The document describes the calculation of power flow analysis on a 3-bus system using the Gauss-Seidel method. It provides the bus data, line impedance values, generator real and reactive power outputs. It then calculates the admittance matrix and performs iterative calculations to determine the voltage phase angle and magnitude at each bus. The results show the voltage values converging with iterations to within the specified tolerance of 0.0001 per unit. It also calculates the real and reactive power flow between buses.

1. Gambar
Slackbus (1)
GeneratorBus (2)
BEban Bus (3)
IndeksPRasisi =0.0001
Base MVA =100 mVA
Kv Dasar =150 Kv
𝛾 =8.5
Kode
Bus
Z Admitansi
1-2 0.017 + j 0.106 0
1-3 0.015 + j0.116 0
2-3 0.05 + j 0.033 0
Kode P Tegangan
Bus
Generator Beban Ket
MW MVA MW MVA
1 1.04+j0.00 - - - - Slack
2 1.0+j0.000 110 - 60 30
3 1.0+j0.000 - - 50 15
 Matriks Admitansi
YPq =
1
𝑌𝑝𝑞
Y12 =
1
𝑍12
Y12 =
1
0.017+𝑗0.106

2. Y12 =
1<0
0.10735<82.63119
= 1.474- j 9.1969
Y13 =
1
𝑍13
Y13 =
1
0.015+𝑗0.116
= 1.0966-j8.4833
Y23 =
1
𝑍23
Y23 =
1
0.05+𝑗0.033
= 13.9332-j9.1953
 Elemensukumatrikadmitansinyaadalah
Y11 = Y12 + Y13
=(1.474-j9.1969)+ (1.0966-j8.4833)
= 2.5706-j17.6802
Y22 = Y21 + Y23
= (1.474-j9.1969)+(13.9332 – j9.1953)
=15.4072-j18.3922
Y33=Y31 + Y23
= (1.0966-j8.4833)+(13.9332-j9.1953)
=15.0298 – j17.678
Y12=Y21 = -Y12 = -1.474+ J9.1969
Y13 = Y31=-Y13 = -1.0966+j8.4833
Y23=Y32 = -Y23 =-13.9332+J9.1953

3. Yrel = [
2.5706 − 𝑗 17.6802 −1.474 + 𝑗9.1969 −1.0966 + 𝑗8.4833
−1.474 + 𝑗9.1969 15.4072 − 𝑗18.3922 −13.9332 + 𝑗9.1953
−1.0966 + 𝑗8.4833 −13.9332 + 𝑗9.1953 15.0289 − 𝑗17.6786
]
PerhitunganKLP
KLP =
𝑃𝑝−𝐽𝑄𝑝
𝑌𝑝𝑝
Untuk P =2 (Bus2)
KL2 =
(1.1−𝑗0.0)−(0.6−𝑗0.3)
15.4072−𝑗18.3922
=
0.5+𝑗0.3
15.4072−𝑗18.3922
= 0.0037 + j0.024
Untuk P = 3
KL2 =
𝑃 𝑝3− 𝐽𝑄 𝑝3
𝑌33
=
(0.0−𝑗0.0)−(0.5−𝑗0.15)
15.0298−𝑗17.6786
=
−0.5+𝑗015
15.0298−𝑗17.6786
= 0.0184+j0.0119
PerhitunganYLPQ
YLPQ = YPQ-LP =
𝑌𝑃𝑄
𝑌𝑃𝑃
YL12 =
𝑌12
𝑌11
=
−1.474+𝑗9.1969
2.5076−𝑗17.6802
=0.52+j0.0075
YL13 =
𝑌13
𝑌11
=
−1.0966+𝑗8.4833
2.5076−𝑗17.6802

4. =0.478-j0.03
YL21=
𝑌21
𝑌22
=
−1.474+𝑗9.1969
15.4072−𝑗18.3922
=0.3332-j0.1990
YL23 =
𝑌23
𝑌22
=
−13.9332+𝑗9.1953
15.4072−𝑗18.3922
=0.6666+j0.19904
YL31 =
𝑌31
𝑌33
=
−1.0966+𝑗8.4833
15.0298−𝑗17.678
= 0.3090711-j0.1982
YL32=
𝑌32
𝑌33
=
−13.9332+𝑗9.1953
15.0298−𝑗17.678
=0.6908+j0.2006
PerhitunganIterasi Gauss-Seidel
1. Iterasi 1 ; K =0
E1 =1.04+j0.00 (SlackBus)
𝐸2
′
=
𝐾𝐿2
( 𝐸2
0 )∗
- 𝑌𝐿21 . 𝐸1
′
- 𝑌𝐿23. 𝐸3
0
=
0.0037+𝑗0.024
1.0+𝑗0.000
- (0.332+j0.1990)(1.04) - (0.6666+j0.19904)(1.0+j0.000)

5. = 0.0037+j0.0240-( 0.346528-j0.20696)- (0.6666+j0.19904)
=-1.009428+j0.03192
=1.0099 | − 17.5460
∆ 𝐸2
1
= 𝐸2
′
- 𝐸2
0
= (-1.009428+j0.03192) – (1.0+j0.000)
=-2.009428 + j0.03192
=2.0096 | − 0.90520
>0.0001
𝐸2(𝑎𝑐𝑐)
1
= 𝐸2
0
+ 𝛾 (∆𝐸2
1
)
=(1.0+j0.000)+8.5 (-2.009428+j0.03192)
= 1.0+j0.000 – 17.0801+ j0.27132
=-16.0801 – j 0.27132
=16.082 |0.9660
𝐸3
′
=
𝐾𝐿3
( 𝐸3
0 )∗
- 𝑌𝐿31 . 𝐸1
′
- 𝑌𝐿32. 𝐸2(𝑎𝑐𝑐)
1
=
0.0184+𝑗0.0119
1.0+𝑗0.000
- (0.3090711 –j0.1982)(1.04) – (0.6908+j0.2006)(-16.0801-j0.27132)
=0.0184+j0.01189 – (0.3214-j0.2061) - (11.053+j3.4137)
=-11.356 – j 3.1957
𝐸3
′
=11.79689 |15.760
∆𝐸3
′
= 𝐸3
′
− 𝐸3
0
=(-11.356 - j3.195) - (1.0+j0.000)
= -12.356 – j3.195
= 12.762 |14.4970
>0.0001

6. 𝐸3(𝑎𝑐𝑐)
1
= 𝐸3
0
+ 𝛾 (∆𝐸3
1
)
=1.0+j0.000+8.5 (-12.356-j3.195)
=1.0+j0.000 – 105.026 – j27.1575
=-104.026 – j27.1575
=107.512 |14.62780
2. Iterasi 2 ; K = 1
𝐸2
2
=
𝐾𝐿2
(𝐸2(𝑎𝑐𝑐)
1 )∗
- 𝑌𝐿21 . 𝐸1
′
- 𝑌𝐿23. 𝐸3
1
=
0.0037+𝑗0.024
−16.0801−𝑗0.2713
- (0.3332- j0.1990) (1.04) – (0.6666 + j 0.19904)(-11.356- j3.19571)
=0.003146+ j0.00179 – (0.3466 – j0.20696) – (6.928 +j4.3989)
=-7.2743 – j4.1016
= 8.394 |29.940
∆𝐸2
1
= 𝐸2
2
− 𝐸2(𝑎𝑐𝑐)
1
= (-7.274 – j4.19016) – (16.0801 – j0.27132)
=8.8061 – j3.917
=9.637 | − 23.9790
>0.0001
𝐸2(𝑎𝑐𝑐)
2
= 𝐸2
2
+ 𝛾 (∆𝐸2
1
)
=8.8061 – j3.917 + 8.5 (8.8061 – j3.917)
=8.8061 – j3.917 +74.85185 – j33.2945
=83.6579 – j37.2115
=91.5605 | − 23.9790

7. 𝐸3
2
=
𝐾𝐿3
( 𝐸3
0 )∗
- 𝑌𝐿31 . 𝐸1
′
- 𝑌𝐿32. 𝐸2(𝑎𝑐𝑐)
2
=
0.0184+𝑗0.0119
1.0+𝑗0.000
- (0.309711 – j0.1982)(1.04) – (0.6908 + j0.2006)(83.6579-j37.2115)
=0.0184+j0.0119 – (0.0614 –j0.2061) –(65.2618 –j8.92)
=-65.3048+j9.138
𝐸3
2
=65.94 | − 7.9650
𝐸3(𝑎𝑐𝑐)
2
= 𝐸3
2
+ 𝛾 (∆𝐸3
2
)
= (-65.3048 + j9.138)+8.5(38.7212+j36.4755)
=263.8254+j319.179
=414.1 |50.4230
3. Iterasi 3 ; K = 2
𝐸2
3
=
𝐾𝐿2
(𝐸2(𝑎𝑐𝑐)
2 )∗
- 𝑌𝐿21 . 𝐸1
′
- 𝑌𝐿23. 𝐸3(𝑎𝑐𝑐)
2
=
0.0037+𝑗0.029
83.6579−𝑗37.215
- (0.3332-j0.1990)(1.04) – (0.6666+j0.19904)(263.8254+j319.179)
=(0.0001436+j0.000219) – (0.346528-j0.20696) – (112.218+j265.334)
=-112.2179 – j265.12682
=287.897 |67.0580
∆𝐸2
3
= 𝐸2
3
− 𝐸2(𝑎𝑐𝑐)
2
= (-112.279-j265.12682) – (83.6579 – j37.2115)
=-195,9369 – j227.915
=300.56 |49.3140
𝐸2(𝑎𝑐𝑐)
3
= 𝐸2(𝑎𝑐𝑐)
2
+ 𝛾 (∆𝐸2
3
)
=(83.6579-j37.2115) + 8.5 (195.9369-j227.915)

8. =83.6579 –j37.2115 -1665.46-j1937.275
=-1581.802+j1900.06
=2472.31 | − 50.220
𝐸3
3
=
𝐾𝐿2
(𝐸3
2 )∗
- 𝑌𝐿31 . 𝐸1
′
- 𝑌𝐿32. 𝐸2(𝑎𝑐𝑐)
3
=
0.0184+𝑗0.0119
38.7212+𝑗36.4755
- (0.3090711 – j0.1982)(1.04) – (0.6908 + j0.2006) (1581.802+j1900.06)
=(1.062.10-6
) –( 0.3214-j0.206) – (1474.09 – j995.186)
=-1474.41 +j995.392
𝐸3
3
= 1778.95 | − 34.0230
∆𝐸2
3
= 𝐸3
3
− 𝐸3(𝑎𝑐𝑐)
2
= (-1474.416+j995.392) – (263.8254+j319.179)
=-1738.26+j676.213
=1865.23 | − 21.2560
𝐸3(𝑎𝑐𝑐)
3
= 𝐸3
2
+ 𝛾 (∆𝐸3
3
)
= (236.8524+j319.179) + 8.5 (-1738.26+j676.213)
=236.8524+j319.179 – 14775.21 +j5747.8
=-14538.357 + j6066.979
=15753.477 | − 22.650
E1 = 1.04 + j0.000 / 1.04 |00
E2acc = -1581.802 +j1900.06 / 2472.31 | − 50.220
E3 = -14538.357+j6066.979 / 15753.477 | − 22.650

9. Daya Pada Bus1-2
P12 – jQ12 = E1 (E1 - E2 ) .Y12
= (1.04+j0.000)(1.04+j0.000) – (-1581.802+j1900.06)(1.474-j9.1969)
= (1.04+j0.0000)(1582.825-j1900.06)(1.474-j9.1969)
= -15745 – j18052.78
P12= - 15745 W
Q12 = -18052.78 Var
 Daya Pada Bus2-1
P21 – j Q21 = E2 (E2 - E1 ) .Y21
= (-1581.802+j1900.06)((-1581.802+j1900.06 – 1.04 + j0.000)) (-1.474+j9.1969)
=(-1581.802+j1900.06)(1582.84+j1900.06)(-1.474+j9.1969)
=784.46 – j4892.097
P21 = 784.46 W
Q21 = -4892.097 Var
Rugi-Rugi
P = P12 + P21
=-15475+784.46
=-14960.54 W
Q = Q12 +Q21
=-18052.78 – 4892.097
=-22944.877 W