Jawaharlal Nehru Engineering College, Aurangabad Power ... · Aim: To determine direct axis...

Power System- II Manual 2019-20

Department of Electrical Engineering | 1

MGM’S

Jawaharlal Nehru Engineering College,

Aurangabad

Power System II Laboratory Manual

For

Third Year Students

Manual made by

Miss.Kanika Chitnavis



LABORATORY MANNUAL CONTENTS

This manual is intended for the third year students of Electrical Engineering in the subject of

Power System - II.

Students are advised to thoroughly go through this manual rather than only topics mentioned in

the syllabus as practical aspects are the key to understanding and conceptual visualization of

theoretical aspects covered in the books.

Good Luck for your Enjoyable Laboratory Sessions

Miss. Kanika Chitnavis



FOREWORD

It is my great pleasure to present this laboratory manual for third year engineering students

for the subject of Power System-II. Keeping in view the vast coverage required for visualization of

concepts of Power System-II components with simple language.

As a student, many of you may be wondering with some of the questions in your mind

regarding the subject and exactly what has been tried is to answer through this manual.

Faculty members are also advised that covering these aspects in initial stage itself, will greatly

relived them in future as much of the load will be taken care by the enthusiasm energies of the

students once they are conceptually clear

Prof. Dr.H.H.Shinde

Principal


Department of Electrical Engineering | JAWAHARLALNEHRUENGINEERINGCOLLEGE 4

JAWAHARLALNEHRUENGINEERINGCOLLEGE AURANGABAD

Department of Electrical Engineering

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Electrical Engineering Department

VISION:

To develop competent Electrical Engineers with human values.

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• To Provide Quality Technical Education To The Students Through Effective Teaching-

learning Process.

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With Ethical Values.



Ex.no Experiment name Page no.

1 Program for Ferranti Effect 8

2 To determine negative & zero sequence reactance of an alternator

10

3 To determine direct axis reactance (xd) & qaudrature axis reactance (xq) of a sailent pole alternator

15

4 To study singular transformation using MATLAB

19

5 Steady state stability of synchronous motor 21

23

6 Study of unsymmetrical fault

7 Use of computers for load flow study

a) Newton-Raphson method

b) Gauss seidal method

24

8 Use of computers for stability study

a)Steady state stability b)Transient stability

34

9 Program for optimum loading of generator 38

10 Program for optimum loading of generator with penalty

factor 40



Experiment no 01 :-

Aim : Program me for Ferranti Effect :-

Theory :

MATLAB PROGRAM :-

clear ;

zprimary = [1100.25

2210.1

3310.1

4200.25

5230.1 ]

[elements columes] = size(zprimary)

zbus = []

currentbus no = 0

for count = 1:elements,

[row cols] = size(zbus)

from=zprimary(count,2)



to=zprimary(count,3)

valu=zprimary(count,4)

newbus=max(from,to)

if newbus>currentbusno & ref = = 0

zbus = [zbus zeros(rows,1)

zeros(1,cols) value]

currentbus no =newbus

continue

end

if newbus>currentbusno & ref ~= 0

zbus=[zbus zbus(:,ref)

zbus(ref,:) value+zbus(ref,ref)]

currentbusno = newbus

continue

end

if newbus<=currentbusno & ref ~= 0

zbus=zbus-1/(value+zbus(from,from)+zbus(to,to)-2*zbus(from,to))*(zbus(:,from)-

zbus(:,to))*((zbus(from,:)-zbus(to,:)))

continue

end

end

Conclusion :



Experiment no. 02:-

Aim: To determine negative and zero sequence reactance of an alternator.

Apparatus Used: Sr.No Items Qty. 1. M.I. Ammeter Portable 0-2.5/5 A 1 2. M.I. Voltmeter Port 300/600v 1 3. M.I. Voltmeter Port 75/1 5 0/3 Oov 1 4. Rheostat 1.4 Amp 230 Ohms 1 5. Rheostat 1.1 A 1800 Ohms 1 6. M.C. Voltmeter Port 150/300 V 1 7. M.C. Ammeter Port 1/2 Amp 1 8. Upf Wattmeter 2.5/5 Amp, 125/250/500 V 1 9. Single Phase Variac 4 A 1 10. M G Set: D C Shunt Motor/3 Phase Alternator 1

THEORY: Direct-axis synchronous reactance and Quadrature axis synchronous reactance are

the steady state reactance of the synchronous machine. These reactance can be measured by

performing, open circuit, short circuit test and the slip test on a synchronous machine.

Direct-axis synchronous reactance, Xd: The Direct-axis synchronous reactance of synchronous

machine in per unit is equal to the ratio of field current, Ifsc at rated armature current from the

short circuit test, to the field current, If at rated voltage on the air gap line. Synchronous

reactance,

Thus Direct-axis synchronous reactance can be found out by performing open circuit and short circuit test on an alternator. Quadrature axis synchronous reactance, Xq by slip test For the slip test the alternator should be driven at a speed, slightly less than the synchronous speed with its field circuit open. 3 phase balanced reduced voltage of same frequency is applied to armature (stator) terminals of the synchronous machine. Applied voltage is to be adjusted, so that the current drawn by the stator winding is full load rated current. Under these conditions of operation, the variation of the current drawn by the stator winding, voltage across the stator winding and the voltage across the field winding. The wave shapes of stator current and stator voltage clearly indicated that these are changing between minimum and maximum value. When the crest of the stator mmf wave coincides with the direct axis of the rotating field the inducted emf in the open field is zero, the voltage across the stator terminals will be maximum and the current drawn by the stator winding is minimum. Thus approximate value of Direct-axis synchronous reactance, Xds is given by,



Circuit Diagram:



For Slip Test:

PROCEDURE :

(a) Open Circuit Test

1. Connect the circuit as per circuit Diagram.

2. Ensure that the external resistance in the field circuit of DC motor acting as a prime mover for

alternator is minimum and the external resistance in the field circuit of alternator is maximum.

3. Switch on DC supply to DC motor and the field of alternator.

4. Start the DC motor with the help of stator. The starter arm should be moved slowly, till the

speed of the motor builds up and finally all the resistance steps are cut out and the starter arm is

held in on position by the magnet of no volt release.

5. Adjust the speed of the DC motor to rated speed of the alternator by varying the external

resistance in the field circuit of the motor.

6. Record the field current of the alternator and its open circuit voltage per phase. 7. Increase the field current of alternator in steps by decreasing the resistance and record the field

current and open circuit voltage of alternator for various values of field current.

8. Field current of alternator is increase till the open circuit voltage of the alternator is 25 to 30

percent higher than the rated voltage of the alternator.



9. Decrease the field current of alternator to minimum by inserting the rheostat fully in the field

circuit.

(b) Short Circuit Test

10. With the DC motor running at rated speed and with minimum field current of alternator close

the switch, thus short-circuiting the stator winding of alternator.

11. Record the field current of alternator and the short circuit current. 12. Increases the field current of alternator in steps till the rated full load short circuit current.

Record the reading of armature in both the circuit at every step. 4 to 5 observations are sufficient

as short circuit characteristics is a straight line.

13. Decrease the field current of alternator to minimum and also decrease the speed of DC motor

by field rheostat of the motor.

14. Switch off the DC supply motor as well as to alternator field.

(c) Slip Test

1. Connect the circuit of alternator as shown in Fig ‘D’ keeping the connections of the DC motor

same.

2. Ensure that the resistance in the field circuit of DC motor is maximum.

3. Switch on the DC supply to the motor.

4. Repeat steps 4 described is (a). 5. Adjust the speed of the DC motor slightly less than the synchronous speed of the alternator by

varying the resistance in the field circuit of the motor. Slip should be extremely low, preferably

less that 4 percent.

6. Ensure that the setting of 3 phase Variac is at zero position.

7. Switch on 3 phase AC supply to the stator winding of alternator.

8. Ensure that the direction of rotation of alternator, when run by the DC motor and when run as

a 3 phase induction motor at reduced voltage (alternator provided with damper winding can be

run as 3 phase induction motor) is the same.

9. Adjust the voltage applied to the stator winding till the current in the stator winding is

approximately full load rated value.

10. Under these conditions the current in the stator winding the applied voltage to the stator

winding and the induced voltage in the open field circuit will fluctuate from minimum values to

maximum values which may be recorded by the meters included in the circuit. For better results,

oscillogram may be take of stator current applied voltage and induced voltage in the field circuit.

11. Reduce the applied voltage to the stator winding of alternator and switch off 3 phase AC

supply.

12. Decrease the speed of DC motor and switch off DC supply.



Observation Table:

Result: We have performed the experiment and determine negative and zero sequence reactance

of an alternator



Experiment No. 03

Aim: To determine direct axis reactance (xd) and quadrature axis reactance (xq) of a salient pole

alternator

.

Apparatus Used:

Theory: The direct-axis subtransient reactance and quadrature-axis subtransient reactance of 3

phase synchronous machine can be measured by applying a reduced single phase voltage to the

two stator phase connected in series, with the field winding short circuited and the machine being

stationary. The rotor is moved by hand, so that the current in the short circuited field winding is

maximum. Under this condition. The reactance offered by the armature is direct-axis subtransient

reactance i.e.

Next the rotor is turned through half a pole pitch, so that q axis coincides with the crest of the

armature mmf and the current in the field winding is minimum. The reactance offered by the

armature under this condition will be quadrature-axis subtransient reactance. This method

necessitates an exact alignment of the rotor with the armature mmf wave, which is not possible.

As such a more convenient method discussed below can be adopted for the measurement of

subtrasient reactances.

Direct-axis subtransient reactance, Xd”

Direct-axis subtransient reactance can be determined by applied voltage method (most

convenient method) in which single phase voltage of reduced magnitude and of rated frequency

is applied across the two terminals of the stator winding the third being left isolated as shown in

Fig ‘A’. The test is repeated for another two combinations of connections of stator terminals i.e.

first voltage applied between terminals A,B, second between B,C and third between terminals

C,A. During this test rotor is stationary and the field winding on the rotor is short circuited



through an armature. The test should be conducted at full load current flowing in the stator

winding as such applied voltage should be adjusted accordingly. Direct-axis subtransient

reactance can now be found out as discussed below.

Circuit diagram:



1. Let the applied voltage across the terminals A,B of the stator winding with terminal C

kept isolated be E volts and the current flowing through the winding in currentbe I

amperes. The ration of voltage across each phase to current is a reactance which can be

represented by a quantity A’ i.e.

2. Similarly the ratio of applied voltage E’/2 across each phase with voltage E’ across the

terminals B,C and the resultant current flowing, I’ can be represented by a quantity B’ i.e.

3. In a similar way the ratio of applied voltage, E”/2across each phase with voltage E”

across the terminals C.A and current flowing I” is represented by a quantity C’ i.e.

4. From the value od A’, B’, and C’ determined from the experimental data, calculate the

values of K and M from the equations given below.



5. Then direct-axis subtransient reactance Xd” = K – M (smaller possible stationary rotor

reactance).

Procedure:

1. Connect the circuit as per the circuit diagram.

2. Ensure that the moving knob of single phase variac is at zero position.

3. Switch on the AC supply.

4. Apply a reduced voltage to the circuit consisting of stator terminals A and B in series, so

that the current flowing in the stator winding is of full load value. Record the voltage applied

and the current flowing in the circuit.

5. Repeat step 4 with stator terminals B and C connected in series.

6. Repeat step 4 with stator terminals C and A connected in series.

7. Repeat step 4, 5 and 6 for a new position of the rotor to confirm that the value of K and M

are same for the both the position of rotor.

8. Switch off the supply.

Observation Table:

Result: We have performed the test and direct-axis sub transient reactance of synchronou



Experiment no.04

Aim: - To study singular transformation using MATLAB

Theory: - Program for Y-Bus matrix

clc

clear all

n=input(‘enter no. of buses:’); % no. of buses excluding reference

nl= input(‘enter no. of lines:’); % no. of transmission lines

sb= input(‘enter starting bus of each line:’); % starting bus of a line

eb= input(‘enter ending bus of each line’); % ending bus of a line

zser= input(‘enter resistance and reactance of each line:’); % line resistance and reactance (R, X)

yshty= input(‘enter shunt admittance of the bus:’); % shunt admittance

i=0;

k=1;

while i<nl

zser1(i+1)=zser(k)+j*zser(k+1); % impedance of a line (R+jX)

i=i+1;

k=k+2; end

zser2=reshape(zser1,nl,1);

yser=ones(nl,1)./zser2;

ypri=zeros(nl+n,nl+n);

ybus=zeros(n,n);

a=zeros(nl+n,n);

for i=1:n

a(i,i)=1;

end

for i=1:nl

a(n+i,sb(i))=1;

a(n+i,eb(i))=-1;

ypri(n+i,n+i)=yser(i);

ypri(sb(i),sb(i))=ypri(sb(i),sb(i))+yshty(i);

ypri(eb(i),eb(i))=ypri(sb(i),eb(i))+yshty(i);

end

at=transpose(a);

ybus=at*ypri*a;

zbus=inv(ybus); Input:

n= 3

nl=3

sb=[1 1 2 ]

eb=[2 3 3 ]

zser=[0.01 0.03 0.08 0.24 0.06 0.18 ]

yshty=[0.01 0.025 0.02]



Conclusion:-



Experiment no.05

Aim:- Steady state stability of synchronous motor.

Theory:-

Steady-State Analysis of a Linear Circuit

Open this ExampleOpen this ExampleThis example shows the use of the Powergui and

Impedance Measurement blocks to analyze the steady-state operation of a linear electrical circuit

G. Sybille (Hydro-Quebec)

Description

A 5th harmonic filter is connected at a bus bar fed by a 60 Hz, 100 V inductive source. A 5th

harmonic (300 Hz, 1 A) current is injected at the bus bar.



This linear system consists of 3 states (2 inductor currents and 1 capacitor voltage), 2 inputs (Vs,

Is) and 2 outputs (Current and Voltage Measurement).

An Impedance Measurement block is used to compute the impedance versus frequency of the

circuit.

Simulation

1. Use the Powergui block to find the steady-state 60Hz and 300 Hz components of voltage and

current phasors. The values of the 3 states (phasors and initial values) can be also obtained from

the powergui block.

2. Open the scope and start the simulation from the Simulation/Start menu. Notice that the

simulation starts in steady-state. Using the Powergui block, select Impedance vs Frequency

Measurement. A new window opens.

3. The measurement will be performed for the specified frequency range vector [0: 2:1000] (0 to

1000 Hz by steps of 2 Hz). Click on the Display button. The impedance is displayed in a graphic

window. Notice the series resonance at 300 Hz corresponding to the tuned frequency of the filter.

Conclusion:-



Experiment no.06:

Aim:- Study of unsymmetrical fault.

Theory:-

Q.1 What are unsymmetrical fault?

Q.2 How to represent unsymmetrical fault using symmetrical sequence components?

Q.3 Derive the expression for L-L fault current?



Experiment no 07:- Use of computers for load flow study

Aim : a) Program for gauss seidal method

Theory: Q. What is the procedure to obtain the solution of power flow problem using

Gauss Seidal method and its advantages and drawbacks ?

MATLAB PROGRAM :

%Program for load flow by Gauss-Seidal Method Clear;

N=4

V=[1.04 1.04 11]

Y= [3-j*9 - 2+j*6 -1+j*3 0

-2+j*6 3.66-j*11 -0.666+j*2 -1+j*3

-1+j*3 -0.666+j*2 3.66-j*11 -2+j*6

0 -1+j*3 -2+j*6 3-j*9 ] Type=ones(n,1) Typechanged=zeros(n,1) Qlimitmax=zeros(n,1) Vmagfixed= zeros(n,1) Type(2)=2 Qlimitmax(2)=1.0 Qlimitmax(2)=0.2 Vmagfixed(2)=1.04 Diff=10;noofiter=1



Vprev=V; While(diff>0.00001/ noofiter==1), Abs(V)

Abs(Vprev)

Vprev=V;

P = [inf 0.5 -1 0.3];

Q = [inf 0.0 0.5 -0.1]; S=[inf+j*inf (0.5-j*0.2) (-1.0+j*0.5) (0.3-j*0.1)]; For i=2:n, If type(i)==2 | typechenged(i)==1, If type(i)>Qlimitmax(i) | Q(i)<Qlimitmin(i), If (Q(i)<Qlimitmin(i),

Q(i)=Qlimitmin(i); Else Q(i)=Qlimitmax(i); End Type(i)=1; Typechanged(i)=1; Else Type(i)=2; Typechanged(i)=0; End

End

End Sumyu=0;



Fork=1:n, If(i~=k) Sumyv=sumyu=y(i,k)*V(k);

end

end V(i)=(1/y(i,i)*( (p(i)-j*Q(i))/conj(V(i))-sumyv); If type(i)==2 & typechanged(i)~=1, V(i)=PolarTorect(Vmagfixed(i),angle(V(i))*180/pi);

End End Diff=max(abs(abs(V(2:n))-abs(Vprev(2:n)))); Noofiter=noofiter=+1;

Output :

Conclusion :



Aim : b)solution of power flow problem using N-R method Theory :

Q. what is the procedure to obtain the solution of power flow problem using N - R

method and its advantages over Gauss Seidal?

Q. write down the difference in Gauss Seidal and N-R method ?

MATLAB PROGRAM :

%Program for load flow by Newton- Raphson Method Clear

n=3;

V=[1.04 1.0 1.04]

Y= [ 5.88228-j*23.50514 -2.9427+j*11.7676 -2.9427+j*11.7676

-2.9427+j*11.7676 5.88228-j*23.50514 -2.9427+j*11.7676

-2.8427+j*11.7676 -2.9427+j*117676 5.88228-j*23.50514 ]

type =ones(n,1);



Typechanged=zeros(n,1): Qlimitmax=zeros(n,1); Qlimitmin=zeros(n,1); Vmagfixed=zero(n,1); Type(3)=2; Qlimitmax(3)=1.5; Qlimitemin(2)=0; Vmagfixed(2)=1.04; Diff=10;noofiter=1.0 Pspec=[inf 0.5 -1.5] Qspec=[inf 1 0]; S=[inf+j*inf(0.5-j*0.2) (-1.0+j*0..5) (0.3-j*0.1)]; While (diff>0.00001 | noofiter==1), Eqcount=1; For i=2:n, Abs(Vprev) Pause Vprev=V; For ceq=1:eqcount,

Am=real(Y(assoeqbus(ceq),assocolbus(ccol))*V(assocolbus(ccol)));

bm =imag(Y(assoeqbus(ceq),assocolbus(ccol))*V(assocolbus(ccol)));

ei= real(V(assoeqbus(ceq))); fi= imag(V(assoeqbus(ceq)));

if assoeqvar(ceq)==’p’ & assocolbus(col)==’d’,

if assoeqbus(ceq)~=assocolbus(ccol),



H=am*fi-bm*ei;

Else

H=Q(assoeqbus(ceq))-imag(Y(assoeqbus(ceq)))^2);

End

Jacob(ceq,ccol)=H End

If assoeqvar(ceq)=’p’ & assocolver(ccol)==’v’, If assoeqbus(ceq)`= assocolver(ccol), N =am*ei+bm*fi Else

N=P(assoeqbus(ceq))+real(Y(assoeqbus(ceq),assocolbus(ceq))*abs(V(assocolbus(ceq)))^2)

; Scal(i)=0;

Sumyv=0; Fork=1:n, Sumyv=sumyv+Y(i,k)*V(k)

; End

Scal(i)=V(i)*conj(sumyv); P(i)=real(scale(i)); Q(i)=image(Scal(i)); If type(i)==2 | typechanged(i)==1, If(Q(i)<Qlimitmax(i) | Q(i)<Qlimitmax(i)),

If(Q(i)<Qlimitmax(i));

Q(i)<Qlimitmax(i); Else Q(i)=Qlimitmax(i); End Type(i)=1;



Typechanged(i)=0; End End If type (i)==1, Assoeqvar(eqcount)=’p’; Assoeqbus(eqcount)=i; Mismatch(eqcount)=Pspec(i)-P(i); Assoceqvar(eqcount+1)=’Q’; Assoeqbus(eqcount+1)=i; Mismatch(eqcount+1)=Qpec(i)-Q(i);

Assoeqvar(eqcount)=’d’; Assoeqbus(eqcount)=i; Assoceqvar(eqcount+1)=’V’; Assoeqbus(eqcount+1)=i; Mismatch=(eqcount+2); Else

Assoeqvar(eqcount)=’p’; Assoeqbus(eqcount)=i; Assoceqvar(eqcount+1)=’Q’; Assoeqbus(eqcount+1)=d; Mismatch(eqcount+1)=Ppec(i)-P(i); Eqcount=eqcount+1: End End Mismatch Eqcount=eqcount-1;



Noofeq=eqcount; Update=Zeros(eqcount,1) Vprwv=V Abs(V); End Jacob(ceq,ccol)=N End If assoeqvar(ceq)==’Q’ & assocolvar (ccol)==’d’, If assoeqbus(ceq)~=assocolbus(ccol), J=am*ei+bm*fi; Else J=p(assoeqbus(ceq))+real(Y(ass0eqbus(ceq),assocolbus(ceq)*abs(V(assoeqbus(ceq)))^2);

End

Jacob(ceq,ccol)=J

End

If assoeqvar(ceq)==’Q’ & ass0colvar (ccol)==’V’, If assoeqbus(ceq)~=assocolbus(ccol),

L=am*fi-bm*ei;

else

L=Q(assoeqbus(ceq))- Imag(Y(assoeqbus(ceq),assocolbus(ceq))*abs(V(assoeqbus(ceq)))^2);

End

Jacob(ceq,ccol)=L;

End



End

End

Jacob;

Pause

Update=inv(Jacob)*mismatch;

Noofeq=1;

For i=2:n,

If i=2:n,

newchinangV=update(noofeq);

newangV=angle(V(i))+newchinangV;

newchinmagV=update(noofeq+1)*abs(V(i));

newangV=abs(V(i))+newchinangV;

V(i)=polartorect(newmagV*180/pi);

Noofeq=noofeq+2;

Else

newchinmagV=update(noofeq);

newangV=angle(V(i))+newchinangV;

V(i)=polartorect(newmagV*180/pi);

Noofeq=noofeq+2;

end

end

clear mismatchJacob update assoeqvar assoeqbus Assoeqlvar assocllous



diff=min(abs(abs(V(2:n))-abs(Vprwv(2:n))));

noofiter=noofiter+1;

Output : Conclusion :



Experiment no.08:- Use of computers for stability study Aim: - A) steady state analysis Theory:- Steady-State Analysis of a Linear Circuit

Open this ExampleOpen this ExampleThis example shows the use of the Powergui and

Impedance Measurement blocks to analyze the steady-state operation of a linear electrical circuit

G. Sybille (Hydro-Quebec)

Description

A 5th harmonic filter is connected at a bus bar fed by a 60 Hz, 100 V inductive source. A 5th

harmonic (300 Hz, 1 A) current is injected at the bus bar.



This linear system consists of 3 states (2 inductor currents and 1 capacitor voltage), 2 inputs (Vs,

Is) and 2 outputs (Current and Voltage Measurement).

An Impedance Measurement block is used to compute the impedance versus frequency of the

circuit.

Simulation

1. Use the Powergui block to find the steady-state 60Hz and 300 Hz components of voltage and

current phasors. The values of the 3 states (phasors and initial values) can be also obtained from

the powergui block.

2. Open the scope and start the simulation from the Simulation/Start menu. Notice that the

simulation starts in steady-state. Using the Powergui block, select Impedance vs Frequency

Measurement. A new window opens.

3. The measurement will be performed for the specified frequency range vector [0: 2:1000] (0 to

1000 Hz by steps of 2 Hz). Click on the Display button. The impedance is displayed in a graphic

window. Notice the series resonance at 300 Hz corresponding to the tuned frequency of the filter.



B) Transient stability Theory:- Transient Analysis of a Linear Circuit

This example shows steady-state and transient simulation of a linear circuit.

H. Le-Huy (Universite Laval, Quebec) and G. Sybille (Hydro-Quebec)

Description

This circuit is a simplified model of a 230 kV three-phase power system. Only one phase of the

transmission system is represented. The equivalent source is modeled by a voltage source (230

kV rms/sqrt(3) or 187.8 kV peak, 60 Hz) in series with its internal impedance (Rs Ls)

corresponding to a 3-phase 2000 MVA short circuit level and X/R = 10. (X = 230e3^2/2000e6 =

26.45 ohms or L = 0.0702 H, R = X/10 = 2.645 ohms). The source feeds a RL load through a 150

km transmission line. The line distributed parameters (R = 0.035ohm/km, L = 0.92 mH/km, C =

12.9 nF/km) are modeled by a single pi section (RL1 branch 5.2 ohm; 138 mH and two shunt



capacitances C1 and C2 of 0.967 uF). The load (75 MW - 20 Mvar per phase) is modeled by a

parallel RLC load block.

A circuit breaker is used to switch the load at the receiving end of the transmission line. The

breaker which is initially closed is opened at t = 2 cycles, then it is reclosed at t = 7 cycles.

Current and Voltage Measurement blocks provide signals for visualization purpose.

Simulation

1. Simulation using a continuous solver (ode23tb)

Start the simulation and observe line voltage and load current transients during load switching

and note that the simulation starts in steady-state. Use the zoom buttons of the oscilloscope to

observe the transient voltage at breaker reclosing.

2. Using the Powergui to obtain steady-state phasors and set initial states

Open the Powergui block and select "Steady State Voltage and Currents" to measure the steady-

state voltage and current phasors. Using the Powergui select now "Initial States Setting" to

obtain the initial state values (voltage across capacitors and current in inductances). Now, reset

all the initial states to zero by clicking the "to zero" button and then "Apply" to confirm changes.

Restart the simulation and observe transients at simulation starting. Using the same Powergui

window, you can also set selected states to specific values.

3. Discretizing your circuit and simulating at fixed steps

The Powergui block can also be used to discretize your circuit and simulate it at fixed steps.

Open the Powergui. Select "Discretize electrical model" and specify a sample time of 50e-6 s.

The state-space model will now be discretized using trapezoidal fixed step integration. The

precision of results is now imposed by the sample time. Restart the simulation and compare

simultion results with the continuous integration method. Vary the sample time of the discrete

system and note the impact on precision of fast transients.

4. Using the phasor simulation method

You will now use a third simulation technique. The "phasor simulation" method consists to

replace the circuit state-space model by a set of algebraic equations evaluated at a fixed

frequency and to replace sinusoidal voltage and current sources by phasors (complex numbers).

This method allows a fast computation of voltage and current phasors at a selected frequency,

disregarding fast transients. It is particularly efficient to study electromechanical transients of

generators and motors involving low frequency oscillation modes. Open the Powergui block and

select "Phasor simulation". Restart the simulation. Observe that the magnitude of 60 Hz voltage

and current is now displayed on the scope. If you double click on the voltage or current

measurement block you can choose to output phasor signals in four different formats: Complex,

Real/Imag, Magnitude/Angle (in degres), or just Magnitude (default value). Notice that you

cannot send a complex signal to an oscilloscope.



Experiment no 09 :-

Aim : Programme for optimum loading of generator :-

MATLAB PROGRAM:-

clear ; n=2 Pd = 231.25 alpha = [0.20 0.25] beta = [40 30] lamda = 20 lamdaprev = lambda esp = 1 deltalambda = 0.25 Pgmin = [20 20] pg = 1008ones(n,1)

while abs (sum (Pg)-Pd)>eps

for i= 1:n,

Pg(i) = (lamda-beta(i)) / alpha(i) ;



if Pg(i)>Pmax(i)

Pg(i) = Pmax (i);

end

if Pg(i)>Pmin(i)

Pg(i) = Pmin (i);

end

end

if(sum (Pg-Pd)) < 0

lamdaprev = lambda

lamda= lamda + deltalamda ; else

lamdaprev = lambda

lamda= lamda - deltalamda ; end

end dis('the final value of lamda is') lamdaprev dis('the distribution of load shared by two units is') Pg Output :

Conclusion :



Experiment no.10

Aim : Programme for optimum loading of generator with penalty factor:-

MATLAB PROGRAM:-

clear

n=2;Pd=237.04;

alpha=[0.020 0.04];

beta=[16 20];

lamda = 20; lamdaprev = lamda ; eps = 1; deltalamda = 0.25;

Pgmax

Pgmax=[200 200];Pgmin=[0 0];

B = [0.0010 0

0 0];

noofiter=0;PL=0;Pg = zeros(n,1);

while abs(sum(Pg)-Pd-PL)>eps

for i=1:n,

sigma=B(i,:)*Pg-B(i,i)*Pg(i);

Pg(i)=(1-beta(i)/(lamda-(2*sigma)))/(alpha(i)/lamda+2*B(i,i));

%PL=Pg'*B*Pg;

if Pg(i)>Pgmax (i)

Pg(i)=Pgmax (i);

end

if Pg(i)<Pgmin(i)

Pg(i)=Pgmin(i);

end

end

PL = Pg'*B*Pg;

if (sum(Pg)-Pd-PL)<0

lamdaprev=lamda;

lamda=lamda+deltalamda;

else



lamdaprev=lamda;

lamda=lamda-deltalamda;

end

noofiter=noofiter + 1;

Pg;

end

Output :

Conclusion :

Jawaharlal Nehru Engineering College, Aurangabad Power ... · Aim: To determine direct axis...

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