LABORATORY MANUAL

EE0405 – SIMULATION LAB

PREPARED BYJ.PREETHA ROSELYN

(AP/Sr.G/EEE)

DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERINGFACULTY OF ENGINEERING & TECHNOLOGY

SRM UNIVERSITY, Kattankulathur – 603 203

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

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.

Syllabus

EE 0405 SIMULATION LAB L T P CPrerequisite 0 0 3 2EE 0302,EE 0308

PURPOSETo 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 following1. Single phase half controlled converter with R and RL load.2. Single phase fully controlled converter with R and RL load3. 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 following1. 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 following1. Load flow solution for GS, NR and FDLF2. Symmetrical and unsymmetrical fault analysis3. Transient stability analysis

TOTAL

REFERENCELaboratory Manual

Course designed by Department of Electrical and Electronics Engineering

Programoutcomes

a B c d e f g h i j kX X X X X

CategoryGeneral

(G)

Basic Sciences

(B)

Engineering Sciences

andTechnical Arts(E)

Professional Subjects(P)

X

Broad area (for ‘P’category)

Electrical Machines

Circuits and

Systems

Electronics Power System

X XStaff responsible for preparing the

syllabusMr.K.Vijayakumar

Date of preparation December 2006

Mapping of Course Outcomes with Instructional Objectives

Mapping of Program Instructional Objectives Vs Program Outcomes

Program Outcomes

Program Instructional objectives

Acquire skills of using computer packages MATLAB coding in Power System studies

Acquire skills of using computer packages MATLAB /SIMULINK in Power Electronics studies.

Acquire skills of using ETAP software for Power System studies

a)An ability to apply knowledge of mathematics, science, and engineering.

X X X

b) An ability to design and conduct experiments, as well as to analyze and interpret results.

X X X

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.

X X

e)An ability to identify, formulate, and solve engineering problems

X X

h)The broad education necessary to understand the impact of engineering solutions in a global perspective

X X X

Mapping of Program Educational Objectives with

Program Outcomes

Mapping of Program Educational Objectives Vs Program Outcomes

PROGRAM EDUCATIONAL OBJECTIVES

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(a–k OUTCOMES)

PROGRAM EDUCATIONAL OBJECTIVES

1 2 3 4

(a) an ability to apply knowledge of mathematics, science, and engineering X

(b) an ability to design and conduct experiments, as well as to analyze and interpret data

X

(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

X

(d) an ability to function on multidisciplinary teams

X X

(e) an ability to identify, formulate, and solve engineering problems

X

(f) an understanding of professional and ethical responsibility

X

(g) an ability to communicate effectively in both verbal and written form.

X

(h) the broad education necessary to understand the impact of engineering solutions in a global perspective.

X

(i) a recognition of the need for, and an ability to engage in life-long learning

X

(j) a knowledge of contemporary issues X

(k) an ability to use the techniques, skills, and modern engineering tools necessary for engineering practice.

X X

Academic Course DescriptionSRM University, Kattankulathur

Faculty of Engineering and Technology

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.

OutcomesStudents 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 System studies.

2. Acquire skills of using ETAP software forPower System Studies.

a)An ability to apply knowledge of mathematics, science, and engineeringb) An ability to design and conduct experiments, 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, and solve engineering problemsh)The broad education necessary to understand the impact of engineering solutions in a global perspective

Text book(s) and/or required materials:

1. P.S.Bimbhra, Power Electronics2. Nagrath and Kothari, Power System Analysis3. B.R.Gupta, Power System Analysis and Design

Web Resources:

www.power-analysis.comwww.4shared.com/power system analysiswww.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 of using computer packages MATLAB /SIMULINK in power electronics.

II Single phase fully controlled converter using R and RL load using MATLAB/ SIMULINK

III Three phase fully controlled converter using R and RL load using MATLAB/ SIMULINK

IV Single phase AC voltage regulator using MATLAB/ SIMULINK

V Formation of Y bus matrix by inspection/analytical method using MATLAB Software

Power system analysis-

Nagrath and Kothari

Acquire skills of using computer packages using MATLAB in power systems.

VI Formation of Zbus matrix using building algorithm using MATLAB Software

VII Gauss Seidal load flow analysis using MATLAB Software

VIII Fast decoupled load flow analysis using MATLAB Software

IX Symmetrical Fault analysis using MATLAB Software

X Economic Dispatch using MATLAB Software

XI Load flow analysis using ETAP Software

Acquire skills of using ETAP software for power system studies

XII Fault analysis using MIPOWER Software

Acquire skills of using MIPOWER software 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 professional presentation 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 lab reports.

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.

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 Lab session.

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 aweightage of 75%. The remaining 25% weightage will be given by conducting an endsemester practical examination for every individual student. Prelab questions will be asked 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 ENGINEERINGSRM UNIVERSITY, Kattankulathur – 603203.

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 52 Preparation of observation/record 103 Pre viva questions 54 Model Calculation 105 Execution 156 Post viva questions 5

TOTAL 50

Signature of the staff

S.NO. 1 Single Phase Half Wave Rectifier with R & RL load

Aim:

To simulate the 1Ø 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) 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 half controlled converter is shown in Figure 1. The output voltage is varied by controlling the firing angle of SCRs. The single phase half 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 is reverse 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 supply via SCR3, load and D4. At the end of negative half cycle, the freewheeling current will flow through the diode D2 and SCR3.

Circuit Diagram:

Model Graph:Resistive Load

Inductive load:

Result:

Thus the Single Phase half controlled Rectifier with R & RL Load circuit is simulated using MATLAB/SIMULINK and the corresponding waveforms are obtained.

S.NO.2Single Phase Full Wave Rectifier with R & RL Load

Aim:

To simulate the 1Ø fully 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) 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 phase fully 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 is resistive, 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, the load current keeps on increasing and becomes maximum at . At , the supply voltage reverses but 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:

Model Graph:Resistive load

Inductive load :

Result:

Thus the Single Phase fully controlled Rectifier with R & RL Load circuit is simulated using MATLAB/SIMULINK and the corresponding waveforms are obtained.

S.NO.3

Three Phase Fully controlled Rectifier with R & RL Load

Aim:

To simulate the 3Ø 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 angle delay 00<α≤900 and as three phase line commutated inverter for 900<α<1800. The numbering of SCRs 1, 3, 5 for the positive group and 2, 4, 6 for negative group. This numbering scheme is adopted here as it agrees with the sequence of gating of six thyristors in a 3-phase full converter.

Here each SCR is conduct for 1200. At any time two SCRs, one from positive group and other from negative group must conduct together and this combination must conduct for 600.this means commutation occurs for every 600. For ABC phase sequence of three phase supply thyristors conduct in pairs: T1 and T2, T2 and T3, T3 and T4, T4 and T5, T5 and T6, T6 and T1.

Period, range of 2π SCR Pair in conduction

α + 30 ° to α + 90 ° S1 and S6

α + 90 °to α + 150 ° S1 and S2

α + 150 ° to α + 210 ° S2 and S3

α + 210 ° to α + 270 ° S3 and S4

α + 270 ° to α + 330 ° S4 and S5

α + 330 ° to α + 360 ° and α + 0 ° to α + 30 ° S5 and S6

Formulae used:

Average output voltage

RMS value of output voltage

RMS of the source current is

Each SCR conducts for 1200 for every 3600. Therefore the RMS value of SCR current is

Circuit Diagram:

Model Graph:Resistive load:

Inductive load:

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:

To simulate the 1Ø 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 in metallurgical 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 these applications 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 isolation between 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) = VR

At this instant t= and hence 2V sin (-) = VR

= - sin- (VR/2V)R max=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

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 the breakdown 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 the reverse 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 Ø AC Voltage regulator with R load circuit is executed with the help of MATLAB software and the graph is plotted.

VS

V0

wt

wt

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. Ybus

matrix is often used in solving load flow problems. It has gained widespread applications owing to 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 generator and 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 be written as,

nnnnn

n

In

nV

V

V

YYY

YYY

YYY

I

I

I

.

.

....

..

..

....

....

.

.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

jij 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

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 the admittances 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

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 by inspection method.

Bus Code Impedance Bus Number Admittance1 - 2 0.06 + j0.18 1 j0.051 – 3 0.02 + j0.06 2 j0.062 - 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

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.

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 for formulating [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 be created (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 Zb between two existing buses.

Type 1 modification:

In type 1 modification, an impedance Zb is added between a new bus p and the reference bus as shown in Figure 1

Let the current through bus p be 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

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 p and an existing bus k as shown in Figure 2. The voltages across the bus k and p can be expressed as,

Vk(new) = Vk + Ip Zkk

Vp = Vk(new) + Ip Zp

= Vk + Ip(Zb + Zkk)

where, Vk is the voltage across bus k before the addition of impedance Zb

Zkk is 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

NetworkZb

k

Figure 2.Type 2 Modification of Zbus

Type 3 Modification:

In this modification, an impedance Zb is added between a existing bus k and a reference bus.Then the following steps are to be followed:

1. Add Zb between 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 as shown in Figure 3. From Figure 3, the relation between the voltages of bus k and j can be written as,

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

Zb

Ib

j

Figure 3.Type 4 Modification of Zbus

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).

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.

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 = j0.05Z12 =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)/j0.8

Z22 =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.

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 day operational 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 the projected 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 day’s 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- linear algebraic 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 independent loop equations. The network equations in terms of the bus admittance matrix can be written as,

busbusbus VYI (1)

For a n bus 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.

At the pth bus, 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

ppp V

jQPI

)( ( for any bus p except slack bus s) (5)

Substituting for Ip 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 these quantities 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

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:

Yes

Read the input data values

Start

Form Y Bus matrix

Set flat voltage profile 1+j0 except slack bus

Set iteration count, p=0

Set the bus count, i = 1

Check for slack bus

Check for Gen 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

min1 QQ p

i Set Qi=Qi min

Check

max1 QQ p

i SetQi=Qi max

A

Y

No

Yes

No

No

No

Yes

Yes

D

E

C

B

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 at the 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 voltage 1 piV

Increment the bus count

Checkni

Check 1p

iV

Print the result

Stop

Increment iteration count P = P+1

Yes

Yes

No

No

B

E

D

C

A

Bus Code Impedance Bus Number Admittance1 – 2 0.06 + j0.18 1 j0.051 – 3 0.02 + j0.06 2 j0.062 – 3 0.04 + j0.12 3 j0.05

Bus Code Assumed Voltage

Generation LoadMW MVAr MW MVAr

1 1.06 + j0 0 0 0 02 1 + j0 0.2 0 0 03 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

qqpqVYV

1

*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 jQ2 = 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

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.

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 system network. 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 of alternative plans for future system expansion to meet increased load demand. Newton-Raphson method is an iterative method that approximates the set of non linear simultaneous equations to a set of linear simultaneous equations using Taylor’s 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 pth bus is given by,

...................,2;1

**

1

*

npVYVVYV

IVS

n

qqpqp

n

qqpqp

ppp

(1)

pqjpqpq

qppq

qjqq

pjpp

eYY

eVV

eVVLet

,

and

In polar co-ordinates, the power on pth bus is given as,

pqjpq

qpjn

qqpppp eYeVVjQPS ||

1

(2)

Separating the Real and Imaginary parts we get,

)cos(1

qpqppq

n

qqpp YVVP

)sin(1

qpqppq

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

nr

nr

n

rn

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

riV 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,

All quantities in

the linear

Equation (4) pertai

n to iterati

on r. The

linear equati

on when solved for , ΔV 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 is repeated. Convergence is tested by the power mismatch criteria. This method converges to high accuracy nearly always in 2 to 5 iterations from a flat start (|V| = 1 p.u. and θ =0 ) for all buses where |V|, θ 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 to a PQ bus. When Q is within limits, then it is switched back to PV bus.

)7()sin(

)6()sin(

:

1

11

qpqppqq

n

pqq

pp

p

qpqppqqpq

p

YVVP

pqYVVP

J

)9()sin(sin2

)8()sin(

:

1

22

qpqppq

n

pqq

qppppp

p

p

qpqppqp

q

p

YVYVV

Q

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

pp

p

qpqppqqpq

p

YVVQ

pqYVVQ

J

Algorithm:

Start

Read bus data, line data, bus power & 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 to

account

Update voltage magnitude and phase angles

VVVoldnew

oldnewat

all buses except slack bus

∆P≤tol∆Q≤tol

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:

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 pV o 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 of pQ 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 and k

pQ for all load buses are within a

specified tolerance, ie, kpP < , k

pQ <

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 2 is 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

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 systemBus code Impedance Line charging

admittance y’pq /21-2 0.06+j0.18 j0.051-3 0.02+j0.06 j0.062-3 0.04+j0.12 j0.05

Table 2: Assumed bus voltages, Generation and loadsBus code Voltages

p.u

GenerationMW MVARp.u p.u

LoadMW MVARp.u p.u

1 1.06 0 0 0 02 1 0.2 0 0 03 1 0 0 0 0.25

Solution:

Formation of Ybus :

Ybus =

333231

232221

131211

YYY

YYY

YYY

Y12 = -18.006.0

1

j=-(1.667-j5)

= 5.27 04.108

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

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 ; δ1 = 00 ; V3 = 1 00

Choose V20 =1+j0 and δ2

0 = δ30 = 0

Calculation of change in real and reactive powers:

Pp = Pp(specified) – Pp(calculated)

Qp = Qp(specified) – Qp(calculated)

n

qqpqppqqpp

n

qqpqppqqpp

YVVQ

YVVP

1

1

)sin(

)cos(

P2(cal) = |V2|

2|Y22|cos θ22 + |V2||V1||Y21|cos(δ2 + θ21 - δ1) + |V2||V3||Y23|cos(δ2 + θ23 – δ3) = 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 θ33 +|V3||V1||Y31|cos(δ3 + θ31 – δ1) + |V3||V2|Y32|cos( θ32 + δ3 - δ2 ) = 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 θ22 + |V2||V1||Y21|sin(δ2 + θ21 – δ1) + |V2||V3||Y23|sin(δ2 + θ23 – δ3)

= 1 13.08 sin(-71.5) + 1 1.06 5.27sin(108.4) + 1 7.906sin(108.4) = 0.4 p.u

Q3(cal) = |V3|2|Y33|sin θ33 + |V3||V1||Y31|sin(δ3 + θ31 – δ1) + |V3||V2||Y32|sin(δ3 + θ32 – δ2)

= 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,

ΔP20 = P2(specified) – P2(calculated) = 0.2 + 0.11 = 0.31 p.u.

ΔP30 = 0 +( - 0.3) = -0.3 p.u.

ΔQ20 = 0- 0.4 = -0.4 p.u.

ΔQ30 = -0.25 -1.02 = -1.27 p.u.

Calculation of Jacobian matrix elements :

Elements of J1:

22

2 V

P|V2||Y22|cos θ22 +|V1||Y21|cos(δ2 + θ21 - δ1) + |V3||Y23|cos(δ3 + θ23 – δ2)

= 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(δ2 + θ23 – δ3)

= 1 7.906 cos(108.4) = -2.5

2

3

V

P |V3||Y32|cos( θ32 + δ2 - δ3 )

= 1 7.906 cos(108.4) = -2.5

3

3

V

P2|V3||Y33|cos θ33 +|V1||Y31|cos(δ1 + θ31 – δ3) + |V2||Y32|cos( θ32 + δ2 - δ3 )

= 2 1 23.61 cos(-71.5) + 1.06 15.81cos(108.4) + 1 7.906cos(108.4) = 7.2

Elements of J2:

2

2

P

|V2||V1||Y21|sin(δ2 + θ21 – δ1) - |V2||V3||Y23|sin(δ3 + θ23 – δ2)

= 1 1.06 5.27sin(108.4) – 1 1 7.906 sin(108.4) = -12.8

3

2

P

- |V2||V3||Y23|sin(δ2 + θ23 – δ3)

= - 1 1 7.906 sin (108.4) = -7.5

2

3

P

- |V3| |V2||Y32|sin(δ3 + θ32 – δ2)

= - 1 1 7.906 sin (108.4) = -7.5

3

3

P

-|V3||V1||Y31|sin(δ3 + θ31 – δ1) - |V3||V2||Y32|sin(δ3 + θ32 – δ2)

= - 1 1.06 15.81sin(108.4)-1 17.906sin(108.4) = - 23.4

Elements of J3:

2

2

V

Q{2|V2||Y22|sin θ22 + |V1||Y21|sin(δ2 + θ21 – δ1) +|V3||Y23|sin(δ2 + θ23 – δ3 ) }

= {2113.08 sin(-71.5) + 1.065.27sin(108.4) + 17.906sin(108.4)} = - 12.01

{3

2

V

Q|V2||Y23|sin(δ2 + θ23 – δ3) }

={ 1 7.906sin(108.4)} = 7.5

{2

3

V

Q|V3||Y32|sin(δ2 + θ32 – δ3)}

= { 1 7.906sin(108.4)} = 7.5

{3

3

V

Q2|V3||Y33|sin θ33 - |V1||Y31|sin(δ3 + θ31 – δ1) - |V2||Y32|sin(δ3 + θ32 – δ2)

= {2 1 23.61 sin(-71.5) +1 1.06 15.81sin(108.4) + 1 7.906 sin (108.4)} = 21.4

Elements of J4:

{2

2 Q

|V2||V1||Y21|cos(δ2 + θ21 - δ1)+ |V2||V3||Y23|cos(δ2 + θ23 – δ3) }

= 11.065.27cos(108.4) + 17.906cos(108.4) = -4.26

{32

Q

|V2||V3||Y23|cos(δ2 + θ23 – δ3) }

= - { 17.906cos(108.4)} = 2.5

{3

3 Q

|V3||V1||Y31|cos(δ3 + θ31 – δ1) + |V3||V2||Y32|cos(δ3 + θ32 – δ2)

= 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 conducted using MATLAB and results was verified.

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 best operation 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 iteration consequently. 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

mkkm E

EPN

m

kkm

QJ

; m

mkkm E

EQL

q

ppq

PH

; q

qP

pq E

EQL

The equation for power flow are again expressed below for calculating elements of Jacobian (ie H & L)

n

pqqppqpqqpppppqpp YEEYEEP

,1

coscos

qppqpqqpppppqpp YEEYEEQ sinsin

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

pPQ

PH

sin

Start

Read the input data

Form the Y bus matrix

Form B’ and B” matrix

Set flat voltage profile except for slack bus

Set convergence criterion є

Set iteration count p=0

Calculate real & reactive power

Calculate

V

Q

V

P,

Are

V

Q

V

P,

Calculate real & reactive line flow, bus powers

Print the results

Stop

VLN

MH

Q

P

Find V & by solving the equations:

Q

V

V 43

21

Uptate voltage magnitude t phase angles

VVVoldnew

oldnew

C

C

No

Yes

Flowchart:

Algorithm:

Step 1: Read the slack bus voltages, real bus powers and reactive bus powers, bus voltage 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|] , [ΔQ/|V|]

Step 9: If ΔP/ |V| & ΔQ/|V| are less than or equal to tolerance limit, solution has convergence and go to step 12 otherwise increase iteration count and go to step 10.

Step 10: Calculate [Δδ] = [B’]-1 [ΔP/ |V|]

[Δ|V|] = [B’’]-1[ΔQ/|V|]

Step 11: Update [δ] & [|V|] for all buses except slack bus. [δ]new = [δ]old + [Δ δ]; [|V|]new = [|V|] old + [Δ|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

Bus specifications:

Bus code

Assumed volt. Generation LoadP Q P Q

1 1 - - - -2 1.1 5.52 0≤QG2≤5.5 0.7 0.13 - - - 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:

003

002

001

00.101

01.101.1

00.101

jV

jV

jV

Calculation of P and Q:

)cos(1

qpqppq

n

qqpp YVVP

0

00,3

0,2

cal

cal

P

P

n

qqpqppqqpp YVVQ

1

)sin(

= 2323333222222221212112 sinsin..sin.. YVVYVVYVV

= -1.65

As 0≤QG2≤5.5Q2= QG2-QD2

QG2=-1.65+0.1 = -1.55 Hence 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 ΔP2, ΔP3, ΔQ3

Let the largest change of ΔE= 4.82 ΔE≤ E; 4.82≤0.01

Find Δδ 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

Δδ2=0.2045

Δδ3=-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

003

03

13

002

02

12

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.

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 are also 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 are caused 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 on the 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 system voltage 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 synchronism

The symmetrical fault occurs when all the three conductors of a three-phase line are brought together simultaneously into a short–circuit 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

AB

C

Figure 1 Symmetrical Fault on Three-Phase system

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

IsI < nb

Print the Result

Stop

Compute Zbus matrix by inverting modified bus

Yes

No

Formula Used:

i) Fault Current, If = ppf

ZZ

V

ii) Fault Voltage, Vf = )1(ppf

bus

ZZ

ZV

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.

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

1j

jjj

Y33 = 83.4508.0

1

03.0

1j

jj

Y12=Y21== 69.713.0

1j

j

Y13=Y31== 5.1208.0

1j

j

Y23=Y32== 33.3303.0

1j

j

We can formulate the bus admittance matrix

If=0

S

1

E1=1.0 E2=1.0

j0.25 j0.2

j0.13

j0.08 j0.03

2

3

Single line diagram

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 = ..462.71343.0

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

VV ff

I13f = ..999.1

1059.0

02114.0

13

31 upjjZ

VV ff

I23f = ..2434.1

115.0

0143.0

23

32 upjjZ

VV ff

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.

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 number of 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 by considering 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 be developed 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 of generation 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 set of schedule which minimizes the system operating cost which is essentially the sum of the production 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 GiGiGiPPP (3)

The unit production cost function is usually approximated by a quadratic function.

iGiiGiiGii

cPbPaPF 2

i=1,2……N (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

the objective function with the constraint function Φ multiplied by Lagrange multiplier λ to obtain the Lagrange function L as,

Min: L(PG1,…..PGN, λ)=

N

i

N

iDGiGii

PPPF1 1

(5)

The necessary conditions for the existence of solution to (5) are given by,

0Gi

P

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 λ 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= λ for PGi,min PGi PGi, max

λ for PGi = PGi,max

λ 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 direct method as given below,

i

ii

dPG

PGdF )(= 2aiPGi + bi = λ; i = 1,2, .......... N (9)

From Equation (9) we obtainPGi = (λ – bi) /2ai ; i = 1,2,............N (10)Substituting Equation (10) in Equation (7) we obtain

N

ii

b1

)( / 2ai = PD

N

i

N

ii

PDaba1 1

11)2/()2/1(

N

ii

N

iii

aabPD11

)2/1(/))2/(( (11)

Flowchart:

Yes

Read the data co-efficients ai,bi,ci and PD

Find Pgi= λ3- bi/2ai

Is∑ PGi< PD No

Find Pgi = λ- bi/2ai ; i= 1,2,…….NG

Assume λ2 > λ1 ValueAssume λ2< λ1 Value

Stop

Start

Find Pgi = λ2-bi/2ai

Calculate λ3 = λ2 + 212

12 PGPdPGPG

Assume initial value of Lagrangian, λ1

Check ∑Pgi = Pd

Print the generator real poweras output

No

Yes

The method of solution involves computing λ 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 λ 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,.........N Then stop, the solution is reached. Otherwise proceed to step 4

Step 4: Fix the schedule of the NV number of violating units whose generation PGi

violates 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 λ using

Equation (11) and the PGi; NRi using equation (10) where

NR is the set of

remaining 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 sets NRNV

, 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

3233

2222

1211

where siP 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

Solution:

Knowing the cost characteristics, incremental cost characteristics are obtained as

MWh / Rs160P1.4IC

MWh / Rs270P2.0IC

MWh / Rs215P1.0IC

33

22

11

Using the equal incremental cost rule

λ160P1.4

λ270P2.0

λ215P1.0

3

2

1

Case i) Total load = 320 MW Since P1 + P2 + P3 = 320 we have

3201.4

160λ

2.0

270λ

1.0

215λ

i.e. 3201.4

160

2.0

270

1.0

215]

1.4

1

2.0

1

1.0

1[λ

i.e. 2.2143 λ = 784.2857

This gives λ = 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 MW

P3 = 138.7093 MW

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 P2

2+ 10P2 + 30 Rs / hr

Assuming both the units are operating at all time then total load varies from 40 to 200 MW and that 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 cost the 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 using direct method and Lambda-iteration method was studied by developing a MATLAB program.

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 and XP 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. Power station 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 will

open 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 bar3. The user information dialog box comes up after you click on ok from the create

project 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

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 elements picture.

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. Now drag 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 generator connections.

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 and in 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 the Cable 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.

Alert View:

After performing a load flow study, you can click on this button to open the Alert View, which lists 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 the different 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.

S.NO 13Fault 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 the power 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 short circuit 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.