Power System Simulation Lab
2022-23
VISVESVARAYA TECHNOLOGICAL UNIVERSITY, BELGAUM.
VII SEMESTER B.E. (EEE)
POWER SYSTEM SIMULATION LAB (18EEL76)
Power system simulation using MATLAB/C or C++ / Sci lab / octave
1. a) Y Bus formation for power systems with and without mutual coupling, by singular
transformation and inspection method.
b) Determination of bus currents, bus power and line flow for a specified system voltage (Bus)
profile.
2. Formation of Z-bus (without mutual coupling), using Z-bus building algorithm.
3. ABCD parameters:Formation for symmetric  /  configuration.Verification of AD – BC =1.
Determination of efficiency and regulation.
4. Determination of power angle diagrams, reluctance power, excitation emf and regulation for
salient and non–salient pole synchronous machines.
5. To obtain Swing curve and to determine critical clearing time and regulation for a single
machine connected to infinity bus through a pair of identical transmission lines under 3phase fault on one of the lines for variation of inertia constant / line parameters / fault location
/ clearing time / pre-fault electrical output.
6. Formation of Jacobian for a system not exceeding 4 buses (no PV buses) in polar coordinates.
7. Write a program to perform load flow analysis using Gauss-seidel method (only PQ bus).
8. To determine fault currents and voltages in a single transmission line system with star-delta
transformers at a specified location for LG, LLG.
9. Load flow analysis using Gauss Seidel method, NR method, Fast decoupled method for both
PQ and PV buses.
10. Optimal Generation scheduling for Thermal power plants.
Note: Questions 1–7: Simulation Experiments using MATLAB/C or C++/Scilab/Octave
Questions 8–10: Use suitable standard software package.
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Power System Simulation Lab
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CYCLE-I: (MATLAB/C EXPERIMENTS)
1. a) Determination of ABCD parameters, efficiency and regulation of a medium
transmission line using Nominal  method. Verification of AD – BC = 1.
b) Determination of ABCD parameters, efficiency and regulation of a medium
transmission line using Nominal  method. Verification of AD – BC = 1.
2. Determination of power angle diagrams, reluctance power, excitation emf and
regulation for non–salient pole synchronous machines and salient pole synchronous
machines
3. a)Formation of YBus for power systems without mutual coupling, by inspection method.
b)Formation of YBus for power systems without mutual coupling, by singular
transformation method.
c)Formation of Y Bus for power systems with mutual coupling, by singular
transformation method.
d)Determination of bus currents, bus power and line flow for a specified system voltage
(Bus) profile.
4. Determination of load flow solution using Gauss-seidel method (only PQ bus).
5. Formation of Jacobian matrix for a power system not exceeding 4 buses (no PV buses) in
polar coordinates.
6. Formation of Z-bus (without mutual coupling), using Z-bus building algorithm.
7. Determination of Swing curve and to determine critical clearing time and regulation for
a single machine connected to infinity bus through a pair of identical transmission lines
under 3-phase fault on one of the lines for variation of inertia constant / line parameters
/ fault location / clearing time / pre-fault electrical output.
CYCLE-II: (Mi-power Experiments)
8. a) Load flow analysis using Gauss Seidel (GS) method for both PQ and PV buses.
b) Load flow analysis using Newton Raphson (NR) method for both PQ and PV buses.
c) Load flow analysis using Fast decoupled method for both PQ and PV buses.
9. Determination of fault currents and voltages in single transmission line systems with stardelta transformers at a specified location for LG, LL and LLG faults.
10. Optimal Generation scheduling for Thermal power plants by using B-coefficients.
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Power System Simulation Lab
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Expt. No.:
Date:
A) PERFORMANCE OF MEDIUM TRANSMISSION LINES
(NOMINAL  METHOD)
AIM:  To determine the ABCD constants of given medium transmission line using nominal 
method.
 To verify AD – BC = 1.
 To determine the efficiency and regulation of the given medium transmission line with the
help of receiving end data.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULAE USED1. The ABCD constants of the medium transmission lines (nominal ) are,
 Y  Z  and D A .
 YZ BZ,
C  Y  1 
A  1
,
4 
2



Where, Z = Total series impedance of the line in = Length in km * series impedance in  / km.
Z = (R + j X) , where R and X are total series resistance and reactance of the line in .
Y = Total shunt admittance of the line in ℧= Length in km * shunt admittance in ℧/ km.
Y = (G + j B)℧, where G and B are total shunt conductance and susceptance in℧.
A) DETERMINATION OF PERFORMANCE USING RECEIVING END DATA (Vr, Ir and cos r):2. Sending end voltage Vs = A * Vr + B * IrVolts
3. Sending end current Is = C * Vr + D * IrAmpere
4. Receiving end power Sr = 3 * Vr * Ir in VA.
Receiving end power Pr = real (Sr) Watts.
5. Sending end power Ss = 3 * Vs * Is in VA.
Sending end power Ps = real (Ss) Watts.
6. % Efficiency =  Receiving end power Pr 
100 %
 Sending end power P 
s 

V
s
7. % Regulation = A  Vr
Vr
100 %
where, Vr, Ir = Receiving end voltage and current of the line in Volts and Amperes respectively.
Cos r = Receiving end power factor (lagging or leading).
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Take Vr as reference phasor. Vr = Vr0 volts.
Ir = Ir - r in case of lagging power factor, Ir = Ir r in case of leading power factor.
B) DETERMINATION OF PERFORMANCE USING SENDING END DATA (Vs and Is):8. Receiving end voltage Vr = A * Vs – B * Is Volts
9. Sending end current Ir = – C * Vs + D * Is Ampere
10. Receiving end power Sr = 3 * Vr * Ir in VA.
Receiving end power Pr = real (Sr) Watts.
11. Sending end power Ss = 3 * Vs * Is in VA.
Sending end power Ps = real (Ss) Watts.
12. % Efficiency =  Receiving end power Pr 
100 %
 Sending end power P 
s 

V
s
13. % Regulation = A  Vr
Vr
100 %
Where, Vs, Is = Receiving end voltage and current of the line in Volts and Amperes respectively.
PROGRAM:A) DETERMINATION OF PERFORMANCE USING RECEIVING END DATA (Vr, Ir and cos r):clc
clear all
length=input('Enter the length of medium line in km= ');
z=input('\nEnter the series impedance of the line per km= ');
y=input('\nEnter the shunt admittance of the line per km= ');
Z=z*length; Y=y*length;
a=1+(Y*Z)/2; d=a; b=Z; c=Y*(1+(Y*Z)/4);
fprintf('\nA,B,C&D contants are:\n');
fprintf('\nA=%15.4f%+15.4fi',real(a),imag(a));
fprintf('\nB=%15.4f%+15.4fi',real(b),imag(b));
fprintf('\nC=%15.4f%+15.4fi',real(c),imag(c));
fprintf('\nD=%15.4f%+15.4fi',real(d),imag(d));
fprintf('\n AD-BC=%f',a*d-b*c);
Vr=input ('\nEnter the receiving end line voltage in kV=');
Ir=input('Enter the receiving end current per phase in A=');
PF=input('Enter the receiving end power factor =');
pf_a=acos(PF);
Vr1=Vr*1e3/sqrt(3);
Ir_p=Ir*cos(pf_a)-i*Ir*sin(pf_a);
Vs=(a*Vr1+b*Ir_p)/1e3;
Is=c*Vr1+d*Ir_p;
fprintf('\nSending end voltage/ph =%15.4f%+15.4fi kV',real(Vs),imag(Vs));
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fprintf('\nSending end current/ph =%15.4f%+15.4fi A',real(Is),imag(Is));
Rec_pow=real(3*Vr1*Ir_p/1e6);
Send_pow=real(3*Vs*Is/1e3);
Efficiency=(Rec_pow/Send_pow)*100;
Regulation=(abs(Vs/a)-abs(Vr))*100/abs(Vr);
fprintf('\nReceiving end power =%0.2f MW',Rec_pow);
fprintf('\nSending end power =%0.2f MW',Send_pow);
fprintf('\nEfficiency=%0.2f%%',Efficiency);
fprintf('\nRegulation=%0.2f%%',Regulation);
B) DETERMINATION OF PERFORMANCE USING SENDING END DATA (Vs and Is):clc
clear all
length=input('Enter the length of short line in km= ');
z=input('\nEnter the series impedance of the line per km= ');
y=input('\nEnter the shunt admittance of the line per km= ');
Z=z*length; Y=y*length;
a=1+(Y*Z)/2; d=a; b=Z; c=Y*(1+(Y*Z)/4);
fprintf('\nA,B,C&D contants are:\n');
fprintf('\nA=%15.4f%+15.4fi',real(a),imag(a));
fprintf('\nB=%15.4f%+15.4fi',real(b),imag(b));
fprintf('\nC=%15.4f%+15.4fi',real(c),imag(c));
fprintf('\nD=%15.4f%+15.4fi',real(d),imag(d));
fprintf('\nproduct of AD-BC=%f,a*d-b*c);
Vs=input ('Enter the sending end line voltage in kV=');
Is=input('Enter the sending end current per phase in A=');
Vs1=Vs*1e3/sqrt(3);
Vr=(a*Vs1-b*Is)/1e3;
Ir=-c*Vs1+d*Is;
fprintf('\nReceiving end voltage/ph =%15.4f%+15.4fi kV',real(Vr),imag(Vr));
fprintf('\nReceiving end current/ph =%15.4f%+15.4fi A',real(Ir),imag(Ir));
Rec_pow=real(3*Vr*Ir/1e3);
Send_pow=real(3*Vs1*Is/1e6);
Efficiency=(Rec_pow/Send_pow)*100;
Regulation=(abs(Vs/a)-abs(Vr))*100/abs(Vr);
fprintf('\nReceiving end power =%0.2f MW',Rec_pow);
fprintf('\nSending end power =%0.2f MW',Send_pow);
fprintf('\nEfficiency=%0.2f%%',Efficiency);
fprintf('\nRegulation=%0.2f%%',Regulation);
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Power System Simulation Lab
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RESULT:Thus, the ABCD constants of the given medium transmission line were found for nominal  configuration.
Also, the efficiency and regulation of it were found for the given receiving end data.
A = _______________________________ B = _____________________________
C= _______________________________
Vr = _____________________
Pr = _______________
D= _____________________________
Vs = ________________ Ir = _____________
Is = ____________
Ps = _________________ Efficiency = __________ Regulation = ___________
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Power System Simulation Lab
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Expt. No.:
Date:
B) PERFORMANCE OF MEDIUM TRANSMISSION LINES
(NOMINAL T METHOD)
AIM:  To determine the ABCD constants of given medium transmission line using nominal T
method.
 To verify AD – BC = 1.
 To determine the efficiency and regulation of the given medium transmission line with the
help of receiving end data.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULAE USED1. The ABCD constants of the medium transmission lines (nominal T) are,
 Y  Z  C  Y and D A .
 YZ
A  1
 , B  Z  1  4 
2




Where, Z = Total series impedance of the line in = Length in km * series impedance in  / km.
Z = (R + j X) , where R and X are total series resistance and reactance of the line in .
Y = Total shunt admittance of the line in ℧= Length in km * shunt admittance in ℧/ km.
Y = (G + j B)℧, where G and B are total shunt conductance and susceptance in℧.
A) DETERMINATION OF PERFORMANCE USING RECEIVING END DATA (Vr, Ir and cos r):2. Sending end voltage Vs = A * Vr + B * Ir Volts
3. Sending end current Is = C * Vr + D * Ir Ampere
4. Receiving end power Sr = 3 * Vr * Ir in VA.
Receiving end power Pr = real (Sr) Watts.
5. Sending end power Ss = 3 * Vs * Is in VA.
Sending end power Ps = real (Ss) Watts.
6. % Efficiency =  Receiving end power Pr 

 100 %
 Sending end power Ps 
V
s
7. % Regulation = A  Vr
Vr
100 %
where, Vr, Ir = Receiving end voltage and current of the line in Volts and Amperes respectively.
Cos r = Receiving end power factor (lagging or leading).
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Power System Simulation Lab
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Take Vr as reference phasor. Vr = Vr0 volts.
Ir = Ir - r in case of lagging power factor, Ir = Ir r in case of leading power factor.
B) DETERMINATION OF PERFORMANCE USING SENDING END DATA (Vs and Is):8. Receiving end voltage Vr = A * Vs – B * Is Volts
9. Sending end current Ir = – C * Vs + D * Is Ampere
10. Receiving end power Sr = 3 * Vr * Ir in VA.
Receiving end power Pr = real (Sr) Watts.
11. Sending end power Ss = 3 * Vs * Is in VA.
Sending end power Ps = real (Ss) Watts.
12. % Efficiency =  Receiving end power Pr 

 100 %
 Sending end power Ps 
V
s
13. % Regulation = A  Vr
Vr
100 %
Where, Vs, Is = Receiving end voltage and current of the line in Volts and Amperes respectively.
PROGRAM:A) DETERMINATION OF PERFORMANCE USING RECEIVING END DATA (Vr, Ir and cos r):clc
clear all
length=input('Enter the length of medium line in km= ');
z=input('\nEnter the series impedance of the line per km= ');
y=input('\nEnter the shunt admittance of the line per km= ');
Z=z*length; Y=y*length;
a=1+(Y*Z)/2; d=a; b=Z*(1+(Y*Z)/4); c=Y;
fprintf('\nA,B,C&D contants are:\n');
fprintf('\nA=%15.4f%+15.4fi',real(a),imag(a));
fprintf('\nB=%15.4f%+15.4fi',real(b),imag(b));
fprintf('\nC=%15.4f%+15.4fi',real(c),imag(c));
fprintf('\nD=%15.4f%+15.4fi',real(d),imag(d));
fprintf('\n AD-BC=%f',a*d-b*c);
Vr=input ('\nEnter the receiving end line voltage in kV=');
Ir=input('Enter the receiving end current per phase in A=');
PF=input('Enter the receiving end power factor =');
pf_a=acos(PF);
Vr1=Vr*1e3/sqrt(3);
Ir_p=Ir*cos(pf_a)-i*Ir*sin(pf_a);
Vs=(a*Vr1+b*Ir_p)/1e3;
Is=c*Vr1+d*Ir_p;
fprintf('\nSending end voltage/ph =%15.4f%+15.4fi kV',real(Vs),imag(Vs));
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fprintf('\nSending end current/ph =%15.4f%+15.4fi A',real(Is),imag(Is));
Rec_pow=real(3*Vr1*Ir_p/1e6);
Send_pow=real(3*Vs*Is/1e3);
Efficiency=(Rec_pow/Send_pow)*100;
Regulation=(abs(Vs/a)-abs(Vr))*100/abs(Vr);
fprintf('\nReceiving end power =%0.2f MW',Rec_pow);
fprintf('\nSending end power =%0.2f MW',Send_pow);
fprintf('\nEfficiency=%0.2f%%',Efficiency);
fprintf('\nRegulation=%0.2f%%',Regulation);
B) DETERMINATION OF PERFORMANCE USING SENDING END DATA (Vs and Is):clc
clear all
length=input('Enter the length of short line in km= ');
z=input('\nEnter the series impedance of the line per km= ');
y=input('\nEnter the shunt admittance of the line per km= ');
Z=z*length; Y=y*length;
a=1+(Y*Z)/2; d=a; b=Z*(1+(Y*Z)/4); c=Y;
fprintf('\nA,B,C&D contants are:\n');
fprintf('\nA=%15.4f%+15.4fi',real(a),imag(a));
fprintf('\nB=%15.4f%+15.4fi',real(b),imag(b));
fprintf('\nC=%15.4f%+15.4fi',real(c),imag(c));
fprintf('\nD=%15.4f%+15.4fi',real(d),imag(d));
fprintf('\nproduct of AD-BC=%f,a*d-b*c);
Vs=input ('Enter the sending end line voltage in kV=');
Is=input('Enter the sending end current per phase in A=');
Vs1=Vs*1e3/sqrt(3);
Vr=(a*Vs1-b*Is)/1e3;
Ir=-c*Vs1+d*Is;
fprintf('\nReceiving end voltage/ph =%15.4f%+15.4fi kV',real(Vr),imag(Vr));
fprintf('\nReceiving end current/ph =%15.4f%+15.4fi A',real(Ir),imag(Ir));
Rec_pow=real(3*Vr*Ir/1e3);
Send_pow=real(3*Vs1*Is/1e6);
Efficiency=(Rec_pow/Send_pow)*100;
Regulation=(abs(Vs/a)-abs(Vr))*100/abs(Vr);
fprintf('\nReceiving end power =%0.2f MW',Rec_pow);
fprintf('\nSending end power =%0.2f MW',Send_pow);
fprintf('\nEfficiency=%0.2f%%',Efficiency);
fprintf('\nRegulation=%0.2f%%',Regulation);
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RESULT:Thus, the ABCD constants of the given medium transmission line were found for nominal T configuration.
Also, the efficiency and regulation of it were found for the given receiving end data.
A = _______________________________ B = _____________________________
C= _______________________________
Vr = _____________________
Pr = _______________
D= _____________________________
Vs = ________________ Ir = _____________
Is = ____________
Ps = _________________ Efficiency = __________ Regulation = ___________
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Power System Simulation Lab
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Expt. No.:
Date:
POWER ANGLE CURVE FOR SALIENT AND NON-SALIENT POLE SYNCHRONOUS
MACHINES
AIM:  To determine the power angle diagram for the given salient and non–salient pole synchronous
machines.

Also, to compute the excitation emf, non-salient power, reluctance power, resultant power and
regulation of it.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULAE USED:-
The excitation current= I= (P-jQ)/(3*Vph)
The Excitation Voltage= E = Vph + (j*I*Xd)
The power angle equation for non salient pole machine is given by,
P = │V││E│* sin δ
Xd
The above equation shows that the power P transmitted from the generator to the motor varies with
the sine of the displacement angle δ between the two rotors.
For salient pole machine,
The power out put P1 = │V││E│* sin δ
Xd
The reluctance powerP2 = │V│2*(Xd-Xq)*sin (2δ)
2XdXq
The total Power =
P = │V││E│* sin δ + │E│2*(Xd-Xq)*sin (2δ)
(Xd)
2XdXq
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Program:
clc
clear all
p=input('Enter power in mw:');
pf=input('Enter power factor:');
vt=input('Enter line to line voltage in kV:');
xd=input('Enter direct axis reactance xd in ohms:');
xq=input('Enter quadrature axis reactance xq in ohms:');
vt_ph=vt*1000/sqrt(3);
pf_a=acos(pf); % power factor angle delta
q=p*tan(pf_a);% reactive power
i=(p-j*q)*1e6/(3*vt_ph); %Current
delta=0:10:180;
% calculation for non salient pole M/c
if xd==xq
e=vt_ph+(j*i*xd);% Excitation voltage
ef=abs(e);
p=ef*vt_ph*sin(delta*pi/180)/xd; % power output
pnet=3*p/1e6;
plot(delta,pnet);
title('Power angle curve for non salient pole synchronous machine');
end
% calculation for salient pole M/c
if xd~=xq
eq=vt_ph+(j*i*xq);
id_mag=abs(i)*sin(angle(eq)-angle(i));
ef=abs(eq)+((xd-xq)*id_mag); % excitation voltage
p1=ef*vt_ph*sin(delta*pi/180)/xd; %real power
p1=3*p1/1e6;
p2=(vt_ph.^2/2)*(1/xq-1/xd)*sin(delta*2*pi/180);% reluctance power
p2=3*p2/1e6;
p=p1+p2;% total power
plot(delta,p1,'r.-',delta,p2,'m.-',delta,p,'b.-');
legend('actual power','relutance power','salient power',2);
title('Power angle curve for a salient pole synchronous machine');
end
reg=(ef-vt_ph)/vt_ph*100;% regulation
xlabel('Delta(in degree)');
ylabel('Three phase power(in MW)');
grid
disp('Percentage regulation:');
disp(reg);
disp('Excitation voltage:');
disp(ef);
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RESULT:Thus, the power angle curves of a given salient pole and non-salient pole synchronous machine
were drawn. Also, the excitation emf, Max. power delivered (Pmax), Non–Salient power, reluctance power,
resultant power and regulation of it were found.
Excitation EMF = ______________ Volts
Regulation = ____________ %
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Power System Simulation Lab
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Expt. No.:
Date:
A) FORMATION OF Y–BUS BY INSPECTION METHOD
AIM:  To determine the Bus Admittance matrix (Y–Bus) for a given power system network by
inspection method.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULAE USED1. Line impedance, which is connecting buses i and k,
zik  R ik  j  X ik  
2. Line Admittance,which is connecting buses i and k, y ik 
1
1
℧.

z ik R ik  j  X ik 
3. The diagonal elements of Bus Admittance Matrix (self admittance or driving point admittance)
n
Yii   y ik
are,
k 1
4. The off-diagonal elements of Bus Admittance Matrix (mutual admittance or transfer admittance)
are,
Yik  Yki   y ik
Where, ‘ i ' and ‘k’ are the buses in the given network and ‘n’ is the no. of buses.
PROGRAM:clc
clear all
Linedata=input('Enter the line data ==>');
fb=Linedata(:,1);
tb=Linedata(:,2);
R=Linedata(:,3);
X=Linedata(:,4);
B= Linedata(:,5);
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z=R+i*X;
y=1./z;
Ysh=i*B;
nbus=max(max(fb),max(tb));
nbranch=length(fb);
ybus=zeros(nbus,nbus);
for k=1:nbranch
ybus(fb(k),tb(k))=-y(k);
ybus(tb(k),fb(k))=ybus(fb(k),tb(k));
end
for m=1:nbus
for n=1:nbranch
if fb(n) == m | tb(n) == m
ybus(m,m)=ybus(m,m)+y(n)+Ysh(n);
end
end
end
ybus
RESULT:Thus, the Bus Admittance matrix (Y–Bus) for a given power system is determined by using
inspection method.
YBus = ______________________.
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Expt. No.:
Date:
B) FORMATION OF Y–BUS BY SINGULAR TRANSFORMATION METHOD
(WITHOUT MUTUAL COUPLING)
AIM:  To determine the Bus Admittance matrix (Y–Bus) for a given power system network without
mutual coupling by using Singular Transformation method.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULA USED: 1. Primitive Impedance matrix [Zprim] = [(R + j X)]
2. Primitive Admittance matrix Y = [Yprim] = 1 / [Zprim]
3. Bus Admittance Matrix YBUS = AT * Y * A
Where, A – Bus incidence matrix and Y – Primitive admittance matrix
PROGRAM:clc
clear all
Linedata=input('Enter the line data ==>');
R=input('Enter the reference node=');
fb=Linedata(:,1);
tb=Linedata(:,2);
r=Linedata(:,3);
x=Linedata(:,4);
z=r+i*x;
nbus=max(max(fb),max(tb));
nbranch=length(fb);
Zprim=zeros(nbranch,nbranch);
Yprim=zeros(nbranch,nbranch);
for i=1:nbranch
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Zprim(i,i)=z(i);
end
Zprim
Yprim=inv(Zprim)
A=zeros(nbranch,nbus);
for i=1:nbranch;
H=Linedata(i,1);
A(i,H)=1;
K=Linedata(i,2);
A(i,K)=-1;
end
A
A(:,R)=[]
At=A'
Ybus=At*(Yprim*A)
RESULT:Thus, the Bus Admittance matrix (Y–Bus) for a given power system without mutual coupling was
determined by using singular transformation method.
YBus = ______________________.
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
Expt. No.:
Date:
C) FORMATION OF Y–BUS BY SINGULAR TRANSFORMATION METHOD
(WITH MUTUAL COUPLING)
AIM:  To determine the Bus Admittance matrix (Y–Bus) for a given power system network with
mutual coupling by using Singular Transformation method.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULA USED: 1. Primitive Impedance matrix [Zprim] = [(R + j X)]
2. Primitive Admittance matrix Y = [Yprim] = 1 / [Zprim]
3. Bus Admittance Matrix YBUS = AT * Y * A
Where, A – Bus incidence matrix and Y – Primitive admittance matrix
PROGRAM:clc
clear all
Linedata=input('Enter the line data ==>');
R=input('Enter the reference node=');
Mc=input('Enter the mutual coupling = >');
fb=Linedata(:,1);
tb=Linedata(:,2);
r=Linedata(:,3);
x=Linedata(:,4);
z=r+i*x;
nbus=max(max(fb),max(tb));
nbranch=length(fb);
Zprim=zeros(nbranch,nbranch);
Yprim=zeros(nbranch,nbranch);
for i=1: nbranch
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
Zprim(i,i)=z(i);
end
m=size(Mc);
for i=1:1:m
p=Mc(i,1);
q=Mc(i,2);
Zprim(p,q)=Mc(i,3);
Zprim(q,p)=Zprim(p,q);
end
Zprim
Yprim=inv(Zprim)
A=zeros(nbranch,nbus);
for i=1:nbranch
H=Linedata(i,1);
A(i,H)=1;
K=Linedata(i,2);
A(i,K) = -1;
end
A
A(:,R)=[ ]
At=A'
Ybus=At*(Yprim*A)
RESULT:Thus, the Bus Admittance matrix (Y–Bus) for a given power system with and without mutual
coupling is determined by using singular transformation method.
YBus = ______________________.
Department of EEE, AIT Bangalore
Page 19
Power System Simulation Lab
2022-23
Expt. No.:
Date:
D) DETERMINATION OF LINE FLOW FOR A POWER SYSTEM NETWORK
AIM:  To determine the line currents, line losses, bus currents, bus power and line flow for a power
system network with a specified system voltage (bus) profile.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULA USED: 1. Line currents
2.
 y  
I ik  Vi Vk  y ik  Vi   sh ik 
 2 
and
 y  
I ki  Vk Vi  y ik  Vk   sh ik 
 2 

Complex power flows S V I  and S V I
ik
3. Line losses
SL ik  Sik  S ki
4. Bus powers S i 
i
ik
ki
k
ki

n
 S ik
k 1
5. Bus currents I i 
n
 I ik
k 1
Where, n is the number of PQ buses in the given network,
Viand Vk arevoltages at the busesi and k respectively,
y ik is the mutual admittance of the line, connecting the buses i and k,
y sh ik is total line charging admittance of the line, connecting buses i and k,
Iik  is the conjugate of line current flowing from bus i to bus k.
Department of EEE, AIT Bangalore
Page 20
Power System Simulation Lab
2022-23
PROGRAM:clc
clear all
linedata=input('\nEnter the line data ----->');
busdata=input('\nEnter the bus data ----->');
fr=linedata(:,1);
to=linedata (:,2);
r=linedata (:,3);
x=linedata (:,4);
b=linedata(:,5);
z=r+i*x;
y=1./z;
ysh=i*b;
nbr=length(fr);
nbus=max(max(fr),max(to));
bus_no=busdata(:,1);
volt_mag=busdata(:,2);
angle_d=busdata(:,3);
%bus voltages in rectangular form
for n=1:nbus
v_bus(n)=(volt_mag(n)*cos(angle_d(n)*pi/180))+i*(volt_mag(n)*sin(angle_d(n)*pi/180));
end
v_bus
%line current
for m=1:nbr
Iik(m)=(v_bus(fr(m))-v_bus(to(m)))*y(m)+ (v_bus(fr(m))*ysh(m));
Iki(m)=(v_bus(to(m))-v_bus(fr(m)))*y(m)+(v_bus(to(m))*ysh(m));
end
Iik
Iki
%complex power flow
for m=1:nbr
Sik(m)=v_bus(fr(m))*conj(Iik(m));
Ski(m)=v_bus(to(m))*conj(Iki(m));
end
Sik
Ski
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
%line losses
for m=1:nbr
Sl(m)=Sik(m)+Ski(m);
end
Sl
%bus power
s_bus=zeros(nbus,1);
for n=1:nbus
s_bus(n)=0;
for m=1:nbr
if fr(m)==n
s_bus(n)=s_bus(n)+Sik(m);
elseif to(m)==n
s_bus(n)=s_bus(n)+Ski(m);
end
end
end
s_bus
%bus current
i_bus=zeros(nbus,1);
for n=1:nbus
i_bus(n)=0;
for m=1:nbr
if fr(m)==n
i_bus(n)=i_bus(n)+Iik(m);
elseif to(m)==n
i_bus(n)=i_bus(n)+Iki(m);
end
end
end
i_bus
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
RESULT:Thus, the line currents, line losses, bus currents, bus power and line flow for a power system
network with a specified system voltage (bus) profile were determined.
Line Currents Iik = ________________________________________________________________
Line Currents Iki =________________________________________________________________
Bus currents Ii =________________________________________________________________
Bus power Si = ________________________________________________________________
Line power Sik =________________________________________________________________
Line power Ski= ________________________________________________________________
Line losses SL = ________________________________________________________________
Department of EEE, AIT Bangalore
Page 23
Power System Simulation Lab
2022-23
Expt. No.:
Date:
LOAD FLOW SOLUTION USING GAUSS–SEIDEL METHOD
AIM:  To perform the load flow solution using Gauss-Seidel (GS) method for a given power system
network with only PQ buses.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULA USED: 1. Bus Voltage at the end of (r+1)th iteration is given by,
Vi
r 1

n
1 
 Pi  j  Q i n
r 1
r
 
  Yik * Vk   Yik  Vk  Volts
Yii  V r 
k 1
k 1



i
 
Where, Yii is the corresponding element in bus admittance matrix (self admittance of bus i),
Yik is the corresponding element in bus admittance matrix (mutual admittance between
buses i and k)
Pi = Net real power injected into the bus i,
Qi = Net reactive power injected into the bus i,
(Vir) = Conjugate of voltage at bus i at the end of rth iteration,
Vkr = Voltage at bus k at the end of rth iteration,
Vk(r+1) = Voltage at bus k at the end of (r+1)th iteration,
n = the number of PQ buses in the given network.
PROGRAM:clc
clear all
ybus=input(‘Enter the Bus Admittance Matrix Ybus=’);
busdata=input('Enter the bus data ==>');
tolerlimit=input('Enter the tolerance limit of Bus Voltages ==>');
busno=busdata(:,1);
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
P=busdata(:,2);
Q=busdata(:,3);
V=busdata(:,4);
del=busdata(:,5);
nbus=max(busno);
Vprev=V;
toler=1;
iteration=1;
while(toler>tolerlimit)
for i=2:nbus
sumyv=0;
for k=1:nbus
if i~=k
sumyv=sumyv+ybus(i,k)* V(k);
end
end
V(i)=(1/ybus(i,i))*((P(i)-j*Q(i))/conj(V(i))-sumyv);
end
if (iteration==1)
iteration
V
else
end
iteration=iteration+1;
toler=max(abs(abs(V)-abs(Vprev)));
Vprev=V;
end
finaliterationcount=iteration
V
Vmag=abs(V)
Vangle=(180/pi)*angle(V)
Department of EEE, AIT Bangalore
Page 25
Power System Simulation Lab
2022-23
RESULT:Thus, the load flow solution using Gauss–Seidel (GS) method for a given network which has
only PQ buses was performed.
Bus Voltage Vi = _______________________________________________________________
Bus Voltage Magnitudes Vi= __________________________________________________
Bus Voltage Angles i = _________________________________________________________
At the end of first iteration,
Bus Voltage V i = ______________________________________________________________
Department of EEE, AIT Bangalore
Page 26
Power System Simulation Lab
2022-23
Expt. No.:
Date:
FORMATION OF JACOBIAN MATRIX
AIM:  To form a Jacobian matrix for a power system which has only PQ buses in polar coordinates.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY: -
FORMULA USED: n
1. Real Power Pi  Vi    Vk  Yik cos ik  k   i 
k 1
n


2. Reactive Power Q i   Vi   Vk  Yik sin ik  k   i
k 1

3. Jacobian elements H ii Q i  B ii  Vi

and L ii  Q i  B ii  Vi
4. Jacobian elements H ij 

, N
ii  Pi

 Gii  Vi
2
, J
ii  Pi 
G  V 
2
ii
i
 Q 
Pi
i 
 L ij  
 Vj   Vi  Vj  Yij sin ij  j   i

 j
 Vj 


 P
N ij  i
V
j

H
5. Jacobian matrix = 
J

2
2



  V   J   Q i  V  V  Y cos    
ij
i
j
ij
ij
j
i
 j
 j




N
L 
Where, i & k are the buses,
Vi is the bus voltage at bus i,
Yik is the corresponding element in bus admittance matrix,
ik is angle of Yikin the corresponding element of bus admittance matrix,
i is the angle of bus voltage at bus i,
Bii is the susceptance of bus i, = Yii* sin ii
Gii is the conductance of bus i, = Yii* cos ii
Department of EEE, AIT Bangalore
Page 27
Power System Simulation Lab
2022-23
PROGRAM:clc
clear all
y_bus=input('Enter the bus admittance matrix ==>');
busdata=input('\nEnter the bus data ==>');
y=abs(y_bus);
th=angle(y_bus);
bus_no=busdata(:,1);
v_mag=busdata(:,2);
v_ang=busdata(:,3);
nbus=max(bus_no);
slackbus=1;
for p=1:nbus
pc(p)=0;
qc(p)=0;
for q=1:nbus
if(p~=slackbus)
pc(p)=pc(p)+(v_mag(p)*v_mag(q)*y(p,q)*cos(th(p,q)+v_ang(q)-v_ang(p)));
qc(p)=qc(p)-(v_mag(p)*v_mag(q)*y(p,q)*sin(th(p,q)+v_ang(q)-v_ang(p)));
end
end
end
for p=1:nbus
p_cal(p)=pc(p);
q_cal(p)=qc(p);
end
p_cal
q_cal
%formation of H N J L
i=1;
for p=1:nbus
j=1;
for q=1:nbus
if(p~=q)&(p~=slackbus)&(q~=slackbus)
H(i,j)=-v_mag(p)*v_mag(q)*y(p,q)*sin(th(p,q)-v_ang(p)+v_ang(q));
N(i,j)=(v_mag(p)*y(p,q))*cos(th(p,q)-v_ang(p)+v_ang(q));
J(i,j)=-N(i,j);
L(i,j)=H(i,j);
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
j=j+1;
end
if(p==q)&(p~=slackbus)&(q~=slackbus)
H(i,i)=-q_cal(p)-(y(p,p)*sin(th(p,p))*v_mag(p)^2);
N(i,i)=p_cal(p)+(y(p,p)*cos(th(p,p))*v_mag(p)^2);
J(i,i)=p_cal(p)-(y(p,p)*cos(th(p,p))*v_mag(p)^2);
L(i,i)=q_cal(p)-(y(p,p)*sin(th(p,p))*v_mag(p)^2);
j=j+1;
end
end
if(p~=q)&(p~=slackbus)&(q~=slackbus)
i=i+1;
end
end
H
N
J
L
Jacobian = [H N; J L]
RESULT:Thus, the Jacobian matrix for a power system which has only PQ buses was formed.
P = _____________________________________________________________________________
Q = _____________________________________________________________________________
Matrix H = _______________________________________________________________________
Matrix N = _______________________________________________________________________
Matrix J = _______________________________________________________________________
Matrix L = _______________________________________________________________________
Jacobian Matrix =_________________________________________________________________
Department of EEE, AIT Bangalore
Page 29
Power System Simulation Lab
2022-23
Expt. No.:
Date:
FORMATION OF Z-BUS USING Z-BUS BUILDING ALGORITHM
AIM:  To determine the Z-bus for a given power system network, using step-by-step Z-bus building
algorithm.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY:-
FORMULA USED:1. Bus impedance matrix in Type-1 modification,

 Z
BUS (Old )


ZBUS ( New ) 


0
0
.
.
0 
0 

. 
. 

Zb 
2. Bus impedance matrix in Type-2 modification,




ZBUS ( New)  



Z j1 Z j2

ZBUS (Old )
.





.

Znj 

Z jj  Zb 

Z1j
Z2 j
.
.
Z jn
3. Bus impedance matrix in Type-3 modification,

 Z1j 

Z 
 1
 2j 
Z BUS ( New )  Z BUS (Old )  
  .   Z j1
 Z jj  Z b  . 
 

Z nj 


4.


Bus impedance matrix in Type-4 modification,



 Z1i

Z

 2i
1
Z BUS ( New )  Z BUS (Old )  

 Z jj  Z b  Z ii  2  Z ij 


 Z ni




Department of EEE, AIT Bangalore


Z j2
 Z1j 
 Z 2j 

.
  Z i1  Z j1

.

 Z nj 


. . Z jn








 Z i2  Z j2 
. .
Z in  Z jn 







Page 30
Power System Simulation Lab
2022-23
PROGRAM:clc
clear all
Z=input('Enter the given line data: ');
[m n]=size(Z);
Zbus=[];
Currentbusno=0;
for i=1:m
[rows cols]=size(Zbus);
fb=Z(i,1);
tb=Z(i,2);
value=Z(i,3);
newbus=max(fb,tb);
ref=min(fb,tb);
%type 1 modification
if newbus>Currentbusno & ref==0
type=1
Zbus=[Zbus zeros(rows,1); zeros(1,cols) value]
Currentbusno=newbus;
continue
end
%type 2 modification
if newbus>Currentbusno & ref~=0
type=2
Zbus=[Zbus Zbus(:,ref); Zbus(ref,:) value+Zbus(ref,ref)]
Currentbusno=newbus;
continue
end
%type 3 modification
if newbus<=Currentbusno & ref==0
type=3
Zbus=Zbus-1/(Zbus(newbus,newbus)+value)*Zbus(:,newbus)* Zbus(newbus,:)
continue
end
%type 4 modification
if newbus<=Currentbusno & ref~=0
type=4
Department of EEE, AIT Bangalore
Page 31
Power System Simulation Lab
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Zbus=Zbus-1/(value+Zbus(fb,fb)+Zbus(tb,tb)-2*Zbus(fb,tb))*((Zbus(:,fb)-Zbus(:,tb))*((Zbus(fb,:)Zbus(tb,:))))
continue
end
end
RESULT:Thus, the Z-bus for a given power system was determined using step-by-step Z-bus building
algorithm.
ZBUS = _____________________.
Department of EEE, AIT Bangalore
Page 32
Power System Simulation Lab
2022-23
Expt. No.:
Date:
DETERMINATION OF SWING CURVE AND CRITICAL CLEARING
ANGLE AND TIME
AIM:  To determine the Swing curve of a single machine connected to an infinite bus.
 To determine the critical clearing angle and time for a single machine connected to an infinite
bus through a pair of identical transmission lines under 3-phase fault on one of the lines.
SOFTWARE REQUIRED:OCTAVE 5.1 and above.
THEORY:-
FORMULA USED:1. Maximum torque angle,

Pm
max    sin 1 
 Pmax aft. fault





2. Critical clearing angle cr  can be found from the following equation.
cos  cr 
Pm   max   0   Pmax dur . fault  cos  0  Pmax aft. fault  cos  max
Pmax aft. fault  Pmax dur . fault
3. Critical clearing time,
t cr
 2  H   cr   0 


   f  Pm

1
2
in sec
4. Solution of swing equation by RK method:
a) Compute the constants k1, l1, k2, l2, k3, l3, k4 and l4 using the following equations.
(i)
(ii)
(iii)
(iv)
b)
c)
 r 
(v)
k1r  r  t
l1r 
Pm  Pmax  sin r  t
M
k 2 r  r  t p  l1r  t
l2r 
(vii)
Pm  Pmax  sin r  t p  k r   t
k1r  2 k 2 r  2 k 3 r  k 4 r 
r 
(vi)
M
(viii)
k 3 r  r  t p  l 2 r  t
l3r 
Pm  Pmax  sin  r  t p  k 2 r   t
M


P  Pmax  sin  r  k 3 r   t
r
 m
k 4 r  r  l 3 r  t
l4
M
rad.
6
l1r  2  l 2r  2  l3 r  l 4 r  rad/sec
6
d)  r 1   r   r rad
e) r 1  r  r rad/sec
Department of EEE, AIT Bangalore
Page 33
Power System Simulation Lab
2022-23
Where, Pm = Mechanical input power under pre-fault condition in pu.
Pmax = Maximum electrical power transfer under pre-fault condition in pu.
Pmax dur. fault = Maximum electrical power transfer during fault condition in pu.
Pmax aft. fault = Maximum electrical power transfer under post-fault condition in pu.
0 = Initial torque angle or load angle or rotor angle.
H = Inertia constant (MJ/MVA) in pu.
M = Inertia constant (s2 / elect deg (or) s2 / elect rad) in pu.
f = frequency in Hz.
 = angular velocity in rad / sec
In constants,
Pmax = Pmax dur. fault when, t < tfc (time of fault clearing)
Pmax = Pmax aft. fault when, t ≥ tfc (time of fault clearing)
∆t = sample time period in sec.
tp = Final time of period of fault analysis in sec.
PROGRAM:clear all
clc
tfc=input('Enter the time of fault clearing: ');
Iner_cons=input('Enter the Inertia constant M in pu: ');
pm=input('Enter the Mechanical input power: ');
pmbf=input('Enter the pre-fault maximum power: ');
pmdf= input('Enter the maximum power during fault: ');
pmaf= input('Enter the post-fault maximum power: ');
d=input('Enter the initial steady state torque angle: ');
w=0; % initial angular velocity omega
h= input('Enter the Inertia constant H in pu: ');
dmax=pi-asin(pm/pmaf);
cosdcr=(pm*(dmax-d)+pmaf*cos(dmax)-pmdf*cos(d))/(pmaf-pmdf);
cc_angle=acos(cosdcr)*180/pi;
cc_time=sqrt(2*h*((cc_angle*pi/180)-d)/(pi*50*pm));
disp('CRITICAL CLEARING ANGLE(IN DEG)');
disp(cc_angle);
disp('CRITICAL CLEARING TIME(IN SEC)');
disp(cc_time);
fprintf('\n\n\t TIME(IN SEC)\t DELTA(IN DEG)\n');
%finding out swing curve by Runge-Kutta Method
a=1;
Department of EEE, AIT Bangalore
Page 34
Power System Simulation Lab
2022-23
for t=0:0.05:0.5
if t<tfc
pmax=pmdf;
elseif t>=tfc
pmax=pmaf;
end
k1=w*0.05;
l1=(pm-pmax*sin(d))*0.05/ Iner_cons;
k2=(w+0.5*l1)*0.05;
l2=(pm-pmax*sin(d+0.5*k1))*0.05/ Iner_cons;
k3=(w+0.5*l2)*0.05;
l3=(pm-pmax*sin(d+0.5*k2))*0.05/ Iner_cons;
k4=(w+l3)*0.05;
l4=(pm-pmax*sin(d+k3))*0.05/ Iner_cons;
deld=(k1+2*k2+2*k3+k4)/6;
delw=(l1+2*l2+2*l3+l4)/6;
d=d+deld;
dg=(d*180)/pi;
fprintf('\t%6.3f\t\t%6.3f\n',t,dg);
w=w+delw;
delta(a)=dg;
a=a+1;
end
t=0:0.05:0.5;
plot(t,delta,'-');
xlabel('Time after fault in sec ----->');
ylabel('Delta in deg.) ----->');
title('Plot: Swing curve for single machine connected to infinite bus system');
legend('SWING CURVE');
RESULT:Thus, the Swing curve of a single machine connected to an infinite bus is determined. Also, the
critical clearing angle and time for a single machine connected to an infinite bus through a pair of
identical transmission lines under 3-phase fault on one of the lines were determined.
Critical clearing angle cr  = _________ rad.
Critical clearing time t cr = _________ sec.
Department of EEE, AIT Bangalore
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Department of EEE, AIT Bangalore
2022-23
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Expt. No.:
Date:
LOAD FLOW ANALYSIS (Mi-Power)
AIM:  To perform the load flow solution for a given power system network using a) Gauss Seidel
method, (b) Newton-Raphson method and (c) Fast decoupled method with the help of MiPower
software package.
SOFTWARE REQUIRED:MiPower Software package.
THEORY:-
PROCEDURE:MiPower – Database configuration:
1. Double click on ‘MiPower’ shortcut icon on the desktop. Double click on ‘Power System Network
Editor’. A ‘Power System Network Editor’ window will appear. In its Menu bar, select ‘Database’ and
then click ‘configure’.
2. A ‘Configure Database’ window will appear. Click ‘browse’ button.An ‘Open’ window will appear.
3. Create a new folder in any disk drives and then open that folder. Now, type ‘Stagg.mdb’ as file name
and then click ‘open’.
4. The ‘Configure Database’ window will reappear with the path chosen. Now, click ‘OK’ button. A
‘Configuration Information’ window will appear now.
5. Uncheck the ‘Power System Libraries’ and ‘Standard Relay Libraries’ which are not required for load
flow analysis in pu base.
6. Click ‘Electrical & Currency Information’ in the menu bar of ‘Configuration Information’ window. In
that menu, enter the given Base MVA and Base Frequency, then, check the pu status. Now, click ‘OK’
to create the database and to return to ‘Power System Network Editor’.
Bus Base VoltageConfiguration:
7. In the ‘Power System Network Editor’, configure the base voltages for the given power system. In its
menu bar, select ‘Configure’ and then click ‘Base Voltage’. A ‘Bus Base Voltage Configuration’ window
will appear. If necessary, change the Base Voltages, Colour and Bus width, then click ‘OK’.
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Drawing of Buses:
8. Click on ‘Bus’ icon provided on the tool bar of the power system which is situated on the right side of
‘Power System Network Editor’. The ‘Bus Base Volt’ window will appear, in which select the given
Bus Base Voltage.
9. Draw the Bus No.1, by right clicking the mouse and dragging it to a straight line. A ‘Bus Data’ window
will appear, in which enter the Bus ID and Bus Name, then click ‘OK’.
10. A ‘MiPower Database – Bus Data’ window will appear, in which enter the minimum and maximum
values of Bus Voltages (usually  5% of rated value). After entering the data, click ‘Save’ icon on this
menu bar.
11. Similarly, by following the above procedure, the remaining buses are drawn with the given data.
Drawing of Transmission Lines:
12. Click on ‘Transmission Line’ icon provided on the tool bar of the power system which is situated on
the right side of ‘Power System Network Editor’.
13. To draw a transmission line, click in between the two buses of that line. If the direction of the line
shown is correct, then connect to the FROM Bus by double clicking LMB (Left Mouse Button) on that
bus, otherwise, click ‘Object(s)’ on the menu bar, then click ‘Rotate’ till it reaches the direction of the
line given. Now, similarly, join the line to the TO Bus by double clicking LMB on that bus.
14. An ‘Element ID’ window will appear, in which enter the Line Number. Now, a ‘Line\Cable Data’
window will appear, in which enter the line name, if any, and enter the Structure Ref. No. as ‘1’. Now,
click ‘Transmission Line Library’, a ‘Line\Cable Library’ window will open.
15. Enter the given Transmission Line Library data for the line 1 as positive sequence resistance, reactance
and susceptance, then click ‘Save’ icon on the menu bar and click close (x).
16. The ‘Line\Cable Data’ window will reappear. Now, click ‘Save’ icon on the menu bar which invokes
the ‘Power System Network Editor’.
17. Similarly, by following the above procedure, the other transmission lines are drawn with the given
data.
18. Take the Structure Ref. Numbers of other transmission lines as 2, 3, etc.
19. If any two of the transmission lines has same data, like, impedance and Line charging admittances,
take the same ‘Structure Ref. No.’ for both lines.
Drawing of Generators:
20. Click on ‘Generator’ icon provided on the tool bar of the power system which is situated on the right
side of ‘Power System Network Editor’.
21. Connect it to respective bus by clicking LMB on the respective bus. The ‘Element ID’ window will
appear, in which, enter the generator number, and then click ‘OK’.
22. A ‘Generator Data’ window will appear, in which enter the given voltage of the generator/bus in
‘Specified Voltage’ column in pu.
23. Now, enter the given MVAR generation of the bus as ‘Reactive Power–Minimum’ and ‘Reactive
Power–Maximum’. Also, enter ‘Manufacturer Ref. No.’ as ‘1’ and then click ‘Library’ which opens
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the ‘Generator Library’ window, in which, enter the ‘MVA Rating’ as ‘ [(MW)2 + (MVAR)2]’ with
the help of given data, kV Rating of the Bus/Generator and the MW rating of the Bus/Generator.
24. After entering the data, click ‘Save’ icon on the menu bar and click close (x) which will reopen the
‘Generator Data’ window. Now, click ‘Save’ icon on the menu bar which invokes the ‘Power System
Network Editor’.
25. Similarly, by following the above procedure, the other generators are drawn with the given data.
26. Take the Structure Ref. Numbers of other generators as 2, 3, etc.
27. If any two of the generators has same data, like, impedance and Line charging admittances, take the
same ‘Structure Ref. No.’ for both generators.
28. At Slack bus, only voltage and its angle are mentioned, but, the Scheduled power, real power
minimum and maximum constraints do not have much importance.
Drawing of Loads:
29. Click on ‘Load’ icon provided on the tool bar of the power system which is situated on the right side
of ‘Power System Network Editor’.
30. Connect it to respective bus by clicking LMB on the respective bus. The ‘Element ID’ window will
appear, in which, enter the Load number, and then click ‘OK’.
31. A ‘Load Data’ window will appear, in which enter the given values of Real Power in MW and Reactive
Power in MVAR. Now, click ‘Save’icon on the menu bar which invokes the ‘Power System Network
Editor’.
32. Similarly, by following the above procedure, the other loads are drawn with the given data.
Solving of Load Flow Analysis:
33. Select ‘Solve’ on the menu bar of the ‘Power System Network Editor’, then click ‘Load Flow Analysis’,
which opens a ‘Load Flow Analysis’ window.
34. Now, click ‘Study Info’ on this window, then ‘Load Flow Studies’ window will appear, in which select
the respective method under ‘Technique’. Also, enter the given values of acceleration factor, Ptolerance and Q-tolerance, then click ‘OK’.
35. The ‘Load Flow Analysis’ window will reappear, in which click ‘Execute’ to execute the Load Flow
Analysis. Then, click on ‘Report’ in the ‘Load Flow Analysis’ window to view the report. After viewing
the report, click ‘Close’ in the ‘Load Flow Analysis’ window to end the experiment.
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RESULT:Thus, the load flow solution for a given power system network using (a) Gauss-Seidel method, (b)
Newton Raphson method and (c) Fast decoupled method were performed with the help of MiPower
software package.
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
Expt. No.:
Date:
UNSYMMETRICAL FAULTS ANALYSIS
AIM:  To determine fault currents and voltages in single transmission line systems with star-delta
transformers at a specified location for LG, LL and LLG faults.
SOFTWARE REQUIRED:MiPower Software package.
THEORY:-
PROCEDURE:MiPower – Database configuration:
1. Double click on ‘MiPower’ shortcut icon on the desktop. Double click on ‘Power System Network
Editor’. A ‘Power System Network Editor’ window will appear. In its Menu bar, select ‘Database’ and
then click ‘configure’.
2. A ‘Configure Database’ window will appear. Click ‘browse’ button.An ‘Open’ window will appear.
3. Create a new folder in any disk drives and then open that folder. Now, type ‘SCS.mdb’ as file name
and then click ‘open’.
4. The ‘Configure Database’ window will reappear with the path chosen. Now, click ‘OK’ button. A
‘Configuration Information’ window will appear now.
5. Uncheck the ‘Power System Libraries’ and ‘Standard Relay Libraries’ which are not required for load
flow analysis in pu base.
6. Click ‘Electrical & Currency Information’ in the menu bar of ‘Configuration Information’ window. In
that menu, enter the given Base MVA and Base Frequency, then, check the pu status. Now, click ‘OK’
to create the database and to return to ‘Power System Network Editor’.
Bus Base VoltageConfiguration:
7. In the ‘Power System Network Editor’, configure the base voltages for the given power system. In its
menu bar, select ‘Configure’ and then click ‘Base Voltage’. A ‘Bus Base Voltage Configuration’ window
will appear. If necessary, change the Base Voltages, Colour and Bus width, then click ‘OK’.
Drawing of Buses:
8. Click on ‘Bus’ icon provided on the tool bar of the power system which is situated on the right side of
‘Power System Network Editor’. The ‘Bus Base Volt’ window will appear, in which select the given
Bus Base Voltage.
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9. Draw the Bus No.1, by right clicking the mouse and dragging it to a straight line. A ‘Bus Data’ window
will appear, in which enter the Bus ID and Bus Name, then click ‘OK’.
10. A ‘BUS’ window will appear, in which enter the minimum and maximum values of Bus Voltages
(usually  5% of rated value). After entering the data, click ‘Save’ icon on this menu bar.
11. Similarly, by following the above procedure, the remaining buses are drawn with the given data.
Drawing of Transmission Lines:
12. Click on ‘Transmission Line’ icon provided on the tool bar of the power system which is situated on
the right side of ‘Power System Network Editor’.
13. To draw a transmission line, click in between the two buses of that line. If the direction of the line
shown is correct, then connect to the FROM Bus by double clicking LMB (Left Mouse Button) on that
bus, otherwise, click ‘Object(s)’ on the menu bar, then click ‘Rotate’ till it reaches the direction of the
line given. Now, similarly, join the line to the TO Bus by double clicking LMB on that bus.
14. An ‘Element ID’ window will appear, in which enter the Line Number. Now, a ‘Line\Cable Data’
window will appear, in which enter the line name, if any, and enter the Structure Ref. No. as ‘1’. Now,
click ‘Transmission Line Library’, a ‘Line\Cable Library’ window will open.
15. Enter the given Transmission Line Library data for the line 1 as positive sequence resistance, reactance
and susceptance, then click ‘Save’ icon on the menu bar and click close (x).
16. The ‘Line\Cable Data’ window will reappear. Now, click ‘Save’ icon on the menu bar which invokes
the ‘Power System Network Editor’.
17. Similarly, by following the above procedure, other transmission lines are drawn with the given data.
18. Take the Structure Ref. Numbers of other transmission lines as 2, 3, etc.
19. If any two of the transmission lines has same data, like, impedance and Line charging admittances,
take the same ‘Structure Ref. No.’ for both lines.
Drawing of Transformers:
20. Click on ‘Two Winding Transformer’ icon provided on the tool bar of the power system which is
situated on the right side of ‘Power System Network Editor’.
21. To draw a transformer, click in between the two buses of that transformer. If the direction of the
transformer shown is correct, then connect to the FROM Bus by double clicking LMB (Left Mouse
Button) on that bus, otherwise, click ‘Object(s)’ on the menu bar, then click ‘Rotate’ till it reaches the
direction of the transformer given. Now, similarly, join the transformer to the TO Bus by double
clicking LMB on that bus.
22. An ‘Element ID’ window will appear, in which enter the Transformer Number. Now, a ‘Two Winding
Transformer Data’ window will appear, in which enter the transformer’s name, if any, and enter the
‘Manufacturer Ref. Number’ as ‘1’. Now, click ‘Transformer Library’, a ‘Two Winding Transformer
Library’ window will open.
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23. Enter the given Transformer Library data for the Transformer 1 as MVA rating, primary and
secondary
voltages,
leakage
reactances
(positive
and
zero
sequence
impedance)
and
windingconfiguration, then click ‘Save’ icon on the menu bar and click close (x).
24. The ‘Two Winding Transformer Data’ window will reappear. Now, click ‘Save’ icon on the menu bar
which invokes the ‘Power System Network Editor’.
25. Similarly, by following the above procedure, the other transformers are drawn with the given data.
26. Take the Structure Ref. Numbers of other transformers as 2, 3, etc.
27. If any two of the transformers has same data, like, primary and secondary voltages, MVA rating and
leakage reactances, take the same ‘Structure Ref. No.’ for both transformers.
Drawing of Generators:
28. Click on ‘Generator’ icon provided on the tool bar of the power system which is situated on the right
side of ‘Power System Network Editor’.
29. Connect it to respective bus by clicking LMB on the respective bus. The ‘Element ID’ window will
appear, in which, enter the generator number, and then click ‘OK’.
30. A ‘Generator Data’ window will appear, in which enter the given voltage of the generator/bus in
‘Specified Voltage’ column in pu.
31. Now, enter ‘Manufacturer Ref. No.’ as ‘1’ and then click ‘Library’ which opens the ‘Generator Library’
window, in which, enter the MVA Rating, MW Rating, kV Rating of the Bus/Generator and the given
reactance of the Generator.
32. After entering the data, click ‘Save’ icon on the menu bar and click close (x) which will reopen the
‘Generator Data’ window. Now, click ‘Save’ icon on the menu bar which invokes the ‘Power System
Network Editor’.
33. Similarly, by following the above procedure, the other generators are drawn with the given data.
34. Take the Structure Ref. Numbers of other generators as 2, 3, etc.
35. If any two of the generators has given same data, take the same ‘Structure Ref. No.’ for both generators.
Solving of Load Flow Analysis:
36. Select ‘Solve’ on the menu bar of the ‘Power System Network Editor’, then click ‘Short Circuit
Analysis’, which opens a ‘Short Circuit Analysis’ window.
37. Now, click ‘Study Info’ on this window, then ‘Short Circuit Studies’ window will appear, in which
select ‘Fault Type’ and check the ‘Fault on Bus’. Also, select the Bus Number and Bus Name at which
the fault appears, then click ‘OK’.
38. The ‘Short Circuit Analysis’ window will reappear, in which click ‘Execute’ to execute the Short
Circuit Analysis. Then, click on ‘Report’ in the ‘Short Circuit Analysis’ window to view the report.
After viewing the report, click ‘Close’ in the ‘Short Circuit Analysis’ window to end the experiment.
RESULT: Thus, the fault currents, voltages and MVAin single transmission line systems with star-delta
transformers at a specified location for LG, LLG and LLG faults were determined.
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Expt. No.:
Date:
OPTIMUM GENERATION SCHEDULING (ECONOMIC LOAD DISPATCH)
AIM:  To determine the economic load dispatch or optimum generation schedule for a given thermal
power plants with the help of B-Coefficients.
SOFTWARE REQUIRED:MiPower Software package.
THEORY:-
CALCULATIONS:Assume the scheduled power in all the generators as P1, P2, P3, …. Pn and value of Lambda.
1) Total Loss with scheduled power PL 
k
P B
m 1
n 1
m
P MW. Where, k = no. of generators.
mn n
1
2
2) Initial cost of generation in ith generator Ci  a i Pi 2  b i Pi  c i Rs./hr
where, Pi = scheduled power in MW and ai, bi and ci are cost coefficients.
3) Initial total cost of generation in all the generators C T 
k
C
i 1
i
Rs./hr
4) The calculated values of power generation in all the generators are,
1
Pi 
k
bi
  2  B ij  Pj

j1
j i
ai
 2  B ii

5) The penalty factors of all the generators are,
Li 
Department of EEE, AIT Bangalore
1
PL
1
Pi
Page 44
Power System Simulation Lab
2022-23
PROCEDURE:1. Double click on ‘MiPower’ shortcut icon on the desktop. On the Main menu bar, click ‘Tools’ and
then, ‘Economic Dispatch by B-coefficient’. A ‘BCoefficient’ window will open.
2. In that, select ‘New’ under File to create a new file. Select a desired location to save the file and give
the desired file name.
3. Now, enter the given values of ‘Total demand’ and ‘No. Of Generators’. Select, the ‘Input Type’ as
given.
4. Select the ‘Gen No’ as ‘1’ under ‘Generator Details’ and enter the corresponding given values of Pmin,
Pmax, Pscheduled (usually slightly less than Pmax) and the cost coefficients C0, C1 and C2. The general form
of equation of fuel cost in Rs./hr is,Ci = C2*Pi2 + C1*Pi + C0, where, Pi and Ci are real power generated
and Cost of unit i respectively.
5. Similarly, enter the given values of all the generator units.
6. Now, enter the given values of B-Coefficients (B11, B12, B13, etc.) by selecting the respective ‘Row’ and
‘Column’ under ‘B Coefficient Values’. Take for all B-coefficients correspond to zeroth row and zeroth
column as zero (Bi0 and B0i are zero, where i = 1, 2… n, No. of generating units). After entering the
respective values of B-Coefficients, click ‘save’ for every value of B-Coefficient.
7. Check ‘B in PU’, if B-Coefficients are given in pu. Now, take the ‘Initial Lamda Value’ as ‘5’and enter
in the respective place, then, click ‘Save’ to save all the values.
8. Now, click ‘Execute’ to run economic dispatch problem. The output (solution) will be given in a new
window‘Notepad’.
RESULT: Thus, the economic load dispatch or optimum generation schedule for a given power system was
determined by using B-Coefficients.
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Viva questions:
1.
2.
3.
4.
5.
What do you mean by a load flow analysis?
Which are the four quantities associated with each bus in a power system?
Name the different types of buses, their specifications and unknowns?
Name the different methods for carrying out load flow solution.
Which are the factors to be considered in the evaluation of the methods for obtaining a load
flow solution.
6. How do you compare Gauss Seidel and Newton- Raphson method.
7. Mention the quantities for which the tolerance is specified in case of Gauss Seidel and
Newton Raphson method?
8. What do you mean by Short Circuit calculations?
9. What is the use of information provided by a short circuit study?
10. Name type of faults involved in majority of short circuit study?
11. Define ABCD parameters.
12. What is the expression for the bad capacity of a specific transmission link?
13. State the criterion for optimum dispatch in a power system without considering transmission
losses?
14. State the criterion for optimum dispatch in a power system considering transmission losses?
15. What do you mean by “Incremental transmission losses?
16. Justify the need for economic operation of a power system?
17. State the significance of swing curve.
18. What is critical clearing angle and critical clearing time?
19. Define swing equation.
20. What is meant by a per unit diagram?
21. What are the advantages of the per unit system?
22. What is a bus? What are the different types of buses in power system?
23. What is the need of base values?
24. What is single line diagram?
25. What is bus impedance matrix?
26. What is bus admittance matrix?
27. Define load flow studies?
28. What is the importance of Load Flow Studies?
29. What are the advantages of iterative methods over Network Analysers?
30. Name the different methods of studying load flows?
31. What are the advantages of Gauss Siedal Methods?
32. What are the advantages of Newton Raphson Method?
33. What are the advantages of Fast Decoupled Method?
34. What is Slack bus?
35. What is the importance of slack bus?
36. How many slack buses can be there in a power system?
37. What is meant by load bus?
38. What are the specifications in the load bus?
39. What is meant by voltage bus?
40. What are the specifications in voltage bus?
41. What is the value of Q if it falls below the constraint?
42. What is the value of Q if it falls above the constraint?
43. What are the advantages of Y bus?
44. What are the static load flow equations?
45. What is meant by Acceleration factor?
46. What are the approximations made in impedance diagram?
47. What is Jacobian Matrix?
Department of EEE, AIT Bangalore
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48. What are the advantages of Jacobian Matrix?
49. What is meant by incidence matrix?
50. What is meant by primitive matrix?
51. What do you mean by flat voltage start?
52. How the convergence of N.R. method is speeded up?
53. What is infinite bus?
54. Mention the various methods of voltage control employed in power system?
55. What is synchronous capacitor?
56. What is regulating transformer and booting transformer?
57. What is meant by a fault?
58. How faults occur in power systems?
59. How the faults are classified?
60. What is meant by fault calculations?
61. What is the reason for transients during short circuits?
62. Which is most severe fault in P.S?
63. Give the sequence diagrams of an unleaded generator for various faults?
64. What is meant by stability?
65. What are the different types of stabilities?
66. What is meant by dynamic stability?
67. What are the methods by which steady state stability is assessed?
68. What are the methods of analyzing transient stability?
69. What is meant by equal area criterion?
70. What is meant by swing equation?
71. Write the general power angle characteristics?
72. What is meant by step-by-step method of obtaining the transient stability? What is meant by
swing curve?
73. List the assumptions made in step-by-step method?
74. Differentiate between symmetrical and unsymmetrical faults?
75. What is meant by short transmission lines?
76. What is meant by medium transmission lines?
77. What are the models of a medium transmission lines?
78. What are ABCD constants?
79. Write the expression for ABCD for a long transmission line?
80. Why Z-bus algorithm is preferred for short circuit studies?
81. Why Z-bus algorithm is preferred for load flow studies?
82. What is meant by Economic load dispatch?
83. What is meant by Penalty factor?
84. Write the quadratic polynomial for the Cost equation?
85. What is meant by Loss co-efficient?
86. What is meant by swing curve?
87. Write down the units of inertia constants M and H and their interrelation ship?
88. How convergence of NR method can be speeded up?
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Power System Simulation Lab
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Typical questions on experiments
1. A three phase, 100 km, transmission line is delivering 50 MW, 0.8 pf lagging, at 132 kV.
Each conductor is having a resistance of 0.1Ω/km, reactance of 0.3Ω/ km and admittance of
3x10-6 mho/km. If load is balanced and leakage current is neglected, Calculate the sending
end voltage, sending end current, sending end power factor, efficiency and the regulation of
the line using nominal – T representation. Write an appropriate MATLAB Code to solve the
problem.
2. A three phase, 100 km, transmission line is delivering 50 MW, 0.8 pf lagging, at 132 kV.
Each conductor is having a resistance of 0.1Ω/km, reactance of 0.3Ω/ km and admittance of
3x10-6 mho/km. If load is balanced and leakage current is neglected, Calculate the sending
end voltage, sending end current, sending end power factor, efficiency and the regulation of
the line using nominal π representation. Write an appropriate MATLAB Code to solve the
problem.
3. A balanced 3 phase load of 30 MW is supplied at 132 kV, 50 Hz and 0.85 pf lagging by means
of a transmission line. The series impedance of a single conductor is (20+52j) and the total
phase to neutral admittance is j315x10-6 Siemen. Use nominal – T method to determine
sending end voltage, sending end current, sending end power factor, efficiency and regulation
of the line. Write an appropriate MATLAB Code to solve the problem.
4. A balanced 3 phase load of 30 MW is supplied at 132 kV, 50 Hz and 0.85 pf lagging by means
of a transmission line. The series impedance of a single conductor is (20+52j) and the total
phase to neutral admittance is j315x10-6 Siemen. Use nominal π method to determine sending
end voltage, sending end current, sending end power factor, efficiency and regulation of the
line. Write an appropriate MATLAB Code to solve the problem.
5. A synchronous generator is rated 645 MVA, 24 kV, 0.9 pf lagging. It has a synchronous
reactance of 1.2 Ω. The generator is feeding a full load at 0.9 pf at rated voltage. Calculate i)
Excitation e.m.f and power angle δ. ii) Reactive power drawn by the load. Write an
appropriate code to calculate the specified parameters and draw the power angle curve using
MATLAB Package.
6. A 34.64 kV, 60 MVA, Synchronous generator has a direct axis reactance of 13.5 Ω and
quadrature axis reactance of 9.333 Ω is operating at 0.8 pf determine the excitation e.m.f,
Regulation, reluctance power and write an appropriate MATLAB code to solve the problem
and to draw power angle characters of a salient pole synchronous machine.
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7. A salient pole machine connected to infinite bus having the rating of P = 48MW, Rated
voltage of 11kV at pf 0.8 for a value of Xq= 10Ω and Xd=15Ω. Find out reluctance power,
power, excitation e.m.f and regulation for a salient pole. Draw the power angle curve of the
machine. Write an appropriate code in MATLAB to generate the necessary calculated
parameters & hence the curve.
8. Program to find critical clearing time for a single machine connected to infinite bus through
a pair of transmission line having three phase fault on one of the line away from the sending
end. Time of fault clearing=0.12, Momentum constant= 0.0331, Mechanical power 0.8 p.u,
maximum power angle before fault= 1.714, Maximum power during the fault= 0.63,
maximum power after fault=1.333 pu, δ=0.485, angular velocity (w)=0 inertia constant =5.2
9. Solve the below stated problem and write an appropriate MATLAB Code to form YBUS by
Inspection method.
STARTIN
G BUS
ENDING BUS
1
1
2
2
3
4
2
3
3
4
4
5
IMPEDANCE
REAL
0
0
0
0
0
0
IMAGINARY
1.00
0.80
0.40
0.20
0.20
0.08
10. Solve the below stated problem and write an appropriate MATLAB Code to form YBUS by
Singular transformation technique.
ELEMENT
NO
1
2
3
4
5
Department of EEE, AIT Bangalore
BUS CODE
1–3
1–4
1–2
4–2
3–2
IMPEDANCE
REAL
IMAGINARY
0
0.15
0
0.10
0
0.20
0
0.15
0
0.10
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Power System Simulation Lab
2022-23
11. Use Mi-Power software package to perform load flow analysis using Gauss Seidel method
for the problem given below. The data shows a simple three bus power system. Assume pf to
be 0.8 and1000 MVA at Generator Bus1.
BUS
NO
1
2
3
BUS
VOLTAGE
1.05+0j
--1.04+0j
GENERATION
MW MVAR
------0
0
200
----
LOAD
MW MVAR
0
0
400
250
0
0
Bus1
(Slack)
Bus2(PQ)
Bus3(PV)
BUS CODE IMPEDANCE
1–2
1–3
2–3
0.02+j0.04
0.01+j0.03
0.0125+j0.025
12. Use Mi-Power software package to perform load flow analysis using NR Method for the
problem given below. The figure shows a simple three bus power system. Line impedances
are marked in per unit on 100 MVA base and line charging susceptances are neglected.
Bus no
1
2
3
Bus
Voltage
1.05+0j
----Bus
Code
1–2
1–3
2–3
Generation
MW MVAR
------0
0
0
0
Impedance
Load
MW
MVAR
0
0
50
20
60
25
Bus1
(Slack)
Bus3
0.08+j0.24
0.02+j0.06
0.06+j0.18
13. Use Mi-Power software package to perform load flow analysis using FDLF Method for the
problem given below. The figure shows a simple three bus power system. Line impedances
are marked in per unit on 100 MVA base & line charging susceptances are neglected.
Bus no
Bus
Voltage
1
2
3
1.05+0j
----Bus
Code
1–2
1–3
2–3
Generation
MW MVAR
------0
0
0
0
Load
MW
MVAR
0
0
50
20
60
25
Bus1
(Slack)
Bus2
Bus3
Impedance
0.08+j0.24
0.02+j0.06
0.06+j0.18
Department of EEE, AIT Bangalore
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Power System Simulation Lab
2022-23
14. Use Mi-Power software package to form Jacobian Matrix for the systemfor the problem
given below. The figure shows a simple three bus power system. Line impedances are marked
in per unit on 100 MVA base and line charging susceptances are neglected.
Bus no
Bus
Generation
Load
Voltage MW MVAR
MW
MVAR
Bus1
1
1.05+0j
------0
0
(Slack)
2
--0
0
50
20
Bus3
3
--0
0
60
25
Bus
Impedance
Code
1–2
0.08+j0.24
1–3
0.02+j0.06
2–3
0.06+j0.18
15. Cost equation and loss coefficients of three units in a plant is given by,
𝐶1 = 0.05𝑃12 + 20𝑃1 + 800; 0 ≤ 𝑃1 ≤ 100
𝐶2 = 0.06𝑃22 + 15𝑃2 + 1000; 0 ≤ 𝑃2 ≤ 100
𝐶3 = 0.07𝑃32 + 20𝑃3 + 900; 0 ≤ 𝑃3 ≤ 100
B11 = 0.0005, B22 = 0.0004, B33 = 0.0005, B12 = B21 = 0.0005, B23 = B32 = 0.00018,
B13 = B31 = 0.0002, Assume λ=5, Pst=90MW
Determine the economic generation using MiPower Software Package for a total demand
of 240 MW, With Transmission losses.
16. For a given circuit find the fault current and voltage at bus 3. Analyze the circuit using
MiPower software Packagefor the case of SLG (Single Line to Ground) Fault.
𝐵𝑢𝑠1(11kV)
𝑋𝑑′ = 0.1
11𝑘𝑉
100𝑀𝑉𝐴
𝑋𝑡 = 0.1
𝐵𝑢𝑠2(220kV)
𝐵𝑢𝑠3(220kV)
𝑋𝐿 = 0.4
G
18. For a given circuit find the fault current and voltage at bus 3. Analyze the circuit using
MiPower software Packagefor the case of LL(Line to Line) Fault &LLG Fault.
𝐵𝑢𝑠1(11kV)
𝑋𝑑′ = 0.1
11𝑘𝑉
100𝑀𝑉𝐴
G
Department of EEE, AIT Bangalore
𝑋𝑡 = 0.1
𝐵𝑢𝑠2(220kV)
𝐵𝑢𝑠3(220kV)
𝑋𝐿 = 0.4
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