author griet-jsd created date 5/9/2017 1:29:36 pm

POWER SYSTEMS SIMULATION LAB

___YEAR __ SEM

EEE

By

Dr J Sridevi

Gokaraju Rangaraju Institute of Engineering amp

Technology

Bachupally

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 _____ sem of ________________

year during the year ________________________

Name

Roll No

Branch EEE

Signature of staff member

INDEX

List of Experiments

SNo Date Name of the Experiment Page no Signature

1 Sinusoidal Voltages and currents 1

2 Computation of line parameters 6

3 Modelling of transmission lines 11

4 Formation of bus admittance matrix 18

5 Load Flow solution using gauss seidel

method 23

6 Load flow solution using Newton

Rapshon method in Polar coordinates 30


Rapshon method in Rectangular coordinates 34

8 Transient stability analysis of single-

machine infinite bus 38

system

9 Power flow solution of 3 ndash bus system 43

10

a)Optimal dispatch neglecting losses b) Optimal dispatch including losses

48 54

11

Three phase short circuit analysis in a

synchronous machine(symmetrical fault

analysis) 58

12 Unsymmetrical fault analysis-LGLLLLG

fault 64

13 Z- Bus building algorithm 73

14

a)Obtain symmetrical components of a set of unbalanced currents b)Obtain the original unbalanced phase voltages from symmetrical components

78 82

15 Short circuit analysis of a power system

with IEEE 9 bus system 85

16 Power flow analysis of a slack bus

connected to different loads 89

17 Load flow analysis of 3 motor systems

connected to slack bus 94

GRIETEEE Power Systems Simulation Lab

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

ExpNo2 Date

COMPUTATION OF LINE PARAMETERS

AIM

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

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

understand modeling and performance of medium lines SOFTWARE REQUIRED MAT LAB THEORY

Transmission line has four parameters namely resistance inductance capacitance and

conductance The inductance and capacitance are due to the effect of magnetic and electric fields

around the conductor The resistance of the conductor is best determined from the manufactures

data the inductances and capacitances can be evaluated using the formula Inductance The general formula

L = 02 ln (Dm Ds)

Where Dm = geometric mean distance (GMD)

Ds = geometric mean radius (GMR) I Single phase 2 wire system GMD = D GMR = re-14 = rprime Where r = radius of conductor II Three phase ndash symmetrical spacing GMD = D GMR = re-14 = rprime Where r = radius of conductor III Three phase ndash Asymmetrical Transposed GMD = geometric mean of the three distance of the symmetrically placed conductors = (DABDBCDCA)13 GMR = re-14 = rprime Where r = radius of conductors Composite conductor lines The inductance of composite conductor X is given by Lx = 02 ln (GMDGMR)


7

where GMD = (Daa Dab )helliphellip(Dna helliphellipDnm ) GMR = n2

(Daa DabhelliphellipDan )helliphellip(DnaDnbhelliphellipDnn) where rrsquoa

= ra e(-1 4) Bundle Conductors The GMR of bundled conductor is normally calculated

GMR for two sub conductor Dsb = (Dstimesd)12

GMR for three sub conductor Dsb = (Dstimesd2)

13 GMR

for four sub conductor Dsb = 109 times (Dstimesd2)14

where Ds is the GMR of each subconductor d is bundle spacing Three phase ndash Double circuit transposed The inductance per phase in mH per km

is L = 02timesln(GMD GMRL) mHkm

where GMRL is equivalent geometric mean radius and is given by

GMRL = (DSADSBDSC)13

where DSADSB and DSC are GMR of each phase group and given by

12 DSA = 4 sb Da1a2)2 = [Dsb Da1a2]

12 DSB = 4 sb Db1b2)2 = [Dsb Db1b2]

DSC = 4 sb Dc1c2)2 = [Dsb Dc1c2]12

where Dsb =GMR of bundle conductor if conductor a1 a2hellip are bundled conductor Dsb = ra1rsquo= rb1= rarsquo2 = rbrsquo2 = rcrsquo2 if a1 a2helliphellip are bundled conductor GMD is the equivalent GMD per phaserdquo amp is given by GMD = [DAB DBC

DCA]13

where DAB DBCampDCA are GMD between each phase group A-B B-C C-A which are given by

DAB = [Da1b1 Da1b2 Da2b1 Da2b2]14

DBC = [Db1c1 Db1c2 Db2c1 Db2c2]14

DCA = [Dc1a1 Dc2a1 Dc2a1 Dc2a2]14

Capacitance A general formula for evaluating capacitance per phase in micro farad per km of a transmission line is given by C = 00556 ln (GMDGMR) Fkm Where GMD is the ldquoGeometric mean distancerdquo which is same as that defined for inductance under various cases


8

PROCEDURE

Enter the command window of the MAT LAB

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

EXERCISE 1A 500kv 3 φ transposed line is composed of one ACSR 1272000-cmil 457 bittern conductor

per phase with horizontal conductor configuration as show in fig1 The conductors have a

diameter of 1345in and a GMR of 05328in Find the inductance and capacitance per phase per

kilometer of the line and justify the result using MAT LAB

a b c

D12 =35rsquo D23 =35rsquo

D13=70rsquo

Fig1

2The transmission line is replaced by two ACSR 636000-cmil 247 Rook conductors which have

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

measured from the centre of the bundle is the same as before and is shown in fig2 The conductors

have a diameter of 0977in and a GMR of 03924inBundle spacing is 18in Find the inductance

and capacitance per phase per kilometer of the line and justify the result using MAT LAB

a b c


D13=70rsquo


9

3A 345- KV double ndashcircuit three- phase transposed line is composed of two ACSR 1431000-

cmil 457 Bobolink conductors per phase with vertical conductor configuration as shown in fig3

The conductors have a diameter of 1427in and a GMR of 0564 in the bundle spacing in 18in

find the inductance and capacitance per phase per kilometer of the line and justify the result using

MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM

[GMD GMRL GMRC] = gmd

L = 02log(GMDGMRL) C = 00556log(GMDGMRC)

OUTPUT


10

RESULT

Thus the positive sequence line parameters L and C per phase per kilometre of a three phase single and double circuit transmission lines for different conductor arrangements were determined and verified with MAT LAB software The value of L and C obtained from MAT LAB program are

Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date

MODELLING OF TRANSMISSION LINES

AIM To understand modeling and performance of Short Medium and Long transmission

lines

SOFTWARE REQUIRED MAT LAB

THEORY

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

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

influenced by the line constants R L and C of the transmission line For instance the voltage drop

in the line depends upon the values of above three line constants Similarly the resistance of

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

the transmission efficiency

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

length of the line The resistance and inductance form the series impedance The capacitance

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

forms a shunt path throughout the length of the line Short Transmission Line

When the length of an overhead transmission line is upto about 50km and the line voltage is comparatively low (lt 20 kV) it is usually considered as a short transmission line Medium Transmission Lines

When the length of an overhead transmission line is about 50-150 km and the line voltage is moderatly high (gt20 kV lt 100 kV) it is considered as a medium transmission line Long Transmission Lines

When the length of an overhead transmission line is more than 150km and line voltage is very high (gt 100 kV) it is considered as a long transmission line Voltage Regulation

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

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

end voltage

Performance of Single Phase Short Transmission Lines

As stated earlier the effects of line capacitance are neglected for a short transmission line Therefore while studying the performance of such a line only resistance and inductance of

6


12

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

shown in Fig (i)Here the total line resistance and inductance are shown as concentrated or lumped

instead of being distributed The circuit is a simple ac series circuit Let I = load current R = loop resistance ie resistance of both conductors XL = loop reactance VR = receiving end voltage cos ϕR = receiving end power factor (lagging) VS = sending end voltage cos ϕS = sending end power factor The phasor diagram of the line for lagging load power factor is shown in Fig (ii) From the right angled traingle ODC we get

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

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

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


13

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

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

ie

OC = OF = OA + AF = OA + AG + GF = OA + AG + BH

VS = VR + I R cos ϕR + I XL sin ϕR Medium Transmission

Line Nominal T Method

In this method the whole line capacitance is assumed to be concentrated at the middle

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

in Fig1 Therefore in this arrangement full charging current flows over half the line In Fig1 one

phase of 3-phase transmission line is shown as it is advantageous to work in phase instead of line-

to-line values

Let IR = load current per phase R = resistance per phase


14

XL = inductive reactance per phase C = capacitance per phase cos ϕR = receiving end power factor (lagging) VS = sending end voltagephase V1 = voltage across capacitor C The phasor diagram for the circuit is shown in Fig2 Taking the receiving end voltage VR as the reference phasor we have Receiving end voltage VR = VR + j 0 Load current IR = IR (cos ϕR - j sin ϕR)

Nominal π Method In this method capacitance of each conductor (ie line to neutral) is divided into two halves

one half being lumped at the sending end and the other half at the receiving end as shown in

Fig3 It is obvious that capacitance at the sending end has no effect on the line drop However

itrsquos charging current must be added to line current in order to obtain the total sending end

current


15

IR = load current per phase R = resistance per phase XL = inductive reactance per phase C = capacitance per phase cos R = receiving end power factor (lagging) VS = sending end voltage per phase The phasor diagram for the circuit is shown in Fig4 Taking the receiving end voltage as the reference phasor we have VR = VR + j 0 Load current IR = IR (cos R - j sin R)


16

EXERCISE

1 A 220- KV 3φ transmission line is 40 km long The resistance per phase is 015 Ω per

km and the inductance per phase is 13623 mH per km The shunt capacitance is negligible

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

regulation and efficiency when the line supplying a three phase load of a) 381 MVA at 08

power factor lagging at 220 KV b) 381 MVA at 08 power factor leading at 220 KV

PROGRAM

VRLL=220 VR=VRLLsqrt(3)

Z=[015+j2pi6013263e-

3]40 disp=((a)) SR=3048+j2286 IR=conj(SR)(3conj(VR)) IS=IR VS=VR+ZIR VSLL=sqrt(3)abs(VS) SS=3VSconj(IS)

REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286

IR=conj(SR)(3conj(VR))

IS=IR VS=VR+ZIR VSLL=sqrt(3)abs(VS) SS=3VSconj(IS) REG=((VSLL-

VRLL)(VRLL))100

EFF=(real(SR)real(SS))100

FF=(real(SR)real(SS))100


17

OUTPUT RESULT Thus the program for modeling of transmission line was executed by using MAT LAB

and the output was verified with theoretical calculation The value of the voltage and power at the

sending end voltage regulation and efficiency obtained from the MAT LAB program are Voltage VS =

Power SS = Voltage regulation REG = Efficiency EFF =



18

Expt No4 Date

FORMATION OF BUS ADMITTANCE MATRIX

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


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 FORMULA USED Yij=ΣYij for i=1 to n Yij= -Yij= -1Zij Yij=Yji Where Yii=yjj= Sum of admittance connected to bus Yij= Negative admittance between buses


19

START

Read the no Of buses no of lines and line data

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) NO Is l =NL YES

l = l+1 Print Y -Bus

Stop


20

Exercise

Line number

Starting Bus

Ending Bus Series Line Line Changing

Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j

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


21


22

OUTPUT

RESULT Thus the bus admittance matrix of the given power system using inspection method

was found and verified by theoretical calculation

Y Bus



23

Expt No5 Date

LOAD FLOW SOLUTION USING GAUSS SEIDEL METHOD

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

method

SOFTWARE REQUIRED POWERWORLD

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 Vo

i - Σ YijVjn- 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


24

During each iteration we compute all the bus voltage and check for convergence is carried out by comparison with the voltages obtained at the end of previous iteration After the solutions is obtained The stack bus real and reactive powers the reactive power generation at other generator buses and line flows can be calculated

ALGORITHM Step1 Read the data such as line data specified power specified voltages Q limits at the generator buses

and tolerance for convergences Step2 Compute Y-bus matrix Step3

Initialize all the bus voltages Step4 Iter=1 Step5 Consider i=2 where 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


25

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) =|Vi

s|δi(r+1)


26

FLOW CHART FOR GAUSS SEIDEL METHOD

Exercise

1 The Fig1 shows the single line diagram of a simple three bus power system with generation at buses 1

The magnitude of voltage at bus 1 is adjusted to 105 pu The scheduled loads at buses 2 amp 3 are marked

on the diagram line impedances are marked in per unit on a 100-MVA base and the line charging charging

susceptances are neglected a) using the gauss-seidal method determine the phasor values of the voltage at the load buses 2 and 3(P-

Q buses) accurate to four decimal places b)find the slack bus real and reactive power c) Determine the line flow and line losses Using software write an algorithm

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

GRIETEEE Power Systems Simulation

Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar

Slack bus v1 = 105L 3

1386 MW 452 Mvar

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 power flow by gauss seidel method

View the results in case information-Bus information

Tabulate the results

POWER WORLD bus diagram


28


29

RESULTS A one line diagram has been developed using POWER WORLD for the given power system by

Gauss Seidal method and the results are verified with model calculation



30

ExpNo 6 Date

LOAD FLOW SOLUTION USING NEWTON RAPSHON METHOD IN

POLAR COORDINATES

AIM

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

polar coordinates

SOFTWARE REQUIRED POWERWORLD 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)


31

Evalute pp=psp-pp 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

EXERCISE 1 For the sample system of Fig the generators are connected at all the four buses while loads are

at buses 2 and 3 Values of real and reactive powers are listed in Table 63 All buses other than

the slack are PQ type

Assuming a flat voltage start find the voltages and bus angles at the three buses using Newton

Raphson Method


32

PROCEDURE



Execute the program in run mode by selecting power flow by Newton Raphson method





33

RESULT A one line diagram has been developed using POWERWORLD for the given power system by

Newton raphson method and the results are verified with model calculation



34

ExpNo 7 Date


RECTANGULAR COORDINATES

AIM


Rectangular coordinates










35




EXERCISE 1 For the sample system of Fig the generators are connected at all the four buses while loads are at buses 2 and 3 Values of real and reactive powers are listed in Table 63 All buses other than the slack are PQ type Assuming a flat voltage start find the voltages and bus angles at the three buses using Newton Raphson Method


36

PROCEDURE








37





38

Expno8 Date

TRANSIENT STABILITY ANALYSIS OF SINGLE-MACHINE INFINITE BUS SYSTEM

AIM To become familiar with various aspects of the transient stability analysis of Single-Machine-Infinite

Bus (SMIB) system OBJECTIVES The objectives of this experiment are 1 To study the stability behavior of one machine connected to a large power system subjected to a

severe disturbance (3-phase short circuit) 2 To understand the principle of equal-area criterion and apply the criterion to study the stability of

one machine connected to an infinite bus 3 To determine the critical clearing angle and critical clearing time with the help of equal-area

criterion 4 To do the stability analysis using numerical solution of the swing equation SOFTWARE REQUIRED MAT LAB 77 THEORY The tendency of a power system to develop restoring forces to compensate for the disturbing forces to

maintain the state of equilibrium is known as stability If the forces tending to hold the machines in

synchronism with one another are sufficient to overcome the disturbing forces the system is said to

remain stable The stability studies which evaluate the impact of disturbances on the behavior of synchronous

machines of the power system are of two types ndash transient stability and steady state stability The

transient stability studies involve the determination of whether or not synchronism is maintained after

the machine has been subjected to a severe disturbance This may be a sudden application of large load

a loss of generation a loss of large load or a fault (short circuit) on the system In most disturbances

oscillations are such magnitude that linearization is not permissible and nonlinear equations must be

solved to determine the stability of the system On the other hand the steady-state stability is concerned

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

linearized version of nonlinear equations In this experiment we are concerned with the transient stability of power systems A method known as the equal-area criterion can be used for a quick prediction of stability of a one-

machine system connected to an infinite bus This method is based on the graphical interpretation of


39

energy stored in the rotating mass as an aid to determine if the machine maintains its stability after a

disturbance The method is applicable to a one-machine system connected to an infinite bus or a two-

machine system Because it provides physical insight to the dynamic behavior of the machine the

application of the method to analyze a single-machine system is considered here Stability Stability problem is concerned with the behavior of power system when it is subjected to

disturbance and is classified into small signal stability problem if the disturbances are small and

transient stability problem when the disturbances are large Transient stability When a power system is under steady state the load plus transmission loss equals

to the generation in the system The generating units run at synchronous speed and system frequency

voltage current and power flows are steady When a large disturbance such as three phase fault loss

of load loss of generation etc occurs the power balance is upset and the generating units rotors

experience either acceleration or deceleration The system may come back to a steady state condition

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

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

unstable

EXERCISE A 60Hz synchronous generator having inertia constant H = 994 MJMVA and a direct axis transient

reactance Xdrsquo= 03 per unit is connected to an infinite bus through a purely resistive circuit as shown

in fig1 Reactances are marked on the diagram on a common system base The generator is delivering

real power of 06 unit 08 pf lagging and the infinite bus at a voltage of 1 per unit Assume the pu

damping power coefficient d=0138 Consider a small disturbance change in delta=100 Obtain the

equation describes the motion of the rotor angle and generation frequency


40

If the generation is operating at steady state at when the input power is increased by

small amount The generator excitation and infinite bus bar voltage are same

as before Obtain a simulink block diagram of state space mode

and simulate to obtain the response


41


42

RESULT



43

ExpNo 9 Date

POWER FLOW SOLUTION OF 3 ndash BUS SYSTEM

Aim To perform power flow solution of a 3-Bus system

Apparatus MATLAB-PSAT

Theory

Slack Bus To calculate the angles θi (as discussed above) a reference angle (θi = 0) needs to

be specified so that all the other bus voltage angles are calculated with respect to this reference

angle Moreover physically total power supplied by all the generation must be equal to the sum

of total load in the system and system power loss However as the system loss cannot be

computed before the load flow problem is solved the real power output of all the generators in

the system cannot be pre-specified There should be at least one generator in the system which

would supply the loss (plus its share of the loads) and thus for this generator the real power

output canrsquot be pre-specified However because of the exciter action Vi for this generator can

still be specified Hence for this generator Vi and θi(= 0) are specified and the quantities Pi and

Qi are calculated This generator bus is designated as the slack bus Usually the largest generator

in the system is designated as the slack bus

In the General Load Flow Problem the Bus with largest generating capacity is taken as the Slack

Bus or Swing Bus The Slack Bus Voltage is taken to be 10 + j 0 PU and should be capable of

supplying total Losses in the System But usually the generator bus are only having station

auxiliary which may be only up to 3 of total generation If the Generation at Slack Bus is

more it can take more load connected to the slack bus

A slack bus is usually a generator bus with a large real and reactive power output It is assumed

that its real and reactive power outputs are big enough that they can be adjusted as required in

order to balance the power in the whole system so that the power flow can be solved A slack

bus can have load on it because in real systems it is actually the bus of a power plant which can

have its own load It also takes care of the Line losses


44

Circuit Diagram

Procedure

Create a new file in MATLAB-PSAT

Browse the components in the library and build the bus system

Save the file and upload the data file in PSAT main window

Execute the program and run powerflow

Get the network visualization


45


46


47

Result



48

ExpNo 10(a) Date

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 + γiPi2

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


49

where Pi = power generated by ith unit

PD = total power demand

Inequality constraint

Each generator should generate power within the limits imposed

Pimin 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


50

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

Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]

Find the optimal dispatch neglecting losses

Procedure



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


Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date

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


55

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

Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]

Find the optimal dispatch including losses

Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date

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


59

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








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date

UNSYMMETRICAL FAULT ANALYSIS

Aim To analyze unsymmetrical faults like LGLLLLG

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


65

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]


66

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

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


67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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

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


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date

Z-BUS BUILDING ALGORITHM

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

Apparatus MATLAB

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


74

Procedure

Enter the command window of the MATLAB


Type and save the program in the editor Window


View the results

Read the no Of buses no of

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


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j

disp(STEP1 Add an element between reference node(0) and node(1))

zBus1 = [z01]

disp(STEP2 Add an element between existing node(1) and new node(2))

zBus2 = [zBus1(11) zBus1(11)

zBus1(11) zBus1(11)+z12]

disp(STEP3 Add an element between existing node(2) and reference node(0))

zBus4 = [zBus2(11) zBus2(12) zBus2(12)

zBus2(21) zBus2(22) zBus2(22)

zBus2(12) zBus2(22) zBus2(22)+z02]

disp(Fictitious node(0) can be eliminated)

zz11 = zBus4(11) - ((zBus4(13) zBus4(31)) zBus4(33))


zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]

disp(STEP5 Add an element between existing nodes (3) and (1))

zBus7 = [zBus6(11) zBus6(12) zBus6(13) zBus6(13)-zBus6(11)

zBus6(21) zBus6(22) zBus6(23) zBus6(23)-zBus6(21)


zBus6(31)-zBus6(11) zBus6(32)-zBus6(12) zBus6(33)-zBus6(13) z23+zBus6(11)+zBus6(33)-

2zBus6(13)]


zzz11 = zBus7(11) - ((zBus7(14) zBus7(41)) zBus7(44))


76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)

zBus = [zzz11 zzz12 zzz13

zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date

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


79

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]


80

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


81

Result



82

ExpNo 14(b) Date

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


83

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


84

Result



85

ExpNo 15 Date

SHORT CIRCUIT ANALYSIS OF IEEE 9 BUS SYSTEM

Aim To obtain the short circuit analysis for a IEEE 9 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

IEEE 9 bus system

Sbase = 100 MVA Vbase = 220KV

Vmax = 106 pu Vmin = 1 pu

NUMBER OF LINES = 8 NUMBER OF BUSES = 9


86

Case Data

Bus Records

Bus No Area PU Volt Volt (kV) Ang(Deg)

Load

MW

Load

Mvar Gen MW Gen Mvar

1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW Cost $Hr IC LossSens Lambda

1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250

Number Gen MW Gen Mvar Min MW Max MW Min Mvar

Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure



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

Select the fault on which bus and calculate


Results

Fault current =

Fault current angle =


88

Result



89

ExpNo 16 Date

POWER FLOW ANALYSIS OF SLACK BUS CONNECTED TO DIFFERENT LOADS

Aim To perform power flow analysis of a slack bus connected to different loads


Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date

LOAD FLOW ANALYSIS OF 3 MOTOR SYSTEMS CONNECTED TO SLACK BUS

Aim To perform load flow analysis of 3 motor systems connected to slack bus


Theory

Slack Bus To calculate the angles θi (as discussed above) a reference angle (θi = 0) needs to be

specified so that all the other bus voltage angles are calculated with respect to this reference angle

Moreover physically total power supplied by all the generation must be equal to the sum of total

load in the system and system power loss However as the system loss cannot be computed before

the load flow problem is solved the real power output of all the generators in the system cannot

be pre-specified There should be at least one generator in the system which would supply the

loss (plus its share of the loads) and thus for this generator the real power output canrsquot be pre-

specified However because of the exciter action Vi for this generator can still be specified

Hence for this generator Vi and θi(= 0) are specified and the quantities Pi and Qi are calculated

This generator bus is designated as the slack bus Usually the largest generator in the system is

designated as the slack bus




auxiliary which may be only up to 3 of total generation If the Generation at Slack Bus is more

it can take more load connected to the slack bus



order to balance the power in the whole system so that the power flow can be solved A slack bus

can have load on it because in real systems it is actually the bus of a power plant which can have

its own load It also takes care of the Line losses


95

Circuit Diagram

Procedure







96


97

RESULT


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 _____ sem of ________________

year during the year ________________________

Name

Roll No

Branch EEE

Signature of staff member

INDEX

List of Experiments







method 23







system


10


48 54

11



analysis) 58


fault 64


14


78 82








1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT


INDEX

List of Experiments







method 23







system


10


48 54

11



analysis) 58


fault 64


14


78 82








1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



1

Date Experiment-1



Apparatus MATLAB


















2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



2

Circuit diagram


Procedure








3

Graph

Calculations


4


5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



3

Graph

Calculations


4


5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



4


5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



5

Result


6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



6


ExpNo2 Date


AIM








L = 02 ln (Dm Ds)




7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



7






13 GMR











DCA]13







8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



8

PROCEDURE





View the results





a b c


D13=70rsquo

Fig1






a b c


D13=70rsquo


9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



9





MAT LAB

a arsquo S11 =11m

H12=7m

b S22=165m brsquo

H23=65m

S33=165m

c crsquo

PROGRAM



OUTPUT


10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



10

RESULT


Case1 L= C=

Case2 L= C=

Case3 L= C=



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



11

ExpNo3 Date



lines


THEORY
















end voltage



6


12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



12




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

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



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



13



ie








to-line values



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



14






current


15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



15



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



16

EXERCISE






PROGRAM


Z=[015+j2pi6013263e-


REG=((VSLL-

VRLL)(VRLL))100 E

disp=((b)) SR=3048-

j2286



VRLL)(VRLL))100




17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



17







18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



18

Expt No4 Date




THEORY












19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



19

START



Consider line l = 1

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



Y(ij) = -Yseries(l)



Stop


20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



20

Exercise

Line number

Starting Bus


Impedance

Admittance

1 1 2 01+04j 015j

2 2 3 015+06j 002j

3 2 4 018+055j 0018j

4 3 4 01+035j 0012j

5 4 1 025+07j 003j









end

end


21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



21


22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



22

OUTPUT



Y Bus



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



23

Expt No5 Date



method


THEORY




PROBLEM FORMULATION





Where

n


k=1

n


k=1


of

n


j=1



n


J=1






24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



24














25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



25

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


Compute Ai

Compute Vi(r+1)


Vi(r+1) =|Vi

s|δi(r+1)


26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



26


Exercise








Is ilt=n

Is

ΔVmaxlt=ε

Advance iteration

count r = r+1


B

A


Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



Lab

27

1 002 + j004 2

2566MW

001+j003 00125+j0025 1102Mvar


1386 MW 452 Mvar

PROCEDURE








28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



28


29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



29





30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



30

ExpNo 6 Date


POLAR COORDINATES

AIM


polar coordinates










31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



31








Raphson Method


32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



32

PROCEDURE








33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



33





34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



34

ExpNo 7 Date



AIM












35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



35






36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



36

PROCEDURE








37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



37





38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



38

Expno8 Date




















39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



39













unstable








40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



40






41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



41


42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



42

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



43

ExpNo 9 Date




Theory























44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



44

Circuit Diagram

Procedure







45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



45


46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



46


47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



47

Result



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



48

ExpNo 10(a) Date




Apparatus MATLAB

Theory




losses









Equality constraint



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



49





Pimin le Pi le Pi












50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



50







Exercise


51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



51

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World diagram


52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



52

Results

Calculations


53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



53

Result



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



54

ExpNo 10(b) Date




Apparatus MATLAB


can 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


55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



55

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




Exercise


56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



56

119872119882 119871119894119898119894119905119904 = [10 8510 8010 70

] 119888119900119904119905 = [200 7 0008180 63 0009140 68 0007

]


Procedure




Run the primal lp



Power World Diagram


57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



57

Results

Result



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



58

ExpNo 11 Date


SYNCHRONOUS MACHINE


Apparatus MATLAB





current is given by





with



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



59











steady state




is given by


Procedure








60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



60

Circuit Diagram


61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



61

Graph

Calculations


62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



62


63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



63

Result



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



64

ExpNo 12 Date



Apparatus MATLAB

Theory







[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





0 1 2

0 1 2

10 0

3a a a

f

I I IZ Z Z Z


65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



65


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]


66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



66

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



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



67

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

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



68

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


0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



69

Circuit Diagram

Procedure







Graph


70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



70

Calculations


71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



71


72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



72

Result



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



73

ExpNo 13 Date



Apparatus MATLAB

Theory










Algorithm












74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



74

Procedure





View the results


lines and line data

START




Y(ij) = -Yseries(l)

Y(ji) =Y(ij)



diagonal elements


modified Y bus




STOP


75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



75

MATLAB Program clc

clear all

z01 = 02j

z02 = 04j

z13 = 04j

z23 = 04j

z12 = 08j


zBus1 = [z01]



zBus1(11) zBus1(11)+z12]




zBus2(12) zBus2(22) zBus2(22)+z02]




zz21 = zz12


zBus5 = [zz11 zz12

zz21 zz22]




zBus5(21) zBus5(22) zBus5(22)+z23]






2zBus6(13)]




76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



76



zzz21 = zzz12



zzz31 = zzz13

zzz32 = zzz23


disp(RESULT)


zzz21 zzz22 zzz23

zzz31 zzz32 zzz33]

OUTPUT

Calculations


77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



77

Result



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



78

ExpNo 14(a) Date



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



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



79

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




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]


80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



80

Procedure


2 Write the program



Result

Vabc =

15588 + 07000i

-00000 + 04000i

-15588 + 07000i

Unbalanced phasors

Magnitude Angle Deg

17088 241825

04000 900000

17088 1558175


81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



81

Result



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



82

ExpNo 14(b) Date


COMPONENTS


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


83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



83

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


84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



84

Result



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



85

ExpNo 15 Date




Theory

ff IZV 44

32144

44 iV

Z

ZIZV f

ifii





by

41144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system





86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



86

Case Data

Bus Records


Load

MW

Load


1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max


1 0 150 5 011 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 00576 0 250 250 250

8 2 No 0 00625 0 250 250 250

3 6 No 0 00586 0 300 300 300

4 5 No 0017 0092 0158 250 250 250

9 4 No 001 0085 0176 250 250 250

5 6 No 0039 017 0358 150 150 150

6 7 No 00119 01008 0209 150 150 150

7 8 No 00085 0072 0149 250 250 250

8 9 No 0032 0161 0306 250 250 250


Max

Mvar

Cost

Model Part Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



87

2 163 600 12 0085 0 10 300 305397 2891 0 2891

3 85 335 1 01225 0 10 270 130506 2183 0 2183

Procedure






Results

Fault current =



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



88

Result



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



89

ExpNo 16 Date




Theory























90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



90

Circuit Diagram


91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



91

Procedure







92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



92


93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



93

Result



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



94

ExpNo17 Date




Theory























95

Circuit Diagram

Procedure







96


97

RESULT



95

Circuit Diagram

Procedure







96


97

RESULT



96


97

RESULT



97

RESULT


author griet-jsd created date 5/9/2017 1:29:36 pm

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