ASSIGNMENT NO:1
Gauss Seidel and Newton Raphson Method for Load Flow Analysis
Submitted To: Professor Asif Gulraiz
Submitted By: Engr. Shahnila Badar (EE-221601)
MAY 17, 2023
ELECTRICAL DEPARTMENT
DHA SUFFA University Karachi
ASSIGNMENT 1
Task: Use MATLAB to apply Gauss Seidel and Newton Raphson method on given
examples. Compare the results between simulation and hand calculated.
During simulation I found the results for Load lines, Load Losses and Power flow
results to be same as hand calculated values.
Load Flow analysis using Gauss Seidel Method
Example 1:
MATLAB Code:
clear all;
clc;
1
y12=10-j*20;
y13=10-j*30;
y23=16-j*32;
V1=1.05+j*0;
iter =0;
S2=-2.566-j*1.102;
S3=-1.386-j*.452;
V2=1+j*0;
V3=1+j*0;
for I=1:10;
iter=iter+1;
V2 = (conj(S2)/conj(V2)+y12*V1+y23*V3)/(y12+y23);
V3 = (conj(S3)/conj(V3)+y13*V1+y23*V2)/(y13+y23);
disp([iter, V2, V3])
end
V2= .98-j*.06;
V3= 1-j*.05;
I12=y12*(V1-V2); I21=-I12;
I13=y13*(V1-V3); I31=-I13;
I23=y23*(V2-V3); I32=-I23;
S12=V1*conj(I12); S21=V2*conj(I21);
S13=V1*conj(I13); S31=V3*conj(I31);
S23=V2*conj(I23); S32=V3*conj(I32);
I1221=[I12,I21]
I1331=[I13,I31]
I2332=[I23,I32]
S1221=[S12, S21 (S12+S13) S12+S21]
S1331=[S13, S31 (S31+S32) S13+S31]
S2332=[S23, S32 (S23+S21) S23+S32]
Simulation Results:
a) Phasor values of Voltage at Load bus 2 and 3 (P-Q Buses) Iterations 1 to 10
1.0000 + 0.0000i 0.9825 - 0.0310i 1.0011 - 0.0353i
2.0000 + 0.0000i 0.9816 - 0.0520i 1.0008 - 0.0459i
3.0000 + 0.0000i 0.9808 - 0.0578i 1.0004 - 0.0488i
4.0000 + 0.0000i 0.9803 - 0.0594i 1.0002 - 0.0497i
5.0000 + 0.0000i 0.9801 - 0.0598i 1.0001 - 0.0499i
6.0000 + 0.0000i 0.9801 - 0.0599i 1.0000 - 0.0500i
7.0000 + 0.0000i 0.9800 - 0.0600i 1.0000 - 0.0500i
2
8.0000 + 0.0000i 0.9800 - 0.0600i 1.0000 - 0.0500i
9.0000 + 0.0000i 0.9800 - 0.0600i 1.0000 - 0.0500i
10.0000 + 0.0000i 0.9800 - 0.0600i 1.0000 - 0.0500i
b) Line Current
I1221 =
1.9000 - 0.8000i -1.9000 + 0.8000i
I1331 =
2.0000 - 1.0000i -2.0000 + 1.0000i
I2332 =
-0.6400 + 0.4800i 0.6400 - 0.4800i
c) Line Flows and Line Losses
S1221 =
1.9950 + 0.8400i -1.9100 - 0.6700i 4.0950 + 1.8900i 0.0850 + 0.1700i
S1331 =
2.1000 + 1.0500i -2.0500 - 0.9000i -1.3860 - 0.4520i 0.0500 + 0.1500i
S2332 =
-0.6560 - 0.4320i 0.6640 + 0.4480i -2.5660 - 1.1020i 0.0080 + 0.0160i
3
Example 2:
MATLAB CODE:
%Power flow analysis using Gauss Seidal Method %
%Engr. Shahnila Badar EE-221601%
clc;
clear all;
y12=10-j*20;
y13=10-j*30;
y23=16-j*32;
y33=y13+y23;
V1=1.05+j*0;
4
format long
iter =0;
S2=-4.0-j*2.5;
P3 = 2;
V2=1+j*0;
Vm3=1.04;
V3=1.04+j*0;
for I=1:10;
iter=iter+1
E2 = V2;
E3=V3;
V2 = (conj(S2)/conj(V2)+y12*V1+y23*V3)/(y12+y23)
DV2 = V2-E2
Q3 = -imag(conj(V3)*(y33*V3-y13*V1-y23*V2))
S3 = P3 +j*Q3;
Vc3 = (conj(S3)/conj(V3)+y13*V1+y23*V2)/(y13+y23)
Vi3 = imag(Vc3);
Vr3= sqrt(Vm3^2 - Vi3^2);
V3 = Vr3 + j*Vi3
DV3=V3-E3
end
format short
I12=y12*(V1-V2); I21=-I12;
I13=y13*(V1-V3); I31=-I13;
I23=y23*(V2-V3); I32=-I23;
S12=V1*conj(I12); S21=V2*conj(I21);
S13=V1*conj(I13); S31=V3*conj(I31);
S23=V2*conj(I23); S32=V3*conj(I32);
I1221=[I12,I21]
I1331=[I13,I31]
I2332=[I23,I32]
S1221=[S12, S21 (S12+S13) S12+S21]
S1331=[S13, S31 (S31+S32) S13+S31]
S2332=[S23, S32 (S23+S21) S23+S32]
Simulation Result:
a) Phasor values of Voltage at Load bus 2 and 3 (P-Q Buses) Iterations 1 to 10
iter = 1
V2 = 0.974615384615385 - 0.042307692307692i
DV2 = -0.025384615384615 - 0.042307692307692i
Q3 = 1.160000000000001
Vc3 = 1.037831858407080 - 0.005170183798502i
V3 = 1.039987148574197 - 0.005170183798502i
DV3 = -0.000012851425803 - 0.005170183798502i
iter = 2
V2 = 0.971057059512953 - 0.043431876337850i
DV2 = -0.003558325102432 - 0.001124184030158i
Q3 = 1.387957731052817
Vc3 = 1.039081476179430 - 0.007300111679686i
V3 = 1.039974378708180 - 0.007300111679686i
5
DV3 = -0.000012769866018 - 0.002129927881184i
iter = 3
V2 = 0.970733708554698 - 0.044791724463619i
DV2 = -0.000323350958254 - 0.001359848125769i
Q3 = 1.429040300785471
Vc3 = 1.039536102030377 - 0.008325001047174i
V3 = 1.039966679445820 - 0.008325001047174i
DV3 = -0.000007699262360 - 0.001024889367487i
iter = 4
V2 = 0.970652437281433 - 0.045329920732880i
DV2 = -8.127127326573724e-05 - 5.381962692614858e-04i
Q3 = 1.448333275594840
Vc3 = 1.039783582412907 - 0.008752000354604i
V3 = 1.039963173621928 - 0.008752000354604i
DV3 = -3.505823891414295e-06 - 4.269993074304240e-04i
iter = 5
V2 = 0.970623655331095 - 0.045554240372625i
DV2 = -2.878195033773068e-05 - 2.243196397443484e-04i
Q3 = 1.456209166612119
Vc3 = 1.039887186964488 - 0.008929007616053i
V3 = 1.039961668920058 - 0.008929007616053i
DV3 = -1.504701870214120e-06 - 1.770072614492024e-04i
iter =
6
V2 = 0.970612037114234 - 0.045646940090561i
DV2 = -1.161821686124220e-05 - 9.269971793655907e-05i
Q3 = 1.459469889628077
Vc3 = 1.039930224431334 - 0.009002221658867i
V3 = 1.039961037734205 - 0.009002221658867i
DV3 = -6.311858531393710e-07 - 7.321404281408414e-05i
iter =
7
V2 = 0.970607253520093 - 0.045685276728252i
DV2 = -4.783594140689296e-06 - 3.833663769042817e-05i
Q3 = 1.460818201396914
Vc3 = 1.039948029840190 - 0.009032502820155i
V3 = 1.039960775170297 - 0.009032502820155i
DV3 = -2.625639083930764e-07 - 3.028116128733424e-05i
iter = 8
6
V2 = 0.970605276281561 - 0.045701131870879i
DV2 = -1.977238532457903e-06 - 1.585514262764792e-05i
Q3 = 1.461375872168914
Vc3 = 1.039955394551153 - 0.009045027392915i
V3 = 1.039960666313617 - 0.009045027392915i
DV3 = -1.088566798923551e-07 - 1.252457276070332e-05i
iter = 9
V2 0.970604458527297 - 0.045707689707255i
DV2 =-8.177542636378377e-07 - 6.557836375535586e-06i
Q3 = 1.461606535170453
Vc3 =1.039958440697760 - 0.009050207830587i
V3 = 1.039960621244008 - 0.009050207830587i
DV3 = -4.506960848971175e-08 - 5.180437672010207e-06i
iter =10
V2 = 0.970604120282796 - 0.045710402176455i
DV2 = -3.382445014077362e-07 - 2.712469200305545e-06i
Q3 = 1.461701943643423
Vc3 = 1.039959700654411 - 0.009052350604469i
V3 = 1.039960602594413 - 0.009052350604469i
DV3 = -1.864959564557012e-08 - 2.142773881465970e-06i
b) Line Current
I1221 = 1.7082 - 1.1308i -1.7082 + 1.1308i
I1331 = 0.3720 - 0.2107i -0.3720 + 0.2107i
I2332 = -2.2828 + 1.6329i 2.2828 - 1.6329i
c) Line Flow and Line Losses
S1221 = 1.7936 + 1.1874i -1.7096 - 1.0195i 2.1841 + 1.4085i 0.0839 + 0.1679i
S1331 = 0.3906 + 0.2212i -0.3887 - 0.2157i 2.0000 + 1.4618i 0.0018 + 0.0055i
S2332 = -2.2903 - 1.4805i 2.3888 + 1.6775i -3.9999 - 2.5000i 0.0985 + 0.1969i
7
Load Flow analysis using Newton Raphson Method
MATLAB Code:
% Newton-Raphson method
V = [1.05; 1.0; 1.04];
d = [0; 0; 0];
Ps=[-4; 2.0];
Qs= -2.5;
YB = [ 20-j*50 -10+j*20
-10+j*30
8
-10+j*20
26-j*52
-10+j*30 -16+j*32
Y= abs(YB); t = angle(YB);
-16+j*32
26-j*62];
iter=0;
pwracur = 0.00025; % Power accuracy
DC = 10;
% Set the maximum power residual to a high value
while max(abs(DC)) > pwracur
iter = iter +1
P=[V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+V(2)^2*Y(2,2)*cos(t(2,2))+ ...
V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
V(3)*V(1)*Y(3,1)*cos(t(3,1)-d(3)+d(1))+V(3)^2*Y(3,3)*cos(t(3,3))+ ...
V(3)*V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2))];
Q= -V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-V(2)^2*Y(2,2)*sin(t(2,2))- ...
V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,1)=V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))+...
V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,2)=-V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,3)=V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+2*V(2)*Y(2,2)*cos(t(2,2))+...
V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(2,1)=-V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
J(2,2)=V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))+...
V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
J(2,3)=V(3)*Y(2,3)*cos(t(3,2)-d(3)+d(2));
J(3,1)=V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+...
V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(3,2)=-V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(3,3)=-V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-2*V(2)*Y(2,2)*sin(t(2,2))-...
V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
DP = Ps - P;
DQ = Qs - Q;
DC = [DP; DQ]
J
DX = J\DC
d(2) =d(2)+DX(1);
d(3)=d(3) +DX(2);
V(2)= V(2)+DX(3);
V, d, delta =180/pi*d;
end
P1= V(1)^2*Y(1,1)*cos(t(1,1))+V(1)*V(2)*Y(1,2)*cos(t(1,2)-d(1)+d(2))+...
V(1)*V(3)*Y(1,3)*cos(t(1,3)-d(1)+d(3))
Q1=-V(1)^2*Y(1,1)*sin(t(1,1))-V(1)*V(2)*Y(1,2)*sin(t(1,2)-d(1)+d(2))-...
V(1)*V(3)*Y(1,3)*sin(t(1,3)-d(1)+d(3))
Q3=-V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-V(3)*V(2)*Y(3,2)*...
sin(t(3,2)-d(3)+d(2))-V(3)^2*Y(3,3)*sin(t(3,3))
Simulation Result:
iter =1
9
DC = -2.8600
1.4384
-0.2200
J = 54.2800 -33.2800 24.8600
-33.2800 66.0400 -16.6400
-27.1400 16.6400 49.7200
DX = -0.0453
-0.0077
-0.0265
V = 1.0500
0.9735
1.0400
d=0
-0.0453
-0.0077
iter = 2
DC =-0.0992
0.0217
-0.0509
J = 51.7247 -31.7656 21.3026
-32.9816 65.6564 -15.3791
-28.5386 17.4028 48.1036
DX = -0.0018
-0.0010
-0.0018
V =1.0500
0.9717
1.0400
d= 0
-0.0471
-0.0087
iter =3
DC = 1.0e-03 *
-0.2166
0.0382
-0.1430
J = 51.5967 -31.6939 21.1474
10
-32.9339 65.5976 -15.3516
-28.5482 17.3969 47.9549
DX = 1.0e-05 *
-0.3856
-0.2386
-0.4412
V = 1.0500
0.9717
1.0400
d= 0
-0.0471
-0.0087
P1 = 2.1842
Q1 =1.4085
Q3 = 1.4618
11