Lanjutan Contoh Penyelesaian Aliran Daya Listrik dengan Metode Newton-Raphson, Decoupled dan Fast Decoupled Load Flow (2)
Melanjutkan tulisan terdahulu, kita sudah mempelajari bagaimana menyelesaikan aliran daya di 3 bus secara analitis, baik dengan metode Newton-Raphson maupun dengan DLF dan Fast Decoupled Load Flow. Kita juga sudah belajar mengenal software PowerWorld yang membantu kita menganalisis STL. Sekarang kita akan mengkonfirmasi hasil-hasil perhitungan kita dengan software PowerWorld ini.
Yang pertama kita simulasikan dengan metode NR, hasilnya sbb:
Bus Records from Powerworld’s Full Newton Load Flow Solution
Name |
PU Volt |
Angle (Deg) |
Load MW |
Load Mvar |
Gen MW |
Gen Mvar |
1 |
1 |
0 |
|
|
99.96 |
25.77 |
2 |
1.03 |
1.32 |
|
|
400 |
418.55 |
3 |
0.93437 |
-5.38 |
500 |
350 |
|
|
Yang kedua kita simulasikan dengan memilih algoritma FDLF, yang hasilnya sbb:
Bus Records from Powerworld’s Fast Decoupled Load Flow Solution
Name |
PU Volt |
Angle (Deg) |
Load MW |
Load Mvar |
Gen MW |
Gen Mvar |
1 |
1 |
0 |
|
|
99.99 |
24.41 |
2 |
1.03 |
1.31 |
|
|
400 |
417.13 |
3 |
0.93485 |
-5.35 |
500 |
350 |
|
|
Ternyata perbandingan hasil kalkulasi tangan/manual dengan hasil perhitungan software sangat dekat, sehingga kita dapat memastikan bahwa kita menggunakan metode NR, DLF dan FDLF dengan benar.
Selain PowerWorld, sebenarnya ada banyak software yang dapat membantu kita menyelesaikan aliran daya STL, termasuk PSS®E dari Siemens yang merupakan software “wajib” yang banyak dipakai oleh designer ataupun operator STL di seluruh dunia.
Untuk para akademisi/pelajar, biasanya juga banyak memakai Matlab untuk menyelesaikannya. Di Matlab kita harus teliti, namun selama kita memahami algoritmanya, hasilnya juga akan sama dengan software2 lain.
Source code Matlab dengan metode NR untuk permasalahan ini :
% % Question #5A Newton-Raphson method
% Modified from Saadat’s Power System Analysis Example 6.10
clear; clc;
V = [1.0; 1.03; 1.0];
d = [0; 0; 0];
Ps=[4 ; -5];
Qs= -3.5;
YB = [ -j*75 j*50 j*25
j*50 -j*75 j*25
j*25 j*25 -j*50];
Y= abs(YB); t = angle(YB);
iter=0;
while iter < 2
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(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-V(3)^2*Y(3,3)*sin(t(3,3))- …
V(2)*V(3)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
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(2)*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(1)*Y(3,1)*cos(t(3,1)-d(3)+d(1))+…
V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2));
J(3,1)=-V(3)*V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2));
J(3,2)=V(2)*V(3)*Y(3,2)*cos(t(3,2)-d(3)+d(2))+…
V(1)*V(3)*Y(3,1)*cos(t(3,1)-d(3)+d(1));
J(3,3)=-V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-2*V(3)*Y(3,3)*sin(t(3,3))-…
V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
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(3)= V(3)+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))
Q2=-V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-V(3)*V(2)*Y(2,3)*…
sin(t(2,3)-d(2)+d(3))-V(2)^2*Y(2,2)*sin(t(2,2))
P_loss = P1+4-5
Q_loss = Q1+Q2-3.5
Kode Matlab dengan FDLF :
% Question #5C Fast decoupled method
% Modified from Saadat’s Power System Analysis Example 6.12
clear; clc;
V1= 1.0; V2 = 1.03; V3 = 1.0;
d1 = 0; d2 = 0; d3=0;
Ps2=4; Ps3 =-5;
Qs3= -3.5;
YB = [ -j*75 j*50 j*25
j*50 -j*75 j*25
j*25 j*25 -j*50];
Y= abs(YB); t = angle(YB);
B1 =[-75 25; 25 -50]
Binv = inv(B1)
iter=0;
while iter < 2
iter = iter +1;
P2= V2*V1*Y(2,1)*cos(t(2,1)-d2+d1)+V2^2*Y(2,2)*cos(t(2,2))+ …
V2*V3*Y(2,3)*cos(t(2,3)-d2+d3);
P3= V3*V1*Y(3,1)*cos(t(3,1)-d3+d1)+V3^2*Y(3,3)*cos(t(3,3))+ …
V3*V2*Y(3,2)*cos(t(3,2)-d3+d2);
Q3=-V3*V1*Y(3,1)*sin(t(3,1)-d3+d1)-V3^2*Y(3,3)*sin(t(3,3))- …
V2*V3*Y(3,2)*sin(t(3,2)-d3+d2);
DP2 = Ps2 – P2; DP2V = DP2/V2;
DP3 = Ps3 – P3; DP3V = DP3/V3;
DQ3 = Qs3 – Q3; DQ3V = DQ3/V3;
DC =[DP2; DP3; DQ3];
Dd = -Binv*[DP2V;DP3V];
DV = -1/B1(2,2)*DQ3V
d2 =d2+Dd(1);
d3 =d3+Dd(2);
V3= V3+DV;
angle2 =180/pi*d2;
angle3 =180/pi*d3;
disp(‘ iter d2 d3 V3 DP2 DP3 DQ3’);
R = [iter d2 d3 V3 DP2 DP3 DQ3]
end
Q2=-V2*V1*Y(2,1)*sin(t(2,1)-d2+d1)-V2^2*Y(2,2)*sin(t(2,2))- …
V2*V3*Y(2,3)*sin(t(2,3)-d2+d3);
P1= V1^2*Y(1,1)*cos(t(1,1))+V1*V2*Y(1,2)*cos(t(1,2)-d1+d2)+ …
V1*V3*Y(1,3)*cos(t(1,3)-d1+d3);
Q1=-V1^2*Y(1,1)*sin(t(1,1))-V1*V2*Y(1,2)*sin(t(1,2)-d1+d2)- …
V1*V3*Y(1,3)*sin(t(1,3)-d1+d3);
S1=P1+j*Q1
Q2
P_loss = P1+4-5
Q_loss = Q1+Q2-3.5
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