LAPORAN

ANILISA SISTEM TENAGA LISTRIK MENGGUNAKAN SIMULASI

POWER WORLD DAN METODE NEWTON RAPHSON

MENGGUNAKAN MATLAB DAN POWER WORLD

ANGGA NUR RAHMAT

S1 PTE

UNIVERSITAS NEGERI MALANG

FAKULTAS TEKNIK

JURUSAN TEKNIK ELEKTRO

PRODI SI PENDIDIKAN TEKNIK ELEKTRO

Desember 2012

1. Tujuan Praktikum

Agar mahasiswa dapat mengetahui sintak-sintak yang ada pada matlab dan simulasi

menggunakan power world yang akan digunakan sebagai perhitungan analisa sitem tenaga

listrik.

Agar mahasiswa mampu menghitung kesalahan dan mengetahui segala letak kesalahan

perhitungan sistem tenaga menggunakan metode Newton Raphson.

1. Dasar teori

Matlab

Matlab merupakan bahasa pemrograman yang hadir dengan fungsi dan karakteristik yang

berbeda dengan bahasa pemrograman lain yang sudah ada lebih dahulu seperti Delphi, Basic

maupun C++. Matlab merupakan bahasa pemrograman level tinggi yang dikhususkan untuk

kebutuhan komputasi teknis, visualisasi dan pemrograman seperti komputasi matematik,

analisis data, pengembangan algoritma, simulasi dan pemodelan dan grafik-grafik

perhitungan. Matlab hadir dengan membawa warna yang berbeda. Hal ini karena matlab

membawa keistimewaan dalam fungsi-fungsi matematika, fisika, statistik, dan visualisasi.

Matlab dikembangkan oleh MathWorks, yang pada awalnya dibuat untuk memberikan

kemudahan mengakses data matrik pada proyek LINPACK dan EISPACK. Saat ini matlab

memiliki ratusan fungsi yang dapat digunakan sebagai problem solver mulai dari simple

sampai masalah-masalah yang kompleks dari berbagai disiplin ilmu.

Current Directory

Window ini menampilkan isi dari direktori kerja saat menggunakan matlab. Kita

dapat mengganti direktori ini sesuai dengan tempat direktori kerja yang diinginkan. Default

dari alamat direktori berada dalam folder works tempat program files Matlab berada.

Command History

Window ini berfungsi untuk menyimpan perintah-perintah apa saja yang sebelumnya

dilakukan oleh pengguna terhadap matlab.

Command Window

Window ini adalah window utama dari Matlab. Disini adalah tempat untuk

menjalankan fungsi, mendeklarasikan variable, menjalankan proses-proses , serta melihat isi

variable.

Workspace

Workspace berfungsi untuk menampilkan seluruh variabel-variabel yang sedang aktif

pada saat pemakaian matlab. Apabila variabel berupa data matriks berukuran besar maka

user dapat melihat isi dari seluruh data dengan melakukan double klik pada variabel tersebut.

Matlab secara otomatis akan menampilkan window “array editor” yang berisikan data pada

setiap variabel yang dipilih user.

Power World

Power word simulator adalah software simulasi yang digunakan untuk mensimulasikan

sistem tenaga listrik. Dengan power word kita dapat membuat sistem tenaga listrik dan

kemudian mensimulasikanya dengan mengubah parameter yang ada dalam setiam komponen

yang digunakan sesuai dengan yang kita inginkan dan kemudian membuat analisisnya.

2. Alat Dan Bahan

Komputer

Software Matlab

Software Power world

3. Cara Kerja

Matlab

1. Buka program matlab.

2. Ketik program sebagai berikut sesuai

dengan program yang akan di run.

3. Jalankan program(tekan run/f8)

4. Amati hasilnya

5. Tulis hasil praktek pada laporan

Power world1. Buka program power world.

2. Mulai membuat sistem tenaga listrik

dengan komponen yang sudah

tersedia pada simulator power word.

3. Jalankan program

4. Amati hasilnya

5. Tulis hasil praktek pada laporan

Simulasi Power World

Simulasi metode NR :

Algoritma FDLF :

Bus Known Unknown(1) Slack V1, δ1 P1, Q1

(2) Generator P2, |V2| Q2, δ2

(3) Load P3, Q3 |V3|, δ3

Matlab scribt

% % 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*25j*50 -j*75 j*25j*25 j*25 -j*50];Y= abs(YB); t = angle(YB);

iter=0;b=input ('masukan perulangan = ');

for a=drange (1:b)iter = iter+1P=[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]JDX = J\DCd(2) =d(2)+DX(1);d(3)=d(3) +DX(2);V(3)= V(3)+DX(3);V, d, delta =180/pi*d;

fprintf ('%f \n', b);

endP1= 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+4-5Q_loss = Q1+Q2-3.5

Perhitungan

YBus = − j75 j 50 j25

j 50 − j 75 j25j 25 j 25 − j50

Merubah matrik bus admittance menjadi bentuk polar

YBus = 75←90 50<90 25<9050<90 75←90 25<9025<90 25<90 50←90

P1 = ∑j=1

n

❑.|Vj|.|Yij| cos (θij – δi + δj) . . . . (1)

Q1= −∑j=1

n

❑.|V1|.|Vj|.|Yij| sin (θij – δi + δj) . . (2)

Dari bentuk (1) dan (2), ekpresi 1 dari tegangan asli pada bus 2 dan 3 dan tegangan reaktiv pada bus 3 adalah . . .P2 = |V2|.|V1|.|Y21| cos (θ21 – δ2 + δ1) +|V2| 2.|Y22| cos (θ22) + |V2|.|V3|.|Y23| cos (θ23 – δ2 + δ3)P3 = |V3|.|V1|.|Y31| cos (θ31 – δ3 + δ1) + |V3|.|V2|.|Y32| cos (θ22 - δ3 + δ2) + |V3|2.|Y33| cos (θ23)Q3 = - |V3|.|V1|.|Y31| sin (θ31 – δ3 + δ1) + |V3|.|V2|.|Y32| sin (θ32 - δ3 + δ2) +

- |V3|2.|Y33| sin (θ33)

Mengganti Yij dan θijP2 = 50.|V2|.|V1| cos (90 – δ2 + δ1) + 25.|V2|.|V3|cos (90 – δ2 + δ3)P2 = 25.|V3|.|V1| cos (90 – δ3 + δ1) + 25.|V1|.|V2|cos (90 – δ3 + δ2)Q3 = - {25.|V3|.|V1| sin (90 – δ3 + δ1) + 25.|V3|.|V2| sin (90 – δ3 + δ2) – 50 |V3|2}

Matrik jacobian dan persamaannya

¿ = [△P2

△P3

△Q3]

Elements of the jacobian matrix are obtained by taking partial derivates of the above equations with respect to δ2, δ3, and |V3|∂ P 2∂ δ 2 = 50.|V2|.|V1| cos (90 – δ2 + δ1) + 25.|V2|.|V3|cos (90 – δ2 + δ3)

∂ P 2∂ δ 2 = - 25.|V2|.|V3| sin (90 – δ2 + δ3)

∂ P 2∂∨V 3∨¿¿ = 25.|V2| cos (90 – δ2 + δ3)

∂ P 3∂ δ 2 = - 25.|V3|.|V2| sin (90 – δ3 + δ2)

∂ P 3∂ δ 3 = 25.|V3|.|V1| sin (90 – δ3 + δ1) + 25.|V3|.|V2| sin (90 – δ3 + δ2)

∂ P3∂∨V 3∨¿¿ = 25.|V1| cos (90 – δ3 + δ1) + 25 |V2| cos (90 – δ3 + δ2)

∂Q 3∂ δ 2 = - 25.|V3|.|V2| cos (90 – δ3 + δ2)

∂Q 3∂ δ 3 = 25.|V3|.|V1| cos (90 – δ3 + δ1) + 25.|V3|.|V2| cos (90 – δ3 + δ2)

∂Q 3∂∨V 3∨¿¿ = - 25.|V1| sin (90 – δ3 + δ1) - 25.|V2| sin (90 – δ3 + δ2) + 100 |V3|

The load and generation expressed in per unit are P2

sch = P2g - P2

d = 4 – 0 = 4 puS3

sch = S3g - S3

d = 0 – (5 + j 3.5) = - 5 – j 3.5 pu

The slack bus voltage is V1 = 1 < 0 pu and the bus 2 voltage magnitude is |V2| = 1.03 pu. Starting with an initial estimate of |V3

(0)| = 1.0, δ3(0) = 0.0 and δ2

(0) = 0.0, the power residual are computed from∆Pi

(K) = Pisch – Pi

(K) ∆Qi

(K) = Qisch – Qi

(K)

Which yields ∆P2

(0) = P2sch – P2

(0) = 4 – 0 = 4∆P3

(0) = P3sch – P3

(0) = -5 – 0 = -5∆Q3

(0) = Q3sch – Q3

(0) = -3.5 – (-0.75) = - 2.75

Evaluating the elements of the jacobian matrix with the initial estimate, the set of linier equations in the first iteration are

[ 4−5

−2.75]=[ 77.25 −25.75 0−25.75 50.75 0

0 0 49.25]¿

Observing the solution of the above matrix equation, the new bus voltages in the first rheration are∆δ2

(0) = 0.0228

∆δ3(0) = - 0.0870

∆ |V3(0)| = - 0.0558

δ2(1) = 0 + 0.0228 = 0.0228

δ3(1) = 0 + (-0.0870) = -0.0870

|V3(1)| = 1 + (-0.0558) = 0.9442

Voltage phase angles are in radians.Elements of the jacobian matrix for the second iteration.∂ P 2∂ δ 2

(1) = 50.(1.03) sin (90 – 0.0228) + 25 (1.03)(0.9442) sin (90 – 0.0228 – 0.087)

∂ P 2∂ δ 3

(1) = - 25.(1.03)(0.9442) sin (90 – 0.0228 – 0.087)

∂ P 2∂∨V 3∨¿¿

(1) = 25.(1.03) cos (90 – 0.0228 – 0.087)

∂ P 3∂ δ 2

(1) = - 25 (0.9442)(1.03) sin (90 + 0.087 + 0.0228)

∂ P 3∂ δ 3

(1) = 25 (0.9442) sin (90 + 0.087) + 25 (0.9442) (1.03) sin (90 + 0.087 + 0.0228)

∂ P3∂∨V 3∨¿¿

(1) = 25 cos (90 + 0.087 + 0) + 25 (1.03) cos (90 + 0.087 + 0.0228)

∂Q 3∂ δ 2

(1) = 25 (0.9442) (1) cos (90 + 0.087) + 25 (0.9442) (1.03) cos (90 + 0.087 + 0.0228)

∂Q 3∂∨V 3∨¿¿

(1) = - 25 sin (90 + 0.087) – 25 (1.03) sin (90 + 0.087 + 0.0228) + 100 (0.9442)

The real power at bus 2 and 3 the reactive power at bus 3 areP2

(1) = (1.03) (50) cos (90 – 0.0228) + (25) (0.9442) (1.03) cos (90 – 0.0228 – 0.087)P3

(1) = (25) (0.9442) cos (90 – 0.087) + (25) (0.9442) (1.03) cos (90 + 0.087 + 0.0228)Q3

(1) = - {(25) (0.9442) sin (90 – 0.087) + (25) (0.9442) (1.03) sin (90 + 0.087 + 0.0228) – (50) (0.9442)2}

The power residuals for second iteration are∆P2

(1) = P2sch – P2

(1) = 4 – 3.84 = 0.16∆P3

(1) = P3sch – P3

(1) = -5 – (- 4.72) = - 0.28∆Q3

(1) = Q3sch – Q3

(1) = -3.5 – (- 3.11) = - 2.39

The of linier equations in the second iteration is

[ 0.16−0.28−0.39]=[ 75.65 −24.17 2.82

−24.17 47.68 −4.992.66 −4.72 43.92 ]¿

∆δ2(1)

= 0.0003∆δ3

(1) = - 0.0067∆ |V3

(1)| = - 0.0096

δ2(2) = 0.0228 + 0.0003 = 0.0231 rad

δ3(2) = -0.0870 + (- 0.0067) = - 0.0937 rad

|V3(2)| = 0.9442 + (-0.0096) = 0.9346 pu

Decoupled power flowBy setting elements J2 and J3 of the Jacobian Matrix to zero, the equation becomes

[△P△Q ]=[ J1 0

0 J 4]¿∆P = J1 . Δδ = (

∂ P∂ δ ) . Δδ

∆Q = J4 – Δ |V| = ( ∂Q

∂∨V ∨¿¿ ) . Δ |V|

The set of linier equations in the first iteration (by using #5A known values)

[△ P1

△ P3]=[ ðP2

ðδ 2

ðP2

ð δ3

ðP3

ðδ 2

ðP3

ð δ3][△ δ2

△ δ3]

[ 4−5]=[ 77.25 −25.75

−25.75 50.75 ][∆ δ2(0 )

∆ δ3(0 )]

∆δ2(0)

= 0.0228∆δ3

(0) = - 0.087

And

[ΔQ3] = [∂Q 3

∂∨V 3∨¿¿] [Δ|V2|]

- 2.75 = (49.25) (Δ|V3|)Δ|V3(0)| = - 0.0558

δ2(1) = 0 + 0.0228 = 0.0228

δ3(1) = 0 + ( -0.0870) = - 0.0870

|V3(1)| = 1 + (-0.0558) = 0.9442

Voltage phase angels are in radiansElements of the J1 for the second iteration are (by using #5A known values)

J4 = [ 75.65 −24.17−24.17 47.68 ]

Elements of the J4 for the second iteration areJ4 = [43.92]

The set of linier euation in the second iteration

[ 0.16−0.28]=[ 75.65 −24.17

−24.17 47.68 ] [∆ δ 2(1)

∆ δ3(1)]

∆δ2(1)

= 0.0003∆δ3

(1) = - 0.0057

[-0.39] = [43.92].[Δ|V3(0)|]

Δ|V3(1) | = 0.0089

δ2(2) = 0.0228 + 0.0003 = 0.0231 rad

δ3(2) = -0.0870 + (- 0.0057) = - 0.0927 rad

|V3(2)| = 0.9442 + (-0.0084) = 0.9531 pu

The original bus admittance matrix of the system is

ybus=[− j75 j 50 j 25j50 − j75 j 25j25 j 25 − j 50]

In this system, bus I is the slack bus and the corresponding bus susceptance matriks for evaluatuion of phase angles ∆δ2 and ∆δ3 is

B’ = [−75 2525 −50 ]

The inverse of the above matrix is

[B’]-1 = [−0.016 −0.008−0.008 −0.024]

From #5A, the expression for real power at bus 2 and 3 and the reactive power at bus 3 areP2 = 50.|V2|.|V1| cos (90 – δ2 + δ1) + 25.|V2|.|V3|cos (90 – δ2 + δ3)P2 = 25.|V3|.|V1| cos (90 – δ3 + δ1) + 25.|V1|.|V2|cos (90 – δ3 + δ2)Q3 = - 25.|V3|.|V1| sin (90 – δ3 + δ1) + - (25).|V3|.|V2| sin (90 – δ3 + δ2) + 50 |V3|2} The load and generation espressed per unit are P2

sch = P2g – P2

(d) = 4 – 0 = 4 pu S3

sch = s3g – S3

(d) = 0– (5 + j 3.3) = -5 – j 3.5 V

The slack bus voltage is V1 = 1< 0 pu, and the bus 2 voltage magnitude is |V2| = 1.03 pu. Starting with an initial estimate of |V3

(1)| = 1.0, δ3(0) = 0.0 and δ2

(0) = 0.0, the power residuals are computed from (3) and (4)∆P2

(0) = P2sch – P2

(0) = 4 – 0 = 4∆P3

(0) = P3sch – P3

(0) = -5 – 0 = - 5∆Q3

(0) = Q3sch – Q3

(0) = -3.5 – (- 0.75) = - 2.75The fast decoupled power flow algorithm given by

Δδ = - [B1]-1 ΔP¿V∨¿¿

Becomes

[∆ δ 2(0)

∆ δ 3(0)] = [−0.016 −0.008

−0.008 −0.024] [ 4103−51 ]=[ 0.0221

−0.0889]

Since bus 2 is a regulated bus, the corresponding row and column of B1 are elimated and we getB11 = [-50]

From ∆ |V| = - [B11]-1 ΔP

¿V∨¿¿ , we have

∆ |V3(0)| = - [

−150

¿ [ −2.75

1 ] = - 0.055

∆δ2(0)

= 0.0221∆δ3

(0) = - 0.0889∆ |V3

(0)| = - 0.055The new bus voltages in the first iteration are δ2

(1) = 0 + 0.0221 = 0.0221δ3

(1) = 0 + ( -0.0889) = - 0.0889

|V3(1)| = 1 + (-0.055) = 0.945

The voltage phase angels are in radians.

The process in continued to second iteration.

[∆ δ 2(1)

∆ δ 3(1)]=−[−0.016 −0.008

−0.008 −0.024 ]¿P2

(1) = (50) (1.03) (1) cos (90 – 0.0221) + (25) (1.03) (0.945) cos (90 – 0.0221 – 0.0889)P3

(1) = (25) (0.945) cos (90 – 0.0889) + (25) (0.945) (1.03) cos (90 + 0.0889 + 0.0221)Q3

(1) = - (25) (0.945) sin (90 – 0.0889) + (25) (0.945) (1.03) sin (90 + 0.08889 + 0.0221) – (50) (0.945)2

The power residuals for second iteration are∆P2

(1) = P2sch – P2

(1) = 4 – 3.8336 = 0.1664∆P3

(1) = P3sch – P3

(1) = -5 – (- 4.7930) = - 0.207∆Q3

(1) = Q3sch – Q3

(1) = -3.5 – (- 3.0645) = - 0.4355

The linier equation for the second iteration

[∆ δ 2(1)

∆ δ 3(1)]=[−0.016 −0.008

−0.008 −0.024 ][ 0.16641.03

−0.2070.945 ]=[−0.0008

−0.004 ]

∆ |V3(1)| = - [

−150

¿ [ −0.4355

0.945 ] = - 0.0092

∆δ2(1)

= 0.0008∆δ3

(1) = - 0.0040∆ |V3

(1)| = - 0.0092

The new bus voltages in the second iteration areδ2

(1) = 0.0221 + 0.0008 = 0.0229 radδ3

(1) = - 0.0889 + (- 0.0040) = - 0.0929 rad|V3

(1)| = 0.945 + (-0.0092) = 0.9358 pu

Using (1) and (2), the reactive power atbus 2 and slack bus real and reactive power areP2 = |V1|2.|Y11| cos (θ11) + |V1|. |V2|.|Y12| cos (θ12 – δ1 + δ2) + |V1|.|V3|.|Y13| cos (θ13 – δ1 + δ3)P2 = - {|V3|2.|Y11| sin (θ11) + |V1|. |V2|.|Y12| sin (θ12 – δ1 + δ2) + |V1|.|V3|.|Y13| sin (θ13 – δ1 + δ3)}Q3 = - {|V2|.|V1.|Y21| sin (θ21 – δ2 + δ1) + |V2|2.|Y22| sin (θ22) + |V2|.|V3|.|Y23| sin (θ23 – δ2 + δ3)}

Subtituting the corresponding Yij and θij yields

P1 = 50.|V1|.|V2| cos (90 – δ1 + δ2) + 25.|V1|.|V3|cos (90 – δ1 + δ3)Q1 = +75.|V1|2 – 50 |V1|.|V2| sin (90 – δ1 + δ2) + 25.|V1|.|V3|cos (90 – δ1 + δ3)Q2 = -50.|V1|.|V2| sin (90 – δ2 + δ1) + 75.|V2|2 – 25 |V2|.|V3| sin (90 – δ2 + δ3)

All of NR, DLF, and FDLF algorithm use the same |V1| = 1, δ1 = 0 from the slack busP1 = 50.|V2| cos (90 + δ2) + 25.|V3|cos (90 + δ3)Q1 = +75 – 50 |V2| sin (90 + δ2) + 25.|V3|cos (90 + δ3)Q2 = -50.|V2| sin (90 – δ2) + 75.|V2|2 – 25 |V2|.|V3| sin (90 – δ2 + δ3)

NRP1

NR = (50) (1.03) cos (90 – 0.0231) + (25) (0.9346) cos (90 – 0.0937) = 0.9966Q1

NR = +75 – (50).(1.03) sin (90 + 0.0231) – (25).(0.9346) sin (90 – 0.0937) = 0.2512Q2

NR = - (50).(1.03) sin (90-0.0231) + (75).(1.03)2 – (25).(1.03).(0.9346) sin (90-0.0231-0.0937) = 4.1793

PLossNR = ΣPg - ΣPd = (P1

NR + P2) – P3

= (0.9966 + 4) – 5 = - 0.0034

QLossNR = ΣQg - ΣQd = (Q1

NR + Q2NR) – Q3

= (0.2512 + 4.1793) – 35 = 0.9305

DLFP1

DLF = (50) (1.03) cos (90 – 0.0231) + (25) (0.9531) cos (90 – 0.0927) = 1.0161Q1

DLF = 75 – (50).(1.03) sin (90 + 0.0231) – (25).(0.9531) sin (90 – 0.0927) = 0.2115Q2

DLF = - (50).(1.03) sin (90-0.0231) + (75).(1.03)2 – (25).(1.03).(0.9346) sin (90-0.0231-0.0927) = 3.7033

PLossDLF = (P1

DLF + P2) – P3

= (1.0161 + 4) – 5 = 0.0161

QLossDLF = (Q1

DLF + Q2) – Q3

= (0.2115 + 3.7033) – 3.5 = 0.4148

FDLFP1

FDLF = (50) (1.03) cos (90 – 0.0229) + (25) (0.9358) cos (90 – 0.0929) = 0.9910Q1

FDLF = 75 – (50).(1.03) sin (90 + 0.0229) – (25).(0.9358) sin (90 – 0.0929) = 0.2194

Q2FDLF = - (50).(1.03) sin (90-0.0229) + (75).(1.03)2 – (25).(1.03).(0.9358) sin (90-0.0229-

0.0929) = 4.1455PLoss

FDLF = (P1FDLF + P2) – P3

= (0.991 + 4) – 5 = - 0.009

QLosFDLF = (Q1

FDLF + Q2) – Q3

= (0.2194 + 4.1455) – 3.5 = 0.8649

Comparison of the results of NR, DLF, and FDLF methodsNR DLF FDLF

Slack bus powerP1 (pu) 0.9966 1.0161 0.991Q1 (pu) 0.2512 0.2115 0.2194

System lossesPLoss (pu) - 0.0034 0.0161 -0.009QLoss (pu) 0.9305 0.4148 0.8649

Kesimpulan Ketiga teori tersebut harus menghilangkan Ploss dalam sistem ketika resisiance tidak ada, mungkin karena jumlah iterasi tidak cukup besar (hanya 2 iterasi per metode), atau kesalahan pembulatan dalam menggunakan kalkulator.NR adalah metode yang paling akurat, dan FDLF adalah yang paling akurat, namun memberikan hasil tercepat.

Daftar Pustaka

PTE2009.Simulator Power Word Untuk Analisa Sistem Tenaga Listrik.Universitas Negeri Malang:Malang

PTE2009. Anilisa Sistem Tenaga Listrik Menggunakan Penyeleseian Metode Newton Raphson Dan Gaus Seidel Menggunakan Matlab.Universitas Negeri Malang:Malang

____.2009.http://imadudd1n.wordpress.com (online diakses pada 25 November 2012)