power system simulation lab

Power Systems Simulation Lab GRIETEEE

1

POWER SYSTEMS SIMULATION LAB

IV YEAR I SEM

EEE

By

Dr J Sridevi

Syed Sarfaraz Nawaz

GSandhya Rani

Gokaraju Rangaraju Institute of Engineering amp

Technology

Bachupally


2

GOKARAJU RANGARAJU INSTITUTE OF

ENGINEERING AND TECHNOLOGY

(Autonomous)

Bachupally Hyderabad-500 072

CERTIFICATE

This is to certify that it is a record of practical work done in

the Power Systems Simulation Laboratory in I sem of IV

year during the year ________________________

Name

Roll No

Branch EEE

Signature of staff member


3

INDEX

SNo Date Topic Page no Signature of

the Faculty

1 Sinusoidal Voltages and Currents 1

2 Equivalent circuit of a Transformer 6

3 Determination of voltage and power at

the sending end voltage regulation using

medium line model

14

4 Determination of line performance when

loaded at receiving end

19

5 Formation of bus Admittance matrix 25

6 Load flow Solution using Gauss Seidel

Method

30

7 Load flow solution using Newton

Raphson method in Rectangular

Coordinates

37

8 a) Optimal dispatch neglecting Losses

b) Optimal dispatch including Losses

41

47

9 Transient Response of an RLC Circuit 51

10 Three phase short circuit analysis in a

Synchronous Machine

57

11 Unsymmetrical Fault Analysis 63

12 Zbus Building Algorithm 73


4

13 a) Obtain Symmetrical components of a

set of Unbalanced currents

b) Obtain the original Unbalanced phase

voltages from Symmetrical Components

78

82

14 Short Circuit Analysis of 14 bus system 86

15 Load Frequency control of a single area

system

90

16 Load frequency control of a two area

system

95

17 Step response of rotor angle and

generator frequency of a Synchronous

Machine

100

18

19

20

21

22

23

24

25

26

27


1

Date Experiment-1

SINUSOIDAL VOLTAGES AND CURRENTS

Aim To determine sinusoidal voltages and currents

Apparatus MATLAB

Theory The RMS Voltage of an AC Waveform

The RMS value is the square root of the mean (average) value of the squared function of the

instantaneous values The symbols used for defining an RMS value are VRMS or IRMS

The term RMS refers to time-varying sinusoidal voltages currents or complex waveforms were

the magnitude of the waveform changes over time and is not used in DC circuit analysis or

calculations were the magnitude is always constant When used to compare the equivalent RMS

voltage value of an alternating sinusoidal waveform that supplies the same electrical power to a

given load as an equivalent DC circuit the RMS value is called the ldquoeffective valuerdquo and is

presented as Veffor Ieff

In other words the effective value is an equivalent DC value which tells you how many volts or

amps of DC that a time-varying sinusoidal waveform is equal to in terms of its ability to produce

the same power For example the domestic mains supply in the United Kingdom is 240Vac This

value is assumed to indicate an effective value of ldquo240 Volts RMSrdquo This means then that the

sinusoidal RMS voltage from the wall sockets of a UK home is capable of producing the same

average positive power as 240 volts of steady DC voltage as shown below

RMS Voltage Equivalent


2

Circuit diagram

Fig Simulink model for voltage and current measurement

Procedure

1 Open Matlab--gtSimulink--gt File ---gt New---gt Model

2 Open Simulink Library and browse the components

3 Connect the components as per circuit diagram

4 Set the desired voltage and required frequency

5 Simulate the circuit using MATLAB

6 Plot the waveforms


3

Graph

Calculations


4


5

Result


6

Signature of the faculty

Date Experiment-2

EQUIVALENT CIRCUIT OF TRANSFORMER

Aim To determine the parameters of equivalent circuit of transformer from OC SC test

data

Apparatus MATLAB

Theory Equivalent Circuit of Transformer

Equivalent impedance of transformer is essential to be calculated because the electrical power

transformer is an electrical power system equipment for estimating different parameters of

electrical power system which may be required to calculate total internal impedance of an

electrical power transformer viewing from primary side or secondary side as per requirement

This calculation requires equivalent circuit of transformer referred to primary or equivalent

circuit of transformer referred to secondary sides respectively

Equivalent Circuit of Transformer Referred to Primary

Let us consider the transformation ratio be

In the figure right the applied voltage to the primary is V1 and

voltage across the primary winding is E1 Total electric current

supplied to primary is I1 So the voltage V1 applied to the

primary is partly dropped by I1Z1 or I1R1 + jI1X1 before it

appears across primary winding The voltage appeared across

winding is countered by primary induced emf E1 So voltage

equation of this portion of the transformer can be written as

From the vector diagram above it is found that the total

primary current I1 has two components one is no - load

component Io and the other is load component I2prime As this


7

primary current has two components or branches so there must be a parallel path with

primary winding of transformer This parallel path of electric current is known as excitation

branch of equivalent circuit of transformer The resistive and reactive branches of the

excitation circuit can be represented as

The load component I2prime flows through the

primary winding of transformer and induced

voltage across the winding is E1as shown in

the figure right This induced voltage

E1 transforms to secondary and it is E2 and

load component of primary current I2prime is

transformed to secondary as secondary

current I2 Current of secondary is I2 So the

voltage E2 across secondary winding is

partly dropped by I2Z2 or I2R2 +

jI2X2 before it appears across load The load

voltage is V2

Now if we see the voltage drop in secondary from primary side then it would be primeKprime times

greater and would be written as KZ2I2

Again I2primeN1 = I2N2

Therefore


8

From above equation secondary impedance of transformer referred to primary is

So the complete equivalent circuit of transformer referred to primary is shown in the figure

below

Approximate Equivalent Circuit of

Transformer

Since Io is very small compared to I1 it

is less than 5 of full load primary

current Io changes the voltage drop

insignificantly Hence it is good

approximation to ignore the excitation

circuit in approximate equivalent circuit

of transformer The winding resistance

and reactance being in series can now

be combined into equivalent resistance

and reactance of transformer referred to

any particular side In this case it is side 1 or primary side

Equivalent Circuit of Transformer Referred to Secondary

In similar way approximate equivalent circuit of transformer referred to secondary can be

drawn

Where equivalent impedance of transformer referred to secondary can be derived as


9

Circuit Diagram


10

Procedure




4 Set the desired voltage and current



Graph

Calculations


11


12


13

Result



14

Date Experiment-3

VOLTAGE REGULATION OF A MEDIUM LINE MODEL

Aim To determine voltage and power at the sending end and to regulate the voltage using

medium line model

Apparatus MATLAB

Theory The transmission line having its effective length more than 80 km but less than

250 km is generally referred to as a medium transmission line Due to the line length being

considerably high admittance Y of the network does play a role in calculating the effective

circuit parameters unlike in the case of short transmission lines For this reason the modelling

of a medium length transmission line is done using lumped shunt admittance along with the

lumped impedance in series to the circuit

These lumped parameters of a medium length transmission line can be represented using two

different models namely-

1)Nominal Π representation

2)Nominal T representation

Letrsquos now go into the detailed discussion of these above mentioned models

Nominal Π Representation of a Medium Transmission Line

In case of a nominal Π representation

the lumped series impedance is placed

at the middle of the circuit where as

the shunt admittances are at the ends

As we can see from the diagram of the

Π network below the total lumped

shunt admittance is divided into 2

equal halves and each half with value

Y frasl 2 is placed at both the sending and the receiving end while the entire circuit impedance is

between the two The shape of the circuit so formed resembles that of a symbol Π and for

this reason it is known as the nominal Π representation of a medium transmission line It is

mainly used for determining the general circuit parameters and performing load flow analysis


15

As we can see here VS and VR is the supply and receiving end voltages respectively and

Is is the current flowing through the supply end

IR is the current flowing through the receiving end of the circuit

I1 and I3 are the values of currents flowing through the admittances And

I2 is the current through the impedance Z

Now applying KCL at node P we get

Similarly applying KCL to node Q

Now substituting equation (2) to equation (1)

Now by applying KVL to the circuit

Comparing equation (4) and (5) with the standard ABCD parameter equations


16

We derive the parameters of a medium transmission line as

Voltage regulation of transmission line is measure of change of receiving end voltage from

no-load to full load condition

Circuit Diagram


17

PROCEDURE






6 Calculate the voltage regulation of medium line model

Calculations


18

Result



19

Date Experiment-4

LINE PERFORMANCE WHEN LOADED AT RECEIVING END

Aim To determine line performance when loaded at receiving end

Apparatus MATLAB

Theory Nominal Π Representation of a Medium Transmission Line



















(1)


(2)


20








21

Circuit Diagram

PROCEDURE






6 Obtain the line performance of a line

Graph


22


23


24

Result



25

Date Experiment-5

FORMATION OF BUS ADMITTANCE MATRIX USING MATLAB

Aim To determine the bus admittance matrix for the given power system Network

Apparatus MATLAB 77

Theory

Formation of Y BUS matrix

Bus admittance matrix is often used in power system studiesIn most of power system

studies it is necessary to form Y-bus matrix of the system by considering certain power system

parameters depending upon the type of analysis For example in load flow analysis it is necessary

to form Y-bus matrix without taking into account the generator impedance and load impedance

In short circuit analysis the generator transient reactance and transformer impedance taken in

account in addition to line data Y-bus may be computed by inspection method only if there is

no natural coupling between the lines Shunt admittance are added to the diagonal elements

corresponding to the buses at which these are connected The off diagonal elements are

unaffected The equivalent circuit of tap changing transformer may be considered in forming[y-

bus] matrix

There are b independent equations (b = no of buses) relating the bus vectors of currents

and voltages through the bus impedance matrix and bus admittance matrix

EBUS = ZBUS IBUS

IBUS = YBUS EBUS

The relation of equation can be represented in the form

IBUS = YBUS EBUS

Where YBUS is the bus admittance matrix IBUS amp EBUS are the bus current and bus

voltage vectors respectively

Diagonal elements A diagonal element (Yii) of the bus admittance matrix YBUS is equal to

the sum total of the admittance values of all the elements incident at the busnode i

Off Diagonal elements An off-diagonal element (Yij) of the bus admittance matrix YBUS is

equal to the negative of the admittance value of the connecting element present between the

buses I and j if any

This is the principle of the rule of inspection Thus the algorithmic equations for the rule of

inspection are obtained as

Yii = Σ yij (j = 12helliphellipn)

Yij = - yij (j = 12helliphellipn)


26

For i = 12hellipn n = no of buses of the given system yij is the admittance of element

connected between buses i and j and yii is the admittance of element connected

between bus i and ground (reference bus)

Read the no Of buses no of

lines and line data

START

Initialize the Y- BUS Matrix

Consider line l = 1

i = sb(1) I= eb(1)

Y(ii) =Y(ii)+Yseries(l) +05Yseries(l)

Y(jj) =Y(jj)+Yseries(l) +05Yseries(l)

Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM

function[Ybus] = ybus(zdata)

nl=zdata(1) nr=zdata(2) R=zdata(3) X=zdata(4)

nbr=length(zdata(1)) nbus = max(max(nl) max(nr))

Z = R + jX branch impedance

y= ones(nbr1)Z branch admittance

Ybus=zeros(nbusnbus) initialize Ybus to zero

for k = 1nbr formation of the off diagonal elements

if nl(k) gt 0 amp nr(k) gt 0

Ybus(nl(k)nr(k)) = Ybus(nl(k)nr(k)) - y(k)

Ybus(nr(k)nl(k)) = Ybus(nl(k)nr(k))

end

end

for n = 1nbus formation of the diagonal elements

for k = 1nbr

if nl(k) == n | nr(k) == n

Ybus(nn) = Ybus(nn) + y(k)

else end

end

end

Calculations


28


29

Result



30

Date Experiment-6

LOAD FLOW ANALYSIS BY GAUSS SEIDEL METHOD

Aim

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

method

Apparatus MATLAB

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 Seidel 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=123helliphelliphelliphellipn) (2)

Where

n

Pi=Re [ Σ ViYik Vk] (3)

k=1

n

Qi= -Im [ Σ ViYik Vk] (4)

k=1

The bus voltage can be written in form of

n

Vi=(10Yii)[Ii- Σ Yij Vj] (5)

j=1

jnei(i=12helliphelliphelliphellipn)amp ineslack bus

Substituting Ii in the expression for Vi we get

n

Vi new=(10Yii)[Pi-JQi Vio - Σ Yij Vio] (6)

J=1

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

Vi new=(10Yii)[Pi-JQi Voi - Σ YijVj

n- Σ Yij Vio] (7) J=1 j=i+1

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


31

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 irsquo 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 QGigtQi max equate QGi = Qimax Then convert it into PQ bus

7)b)If QGiltQi 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=(10Yii) [(Pi-j Qi)vi0- Σ Yij Vj- Σ 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=12hellipn) ine slackbus α is the acceleration factor=14

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


32

START

Read

1 Primitive Y matrix

2 Bus incidence matrix A

3 Slack bus voltages

4 Real and reactive bus powers Piamp Qi

5 Voltage magnitudes and their limits

Form Ybus

Make initial assumptions

Compute the parameters Ai for i=m+1hellipn and Bik for i=12hellipn

k=12hellipn

Set iteration count r=0

Set bus count i=2 and ΔVmax=0

Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi

(r+1) = Qimin Compute Ai

(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33

FLOW CHART FOR GAUSS SEIDEL METHOD

Procedure

Enter the command window of the MATLAB

Create a new M ndash file by selecting File - New ndash M ndash File

Type and save the program in the editor Window

Execute the program by pressing Tools ndash Run

View the results

MATLAB program

clear

basemva=100

accuracy=0001 maxiter=100

no code mag degree MW Mvar MW Mvar Qmin Qmax Mvar

busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]

bus bus R X 12B =1 for lines

linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and

advance bus count i = i+1

Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1

Compute slack bus power P1+jQ1 and all line flows

B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7

LOAD FLOW ANALYSIS BY NEWTON RAPSHON METHOD

Aim

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

Apparatus MATLAB 77

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

Taylorrsquos 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=ΣvpqYpqcos(Qpq+εp-εq)

Qp=ΣVpqYpasin(qpq+εp-εa)

Evalute pp=psp-pp


38

Step5 If the bus is generator (PV) bus check the value of Qpis within the limitsIf it Violates

the limits then equate the violated limit as reactive power and treat it as PQ bus If limit is not

isolated 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 eph+ 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

Procedure





View the results

MATLAB program

clear

basemva=100accuracy=0001maxiter=100


busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41

Date Experiment-8(a)

OPTIMAL DISPATCH NEGLECTING LOSSES

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

Apparatus MATLAB

Theory

As the losses are neglected the system model can be understood as shown in Fig here n number

of generating units are connected to a common bus bar collectively meeting the total power

demand PD It should be understood that share of power demand by the units does not involve

losses

Since transmission losses are neglected total demand PD is the sum of all generations of n-

number of units For each unit a cost functions Ci is assumed and the sum of all costs computed

from these cost functions gives the total cost of production CT

Fig System with n-generators

where the cost function of the ith unit from Eq (11) is

Ci = αi + βiPi + γ

iPi

2

Now the ED problem is to minimize CT subject to the satisfaction of the following equality and

inequality constraints

Equality constraint

The total power generation by all the generating units must be equal to the power demand


42

where Pi = power generated by ith unit

PD = total power demand

Inequality constraint

Each generator should generate power within the limits imposed

Pi

min le Pi le Pi

max i = 1 2 hellip n

Economic dispatch problem can be carried out by excluding or including generator power

limits ie the inequality constraint

The constrained total cost function can be converted into an unconstrained function by using

the Lagrange multiplier as

The conditions for minimization of objective function can be found by equating partial

differentials of the unconstrained function to zero as

Since Ci = C1 + C2+hellip+Cn

From the above equation the coordinate equations can be written as

The second condition can be obtained from the following equation


43

Equations Required for the ED solution

For a known value of λ the power generated by the ith unit from can be written as

which can be written as

The required value of λ is

The value of λ can be calculated and compute the values of Pi for i = 1 2hellip n for optimal

scheduling of generation

POWER WORLD bus diagram


44

Procedure

Create a new file in edit mode by selecting File - New File

Browse the components and build the bus sytem

Execute the program in run mode by selecting tools-opf areas-select opf

Run the primal lp

View the results in case information-Generator fuel costs

Tabulate the results

Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893

Total Cost = 5384 79$hr


45

Calculations


46

Result



47

Date Experiment-8(b)

OPTIMAL DISPATCH INCLUDING LOSSES

Aim To develop a program for solving economic dispatch problem including transmission


Apparatus MATLAB

Theory When the transmission losses are included in the economic dispatch problem we can

modify (54) as

LOSSNT PPPPP 21

where PLOSS is the total line loss Since PT is assumed to be constant we have

LOSSN dPdPdPdP 210

In the above equation dPLOSS includes the power loss due to every generator ie

N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1

Also minimum generation cost implies dfT = 0 as given Multiplying by and combing we get

N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0

Adding with we obtain

N

i

i

i

LOSS

i

T dPP

P

P

f

1

0

The above equation satisfies when

NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1

where Li is called the penalty factor of load-i and is given by

NiPP

LiLOSS

i 11

1

Consider an area with N number of units The power generated are defined by the vector

TNPPPP 21

Then the transmission losses are expressed in general as

BPPP T

LOSS

where B is a symmetric matrix given by

NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211

The elements Bij of the matrix B are called the loss coefficients These coefficients are not

constant but vary with plant loading However for the simplified calculation of the penalty factor

Li these coefficients are often assumed to be constant


49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9

TRANSIENT RESPONSE OF A RLC CIRCUIT

Aim To obtain the transient response of an RLC circuit with its damping frequency and

damping ratio using MATLAB

Apparatus MATLAB

Theory RLC circuits are widely used in a variety of applications such as filters in

communications systems ignition systems in automobiles defibrillator circuits in biomedical

applications etc The analysis of RLC circuits is more complex than of the RC circuits we have

seen in the previous lab RLC circuits have a much richer and interesting response than the

previously studied RC or RL circuits A summary of the response is given below

Lets assume a series RLC circuit The discussion is also applicable to other RLC circuits such as

the parallel circuit

Figure 1 Series RLC circuit

By writing KVL one gets a second order differential equation The solution consists of two parts

x(t) = xn(t) + xp(t)

in which xn(t) is the complementary solution (=solution of the homogeneous differential equation

also called the natural response) and a xp(t) is the particular solution (also called forced

response) Lets focus on the complementary solution The form of this solution depends on the

roots of the characteristic equation

in which is the damping ratio and is the undamped resonant frequency The roots of the

quadratic equation are equal to

For the example of the series RLC circuit one has the following characteristic equation for the

current iL(t) or vC(t)

s2 + RLs + 1LC =0

Depending on the value of the damping ratio one has three possible cases

Case 1 Critically damped response two equal roots s= s1= s2


54

The total response consists of the sum of the

complementary and the particular solution The case of

a critically damped response to a unit input step

function

Case 2 Overdamped response two real and unequal roots s1 and

s2

Case 3 Underdamped response two complex roots

Procedure





View the results

MATLAB Program

wn = input(Enter value of undamped natural frequency)

z = input(Enter value of damping ratio)

n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result

Enter value of undamped natural frequency 16

wn = 16

Enter value of damping ratio05

z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10

THREE PHASE SHORT CIRCUIT ANALYSIS OF A SYNCHRONOUS

MACHINE

Aim To Analyze symmetrical fault

Apparatus MATLAB

Theory The response of the armature current when a three-phase symmetrical short circuit

occurs at the terminals of an unloaded synchronous generator

It is assumed that there is no dc offset in the armature current The magnitude of the current

decreases exponentially from a high initial value The instantaneous expression for the fault

current is given by

where Vt is the magnitude of the terminal voltage α is its phase angle and

is the direct axis subtransient reactance

is the direct axis transient reactance

is the direct axis synchronous reactance

with

The time constants are


60

is the direct axis subtransient time constant

is the direct axis transient time constant

we have neglected the effect of the armature resistance hence α = π2 Let us assume that the

fault occurs at time t = 0 From (69) we get the rms value of the

current as

which is called the subtransient fault current The duration of the subtransient current is

dictated by the time constant Td As the time progresses and Td´´lt t lt Td´ the first

exponential term will start decaying and will eventually vanish However since t is still nearly

equal to zero we have the following rms value of the current

This is called the transient fault current Now as the time progress further and the second

exponential term also decays we get the following rms value of the current for the sinusoidal

steady state

In addition to the ac the fault currents will also contain the dc offset Note that a symmetrical

fault occurs when three different phases are in three different locations in the ac cycle

Therefore the dc offsets in the three phases are different The maximum value of the dc offset

is given by

where TA is the armature time constant

Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11

UNSYMMETRICAL FAULT ANALYSIS

Aim To analyze unsymmetrical fault

Apparatus MATLAB

Theory

Single Line-to-Ground Fault

The single line-to-ground fault is usually referred as ldquoshort circuitrdquo fault and occurs when one

conductor falls to ground or makes contact with the neutral wire The general representation

of a single line-to-ground fault is shown in Figure 310 where F is the fault point with

impedances Zf Figure 311 shows the sequences network diagram Phase a is usually

assumed to be the faulted phase this is for simplicity in the fault analysis calculations [1]

a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf

General representation of a single Sequence network diagram of a

line-to-ground fault single line-to-ground fault

Since the zero- positive- and negative-sequence currents are equals as it can be observed

in Figure 311 Therefore

0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66

With the results obtained for sequence currents the sequence voltages can be obtained

from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I

If the single line-to-ground fault occurs on phase b or c the voltages can be found by the

relation that exists to the known phase voltage components

0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault

A line-to-line fault may take place either on an overhead andor underground

transmission system and occurs when two conductors are short-circuited One of the

characteristic of this type of fault is that its fault impedance magnitude could vary over a wide

range making very hard to predict its upper and lower limits It is when the fault impedance is

zero that the highest asymmetry at the line-to-line fault occurs

The general representation of a line-to-line fault is shown in Figure 312 where F is the

fault point with impedances Zf Figure 313 shows the sequences network diagram Phase b and

c are usually assumed to be the faulted phases this is for simplicity in the fault analysis

calculations [1]


67

a

c

b

F

Iaf = 0 Ibf Icf = -Ibf

Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-

Sequence network diagram of a line-to-line fault Sequence network diagram of a single

line-to-line fault

It can be noticed that

0afI

bf cfI I

bc f bfV Z I

And the sequence currents can be obtained as

0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z

The fault currents for phase b and c can be obtained as

13 90bf cf aI I I

The sequence voltages can be found as

0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68

Finally the line-to-line voltages for a line-to-line fault can be expressed as

ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V

Double Line-to-Ground Fault

A double line-to-ground fault represents a serious event that causes a significant

asymmetry in a three-phase symmetrical system and it may spread into a three-phase fault when

not clear in appropriate time The major problem when analyzing this type of fault is the

assumption of the fault impedance Zf and the value of the impedance towards the ground Zg

The general representation of a double line-to-ground fault is shown in Figure 314 where

F is the fault point with impedances Zf and the impedance from line to ground Zg Figure 315

shows the sequences network diagram Phase b and c are assumed to be the faulted phases this is

for simplicity in the fault analysis calculations

a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-

General representation of a Sequence network diagram

double line-to-ground fault of a double line-to-ground fault

It can be observed that

0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I

The positive-sequence currents can be found as

12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z

An alternative method is

0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I

If Zf and Zg are both equal to zero then the positive- negative- and zero-sequences can

be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z

The current for phase a is

0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I

The total fault current flowing into the neutral is

03n a bf cfI I I I

The resultant phase voltages from the relationship given in Equation 378 can be

expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70

And the line-to-line voltages are

0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12

FORMATION OF BUS IMPEDANCE MATRIX USING MATLAB

Aim To determine the bus impedance matrix for the given power system Network

Apparatus MATLAB 77

Theory

Formation of Z BUS matrix

Z-bus matrix is an important matrix used in different kinds of power system study such as

short circuit study load flow study etc In short circuit analysis the generator uses transformer

impedance must be taken into account In quality analysis the two-short element are neglected by

forming the z-bus matrix which is used to compute the voltage distribution factor This can be

largely obtained by reversing the y-bus formed by inspection method or by analytical method

Taking inverse of the y-bus for large system in time consuming Moreover modification in the

system requires whole process to be repeated to reflect the changes in the system In such cases

is computed by z-bus building algorithm

Algorithm

Step 1 Read the values such as number of lines number of buses and line data generator data

and transformer data

Step 2 Initialize y-bus matrix y-bus[i] [j] =complex(0000)

Step 3 Compute y-bus matrix by considering only line data

Step 4 Modifies the y-bus matrix by adding the transformer and the generator admittance to the

respective diagonal elements of y-bus matrix

Step 5 Compute the z-bus matrix by inverting the modified y-bus matrix

Step 6 Check the inversion by multiplying modified y-bus and z-bus matrices to check whether

the resulting matrix is unit matrix or not

Step 7 Print the z-bus matrix


76

Procedure






lines and line data

START

Form Y bus matrix using the algorithm



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Modifying the y bus by adding generator

and transformer admittances to respective

diagonal elements

Compute Z bus matrix by inverting

modified Y bus

Multiply modified Y bus and Z bus and check

whether the product is a unity matrix

Print all the results

STOP


77

View the results

MATLAB Program


zdata(1)) nbus = max(max(nl) max(nr))








end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80


SYMMETRICAL COMPONENTS

Aim To obtain symmetrical components of set of unbalanced currents

Apparatus MATLAB

Theory

Before we discuss the symmetrical component transformation let us first define the a-

operator

2

3

2

10120 jea j

Note that for the above operator the following relations hold

on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j

Also note that we have

02

3

2

1

2

3

2

111 2 jjaa

Using the a-operator we can write from Fig 71 (b)

111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV

The symmetrical component transformation matrix is then given by


81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

Defining the vectors Va012 and Vabc as

c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]

rankV012=length(V012(1))

if rankV012 == 2

mag= V012(1) ang=pi180V012(2)

V012r=mag(cos(ang)+jsin(ang))

elseif rankV012 ==1

V012r=V012

else

fprintf(n Symmetrical components must be expressed in a one column array in rectangular

complex form n)

fprintf( or in a two column array in polar form with 1st column magnitude amp 2nd column

n)

fprintf( phase angle in degree n)

return end

a=cos(2pi3)+jsin(2pi3)

A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r

Vabcp= [abs(Vabc) 180piangle(Vabc)]

fprintf( n Unbalanced phasors n)

fprintf( Magnitude Angle Degn)

disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure

1 Open Matlab--gt File ---gt New---gt Script

2 Write the program

3 Enter F5 to run the program

4 Observe the results in MATLAB command window

Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84


UNBALANCED VOLTAGES FROM SYMMETRICAL COMPONENTS

Aim To obtain the original unbalanced phase voltages from symmetrical components

Apparatus MATLAB

Theory

abca CVV 012

where C is the symmetrical component transformation matrix and is given by

aa

aaC2

2

1

1

111

3

1

The original phasor components can also be obtained from the inverse symmetrical

component transformation ie

012

1

aabc VCV

Inverting the matrix C we get

2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2

10 cccaaac VVVVaaVVV

Finally if we define a set of unbalanced current phasors as Iabc and their symmetrical

components as Ia012 we can then define

012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]

rankIabc=length(Iabc(1))

if rankIabc == 2

mag= Iabc(1) ang=pi180Iabc(2)

Iabcr=mag(cos(ang)+jsin(ang))

elseif rankIabc ==1

Iabcr=Iabc

else

fprintf(n Three phasors must be expressed in a one column array in rectangular complex

form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]

fprintf( n Symmetrical components n)


disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i

Symmetrical components

Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14

SHORT CIRCUIT ANALYSIS OF 14 BUS SYSTEM

Aim To obtain the short circuit analysis for a IEEE 14 bus system

Apparatus POWERWORLD

Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii

We further assume that the system is unloaded before the fault occurs and that the magnitude and

phase angles of all the generator internal emfs are the same Then there will be no current

circulating anywhere in the network and the bus voltages of all the nodes before the fault will be

same and equal to Vf Then the new altered bus voltages due to the fault will be given from by

41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure



Execute the program in run mode by selecting tools-fault analysis

Select the fault on which bus and calculate


Results 3-phase fault is applied on bus 3

Bus

Number Va Vb Vc Ang A Ang B Ang C

1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087

Fault current = 445 pu

= 186167 A

Fault current angle = -7774 deg


91

Result



92

Date Experiment-15

LOAD FREQUENCY CONTROL OF A SINGLE AREA POWER SYSTEM

Aim To obtain the frequency response of single and two area power system using MATLAB

Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where

D = damping coefficient

GG ndash gain of generator GT - gain of turbine

GP - gain of power

KP ndash power system constant

KT- turbine constant

KG- generator constant

TP ndash power system time constant

TG- generator time constant

Procedure








93

SIMULINK RESULTS

Single Area Power System Block Diagram

Graph


94


95


96

RESULT



97

Date Experiment-16

LOAD FREQUENCY CONTROL OF A TWO AREA POWER SYSTEM

Aim To obtain the frequency response of two area power system using MATLAB

Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17

STEP RESPONSE OF A SYNCHRONOUS MACHINE

Aim To obtain step response of rotor angle and generator frequency of a synchronous

machine

Apparatus MATLAB

Circuit Diagram

Procedure

1 Open File ---gt New---gt Model







103

Graph


104


105

Result



2

GOKARAJU RANGARAJU INSTITUTE OF

ENGINEERING AND TECHNOLOGY

(Autonomous)

Bachupally Hyderabad-500 072

CERTIFICATE

This is to certify that it is a record of practical work done in

the Power Systems Simulation Laboratory in I sem of IV

year during the year ________________________

Name

Roll No

Branch EEE

Signature of staff member


3

INDEX


the Faculty





medium line model

14



19



Method

30



Coordinates

37



41

47



Synchronous Machine

57




4





78

82



system

90


system

95



Machine

100

18

19

20

21

22

23

24

25

26

27


1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



3

INDEX


the Faculty





medium line model

14



19



Method

30



Coordinates

37



41

47



Synchronous Machine

57




4





78

82



system

90


system

95



Machine

100

18

19

20

21

22

23

24

25

26

27


1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



4





78

82



system

90


system

95



Machine

100

18

19

20

21

22

23

24

25

26

27


1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



3

Graph

Calculations


4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



4


5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



5

Result


6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



6


Date Experiment-2



data

Apparatus MATLAB





















7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



7
















voltage is V2




Therefore


8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



8



below


Transformer














drawn



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



9

Circuit Diagram


10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



10

Procedure







Graph

Calculations


11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



11


12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



12


13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



13

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



14

Date Experiment-3



medium line model

Apparatus MATLAB


























15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



15












16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



16




Circuit Diagram


17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



17

PROCEDURE







Calculations


18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



18

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



19

Date Experiment-4



Apparatus MATLAB




















(1)


(2)


20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



20








21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



21

Circuit Diagram

PROCEDURE







Graph


22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



22


23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



23


24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



24

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



25

Date Experiment-5



Apparatus MATLAB 77

Theory











bus] matrix



EBUS = ZBUS IBUS

IBUS = YBUS EBUS


IBUS = YBUS EBUS













26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



26





lines and line data

START


Consider line l = 1

i = sb(1) I= eb(1)



Y(ij) = -Yseries(l)

Y(ji) =Y(ij)

Is l =NL

l = l+1

Stop

NO YES

Print Y -Bus


27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



27

MATLAB PROGRAM











end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Calculations


28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



28


29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



29

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



30

Date Experiment-6


Aim


method

Apparatus MATLAB

Theory




Problem Formulation






Where

n


k=1

n


k=1


n


j=1



n


J=1







31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



31




Algorithm





Step4 Iter=1



step








step6


step12





bus angles

Step13 Stop


32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



32

START

Read






Form Ybus



k=12hellipn



Test for

type of bus

Qi(r+1)

gtQi max Qi(r+1)

lt Qimin

Compute Qi(r+1)

Qi(r+1)

= Qimax Qi


(r+1)

Compute Ai

Compute Vi

(r+1)

Compute δi(r+1)

and

Vi(r+1)

=|Vis|δi

(r+1)


33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



33


Procedure





View the results

MATLAB program

clear

basemva=100



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

Replace Vir by Vi

(r+1) and


Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



34

2 3 00125 0025 00 1]

lfybus

lfgauss

busout

lineflow

Calculations


35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



35


36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



36

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



37

Date Experiment-7


Aim


Apparatus MATLAB 77

Theory









Algorithm








Evalute pp=psp-pp


38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



38









Fp^k+1=fpK+fpK



Procedure





View the results

MATLAB program

clear



busdata=[1 1 105 00 00 00 00 00 0 0 0

2 0 10 00 25666 1102 00 00 0 0 0

3 0 10 00 1386 452 00 00 0 0 0]


linedata=[1 2 002 004 00 1

1 3 001 003 00 1

2 3 00125 0025 00 1]

lfybus

lfnewton


39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



39

busout

lineflow

Calculations


40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



40

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



41





Apparatus MATLAB

Theory




losses







iPi

2



Equality constraint



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



42





Pi

min le Pi le Pi












43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



43









44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



44

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC

Min

MW

Max

MW

Cost

$Hr Lambda

1 7952 150 5 011 10 250 124312 2249

2 126 600 12 0085 10 300 210066 2262

3 114 335 1 01225 10 270 204101 2893



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



45

Calculations


46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



46

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



47





Apparatus MATLAB


modify (54) as

LOSSNT PPPPP 21


LOSSN dPdPdPdP 210


N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1


N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0


N

i

i

i

LOSS

i

T dPP

P

P

f

1

0


NiP

P

P

f

i

LOSS

i

T 10

Again since

NiP

df

P

f

i

T

i

T 1


48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



48

we get

N

N

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1


NiPP

LiLOSS

i 11

1


TNPPPP 21


BPPP T

LOSS


NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211





49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



49

Circuit Diagram

`


50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



50

Procedure




Run the primal lp



Results

Generator

Name

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW

Cost

$Hr Lambda

1 10638 150 5 011 0 10 250 192666 284

2 126 600 12 0085 0 10 300 210066 2262

3 8622 335 1 01225 0 10 270 133184 2212

Total Cost =

535915$hr


51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



51


52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



52

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



53

Date Experiment-9




Apparatus MATLAB



















s2 + RLs + 1LC =0




54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



54




function


s2


Procedure





View the results

MATLAB Program



n = [wnwn]

p = sqrt(1-z^2)


55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



55

wd = wnp

h = [pz]

k = atan(h)

m = pi-k

tr = [mwd]

tp = [piwd]

q = zwn

ts = [4q]

r = zpi

f = [rp]

mp = exp(-f)

num = [0 0 n]

den = [1 zzwn n]

s = tf(numden)

hold on

step(s)

impulse(s)

hold off

Result


wn = 16


z = 05000

n = 256


56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



56

p = 08660

k = 10472

tr = 01511

tp = 02267

q = 8

ts = 05000

r = 15708

f = 18138

mp = 01630

num = 0 0 256

den = 1 4 256

Transfer function

256

---------------

s^2 + 4 s + 256

Graph


57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



57

Calculations


58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



58

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



59

Date Experiment-10


MACHINE


Apparatus MATLAB





current is given by





with



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



60





current as







steady state




is given by


Procedure








61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



61

Circuit Diagram

Graph


62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



62

Calculations


63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



63


64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



64

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



65

Date Experiment-11



Apparatus MATLAB

Theory







a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

10

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



66


from

0 0

2

1 1

2

22

0 1 1 1

10 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

10

a a

a a

a a

V Z I

V Z I

V Z I



0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault









calculations [1]


67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



67

a

c

b

F


Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-


line-to-line fault


0afI

bf cfI I

bc f bfV Z I


0 0aI

1 2

1 2

10 0a a

f

I IZ Z Z

If Zf = 0

1 2

1 2

10 0a aI I

Z Z


13 90bf cf aI I I


0

1 1 1

2 2 2 2 1

0

10 -

a

a a

a a a

V

V Z I

V Z I Z I


68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



68


ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V










a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

10 0o

-




0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I


12 0

1

2 0

10 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z


69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



69

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z


0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I


be obtained from

12 0

1

2 0

10 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z


0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I


03n a bf cfI I I I


expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V


70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



70


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

Circuit Diagram

Procedure








71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



71

Graph


72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



72


73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



73


74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



74

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



75

Date Experiment-12



Apparatus MATLAB 77

Theory










Algorithm












76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



76

Procedure






lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



77

View the results

MATLAB Program










end

end


for k = 1nbr

if nl(k) == n | nr(k) == n


else end

end

end

Zbus= inv(Y)

Ibus=[-j11-j12500]

Vbus=ZbusIbus

OUTPUT


78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



78


79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



79

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



80




Apparatus MATLAB

Theory


operator

2

3

2

10120 jea j


on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j


02

3

2

1

2

3

2

111 2 jjaa


111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



81

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V

2

1

0

012

Program

V012 = [06 90

10 30

08 -30]


if rankV012 == 2

mag= V012(1) ang=pi180V012(2)


elseif rankV012 ==1

V012r=V012

else


complex form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

Vabc= AV012r




disp(Vabcp)

Vabc0=V012r(1)[1 1 1]

Vabc1=V012r(2)[1 a^2 a]

Vabc2=V012r(3)[1 a a^2]


82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



82

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



83

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



84




Apparatus MATLAB

Theory

abca CVV 012


aa

aaC2

2

1

1

111

3

1



012

1

aabc VCV


2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2




012

1

012

aabc

abca

ICI

CII


85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



85

Program

Iabc = [16 25

10 180

09 132]


if rankIabc == 2



elseif rankIabc ==1

Iabcr=Iabc

else


form n)


n)


return end


A = [1 1 1 1 a^2 a 1 a a^2]

I012=inv(A)Iabcr

symcomp= I012

I012p = [abs(I012) 180piangle(I012)]



disp(I012p)

Iabc0=I012(1)[1 1 1]

Iabc1=I012(2)[1 a^2 a]

Iabc2=I012(3)[1 a a^2]

Result

symcomp =

-00507 + 04483i

09435 - 00009i

05573 + 02288i


Magnitude Angle Deg

04512 964529

09435 -00550

06024 223157


86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



86


87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



87

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



88

Date Experiment-14




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





41144

4

iV

Z

ZVVV f

iifi

14 bus system


89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



89

Procedure







Bus


1 041937 041937 041937 1964 -10036 13964

2 034862 034862 034862 1087 -10913 13087

3 0 0 0 0 0 0

4 029602 029602 029602 -225 -12225 11775

5 032956 032956 032956 058 -11942 12058


90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



90

6 047538 047538 047538 -1063 -13063 10937

7 042742 042742 042742 -844 -12844 11156

8 055864 055864 055864 -1006 -13006 10994

9 041373 041373 041373 -95 -1295 1105

10 042145 042145 042145 -994 -12994 11006

11 044639 044639 044639 -1037 -13037 10963

12 046474 046474 046474 -1144 -13144 10856

13 045879 045879 045879 -1127 -13127 10873

14 042532 042532 042532 -113 -1313 1087


= 186167 A



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



91

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



92

Date Experiment-15



Apparatus MATLAB 61

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



93

SIMULINK RESULTS


Graph


94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



94


95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



95


96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



96

RESULT



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



97

Date Experiment-16



Apparatus MATLAB

Formula used

1)

2)

3)

4)

Where



GP - gain of power






Procedure








98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



98

SIMULINK RESULTS


Graphs


99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



99


100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



100


101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



101

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



102

Date Experiment-17



machine

Apparatus MATLAB

Circuit Diagram

Procedure








103

Graph


104


105

Result



103

Graph


104


105

Result



104


105

Result



105

Result


power system simulation lab

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