INTRODUCTION TO MATLABExpt.No:Date :

OBJECTIVES

(i) To procure sufficient knowledge in MATLAB to solve the power systemProblems.

(ii) To write a MATLAB program.

SOFTWARE REQUIRED

(i) MATLAB

1. INTRODUCTION TO MATLAB

MATLAB is a high performance language for technical computing. It integratescomputation, visualization and programming in an easy-to-use environment where problems andsolutions are expressed in familiar mathematical notation.

MATLAB is numeric computation software for engineering and scientific calculations.MATLAB is primary tool for matrix computations. MATLAB is being used to simulate randomprocess, power system, control system and communication theory.

MATLAB comprising lot of optional tool boxes and block set like control system,optimization, and power system and so on.

1.1. TYPICAL USES INCLUDE

Math and computation.

Algorithm development.

Modeling, simulation and prototype.

Data analysis, exploration and Visualization.

Scientific and engineering graphics.

Application development, including graphical user interface building.

MATLAB is a widely used tool in electrical engineering community. It can be used forsimple mathematical manipulation with matrices for understanding and teaching basicmathematical and engineering concepts and even for studying and simulating actual power systemand electrical system in general. The original concept of a small and handy tool has evolved to

become an engineering work house. It is now accepted that MATLAB and its numerous tool boxesreplace and/or enhance the usage of traditional simulation tool for advanced engineeringapplications.

Engineering personnel responsible for studies of electrical power system, control systemand power electronics circuits will benefit from the MATLAB. To expertise in Electrical SystemSimulation one should have a basic understanding of electric circuits, power system and powerelectronics.

1.2. GETTING STARTED WITH MATLAB

To open the MATLAB applications double click the Matlab icon on the desktop. This willopen the MATLAB window space with Matlab prompt as shown in the fig.1.

Fig-1: MATLAB window space

To quit from MATLAB type…

>> quit(Or)

>>exitTo select the (default) current directory click ON the icon […] and browse for the folder

named “D:\SIMULAB\xxx”, where xxx represents roll number of the individual candidate inwhich a folder should be created already.

When you start MATLAB you are presented with a window from which you can entercommands interactively. Alternatively, you can put your commands in an M- file and execute it atthe MATLAB prompt. In practice you will probably do a little of both. One good approach is toincrementally create your file of commands by first executing them.

M-files can be classified into following 2 categories,

i) Script M-files – Main file contains commands and from whichfunctions can also be called.

ii) Function M-files – Function file that contains function command at thefirst line of the M-file.

M-files to be created by you should be placed in your default directory. The M-filesdeveloped can be loaded into the work space by just typing the M-file name.

To load and run a M-file named “ybus.m” in the workspace.

>>ybus

These M-files of commands must be given the file extension of “.m”. However M-files arenot limited to being a series of commands that you don’t want to type at the MATLAB window,they can also be used to create user defined function. It turns out that a MATLAB tool box isusually nothing more than a grouping of M-files that someone created to perform a special type ofanalysis like control system design and power system analysis. Any of the matlab commands (eg:sqrt) is really an M-file.

One of the more generally useful matlab tool boxes is simulink – a drag and-drop dynamicsystem simulation environment. This will be used extensively in laboratory, forming the heart ofthe computer aided control system design (CACSD) methodology that is used.

>>simulinkAt the matlab prompt type simulink and brings up the “Simulink Library Browser”. Each of

the items in the Simulink Library Browser are the top level of a hierarchy of palette of elementsthat you can add to a simulink model of your own creation. At this time expand the “simulink”pallete as it contains the majority of the elements you will use in this course. Simulink has builtinto it a variety of integration algorithm for integrating the dynamic equations. You can place thedynamic equations of your system into simulink in four ways.

1 Using integrators2. Using transfer functions.3. Using state space equations.4. Using S- functions (the most versatile approach)

Once you have the dynamics in place you can apply inputs from the “sources” palettes andlook at the results in the “sinks” palette.

Finally the most important MATLAB features are its help. At the MATLAB Promptsimply typing helpdesk gives you access to searchable help as well as all the MATLAB manuals.

>>helpdesk

To get the details about the command name sqrt, just type…

>>help sqrt

Where sqrt is the command name and you will get pretty good description in the MATLABwindow as follows./SQRT Square root.

SQRT(X) is the square root of the elements of X. Complexresults are produced if X is not positive.

See also SQRTM.

Overloaded methodshelp sym/sqrt.m

1.3 MATLAB WORKSPACEThe workspace is the window where you execute MATLAB commands (Ref. figure-1). The

best way to probe the workspace is to type whos. This command shows you all the variables thatare currently in workspace. You should always change working directory to an appropriatelocation under your user name.

Another useful workspace-like command is>>clear all

It eliminates all the variables in your workspace. For example start MATLAB and executethe following sequence of commands

>>a=2;>>b=5;>>whos>>clear all

The first two commands loaded the two variables a and b to the workspace and assignedvalue of 2 and 5 respectively. The clear all command clear the variables available in the workspace. The arrow keys are real handy in MATLAB. When typing in long expression at thecommand line, the up arrow scrolls through previous commands and down arrow advances theother direction. Instead of retyping a previously entered command just hit the up arrow until youfind it. If you need to change it slightly the other arrows let you position the cursor anywhere.

Finally any DOS command can be entered in MATLAB as long as it is preceded by anyexclamination mark.

>>!dir1.4 MATLAB Data Types

The most distinguishing aspect of MATLAB is that it allows the user to manipulate vectors(like 5+j8) and matrices with the same ease as manipulating scalars (like5,8). Before diving intothe actual commands everybody must spend a few moments reviewing the main MATLAB datatypes. The three most common data types you may see are,

1) arrays 2) strings 3) structures.As for as MATLAB is concerned a scalar is also a 1 x 1 array. For example clear your

workspace and execute the commands.

>>a=4.2:>>A=[1 4;6 3];>>whos

Two things should be evident. First MATLAB distinguishes the case of a variable nameand that both a and A are considered arrays. Now let’s look at the content of A and a.

>>a>>A

Again two things are important from this example. First anybody can examine thecontents of any variables simply by typing its name at the MATLAB prompt. Second, when typingin a matrix space between elements separate columns, whereas semicolon separate rows. Forpractice create the matrix in your workspace by typing it in all the MATLAB prompt.

>>B= [3 0 -1; 4 4 2;7 2 11];(use semicolon(;) to represent the end of a row)

>>BArrays can be constructed automatically. For instance to create a time vector where the

time points start at 0 seconds and go up to 5 seconds by increments of 0.001

>>mytime =0:0.001:5;Automatic construction of arrays of all ones can also be created as follows,

>>myone=ones (3,2)Note:

Any MATLAB command can be terminated by a semicolon, which suppressed anyecho information to the screen.

1.5 Scalar versus Array Mathematical Operation

Since MATLAB treats everything as an array, you can add matrices as easily asscalars.Example:

>>clear all>> a=4;>> A=7;>>alpha=a+A;>>b= [1 2; 3 4];>>B= [6 5; 3 1];>>beta=b+B

Of course cannot violate the rules of matrix algebra which can be understood from the followingexample.

>>clear all>>b=[1 2;3 4];>>B=[6 7];>>beta=b*B

In contrast to matrix algebra rules, the need may arise to divide, multiply, raise to apower one vector by another, element by element. The typical scalar commands are used for this

“+,-,/, *, ^” except you put a “.” in front of the scalar command. That is, if you need to multiply theelements of [1 2 3 4] by [6 7 8 9], just type...

>>[1 2 3 4].*[6 7 8 9]

1.6 Conditional Statements

Like most Programming languages, MATLAB supports a variety of conditional statementsand looping statements. To explore these simply type

>>help if>>help for>>help while

Example :>>if z=0>>y=0>>else>>y=1/z>>end

Looping :>>for n=1:2:10>>s=s+n^2>>end

- Yields the sum of 1^2+3^2+5^2+7^2+9^21.7 PLOTTING

MATLAB’s potential in visualizing data is pretty amazing. One of the nice features is thatwith the simplest of commands you can have quite a bit of capability.Graphs can be plotted and can be saved in different formulas.

>>clear all>>t=0:10:360;>>y=sin (pi/180 * t);

To see a plot of y versus t simply type,>>plot(t,y)

To add label, legend, grid and title use

>>xlabel (‘Time in sec’);>>ylabel (‘Voltage in volts’)>>title (‘Sinusoidal O/P’);>>legend (‘Signal’);

The commands above provide the most plotting capability and represent several shortcutsto the low-level approach to generating MATLAB plots, specifically the use of handle graphics.The helpdesk provides access to a pdf manual on handle graphics for those really interested in it.1.8 Functions

As mentioned earlier, a M-file can be used to store a sequence of commands or a user-defined function. The commands and functions that comprise the new function must be put in afile whose name defines the name of the new function, with a filename extension of '.m'.A functionis a generalized input/output device. That is you can give some input.(arguments) and providessome output. MATLAB functions allow you much capability to expand MATLAB’s usefulness.We will just touch on function here as you may find them beneficial later.We will start by looking at the help on functions :

>>help functionWe will create our own function that given an input matrix returns a vector containing theadmittance matrix(y) of given impedance matrix(z)’

z=[5 2 4;1 4 5] as input, the output would be,

y=[0.2 0.5 0.25;1 0.25 0.2] which is the reciprocal of each elements.

To perform the same name the function “admin” and noted that “admin” must bestored in a function M-file named “admin.m”. Using an editor, type the following commands andsave as “admin.m”.admin.m :

function y = admin(z)y = 1./z

returnSimply call the function admin from the workspace as follows,

>>z=[5 2 4;1 4 5]

>>admin(z)The output will be,

ans = 0.2 0.5 0.251 0.25 0.2

Otherwise the same function can be called for any times from any script file provided the functionM-file is available in the current directory.

With this introduction anybody can start programming in MATLAB and can beupdated themselves by using various commands and functions available. Concerned with the“Power System Simulation Laboratory”, initially solve the Power System Problems manually, listthe expressions used in the problem and then build your own MATLAB program or function.

RESULT:

Expt. No: Date:

COMPUTATION OF LINE PARAMETERS USING MATLAB AIM

To determine the positive sequence line parameters L and C per phase per kilometre

of a three phase single and double circuit transmission lines for different conductor

arrangements and to understand modeling and performance of medium lines.

SOFTWARE REQUIRED: MATLAB 7.7

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THEORETICAL BACK GROUND:-

Line Parameters

Transmission line has four electrical parameters - resistance, inductance, capacitance and conductance. The inductance and capacitance are due to the effect of magnetic and electric fields around the conductor. The shunt conductance characterizes the leakage current through insulators, which is very small and can be neglected. The parameters R, L and C are essential for the development of the transmission line models to be used in power system analysis both during planning and operation stages.

While the resistance of the conductor is best determined from manufactures data, the inductances and capacitances can be evaluated using formula. The student is advised to read any other text book before taking up the experiment.

INDUCTANCE

The inductance is computed from flux linkage per ampere. In the case of the three phase lines, the inductance of each phase is not the same if conductors are not spaced equilaterally. A different inductance in each phase results in unbalanced circuit. Conductors are transposed in order to balance the inductance of the phases and the average inductance per phase is given by simple formulas, which depends on conductor configuration and

PROCEDURE

Enter the command window of the MATLAB.

Create a new M – file by selecting File - New – M – File.

Type and save the program in the editor window.

Execute the program by pressing Tools – Run.

View the results.

EXERCISE

1.A 500kv 3φ transposed line is composed of one ACSR 1,272,000-cmil, 45/7 bittern

conductor per phase with horizontal conductor configuration as show in fig.1. The conductors

have a diameter of 1.345in and a GMR of 0.5328in. Find the inductance and capacitance per

phase per kilometer of the line and justify the result using MATLAB.

A D12=35’ B D 23=35’ C

D =70’

Fig.1

2.The transmission line is replaced by two ACSR 636,000-cmil, 24/7 Rook conductors which

have the same total cross-sectional area of aluminum as one bittern conductor. The line

spacing as measured from the centre of the bundle is the same as before and is shown in fig.2.

The conductors have a diameter of 0.977in and a GMR of 0.3924in.Bundle spacing is 18in .Find the inductance and capacitance per phase per kilometer of the line and justify the result

using MATLAB.

18” A B C

D12=35’ D 23=35’

D13=70’

Fig.2

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3.A 345- KV double –circuit three- phase transposed line is composed of two ACSR,

1,431,000-cmil, 45/7 Bobolink conductors per phase with vertical conductor configuration as

shown in fig.3. The conductors have a diameter of 1.427in and a GMR of 0.564 in .the

bundle spacing in 18in. find the inductance and capacitance per phase per kilometer of the

line and justify the result using MATLAB.

a S11=11m

c’ H12 = 7m

b S22=16.5m b’

H12 = 6.5m

S33=12.5m

c a’ Fig.3

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RESULT

Thus the positive sequence line parameters L and C per phase per kilometre of a three

phase single and double circuit transmission lines for different conductor arrangements were

determined and verified with MATLAB software. The value of L and C obtained from MATLAB program are: Case1: L= C= Case2: L= C= Case3: L= C=

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Expt. No: Date:

MODELING OF TRANSMISSION LINES

AIM

To understand modeling and performance of Short, Medium and Long transmission lines.

SOFTWARE REQUIRED: MATLAB 7.7 THEORY

The important considerations in the design and operation of a transmission line are the

determination of voltage drop, line losses and efficiency of transmission. These values are

greatly influenced by the line constants R, L and C of the transmission line. For instance, the

voltage drop in the line depends upon the values of above three line constants. Similarly, the

resistance of transmission line conductors is the most important cause of power loss in the line

and determines the transmission efficiency.

A transmission line has three constants R, L and C distributed uniformly along the

whole length of the line. The resistance and inductance form the series impedance. The

capacitance existing between conductors for 1-phase line or from a conductor to neutral for a 3-

phase line forms a shunt path throughout the length of the line.

Short Transmission Line:

When the length of an overhead transmission line is upto about 50km and the line

voltage is comparatively low (< 20 kV), it is usually considered as a short transmission line.

Medium Transmission Lines:

When the length of an overhead transmission line is about 50-150 km and the line

voltage is moderatly high (>20 kV < 100 kV), it is considered as a medium transmission line.

Long Transmission Lines:

When the length of an overhead transmission line is more than 150km and line voltage

is very high (> 100 kV), it is considered as a long transmission line.

Voltage Regulation:

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The difference in voltage at the receiving end of a transmission line between conditions of

no load and full load is called voltage regulation and is expressed as a percentage of the receiving

end voltage.

Performance of Single Phase Short Transmission Lines As stated earlier, the effects of line capacitance are neglected for a short transmission line.

Therefore, while studying the performance of such a line, only resistance and inductance of

the line are taken into account. The equivalent circuit of a single phase short transmission line

is shown in Fig. (i).Here, the total line resistance and inductance are shown as concentrated

or lumped instead of being distributed. The circuit is a simple a.c. series circuit. Let I = load current R = loop resistance i.e., resistance of both conductors XL = loop reactance VR = receiving end voltage cos ϕR = receiving end power factor (lagging) VS = sending end voltage cos ϕS = sending end power factor The phasor diagram of the line for lagging load power factor is shown in Fig. (ii). From the

right angled traingle ODC, we get,

(OC)2 = (OD)2 + (DC)

2

VS2 = (OE + ED)2 + (DB + BC)

2

= (VR cos ϕR + IR)2 + (VR sinϕR + IXL)

2

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An approximate expression for the sending end voltage VS can be obtained as follows. Draw

perpendicular from B and C on OA produced as shown in Fig. 2. Then OC is nearly equal to

OF i.e., OC = OF = OA + AF = OA + AG + GF

= OA + AG + BH VS = VR + I R cos ϕR + I XL sin ϕR Medium Transmission Line

Nominal T Method In this method, the whole line capacitance is assumed to be concentrated at the middle

point of the line and half the line resistance and reactance are lumped on its either side as

shown in Fig.1, Therefore in this arrangement, full charging current flows over half the line.

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In Fig.1, one phase of 3-phase transmission line is shown as it is advantageous to work in

phase instead of line-to-line values.

Fig.1 Let IR = load current per phase

R = resistance per phase

XL = inductive reactance per phase

C = capacitance per phase

cos ϕR = receiving end power factor (lagging)

VS = sending end voltage/phase

V1 = voltage across capacitor C

The phasor diagram for the circuit is shown in Fig.2. Taking the receiving end voltage VR as the reference phasor, we have, Receiving end voltage, VR = VR + j 0

Load current, IR = IR (cos ϕR - j sin ϕR)

Fig.2

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Nominal π Method

In this method, capacitance of each conductor (i.e., line to neutral) is divided into two

halves; one half being lumped at the sending end and the other half at the receiving end as

shown in Fig.3. It is obvious that capacitance at the sending end has no effect on the line

drop. However, it’s charging current must be added to line current in order to obtain the total

sending end current.

Fig.3 IR = load current per phase R = resistance per phase XL = inductive reactance per phase C = capacitance per phase cos R = receiving end power factor (lagging) VS = sending end voltage per phase The phasor diagram for the circuit is shown in Fig.4. Taking the receiving end voltage as the reference phasor, we have, VR = VR + j 0 Load current, IR = IR (cos R - j sin R)

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and the inductance per phase is 1.3623 mH per km. The shunt capacitance is negligible. Use

the short line model to find the voltage and power at the sending end and the voltage

regulation and efficiency when the line supplying a three phase load of

b) 381 MVA at 0.8 power factor leading at 220 KV.

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EXERCISE ( Short Transmission Line)

A 220- KV, 3φ transmission line is 40 km long. The resistance per phase is 0.15 Ω per km

a) 381 MVA at 0.8 power factor lagging at 220 KV.

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RESULT

Thus the program for modeling of transmission line was executed by using

MATLAB and the output was verified with theoretical calculation.

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Voltage regulation REG =

FOR LAGGING PF: FOR LEADING PF:

Efficiency EFF =

Voltage regulation REG =

Efficiency EFF =

Voltage regulation REG =

Efficiency EFF =

The value of the voltage regulation and efficiency obtained from the MATLAB

program are

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Expt. No: Date:

1919

FORMATION OF BUS ADMITTANCE & BUS IMPEDANCE

MATRIX USING MATLAB

AIM To determine the bus admittance & bus impedance matrix for the given power system Network.

OBJECTIVES

SOFTWARE REQUIRED

THEORETICAL BACKGROUND

Network Description of a Multimode Power System

The bus admittance matrix Y and bus impedance matrix Z are the two important network descriptions of interconnected power system. The injected bus currents and bus voltages of a power system under steady state condition can be related through these matrices as

Y V = I

MATLAB.

To obtain network solution using these matrices.

To determine the bus admittance and impedance matrices for the given power system network.

To obtain certain specified columns of the bus impedance matrix Z

or the full matrix Z using the factors of Y or the inverse of Y.

Z I = V

SOFTWARE REQUIRED: MATLAB 7

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MANUAL CALCULATION:(Problem-1)-YBUS MATRIX

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MANUAL CALCULATION:(Problem-2)-Z BUS MATRIX

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MANUAL CALCULATION:(Problem-3)- Z BUS MATRIX-Building Algorithm

RESULT

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Expt. No: Date:

LOAD FLOW ANALYSIS BY GAUSS SEIDAL METHOD USING MATLAB AIM

To carry out load flow analysis of the given power system network by Gauss Seidal

method.

SOFTWARE REQUIRED: MATLAB 7.7

THEORY

Load flow analysis is the study conducted to determine the steady state operating

condition of the given system under given conditions. A large number of numerical

algorithms have been developed and gauss seidal method is one of such algorithm.

PROBLEM FORMULATION

The performance equation of the power system may be written of

[I bus]=[Y bus][V bus] (1)

Selecting one of the buses as the reference bus, we get (n-1) simultaneous equations.

The bus loading equations can be written as

Ii = Pi-jQi / Vi* (i=1,2,3,…………..n) (2)

Where,

n

Pi=Re [ ∑ Vi*Yik Vk.] (3)

K=1

n

Qi= -Im [ ∑ Vi*Yik Vk]. (4)

k=1

The bus voltage can be written in form of

n

Vi=(1.0/Yii)[Ii- ∑ Yij Vj] (5)

J=1

j≠i(i=1,2,…………n)& i ≠slack bus

Substituting Ii in the expression for Vi, we get

n

Vi new

=(1.0/Yii)[Pi-JQi / Vio* - ∑ Yij Vi

o] (6)

J=1

The latest available voltages are used in the above expression,we get n n

Vi new

=(1.0/Yii)[Pi-JQi / Vio* - ∑ YijVj

n - ∑ Yij Vi

o] (7)

J=1 j=i+1

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The above equation is the required formula .this equation can be solved for voltages in

interactive manner. During each iteration, we compute all the bus voltage and check for

convergence is carried out by comparison with the voltages obtained at the end of previous

iteration. After the solutions is obtained. The stack bus real and reactive powers, the reactive

power generation at other generator buses and line flows can be calculated.

ALGORITHM

Step1:Read the data such as line data, specified power ,specified voltages, Q limits at the

generator buses and tolerance for convergences Step2: Compute Y-bus matrix. Step3: Initialize all the bus voltages. Step4: Iter=1 Step5: Consider i=2, where i’ is the bus number. Step6:check whether this is PV bus or PQ bus . if it is PQ bus goto step 8 otherwise go to

next step. Step7: Compute Qi check for q limit violation. QGi=Qi+QLi.

7).a).If QGi>Qi max ,equate QGi = Qimax. Then convert it into PQ bus.

7).b).If QGi<Qi min, equate QGi = Qi min. Then convert it into PQ bus. Step8: Calculate the new value of the bus voltage using gauss seidal formula.

i=1 n

Vi=(1.0/Yii) [(Pi-j Qi)/vi0*- ∑ Yij Vj- ∑ YijVj0]

J=1 J=i+1

Adjust voltage magnitude of the bus to specify magnitude if Q limits are not violated.

Step9: If all buses are considered go to step 10 otherwise increments the bus no. i=i+1 and

Go to step6. Step10: Check for convergence. If there is no convergence goes to step 11 otherwise go to

step12. Step11: Update the bus voltage using the formula.

Vinew

=Vi old

+ α(vinew

-Viold

) (i=1,2,…..n) i ≠ slackbus ,α is the acceleration factor=1.4 Step12: Calculate the slack bus power, Q at P-V buses real and reactive give flows real and

reactance line losses and print all the results including all the bus voltages and all the

bus angles. Step13: Stop.

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FLOW CHART:

FLOW CHART FOR GAUSS SEIDAL METHOD

PROCEDURE

Enter the command window of the MATLAB. Create a new M – file by selecting File - New – M – File.

Type and save the program in the editor Window.

Execute the program by pressing Tools – Run. View the results.

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MANUAL CALCULATION

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A program has been developed using MATLAB for the given power system by Gauss

Seidal method and the results are verified with model calculation.

EXERCISE

Expt. No: Date:

LOAD FLOW ANALYSIS BY NEWTON RAPSHON METHOD USING MATLAB

AIM

To carry out load flow analysis of the given power system by Newton Raphson

method.

SOFTWARE REQUIRED: MATLAB 7.7

THEORY The Newton Raphson method of load flow analysis is an iterative method which

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 order

approximation. The load flow equations for Newton Raphson method are non-linear

equations in terms of real and imaginary part of bus voltages.

where, ep = Real part of

Vp fp = Imaginary part of

Vp

Gpq, Bpq = Conductance and Susceptances of admittance Ypq respectively.

ALGORITHM

Step1: Input the total number of buses. Input the details of series line impendence and line

charging admittance to calculate the Y-bus matrix. Step2: Assume all bus voltage as 1 per unit except slack bus. Step3: Set the iteration count as k=0 and bus count as p=1. Step4: Calculate the real and reactive power pp and qp using the formula

P=∑vpqYpq*cos(Qpq+εp-εq)

Qp=∑VpqYpa*sin(qpq+εp-εa)

Evalute pp*=psp-pp* Step5: If the bus is generator (PV) bus, check the value of Qp*is within the limits.If it violates

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the limits, then equate the violated limit as reactive power and treat it as PQ bus. If limit is

not violated then calculate,

|vp|^r=|vgp|^rspe-|vp|r

Qp*=qsp-qp*

Step6: Advance bus count by 1 and check if all the buses have been accounted if not go to step5. Step7: Calculate the elements of Jacobean matrix. Step8: Calculate new bus voltage increment pk and fpk Step9: Calculate new bus voltage ep*h+ ep*

Fp^k+1=fpK+fpK Step10: Advance iteration count by 1 and go to step3. Step11: Evaluate bus voltage and power flows through the line .

Enter the command window of the MATLAB. Create a new M – file by selecting File - New – M – File.

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PROCEDURE

Type and save the program in the editor Window.

Execute the program by pressing Tools – Run. View the results.

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RESULT

A program has been developed using MATLAB for the given power system by Newton

Raphson method and the results are verified with model calculation

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EXERCISE

Expt. No: Date:

LOAD FLOW ANALYSIS BY FAST DECOUPLED METHOD USING MATLAB

AIM

To carry out load flow analysis of the given power system by Fast Decoupled method.

SOFTWARE REQUIRED: MATLAB 7.7

THEORY

Due to the weak coupling between PV and q-δ half of the elements of jacobian matrix

are neglected further the assumptions made are Cos δij = 1 Sin δij = 0 Qi << Bij

|V|2

The simplified FDLF equations are (∆P / |V|) = [B’][ ∆δ]

(∆Q / |V|) = [B’’][

∆|V|] One iteration implies one solution for [∆δ] to update [δ] and one solution for [∆P/|V|] to

update [|V|] and is termed as 1-δ and 1-V iteration. The convergence for the real and reactive

power is achieved when max [∆P] < ∑P; max [∆Q]<=∑Q. The main advantage of the decoupled load flow as compared to Newton Raphson method is

its reduced memory is storing Jacobian.

ALGORITHM Step1: Input the total number of buses. Step2: Input the Y-bus matrix of order n X n. Step3: Assume all the bus voltages are 1 pu except slack bus. Step4: Form susceptance matrix B’ and B’’. Step5: Set the iteration count as K=0. Step6: Set Vs=0; rv=0. Step7: Calculate the real and reactive power Pi using the formula

∆Pir = ∑VpqYpq*cos(Qpq + ξp - ξq)

Step8: Evaluate ∆Pir = Pispec – P i

r.

Step9: Test for convergence if max (∆ Pir < ∑P) then set rs = 1 and go to next step otherwise go

to step 2.

Step10: Check if rv = 1 then calculate the slack bus power and all line flows and print the result

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d2 =d2+Dd(1); d3 =d3+Dd(2); V2= V2+DV; angle2 =180/pi*d2; angle3 =180/pi*d3; R = [iter d2 d3 V2 DP2 DP3 DQ2]; disp(R) end Q3=-V3*V1*Y(3,1)*sin(t(3,1)-d3+d1)-V3^2*Y(3,3)*sin(t(3,3))- ... V3*V2*Y(3,2)*sin(t(3,2)-d3+d2); P1= V1^2*Y(1,1)*cos(t(1,1))+V1*V2*Y(1,2)*cos(t(1,2)-d1+d2)+ ... V1*V3*Y(1,3)*cos(t(1,3)-d1+d3); Q1=-V1^2*Y(1,1)*sin(t(1,1))-V1*V2*Y(1,2)*sin(t(1,2)-d1+d2)- ... V1*V3*Y(1,3)*sin(t(1,3)-d1+d3); S1=P1+j*Q1 Q3

otherwise go to step 13.

Step11: Set rv = 0 then calculate ∆Sir using (∆P / |V|) = [B’][∆δ]

Step12: Calculate δir+1

= δir + ∆δi

r.

Step13: Calculate ∆Qir using ∆Qispec – Q i

r.

Step14: Test for convergence if max (∆Qir)<∑Q then set rv = 1 and do next step otherwise go to

step 17.

Step15: Check if rs=1 then go to step 19. Step16: Otherwise advance the iteration count by 1 and go to step 8.

Step17: Set rs = 0 and calculate ∆|Vi|r using (∆Q / |V|) = [B’’][ ∆|V|]

Step18: Calculate |Vi|r+1 = |Vi|

r + ∆|Vi|

r and go to step 16.

Step19: Calculate slack bus power and all line flows and print the result.

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PROCEDURE

Enter the command window of the MATLAB. Create a new M – file by selecting File - New – M – File.

Type and save the program in the editor Window.

Execute the program by pressing Tools – Run. View the results.

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RESULT

A program has been developed using MATLAB for the given power system by fast

decoupled method and the results are verified with model calculation.

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Expt. No: Date: TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS OF SINGLE-MACHINE

INFINITE BUS SYSTEM AIM

To become familiar with various aspects of the transient and small signal stability

analysis of Single-Machine-Infinite Bus (SMIB) system.

OBJECTIVES

The objectives of this experiment are:

1. To study the stability behavior of one machine connected to a large power system

subjected to a severe disturbance (3-phase short circuit)

2. To understand the principle of equal-area criterion and apply the criterion to study

the stability of one machine connected to an infinite bus

3. To determine the critical clearing angle and critical clearing time with the help of

equal-area criterion

4. To do the stability analysis using numerical solution of the swing equation.

SOFTWARE REQUIRED: MATLAB 7.7

THEORY

The tendency of a power system to develop restoring forces to compensate for the

disturbing forces to maintain the state of equilibrium is known as stability. If the forces

tending to hold the machines in synchronism with one another are sufficient to overcome the

disturbing forces, the system is said to remain stable.

The stability studies which evaluate the impact of disturbances on the behavior of

synchronous machines of the power system are of two types – transient stability and steady

state stability. The transient stability studies involve the determination of whether or not

synchronism is maintained after the machine has been subjected to a severe disturbance. This

may be a sudden application of large load, a loss of generation, a loss of large load, or a fault

(short circuit) on the system. In most disturbances, oscillations are such magnitude that

linearization is not permissible and nonlinear equations must be solved to determine the

stability of the system. On the other hand, the steady-state stability is concerned with the

system subjected to small disturbances wherein the stability analysis could be done using the

linearized version of nonlinear equations. In this experiment we are concerned with the

transient stability of power systems.

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MANUAL CALCULATION:

A method known as the equal-area criterion can be used for a quick prediction of

stability of a one- machine system connected to an infinite bus. This method is based on the

graphical interpretation of energy stored in the rotating mass as an aid to determine if the

machine maintains its stability after a disturbance. The method is applicable to a one-

machine system connected to an infinite bus or a two- machine system. Because it provides

physical insight to the dynamic behavior of the machine, the application of the method to

analyze a single-machine system is considered here.

Stability: Stability problem is concerned with the behavior of power system when it is

subjected to disturbance and is classified into small signal stability problem if the

disturbances are small and transient stability problem when the disturbances are large.

Transient stability: When a power system is under steady state, the load plus transmission

loss equals to the generation in the system. The generating units run at synchronous speed

and system frequency, voltage, current and power flows are steady. When a large

disturbance such as three phase fault, loss of load, loss of generation etc., occurs the power

balance is upset and the generating units rotors experience either acceleration or

deceleration. The system may come back to a steady state condition maintaining

synchronism or it may break into subsystems or one or more machines may pull out of

synchronism. In the former case the system is said to be stable and in the later case it is said

to be unstable.

Small signal stability: When a power system is under steady state, normal operating

condition, the system may be subjected to small disturbances such as variation in load and

generation, change in field voltage, change in mechanical toque etc., the nature of system

response to small disturbance depends on the operating conditions, the transmission system

strength, types of controllers etc. Instability that may result from small disturbance may be

of two forms,

(i) Steady increase in rotor angle due to lack of synchronizing torque.

(ii) Rotor oscillations of increasing magnitude due to lack of sufficient damping torque.

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Thus the stability analysis of single machine infinite bus system is verified using the

MATLAB program.

Expt. No: Date:

LOAD FREQUENCY DYNAMICS OF SINGLE AND TWO AREA POWER SYSTEM

AIM

To obtain the frequency response and steady state frequency deviation of single and two

area power system using MATLAB.

SOFTWARE REQUIRED: MATLAB 7.7

THEORY

Active power control is one of the important control actions to be performed to be

normal operation of the system to match the system generation with the continuously changing

system load in order to maintain the constancy of system frequency to a fine tolerance level.

This is one of the foremost requirements in proving quality power supply. A change in system

load cases a change in the speed of all rotating masses (Turbine – generator rotor systems) of

the system leading to change in system frequency. The speed change form synchronous speed

initiates the governor control (primary control) action result in the entire participating generator

– turbine units taking up the change in load, stabilizing system frequency. Restoration of

frequency to nominal value requires secondary control action which adjusts the load - reference

set points of selected (regulating) generator – turbine units. The primary objectives of automatic

generation control (AGC) are to regulate system frequency to the set nominal value and also to

regulate the net interchange of each area to the scheduled value by adjusting the outputs of the

regulating units. This function is referred to as load – frequency control (LFC).

PROCEDURE

Enter the command window of the MATLAB.

Create a new Model by selecting File - New – Model.

Pick up the blocks from the Simulink library browser and form a block

diagram. After forming the block diagram, save the block diagram.

Double click the scope and view the result.

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EXERCISE 1

An isolated power station has the following

parameters Turbine time constant, τT = 0.5sec

Governor Time constant, τg = 0.2sec

Generator inertia constant, H = 5sec

(c) The governor speed regulation is set to R = 0.05 per unit. The turbine rated output is 250MW at

nominal frequency of 60Hz. A sudden load change of 50 MW (∆PL = 0.2 per unit) occurs.

(i) Find the steady state frequency deviation in Hz. (ii) Use MATLAB to obtain the time domain performance specifications and the frequency

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deviation step response.

Governor speed regulation = R per unit

The load varies by 0.8 percent for a 1 percent change in frequency, i.e., D=0.8

(a) Use the Routh – Hurwitz array to find the range of R for control system stability.

(b) Use MATLAB to obtain the root locus plot.

EXERCISE 2 A two area system connected by a tie line has the following parameters on a 1000MVA common base.

Area 1 2

Speed Regulation R1=0.05 R2=0.0625

Frequency – sens. load D1=0.6 D2=0.9

Inertia Constant H1=5 H2=4

Base Power 1000MVA 1000MVA

Governor Time Constant τg1= 0.2sec τg2= 0.3sec

Turbine Time Constant τT1=0.5sec τT2=0.6sec

The units are operating in parallel at the nominal frequency of 60Hz. The synchronizing power

coefficient is computed from the initial operating condition and is given to be Ps = 2 p.u. A load change of 187.5 MW occurs in area1. (a) Determine the new steady state frequency and the change in the tie-line flow. (b) Construct the SIMULINK block diagram and obtain the frequency deviation response for the

condition in part (a).

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RESULT

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1 G1 G1C 0.004P2 5.3P 500Rs / hr

C2 0.006PG22 5.5PG2 400Rs/ hr

C3 0.009PG23 5.8P G3 200Rs / hr

where PGi ’s are in MW. Find the scheduling for a load of 975 MW.

PROBLEM1a: (WITHOUT LOSS &no GENERATING LIMITS)

A power plant has three units with the following cost characteristics:

ECONOMIC LOAD DISPATCH IN POWER SYSTEM

AIM:

To understand the fundamentals of economic dispatch and solve the problem using classical method without line losses.

SOFTWARE REQUIRED: MATLAB 7 & above

THEORY:

Power is allowed to vary within certain limits so as to meet a particular load demand with minimum fuel cost. This is called the optimal power flow (OPF) problem. The OPF is used to optimize the power flow solution of large scale power system. This is done by minimizing selected the same distance from the centre of loads and their fuel costs are different. Also under normal operating conditions, the generation capacity is more than the total load demand and losses. Thus, there objective functions while maintaining an acceptable system performance in terms of generator capability limits and the output of the compensating devices.

The simplest economic dispatch problem is the case when transmission line losses are neglected. This is the problem model does not consider system configuration and line impedances. In essence, the model assumes that the system is only one bus with all generation and loads connected to it as shown in figure below.

Since transmission losses are neglected, the total demand PD is the sum of all generation. A cost function Ci is assumed to be known for each plant. The problem is to find the real power generation for each plant such that the objective function as defined by the equation,

2

1 1iiii

ng

i

n

iiit PvPCC ++==

= =

βα

is minimum, subject to the constraint,

C1

P1

C2

P2

Cn

Png

PD (Plants connected to a common bus)

EXPT. NO : DATE :

8989

9090

=

=ng

iDi PP

1

where, Ct is the total production cost, Ci is the production cost of ith plant, Pi is generation of the ith plant, PD is the total demand and ng is the total number of despicable generating plants.

A rapid solution is obtained by the use of the gradient method (Lambda iteration). Let f(2)= PD

Expanding the left hand side of the above equation in Taylor’s series about an operating point 2k and neglecting the higher-order terms result in,

=

+

−=∆

∆+=∆

∆=∆

∆=

∆=∆

=∆⋅

+

ng

i

kiD

k

kkk

i

kk

ki

k

k

kk

Dk

kk

PPP

v

Por

ddP

P

ddf

P

Pd

dff

1

1

where,

, thereforeand

21

)()(

)()(

λλλ

λ

λλλ

λ

λλλλ

ALGORITHM:

1. Start the program. 2. Read the input data values 3. Start the iteration counter. 4. Check the test for convergence. 5. iter=iter+1 6. Calculate P,delp, J, dellambda and lambda. 7. Display the above values. 8. To find the total cost=sum(alpha+beta.*P+gamma.*P.2) 9. Print the value of total cost. 10. Stop the program.

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MANUAL CALCULATION

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MANUAL SOLUTION (Problem-2):

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AIM:

fault on electrical distribution systems. The program calculates the total short circuit current as well as the distributions of individual motors, generators and utility in the system. Fault duties are in competence with the latest conditions which calculates.

• Momentary symmetrical fault current in kA • Interrupting symmetrical fault current in kA

Unsymmetrical Faults:

• Line to ground faults • Double line faults

The MATLAB short circuit analysis program also analyzes the effect of 3 phase unsymmetrical faults like,

SOFTWARE REQUIRED: MATLAB

THEORY:

Symmetrical Faults:

The MATLAB Short circuit analysis program analyzer the effect of 3 phase

symmetrical

To conduct short circuit analysis of the given power system using MATLAB software package and verify the same.

FAULT ANALYSIS

EXPT NO : DATE :

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99

ZERO SEQUENCE NETWORK:

100

101

Thus the fault analysis of the given power system using MATLAB software

package was experimented and the output was verified.

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STUDY THE ELECTROMAGNETIC TRANSIENTS IN POWER

SYSTEMS AIM:

To study and understand the electromagnetic transient phenomena in power systems caused due to switching and faults.

THEORETICAL BACKGROUND

Solution Method for Electromagnetic Transients Analysis

Intentional and inadvertent switching operations in EHV systems initiate over voltages, which might attain dangerous values resulting in destruction of apparatus. Accurate computation of these over voltages is essential for proper sizing, coordination of insulation of various equipment’s and specification of protective devices. Meaningful design of EHV systems is dependent on modeling philosophy built into a computer program. The models of equipment’s must be detailed enough to reproduce actual conditions successfully – an important aspect where a general purpose digital computer program scores over transient network analyzers.

The program employs a direct integration time-domain technique evolved by Dommel. The essence of this method is discretization of differential equations associated with network elements using trapezoidal rule of integration and solution of the resulting difference equations for the unknown voltages. Any network which consists of interconnections of resistances, inductances, capacitances, single and multiphase π circuits, distributed parameter lines, and certain other elements can be solved. To keep explanations sample, however, single phase network elements will be used, rather than the more compels multiphase network elements.

Figure 1. Part of the Network around a Node of Large System

Modeling Details:

Fig. 1 shows the details for the region around node 1 of a large system. Suppose that voltages and currents have already been computed in steps of time (∆t) for t=0, ∆t, 2∆t etc., upto t-∆t, and that the solution must now be found for time t. At any distance of time the sum of the currents flowing away from node 1 through the branches must be equal to the injected current i1.

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EXPT NO : 1DATE :

)1()()()()()( 115141312 →=+++ tititititi

Since node voltages are used as state variables, it is necessary to express the branch currents, i12 etc., as function of node voltages. For the resistance:

)2()()(

)( 2112 →−=

Rtvtv

ti

For the inductance, a simple relationship is obtained by replacing the differential equation

dtdi

Lv = with a central difference equation:

∆∆−−=∆−+

tttiti

Lttvtv )()(

2)()(

Applying the above relationship for the case of above figure,

∆∆−−=∆−+

tttiti

Lttvtv )()(

2)()( 13131313

Rewriting the above equation, we get

( )

steps. preceding theof values thefrom

computedi with rmhistory tepast the, )()(2

)(,

)()()(2

)(

)()(2

)(2

)(

)(2

)(2

)()(

13131313

133113

13131313

13131313

ttittvLt

ttIwhere

ttItvtvLt

ti

ttittvLt

tvLt

ti

ttvLt

tvLt

ttiti

∆−+∆−⋅

∆=∆−

∆−+−⋅

∆=

∆−+∆−⋅

∆+⋅

∆=

∆−⋅

∆+⋅

∆=∆−−

( ))()(2

)(13)( 3113 ttvttvLt

ttitI ∆−−∆−⋅

∆+∆−=

The derivation for the branch equation of the capacitance is analogous, and leads to

( ) )()()(2

)( 144114 ttItvtvt

Cti ∆−+−⋅

∆=

with ( ))()(2

)(14)( 4114 tvtvt

CttittI −⋅

∆−∆−−=∆−

For the transmission line between nodes 1 and 5, losses are first ignored, then reintroduced later on. Then the wave equations

107

xv

Cxi

xi

Lxv

∂∂=

∂∂−

∂∂=

∂∂−

'

'

where, L’ &C’=inductance and capacitance per unit length, x=distance from sending end.

The well known solution due to de’ Alembert:

)()()5()()(

ctxzfctxzFv

actafctxFi

+−−=→++−=

Where, F(x-ct) and f(x+ct) are functions of the composite variables x-ct and x+ct, z is the surge impedance and ‘c’ is the velocity of propagation.

If the current in equation (5a) is multiplied by z and added to the voltage, then v+zi=2zF(x-ct) (5b).

Note that the composite expression v+zi does not change if x-ct does not change. Imagine a fictitious observer travelling on the line with wave velocity c. The distance travelled is x=x0+ct (x0 – location of the starting point), or x-ct = constant. If x-ct is constant, then the value v+zi, measured by the observer, must also remain constant. With travel time

cLength Line=τ

An observer leaving node 5 at time t-ττττ will measure the value v5(t-ττττ)+zi15(t-ττττ), and upon arrival at node 1 (after the elapse of travel time ττττ), will measure the value v1(t)+zi15(t)(negative sign because i15 has opposite direction of i51 ). But the value as measured by the observer must remain constant, so these values must be equal, giving, after rewriting,

)6()()(

)( 515

15 atIz

tvti →−−−= ττ

Where the term I15 is again known from previously computed values

)6()()(

)( 515

51 btiz

tvtI →−−−−=− τττ

RESULT:

Thus the electromagnetic transient phenomena in power systems caused due to switching and faults are studied.

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