lab manual-ps lab

8/8/2019 Lab Manual-ps Lab

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REGISTER

NUMBER

Certified that this is the bonafide record of work

done by Selvan / Selvi_____________________________________ of the

______________________

________________________________________________________ branch during

the year ___________ in the

_______________________________________________________________

laboratory

STAFF IN CHARGE HEAD OF THE

DEPARTMENT

Submitted for the university Practical Examinations on

_________________________



INTERNAL EXAMINER EXTERNAL

EXAMINER



S.N

oDATE NAME OF THE EXPERIMENT

PAG

E No

MARKS

AWARDE

D

SIGNATURE



AIM:

To develop a program to compute bus admittance matrix for the given powersystem network by inspection method

ALGORITHM

1. Intially Y-Bus matrix i.e. replace all entries as zeros. It means that Yij=Yij-

Yij=Yii

2. Read the number of buses [NB], Number of Lines [NL] and line data

3. Consider line l=1

n

4. Compute Yii= ∑ Yij = diagonal element

J=1

5. Let Y(I,j)= Y(i,i)+Y series (l)+ 0.5 Ysh(l)

Y(j,i)= Y(i,i)+Yseries (l)+0.5 Ysh(l)

Y(i,j)=Yseries (i,j)

Y(j,j)=Y(i,j)

6. Is l=NL?

If Yes , Print Y-bus

If no, l=l+1

7. Stop the program

EX. No: 01 FORMATION OF Y-BUS MATRIX BY THE METHOD OFINSPECTIONDATE:



Line Data

LINENo

START

BUS

ENDBUS

SERIESIMPENDNCE (P.u)

LINE CHARGINGADMITTANCE (P.U)

1 1 2 0.1+j0.3 0.0+j0.022 2 3 0.15+j0.5 0.0+j0.01253 3 1 0.2+j0.6 0.0+j0.028

One Line Diagram

Impedance Diagram



Program:

#include<conio.h>

#include<complex.h>

#include<iostream.h>

#include<fstream.h>

#include<stdio.h>

#include<math.h>#define NB 3+1

#define NL 3+1

complex y_bus[NB][NB],line_Z[NL],halfline_y[NL];

int i,j,k1,k2,line[NL],sb[NL],eb[NL],nl,nb;

void main()

{

ifstream infile;

infile.open("ybus.dat");

ofstream outfile;outfile.open("ybus.out");

outfile.precision(4);

outfile.setf(ios::showpoint);

outfile.fill('0');

outfile<<"\t\t\t Y-BUS FORMATION BY INSPECTION METHOD \n";

infile>>nl>>nb;

outfile<<"\t"<<"NUMBER OF LINES:"<<nl<<" \n";

outfile<<"\t"<<"NUMBER OF BUSES:"<<nb<<"\n";

outfile<< "LINE No\t\tse\teb \t\tline_Z \t\t\t halfline_y\n";for(i=1;i<=nl;i++)

{

infile>>line[i]>>sb[i]>>eb[i]>>line_Z[i]>>halfline_y[i];

outfile<<line[i]<<"\t\t"<<sb[i]<<"\t"<<eb[i]<<"\t\t"<<line_Z[i]<<"\t"<<halfline

_y[i]<<"\n";

k1=sb[i];

k2=eb[i];

y_bus[k1][k1]+= (1.0/line_Z[i])+halfline_y[i];

y_bus[k2][k2]+= (1.0/line_Z[i])+halfline_y[i];

y_bus[k1][k2]+= -1.0/line_Z[i];

y_bus[k2][k1] = y_bus[k1][k2];

}

outfile<<"\n\t\t\t Y-BUS MATRIX \n";

for(i=1;i<=nb;i++)

{

outfile<<"\n";

for(j=1;j<=nb;j++)

{

outfile<<y_bus[i][j]<<"\t";}

}

}



INPUT

File Name: [“YBUS.DAT”]

3

3

1 1 2 (0.1,0.3)(0.0,0.01)

2 2 3 (0.15,0.5)(0.0,0.0625)

3 3 1 (0.2,0.6)(0.0,0.014)



OUTPUT

File Name: [“YBUS.OUT”]

Y-BUS FORMATION BY INSPECTION METHOD

NUMBER OF LINES:3

NUMBER OF BUSES:3

LINE No se eb line_Z halfline_y

1 1 2 (0.1000, 0.3000) (0.0000, 0.0100)

2 2 3 (0.1500, 0.5000) (0.0000, 0.0625)

3 3 1 (0.2000, 0.6000) (0.0000, 0.0140)

Y-BUS MATRIX

(1.5000, -4.4760) (-1.0000, 3.0000) (-0.5000, 1.5000)

(-1.0000, 3.0000) (1.5505, -4.7624) (-0.5505, 1.8349)

(-0.5000, 1.5000) (-0.5505, 1.8349) (1.0505, -3.2584)

Result



AIM:

To develop a program to obtain bus impendance matrix for the given power

system network

ALGORITHM

1. Read the values of such as no of lines, no of buses & line data, generator data

and Transformer data

2. Intialize Y-Bus matrix, Y bus[i][j]=complex (0.0,0.0)

3. Compute Y-bus matrix, by considering only line data

4. Modify Y-Bus matrix by adding the transformer and generator admittance to

respective diagonal element of Y-Bus matrix

5. Compare Z-bus matrix by inverting the modified Y-bus matrix

6. Check the inversion by multiplying modified y-bus & z-bus matrix to see

whether the resultingmatrix is unity or not

7. Print the Z-bus matrix

EX. No: 02FORMATION OF Z-BUS MATRIX

DATE:



Data

No of lines

No of Buses

No of generator

No of Transformer

5 4 2 2Generator1: 0.0+j0.02

Generator 2: 0.0+0.08

Transformer 1: 0.0+j0.25

Transformer2: 0.0+j0.1

LINENo

STARTBUS

ENDBUS

SERIES IMPENDNCE(P.u)

LINE CHARGINGADMITTANCE (P.U)

1 1 2 0.0+j0.4 0.0+j0.00752 2 3 0.15+j0.6 0.0+j0.01

3 3 4 0.18+j0.55 0.0+j0.0094 4 1 0.1+j0.35 0.0+j0.0065 4 2 0.25+j0.7 0.0+j0.015



Program

#include<complex.h>

#include<conio.h>

#include<fstream.h>

#include<math.h>

#define NL 5+1

#define NB 4+1

#define NG 2+1

#define NT 2+1

complex zg[NG],zt[NT],linez[NL],halfliney[NL];

complex ybus[NB][NB],zbus[NB][NB],check[NB][NB];

int nl,line[NL],sb[NL],eb[NL],nb,i,j,m,k1,k2,k,ng,nt;

void cinver(complex [NB][NB],int);

void main()

{

ifstream infile;

infile.open("ZBUS.dat");

ofstream outfile;

outfile.open("ZBUS.out");



outfile.fill('0');

outfile<<"\t\t ZBUS FORMATION AFTER Y BUS MODIFICATION \n";

infile>>nl>>nb>>ng>>nt;

outfile<<"----------------------------------------------------\n";

outfile<<"\t\t Number of lines :"<<nl<<"\n";

outfile<<"\t\t number of buses :"<<nb<<"\n";

outfile<<"\t\t number of gen :"<<ng<<"\n";

outfile<<"\t\t number of trans :"<<nt<<"\n";

for(i=1;i<=ng;i++)

infile>>zg[i]>>zt[i];

outfile<<" line starting ending line halfline\n";

outfile<<" No. bus bus imp adm \n\n";

for(i=1;i<=nl;i++)

{



infile>>line[i]>>sb[i]>>eb[i]>>linez[i]>>halfliney[i];

outfile<<line[i]<<"\t"<<sb[i]<<"\t"<<eb[i]<<"\t"<<linez[i]<<"\t"<<halfliney[i]

<<"\n";

k1=sb[i];

k2=eb[i];

ybus[k1][k1]+=(1.0/linez[i])+halfliney[i];

ybus[k2][k2]+=(1.0/linez[i])+halfliney[i];

ybus[k1][k2]+=-1.0/linez[i];

ybus[k2][k1]+=ybus[k1][k2];

}

for(i=1;i<=ng;i++)

ybus[i][i]+=1.0/(zg[i]+zt[i]);

outfile<<"\n modified y bus matrix: \n\n";for(i=1;i<=nb;i++)

{

for(j=1;j<=nb;j++)

{

outfile<<ybus[i][j]<<"";

zbus[i][j]=ybus[i][j];

}outfile<<"\n";

}

cinver(zbus,nb);

outfile<<"\n Zbus matrix:\n\n";

for(i=1;i<=nb;i++)

{

for(j=1;j<=nb;j++)outfile<<zbus[i][j]<<"";

outfile<<"\n";

}

for(i=1;i<=nb;i++)

{

for(j=1;j<=nb;j++)

{check[i][j]=complex(0.0,0.0);

for(k=1;k<=nb;k++)

check[i][j]+=zbus[i][k]*ybus[k][j];



}

}

outfile<<"\n check matrix\n\n";

for(i=1;i<=nb;i++)

{

for(j=1;j<=nb;j++)

outfile<<check[i][j]<<"";

outfile<<"\n";

}

}

void cinver(complex xxyy[NB][NB],int nnn)

{

complex x;m=nnn+1;

for(i=1;i<=nnn;i++)

{

for(j=1;j<=nnn;j++)

xxyy[j][m]=complex(0.0,0.0);

xxyy[i][m]=complex(1.0,0.0);

x=xxyy[i][i];for(j=1;j<=m;j++)

xxyy[i][j]=xxyy[i][j]/x;

for(k=1;k<=nnn;k++)

{

if(k==i)goto zz;

x=xxyy[k][i];

for(j=1;j<=m;j++)xxyy[k][j]=xxyy[k][j]-x*xxyy[i][j];

zz:;

}

for(j=1;j<=nnn;j++)

xxyy[j][i]=xxyy[j][m];

}

}



INPUT

[“ZBUS.DAT”]

5 4 2 2

(0.0,0.2) (0.0,0.08) (0.0,0.25) (0.0,0.1)

1 1 2 (0.1,0.4) (0.0,0.0075)

2 2 3 (0.15,0.6)(0.0,0.01)

3 3 4 (0.18,0.55)(0.0,0.009)

4 4 1 (0.1,0.35)(0.0,0.006)

5 4 2 (0.25,0.7)(0.0,0.015)



OUTPUT

[“ZBUS.OUT”]

ZBUS FORMATION AFTER Y BUS MODIFICATION

---------------------------------------------------- Number of lines :5

number of buses :4

number of gen : 2

number of trans :2

line starting ending line halfline

No. bus bus imp adm

1 1 2 (0.1000, 0.4000) (0.0000, 0.0075)

2 2 3 (0.1500, 0.6000) (0.0000, 0.0100)

3 3 4 (0.1800, 0.5500) (0.0000, 0.0090)

4 4 1 (0.1000, 0.3500) (0.0000, 0.0060)

5 4 2 (0.2500, 0.7000) (0.0000, 0.0150)

modified y bus matrix:

(1.3430, -8.5524)(-0.5882, 2.3529)(0.0000, 0.0000)(-0.7547, 2.6415)

(-0.5882, 2.3529)(1.4329, -8.0132)(-0.3922, 1.5686)(-0.4525, 1.2670)

(0.0000, 0.0000)(-0.3922, 1.5686)(0.9296, -3.1919)(-0.5375, 1.6423)

(-0.7547, 2.6415)(-0.4525, 1.2670)(-0.5375, 1.6423)(1.7447, -5.5208)

Zbus matrix:

(0.0069, 0.1950)(-0.0087, 0.1111)(-0.0043, 0.1367)(0.0012, 0.1589)

(-0.0087, 0.1111)(0.0110, 0.2171)(0.0064, 0.1881)(-0.0007, 0.1595)

(-0.0043, 0.1367)(0.0064, 0.1881)(0.1008, 0.5174)(0.0282, 0.2633)

(0.0012, 0.1589)(-0.0007, 0.1595)(0.0282, 0.2633)(0.0608, 0.3557)

check matrix

(1.0000, 4.1633e-17)(1.3878e-16, 4.1633e-17)(0.0000, -1.3878e-17)(0.0000, 0.0000)

(2.7756e-16, 2.7756e-17)(1.0000, 2.7756e-17)(5.5511e-17, -1.3878e-17)(-1.1102e-16, 0.0000)

(2.2204e-16, 5.5511e-17)(2.2204e-16, 4.1633e-17)(1.0000, -5.5511e-17)(0.0000, -5.5511e-17)

(2.2204e-16, 5.5511e-17)(2.2204e-16, 0.0000)(0.0000, 0.0000)(1.0000, 0.0000)

RESULT



AIM:

To find The Load Frequency Dynamics of Single & Two Area Power Systems

ALGORITHM

1. Start the program

2. Read all the variables

3. Chech whether the system is single area or two area systems

4. If it is single area system calculate the change in frequency by using the

following formulas

D=Pd/(f*Pr )

K p=1/D

Tp=2h/(f*D)

B=D+(1/R)

Df=-M/B

5. If it is two area system, calculate the change in frequency and tie line power

by using the following formulaes

R1=(r1+100)*(f/(g1*sr1)

R2=(r2+100)*(f/(g2*sr2)

D1=d1(l1-m1)/f

D2=d2(l2-m2)/f

B1=D1+(1/R1)

B2=D2+(1/R2)

Df=(M1+M2)/(B1+B2)

Static Frequency = f+Df

6. Print all the values

EX. No: 03 LOAD FREQUENCY DYNAMICS OF SINGLE AREA & TWOAREA SYSTEMSDATE:



7. Stop the program



Problem for Single Area system

For an isolated single area system considered the following data

Total rated area Capacity P r=2000 MW

Normal operating load Pd=1000MW

Inertia constant H=5 sec

Regulation R=2.4 Hz/P.u.M.w

Normal frequency F=60 Hz

assume that the load frequency charcteristic is linear meaning that the load willincrease 1% for 1% frequency increase . Find 1.Gain & Time constant of power

system 2. Change in frequency under static condition . Also verify the obtained

result with the calculated value

Problem for two area system

Considering the two area system, find the new steady state frequency and

change in tie line flow for a load change of area 2. For both areas each percent

change in frequency causes 1% change in load. Also verify the obtained result.

Assume the following data

Value of Generator 1& 2 =19000 & 41000

Value of generator 1& 2 = 2000 & 40000

Spinning reserves area 1 & 2 = 5 & 5

Regulation of Area 1 & 2 = 1000 & 60

Exchange in load area = 1.0000 & 1.0000

Frequency = 60 Hz



Program

#include<conio.h>

#include<stdio.h>

#include<complex.h>

#include<fstream.h>

#include<math.h>

float pg1,pg2,load1,load2,pd1,pd2,gen1,gen2,tie,x1,x2,c,pr,pd,r,l,h,f,x,D;float D1,D2,b,b1,b2,r1,r2,m,m1,m2,kp,tp,df,g1,g2,l1,l2,sr1,sr2,R,R1,R2,d1,d2;

void main()

{

ifstream infile;

infile.open("load.dat");

ofstream outfile;

outfile.open("load.out");


outfile.setf(ios::showpoint);outfile.fill('0');

outfile<<"\n\n\n load frequency dyanmics of single area and two area system

\n\n";

outfile<<"\n------------------------------------------------";

outfile<<"\n\t\t SINGLE AREA SYSTEM";

outfile<<"\n------------------------------------------------";

infile>>pr;

outfile<<"\n\t AREA CAPACITY :"<<pr;

infile>>pd;outfile<<"\n\t Normal Operating system :"<<pd;

infile>>h;

outfile<<"\n\t Inertia constant :"<<h;

infile>>R;

outfile<<"\n\t Regulation :"<<R;

infile>>f;

outfile<<"\n\t operating frequency(given/assume) :"<<f;

infile>>m;

outfile<<"\n\t Increase in total load :"<<m;

D=pd/(f*pr);

kp= 1/D;

tp=(2*h)/(f*D);

outfile<<"\n\t kp :"<<kp;

outfile<<"\n\t tp :"<<tp;

b=D+(1/R);

df=-(m/b);

outfile<<"\n\t change in frequency :"<<df;

outfile<<"\n--------------------------------------------------";

outfile<<"\n\t\t TWO AREA SYSTEM :";outfile<<"\n --------------------------------------------------";

infile>>g1>>g2;

outfile<<"\n\t value of gen1 & gen 2 :"<<g1<<"\t\t"<<g2;



infile>>l1>>l2;

outfile<<"\n\t value of load 1 & 2 :"<< l1<<"\t\t"<<l2;

infile>>sr1>>sr2;

outfile<<"\n\t the spring reverse of area2 :"<< sr1<<"\t\t"<<sr2;

infile>>r1>>r2;

outfile<<"\n\t the regulation of area2 :"<<r1<<"\t\t"<< r2;

infile>>m1>>m2;

outfile<<"\n\t the exchange in the load of area :"<< m1<<"\t\t"<<m2;

infile>>f;

outfile<<"\n\t frequency :"<<f;

infile>>d1>>d2;

outfile<<"\n\t the percentage of load :"<<d1<<"\t\t"<<d2;

R1=(r1/100)*(f/(g1+sr1));

outfile<< "\n\t R1 :"<<R1;

R2=(r1/100)*(f/(g2+sr2));

outfile<< "\n\t R2 :"<<R2;

D1=d1*((l1-m1)/f);D2=d2*((l2-m2)/f);

outfile<<"\n\t D1 :"<<D1;

outfile<<"\n\t D2 :"<<D2;

b1=D1+(1/R1);

b2=D2+(1/R2);

outfile<<"\n\t b1 :"<<b1;

outfile<<"\n\t b2 :"<<b2;

df=(m1+m2)/(b1+b2);

x=f+df;outfile<<"\n\t DF :"<<df;

outfile<<"\n\t Static frequency :"<<x;

pg1=(-df)/R1;

pg2=(-df)/R2;

outfile<<"\n\t pg1 :"<<pg1;

outfile<<"\n\t pg2 :"<<pg2;

pd1=D1*df;

pd2=D2*df;

outfile<<"\n\t pd1 :"<<pd1;outfile<<"\n\t pd2 :"<<pd2;

gen1=g1+pg1;

gen2=g2+pg2;

load1=(11-m1)+pd1;

load2=(12-m2)+pd2;

tie=gen2-load2;

outfile<<"\n\t Tie Line power :"<<tie;

}



Input data

[“Load.DAT”]

2000

1000

52.4

60

0.01

19000 41000

2000 40000

5 5

1000

60

1 1



OUTPUT

[“Load.OUT”]

Load Frequency Dynamics of Single Area And Two Area System

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

SINGLE AREA SYSTEM

------------------------------------------------AREA CAPACITY :2000.0000

Normal Operating system :1 000.0000

Inertia constant :5 .0 000

Regulation : 2. 4 000

operating frequency(given/assume) :6 0.0000

Increase in total load : 0. 0100

kp :120.0000

tp :20.0000

change in frequency :-0.0235--------------------------------------------------

TWO AREA SYSTEMS:

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

value of gen1 & gen 2 :19000.0000 41000.0000

value of load 1 & 2 :2000.0000 40000.0000

the spring reverse of area2 :5 .0 000 5.0000

the regulation of area2 : 1 00 0 .0 000 60.0000

the exchange in the load of area :1.0000 1.0000

frequency :6 0 . 0 000the percentage of load :0 .0 00 0 0.0000

R1 :0 . 0 3 16

R2 : 0 . 0 1 4 6

D1 :0 . 0000

D2 :0 . 0000

b1 : 31 .6750

b2 : 68 .3417

DF : 0 .0 200

Static frequency :6 0 . 0 200

pg1 : -0 .6 334

pg2 : -1 .3 666

pd1 : 0 .0 000

pd2 : 0 .0 000

Tie Line power :999.6328

Result



AIM

To find the performance parameters of short transmission line

ALGORITHM


2. Get the power delivered, reactive end voltage, total line impendance, power

factor & total resistance of short transmission line

3. Calculate the follwing

4. Line current I1= (Pr)/(Vr*Pf)

5. Receiving end current Ir= (I1*fi)

6. Sending end voltage Vs=(Vr*1000)+(Ir*Z)

7. Magnitude of sending end voltage Vsm=(abs(Vs)/1000)

8. Line losses L1=((I1*I1*R)/1000)

9. Power sent Ps=Pr+L1

10. Transmission Efficiency Ty=((Pr/Ps)*100)

11. %Regulation =(Vsm-Vr)/Vr)*100

12. Print all the calculated parameters

13. Stop the program

EX. No: 04PERFORMANCE OF SHORT TRANSMISSION LINE

DATE:



PROGRAM

#include<stdio.h>

#include<fstream.h>

#include<complex.h>

#include<conio.h>

#include<math.h>

float I1,L1,Vr,Vsm,Ps,Ty,PF,Reg;

complex Z,Ir,Vs,FI;

int pr,R;

void main()

{

ifstream infile;

infile.open("ST.dat");

ofstream outfile;

outfile.open("ST.out");



outfile.fill('0');

outfile<<"\t\t PERFORMANCE OF SHORT TRANSMISSION LINE \n";

infile>>pr>>Vr>>FI>>Z>>PF>>R;

outfile<<"\n power delivered ="<<pr<<"KW";

outfile<<"\n the receiving end voltage ="<<Vr<<"KV";

outfile<<"\n FI ="<<FI<< "ohm";

outfile<<"\n total line impendance ="<<Z<<"ohm";

outfile<<"\n power factor ="<<PF<<"\t Lagging";

outfile<<"\n the total resistance ="<<R<<"ohm";

outfile<<"\n calculation for Ir,Vs,Trans.eff1,%Reg"<<"\n";

I1=(pr)/(Vr*PF);

o utfile<<"\n Line current = " <<I1<<"Amperes";

Ir=(I1*FI);

outfile<<"\n Receiving End Current ="<<Ir<<"Amperes";

Vs=(Vr*1000)+(Ir*Z);

outfile<<"\n Sending End Voltage ="<<Vs<<"Volt";

Vsm=(abs(Vs)/1000);

outfile<<"\n The magnitude of sending end voltage ="<<Vsm;



L1=((I1*I1*R)/1000);

o utfile<<"\n Line Losses = "<<L1<<"KV";

Ps=pr+L1;

o utfile<<"\n Power sent =" < <P s<<"KW";

Ty=((pr/Ps)*100);

outfile<<"\n The transmission Efficiency ="<<Ty<<"%";

Reg=((Vsm-Vr)/Vr)*100;

outfile<<"\n P ercentage Regulation ="<<Reg<<"%";

}



INPUT DATA

[“ST.DAT”]

1100

11

(0.8,-0.6)

(8,16)

0.8

8



OUTPUT

[“ST.OUT”]

PERFORMANCE OF SHORT TRANSMISSION LINE

Power delivered =1100KW

The receiving end voltage =11.00KV

F I =(0.80, -0.60)ohm

Total line impendance =(8.00, 16.00)ohm

Power factor =0.80 Lagging

The total resistance =8ohm

calculation for Ir,Vs,Trans.eff1,%Reg

Line current = 1 25 .00Amperes

Receiving End Current = (100.00, -75.00) Amperes

Sending End Voltage = (1.30e+04, 1000.00) Volt

The magnitude of sending end voltage =13.04

Line Losses =125.00KV

Power sent = 1. 23 e+0 3KW

The transmission Efficiency =89.80%

Pe rc entage Regulation =18.53%



RESULT

AIM

To find the optimum generation by considering with & without losses.

ALGORITHM


2. Get the cost inputs of generating units

3. Get the value of load demand and incremental cost transmission loss

coefficient

4. Calculate the optimum generator P1 & p2

5. P1=(λ-b1)/2a1; P2=(λ-b2)/2a2;

6. Calculate the toalpower generation P=p1+p2

7. Check whether total power generation is equal to demand

8. If the power & demand value not equal , then the incremental cost or

decremental cost value

9. If they are equal , then display P1 &P2

10. When there is Presence of loss , then get the Value of transmission loss and

co-efficient & incremental cost (λ)

11. Calculate PL

12. PL= P2B11 + P2B22+ P1P2B12

13. Calculate the total generation and the received power Ptg=P1+P2

14. Total power demand Pd=Pg-P1

EX. No: 05ECONOMIC DISPTACH WITH AND WITHOUT LOSSES

DATE:



15. Print all the values

16. Stop the Program



Problem

The fuel cost of the two units are given by

F1=0.05 P12+2 P1+100 Rs/Hr

F2=0.05 P22+1.5 P2+120 Rs/Hr

Without Losses

Determine the economic schedule for the incremental cost of received power is

Rs.220 /MWhr. Also find the total generation and demand. Also verify the

calculated result

With Losses

Determine the economic schedule for the incremental cost of received power is

Rs.220 /MWhr. Also find the total generation & losses and demand. Also verify thecalculated result

The Transmission co-effficient B11=0.0015, B12=-0.00050,B22=0.0025

Incremental cost e=10



PROGRAM

#include<conio.h>

#include<complex.h>


#include<fstream.h>

#include<stdio.h>

#include<math.h>

float l,m,i,j,n,td,x1,x2,y1,y2,z1,z2,e,pg,pd,p,p1,p2,b11,b12,b22;

void main()

{

ifstream infile;

infile.open("Economic.dat");

ofstream outfile;

outfile.open("Economic.out");



outfile.fill('0');

outfile<<"\n\t\t Calcultion of optimum generation with and without losses";

outfile<<"\n the fuel cost function-1 :";

infile>>x1>>y1>>z1;

outfile<<"\n f1="<<x1<<"p1(pow)2+"<<y1<<"p1+"<<z1;

o u tfile<<"\n the fuel cost function-2 :";

infile>>x2>>y2>>z2;

outfile<<"\n f2="<<x2<<"p2(pow)2+"<<y2<<"p2+"<<z2;

outfile<<"\n-----------------------------------------------------------";

outfile<<"\n Calculation of p1 & p2 with and without losses";

outfile<<"\n-----------------------------------------------------------";

infile>>td;

outfile<<"\n total demand ="<<td;

infile>>e;

outfile<<"\n the incremental cost ="<< e;

do

{

if(p>td)

e--;



else

e++;

p1=(e-y1)/(2*x1);

p2=(e-y2)/(2*x2);

p=p1+p2;

}while(p!=td);

outfile<<"\n optimum generation\n p1="<<p1<<"\n p2="<<p2;

outfile<<"\n the total demand =" <<td<<"\n incremental cost ="<<e;

outfile<<"\n-----------------------------------------------------------";

outfile<<"\n\t\t calculation of p1 & p2 with losses";

outfile<<"\n-----------------------------------------------------------";

infile>>b11>>b12>>b22;

outfile<<"\n The transmission coefficient:"<<b11<<b12<<b22;infile>>e;

outfile<<"\n the incremental cost:"<<e;

i=((2*x1)+(e*b11*2));

l=(2*x2)+(e*b22*2);

j=(2*e*b12);

m=e-y1;

n=e-y2;outfile<<"\n"<<i<<"p1"<<j<<"p2="<<m;

outfile<<"\n"<<j<<"p1+"<<i<<"p2="<<n;

p1=(((m*l)-(n*j))/((i*l)-(j*j)));

outfile<<"\n p1="<<p1;

p2=(((n*i)-(m*j))/((i*l)-(j*j)));

outfile<<"\n p2="<<p2;

pg=p1+p2;outfile<<"\n the total generation pg="<<pg<<"mw";

p1=((p1*p1*b11)+(p1*p2*b12*2)+(p2*p2*b22));

outfile<<"\n the transmission losses p1="<<p1<<"mw";

pd=pg-p1;

outfile<<"\n the demand pd="<<pd<<"mw";

}



INPUT

[“ECONOMIC.DAT”]

0.05 2 100

0.075 1.5 120

220

50

0.0015 -0.00050 0.0025

10



AIM

To develop a computer program to carryout simulation study of symmetrical three

phase short circuit on a given three phase system

ALGORITHM


2. Read line data, machine & Transformer data & fault impendence etc

3. Compute y-bus & calculate modified y-bus matrix

4. Compute y-bus & calcuate modified y-bus matrix

5. Form z-bus by inverting the modified y-bus matrix

6. Intialize count i=0

7. i=i+1 this means fault occurs at ‘I’th bus

8. compute fault current at faulted bus and bus voltage at all buses

9. compute fault current at faulted bus and bus volatges at all buses

10. compute all line currents & generator currents

11. check whether I it is less than the number of buses, go to step 6,

otherwise go to the next step

12. print the result

13. stop the program

EX. No: 06SYMMETRICAL SHORT CIRCUIT ANALYSIS

DATE:



PROGRAM

#include<stdio.h>

#include<complex.h>


#include<fstream.h>

#include<math.h>

#include<stdio.h>

#define NB 3+1

#define NL 4+1

int nl,nb,ng,nt,i,j,k,l,m,ln[NL],sb[NB],eb[NL],p,tag[NB];

complex y[NB][NB],z[NB][NB],chk[NB]

[NB],lz[NL],g_z[NB],t_z[NB],voa,lf,zf,v[NB],genr_cur,lc;

void cinver(complex[NB][NB],int nnn);

void main()

{

ifstream infile;

infile.open("sca.dat");

ofstream outfile;

outfile.open("sca.out");



outfile<<" symmetrical short circuit analysis \n";

infile>>nb>>nl>>ng>>nt>>zf;

outfile<<"\n NB \t NL \t NG \t NT \t ZF \n";

outfile<<nb<<"\t"<<nl<<"\t"<<ng<<"\t"<<nt<<"\t"<<zf;

outfile<<"\n line data\n";

outfile<<"LNo\tSB\t EB \t LINE IMP \n";

for(i=1;i<=nl;i++)

{

infile>>ln[i]>>sb[i]>>eb[i]>>lz[i];

outfile<<"\n"<<ln[i]<<"\t"<<sb[i]<<"\t"<<eb[i]<<"\t"<<lz[i];

l=sb[i];

m=eb[i];

y[l][l]+=1.0/lz[i];y[m][m]+=1.0/lz[i];

y[l][m]+=-1.0/lz[i];



y[m][l]=y[l][m];

}

outfile<<"\n Y BUS MATRIX \n";

for(i=1;i<=nb;i++)

{

outfile<<"\n";

for(j=1;j<=nb;j++)

outfile<<y[i][j]<<"\t";

}

outfile<<"\n BUS no \t gen_z \t tran_z \t tag \n";

for(i=1;i<=nb;i++)

{

infile>>g_z[i]>>t_z[i]>>tag[i];if(tag[i]==1)

{

outfile<<"\n"<<i<<"\t"<<g_z[i]<<"\t"<<t_z[i]<<"\t"<<tag[i];

y[i][i]+=1.0/(g_z[i]+t_z[i]);

}

}

outfile<<"\n modified bus matrix:\n";for(i=1;i<=nb;i++)

{

outfile<<"\n";

for(j=1;j<=nb;j++)

{

outfile<<y[i][j]<<"\t";

z[i][j]=y[i][j];}

}

cinver(z,nb);

outfile<<"\n zbus matrix";

for(i=1;i<=nb;i++)

{

outfile<<"\n";for(j=1;j<=nb;j++)

outfile<<z[i][j]<<"\t";

}



outfile<<"\n check matrix \n";

for(i=1;i<=nb;i++)

{

outfile<<"\n";

for(j=1;j<=nb;j++)

{

chk[i][j]=complex(0.0,0.0);

for(k=1;k<=nb;k++)

chk[i][j]+=y[i][k]*z[k][j];

outfile<<chk[i][j]<<"\t";

}

}

voa=complex(1.0,0.0);for(p=1;p<=nb;p++)

{

lf=voa/(z[p][p]+zf);

outfile<<"\n fault current due to the fault bus"<<p<<"is :"<<lf;

outfile<<"\n BUs voltage under faulted condition";

for(i=1;i<=nb;i++)

{v[i]=voa-(lf*z[i][p]);

outfile<<"\n voltage at bus"<<i<<" due to fault bus"<<p<<"is:"<<v[i];

}

outfile<<"\n line currents under faulted condition \n:";

outfile<< "LN \t SB \t EB \t LC \n";

for(i=1;i<=nl;i++)

{l=sb[i];

m=eb[i];

lc=(v[l]-v[m])/lz[i];

outfile<<"\n"<<i<<"\t"<<l<<"\t"<<m<<"\t"<<lc;

}

outfile<<" \n generator current under faulted condition: \n";

for(i=1;i<=nb;i++){

if(tag[i]==1)

{



genr_cur=(voa-v[i])/(g_z[i]+t_z[i]);

outfile<<"\n generator current due to fault at bus"<<p<<"is:"<<genr_cur;

}

}

}

}

void cinver(complex xxx[NB][NB],int n)

{

int m;

complex yyy;

m=n+1;

for(i=1;i<=n;i++)

{for(j=1;j<=n;j++)

xxx[j][m]=complex(0.0,0.0);

xxx[i][m]=complex(1.0,0.0);

yyy=xxx[i][i];

for(j=1;j<=m;j++)

xxx[i][j]=xxx[i][j]/yyy;

for(j=1;j<=n;j++){

if(i==j)goto abcd;

yyy=xxx[j][i];

for(k=1;k<=m;k++)

xxx[j][k]=xxx[j][k]-yyy*xxx[i][k];

abcd:;

}for(j=1;j<=n;j++)

xxx[j][i]=xxx[j][m];

}

}



Input

3 4 2 2 (0.1,0.05)

1 1 2 (0.0,0.05)

2 1 2 (0.0,0.05)

3 2 3 (0.0,0.06)

4 1 3 (0.0,0.1)

(0.0,0.25) (0.0,0.1) 1

(0.0,0.0) (0.0,0.0) 0

(0.0,0.2) (0.0,0.08) 1



OUTPUT

Symmetrical Short Circuit Analysis

NB NL NG NT ZF

3 4 2 2 (0.1000, 0.0500)

line data

LNo SB EB LINE IMP

1 1 2 (0.0000, 0.0500)

2 1 2 (0.0000, 0.0500)

3 2 3 (0.0000, 0.0600)

4 1 3 (0.0000, 0.1000)

Y BUS MATRIX

(0.0000, -50.0000) (0.0000, 40.0000) (0.0000, 10.0000)

(0.0000, 40.0000) (0.0000, -56.6667) (0.0000, 16.6667)

(0.0000, 10.0000) (0.0000, 16.6667) (0.0000, -26.6667)

BUS no gen_z t ran_z tag

1 (0.0000, 0.2500) (0.0000, 0.1000) 1

3 (0.0000, 0.2000) (0.0000, 0.0800) 1

modified bus matrix:

(0.0000, -52.8571) (0.0000, 40.0000) (0.0000, 10.0000)

(0.0000, 40.0000) (0.0000, -56.6667) (0.0000, 16.6667)

(0.0000, 10.0000) (0.0000, 16.6667) (0.0000, -30.2381)

zbus matrix

(0.0000, 0.1688) (0.0000, 0.1618) (0.0000, 0.1450)

(0.0000, 0.1618) (0.0000, 0.1761) (0.0000, 0.1506)

(0.0000, 0.1450) (0.0000, 0.1506) (0.0000, 0.1640)

check matrix

(1.0000, 0.0000) (-1.1102e-15, 0.0000) (-8.8818e-16, 0.0000)

(-1.7764e-15, 0.0000) (1.0000, 0.0000) (4.4409e-16, 0.0000)

(-8.8818e-16, 0.0000) (0.0000, 0.0000) (1.0000, 0.0000)

fault current due to the fault bus1is :(1.7283, -3.7810)

BUs voltage under faulted condition

voltage at bus1 due to fault bus1is:(0.3619, -0.2917)

voltage at bus2 due to fault bus1is:(0.3883, -0.2796)voltage at bus3 due to fault bus1is:(0.4518, -0.2506)

line currents under faulted condition



LN SB EB LC

1 1 2 (-0.2419, 0.5291)

2 1 2 (-0.2419, 0.5291)

3 2 3 (-0.4837, 1.0582)

4 1 3 (-0.4112, 0.8995)

generator current under faulted condition:

generator current due to fault at bus1is:(0.8334, -1.8232)






voltage at bus3 due to fault bus2is:(0.4430, -0.2463)line currents under faulted condition

LN SB EB LC

1 1 2 (0.4696, -1.0619)

2 1 2 (0.4696, -1.0619)

3 2 3 (-0.6965, 1.5751)

4 1 3 (-0.1831, 0.4141)

generator current under faulted condition:generator current due to fault at bus2is:(0.7561, -1.7097)






voltage at bus3 due to fault bus3is:(0.3710, -0.2939)line currents under faulted condition

LN SB EB LC

1 1 2 (0.2006, -0.4294)

2 1 2 (0.2006, -0.4294)

3 2 3 (0.4013, -0.8587)

4 1 3 (0.3411, -0.7299)

generator current under faulted condition:generator current due to fault at bus3is:(0.7423, -1.5887)




RESULT



AIM

To obtain the bus incidence and loop incidence matrix by using C++program.

ALGORITHM


2. Intilaize the variables

3. Enter the number of nodes, number of elements and a number of branches

4. Check whether there is connection or not. If there is no connection go to

loop and enter the direction and bus incidence mtrix is calculated.

5. Calculate the number of loops

6. Check whether is connection or not. If there is connection means go inside

the loop . if the direction is same then C[i][j]=1, else C[i][j]=-1

7. Loop incidence matrix is calculated

8. Stop the program

EX. No: 07 FORMATION OF BUS INCIDENCE AND LOOP INCIDENCEMATRIXDATE:



PROGRAM

#include<stdio.h>

#include<conio.h>

#include<fstream.h>


int e,n,b,l,m,r,s,lo,i,j;

float a[10][10],lp[10][10];

void main()

{

ifstream infile;

infile.open("loop.dat");

ofstream outfile;

outfile.open("loop.out");

outfile<<"\n\n/* formation of buses incidence & loop incidence matrix */";

infile>>e;

outfile<<"\n No of elements:"<<e;

infile>>n;

outfile<<"\n No of Nodes:"<<n;

infile>>b;

outfile<<"\n No of branches:"<<b;

for(i=1;i<=e;i++)

{

for(j=1;j<=n;j++)

{

infile>>l;

outfile<<"\n\n the elements"<<i<<"is connected to node"<<j<<"connected to

enter node"<<j<<"connected to enter 1 (or) Not connected enter 0:"<<l;

if(l==0)

a[i][j]=0;



else

{

infile>>m;

outfile<<"\n whether the direction is towards 1 (pr) outwards 0:"<<m;

if(m==1)

a[i][j]=-1;

else

a[i][j]=1;

}

}

}

outfile<<"\n\n Bus Incidence matrix:\n";

for(i=1;i<=e;i++)

{

for(j=1;j<=n;j++)

{

outfile<<"\t"<<a[i][j];

}

outfile<<"\n";

}

lo=e-b;

outfile<<"\n No of loops="<<lo;

for(i=1;i<=lo;i++)

{

for(j=1;j<=e;j++)

{

infile>>r;

outfile<<"\n\n The Loop"<<i<<"The element"<<j<<"is connection enter 1 or

not 0:"<<r;

if(r==0)



lp[i][j]=0;

else

{

infile>>s;

outfile<<"\n whether the direction is towards (or) outwards 0:"<<s;

if(s==1)

lp[i][j]=1;

else

lp[i][j]=-1;

}

}

}

outfile<<"\n \n The loop incidence matrix:\n";

for(i=1;i<=lo;i++)

{

for(j=1;j<=e;j++)

{

outfile<<"\n"<<lp[i][j];

}

outfile<<"\n";

}

}



INPUT

4 3 3

1 0

0

0

0

0

1 0

0

1 0

1 1

1 0

1 1

0

1 0

1 1

1 1

1 1



OUTPUT

/* formation of buses incidence & loop incidence matrix */

No of elements:4

No of Nodes:3

No of branches:3

the elements1is connected to node1connected to enter node1connected to enter

1 (or) Not connected enter 0:1

whether the direction is towards 1 (pr) outwards 0:0









the elements2is connected to node3connected to enter node3connected to enter1 (or) Not connected enter 0:1




















Bus Incidence matrix:

1 0 0

0 0 1

0 1 -1

1 -1 0

No of loops=1

The Loop1The element1is connection enter 1 or not 0:1

whether the direction is towards (or) outwards 0:0







AIM

To conduct the load flow study on the given power system using Fast decoupled

method

ALGORITHM

1. Read the slack bus voltage , real bus powers & reactive bus powers , bus

voltage magnitudes & reactive power limits

2. Form the Ybus matrix without line charging admittances & shunt admittances

3. Form B’ matrix from Ybus matrix obtained in step 2

4. Form Y bus matrix with double the line charging admittance

5. Form B’’ matrix from Y bus matrix obtained in step 4

6. Calculate the inverse of B’ and B’’ matrices

7. Calculate [ΔP / |V| ] & [ΔQ / |V| ]

8. If [ΔP / |V|] and [ΔQ / |V| ] are less than or equal to tolerance limit , solution

has converged and go to step 12 . Otherwise, increase iteration count and go

to step 10

9. Calculate [Δδ] =[B’]-1 [ΔP / |V| ] and [Δ|V|] = [B’’]-1[ΔQ / |V| ]

10. Update [δ], [|v|]

11. [δ}new=[δ]old+[Δδ] at all the buses except slack bus

12. [|V|]new

=|V|old

+Δ|V| at all PQ buses

13. Then go to step -8

14. Compute the slack bus power, line flows, real power loss, reactive power loss

etc.

EX. No: 08LOAD FLOW ANALYSIS USING FAST DECOUPLED METHOD

DATE:



One Line Diagram

Bus data

Line data



Outfile<<vsp[i]«" ";}Outfile<<"\nSpecified Real Bus Power\n";for(i=2;i<=nb; i++ ){Infile>>psp[i];Outfile<<psp[i]«" ";}outfile«"\nSpecified Reactive Bus Power\n";for(i=mb+ 1 ;i<=nb; i++){Infile>>qsp[i];Outfile<< qsp[i]«" ";}outfile< <"\n initial Angles\n";for(i=2;i<=nb; i++ ){infile> >deltanew[ i];

outfile<<deltanew[i]<<" ";}outfile«"\nB'Matrix\n";for(i=2;i<=nb; i++ ){l=i- l;outfile<<"\n";for(j=2;j<= nb; j++ ){m=j- l;b[l][ m ]= - imag (y[i] [j]);

bl [l][m]=b[l][m];outfile<<b[l][m]<<" ";}}float_inverse (b1, nb-l);outfile<<"\n inverse of B' Matrix\n";for(i=1 ;i<=nb-l ;i++){outfile< <"\n ";for(j=1 ;j<=nb-l ;j++)

outfile<<bl[i][j]<<" ";}Outfile<<"\nCheck Matrix\n";for(i= 1 ;i<= nb-l ; i++){outfile< <"\n";for(j= 1 ;j<=nb-l ;j++){chk[i] [j]=O.O;for(k= 1 ;k<=nb-l ;k++ )chk[i] [j]+= b[i] [k] *b1 [k] [j];

outfile<<chk[i][j]<<" ";}}Outfile<<"\nB" Matrix \n";for(i=mb+ 1 ;i<=nb; i++)



{outfile<<"\n";for(j= 1 ;j<= nb-mb; j+-+)outfile<<b3[i][j]<<" ";}outfile<<"\nCheck Matrix\n";for(i= 1 ;i<=nb-mb;i++){Outfile<<\n";17

For(j = 1 ;j<=nb-mb; j++){chk[i] [j]=0.0;for(k= 1 ;k<=nb-mb; k++)chk[i] [j]+=b2[i] [k] *b3 [k][j];outfile<<chk[i][j]<<" ";}}

//intialization of bus voltagesfor(i= 1 ;i<=mb; i++)vnew[i ]=complex( vsp [i], 0. 0);for(i=mb+ 1 ;i<=nb; i++)vnew[i]=complex( 1.0,0.0);iter=0;ll: ;iter+= 1;outfile< <"\niter: "<<iter;for(i= 1 ;i<=nb: i++){

vold[i]=vnew[i];deltaold[i]=deltanew[i] ;}outfile< <"\nDelP values";for(i=2;i<=nb; i++ ){il=i-l;sum=complex(0.0,0.0);for(j= 1 ;j<=nb; j++)sum+=y[i] [j] *vold[j];

pcal[i]=real( conj(vold[i])*sum);delp[il]= (psp[i]-pcal[i])/abs(vold[i]);outfile<<"\n"<<delp[i1];}Outfile<<"\nDelQ values\n";for(i=mb+ 1 ;i<=nb; i++){il =i-mb;sum=complex(0.0,0.0);for(j= 1 ;j<=nb; j++)sum+=y(i] [j] *vold[j];

qcal (i ]=- image (sum * conj( vold[ i]));delq(i1]=(qsp(i]-qcal[i])/abs(vold[i]);outfile<<"\n"<<delq(i1];}for(i= 1 ;i<=nb-l ;i++)



{g=fabs( delp[i]);if(g>tol) goto jj;}for(i=mb+ 1 ;i<=nb; i++){g=fabs( delq[i]);if(g>tol) goto jj;}goto hh; jj:;outfile< <"\nDelDelta \n";for(i=2;i<=nb; i++ ){il= i-l;deldelta[i]=0.0;for(j=2;j<=nb;j++){

j1 = j - 1;deldelta[i]+=b1[i1] [j1] *delp[j1];}outfile< <"\n "< <deldelta[i];}Outfile<<"\nDEL V\n";for(i=mb+ 1 ;i<=nb;i++){i1 =i-mb;delv(i]=0.0;for(j=2;j<=nb; j++ )

{ j1 = j-mb;delv[i]+=b3 [i1] [j1] *delq[j1];}outfile< <"\n "< <delv(i];}Outfile<<"\nDelta values (new)\n";for(i=2;i<=nb; i++ ){deltanew[ i ]=deltaold[i]+deldelta[ i];

outfile< <"\n "< <deltanew[i];}Outfile<<\nvoltage values(new)\n";for(i=2;i<=nb;i++ ){vnewl =abs(vold(i))+delv[i];vnew[i]=polar( vnew1 ,deltanew[i]);outfile< <"\n "< <vnew[i];}goto 11;hh:;

outfile<<"\nSolution is obtained at: "<<iter<<"-th iteration\n";outfile< <"\nBusNo\tV new(mag)\t Vnew angle(radians )\tReal power\tReactivePower\n";for(i= 1;i<=nb;i++){



sum=complex(0.0,0.0);for(j=1 ;j<=nb;j++)sum+=conj( vnew[i])*y[i] [j] *vnew[j];p[i]=real(sum);q[i]=-imag(sum);outfile< <"\n" < <i < <"\t" < <abs( vnew[i])< <"\t\t" < <deltanew[i]< <"\t\t\t" <<p[i]«"\t\t"«q[i] ;}outfile< <"\nPl oss\tQ loss\n";ploss=0.0;qloss=0.0;for(i=1 ;i<=nb; i++){ploss+=p[i];qloss+=q[i];}outfile<<ploss<<"\t"<<qloss;

outfile<<"\nLine flows";outfile<<"\nLNo\tSB\tEB\tReal power\tReactive Power\n";for(i=l;i<=nl;i++){ j=sb[i];k=eb[i];ctemp=vnew[j]-vnew[k] ;ctemp=ctemp*( -y[j] [k])+(hly[i]*vnew[j]);ctemp=ctemp * conj (vnew[j]);temp 1 =real( ctemp);temp2=-imag( ctemp);

outfile< <"\n"< <i< <"\t"< <j< <"\t"< <k< <"\t"< <temp1 < <"\t\t"< <temp2;}}void float_ inverse(float x[NB] [NB],int n){int m;float y;m=n+ 1 ;for(i=1 ;i<=n;i++){

for(j= 1 ;j<=n;j++)x[j] [m]=0.0;x[i][m]=1.0;y=x[i][i];for(j= 1 ;j<=m; j++)x[i] [j]=x[i][j]/y;for(j= 1 ;j<=n; j++){if(i= = j)goto a;y=x[j] [i];for(k=1 ;k<=m;k++)

x[j] [k]=x[j] [k ]-y*x[i] [k];a:;}for(j=1;j<=n; j++)x[j][i]=x[j][m];



}}

RESULT



AIM

To carry out the load flow analysis of the given power system by gauss seidel

method

ALGORITHM

1. Read the line data , specified power & voltage

2. Compute Y-bus matrix

3. Intialize all bus voltages

4. Iteration=1

5. Consider i=2

6. Check the bus is PV or PQ. If PQ go to step 8 , otherwise go to next step

7. Compute Qi and check Q limits

8. Calculate the new value of bus voltage

9. If all the buses are considered go to step 10, otherwise go to step 6

10. Check for the convergence , if no go to step 11, else step 12

11. Update the bus voltage with acceleration factor α=1.4 & iteration

=iteration+1

12. Calculate slack bus power Q at PV buses, real & reactive power flow &

line losses

13. Print all the result

14. Stop the program

Ex. No: 09GAUSS SEIDEL LOAD FLOW ANALYSIS

DATE:



ONE LINE DIAGRAM

Acceleration factor=1.2 AND TOLERANCE=0.0001

BUS SPECIFICATION

LINE DATA

BU

S

BUS

TYPEVSPECIFIED

GENERATING LOAD IN Qmi

n

Qma

xP Q P Q1 SLACK 1.06 - - - - - -2 P-Q 1.00 0.6 - 0.0 0.0 -1.0 1.53 P-V - - - 0.45 0.15 - -4 P-V - - - 0.40 0.05 - -5 P-V - - - 0.60 0.1 - -

LINE No STARTING

BUS

ENDING

BUS

SERIES IMPEDANCE HALF LINE CHARGING

ADMITTANCE (P.U)

1 1 2 (0.02,0.06) (0.0,0.03)

2 1 3 (0.08,0.24) (0.0,0.025)

3 2 3 (0.06,0.18) (0.0,0.02)

4 2 4 (0.02,0.08) (0.0,0.02)

5 2 5 (0.04,0.12) (0.0,0.015)

6 3 4 (0.01,0.03) (0.0,0.01)

7 4 5 (0.08,0.24) (0.0,0.035)



outfile<<"\t\t\t Program For Gauss Seidal Method \n";

outfile<<"\n NB \t NL \t MB \t ALPHA \t TOL \n";

infile>>nb>>nl>>mb>>alpha>>tol;

outfile<<nb<<"\t"<<nl<<"\t"<<mb<<"\t"<<alpha<<"\t"<<tol<<"\n";

outfile<<"\t Line No. \t SB \t Eb \t LINE_Z \t Halfline_y\n\n";

for(i=1;i<=nl;i++)

{

infile>>line[i]>>sb[i]>>eb[i]>>line_z[i]>>halfline_y[i];

outfile<<line[i]<<"\t"<<sb[i]<<"\t"<<eb[i]<<"\t"<<line_z[i]<<"\t"<<halfline_y[i]<<"\n";

l=sb[i];

m=eb[i];

y_bus[l][l]+=(1.0/line_z[i])+halfline_y[i];

y_bus[m][m]+=(1.0/line_z[i])+halfline_y[i];

y_bus[l][m]+=-1.0/line_z[i];

y_bus[m][l] =y_bus[l][m];

}

outfile<<"\t\t\t YBus Matrix \n";

for(i=1;i<=nb;i++)

{

outfile<<"\n";

for(j=1;j<=nb;j++)

{

outfile<<y_bus[i][j]<<"\t";

}



}

outfile<<"\n Specified Voltages \n";

for(i=1;i<=mb;i++)

{

infile>>vsp[i];

outfile<<vsp[i]<<"\t";

}

outfile<<"\n Specified Real Bus Power \n";

for(i=2;i<=nb;i++)

{

infile>>psp[i];

outfile<<psp[i]<<"\t";

}

outfile<<"\n Specified Reactive Bus Power \n";

for(i=mb+1;i<=nb;i++)

{

infile>>qsp[i];

outfile<<qsp[i]<<"\t";

}

outfile<<"\n Local Load \n";

for(i=2;i<=mb;i++)

{

infile>>ql[i];

outfile<<ql[i]<<"\t";

}



vold[i]=complex(1.0,0.0);

vnew[i]=vold[i];

}

iter=0;

ll:;

iter+=1;

outfile<<"\n ITER \t"<<iter<<"\n";

outfile<<"\n GENERATOR REACTIVE POWER \n";

for(i=2;i<=nb;i++)

{

if(i>mb) goto dd;

sum=complex(0.0,0.0);

for(j=1;j<=nb;j++)

sum+=y_bus[i][j]*vold[j];

qbus[i]=-imag(conj(vold[i])*sum);

qg[i]=qbus[i]+ql[i];

outfile<<qg[i]<<"\n";

if(qg[i]<qmin[i]) goto bb;

if(qg[i]>qmax[i]) goto cc;

qsp[i]=qbus[i];

goto dd;

bb:;

qsp[i]=qmin[i]-ql[i];

tag[i]=1.0;

goto dd;



cc:;

qsp[i]=qmax[i]-ql[i];

tag[i]=1.0;

dd:;

a=complex(psp[i],-qsp[i])/conj(vold[i]);

b=complex(0.0,0.0);

for(j=1;j<=i-1;j++)

b+=y_bus[i][j]*vnew[j];

c=complex(0.0,0.0);

for(j=i+1;j<=nb;j++)

c+=y_bus[i][j]*vold[j];

vnew[i]=(a-b-c)/y_bus[i][i];

}

for(i=2;i<=mb;i++)

{

if(tag[i]==1) goto gg;

vnew[i]=vnew[i]*vsp[i]/abs(vnew[i]);

gg:;

}

outfile<<"\n DEL V VALUES \n";

for(i=2;i<=nb;i++)

{

xx=abs(vnew[i]-vold[i]);

outfile<<xx<<"\n";

}



for(i=2;i<=nb;i++)

{

xx=abs(vnew[i]-vold[i]);

if(xx>tol) goto ff;

}

goto hh;

ff:;

for(i=2;i<=nb;i++)

{

vnew[i]=vold[i]+(alpha*(vnew[i]-vold[i]));

vold[i]=vnew[i];

}

for(i=2;i<=mb;i++)

tag[i]=0.0;

goto ll;

hh:;

outfile<<"\n Solution Is Obtained At \t"<<iter<<"_th"<<"\t Iteration \n";

outfile <<"\n VOLTAGE AT ALL THE BUSES \n";

for(i=1;i<=nb;i++)

outfile<<vnew[i]<<"\n";

outfile<<"\n Voltage Magnitude At All Buses \n";

for(i=1;i<=nb;i++)

{

vabs[i]=abs(vnew[i]);

outfile<<vabs[i]<<"\n";



}

outfile<<"\n Delta Values In Degrees \n";

for(i=1;i<=nb;i++)

{

delta[i]=atan(imag(vnew[i])/real(vnew[i]))*180/3.14;

outfile<<delta[i]<<"\n";

}

outfile<<"\n Bus No Real \t Reactive \n";

for(i=1;i<=nb;i++)

{

sum=complex(0.0,0.0);

for(j=1;j<=nb;j++)

sum+=conj(vnew[i])*y_bus[i][j]*vnew[j];

p[i]=real(sum);

q[i]=-imag(sum);

outfile<<i<<"\t"<<p[i]<<"\t"<<q[i]<<"\n";

}

outfile<<"\n Ploss \t Qloss \n";

ploss=0.0;

qloss=0.0;

for(i=1;i<=nb;i++)

{

ploss+=p[i];

qloss+=q[i];

}



outfile<<ploss<<"\t"<<qloss<<"\n";

outfile<<"\n Line Flows \n";

outfile<<"\n"<<"Line\t Sb \tEb\t Real Reactive \n";

outfile<<"No\t\t\t Power\t Power\n";

for(i=1;i<=nl;i++)

{

j=sb[i];

k=eb[i];

ctemp=vnew[j]-vnew[k];

ctemp=ctemp*(-(y_bus[j][k]))+(halfline_y[i]*vnew[j]);

ctemp=ctemp*conj(vnew[j]);

temp1=real(ctemp);

temp2=-imag(ctemp);

outfile<<i<<"\t"<<sb[i]<<"\t"<<eb[i]<<"\t"<<temp1<<"\t"<<temp2<<"\n";

}

}



INPUT

5 7 2 1.2 0.0001

1 1 2 (0.02,0.06) (0.0,0.03)

2 1 3 (0.08,0.24) (0.0,0.025)

3 2 3 (0.06,0.18) (0.0,0.02)

4 2 4 (0.02,0.08) (0.0,0.02)

5 2 5 (0.04,0.12) (0.0,0.015)

6 3 4 (0.01,0.03) (0.0,0.01)

7 4 5 (0.08,0.24) (0.0,0.035)

1.06 1.0

0.6 -0.45 -0.4 -0.6

-0.15 -0.05 -0.1

0.1

-1.0 1.5

0.0



OUTPUT

Program for Gauss Seidel Method

NB NL MB ALPHA TOL5 7 2 1.2000 1.0000e-04

Line No. SB Eb LINE_Z Halfline_y

1 1 2 (0.0200, 0.0600) (0.0000, 0.0300)2 1 3 (0.0800, 0.2400) (0.0000, 0.0250)3 2 3 (0.0600, 0.1800) (0.0000, 0.0200)4 2 4 (0.0200, 0.0800) (0.0000, 0.0200)5 2 5 (0.0400, 0.1200) (0.0000, 0.0150)6 3 4 (0.0100, 0.0300) (0.0000, 0.0100)7 4 5 (0.0800, 0.2400) (0.0000, 0.0350)

YBus Matrix

(6.2500, -18.6950) (-5.0000, 15.0000) (-1.2500, 3.7500) (0.0000, 0.0000) (0.0000,

0.0000)(-5.0000, 15.0000) (12.1078, -39.1797) (-1.6667, 5.0000) (-2.9412, 11.7647) (-2.5000, 7.5000)(-1.2500, 3.7500) (-1.6667, 5.0000) (12.9167, -38.6950) (-10.0000, 30.0000)

(0.0000, 0.0000)(0.0000, 0.0000) (-2.9412, 11.7647) (-10.0000, 30.0000) (14.1912, -45.4497)

(-1.2500, 3.7500)(0.0000, 0.0000) (-2.5000, 7.5000) (0.0000, 0.0000) (-1.2500, 3.7500) (3.7500,-11.2000)Specified Voltages

1.0600 1.0000

Specified Real Bus Power0.6000 -0.4500 -0.4000 -0.6000Specified Reactive Bus Power

-0.1500 -0.0500 -0.1000Local Load

0.1000Q_Min

-1.0000Q_Max

1.5000TAG

0.0000

ITER 1

GENERATOR REACTIVE POWER-0.8850

DEL V VALUES0.02080.00710.00750.0386

ITER 2


DEL V VALUES



0.01570.00710.01110.0066

ITER 3

GENERATOR REACTIVE POWER

-0.9355

DEL V VALUES0.00270.00970.00720.0062

ITER 4


DEL V VALUES0.00490.00550.00550.0046

ITER 5


DEL V VALUES0.00310.00450.00410.0034

ITER 6


DEL V VALUES0.00240.00330.00300.0025

ITER 7


DEL V VALUES

0.00170.00240.00220.0018



ITER 8


DEL V VALUES0.00130.0018

0.00160.0013

ITER 9


DEL V VALUES0.00090.00130.0012

0.0010

ITER 10


DEL V VALUES0.00070.00100.00090.0007

ITER 11


DEL V VALUES0.00050.00070.00060.0005

ITER 12


DEL V VALUES0.00040.00050.00050.0004

ITER 13




DEL V VALUES0.00030.00040.00030.0003

ITER 14


DEL V VALUES0.00020.00030.00020.0002

ITER 15

GENERATOR REACTIVE POWER

-0.7701

DEL V VALUES0.00010.00020.00020.0001

ITER 16


DEL V VALUES0.00010.00010.00010.0001

ITER 17


DEL V VALUES7.3310e-050.00019.6077e-057.8040e-05

ITER 18


DEL V VALUES5.3362e-057.6210e-056.9938e-055.6806e-05



Solution Is Obtained At 18_th Iteration

VOLTAGE AT ALL THE BUSES(1.0600, 0.0000)(0.9999, -0.0170)(0.9919, -0.0537)(0.9902, -0.0544)

(0.9723, -0.0755)

Voltage Magnitude At All Buses1.06001.00000.99340.99170.9752

Delta Values In Degrees0.0000-0.9768

-3.0990-3.1436-4.4436

Bus No Real Reactive1 0.8932 1.00372 0.6018 -0.86933 -0.4480 -0.15094 -0.3999 -0.05025 -0.6001 -0.1001

Ploss Qloss0.0471 -0.1669

Line Flows

Line Sb Eb Real ReactiveNo Power Power1 1 2 0.5897 0.83232 1 3 0.3036 0.17143 2 3 0.1960 -0.04484 2 4 0.4674 -0.02415 2 5 0.5085 0.0370

6 3 4 0.0398 0.03307 4 5 0.1030 0.0004

lab manual-ps lab

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