Analisis Sistem Tenaga 1

Buku Ajar Analisa Sistem Tenaga 1

L66 = |V6|

= |V6| {- |V1|B61 Cos (θ6 – θ1) 1 – |V6||V2| B62 Cos (θ6 – θ2)
1 – |V6||V3| B63 Cos (θ6 – θ3) 1-|V6||V5| B65 Cos (θ6 – θ5) 1
– 2 |V6|B66 1

= (1,0) {- (1,05) (19,672131) 1 – (1,0) (1,0) (13,736853) 1
– (1,0) (1,1) (86,557377) 1 - (1,0) (1,0) (15,384615) 1– 2
(1,0) (-135,350976) 1

= (-20,6557) - (13,736853) – (95,2131) – (15,3846) – (-
270,7019)

= 141,0962
N66 = |V6|

= |V6| {|V1|G61 Cos (θ6 – θ1) 1 + |V6||V2| G62 Cos (θ6 – θ2) 1
+ |V6||V3| G63 Cos (θ6 – θ3) 1+ |V6||V5| G65 Cos (θ6 – θ5) 1
+ 2 |V6|G66 1

= (1,0) {(1,05) (-16,393442) 1+ (1,0) (1,0) (-9,873363) 1
+ (1,0) (1,1) (-55,081967) 1 + (1,0) (1,0) (-10,256410)
1+ 2 (1,0) (91,605182) 1

= (-17,2131) + (-9,8733) + (-60,5901) + (-10,2564) +
183,2103

= 85,2774

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Buku Ajar Analisa Sistem Tenaga 1

Sedangkan
ΔP2 = P2 - ((V2)2.G22) + V2V3 { G23 Cos (θ2 – θ3) + B23 Sin (θ2 –

θ3)} + V2V6 { G26 Cos (θ2 – θ6) + B26 Sin (θ2 – θ6)}
= 0,75 – {(1,0) (1,0) (40,642593) + (1,0) (1,1) (-30,769230)

+ (1,0) (1,0) (-9,873363)}
= 0,75 – {(40,6425) + (-33,8461) + (-9,8733)}
= 0,75 – (-3,0769)
= 3,8269

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Buku Ajar Analisa Sistem Tenaga 1

ΔQ2 = Q2- ((-V2)2 . B22) + V2V3 {G23 Sin (θ2 – θ3) - B23 Cos (θ2 –
θ3)} + V2V6 {G26 Sin (θ2 – θ6) - B26 Cos (θ2 – θ6)}

= 0,15 – {– (1.0) (1.0) (-59,890699) – (1.0)(1.1) (-
46,153846) – (1.0)(1.0) (13,736853)}

= 0,15 – {(-59,8906) – (-50,7692) - (13,7368)}
= 0,15 – (-22,8582)
= 23,0082
ΔP3 = P3 - V3 V2 {G32 Cos (θ3 – θ2) + B32 Sin (θ3 – θ2)} + ((V3)2.

G33) + V3V4 {G34 Cos (θ3 – θ4) + B34 Sin (θ3 – θ4)} + V3V6
{G36 Cos (θ3 – θ6) + B36 Sin (θ3 – θ6)}
= 0,75 – {(1,1) (1,0) (-30,769230) + (1,1) (1,1) (99,576687)
+ (1,1) (1,0) (-13,725490) + (1,1) (1,0) (-55,081967)}
= 0,75 – {(-33,8461) + (120,4877) + (-15,0980) + (-
60,5901)}
= 0,75 – 13,9535
= -13,2035
ΔP4 = P4 - V4V3 {G43 Cos (θ4 – θ3) + B43 Sin (θ4 – θ3)} + ((V4)2.
G44) + V4V5 {G45 Cos (θ4 – θ5) + B45 Sin (θ4 – θ5)}
= 0,9 – {(1,0) (1,1) (-13,725490) + (1,0) (1,0) (24,754901) +
(1,0) (1,0) (-11,029411)}
= 0,9 – {(-15,0980) + (24,7549) + (-11,0294)}
= 0,9 – (-1,3725)
= 2,2725

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Buku Ajar Analisa Sistem Tenaga 1

ΔQ4 = Q4 - V4V3 {G43 Sin (θ4 – θ3) - B43 Cos (θ4 – θ3)} - ((V4)2.
B44) + V4V5 {G45 Sin (θ4 – θ5) - B45 Cos (θ4 – θ5)}
= 0,35 – {-(1,0) (1,1) (21,568627) – (1,0) (1,0)

(39,950979) – (1,0) (1,0) (18,382352)}
= 0,35 – {- (23,7254) - (39,9509) - (18,3823)}
= 0,35 – (-82,0586)
= 82,4086

ΔP5 = P5 - V5V4 {G54 Cos (θ5 – θ4) + B54 Sin (θ5 – θ4)} + ((V5)2 .
G55) + V5V6 {G56 Cos (θ5 – θ6) + B56 Sin (θ5 – θ6)}
= 1 – {(1,0) (1,0) (-11,029411) + (1,0) (1,0) (21,285821) +
(1,0) (1,0) (-10,256410)}
= 1- {(-11,0294) + (21,2858) + (-10,2564)}
= 1- (0)
=1

ΔQ5 = Q5 - V5V4 {G54 Sin (θ5 – θ4) - B54 Cos (θ5 – θ4)} - ((V5)2.
B55) + V5V6 {G56 Sin (θ5 – θ6) - B56 Cos (θ5 – θ6)}
= 0,5 – {-(1,0) (1,0) (18,382352) – (1,0) (1,0) (33,766967)
– (1,0) (1,0) (15,384615)}
= 0,5 – {-(18,3823) - (33,7669) - (15,3846)}
= 0,5 – (-67,5338)
= 68,0338

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Buku Ajar Analisa Sistem Tenaga 1

ΔP6 = V6V1 {G61 Cos (θ6 – θ1) + B61 Sin (θ6 – θ1)} + V6V2 {G62
Cos (θ6 – θ2) + B62 Sin (θ6 – θ2)} + V6V3 {G63 Cos (θ6 – θ3)
+ B63 Sin (θ6 – θ3)} + V6V5 {G65 Cos (θ6 – θ5) + B65 Sin
(θ6 – θ5)} + ((V6)2 . G66)
= 0,9-{(1,0)(1,05)(-16,393442) + (1,0)(1,0)(-9,873363) +
(1,0)(1,1)(-55,081967) + (1,0)(1,0) (-10,256410) +
(1,0)(1,0)(91,605182) }
= 0,9– {(-17,2131) + (-9,8733) + (-60,5901) + (-10,2564)
+ (91,6051)}
= 0,9-(-6,3278)
= 7,2278

ΔQ6 = V6V1 { G61 Sin (θ6 – θ1) – B61 Cos (θ6 – θ1)} + V6V2
{ G62 Sin (θ6 – θ2) -B62Cos (θ6 – θ2)} + V6V3 { G63 Sin (θ6
– θ3) -B62Cos (θ6 – θ3)} + V6V5 { G65 Sin (θ6 – θ5) – B65
Cos (θ6 – θ5)} - ((V6)2 . B66 )
= 0,35-{-(1,0)(1,05)(19,672131) - (1,0)(1,0)(13,736853) -
(1,0)(1,1)( 86,557377) -(1,0)(1,0)( 15,384615) - (1,0)(1,0)
(135,350976)}
= 0,35– (20,6557) - (13,7368) - (95,2131) - (15,3846) -
(135,3509)
= 0,35– (-239,0297)
= 239,3797

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Buku Ajar Analisa Sistem Tenaga 1
147

Buku Ajar Analisa Sistem Tenaga 1
Dapat dihitung dengan menggunakan inverse matrik, diperoleh
hasilnya:
[A]. [X] = [H]
[X] = [Inverse matrik A]. [H]
Atau
[X] = [A]-1. [H]

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Buku Ajar Analisa Sistem Tenaga 1

Δθ2 = -0,27613
= -0,05105

Δθ3 = -0,14107
Δθ4 = -2,00482

= 3,58808
Δθ5 = -2,46068

= 4,15134
Δθ6 = -1,02366

= 1,68700

Nilai awal (Iterasi ke 0) adalah :
= 1,05∠ =1,05 + j0
= 1,0∠ = 1,0 + j0
= 1,1∠ = 1,1 + j0
= 1,0∠ = 1,0 + j0
= 1,0∠ = 1,0 + j0
= 1,0∠ = 1,0 + j0

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Buku Ajar Analisa Sistem Tenaga 1

Perhitungan tegangan dan sudut tegangan masing-masing bus
pada iterasi ke 1:
1. V1’ = 1,05 ∠00 ( Nilai tetap karena Sluck Bus )
2. V2’ = ( 1,0 + Δ|V2|∠ (00 + Δθ2)

= (1,0 + (-0,05105))∠ (00 + (-0,27613))
= -0,05105 ∠-0,27613
3. V3’ = 1,1 - ∠ (00 + Δθ3)
= 1,1 - ∠ (00 + (-0,14107))
=1,1 - ∠(-0,14107)
4. V4’ = ( 1,0 + Δ|V4|∠ (00 + Δθ4)
= (1,0 + (3,58808)∠ (00 + (-2,00482))
=3,58808 ∠-2,00482
5. V5’ = ( 1,0 + Δ|V5|∠ (00 + Δθ5)
= (1,0 +4,15134)∠ (00 + (-2,46068))
= 4,15134 ∠-2,46068
6. V6’ = ( 1,0 + Δ|V6|∠ (00 + Δθ6)
= (1,0 +1,68700)∠ (00 + (-1,02366))
= 1,68700 ∠-1,02366

150

Buku Ajar Analisa Sistem Tenaga 1

Bila dan Δθi masih lebih besar dari angka toleransi maka

perhitungan dilanjutkan pada iterasi berikutnya dan sebaliknya

bila dan Δθi lebih kecil dari angka toleransi maka proses

iterasi selesai.

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Buku Ajar Analisa Sistem Tenaga 1
DAFTAR PUSTAKA

1. Budiono Mismail, 1995, Rangkaian Listrik Jilid Pertama,
Penerbit ITB-Bandung.

2. Pabla A.S, Alih Bahasa Abdul Hadi, 1994, Sistem Distribusi
Daya Listrik, Penerbit Erlangga, Jakarta.

3. R. Soegeng, 1993, Komputasi Numerik Dengan Turbo Pascal,
Penerbit Andi Offset Yogyakarta.

4. T.S. Hutauruk., Ir., M.Sc, 1985, Transmisi Daya Listrik,
Penerbit Erlangga, Jakarta.

5. William D. Stevenson, Jr, Alih Bahasa, Kamal Idris, 1990,
Analisis Sistem Tenaga Listrik, Penerbit Erlangga.

6. William D. Stevenson, Jr, 1975, Elements Of Power System
Analysis, Mc. Graw-Hill.

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