Crct Pss Recrd

1

PARK COLLEGE OF ENGINEERING AND TECHNOLOGY

KANIYUR, COIMBATORE – 641 659.

DEPARTMENT OF ELECTRICAL AND ELECTRONICS

ENGINEERING

RECORD NOTE BOOK

REGISTER NO : _________________________________

SUBJECT : _________________________________

SUBJECT CODE: _________________________________

Certified that this is the bonafide record of work done by Mr/Miss

_____________________________ of VII semester of ELECTRICAL AND

ELECTRONICS ENGINEERING during the year 2014.

Date: …………… Staff In-charge

Submitted for the University Practical Examination held on

………………… at Park College of Engineering and Technology.

H.O.D. Internal Examiner External Examiner

2

CONTENTS

Sl. No. Date Name Of The Experiments Page

No. Marks Signature

1a)

COMPUTATION OF LINE

PARAMETERS 4

1b)

MODELING AND

PERFORMANCE OF

TRANSMISSION LINES

10

2a).

FORMATION OF BUS

IMPEDANCE MATRICES 17

2b)

FORMATION OF BUS

ADMITTANCE

MATRICES

25

3)

SOLUTION OF LOAD FLOW

ANALYSIS USING GAUSS

SEIDAL METHOD

32

4a)


ANALYSIS USING NEWTON

RAPHSON METHOD

43

4b)


ANALYSIS USING FAST

DECOUPLED METHOD

54

5a)

FAULT ANALYSIS-

SYMMETRICAL FAULT

ANALYSIS

65

5b)

UNSYMMETRICAL FAULT

ANALYSIS 70

6)

TRANSIENT STABILITY

ANALYSIS OF SINGLE

MACHINE CONNECTED TO

INFINITE BUS

76

7)

TRANSIENT STABILITY

ANALYSIS OF POWER SYSTEM

(STEADY STATE STABILITY)

88

8)

ELECTRO MAGNETIC

TRANSIENTS 93

3

Sl. No. Date Name Of The Experiments Page

No. Marks Signature

9a)

LOAD-FREQUENCY

DYNAMICS OF SINGLE AREA

POWER SYSTEM

96

9b)

LOAD-FREQUENCY

DYNAMICS OF STWO AREA

POWER SYSTEM

102

10)

ECONOMIC DISPATCH IN

POWER SYSTEM 107

4

Ex. No: 1 a) COMPUTATION OF LINE PARAMETERS AND MODELLING OF

TRANSMISSSION LINES

Date :

AIM:

To determine the positive sequence line parameters L & C per phase/KM of a 3phase and

single circuit transmission line for different conductor arrangement.

ALGORITHM:

Step1: Select the type of circuit Single phase two wire system

Step2: For all the cases, get the required input.

Step3: Perform necessary calculations.

Step4: Print the output.

THEORY:

(i)Single Phase Two Wire system

L=0.5×10-7(1+4ln(D/R’)) H/m

Can=2 𝜋휀o/ln(D/R) F/m

Where,

L= Inductance of conductor

Can=Capacitance of conductor ‘a’ w.r.t neutral

D= Distance between the conductors (m)

R’=0.7788R

R=Radius of the conductors (m),

𝜖o= Absolute permittivity=8.854×10-12

5

EXCERSISE:

A 3 phase transposed line composed of one ACSR, 636000cmill, 54/7 rook conductor

per phase with horizontal spacing of 11m between phases A and B and between phase B and C

Diameter=3.625,determine the inductance and capacitance?

D12 11cm D23 11cm

22 cm

D33

D1 D2 D3

6

PROGRAM:

%3 phase single circuit

D12=input ('enter the distance b/w D12 in cm:');

D23=input ('enter the distance b/w D23 in cm:');

D31=input('enter the distance b/w D31 in cm:');

d=input ('enter the value of d:');

r=d/2;

Ds=0.7788*r;

X=D12*D23*D31;

Deq=nthroot(X,3);

y=log(Deq/Ds);

Inductance=0.2*y

Capacitance=0.0556/(log(Deq/r))

fprintf ('\n the inductance per phase per km is %f mh/ph/km \n',inductance);

fprintf ('\n the capacitance per phase per km is %f mh/ph/km \n',capacitance);

7

OUTPUT:

Enter the distance b/w D12 in cm:11



Enter the value of d:3.625

Inductance =

0.4568

Capacitance =

0.0273

the inductance per phase per km is 0.456848 mh/ph/km

the capaitance per phase per km is 0.027332 mh/ph/km

8

MANUAL CALCULATION:

9

RESULT:

Thus the line parameters of single phase and transmission lines were determined.

10

Ex. No: 1 b) MODELLING OF TRANSMISSIONLINES PARAMETERS

Date :

AIM:

To determine the positive sequence line parameters L & C per phase/KM of a 3phase and

double circuit transmission line for different conductor arrangement.

ALGORITHM:

Step1: Select the type of circuit Single phase two wire system

Step2: For all the cases, get the required input.

Step3: Perform necessary calculations.

Step4: Print the output.

THEORY:

(I)Three Phase Two Wire system

Inductance:

General formula:

L=0.2*Ln[Dm/Ds] mH/Km

Dm=geometric mean distance[GMD]

Ds=geometric mean radius[GMR]

3phase-symmertical sapcing

GMD=D

GMR=re-1/4=r’

r’=0.7788r,

CAPACITANCE:

C=0.0556/Ln(Deq/r)uF/Km

11

EXCERSISE:

A 345KV double circuit 3phase transposed line is composed of two ACSR 1,431,000

45/7 bobolink conductors per phase with vertical conductor configuration as shown in the

figure, the conductors have a diameter of 1.427inch and a GMR of a 0.456 inch.the bundle

spacing in 18 inch.find the inductance and capacitance per phase kilometer of the line?

a a’

S11=11m

b h13=11m b’

s22=16.5m

h23=6.5m

c c’

S33=12m

12

PROGRAM:

%3 phase double circuit

s=input('enter row vector[s11,s22,s33]=');

H=input('enter row vector[h12,h23]=');

d=input('bundle spacing in inch=');

dia=input('conductor diameter in inch='); r=dia/2;

Ds=input('geometric mean radius in inch=');

s11=s(1);s22=s(2);s33=s(3);H12=H(1);H23=H(2);

a1=-s11/2+1i*H12;

b1=-s22/2+1i*0;

c1=-s33/2-1i*H23;

a2=s11/2+1i*H12;

b2=s33/2+1i*0;

c2=s33/2-1i*H23;

Da1b1=abs(a1-b1); Da1b2=abs(a1-b2);

Da1c1=abs(a1-c1); Da1c2=abs(a1-c2);

Db1c1=abs(b1-c1); Db1c2=abs(b1-c2);

Da2b1=abs(a2-b1); Da2b2=abs(a2-b2);

Da2c1=abs(a2-c1); Da2c2=abs(a2-c2);

Db2c1=abs(b2-c1); Db2c2=abs(b2-c2);

Da1a2=abs(a1-a2);

Db1b2=abs(b1-b2);

Dc1c2=abs(c1-c2);

DAB=(Da1b1*Da1b2*Da2b1*Da2b2)^0.25;

DBC=(Da1c1*Db1c2*Db2c1*Db2c2)^025;

DCA=(Da1c1*Da1c2*Da2c1*Da2c2)^0.25;

GMD=(DAB*DBC*DCA)^(1/3);

Ds=2.54*Ds/100;r=2.54*r/100;d=2.54*d/100;

Dsb=(d*Ds)^(1/2);rb=(d*r)^(1/2);

DSA=sqrt(Dsb*Da1a2); rA=sqrt(rb*Da1a2);

DSB=sqrt(Dsb*Db1b2); rB=sqrt(rb*Db1b2);

DSC=sqrt(Dsb*Dc1c2); rC=sqrt(rb*Dc1c2);

GMRL=(DSA*DSB*DSC)^(1/3);

GMRC=(rA*rB*rC)^(1/3);

L=0.2*log(GMD/GMRL) %mH/km

C=0.0556/log(GMD/GMRC) %microF/km

13

OUTPUT:

enter row vector[s11,s22,s33]=[11 16.5 12]

enter row vector[h12,h23]=[11 6.5]

bundle spacing in inch=18

conductor diameter in inch=1.427

geometric mean radius in inch=0.456

L =

17.2140

C =

6.4683e-004

14

MANUAL CALCULATION:

15

16

RESULT:

Thus the modeling of transmission line parameters found successfully using mat lab.

17

Ex. No: 2a) FORMATION OF BUS IMPEDANCE MATRIX USING MATLA

Date :

AIM:

To develop a MATLAB program to obtain the bus impedance matrix for the given power system

network

ALGORITHM:

Step 1: Obtain the system data.

Step 2: Obtain the impedance of the lines.

Step 3: Check for the following four cases and obtain the Z bus matrix using the calculations

specified for each case.

Step 3: Adding branch impedance Zb, from a new bus-p to reference bus,

Zbus,new= b

orig

Z

Z

0

0

Step 4: Adding Zb from a new bus-p to existing bus-q,

Zbus,new = bqqq ZZZ 1

1qorig ZZ

Step 5: Adding Zb from an existing bus-q to the reference bus,

i) Form the matrix Zbus,new as in step 4.

ii) Eliminate the last row and last column by node elimination method.

Step 6: Adding Zb between two existing buses h and q,

i) Zbus,new= hqqqhhbqh

qhorig

ZZZZZZ

ZZZ

211

11

ii) Eliminate the last row and last column by node elimination method.

Step 7: Z bus matrix are formed.

18

SINGLE LINE DIAGRAM:

0.12+j0.35

DATA FOR THE SYSTEM:

NO. OF BUSES: 4

NO. OF LINES: 5

BASE MVA : 100

TRANSMISSION LINE DATA:

LINE

NO.

START

BUS

END

BUS

RESISTANCE

(P.U)

REACTANCE

(P.U)

SUSCEPTANCE

(P.U)

RATED

MVA 1

2

3

4

5

1

2

3

4

2

2

3

4

1

4

0.1

0.15

0.12

0.1

0.25

0.4

0.6

0.35

0.35

0.7

0.015

0.082

0.018

0.012

0.030

100

100

100

100

100

0.1+j0.4

2 1

3 4

0.1+j0.3

5 0.25+j0.7

0.15+j0.6

19

FLOWCHART:

START

OBTAIN THE SYSTEM DATA

OBTAIN THE LINE IMPEDANCE

ADDING BRANCH IMPEDANCE Zb, FROM A NEW BUS-p TO

REFERENCE BUS,

Zbus,new=

CHECK FOR THE FOLLOWING CASES AND OBTAIN THE Z

BUS MATRIX USING THE CALCULATIONS SPECIFIED FOR

EACH CASE

ADDING Zb FROM NEW BUS-p TO EXISTING BUS-q,

Zbus,new =

ADDING Zb FROM AN EXISTING BUS-q TO THE REFERENCE

BUS,

i) FORM THE MATRIX Zbus,new AS ABOVE.

ii) ELIMINATE THE LAST ROW AND LAST COLUMN BY NODE

ELIMINATION METHOD

ADDING Zb BETWEEN TWO EXISTING BUSES h and q,

i) Zbus,new=

ii) ELIMINATE THE LAST ROW AND LAST COLUMN BY NODE

ELIMINATION METHOD

A

20

A

Z BUS MATRIX IS FORMED

STOP

21

PROGRAM:

clear all;

clc;

n=input('ENTER THE NUMBER OF BUSES:');

z=zeros(n);

for i=1:n

for j=i:n

if i~=j

disp(i);

disp(j);

z(i,j)=input('ENTER THE IMPEDANCE VALUE:');

z(j,i)=-z(i,j);

z(i,j)=z(j,i);

end

end

end

for i=1:n

for j=1:n

if i~=j

z(i,i)=z(i,i)-z(i,j);

end

end

end

'output'

disp(z);

22

OUTPUT:

ENTER THE NUMBER OF BUSES:4

1

2

ENTER THE IMPEDANCE VALUE:0.1+0.4i

1

3

ENTER THE IMPEDANCE VALUE:0

1

4


2

3


2

4


3

4


ans =

OUTPUT

Columns 1 through 3

0.2000 + 0.7500i -0.1000 - 0.4000i 0

-0.1000 - 0.4000i 0.5000 + 1.7000i -0.1500 - 0.6000i

0 -0.1500 - 0.6000i 0.2700 + 0.9500i

-0.1000 - 0.3500i -0.2500 - 0.7000i -0.1200 - 0.3500i

Column 4

-0.1000 - 0.3500i

-0.2500 - 0.7000i

-0.1200 - 0.3500i

0.4700 + 1.4000i

23

MANUAL CALCULATION:

24

RESULT:

Thus the MATLAB program to obtain the bus impedance matrix for the given power

system network was executed and the output was obtained

25

Ex. No :2b) FORMATION OF BUS ADMITTANCE MATRIX USING MATLAB

Date :

AIM:

To develop a MATLAB program to obtain the bus admittance matrix for the given power system

network

ALGORITHM:

STEP1: Start the Program.

STEP2: Initialize the variable and data type.

STEP 3: Initialize the bus admittance matrix.

STEP 4: Read the total number of buses and total no of transmission line of a given power

System.

STEP 5: From Ybus Matrix using the formula

Ybus[i][j] = Ybus[i][j] + 1 + shunt Y[i] Series Z[i]

STEP 6: If the admittance matrix is mutual the inverse of that line impedance is tabulated.

STEP 7: print the Ybus matrix.

STEP 8: Invert the Ybus matrix from Z bus

STEP 9: Print the Z bus matrix.

STEP 10: Stop the Program

26



NO. OF BUSES: 3

NO. OF LINES: 3

BASE MVA : 100


LINE

NO.

START

BUS

END

BUS

SERIES

IMPEDENCE

LINE

CHARGING

ADMITTANCE

1

2

3

1

2

3

2

3

1

0.1+0.3j

0.15+0.5j

0.2+0.8j

0.0+0.2j

0.0+0.01j

0.0+0.28j

2

G2

1

G

1

3

27

FLOW CHART:

START

DIAGNAL ELEMENT OF Y MATRIX IS

Y[i][j]=Y[i][j]+[1/Z[i][j]

READ NUMBER OF LINES GENERATORTRANSFORMERS IF

ANY

OTHER ELEMENT OF Y MATRIX IS Y[i][j]= -[i][j]

STOP

IS THERE

ANY

GENERATOR

OR

TRANSFORME

R CONNECT

TO THEM

Y[i][j]=Y[i][j]

Y BUS MATRIX IS FORMED

28

PROGRAM:

clear all;

clc;

n=input('no of buses:');

y=zeros(n);

for i=1:n

for j=1:n

if i~=j

disp(i);

disp(j);

z(i,j)=input('enter the impedence value:');

y(i,j)=input('enter the line charging admittance:');

z(j,i)=z(i,j);

end

end

end

'input';

disp(z);

for i=1:n

for j=1:n

if z(i,j)~=0

y(i,j)=-1/z(i,j);

end

y(i,j)=y(i,j);

if i~=j

y(i,i)=y(i,i)-y(i,j);

end

end

end

'output';

disp(y);

29

OUTPUT

no of buses:3

1

2

entner the impedance value:0.1+0.3i

enter the line charging admittance:0.2i

1

3

enter the impedence value:0.2+0.8i


2

1



2

3



3

1



3

2



0 0.1000 + 0.3000i 0.2000 + 0.8000i

0.1000 + 0.3000i 0 0.1500 + 0.5000i

0.2000 + 0.8000i 0.1500 + 0.5000i 0

1.2941 - 4.1765i -1.0000 + 3.0000i -0.2941 + 1.1765i

-1.0000 + 3.0000i 1.5505 - 4.8349i -0.5505 + 1.8349i

-0.2941 + 1.1765i -0.5505 + 1.8349i 0.8446 - 3.0113i

30

MANUAL CALCULATION:

31

RESULT:

Thus the MATLAB program to obtain the bus admittance matrix for the given power

system network was executed and the output was obtained

32

Ex. No :03 LOAD FLOW ANALYSIS BY GAUSS SEIDAL METHOD USING

Date : MATLAB

AIM:

To develop a MATLAB program for performing the power flow analysis using Gauss-Seidal

method.

ALGORITHM:

Step1: Assume a flat voltage 1+j0 for all except the slack bus. The voltage of slack bus is

the specified voltage and it is not modified in any iteration.

Step2: Assume a suitable value of convergence factor. It is used to compare the actual

change in bus voltage between kth and (k+1) th iteration.

Step3: Set iteration constant k=0 and the voltage profile of the buses are denoted as V10,

V20, V3

0….Vn0 except slack bus. Set bus count p=1.

Step4: Check for slack bus, if it is slack bus then go to step 11 otherwise go to next step.

Step5: Check for generator bus, if it is generator bus then go to next step, otherwise go to

Step 8.

Step6: Calculate reactive power by using,

k+1 p-1 n

Qpcal = (-1)Im (Vpk)* [ Ypq Vq

k+1 + Ypq Vqk ]

q=1 q=p

k+1

If Qpcal < Qpmin then Qp=

Qpmin.

k+1

If Qpcal > Qpmax then Qp= Qpmax

Step7: The magnitude and phase of the bus voltage is calculated by,

k+1 p-1 n

Vptemp = 1/Ypp (Pp – jQp) / (Vpk)* - Ypq Vq

k+1 - Ypq Vqk

q=1 q=p+1

k+1 k+1

p(k+1) =tan-1

Imaginary part of Vptemp / Real part of Vptemp

Vp(k+1) =│Vp│spec p

(k+1).

After calculating Vp(k+1) for generator bus go to step10.

Step8: (k+1)th iteration of load bus P- voltage Vpk+1 is calculated by,

k+1 p-1 n

Vp = 1/Ypp (Pp – jQp) / (Vpk)* - Ypq Vq

k+1 - Ypq Vqk

q=1 q=p+

Step9: Calculate the voltage by acceleration factor,

33

k+1 k k+1 k

Vpacc = Vp + Vp - Vp

k+1 k+1 k

Step10: Calculate the change in the bus voltage i.e., Vp = Vp – Vp

Step11: Repeat the steps 4 to 10 until all the bus voltages are calculated.

Step12: Find the largest of the absolute value of the change in voltage.

Step13: Calculate the line flows and slack bus power using the bus voltages.



NO. OF BUSES : 4

NO. OF LINES : 5

BASE MVA : 100


LINE

NO.

START

BUS

END

BUS

RESISTANCE

(P.U)

REACTANCE

(P.U)

RATED

MVA

1

2

3

4

5

1

1

2

2

3

2

3

3

4

4

0.03

0.10

0.15

0.10

0.05

0.15

0.30

0.45

0.30

0.15

55

63

30

55

40

0.03+0.15j

0.1+0.3j 0.1+0.3j

0.05+0.15j

0.15+0.45j

1 2

3 4

34

BUS DATA:

P Q V TYPE

-

0.5

-1.0

0.3

-

-0.2

0.5

-0.1

1.04+0j

-

-

-

SLACK

PQ

PQ

PQ

35

FLOW CHART:

START

READ SYSTEM DATA AND FORM Y BUS

GET SLACK BUS NUMBER, Qmin,

Qmax, , & SET K=0

SET P=1, FLAG=0

CHECK

SLACK

BUS

CHECK PV

BUS

n

CALCULATE Qp = -(Im [Ep* Ypq Eq])

q=1

IF

QP<Qmin

A

C

B

B

A

IF

Qp>Qmax

Qp = Qmin

Qp = Qmax

YES

NO

YES

NO

NO

YES

NO

YES

x

SET FLAG = 1

36

CALCULATE p-1 n

Epk+1= 1/Ypp (Pp – jQp) / (Ep

k)* - Ypq Eqk+1 - Ypq Eq

k

q=1 q=p+1

Epk = Ep

k+1 - Epk

k+1 k

E pacc = E p + Epk

CHECK

FLAG = 1

NO

YES

x

Epk+1 = Ep Cos p

+ Ep Sin p

P=P+1

IF

P < = n

EVALUATE E max

IF

E max<

EVALUATE LINE FLOW AND

SLACK BUS POWER

STOP

K = K+1 C

B

A

S NO

YES

YES

NO

37

PROGRAM:

clear all;

clc;

n=input('ENTER THE NUMBER OF BUSES:');

y=zeros(n);

for i=1:n

for j=i:n

if i~=j

disp(i);

disp(j);

z(i,j)=input('ENTER THE IMPEDANCE VALUE:');

z(j,i)=z(i,j);

end

end

end

'input'

disp(z);

for i=1:n

for j=1:n

if z(i,j)~=0

y(i,j)=-1/(z(i,j));

end

y(j,i)=y(i,j);

if i~=j

y(i,i)=y(i,i)-y(i,j);

end

end

end

disp(y);

m=input('ENTER THE NUMBER OF PV BUSES:');

for i=1:m

v(i+1)=input('ENTER THE PV BUS VOLTAGE:');

qmax(i)=input('ENTER THE MAXIMUM LIMIT OF Q FOR PV BUS:');

qmin(i)=input('ENTER THE MINIMUM LIMIT OF Q FOR PV BUS:');

end

p=input('ENTER THE REAL POWER:');

q=input('ENTER THE REACTIVE POWER:');

v(1)=input('ENTER THE SLACK BUS VOLTAGE:');

itrn=input('ENTER THE NUMBER OF ITERATIONS:');

for i=m+2:n

v(i)=1;

end

for i=2:m+1

yv=0;

for i=1:n

yv=yv+(y(i,j)*v(j));

end

yv=(conj(v(i)))*yv;

q(i)=-imag(yv);

if q(i)<qmin(i-1)

38

q(i)=qmin(i-1);

end

if q(i)>qmax(i-1)

q(i)=qmax(i-1);

end

yv=0;

for j=1:n

if i~=j


end

end

v(i)=(complex(p(i),-q(i))/(conj(v(i)-yv))/y(i,i));

del(i)=angle(v1);

[re,im]=pol2cart(del(i),v(i));

v(i)=complex(re,im);

end

for i=m+2:n

yv=0;

for j=1:n

if i~=j


end

end

v(i)=(complex(-p(i),q(i))/(conj(v(j)-yv))/y(i,i));

end

for k=2:itrn

for i=2:n

yv=0;

for j=1:n

if i~=j

yv=yv+y(i,j)*v(i);

end

end

v(i)=(complex(-p(i),q(i))/(conj(v(i)-yv))/y(i,i));

end

end

disp('THE FINAL VALUE OF VOLTAGE AFTER THE REQUIRED ITERATIONS IS:');

disp(v);

39

OUTPUT:

ENTER THE NUMBER OF BUSES:4

1

2


1

3


1

4

ENTER THE IMPEDANCE VALUE:0

2

3


2

4


3

4


ans =

input

0 0.0300 + 0.1500i 0.1000 + 0.3000i 0

0.0300 + 0.1500i 0 0.1500 + 0.4500i 0.1000 + 0.3000i

0.1000 + 0.3000i 0.1500 + 0.4500i 0 0.0500 + 0.1500i

0 0.1000 + 0.3000i 0.0500 + 0.1500i 0

2.2821 - 9.4103i -1.2821 + 6.4103i -1.0000 + 3.0000i 0

-1.2821 + 6.4103i 2.9487 -11.4103i -0.6667 + 2.0000i -1.0000 + 3.0000i

-1.0000 + 3.0000i -0.6667 + 2.0000i 3.6667 -11.0000i -2.0000 + 6.0000i

0 -1.0000 + 3.0000i -2.0000 + 6.0000i 3.0000 - 9.0000i

ENTER THE NUMBER OF PV BUSES:0

ENTER THE REAL POWER:[0 0.5 -1 0.3]

ENTER THE REACTIVE POWER:[0 -0.2 0.5 -0.1]

ENTER THE SLACK BUS VOLTAGE:1.04

ENTER THE NUMBER OF ITERATIONS:1

THE FINAL VALUE OF VOLTAGE AFTER THE REQUIRED ITERATIONS IS:

1.0400 -0.0033 - 0.0016i 0.0082 + 0.0051i -0.0009 - 0.0321i

40

MANUAL CALCULATION:

41

42

RESULT:

Thus the MATLAB program for performing the power flow analysis for the given power

system using Gauss-Seidal method was executed and the output was obtained.

43

Ex. No: 4a) LOAD FLOW ANALYSIS BY NEWTON RAPHSON METHOD

Date :

AIM:

To write a MatLab program to determine the power flow solution by Newton raphson

method for a given problem.

ALGORITHM:

Step1: Start the program

Step2: Clear the screen

Step3: Give values to v1, v2,v3,d1,d2,d3,ps1,ps2&qs

Step4: Form the Y bus matrix for the given problem

Step5: Determine the magnitude &angle of Y bus matrix

Step6: Set iter as 0

Step7:Set power accuracy value as 0.00025

Step8: Check maximum of PQ is greater than power accuracy value

Step9: iter=iter+1

Step10: Find the p1,p2and q using formula

Step11: Form the Jacobian matrix

Step12: Find the dp and dq

Step13: dc=[dp;dq;]

Step14: dv=inv(j)*dc

Step15: d(2)=d(2)+dv(1)

d(3)=d(3)+dv(2)

v(2)=v(2)+dv(3)

Step16: Stop the program.

44

FORMULA:

1.

2.

4.

5.

6.

7.

8.

45

EXCERSISE:

1. Figure shows the one line diagram of simple three bus power system with generators at

buses at 1 and 3. The magnitude of voltage at bus 1 is adjusted to1.05pu. Voltage magnitude

at bus 3 is fixed at 1.04 pu with a real power generation of 200 MW. A load consisting of 400

MW and 250 Mvar is taken from bus 2. Line impedances are marked on per unit on a 100

MVA base, and the line charging susceptances are neglected. Obtain the power flow solution

by fast decoupled method including line flows and line losses.

1 2

0.02+j0.04 400MW 0.01+jo.o3 0.0125+j0.025 250Mvar 3

Slack bus V1=1.05

200 MW |v3|=1.04

46

PROGRAM:

clear all;

clc;

y=[20-50i -10+20i -10+30i;-10+20i 26-52i -16+32i;-10+30i -16+32i 26-62i;];

Y=abs(y)

th=angle(y)

p2s=-4;

q2s=-2.50;

p3s=2;

v1=1.05;

v2=1;

v3=1.04;

del1=0;

del2=0;

del3=0;

iter=0

PQ=[1;1;1]

pwrctr=0.00025;

while(max(abs(PQ))>pwrctr)

iter=iter+1

p2=abs(v2)*abs(v1)*Y(2,1)*cos(th(2,1)-

del2+del1)+abs(v2^2)*Y(2,2)*cos(th(2,2))+abs(v2)*abs(v3)*Y(2,3)*cos(th(2,3)-

del2+del3);


del3+del1)+abs(v3^2)*Y(3,3)*cos(th(3,3))+abs(v3)*abs(v2)*Y(3,2)*cos(th(3,2)-

del3+del2);

q2=-(abs(v2)*abs(v1)*Y(2,1)*sin(th(2,1)-del2+del1))-

(abs(v2^2)*Y(2,2)*sin(th(2,2)))-(abs(v2)*abs(v3)*Y(2,3)*sin(th(2,3)-del2+del3));

Dp2=p2s-p2;

Dp3=p3s-p3;

Dq2=q2s-q2;

j11=(abs(v2)*abs(v1)*Y(2,1)*sin(th(2,1)-

del2+del1))+(abs(v2)*abs(v3)*Y(2,3)*sin(th(2,3)-del2+del3));

j12=-(abs(v2)*abs(v3)*Y(2,3)*sin(th(2,3)-del2+del3));

j13=(abs(v1)*Y(2,1)*cos(th(2,1)-

del2+del1))+(2*abs(v2)*Y(2,2)*cos(th(2,2)))+(abs(v3)*Y(2,3)*cos(th(2,3)-

del2+del3));

j21=-(abs(v3)*abs(v2)*Y(3,2)*sin(th(3,2)-del3+del2));

j22=(abs(v3)*abs(v1)*Y(3,1)*sin(th(3,1)-

del3+del1))+(abs(v3)*abs(v2)*Y(3,2)*sin(th(3,2)-del3+del2));

j23=(abs(v3)*Y(3,2)*cos(th(3,2)-del3+del2));

j31=(abs(v2)*abs(v1)*Y(2,1)*cos(th(2,1)-

del2+del1))+(abs(v2)*abs(v3)*Y(2,3)*cos(th(2,3)-del2+del3));

j32=-(abs(v2)*abs(v3)*Y(2,3)*cos(th(2,3)-del2+del3));

j33=-(abs(v1)*Y(2,1)*sin(th(2,1)-del2+del1))-(2*abs(v2)*Y(2,2)*sin(th(2,2)))-

(abs(v3)*Y(2,3)*sin(th(2,3)-del2+del3));

J=[j11 j12 j13;j21 j22 j23;j31 j32 j33]

PQ=[Dp2;Dp3;Dq2]

DV=inv(J)*PQ

47

del2=DV(1,1)+del2

del3=DV(2,1)+del3

v2=DV(3,1)+v2

end

p1=(abs(v1^2)*Y(1,1)*cos(th(1,1)))+(abs(v1)*abs(v2)*Y(1,2)*cos(th(1,2)-

del1+del2))+(abs(v1)*abs(v3)*Y(1,3)*cos(th(1,3)-del1+del3))

q1=-(abs(v1^2)*Y(1,1)*sin(th(1,1)))-(abs(v1)*abs(v2)*Y(1,2)*sin(th(1,2)-del1+del2))-

(abs(v1)*abs(v3)*Y(1,3)*sin(th(1,3)-del1+del3))

q3=-(abs(v3)*abs(v1)*Y(3,1)*sin(th(3,1)-del3+del1))-(abs(v3)*abs(v2)*Y(3,2)*sin(th(3,2)-

del3+del2))-(abs(v3^2)*Y(3,3)*sin(th(3,3)))

48

OUTPUT:

Y = 53.8516 22.3607 31.6228

22.3607 58.1378 35.7771

31.6228 35.7771 67.2309

th = -1.1903 2.0344 1.8925

2.0344 -1.1071 2.0344

1.8925 2.0344 -1.1737

iter = 0

PQ = 1

1

1

Iter = 1

J = 54.2800 -33.2800 24.8600

-33.2800 66.0400 -16.6400

-27.1400 16.6400 49.7200

PQ =-2.8600

1.4384

-0.2200

DV = -0.0453

-0.0077

-0.0265

del2 = -0.0453

del3 =-0.0077

v2 =0.9735

iter = 2

J =-0 51.7247 -31.7656 21.3026

-32.9816 65.6564 -15.3791

-28.5386 17.4028 48.1036

PQ =-0.0992

0.0217

-0.0509

DV =-0.0018

-0.0010

-0.0018

del2 = -0.0471

del3 =-0.0087

v2 = 0.971

49

iter = 3

J = 51.5967 -31.6939 21.1474

-32.9339 65.5976 -15.3516

-28.5482 17.3969 47.9549

PQ = 1.0e-003 *

-0.2166

0.0382

-0.1430

DV = 1.0e-005 *

-0.3856

-0.2386

-0.4412

del2 = -0.0471

del3 =-0.0087

v2 = 0.9717

p1 = 2.1842

q1 =1.4085

q3 = 1.4618

50

MANUAL CALCULATION :

51

52

53

RESULT:

Thus the power flow solution is obtained by newton raphson method with one slack

bus, one load bus and one generated bus

54

Ex. No: 4b) LOAD FLOW ANALYSIS BY FAST DECOUPLED METHOD

Date :

AIM:

To write a Mat lab program to determine the load flow solution of the given bus

condition using Fast decoupled method.

ALGORITHM:

Step1: Start the program.

Step2: Get the values of the Ybus matrix and the variables.

Step3: Calculate the magnitude and Angles of the Ybus.

Step4: Form B’ matrix by imaginary part of Ybus matrix eliminating row &

column of Slack bus.

Step5:Form B’’ matrix by imaginary part of Ybus matrix eliminating row & column

of Slack bus and generator bus.

Step6: Get the values of Real & reactive power, voltage and magnitude of all the

three buses. Intialize iterative value as zero and pwrctr value as 0.00025

Set PQ matrix as[1;1;1]

Step7:Check whether the PQ matrix elements are greater than the pwrctr value. If yes

proceed with calculating values of P2,Q2 and P3.Also calculate the DV and PQ

matrices using relevant formulae.

Step8: If the condition is not satisfied, then continue calculating the values of P1,Q1

and Q3 with the available input values.

Step9: Stop the program.decoupled method by MATLAB application.

55

EXCERSISE:

1. Figure shows the one line diagram of simple three bus power system with generators at

buses at 1 and 3. The magnitude of voltage at bus 1 is adjusted to1.05pu. Voltage magnitude

at bus 3 is fixed at 1.04 pu with a real power generation of 200 MW. A load consisting of 400

MW and 250 Mvar is taken from bus 2. Line impedances are marked on per unit on a 100

MVA base, and the line charging susceptances are neglected. Obtain the power flow solution

by fast decoupled method including line flows and line losses.

1 2

0.02+j0.04 400MW 0.01+jo.o3 0.0125+j0.025 250Mvar 3

Slack bus V1=1.05

200 MW |v3|=1.04

-52 32 B’= 32 -62

56

FORMULA:

FAST DECOUPLED METHOD

1.𝑷𝒊 = ∑ 𝑽𝒊𝑽𝒋𝒀𝒊𝒋𝑪𝒐𝒔(Ѳ𝒊𝒋𝒏𝒋=𝟏 − 𝜹𝒊+𝜹𝒋)

2. 𝑸𝒊 = − ∑ 𝑽𝒊𝑽𝒋𝒀𝒊𝒋𝑺𝒊𝒏(Ѳ𝒊𝒋𝒏𝒋=𝟏 − 𝜹𝒊+𝜹𝒋)

3. ∆𝑷𝒊(𝒌)

= 𝑷𝒊𝒔𝒄𝒉 − 𝑷𝒊

(𝒌)

4. ∆𝑸𝒊(𝒌)

= 𝑸𝒊𝒔𝒄𝒉 − 𝑸𝒊

(𝒌)

5. ∆ 𝜹 = −[𝑩′]−𝟏 ∆𝑷

𝑽

6. ∆ 𝑽 = −[𝑩′′]−𝟏 ∆𝑸

𝑽

7. 𝜹𝒊(𝒌+𝟏)

= 𝜹𝒊(𝒌)

+ ∆𝜹𝒊(𝒌)

8. 𝑽𝒊(𝒌+𝟏)

= 𝑽𝒊(𝒌)

+ ∆𝑽𝒊(𝒌)

57

PROGRAM:

%Fast decoupled method using B matrix

clear all;

clc;

y=[20-50i -10+20i -10+30i; -10+20i 26-52i -16+32i; -10+30i -16+32i 26-62i]

Y=abs(y) %Magnitude of the Ybus elements

th=angle(y) %Angle of the ybus elements

B1=[-52 32;32 -62] % imaginary part of Ybus matrix eliminating row & column of Slack

bus

B11=[-52] % imaginary part of Ybus matrix eliminating row & column of Slack

p2s=-4; bus & generator bus

q2s=-2.50;

p3s=2;

v1=1.05;

v2=1;

v3=1.04;

del1=0;

del2=0;

del3=0;

iter=0

PQ=[1;1;1]

pwrctr=0.00025; %Power counter value for iterative condition cheching

while(max(abs(PQ))>pwrctr)

58

iter=iter+1


del2+del1)+abs(v2^2)*Y(2,2)*cos(th(2,2))+abs(v2)*abs(v3)*Y(2,3)*cos(th(2,3)-del2+del3);


del3+del1)+abs(v3^2)*Y(3,3)*cos(th(3,3))+abs(v3)*abs(v2)*Y(3,2)*cos(th(3,2)-del3+del2);

q2=-(abs(v2)*abs(v1)*Y(2,1)*sin(th(2,1)-del2+del1))-(abs(v2^2)*Y(2,2)*sin(th(2,2)))-

(abs(v2)*abs(v3)*Y(2,3)*sin(th(2,3)-del2+del3));

Dp2v=(p2s-p2)/v2;

Dp3v=(p3s-p3)/v3;

Dq2v=(q2s-q2)/v2;

Ddel=-inv(B1)*[Dp2v;Dp3v]

DV=-inv(B11)*[Dq2v]

PQ=[Dp2v;Dp3v;Dq2v]

del2=del2+Ddel(1,1)

del3=del3+Ddel(2,1)

v2=v2+DV

end

p1=(abs(v1^2)*Y(1,1)*cos(th(1,1)))+(abs(v1)*abs(v2)*Y(1,2)*cos(th(1,2)-

del1+del2))+(abs(v1)*abs(v3)*Y(1,3)*cos(th(1,3)-del1+del3))

q1=-(abs(v1^2)*Y(1,1)*sin(th(1,1)))-(abs(v1)*abs(v2)*Y(1,2)*sin(th(1,2)-del1+del2))-

(abs(v1)*abs(v3)*Y(1,3)*sin(th(1,3)-del1+del3))

q3=-(abs(v3)*abs(v1)*Y(3,1)*sin(th(3,1)-del3+del1))-(abs(v3)*abs(v2)*Y(3,2)*sin(th(3,2)-

del3+del2))-(abs(v3^2)*Y(3,3)*sin(th(3,3)))

59

OUTPUT:

y = 20.0000 -50.0000i -10.0000 +20.0000i -10.0000 +30.0000i

-10.0000 +20.0000i 26.0000 -52.0000i -16.0000 +32.0000i

-10.0000 +30.0000i -16.0000 +32.0000i 26.0000 -62.0000i

Y = 53.8516 22.3607 31.6228

22.3607 58.1378 35.7771

31.6228 35.7771 67.2309

th = -1.1903 2.0344 1.8925

2.0344 -1.1071 2.0344

1.8925 2.0344 -1.1737

B1 = -52 32

32 -62

B11 = -52

iter =0

PQ = 1 1 1

iter = 1

Ddel = -0.0605

-0.0089

DV = -0.0042

PQ = -2.8600

1.3831

-0.2200

del2 = -0.0605

del3 = -0.0089

v2 =0.9958

.

.

.

.

.

.

60

iter =15

Ddel = 1.0e-005 *

0.2016

-0.0145

DV =-8.5243e-007

PQ = 1.0e-003 *

0.1094

-0.0735

-0.0443

del2 = -0.0471

del3 = -0.0087

v2 =0.9717

p1 = 2.1842

q1 = 1.4085

q3 = 1>>

61

MANUAL CALCULATION:

62

63

64

RESULT:

Thus the power flow solution is obtained by fast decoupled method with one slack

bus, one load bus and one generated bus

65

Ex. No: 5a) SYMMETRICAL FAULT ANALYSIS Date :

AIM:

To develop a Matlab program for performing symmetrical fault analysis for the given power

system network.

ALGORITHM:

1. Start the program.

2. Read the available input datas.

3. Find the per unit reactance for all the power system components.

4. Find the subtransient reactance or transient or synchronous emf of the

generator or motor.

5. Find the subtransient current.

6. Stop the program.

FORMULAE:

EMF behind the Subtransient reactance of the generator

Eg” = Vt + ILXd”

EMF behind the transient reactance of the generator

Eg’ = Vt + ILXd’

EMF behind the synchronous reactance of the generator

Eg = Vt + ILXd

Where,

Vt is the terminal voltage in p.u

IL is the load current in p.u

Xd” is the subtransient reactance of the generator in p.u

Xd’ is the transient reactance of the generator in p.u

Xd is the synchronous reactance of the generator in p.u

Fault Current,

If “ = Eg” / X” from generator to fault

If ‘ = Eg ‘ / X’ from generator to fault

If = Eg / X from generator to fault

66

PROGRAM:

clear all;

clc;

baseMVA=25;

baseKVgen=11;

Xg=0.15i;

Xt1=0.1i;

baseKVline=66;

Xline=0.1i;

baseKVt2=66;

Xt2=0.1i;

baseKVmotor=11;

Xmotor=0.15i;

Vt=10.6;

Pout=15;

pf=0.8;

'Load voltage in p.u'

Vpu=Vt/baseKVmotor

'Load MVA in p.u'

LMVA=(Pout/pf)/baseMVA

'Load current in p.u'

I=LMVA/Vpu

IL=I*(pf-j*asind(pf))

'EMF behind subtransient reactance of the generator'

Xdg=Xg+Xt1+Xline+Xt2

Eg=Vpu+(IL*Xdg)

'EMF behind subtransient reactance of the motor'

Xdm=Xmotor

Em=Vpu-(IL*Xdm)

'subtransient current of the generator'

Ig=Eg/Xdg

'subtransient current of the motor'

Im=Em/Xmotor

'Total Fault current

If=Ig+Im

67

OUTPUT:

Load voltage in p.u

Vpu =

0.9636

Load MVA in p.u

LMVA =

0.7500

Load current in p.u

I =

0.7783

IL =

0.6226 -41.3513i

EMF behind subtransient reactance of the generator

Xdg =

0 + 0.4500i

Eg =

19.5717 + 0.2802i

EMF behind subtransient reactance of the motor

Xdm =

0 + 0.1500i

Em =

-5.2391 - 0.0934i

subtransient current of the generator

Ig =

0.6226 -43.4927i

subtransient current of the motor

Im =

-0.6226 +34.9270i

Total Fault current

If =

0 - 8.5657i

68

MANUAL CALCULATION:

69

RESULT:

Thus develop a mat lab program for performing symmetrical fault analysis for the

given power system network.

70

Ex. No: 5b) UN SYMMETRICAL FAULT ANALYSIS

Date :

AIM:

To develop a Mat lab program for performing unsymmetrical fault analysis for the given power

system network.

ALGORITHM:

STEP1:Start the program.

STEP2:Input the available datas.

STEP3:Do the Fault analysis based on the type of fault.

STEP4:Find the respective fault current.

STEP5:Stop the program

FORMULAE:

Line-Ground Fault

Fault current, If = 3 x Ea/ (Z0+Z1+Z2)

Line-Line Fault

Fault current, If = √3x Ea /(Z1+Z2)

Line-Line-Ground Fault

Fault Current, If = 3x Ia0

Z0,Z1, Z2 are the zero, positive and negative sequence impedances respectively.

71

PROGRAM:

clear all;

clc;

f=50;

'Base MVA'

MVA=50*10^6

'Base KV'

KV=13.2*10^3

'Positive, negative and zero sequence impedances of alternator'

Xg1=0.1i

Xg2=0.1i

Xg0=0.05i

Eg=1+0i

'Base Current'

Ibase=MVA/(sqrt(3)*KV)

'Line-Ground Fault'

Ia1=Eg/(Xg1+Xg2+Xg0)

If=3*Ia1

Ifactual=abs(If*Ibase)

'--------------------'

'Line-Line Fault'

Ia1=Eg/(Xg1+Xg2)

If=sqrt(3)*Ia1


'---------------------'

'Line-Line-Ground'

Ia1=Eg/(Xg1+(Xg2*Xg0)/(Xg0+Xg2))

Ia0=-Ia1*(Xg2/(Xg2+Xg0))

If=3*Ia0


'--------------------'

72

OUTPUT:

Base MVA

MVA =

50000000

Base KV

KV =

13200

Positive, negative and zero sequence impedances of alternator

Xg1 =

0 + 0.1000i

Xg2 =

0 + 0.1000i

Xg0 =

0 + 0.0500i

Eg =

1

Base Current

Ibase =

2.1869e+003

Line-Ground Fault

Ia1 =

0 - 4.0000i

If =

0 -12.0000i

Ifactual =

73

2.6243e+004

Line-Line Fault

Ia1 =

0 - 5.0000i

If =

0 - 8.6603i

Ifactual =

1.8939e+004

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

Line-Line-Ground

Ia1 =

0 - 7.5000i

Ia0 =

0 + 5.0000i

If =

0 +15.0000i

Ifactual =

3.2804e+004

74

MANUAL CALCULATION :

75

RESULT:

Thus the Mat lab program was developed to analyze the unsymmetrical fault for the

given network.

76

Ex. No : 6) TRANSIENT STABILITY ANALYSIS OF POWER SYSTEMS

Date :

AIM:

To write a Mat Lab program for solving Transient Stability Analysis of power systems

and to obtain corresponding swing curves.

ALGORITHM:

STEP1: Input the time interval, final and initial time.

STEP2:Input the value of kinetic energy.

STEP3:Calculate the moment of inertia and maximum power before fault,

maximum power after fault and minimum power during fault.

m = Gh/180

Pmax1 = |E||V| /X1 - Prefault condition

Pmax2 = |E||V| /X2 - During fault

Pmax3 = |E||V| /X3 - Post fault

STEP4. Find the accelerating power for each time interval.

STEP5. Calculate the change in δ and

Pa(n-1)= Pm-Pmax Sinδ

Δδn = (Δδn-1+ (Δt)²)/m Pa(n-1)

δn = δn-1+Δδn

STEP6. Increase the time value by step.

STEP7. Plot the δ vs time curve.

FORMULA:

Consider the swing equation 𝒅𝟐𝜹

𝒅𝒕𝟐=

𝟏

𝒎(Pm-Pmax sin𝛿)

= 𝑷𝒂

𝒎 where m=Gh/pi

The solution δ(t) is obtained at discrete intervals at time interval spread at uniform

throughout. accelerating power and change in speed which are continuous functions of time.

1) Pa computed at the beginning of an interval to remain constant from the middle of the

preceding interval to the middle of the interval being considered.

2) The angular velocity ω=dδ/dt is assumed constant. Any interval at the value computed for

the middle of the interval.

The clearing angle can be found by using the following formula.

Pa(n-1) = Pm-Pmax Sin δn-1

Δδn = Δδn-1+ (Δt)²/m Pan-1

δn = δn-1+Δδn

Greater accuracy of solution can be achieved by reducing the time duration of intervals.

77

PROBLEM:

A 20 MVA, 50 Hz generator delivers 18 MV over a double circuit line to an infinite bus. The

generator has a kinetic energy of 2.52 MJ / MVA at rated speed. The generator transient reactance is

Xd’=0.35 p.u. Each transient circuit has R=0 under reactance of 0.2 p.u. on a 20 MVA base. |E’|=1.1

p.u., infinite bus voltage, V=10o.

A three phase short circuit occurs at the midpoint of one of the transmission lines. Plot Swing curves

with fault cleared by simultaneous opening of circuit breakers at both ends of the line at 2.5 cycles

and 6.25 cycles after the occurrence of the fault.

Also plot the Swing curve over the period of 0.5 seconds if the fault is sustained.

78

FLOWCHART:

IF

t > tc

START

ENTER TIME IN MATRIX FORM,

ENTER KINETIC ENERGY

CALCULATE M, Pmaxbf, Pmaxdf, Pmaxaf

IF

t = tf

IF

t = tc

Paminus=0.9 - Pmaxbf * sin(delta)

Paplus=0.9 - Pmaxdf * sin(delta)

Pa=(Paminus+Paplus) / 2

A

B

Paminus=0.9 - Pmaxdf * sin(delta)

Paplus=0.9 - Pmaxaf * sin(delta)

Pa=(Paminus+Paplus) / 2

A

IF

t > tf

t < tc

Pa = Pm - Pmaxdf * sin(delta)

Pa = Pm - Pmaxaf * sin(delta)

C

79

C

Ddelta = ddelta + tstep

Delta = delta * d.delta

t = t+step

IF

t<tfinal B

PLOT THE GRAPH A

STOP

80

PROGRAM:

clear all;

clc;

h=input('Enter kinetic Energy');

dpower=input('Enter delivered power');

base=input('Enter base power');

E=input('Enter E value');

V=input('Enter V value');

X=input('Enter Gen Reactance,line1 reactance');

Xmid=X(2)/2;

pm=input('Enter mechanical input');

tinitial=0.0;

M=h/(180*50);

%UNDER PREFAULT CONDITION

preX= X(1)+X(2)/2;

pmaxbf = (E*V)/preX;

%UNDER FAULT CONDITION

faultX=(X(1)*Xmid+X(2)*Xmid+X(1)*X(2))/Xmid;

pmaxdf = (E*V)/faultX;

%POST FAULT CONDITION

postX = X(1)+X(2);

pmaxaf = (E*V)/postX;

n=3;

disp(['Prefault maximum power limit',num2str(pmaxbf)]);

disp(['Maximum power limit during fault',num2str(pmaxdf)]);

disp(['Postfault maximum power limit',num2str(pmaxaf)]);

for k=1:n

time= input('Enter step time:Critical time:Final time:in matrix form:');

delta =asin((dpower/base)/(pmaxbf));

ang(1)=(delta*180)/pi;

tinitial=0.0;

delchange=0.0;

tim(1)=0;

i=2;

t=0;

j=0;

while t<time(3)

if (t==tinitial)

paminus=pm-pmaxbf*sin(delta);

paplus=pm-pmaxdf*sin(delta);

paavg=(paplus+paminus)/2;

pa=paavg;

end

if(t==time(2))

paminus=pm-pmaxdf*sin(delta);

paplus=pm-pmaxaf*sin(delta);

paavg=(paminus+paplus)/2;

pa=paavg;

end

if(t>tinitial & t<time(2))

81

pa=pm-pmaxdf*sin(delta);

end

if(t>time(2))

pa= pm-pmaxaf*sin(delta);

end

delchange=delchange+((time(1)*time(1)*pa)/M);

delta=(delta*180/pi+delchange)*pi/180;

deldeg=delta*180/pi;

if(j==0)

disp('time acclerating power delchange deldegree');

end

j=j+1;

disp([num2str(t),' ',num2str(pa),' ',num2str(delchange),' ',num2str(deldeg)]);

t=t+time(1);

tim(i)=t;

ang(i)=deldeg;

i=i+1;

end

hold on

axis([0 0.6 0 160])

xlabel('Time in sec')

ylabel('Torque angle,deg')

plot(tim,ang,'-')

end

82

INPUT:

enter kinetic energy2.52

enter delivered power18

enter base power20

enter e value1.1

enter v value1

enter gen reactance,line 1 reactance[0.35 .2]

enter mechanical input0.9

OUTPUT:

prefault maximum power limit2.4444

maximum power limit during fault0.88

postfault maximum power limit2

enter step time:critical time:final time:in matrix form:[0.05 .38 .5]

time acclerating power delchange deldegree

00.451644.032525.636

0.050.519278.668834.3048

0.10.4040412.276246.581

0.150.2608114.604961.186

0.20.1289515.756376.9423

0.250.04275416.138193.0803

0.30.02127116.328109.4083

0.350.07000616.953126.3613

0.4-0.7105910.6085136.9698

0.45-0.464776.4588143.4286

0.5-0.291653.8548147.2834

enter step time:critical time:final time:in matrix form:[.05 .05 .5]


00.451644.032525.636

0.050.276986.505532.1415

0.1-0.164025.04137.1825

0.15-0.308712.284739.4672

0.2-0.37127-1.030338.4369

0.25-0.3433-4.095534.3414

0.3-0.22825-6.133428.208

0.35-0.045347-6.538321.6697

0.40.16149-5.096416.5733

0.450.32952-2.154314.419

0.50.401981.434815.8538

enter step time:critical time:final time:in matrix form:[.05 .125 .5]


00.451644.032525.636

0.050.519278.668834.3048

0.10.4040412.276246.581

0.15-0.552697.341553.9225

0.2-0.716440.9446754.8672

0.25-0.73564-5.623549.2436

0.3-0.61498-11.114538.1291

0.35-0.33487-14.104424.0247

83

Graph:

84

MANUAL CALCULATION:

85

86

87

RESULT: Thus the transient analysis of given power systems was studied and corresponding swing

curve was obtained.

88

Ex. No: 7) STEADY STATE STABILITY

Date :

AIM:

To find the frequency and oscillations for given performance of its maximum power.

FORMULA:

M=H/πf in PU

p.e= [|E||V|]/xd*sinδ=Pmax sinδ

dPe/dδ=Pmaxsinδ

PROBLEM:

A 20 MVA, 50 Hz generator delivers 18 MV over a double circuit line to an infinite bus. The

generator has a kinetic energy of 2.52 MJ / MVA at rated speed. The generator transient reactance is

Xd’=0.35 p.u. Each transient circuit has R=0 under reactance of 0.2 p.u. on a 20 MVA base. |E’|=1.1

p.u., infinite bus voltage, V=10o.

A three phase short circuit occurs at the midpoint of one of the transmission lines. Plot Swing curves

with fault cleared by simultaneous opening of circuit breakers at both ends of the line at 2.5 cycles

and 6.25 cycles after the occurrence of the fault.

Also plot the Swing curve over the period of 0.5 seconds if the fault is sustained.

89

PROGRAM:

clear all; clc;

EV=input('enter the voltage for generator and infinite bus:');

X=input('enter the reactance in matrix:');

H=input('enter the inertia const:');

i=0;

n=input('enter the number of generator load:');

Pmax=(EV(1)*EV(2))/(X(1)+X(2));

for i=1:n;

load=input('enter the load value of its maximum power limit:');

del=asin(load);

Ddel=Pmax*cos(del);

freq=sqrt((Ddel*50*pi)/H);

freq=freq/(2*pi);

disp(['frequency of oscillation:',num2str(freq),'hz']);

disp(['maximum power limit:',num2str(Pmax),'mw']); end

90

OUTPUT:

enter the voltage for generator and infinite bus:[1.1 1]

enter the reactance in matrix:[0.35 0.2]

enter the inertia const:2.52

enter the number of generator load:1

enter the load value of its maximum power limit:0.5

frequency of oscillation:1.6537hz

maximum power limit:2mw

91

MANUAL CALCULATION:

92

RESULT: Thus the steady state stability analysis of given power systems was studied and

corresponding swing curve was obtained.

93

: Ex. No: 8 ELECTROMAGNETIC TRANSIENTS

Date :

AIM:

To study the effect in electromagnetic transients in power supply

THEORY:

Transient phenomena is a periodic function of time and does not-last longer than the

operator for which they lost in very insurant compared with operator time of system.

They are very important. The power system can be considerably made up of lines.

Reduces elements of RF and the circuits is normally organized and carrier load until OC fault

occurs, the fault corresponds to closing the switch. The redistribution is accomplished in

general by ac transient period during which the resultant to relic that these redistribution of I

and V cannot take place instantly for the following reason.

Transient in single circuits with DC source

. R-only

As soon as the switch is closed the current is determined by using ohm’s law.

I=V/R

No transients will be there

. L-only

When switch is closed the current is given by

I(s)=V(S)/z(s)={VI/s2L}=(V/l)*t

It is shown that when a pure inductance is switched on to a DC source the current at t=0 and

increases linearly for infinity.C-only

When switch is closed current is circuit is given by

I(s)=V(s)/Z(s)={V/S}C s=VC

Therefore to have transients in an electrical system the following requirements should be

used. Either inductance or capacitance of both should be present.

1. Fundamental frequency

2. Natural frequency

94

3. There is also another component called thermionic due to unbalanced current. Natural

frequency occurs after the occurrence of certain fault which is added to fundamental

frequency which consist transient voltage.

4. Transient which are in form of energy storage magnetic or electric is consider is

called single energy transient. If those are both in magnetic and electric energies it is

called double energy transients. The electromagnetic energy standby an inductance L

is (1/2) LI^2 where I is instantaneous value of current assuming L to be constantly

changing in current which an inductance is allowed.

5. There are only to two components of which store energy and redistribution of energy

following in a circuit change takes a finite time.

1. I can’t change instantaneously through induction

2. V across the capacitor can’t change instantaneous

3. How of conservation of energy must hold.

Which an impulse of strength where s is closed

|Vm*ωcosФ/ L S^2+ω^2 + Vm*s.sinФ/ L S^2+ω^2|

V= Vm sin(ωt +Ф )

I(S)= Vm * ωcos Ф/ (s+ω)(s2+ω2)

Now

L-I I(s) = (Vm/(L2+ω2))( ωcos Ф){e-at+(a/ω)sinωt-cosωt}+sin Ф{a cosωt +ωsinωt- e-∆t }

It can be simplified to

{Vm/(R2+ ω2 L2)V}{sin (ω+0-0)-( sin(0-0) e-at)}

The first term in above can is steady state sinusoidal vibration and second term in the

transient part after infinite line.

95

RESULT:

The effect of transients in power system is studied.

96

EXP.NO : 9a) LOAD FLOW ANALYSIS OF SINGLE AREA SYSTEM

DATE :

AIM:

To study the time response of area frequency deviation following a single load change in a

single area power system provided with an integral frequency controller.

ALGORITHM:

STEP1:Double click the mat lab icon

STEP2: Open a new model file.

STEP3: Open the Simulink library from menu.

STEP4: Click on the required components, drag and place one by one in new model file.

STEP5: Connect them and enter the data to form the model of given single area power

system.

STEP6:Save and run the model to view the output in the scope. Repeat the same procedure

for the given to area power system model.

EXERCISE:

An isolated power system has following parameters

Turbine time constant=0.5sec

Governor time constant=0.2sec

Generator inertia speed constant, H=5sec

Governor speed regulation, R=0.05pu

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

rated output is 250Mw at normal frequency of 60Hz. A sudden load change of 50 MW

(ΔPL=0.2pu) occurs.set the integral controller gainKi=7.

Construct a SIMULINK block and obtain frequency deviation response.

97

BLOCK DIAGRAM:

SINGLE AREA SYSTEM:

SIMULINKMOEDEL:

SINGLE AREASYSTEM:

98

OUTPUT:

SINGLE AREA SYSTEM:

Change in frequency deviation Vs time

Change in turbine power deviation

99

MANUAL CALCULATION:

100

101

RESULT:

Thus the time response of single area and two area system was simulated.

102

Ex. No: 9b) LOAD FLOW ANALYSIS OF TWO AREA SYSTEM

Date :

AIM:

To study the time response of area frequency deviation and net interchange deviation

following a small load change in any one of areas of an interconnected two area power system.

ALGORITHM:

STEP1: Double click the mat lab icon

STEP2: Open a new model file.

STEP3; Open the simulink library from menu.

STEP4 Click on the required components, drag and place one by one in new model file.

STEP5: Connect them and enter the data to form the model of given single area power

system.

STEP6: Save and run the model to view the output in the scope. Repeat the same procedure

for the given to area power system model.

EXCERSISE:

A two area system connected by a tie line has the following parameters on a 1000 MVA

common base.

Area 1 2

Speed regulation, R 0.05 0.0625

Freq-sens. Load co-eff, D 0.6 0.9

Inertia constant, H 5 4

Base power 1000MVA 1000MVA

Governor time constant, TH 0.2sec 0.3sec

Turbine time constant, TT 0.5sec 0.6ec

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

coefficient is computed from the initial operating condition and is given to be PS=2 pu. A

load change of 187.5Mw occurs in area 1.

Construct a SIMULINK block and obtain frequency deviation response.

103

BLOCK DIAGRAM:

TWO AREA SYSTEM:

SIMULINK MODEL

TWO AREA SYSTEM:

104

OUTPUT:

TWO AREA SYSTEMS:

Frequency deviation response

Power deviation response

105

MANUALCALCULATION:

106

RESULT:

Thus the time response of single area and two area system was simulated.

107

Ex. No :10 ECONOMIC DISPATCH WITHOUT LOSSES

Date :

AIM:

To find out the economic schedule for the given lossless system.

ALGORITHM:

Step1: Obtain the data for the cost function of the ‘n’ units, and the total load of the

system if any.

Step2: To find out the incremental cost for each unit and find out the coordination

equation for the Lambda iteration method.

Step3: In the lossless system incremental cost of all the units are same.

Step4: By solving the equations, calculate the power generation for n units (PG1,

PG2,…………. PGn. )

Step5: If the generation limits (minimum or maximum) are violated for any one

generator, then fix the violating limits as the generation value of that generator.

Step6: The economic dispatch schedule was calculated.

PROBLEM:

The input – output characteristics of 3 units are,

F1 = 940 + 5.46 PG1 + 0.0016 PG12

F2 = 820 + 5.35 PG2 + 0.0019 PG22

F3 = 99 + 5.65 PG3 + 0.0032 PG32

150 < = PG1 < = 300

150 < = PG2 < = 250

50 < = PG3 < = 100

Find out the economic scheduling for the total load of 600 MW.

108

PROGRAM:

clear all;

pd=input('ENTER THE POWER DEMAND:');

f1=input('ENTER THE INTERCEPT OF UNIT-1:');



ff1=input('ENTER THE COEFFICIENT OF SLOPE OF UNIT-1:');



if(f1>=f2&f1>=f3)

lambda=f1+rand(1);

elseif(f2>=f3)

lambda=f2+rand(1);

else

lambda=f3+rand(1);

end

for k=1:1000

p1=(lambda-f1)/ff1;

p2=(lambda-f2)/ff2;

p3=(lambda-f3)/ff3;

pc=p1+p2+p3;

delpd=abs(pd-pc);

if delpd<=1

break;

end

if(pc>pd)

lambdanew=lambda+(-rand(1)*0.01);

else

lambdanew=lambda+(rand(1)*0.01);

end

lambda=lambdanew;

end

disp(p1);

disp(p2);

disp(p3);

disp(lambda);

disp(delpd);

109

OUTPUT:

ENTER THE POWER DEMAND:600

ENTER THE INTERCEPT OF UNIT-1:5.46



ENTER THE COEFFICIENT OF SLOPE OF UNIT-1:0.0032



256.8101

245.2085

98.7176

6.2818

0.7363

110

MANUAL CALCULATION:

111

112

RESULT:

Thus the economic dispatch schedule is obtained for the given system by using the

Mat lab.

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