y11, Y12, Y1n are called the self-admittances at the nodes and all other
admittances are called the mutual admittances of the nodes.
Formulae Used:
Main diagonal element in Y-bus matrix = ij
n
j
ij BY 1
where Bij is the half line shunt admittance in mho.
Yij is the series admittance in mho.
Off-diagonal element in Y-bus matrix , Yij = -Yij
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where Yij is the series admittance in mho.
Flowchart:
START
Enter the mutual admittance
between the buses
Calculate the diagonal term,
Yii = sum of all admittances
connected to bus i.
STOP
Calculate the off-diagonal
term, Yij=Negative sum of theadmittances connected from
bus i to bus j.
Enter the number of buses,n
and lines
Set the bus count i =1
Is i = n
i = i +1
Print Y bus and Z bus matrices
Compute Z bus matrix by
inverting Y bus matrix
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Algorithm:
Step 1: Read the values of number of buses and the number of lines of the given
system.
Step 2: Read the self-admittance of each bus and the mutual admittance between the
buses.
Step 3: Calculate the diagonal element term called the bus driving point admittance, Yij
which is the sum of the admittance connected to bus i.
Step 4: The off-diagonal term called the transfer admittance, Yij which is the negative
of the admittance connected from bus i to bus j.
Step 5: Check for the end of bus count and print the computed Y-bus matrix.Step 6: Compute the Z-bus matrix by inverting the Y-bus matrix.
Step 7: Stop the program and print the results.
Sample Problem:
The bus and branch datas for a 3 bus system is given in table below. Form Y bus matrix byinspection method.
Bus Code Impedance Bus Number Admittance
1 - 2 0.06 + j0.18 1 j0.051 3 0.02 + j0.06 2 j0.06
2 - 3 0.04 + j0.12 3 j0.05
Solution:
Formation of Y bus:
06.002.0
105.0
12.004.0
1
12.004.0
1
06.002.0
1
12.004.0
106.0
12.004.0
1
18.006.0
1
18.006.0
1
06.002.0
1
18.006.0
105.0
06.002.0
1
18.006.0
1
jj
jjj
jj
jjj
jjj
jj
Ybus
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Theoretical output:
45.225.75.75.2155
5.75.244.1216.4566.1
155566.195.1966.6
jjj
jjj
jjj
Ybus
Result:
The Y bus matrix was formed for the given system by direct inspection method and the results
were verified using MATLAB program.
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S.NO.6
Z-bus Building Algorithm using MATLAB Software
Aim:
To develop a computer program to obtain the building algorithm for bus impedance matrix of the
given power system.
Theory:
The Ybus /Zbus matrix constitutes the models of the passive portions of the power network. The
impedance matrix is a full matrix and is most useful for short circuit studies. An algorithm forformulating [Zbus] is described in terms of modifying an existing bus impedance matrix
designated as [Zbus]old. The modified matrix is designated as [Zbus]new. The network consists of a
reference bus and a number of other buses. When a new element having self impedance Zb is
added, a new bus may be created (if the new element is a tree branch) or a new bus may not becreated (if the new element is a link). Each of these two cases can be subdivided into two cases
so that Zb may be added in the following ways:1. Adding Zb from a new bus to reference bus.
2. Adding Zb from a new bus to an existing bus.
3. Adding Zb from an existing bus to reference bus.
4. Adding Zbbetween two existing buses.
Type 1 modification:
In type 1 modification, an impedance Zb is added between a new bus pand the reference bus asshown in Figure 1
Let the current through bus pbe Ip, then the voltage across the bus p is given by,
Vp = Ip Zb
Vp
Ref. Bus
p
n
1
Network
Zb
Figure 1. Type 1 modification of Zbus
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The potential at other buses remains unaltered and the system equations can be written as,
p
n
b
oldbus
p
n
I
I
I
I
Z
Z
V
V
V
V
2
1
2
1
00000
0
0
0
0
0
Type 2 modification:
In type 2 modification, an impedance Zb is added between a new bus pand an existing bus kasshown in Figure 2. The voltages across the bus kand pcan be expressed as,
Vk(new) = Vk+ Ip Zkk
Vp = Vk(new) + Ip Zp
= Vk+ Ip(Zb + Zkk)
where, Vkis the voltage across bus kbefore the addition of impedance ZbZkkis the sum of all impedance connected to bus k.
The system of equations can be expressed as,
p
n
bkkkk
oldbus
k
k
p
n
I
I
I
I
ZZZZ
Z
Z
Z
V
V
V
V
2
1
21
2
1
2
1
Ip
Ik+ Ip
Ref. Bus
p
n
1
Network
Z
b
k
Figure 2.Type 2 Modification of Zbus
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Type 3 Modification:
In this modification, an impedance Zb is added between a existing bus kand a reference bus.Then the following steps are to be followed:
1. Add Zbbetween a new bus p and the existing bus k and the modifications are done as in
type 2.2. Connect bus p to the reference bus by letting Vp = 0.To retain the symmetry of the Bus Impedance Matrix, network reduction technique can be
used to remove the excess row or column.
Type 4 Modification:
In this type of modification, an impedance Zb is added between two existing buses j and k asshown in Figure 3. From Figure 3, the relation between the voltages of bus kandj can be writtenas,
Vk Vj = IbZb (3)
The voltages across all the buses connected to the network changes due to the addition of
impedance Zb and they can be expressed as,V1 = Z11I1 + Z12I2 + - - - - - - - - + Z1j(Ij + Ib) + Z1k(Ik Ib)+- - -
V2 = Z21I1 + Z22I2 + - - - - - - - - + Z2j(Ij + Ib) + Z2k(Ik Ib)+ - - -
Vj = Zj1I1 + Zj2I2 + - - - - - - - - + Zjj(Ij + Ib) + Zjk(Ik Ib) + - - - (4)
Vk= Zk1I1 + Zk2I2 + - - - - - - - - + Zkj(Ij + Ib) + Zkk(Ik Ib) + - - -
Vn = Zn1I1 + Zn2I2 + - - - - - - - - + Znj(Ij + Ib) + Znk(Ik Ib) + - - -
On solving the Equations (3) and (4), the system of equations can be rewritten as,
Ik- Ib
Ij + Ib
Ref. Bus
k
n
1
Network
Z
b
Ib
j
Figure 3.Type 4 Modification of Zbus
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p
n
bbkkjkkj
kkkj
oldbus
kj
p
n
I
I
I
I
ZZZZZ
ZZ
Z
ZZ
V
V
V
V
2
1
11
11
2
1
)()(
)(
)(
(5)
where,
Zbb = Zjj + Zkk 2 Zjk+ Zb
Procedure for formation of Zbus matrix:
Step1: Number the nodes of the given network, starting with those nodes at the ends
of branches connected to the reference node.
Step2: Start with a network composed of all those branches connected to the
reference node.
Step3: Add a new node to the ith
node of the existing network.
Step4: Add a branch between ith
and jth
nodes. Continue until all the remaining
branches are connected.
Sample problem:
Form bus impedance matrix using building algorithm:
Solution:
Step1: Add an element between ref (0) bus and a new bus (1).
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Z = [j0.2]
Step2: Add an element between existing bus (1) to a new bus (2).
Z =
6.02.0
2.02.0
jj
jj
Step3: Add an element between existing (2) Bus to a ref (0) Bus.
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Z=
8.06.02.0
6.06.02.0
2.02.02.0
jjj
jjj
jjj
New Z Bus:
Z11 = Z11-(Z31*Z13)/Z33= j0.2 (j0.2*j0.2)/j0.8
Z11 = j0.05
Z12 =Z21= Z12-(Z32*Z13)/Z33
= j0.2 - (j0.6*j0.2)/j0.8= j0.05
Z22 =Z22-(Z32*Z23)/Z33
=J0.6-(j0.6*j0.6)/j0.8Z22 =j0.15
Z Bus =
15.005.0
05.005.0
jj
jj
Result:
The bus impedance matrix using building algorithm for the given system was formed and the
results were verified using MATLAB program.
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S.NO.7
Gauss Seidal Load flow analysis using MATLAB software
Aim:
To develop a computer program to solve the set of non linear load flow equations using Gauss-seidal load flow algorithm.
Theory:
Load flow analysis is the most frequently performed system study by electric utilities. This
analysis is performed on a symmetrical steady-state operating condition of a power system under
normal mode of operation and aims at obtaining bus voltages and line/transformer flows for a
given load condition. This information is essential both for long term planning and next dayoperational planning. In long term planning, load flow analysis helps in investigating the
effectiveness of alternative plans and choosing the best plan for system expansion to meet theprojected operating state. In operational planning, it helps in choosing the best unit
commitment plan and generation schedules to run the system efficiently for them next days load
condition without violating the bus voltage and line flow operating limits.
The Gauss seidal method is an iterative algorithm for solving a set of non- linearalgebraic equations. The relationship between network bus voltages and currents may be
represented by either loop equations or node equations. Node equations are normally preferred
because the number of independent node equation is smaller than the number of independentloop equations.
The network equations in terms of the bus admittance matrix can be written as,
busbusbus VYI (1)
For a nbus system, the above performance equation can be expanded as,
n
p
nnnpnn
pnpppp
np
np
n
p
V
V
V
V
YYYY
YYYY
YYYY
YYYY
I
I
I
I
2
1
21
21
222212
111211
2
1
(2)
where n is the total number of nodes.Vp is the phasor voltage to ground at node p.Ip is the phasor current flowing into the network at node p.
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At the pthbus, current injection:
n
pqq
qpqppp
n
qqpq
npnpppppp
VYVYVY
VYVYVYVYI
11
2211 .........................
(3)
npVYIY
Vn
pqq
qpqppp
p ,....2;1
1
(4)
At bus p , we can write Pp jQp = pp IV
Hence, the current at any node p is related to P, Q and V as follows:
p
pp
pV
jQPI
)(( for any bus p except slack bus s) (5)
Substituting forIp in Equation (4),
npVYV
jQP
YV
n
pqq
qpqp
pp
ppp .....,2;
1
1*
(6)
Ip has been substituted by the real and reactive powers because normally in a power system thesequantities are specified.
Algorithm:
Step 1: Read the input data.
Step 2: Find out the admittance matrix.
Step 3: Choose the flat voltage profile 1+j0 to all buses except slack bus.
Step 4: Set the iteration count p = 0 and bus count i = 1.Step 5: Check the slack bus, if it is the generator bus then go to the next step otherwise go to
next
step 7.Step 6: Before the check for the slack bus if it is slack bus then go to step 11 otherwise go to
next
step.Step 7: Check the reactive power of the generator bus within the given limit.
Step 8: If the reactive power violates a limit then treat the bus as load bus.
Step 9: Calculate the phase of the bus voltage on load bus
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Step 10: Calculate the change in bus voltage of the repeat step mentioned above until all the bus
voltages are calculated.
Step 11: Stop the program and print the results
Flowchart:
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Yes
Read the in ut data values
Start
Form Y Bus matrix
Set flat voltage profile 1+j0 except slack bus
Set iteration count, =0
Set the bus count, i = 1
Check forslack bus
Check forGen bus
It is a load bus
calculate
n
jkik
j
kkik
i
ii
ii
pical VYVY
V
jQP
YV
1
1
1*
1 1
Calculate
p
k
n
ikik
p
k
i
kik
p
iVYVYQ ipV
11
1
*1 Im
Check
in
1 QQpi
SetQi=Qi min
Check
ax
1 QQpi
Set
Qi=Qi max
A
No
Yes
No
No
No
Yes
Yes
D
E
C
B
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Sample Problem:
The load flow data for a 3 bus system is given in tables below. The voltage magnitude at bus 2
is to be maintained at 1.04 p.u. The maximum and minimum reactive power limits for bus 2 are
0.5 to 0.2 respectively. Taking bus 1 as slack bus, determine voltages of the various buses atthe end of first iteration starting with flat voltage profile for all buses except slack bus using
Gauss-Seidal method with acceleration factor of 1.6.
Treat this as gen bus & calculate Vpi
n
ik
pkik
i
k
pkik
i
i
ii
pi VYVY
V
jQP
YV
1
1
1
1
*
1 1
Calculate the change in voltage1 piV
Increment the bus count
Check
ni
Check
1piV
Print the result
Sto
Increment
iteration count
P = P+1
Yes
Yes
No
No
B
E
D
C
A
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Bus Code Impedance Bus Number Admittance
1 2 0.06 + j0.18 1 j0.05
1 3 0.02 + j0.06 2 j0.06
2 3 0.04 + j0.12 3 j0.05
Bus Code AssumedVoltage
Generation Load
MW MVAr MW MVAr
1 1.06 + j0 0 0 0 0
2 1 + j0 0.2 0 0 0
3 1 + j0 0 0 0.6 0.25
Solution:
Formation of Ybus:
45.225.75.75.2155
5.75.244.1216.4566.1
155566.195.1966.6
jjj
jjj
jjj
Ybus
Calculation of Q2:
Q2 =
n
q
qpqVYV1
*
2Im
= )5.75.2(04.1)5.1216.4()06.1)(566.1(04.1Im jjj = )5.75.2(04.1)5.1216.4()30.5763.1(04.1Im jjj = 14.007.0Im j
Q2 = 0.14, it violates the limits of the reactive power.
Q2 = Q min = 0.2 as min2 QQ
[ If suppose, Q2 Qmax then Q2 = Qmax]
Calculation of Bus voltages:
)1(
2V = 0.075
))01)(5.75.2()06.1)(566.1((04.1
2.02.063.71 jj
jVolts
= 0.075 99.12452.463.71 j)1(
2V = 1.047+j 0.007 volts
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Accelerated voltage,)1(
2V = 1.04+ 1.6(1.047+j0.007 -1.04)
= 1+0.048-j0.048)1(
2V =1.0512+j0.0112 Volts
)1(
3V =0.0423 ))0112.00512.1)(5.75.2()06.1)(155((25.06.049.71 jjjj )1(
3V = 1.041 j 0.017 Volts
Accelerated voltage,
)1(
3V = 1+1.6(1.041 j 0.17- 1 )
)1(3V = 1.0656-j0.272 Volts
Theoretical Output:
V1=1.06+j0 Volts,)1(
2V =1.0512+j0.0112 Volts,)1(
3V = 1.0656-j0.272 Volts
Result:
The given set of load flow equations for a given power system were solved using Gauss-Seidal
method.
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S.NO.8
Newton Rapshson load flow analysis using MATLAB software
Aim:
To develop a software program to obtain real and reactive power flows, bus voltage magnitude
and angles by using N R method.
Theory:
Load flow study in power system parlance is the steady state solution of the power systemnetwork. The main information obtained from this study comprises the magnitudes and phase
angles of load bus voltages, reactive powers at generator buses, real and reactive power flow on
transmission lines, other variables being specified. This information is essential for the
continuous monitoring of current state of the system and for analyzing the effectiveness ofalternative plans for future system expansion to meet increased load demand.
Newton-Raphson method is an iterative method that approximates the set of non linearsimultaneous equations to a set of linear simultaneous equations using Taylors series expansion
and the terms are limited to first approximation. The rate of convergence is fast as compared to
the FDLF program and also it is suitable for large size system. So we go for N-R method.
The non-linear equations governing the power system network are,
qp
ppqp pallforVYI
where Ip is the current injected into bus p.The complex power in pthbus is given by,
...................,2;1
**
1
*
npVYVVYV
IVSn
q
qpqp
n
q
qpqp
ppp
(1)
pqjpqpq
qppq
qjqq
pjpp
eYY
eVV
eVVLet
,
and
In polar co-ordinates, the power on pthbus is given as,
pqjpq
qpjn
q
qpppp eYeVVjQPS
||1
(2)
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Separating the Real and Imaginary parts we get,
)cos(
1
qpqppq
n
q
qpp YVVP
)sin(
1qpqppq
n
qqpp YVVQ
(3)
The Newton Raphson method requires that a set of linear equations be formed expressing the
relationship between the changes in real and reactive powers and the components of the bus
voltages as follows:
)4(
|
|
|
|
|
|
|
)(
)(
2
)(
)(
2
)()(
2
)(
2
)(
2
)(
2
)(
2
2
)(
2
)(
2
2
)()(
2
2)()(
2
)(
2
)(
2
2
)(
2
)(
2
2
)(
)(
2
)(
)(
2
rn
r
rn
r
r
n
n
r
n
r
n
r
n
r
n
rr
n
r
r
n
nrr
n
nr
n
r
n
rr
n
r
rn
r
rn
r
V
V
V
Q
V
QQQ
V
Q
V
QQQ
V
P
V
PPP
V
P
V
PPP
Q
Q
P
P
where, the coefficient matrix is known as Jacobian matrix.
In the above equation, bus 1 is assumed to be the slack bus. The Jacobian matrix gives
the linearized relationship between small changes in voltage angle )(ri and voltage magnitude r
iV with the small changes in real and reactive power r
iP and riQ . Elements of the
Jacobian matrix are the partial derivatives of (2) and (3) evaluated at ri and r
iV .
The above relationship can be written in a compact form as,
VJJ
JJ
Q
P
2221
1211(5)
The elements of Jacobian matrix are defined as,
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All
quantit
ies inthe
linear
Equation (4)
pertai
n toiterati
on r.The
linearequati
on when solved for, V gives the correction to be applied to |V| and , i.e.
|V|
(r+1)
= |V|
(r)
+
|V|
(r)
(14) (r+1) = (r) + (r) (15)
Next we get a new set of linear equations evaluated at (r+1)th iteration and the process isrepeated. Convergence is tested by the power mismatch criteria. This method converges to highaccuracy nearly always in 2 to 5 iterations from a flat start (|V| = 1 p.u. and =0 ) for all buses
where |V|, are unknown, independent of system size.
At PV bus at the end of an iteration and if it violates the limits, the PV bus is switched toa PQ bus. When Q is within limits, then it is switched back to PV bus.
)7()sin(
)6()sin(
:
1
11
qpqppqq
n
pqq
p
p
p
qpqppqqp
q
p
YVVP
pqYVVP
J
)sin(sin2
)sin(
:
1
22
qpqppq
n
pqq
qppppp
p
p
qpqppqp
q
p
YVYVVQ
pqYVV
Q
J
)11()cos(cos2
)10()cos(
:
1
12
qpqppq
n
pqq
qppppp
p
p
qpqppqp
q
p
YVYVV
P
pqYVV
P
J
)13()cos(
)12()cos(
:
1
21
qpqppqq
n
pqq
p
p
p
qpqppqqp
q
p
YVVQ
pqYVVQ
J
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Algorithm:
Start
Read bus data, line data, bus ower & tolerance
Form Y bus matrix
Initialize all bus voltages
Set iter count = 0
Iter = Iter +1
Calculate real power & reactive power
mismatch [P] [Q] using the current values
of V& taking Q limit violations in toaccount
Update voltagemagnitude and phase
angles
VVV oldnew oldnew atall buses except slackbus
Ptol
Qtol
Calculate real & reactive
line flows in all the lines
Print the result
Stop
Solve the equation
Q
P
V 43
21
Solve the equation
VLN
MH
Q
P
To find V &
Yes
No
Flowchart:
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The computational procedure for Newton-Raphson method using polar coordinate is as follows:
Step 1: Form Ybus matrix.
Step 2: Assume initial values of bus voltages pVo
and phase angles op for load buses
and phase angles for PV buses. Normally we set the assumed bus voltage
magnitude and its phase angle equal slack bus quantities 1V = 1.0, 1 =0o.
Step 3: Compute Pp and Qp for each load bus from the Equations (2) and (3).
Step 4: Compute the scheduled errors pP and pQ for each load bus from the
following relations:
npQQQ
npPPP
kcalpspp
kp
kcalpspp
kp
.....3,2
.....3,2
For PV buses, the exact value ofp
Q is not specified, but its limits are known. If
the calculated value of pQ is within limits, only pP is calculated. If the
calculated value of pQ is beyond the limits, then an appropriate limit is imposed
and pQ is also calculated by subtracting the calculated value of pQ from the
appropriate limit. The bus under consideration is now treated as a load on
(PQ) bus.
Step 5: Compute the elements of the Jacobian matrix using the estimated pV and p
from step2.
Step 6: Obtain and pV from Equations (4) and (5).
Step 7: Using the values of p and pV calculated in step 6, modify the voltage
magnitude and phase angle at all loads by the Equations (14) and (15). Start the
next iteration cycle at step 2 with these modified pV and p .
Step 8: Continue until scheduled errors kpP andkpQ for all load buses are within a
specified tolerance, ie, kpP < ,kpQ <
where, denotes the tolerance level for load buses.Step10: Calculate line flows and power at the slack bus exactly in the same manner as in
the Gauss Seidal method.
Sample Problem:
The load flow data for a 3-bus system is given in tables 1 and 2. The voltage magnitude at bus 2is to be maintained at 1.0 p.u. The maximum and minimum reactive power limits for bus 2 are
0.3 and 0 p.u. respectively. Taking bus 1 as slack bus, determine the voltages of the various
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buses at the end of first iteration starting with a flat voltage profile for all buses except slack bus
using N-R method.
Table 1: Impedance for sample system
Bus code Impedance Line charging
admittance ypq /21-2 0.06+j0.18 j0.05
1-3 0.02+j0.06 j0.06
2-3 0.04+j0.12 j0.05
Table 2: Assumed bus voltages, Generation and loads
Bus code Voltages
p.u
Generation
MW MVARp.u p.u
Load
MW MVARp.u p.u
1 1.06 0 0 0 0
2 1 0.2 0 0 0
3 1 0 0 0 0.25
Solution:
Formation of Ybus :
Ybus =
333231
232221
131211
YYY
YYY
YYY
Y12 = -18.006.0
1j
=-(1.667-j5)
= 5.270
4.108
Y13 = -06.002.0
1
j=-(5-j15)
= 15.81 04.108
Y23 = -12.004.0
1
j=-(2.5-j7.5)
= 7.906 04.108
Y11 =18.006.0
1
j+
06.002.0
1
j+ j0.05+j0.06 =6.667-j19.89
=21.97 05.71
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Y22 =18.006.0
1
j+
12.004.0
1
j+ j0.05+j0.05 =4.167-j12.4
=13.08 05.71
Y33 = 06.002.0
1
j + 12.004.0
1
j + j0.06+j0.05=7.5-j22.39
=23.61 05.71
Ybus =
000
000
000
5.7161.234.108906.74.10881.15
4.108906.75.7108.134.10827.5
4.10881.154.10827.55.7197.21
Flat start profile:
Given V1 = 1.06+ j0 ; 1 = 00 ; V3 = 1
00
Choose V20
=1+j0 and 20
= 30
= 0
Calculation of change in real and reactive powers:
Pp = Pp(specified) Pp(calculated)
Qp = Qp(specified) Qp(calculated)
n
q
qpqppqqpp
n
qqpqppqqpp
YVVQ
YVVP
1
1
)sin(
)cos(
P2(cal) = |V2|2|Y22|cos 22 + |V2||V1||Y21|cos(2 + 21 - 1) + |V2||V3||Y23|cos(2 + 233)
= 1 13.08 cos(-71.5) + 1 1.06 5.27cos(108.4) + 1 7.906cos(108.4)= -0.11p.u
P3(cal) = |V3|2
|Y33|cos 33 +|V3||V1||Y31|cos(3 + 311) + |V3||V2|Y32|cos( 32 + 3 - 2 )= 1 23.61 cos(-71.5) + 1 1.06 15.81cos(108.4) + 1 7.906cos(108.4)=- 0.3 p.u
Q2(cal) = |V2|2|Y22|sin 22 + |V2||V1||Y21|sin(2 + 211) + |V2||V3||Y23|sin(2 + 233)
= 1 13.08 sin(-71.5) + 1 1.06 5.27sin(108.4) + 1 7.906sin(108.4)= 0.4 p.u
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Q3(cal) = |V3|2|Y33|sin 33 + |V3||V1||Y31|sin(3 + 311) + |V3||V2||Y32|sin(3 + 322)
= 1 23.61 sin(-71.5) + 1 1.06 15.81sin(108.4) + 1 7.906sin(108.4)= 1.02 p.u.
Calculation of specified quantities :
P2(specified) = PG2 - PD2 = 0.2 0.0 = 0.2 p.u
Q2(specified) = QG2 - QD2 = 0 p.u
P3(specified) = PG3 - PD3 = 0.0 p.u
Q3(specified) = QG3 - QD3 = -0.25 p.u
The change in real and reactive powers are,
P20
= P2(specified) P2(calculated) = 0.2 + 0.11 = 0.31 p.u.P3
0= 0 +( - 0.3) = -0.3 p.u.
Q20
= 0- 0.4 = -0.4 p.u.
Q30
= -0.25 -1.02 = -1.27 p.u.
Calculation of Jacobian matrix elements :
Elements of J 1:
22
2
V
P|V2||Y22|cos 22 +|V1||Y21|cos(2 + 21 - 1) + |V3||Y23|cos(3 + 232)
= 2 1 13.08 cos(-71.5) + 1.06 5.27cos(108.4) + 1 7.906cos(108.4)= 4.04
3
2
V
P|V2||Y23|cos(2 + 233)
= 1 7.906 cos(108.4)= -2.5
2
3
V
P|V3||Y32|cos( 32 + 2 - 3 )
= 1 7.906 cos(108.4)= -2.5
3
3
V
P2|V3||Y33|cos 33 +|V1||Y31|cos(1 + 313) + |V2||Y32|cos( 32 + 2 - 3 )
= 2 1 23.61 cos(-71.5) + 1.06 15.81cos(108.4) + 1 7.906cos(108.4)= 7.2
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Elements of J 2:
2
2
P|V2||V1||Y21|sin(2 + 211) - |V2||V3||Y23|sin(3 + 232)
= 1 1.06 5.27sin(108.4) 1 1 7.906 sin(108.4)
= -12.8
3
2
P- |V2||V3||Y23|sin(2 + 233)
= - 1 1 7.906 sin (108.4)= -7.5
2
3
P- |V3| |V2||Y32|sin(3 + 322)
= - 1 1 7.906 sin (108.4)= -7.5
3
3
P -|V3||V1||Y31|sin(3 + 311) - |V3||V2||Y32|sin(3 + 322)
= - 1 1.06 15.81sin(108.4)-1 17.906sin(108.4)= - 23.4
Elements of J 3:
2
2
V
Q{2|V2||Y22|sin 22 + |V1||Y21|sin(2 + 211) +|V3||Y23|sin(2 + 233 ) }
= {2 113.08 sin(-71.5) + 1.065.27sin(108.4) + 17.906sin(108.4)}
= - 12.01
{3
2
V
Q|V2||Y23|sin(2 + 233) }
={ 1 7.906sin(108.4)}= 7.5
{2
3
V
Q|V3||Y32|sin(2 + 323)}
= { 1 7.906sin(108.4)}
= 7.5
{3
3
V
Q2|V3||Y33|sin 33 - |V1||Y31|sin(3 + 311) - |V2||Y32|sin(3 + 322)
= {2 1 23.61 sin(-71.5) +1 1.06 15.81sin(108.4) + 1 7.906 sin (108.4)}= 21.4
Elements of J 4:
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{2
2
Q|V2||V1||Y21|cos(2 + 21 - 1)+ |V2||V3||Y23|cos(2 + 233) }
= 11.06 5.27cos(108.4) + 17.906cos(108.4)= -4.26
{3
2
Q|V2||V3||Y23|cos(2 + 233) }
= - { 17.906cos(108.4)}= 2.5
{3
3
Q|V3||V1||Y31|cos(3 + 311) + |V3||V2||Y32|cos(3 + 322)
= 11.0615.81cos(108.4) + 17.906cos(108.4)= - 7.8
The general matrix form of load flow equation is,
V
JJ
JJ
Q
P
2221
1211
3
2
3
2
8.75.24.215.7
5.226.45.701.12
4.235.72.75.2
5.78.125.204.4
27.1
4.0
3.0
31.0
v
v
Result:
The load flow study of the given power system using Newton-Raphson method was conductedusing MATLAB and results was verified.
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S.NO.9
Fast Decoupled Load Flow Analysis using MATLAB Software
Aim:
To become proficient in the usage of software in solving load flow problems using Fast
decoupled load flow method.
Theory:
Load flow study is useful in planning the expansion of power system as well as determining bestoperation of the system. The principle obtained from load flow study is the magnitude and phase
angle of the voltage at each bus and real and reactive power flowing in each line. Load flow
analysis may be performed using A.C. network analyzer and also by digital computer. But now-
a-days digital computer oriented load flow analysis is a standard practice.The fast decoupled load flow method is a very fast method of obtaining load flow
solutions. This method requires less number of arithmetic operations to complete an iterationconsequently. This method requires less time per iterations. In N-R method, the elements of
Jacobian are to be computed in each iteration .So the time per iteration is considerably more in
N-R method than in FDLF. The rate of convergence in FDLF method is slow requiring
considerably more number of iterations to obtain a solution than in the case of N-R method.However accuracy is same in both the cases. In this method both the speeds as well as the
sparsity are exploited. This is an extension of N-R method formulated in polar co-ordinates with
certain approximation which results into a fast algorithm for load flow solution.In practice, transmission system operating under steady state possesses strong
interdependence between active powers and bus voltages, angles, similarly there is strong
interdependence between bus voltage and reactive power
m
kkm
PH
;
m
mk
kmE
EPN
m
kkm
QJ
;
m
mk
kmE
EQL
q
p
pq
PH
;
q
qP
pqE
EQL
The equation for power flow are again expressed below for calculating elements of Jacobian (ieH & L)
n
pq
qppqpqqpppppqpp YEEYEEP,1
coscos
qppqpqqpppppqpp YEEYEEQ sinsin
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Therefore the elements of Jacobian (ie H & L) can be calculated as from the equations above of
power. OFF diagonal element of H is
qppqq
p
PQ
PH
sin
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Start
Read the in ut data
Form the Y bus matrix
Form B and B matrix
Set flat voltage profile except for slack bus
Set convergence criterion
Set iteration count p=0
Calculate real & reactive power
Calculate
V
Q
V
P,
Are
V
Q
V
P,
Calculate real & reactive lineflow, bus powers
Print the results
Stop
VLN
MH
Q
P
Find V & by solving theequations:
Q
V
V
21
Uptate voltage magnitude t phaseangles
VVV oldnew oldnew
C
C
No
Yes
Flowchart:
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Algorithm:
Step 1: Read the slack bus voltages, real bus powers and reactive bus powers, busvoltage magnitudes and reactive power limits.
Step 2: Form the Y bus matrix without line charging admittance and shunt admittance.
Step 3: Form B matrix, form Y bus matrix obtained in step 2.
Step 4: Form Y bus matrix with double the line charging admittance.
Step 5: Form B matrix from Y bus matrix obtained in step 4.
Step 6: Calculate the inverse of B & B matrices.
Step 7: Initialize the bus voltage.
Step 8: Calculate [P/|V|] , [Q/|V|]
Step 9: If P/ |V| & Q/|V| are less than or equal to tolerance limit, solution has convergenceand go to step 12 otherwise increase iteration count and go to step 10.
Step 10: Calculate [] = [B]-1
[P/ |V|]
[|V|] = [B]-1
[Q/|V|]
Step 11: Update [] & [|V|] for all buses except slack bus.
[]new
= []old
+ [ ]; [|V|]new
= [|V|]old
+ [|V|]
Step 12: Compute slack bus power, line flows, real power loss, reactive power loss etc.
Sample Problem:
For the system shown in Figure 4.4 determine the voltage at the end of the 1st
iteration by FDLF
method. The line reactances are marked in the figure.
j 0.2
j 0.2
j 0.11 2
3
Figure 4.4
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Bus specifications:
Bus
code
Assumed volt. Generation Load
P Q P Q1 1 - - - -
2 1.1 5.52 0QG25.5 0.7 0.1
3 - - - 3.65 0.53
Solution:
P2 = PG2-PD2 = 4.82 p.u
P3=PG3-PD3 = -3.65 p.u
Q3= QG3-QD3= -0.53 p.u
10
105
515
9010905905
90590159010
90590109015
1055
51510
51015
''
'
B
B
jjj
jjj
jjj
Ybus
Flat voltage profile:
00
3
00
2
00
1
00.101
01.101.1
00.101
jV
jV
jV
Calculation of P and Q:
)cos(
1
qpqppq
n
qqpp YVVP
00
0
,3
0
,2
cal
cal
PP
n
q
qpqppqqpp YVVQ1
)sin(
= 2323333222222221212112 sinsin..sin.. YVVYVVYVV
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= -1.65
As 0QG25.5Q2= QG2-QD2QG2=-1.65+0.1
= -1.55Hence it is not within the specified limits.
Q3= 3333333232322331313113 sinsin..sin.. YVVYVVYVV = 0.5
Calculation of change in power:
03.15.053.0
65.3065.3
82.4082.4
.3.33
.3.33
.2.22
calspec
calspec
calspec
QQQ
PPP
PPP
Find the largest value of P2, P3, Q3
Let the largest change of E= 4.82
E E; 4.820.01
Find and V :
V
QBV
V
PB
1''
1'
1
1'
105
515
B
=
12.004.0
04.008.0
1.010/11'' B
V
P
12.004.0
04.008.0
65.3
38.4
12.004.0
04.008.0
3
2
2=0.2045
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3=-0.2627
V
QV 1.0
103.03 V
Find the new values of phase angle and magnitude of the voltage:
p
i
p
i
p
i
pi
pi
pi
VVV
1
1
p =0; i=2,3
00
3
0
3
1
3
00
2
0
2
1
2
05.152627.0
71.112045.0
rad
rad
p =0; i=3
897.0103.010
3
0
3
1
3 VVV
New values are:
)(05.15897.0
)(01.1
)(01
0'
3
'
3
'
3
0'
2
'
2
'
2
0'
1
'
1
'
1
busLoadVV
busGeneratorVV
busSlackVV
Result:
The load flow study on the given power system using Fast decoupled method was conducted
using MATLAB and results was verified.
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S.NO.10
Symmetrical Fault Analysis using MATLAB Software
Aim:
To develop a computer program to carry out simulation study of a symmetrical three phase short
circuit on a given power system.
Theory:
Short circuits and other abnormal conditions often occur on a power system. Short circuits are
usually called faults by power system engineers. Some defects, other than short circuits arealso termed as faults.
Faults are caused either by insulation failures or by conducting path failures. The failure
of insulation results in short circuits which are very harmful as they may damage some
equipment of the power system. Most of the faults in transmission and distribution lines arecaused by over voltages due to lightning or switching surges, or by external conducting objects
falling on overhead lines. Overvoltages due to lightning or switching surges cause flashover onthe surface of insulators resulting in short circuits. Short circuits are also caused by tree branches
or other conducting objects falling on the overhead lines.
The fault impedance being low, the fault currents are relatively high. The fault currents
being excessive, they damage the faulty equipment and the supply installation. Also, the systemvoltage may reduce to a low level, windings and busbars may suffer mechanical damage due to
high magnetic forces during faults and the individual generators in a power station or group of
generators in different power stations may loose synchronismThe symmetrical fault occurs when all the three conductors of a three-phase line
are brought together simultaneously into a shortcircuit condition as shown in Figure 1.
This type of fault gives rise to symmetrical currents i.e. equal fault currents with 1200
displacement. Thus referring to Figure 5.1, fault currents IA, IB and IC will be equal in magnitude
with 1200
displacement among them. Because of balanced nature of fault, only one phase needs to
be considered in calculations since condition in the other two phases will also be similar.
IA IB IC
Short circuit
A
B
C
Figure 1 Symmetrical Fault on Three-Phase system
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A three-phase short circuit occurs rarely but it is most severe type of fault involving largest
currents. For this reason, the balanced short-circuit calculations are performed to determine these
large currents to be used to determine the rating of the circuit breakers.
Flowchart:
Start
Read line data, Bus data,fault impedance etc
Compute Ybus matrix & modified Ybus matrix
I = 0
Find the bus at which fault occurs I = I+1
Compute fault current at faulted bus and
bus voltage at all buses
Compute all line current at unfaulty
area & gen currents
Is
I < nb
Print the Result
Stop
Compute Zbus matrix by inverting modified bus
Yes
No
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Formula Used:
i) Fault Current, If=
ppfZZ
V
ii) Fault Voltage, Vf= )1(ppf
bus
ZZZV
where Zf Fault impedance
Zpp Line impedance
Algorithm:
Step 1: Read line data, machine data, transformer data, fault impedance etc.
Step 2: Compute [Ybus] matrix and calculate [Ybus]modi.
Step 3: Form [Zbus] by inverting the [Ybus] modified.
Step 4: Initialize count I = 0.
Step 5: Find the bus at which fault occurs I=I+1.
Step 6: Compute fault current at faulted bus and bus voltage at all buses.
Step 7: Compute all line and generator currents.
Step 8: Check if I< number of buses, if yes go to step 5 else go to step 9.
Step 9: Print the results and stop the program.
Sample problem:
For a simple power system as shown in figure, find with the help of bus-impedance matrix
method the post fault currents in all the branches and post-fault voltages at all buses, if a three
phase dead short circuit occurs at bus-3. The pre- fault currents are neglected.
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Solution:
Formation of bus impedance matrix:
Y11 = 19.2408.0
1
13.0
1
25.0
1j
jjj
Y22 = 02.4603.0
1
13.0
1
20.0
1
jjjj
Y33 = 83.4508.0
1
03.0
1j
jj
Y12=Y21== 69.713.0
1j
j
Y13=Y31== 5.12
08.0
1j
j
Y23=Y32== 33.3303.0
1j
j
We can formulate the bus admittance matrix
I =0
S
1
E1=1.0 E2=1.0
j0.25 j0.2
j0.13
j0.08 j0.03
2
3
Single line diagram
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Ybus =
83.4533.335.12
33.3302.4669.7
5.1269.719.24
jjj
jjj
jjj
By inversion of Ybus , we get
Zbus =
1343.01150.01059.0
1150.01214.00979.0
1059.00979.01270.0
jjj
jjj
jjj
Fault current, If= ..462.7
1343.0
0.1
33
0
upjjZ
Vk
Bus voltages during the fault are,
V1f= 0
33
1333kV
Z
ZZ
= ..2114.01343.0
1059.01 up
j
j
V2f= 0
33
2333kV
Z
ZZ
= ..143.01343.0
1150.01 up
j
j
V3f
= 0
Short circuit currents in the lines are,
I12f= ..69.0
0979.0
143.02114.0
12
21 upjjZ
VVff
I13f= ..999.1
1059.0
02114.0
13
31 upjjZ
VVff
I23f= ..2434.1
115.00143.0
23
32 upjjZ
VVff
Result:
The program to carry out the simulation study of a symmetrical three phase short circuit on a
given power system was developed and the results were verified.
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S.NO.11
Economic Dispatch using MATLAB Software
Aim:
To develop a program for solving economic dispatch problem without transmission losses for a
given load condition using direct method and Lambda-iteration method.
Theory:
A modern power system is invariably fed from a number of power plants. Research and
development has led to efficient power plant equipment. A generating unit added to the system
today is likely to be more efficient than the one added some time back. With a very large numberof generating units at hand, it is the job of the operating engineers to allocate the loads between
the units such that the operating costs are the minimum. The optimal load allocation is byconsidering a system with any number of units. The loads should be so allocated among the
different units that every unit operates at the same incremental cost. This criterion can bedeveloped mathematically by the method of lagrangian multiplier.
Statement of Economic Dispatch Problem:
In a power system, with negligible transmission losses and with N number of spinning thermal
generating units the total system load PD at a particular interval can be met by different sets ofgeneration schedules.
PG1(K)
, PG2(K)
. PGN(K)
; k =1,2,.NS
Out of these NS sets of generation schedules, the system operator has to choose that setof schedule which minimizes the system operating cost which is essentially the sum of theproduction costs of all the generating units. This economic dispatch problem is mathematically
stated as an optimization problem. Given the number of available generating units Ns their
production cost function, their operating limits and the system load PD.To determine the set of generating schedule PG,
Min FT =
N
iGii
PF1
. (1)
N
iDGi
PP1
=0 (2)
maxmin GiGiGi PPP (3)The unit production cost function is usually approximated by a quadratic function.
iGiiGiiGiicPbPaPF
2
i=1,2N (4)
where ai, bi and ci are constants.
The ED problem is given by the equations (1) to (4). By omitting the inequality constraint the
reduced ED problem may be restated as an unconstrained optimization problem by augmenting
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the objective function with the constraint function multiplied by Lagrange multiplier to
obtain the Lagrange function L as,
Min: L(PG1,..PGN, )=
N
i
N
iDGiGii
PPPF1 1
(5)
The necessary conditions for the existence of solution to (5) are given by,
0
GiP
L
Gi
Gii
dP
PdF; i=1,2..N (6)
N
iDGi
PPL
1
0
(7)
The solution to ED problem can be obtained by solving simultaneously the necessary conditions(6) and (7) which state that the economic generation schedules not only satisfy the system power
balance equation (8) but also demand that the incremental cost rates of all the units be equal to
which can be interpreted as incremental cost of received power when the inequality constraints(3) are included in the ED problem the necessary condition (6) gets modified as
i
ii
dPG
PGdF= for PGi,min PGi PGi, max
for PGi = PGi,max
for PGi = PGi, min (8)
Methods of Solution for ED without Loss
The solution to the ED problem with the production cost function assumed to be a quadratic
function, equation (4), can be obtained by simultaneously solving (6) and (7) using a directmethod as given below,
i
ii
dPG
PGdF )(= 2aiPGi + bi = ; i = 1,2, .......... N (9)
From Equation (9) we obtain
PGi = ( bi) /2ai ; i = 1,2,............N (10)Substituting Equation (10) in Equation (7) we obtain
N
i ib1 )( / 2ai = PD
N
i
N
ii
PDaba1 1
11)2/()2/1(
N
ii
N
iii
aabPD11
)2/1(/))2/(( (11)
Flowchart:
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Yes
Read the data co-efficients ai,bi,ci and PD
Find Pgi= 3- bi/2ai
Is PGi 1ValueAssume 2< 1Value
Sto
Start
Find Pgi = 2-bi/2ai
Calculate 3 = 2 + 212
12 PGPdPGPG
Assume initial value of Lagrangian, 1
Check
Pgi = Pd
Print the generator real power
as output
No
Yes
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The method of solution involves computing using equation (11) and than computing the
economic schedules PGi; i=1,2,........N using equation (10). In order to satisfy the operating limits
(3) the following iterative algorithm is to be used.
Algorithm for ED without loss (For quadratic production cost function)
Step 1: Compute using Equation (11)
Step 2: Compute using Equation (10) the economic schedules, PGi ; i = 1,2,........N
Step 3: If the computed PGi satisfy the operating limits
PGi, min PGi PGi, max ; i = 1,2,.........NThen stop, the solution is reached. Otherwise proceed to step 4
Step 4: Fix the schedule of the NV number of violating units whose generation PGiviolates the operating limits (12) at the respective limit, either PGi,max or PGi,min
Step 5: Distribute the remaining system load PD minus the sum of the fixed generation
schedules to the remaining units numbering NR (= N-NV) by computing using
Equation (11) and the PGi; NRi using equation (10) where NR is the set ofremaining units.
Step 6: Check whether optimality condition (8) is satisfied. If yes, stop the solution
Otherwise, release the generation schedule fixed at PGi,max or PGi,min of
those generators not satisfying optimality condition (8), include these units in the
remaining units, modify the setsNRNV
, and the remaining load. Go to step 5.
Sample Problem:Economic Dispatch without loss:
A power plant has three units with the following cost characteristics:
Rs/h9000P160P0.7C
h/Rs5000P270P1.0C
h/Rs5000P215P0.5C
3
2
33
2
2
22
1
2
11
where
s
iP are the generating powers in MW. The maximum and minimum loads allowable on
each unit are 150 and 39 MW. Find the economic scheduling for a total load of i) 320 MW
ii) 200 MW
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Solution:
Knowing the cost characteristics, incremental cost characteristics are obtained as
MWh/Rs160P1.4ICMWh/Rs270P2.0IC
MWh/Rs215P1.0IC
33
22
11
Using the equal incremental cost rule
160P1.4
270P2.0
215P1.0
3
2
1
Case i) Total load = 320 MW Since P1 + P2 + P3 = 320 we have
3201.4
160
2.0
270
1.0
215
i.e. 3201.4
160
2.0
270
1.0
215]
1.4
1
2.0
1
1.0
1[
i.e. 2.2143 = 784.2857
This gives = 354.193 RM / MWh
Thus P1 = ( 354.193 - 215 ) / 1.0 = 139.193 MW
P2 = ( 354.193 - 270 ) / 2.0 = 42.0965 MW
P3 = ( 354.193 -16.0 ) / 1.4 = 138.7093 MW
All sP'
i lie within maximum and minimum limits. Therefore, economic scheduling is
P1 = 139.193 MW
P2 = 42.0965 MWP3 = 138.7093 MW
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Economic Dispatch without loss using Lamda Iteration method:
The fuel cost equations are given by,F1 = 0.035 P1
2+ 15P1 + 20 Rs / hr
F2= 0.04 P22+ 10P2 + 30 Rs / hr
Assuming both the units are operating at all time then total load varies from 40 to 200 MW andthat the maximum and minimum load of each unit is 100 and 20 MW respectively. Find the
Incremental Production Cost and allocation of load between two units for minimum overall costthe given load.
P1 = 20 MW
P2 = 80 MW
4.1610)80(08.0
4.1615)20(07.0
2
2
1
1
dP
dF
dP
dF
2
2
1
1
dP
dF
dP
dF= 16.4
Solving the above Equations, we get
F1 = 0.035 (20)2
+ 15(20)+ 20 = 334 Rs / hr
F2 = 0.04 (80)2+ 10(80)+ 30 = 1086 Rs / hr
FT = F1 + F2 =1420 Rs / hr
Result:
The economic dispatch problem without transmission losses for a given load condition usingdirect method and Lambda-iteration method was studied by developing a MATLAB program.
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S.NO 12
Load Flow analysis using ETAP Software
Aim:To conduct the load flow analysis by using an ETAP power station.
Theory:
ETAP power station is a fully graphical electrical transient analyzer program that can operate
under the Microsoft windows 98, NT4-0, 2000 and XP environments. The Windows 2000 andXP Professional platforms provide the highest performance level for demanding applications,
such as large network analysis requiring intensive computation and online monitoring and
control applications. PowerStation allows us to work directly with graphical one-line diagrams,
underground cable raceway systems, ground grid systems and cable pulling systems. Powerstation combines the electrical, logical, mechanical and physical attributes of system elements in
the same data base.ETAP can simulate various power system problems like load flow analysis, short circuit
analysis, Harmonic analysis, Transient Stability analysis, Optimal power flow analysis, motor
acceleration analysis, Battery sizing discharge, DC load flow and DC short circuit analysis.
Power station organizers and accesses its database using Microsoft open database connectivity(ODBC)
General steps for ETAP Simulation:
Step 1: (To create a new project)
1. To start power station, double click on the power station icon on desktop. This willopen the window.
2. To create a new project, select the file menu option from the start up menu Bark click
on the first button on the project tool bar
3. The user information dialog box comes up after you click on ok from the createproject file.
4. Enter the user name, full name and description and password click on ok in dialogue
box.
Step 2: (Project Properties)
Under the project menu there are some options as follows to give or edit the properties. The
information and standard of the projects can be edited from this menu.
Step 3: (Edit a one line diagram)
One line diagram menu bar contains a comprehensive collection of menu options. This menu bar
is displayed when a one line diagram is active. In the one line diagram presentation (OLV1), we
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can graphically construct our electrical system by connecting the buses, branches motors etc.
from the one line diagram Edit tool bar.
Step 4: (For adding Components)
Click on the required symbol on the edit tool bar which changes the cursor shape to the elementspicture.
Step 5: (Rotation)
For this right click to bring up the menu and select one of the orientation
Edit Properties of the elements:
To change or edit properties of an element right click and select the properties to get the editor.
Relocate elements:
Select an element and move the cursor on top of it, the cursor becomes a move symbol. Nowdrag the element to a new position and release the left button.
Load Flow Analysis:
The PowerStation Load Flow Analysis program calculates the bus voltages, branch power
factors, currents, and power flows throughout the electrical system. The program allows for
swing, voltage regulated, and unregulated power sources with multiple utility and generatorconnections.
Run Load Flow Studies:
Select a study case from the Study Case Editor. Then click on the Run Load Flow Study icon to
perform a load flow study. A dialog box will appear to specify the output report name if the
output file name is set to Prompt. The study results will then appear on the one-line diagram andin the output report.
Update Cable Load Current:
Selecting the Update Cable Load Current icon will transfer cable load current data from the
previously run load flow study. The data is transferred to the Operating Load Current in theCable Editor for each cable associated with the load flow study.
Load Flow Display Options
The results from load flow studies are displayed on the one-line diagram. To edit how these
results look, click on the Load Flow Display Options icon.
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Alert View:
After performing a load flow study, you can click on this button to open the Alert View, whichlists all equipment with critical and marginal violations based on the settings in the study case.
Load Flow Report Manager:
Load flow output reports are provided in two forms: ASCII text files and Crystal Reports. The
Report Manager provides four pages (Complete, Input, Result, and Summary) for viewing thedifferent parts of the output report for both text and Crystal Reports. Available formats for
Crystal Reports are displayed in each page of the Report Manager for load flow studies.
Choosing any format other than Text Report in the Report Manager activates the crystal reports.
Result:
The results obtained for load flow analysis using ETAP power station was verified.
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S.NO 13
Fault analysis using MiPower Software
Aim:
To conduct fault analysis using Mipower software.Theory:POWERSCS module is designed to perform the short circuit study for the given system. Short
circuit studies are performed to determine the magnitude of the currents flowing throughout thepower system at various time intervals after a fault occurs. The magnitudes of current flowing
through the power system after a fault vary with time until they reach steady state condition. This
behavior is due to system characteristics and dynamics. The short circuit information is used to
select fuses, breakers and switchgear ratings in addition to setting protective relays. The shortcircuit program computes the steady state fault current for the impedance considered.
Procedure to enter data for performing studies using Mipower:
1. Draw single line diagram and enter data simultaneously in database manager.
2. Open power system network editor. Select menu option Database-configure.Configure database dialog box is popped up. Click browse button.
3. The elements can be selected from the power system tool bar.
4. The element ID can be selected by double click the element in the file. Enter the
details of the elements in detailed form.
5. Save and close the library screen.
6. To solve short circuit studies choose menu option solve-short circuit analysis.
7. Select the suitable fault in the fault type and select the bus no.
8. Click execute and short circuit study will be executed.
9. Click on report to view the report.
Result:The results obtained for fault analysis using Mipower software was verified.
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