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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
_________________________
8/8/2019 Lab Manual-ps Lab
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INTERNAL EXAMINER EXTERNAL
EXAMINER
8/8/2019 Lab Manual-ps Lab
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S.N
oDATE NAME OF THE EXPERIMENT
PAG
E No
MARKS
AWARDE
D
SIGNATURE
8/8/2019 Lab Manual-ps Lab
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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:
8/8/2019 Lab Manual-ps Lab
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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
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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";}
}
}
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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)
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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
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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:
8/8/2019 Lab Manual-ps Lab
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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
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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.precision(4);
outfile.setf(ios::showpoint);
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++)
{
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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];
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}
}
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];
}
}
8/8/2019 Lab Manual-ps Lab
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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)
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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
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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:
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7. Stop the program
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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
8/8/2019 Lab Manual-ps Lab
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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.precision(4);
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;
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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;
}
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Input data
[“Load.DAT”]
2000
1000
52.4
60
0.01
19000 41000
2000 40000
5 5
1000
60
1 1
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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
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AIM
To find the performance parameters of short transmission line
ALGORITHM
1. Start the program
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:
8/8/2019 Lab Manual-ps Lab
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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.precision(2);
outfile.setf(ios::showpoint);
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;
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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<<"%";
}
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INPUT DATA
[“ST.DAT”]
1100
11
(0.8,-0.6)
(8,16)
0.8
8
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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%
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RESULT
AIM
To find the optimum generation by considering with & without losses.
ALGORITHM
1. Start the program
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:
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15. Print all the values
16. Stop the Program
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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
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PROGRAM
#include<conio.h>
#include<complex.h>
#include<iostream.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.precision(4);
outfile.setf(ios::showpoint);
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--;
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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";
}
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INPUT
[“ECONOMIC.DAT”]
0.05 2 100
0.075 1.5 120
220
50
0.0015 -0.00050 0.0025
10
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AIM
To develop a computer program to carryout simulation study of symmetrical three
phase short circuit on a given three phase system
ALGORITHM
1. Start the program
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:
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PROGRAM
#include<stdio.h>
#include<complex.h>
#include<iostream.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.precision(4);
outfile.setf(ios::showpoint);
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];
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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";
}
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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)
{
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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];
}
}
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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
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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
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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)
generator current due to fault at bus1is:(0.8949, -1.9578)
fault current due to the fault bus2is :(1.6357, -3.6989)
BUs voltage under faulted condition
voltage at bus1 due to fault bus2is:(0.4016, -0.2646)
voltage at bus2 due to fault bus2is:(0.3485, -0.2881)
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)
generator current due to fault at bus2is:(0.8797, -1.9892)
fault current due to the fault bus3is :(1.7920, -3.8352)
BUs voltage under faulted condition
voltage at bus1 due to fault bus3is:(0.4440, -0.2598)
voltage at bus2 due to fault bus3is:(0.4225, -0.2698)
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)
generator current due to fault at bus3is:(1.0497, -2.2465)
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RESULT
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AIM
To obtain the bus incidence and loop incidence matrix by using C++program.
ALGORITHM
1. Start the program
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:
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PROGRAM
#include<stdio.h>
#include<conio.h>
#include<fstream.h>
#include<iostream.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;
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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)
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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";
}
}
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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
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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 elements1is connected to node2connected to enter node2connected to enter
1 (or) Not connected enter 0:0
the elements1is connected to node3connected to enter node3connected to enter
1 (or) Not connected enter 0:0
the elements2is connected to node1connected to enter node1connected to enter
1 (or) Not connected enter 0:0
the elements2is connected to node2connected to enter node2connected to enter
1 (or) Not connected enter 0:0
the elements2is connected to node3connected to enter node3connected to enter1 (or) Not connected enter 0:1
whether the direction is towards 1 (pr) outwards 0:0
the elements3is connected to node1connected to enter node1connected to enter
1 (or) Not connected enter 0:0
the elements3is connected to node2connected to enter node2connected to enter
1 (or) Not connected enter 0:1
whether the direction is towards 1 (pr) outwards 0:0
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the elements3is connected to node3connected to enter node3connected to enter
1 (or) Not connected enter 0:1
whether the direction is towards 1 (pr) outwards 0:1
the elements4is 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 elements4is connected to node2connected to enter node2connected to enter
1 (or) Not connected enter 0:1
whether the direction is towards 1 (pr) outwards 0:1
the elements4is connected to node3connected to enter node3connected to enter
1 (or) Not connected enter 0:0
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
The Loop1The element2is connection enter 1 or not 0:1
whether the direction is towards (or) outwards 0:1
The Loop1The element3is connection enter 1 or not 0:1
whether the direction is towards (or) outwards 0:1
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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:
8/8/2019 Lab Manual-ps Lab
http://slidepdf.com/reader/full/lab-manual-ps-lab 55/80
One Line Diagram
Bus data
Line data
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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++)
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{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++)
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{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++){
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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];
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}}
RESULT
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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:
8/8/2019 Lab Manual-ps Lab
http://slidepdf.com/reader/full/lab-manual-ps-lab 64/80
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)
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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";
}
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}
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";
}
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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;
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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";
}
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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";
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}
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];
}
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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";
}
}
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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
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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
GENERATOR REACTIVE POWER-1.1861
DEL V VALUES
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0.01570.00710.01110.0066
ITER 3
GENERATOR REACTIVE POWER
-0.9355
DEL V VALUES0.00270.00970.00720.0062
ITER 4
GENERATOR REACTIVE POWER-0.8987
DEL V VALUES0.00490.00550.00550.0046
ITER 5
GENERATOR REACTIVE POWER-0.8303
DEL V VALUES0.00310.00450.00410.0034
ITER 6
GENERATOR REACTIVE POWER-0.8066
DEL V VALUES0.00240.00330.00300.0025
ITER 7
GENERATOR REACTIVE POWER-0.7910
DEL V VALUES
0.00170.00240.00220.0018
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ITER 8
GENERATOR REACTIVE POWER-0.7831
DEL V VALUES0.00130.0018
0.00160.0013
ITER 9
GENERATOR REACTIVE POWER-0.7782
DEL V VALUES0.00090.00130.0012
0.0010
ITER 10
GENERATOR REACTIVE POWER-0.7752
DEL V VALUES0.00070.00100.00090.0007
ITER 11
GENERATOR REACTIVE POWER-0.7733
DEL V VALUES0.00050.00070.00060.0005
ITER 12
GENERATOR REACTIVE POWER-0.7721
DEL V VALUES0.00040.00050.00050.0004
ITER 13
GENERATOR REACTIVE POWER-0.7712
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DEL V VALUES0.00030.00040.00030.0003
ITER 14
GENERATOR REACTIVE POWER-0.7706
DEL V VALUES0.00020.00030.00020.0002
ITER 15
GENERATOR REACTIVE POWER
-0.7701
DEL V VALUES0.00010.00020.00020.0001
ITER 16
GENERATOR REACTIVE POWER-0.7698
DEL V VALUES0.00010.00010.00010.0001
ITER 17
GENERATOR REACTIVE POWER-0.7696
DEL V VALUES7.3310e-050.00019.6077e-057.8040e-05
ITER 18
GENERATOR REACTIVE POWER-0.7694
DEL V VALUES5.3362e-057.6210e-056.9938e-055.6806e-05
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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
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