Stabilitas Hand Out 1
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Transcript of Stabilitas Hand Out 1
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Electric Power Systems L8 - OlofSamuelsson
1
STABILITAS
SISTEM TENAGA LISTRIK
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2
MASALAH STABILITAS DALAM
SISTEM TENAGA LISTRIK
DALAM KEADAAN OPERASI YANG STABIL DARI
SISTEM TENAGA LISTRIK, TERDAPAT
KESEIMBANGAN ANTARA DAYA INPUT MEKANISPADA PRIME MOVER DENGAN DAYA OUTPUT
LISTRIK (BEBAN LISTRIK) PADA SISTEM.
DALAM KEADAAN INI SEMUA GENERATORBERPUTAR PADA KECEPATAN SINKRON.
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☻ TERUTAMA JIKA TERJADI GANGGUAN, MAKA
SESAAT AKAN TERJADI PERBEDAAN YANG BESAR
ANTARA DAYA INPUT MEKANIS DAN DAYA OUTPUTLISTRIK DARI GENERATOR.
☻ KELEBIHAN DAYA MEKANIS TERHADAP DAYA
LISTRIK MENGAKIBATKAN PERCEPATAN PADAPUTARAN ROTOR GENERATOR ATAU SEBALIKNYA.
☻ BILA GANGGUAN TIDAK DIHILANGKAN
DENGAN SEGERA, MAKA PERCEPATAN /
PERLAMBATAN PUTARAN ROTOR
GENERATOR
AKAN MENGAKIBATKAN HILANGNYA
SINKRONISASI DALAM SISTEM.
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STABILITAS SISTEM TENAGA LISTRIK
KEMAMPUAN SUATU SISTEM TENAGA LISTRIK
ATAU BAGIAN KOMPONENNYA UNTUK
MEMPERTAHANKAN SINKRONISASI DAN
KESEIMBANGAN DALAM SISTEM.
BATAS STABILITAS SISTEM
DAYA MAKSIMUM YANG DAPAT MENGALIR
MELALUI SUATU TITIK DALAM SISTEM TANPAMENYEBABKAN HILANGNYA STABILITAS.
BERDASARKAN SIFAT DAN BESARNYA
GANGGUAN,
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GANGGUAN TERHADAP STABILITAS :
GANGGUAN KECIL : FLUKTUASI BEBAN
GANGGUAN BESAR (BERSIFAT MENDADAK) :
HUBUNG SINGKAT, PELEPASAN BEBAN
MENDADAK, DSB.
MASALAH STABILITAS DALAM SISTEM
TENAGA LISTRIK DIBEDAKAN ATAS:
STABILITAS STEADY-STATE
STABILITAS TRANSIENT
STABILITAS DINAMIS
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STABILITAS STEADY-STATE :
KEMAMPUAN DARI SUATU SISTEM TENAGA
MEMPERTAHANKAN SINKRONISASI ANTARA MESIN-
MESIN DALAM SISTEM SETELAH MENGALAMI
GANGGUAN KECIL.
STABILITAS TRANSIENT :
KEMAMPUAN DARI SUATU SISTEM TENAGA
MEMPERTAHANKAN SINKRONISASI SETELAH
MENGALAMI GANGGUAN BESAR YANG BERSIFAT
MENDADAK SELAMA SEKITAR SATU “SWING” (YANG
PERTAMA) DENGAN ASUMSI BAHWA PENGATURTEGANGAN OTOMATIS (AVR) DAN GOVERNOR
BELUM
BEKERJA.
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STABILITAS DINAMIS :
BILA SETELAH SWING PERTAMA (PERIODE
STABILITAS TRANSIENT) SISTEM BELUM
MAMPU MEMPERTAHANKAN SINKRONISASI
SAMPAI SISTEM MENCAPAI KEADAANSEIMBANG YANG BARU. (ADALAH STABILITAS
TRANSIENT BILA AVR DAN GOVERNOR
BEKERJA CEPAT DAN DIPERHITUNGKAN
DALAM ANALISIS).
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PENGERTIAN HILANGNYA SINKRONISASI
KETIDAKSEIMBANGAN ANTARA DAYA PEMBANGKIT DANBEBAN MENIMBULKAN SUATU KEADAAN TRANSIENT
YANG MENYEBABKAN ROTOR DARI MESIN SINKRON
BERAYUN KARENA ADANYA TORSI YANG
MENGAKIBATKAN PERCEPATAN ATAU PERLAMBATAN
PADA ROTOR TERSEBUT.
BILA TORSI TERSEBUT CUKUP BESAR, MAKA SATU
ATAU LEBIH DARI MESIN SINKRON TERSEBUT AKAN
KEHILANGAN SINKRONISASINYA.
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MISAL TERJADI KETIDAKSEIMBANGAN YANG
DISEBABKAN OLEH ADANYA PEMBANGKITAN YANG
BERLEBIHAN, MAKA SEBAGIAN BESAR DARI
ENERGI YANG BERLEBIHAN AKAN DIUBAH
MENJADI ENERGI KINETIK YANG MENGAKIBATKAN
SUDUT ROTOR BERTAMBAH BESAR.
WALAUPUN KECEPATAN ROTOR BERTAMBAH
BESAR, TIDAK BERARTI BAHWA
SINKRONISASI DARI MESIN TERSEBUT AKAN
HILANG. FAKTOR YANG MENENTUKAN
ADALAH PERBEDAAN SUDUT ROTOR / DAYA ANTARA MESIN-MESIN DALAM SISTEM,
DIMANA SUDUT ROTOR / DAYA TERSEBUT
DIUKUR TERHADAP REFERENSI PUTARAN
SINKRON.
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PERHATIKAN GAMBAR DIBAWAH INI YANG MENUNJUKKAN
SUDUT ROTOR/DAYA MESIN DALAM SISTEM 2 MESIN
SEBAGAI FUNGSI WAKTU SELAMA KEADAAN TRANSIENT.
SEMUA SUDUT ROTOR
MENINGKAT TETAPI
PERBEDAAN SUDUT ROTORDARI SEMUA MESIN KECIL
DAN SUDUT-SUDUT
TERSEBUT MENUJU POSISI
YANG BARU.
SISTEM STABIL0 0.5 1 1.5
-100
-50
0
50
100
150Phase angle difference (fault cleared at 0.4s)
t, sec
D e l t a ,
d e g r e e
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SEMUA MESIN TERPISAHMENJADI DUA KELOMPOK
TANPA ADANYA
KEMUNGKINAN BERTEMU
PADA SUATU TITIK.
SISTEM TAK STABIL0 0.5 1 1.5-200
0
200
400
600
800
1000
1200
1400Phase angle difference (fault cleared at 0.5s)
t, sec
D e l t a ,
d e g r e e
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P = fs (δ)
DARI GENERATOR SEREMPAK DENGAN ROTOR
BULAT (NON SALIENT POLE)
T ~
E
Xs V
I Infinite bus
E = V + j I Xs
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E
V
IXs
I
d
f
d f sin)90sin(
s IX E
cosIVP
sincos
f
d f
s X
E I
Daya yang dibangkitkan generator :
d sin s X
VE P
P
Pm
Pm= P mekanis turbin
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P = fs (δ)
DARI GENERATOR SEREMPAK DENGAN ROTOR
KUTUB MENONJOL (SALIENT POLE)
)()( qqd d jX jI jX I V E
qd qd qqd d X jI X jI jX jI X jI V
)())(( qd d qqd X X jI jX jI I V
)( qd d q X X jI jIX V E
qd
jI I I dimana
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d
qd X jI
q jIX
d d X jI
V I
d I
qd d X X jI
f
E
qq X I
q I
d I I V
O
q I
f
d
y
a
x
c
b
d d f sincoscosd q
I I I
ob I f cos aboa
d cosq I oa
cbacab d d sinsin y x d sin y x
d sind I
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d
qd X jI
q jIX
d d X jI
V I
d I
qd d X X jI
f
E
qq X I
q I
qq X I V d sin
d d X I E V d cos
V sinδ
V cosδ
q
q X
V I
d sin
d
d X
V E I d cos
d d f sincoscos d q I I I
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Pf cosVI P
d
d 2sin
2
sin
qd
qd
d X X
X X V
X
E V
d d 2sin2
sin
2
qd
qd
d X X X X V
X VE P
d sin
Xd
VE
d 22
)(2
Sin X X
X X V
qd
qd
d d 2sin2
)(sin
2
qd
qd
d x X
x X V
X
VE
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CONTOH SOAL :
GS
Eg
Eg’
X genXe
Trafo
Transmisi dll
Infinite Bus
E~ = 1.0 pu
SUATU ALTERNATOR TURBO 2 KUTUB TERHUBUNG
SISTEM YANG BESAR DENGAN NAME PLATE DATA
SEBAGAI BERIKUT:
100 MVA, 2 POLE, 50 HZ, 85%P.F., 13.2 kV (L-L)
Xd = 100%, Xq = Xq’ = 96%, Xd’ = 20% Xe = 50%
E 1
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UNTUK KEADAAN KERJA DIMANA ARUS YANG MENGALIR
NOMINAL, DENGAN FAKTOR KERJA 1.0 PADA INFINITE BUS,
TENTUKAN P vs d UNTUK KEADAAN STEADY STATE DANTRANSIENT.
STEADY STATE :
TRANSIENT :
d d 2sin2sin
2
eqed
eqed
ed
g
X X X X
X X X X E
X X
E E
P
d d 2sin
2sin
''
''2
'
'
eqed
eqed
ed
g
X X X X X X X X E
X X E E P
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STEADY STATE :
Id(Xd- Xq) Eg
Iq
Id
Ia = 1.0 pu
E~ = 1.0 pu
IaXe
E1
IaXq
d d 2sin
46.15.12
96.00.10.1sin
5.1
0.181.1 2 P
P = 1.20 sin d + 0.0091 sin 2d
1.2 sin dδ
=IaSinδ
Eg
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TRANSIENT :
E’g
IaXe
E1Iq
Id
Ia = 1.0 pu
E~ = 1.0 pu
d d 2sin46.170.02
96.020.00.1sin
50.020.0
0.115.1 2
P
P = 1.65 sin d - 0.372 sin 2d
Id(Xq’ - Xd’)
IaXq’
Eg’
δ
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0 20 40 60 80 100 120 140 160 180-0.5
0
0.5
1
1.5
2
2.5Analisa Stabilitas yang ditunjukkan Oleh Kurva P Vs d
d
P
Steadystate
Transient
P = 1.65 sin d - 0.372 sin 2d
P = 1.20 sin d
1.65 sin d
- 0.372 sin 2d
(transient)
(steady-state)
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TUGAS :KERJAKAN LAGI CONTOH SOAL BILA,
a/ FAKTOR DAYA = 0.75 LAGGING
b/ FAKTOR DAYA = 0.75 LEADING
c/ ARUS Ia = 75% NOMINAL, FAKTOR DAYA 0.75
LAGGING
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STABILITAS STEADY STATE
Y - A x
i s
d
PMAX
PM
rates RotorDecle
P P M E
d d
M E P P
oOperatingP
int
erates RotorAccel
PM PE
d d
d o90 o180
M
E
P Power
Mechanical
X
E E P
Power
Electrical
d sin21
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TITIK KERJA STEADY STATE ADALAH SUATU
KEADAAN DIMANA DAYA LISTRIK YANGDIBANGKITKAN GENERATOR (PE) SEIMBANG
DENGAN DAYA MEKANIS DARI TURBIN (PM).
PERUBAHAN SUDUT TITIK KERJA TERSEBUT (d0) AKAN MENGAKIBATKAN KETIDAKSEIMBANGAN
DAYA YANG AKAN MEMPERCEPAT /
MEMPERLAMBAT KECEPATAN ROTOR KE TITIK
KERJA YANG BARU.
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Y - A x i s
d
P MAX
P M
o90 o180δ1
δ1’ δ
1” δ
2’
δ2
δ2”
STABIL TAK STABIL
BATAS STABILITAS STEADY STATE
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Eg v t Em
Xm
Xg
Ig
Im
g d
md Ig
Eg
IgX
E
ImX
m
Em
m g
total
m g
X
E E P d d sin
total
m g
X
E E P max
UNTUK MENAIKKAN BATAS STABILITAS :
1. MEMPERBESAR Eg2. MEMPERBESAR Em
3. MEMPERKECIL REAKTANSI TOTAL SISTEM
ANTARA Eg& Em
MENGGUNAKAN DUA SALURAN PARALEL.
MENGGUNAKAN KAPASITOR SERI
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STABILITAS TRANSIENT
Y - A x i s
d
P
MAX
PM
d o90 o180
Power
1
2
3 4
PDECEL
P ACCEL
Stable
Turbine
Power
Unstable
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GAMBAR DIATAS MENUNJUKKAN PERILAKU SUATU
GENERATOR DALAM KEADAAN GANGGUAN.
1. TITIK KERJA AWAL(SEBELUM TERJADI GANGGUAN)
2. TIMBUL GANGGUAN YANG MENGAKIBATKAN DAYA
OUTPUT GENERATOR TURUN SECARA DRASTIS.
SELISIH ANTARA DAYA OUTPUT LISTRIK TERSEBUT
DAN DAYA MEKANIS TURBIN MENGAKIBATKAN ROTOR
GENERATOR MENGALAMI PERCEPATAN, SEHINGGASUDUT ROTOR / DAYA BERTAMBAH BESAR.
3. PADA SAAT GANGGUAN HILANG, DAYA OUTPUT
GENERATOR PULIH KEMBALI PADA HARGA YANG
SESUAI DENGAN KURVA P - d DI ATAS.
4. SETELAH GANGGUAN HILANG, DAYA OUTPUT
GENERATOR MENJADI LEBIH BESAR DARIPADA DAYA
MEKANIS TURBIN. HAL INI MENGAKIBATKAN
PERLAMBATAN PADA ROTOR GENERATOR.
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BILA TERDAPAT TORSI LAWAN YANG CUKUP
SETELAH GANGGUAN HILANG UNTUK
MENGIMBANGI PERCEPATAN YANG TERJADISELAMA TERJADINYA GANGGUAN, GENERATOR
AKAN STABIL SETELAH AYUNAN (SWING) YANG
PERTAMA DAN KEMBALI KE TITIK KERJANYA
DALAM WAKTU KIRA-KIRA 0.5 DETIK
BILA KOPEL LAWAN TERSEBUT TIDAK CUKUP
BESAR, SUDUT ROTOR / DAYA AKAN TERUS
BERTAMBAH BESAR SAMPAI SINKRONISASIDENGAN SISTEM HILANG.
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PERSAMAAN AYUNAN ROTOR
(ROTOR SWING EQUATION)
UNTUK GERAK ROTASI BERLAKU
TurbinLoadRotor
Tshaft
Tmaxsin δ
Redaman
TD.ω = TD.d
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.α = Σ T
Hk. Newton :
= Tshaft – TD. – Tmaxsin δ
J
d J d
DIMANA :
: MOMEN INERSIA
α : PERCEPATAN SUDUT
TD
: KOEFISIEN REDAMAN
Tmax sin δ : ELECTROMAGNETIC TORQUE
YANG DIBANGKITKAN
δ : TORQUE/POWER/ROTOR ANGLE
J
ROTOR
SWING EQ.
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BILA REDAMAN DIABAIKAN, MAKA PERSAMAAN
DIATAS MENJADI,
+ Tmax sin δ = Tshaft
d M
d J
PERSAMAAN AYUNAN ROTOR :
(Dinyatakan dlm Torque)
+ Pmax
sin δ = Pshaft
(Dinyatakan dlm Power)
0 0.5 1 1.5 2 2.5 310
15
20
25
30
t, sec
d
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HUBUNGAN ANTARA M (momentum sudut) DAN H
(konstanta inertia)
MegaJoules M
MegaJoules J E K
2
1
2
1. 2
(Tersedia)
Tenaga Kinetis Rotasi :
g
MegaJoules
MVAGenerator Rating
MegaJoules H
)(
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MegaJoules gH M 2
1
f
gH M
180
H BIASA DIGUNAKAN DALAM PERSAMAAN AYUNAN ROTOR
(SWING) YANG DINYATAKAN DALAM TORQUE (KOPEL)
DENGAN SATUAN NEWTON – METERS.
M BIASA DIGUNAKAN DALAM PERSAMAAN AYUNAN
ROTOR (SWING) YANG DINYATAKAN DALAM P DENGAN
SATUAN MEGA WATTS.
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KRITERIA LUAS SAMA
(EQUAL AREA CRITERION)
shaft em P P M d
d sinmax P P em
PERSAMAAN AYUNAN ROTOR :
DIMANA,
PENYELESAIAN DARI PERSAMAAN DIATAS MERUPAKAN
BENTUK OSILASI (TEREDAM/DAMPED - UNTUK KEADAAN
STABIL).
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BILA UNTUK SUATU PERSAMAAN AYUNAN ROTOR DAPAT
DITUNJUKKAN BAHWA NILAI DARI PADA δ MENCAPAI
MAKSIMUM DAN KEMUDIAN BERKURANG (ATAU SEBALIKNYA) ,
MAKA DIKATAKAN SISTEM STABIL (MAMPU MEMPERTAHANKANKESTABILANNYA).
MAKA, AGAR SISTEM STABIL (PERSYARATAN
UNTUK STABIL) HARUS DIPENUHI :
0dt
d d
UNTUK MAKSIMUM/MINIMUM YANG PERTAMA
BILA PERSYARATAN STABIL DIATAS DIAPLIKASIKAN PADA
PERSAMAAN AYUNAN ROTOR DARI SUATU SISTEM
KRITERIA LUAS SAMA UNTUK MENENTUKAN SISTEM
STABIL ATAU TIDAK STABIL
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shaft em P P M d
d d x em sha ft P P M
M P P M
em sha ft
2 x)( d d d
)(2
2 d d d em shaft P P
M
dt dt
d P P
M dt
d em shaft x)(
22 d
d
d d d P P M
d em shaft )(22
K d P P M
em sha ft d d )(22
d
d
d d
0
)(22 d P P
M
em shaft
d
d
d d
0
)(2
d P P M
em shaft
0
d
0)(
0
d
d
d d P P em shaft
d
d
d
d
d d 00
d P d P em shaft
LUASAN DIBAWAH Pshaft SAMA DENGAN LUASAN DIBAWAH Pem
DENGAN BATAS DARI δ0 S/D δ
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Y - A x i s
d
P MAX
P M
o90 o180δ
0 δ
Pshaft
Pem
: Luasan dibawah Pshaft
: Luasan dibawah Pem
Luasan dibawah Pshaft = Luasan dibawah Pem
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Y - A x i s
d
P MAX
P M
o90 o180δ
0 δ
Pshaft
Pem
BerimpitLuasan dibawah
Pshaft
Luasan diatas
Pshaft
Luasan diatas Pshaft = Luasan dibawah Pshaft
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Electric Power Systems L8 - OlofSamuelsson 41
PENERAPAN
KRITERIA LUAS SAMA
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X12
=0.3
X12
=0.3
Xt=0.2
X’d=0.3
E’d
1 2
Inf.
V=1.0
Daya yang dibangkitkan generator pada Infinite bus :
P = 0.6 pu. dengan pf. 0.8 lagging. Tegangan infinitebus 1.0 pu
Tentukan :
a. Daya input maksimum yang bisa diberikan padagenerator dan tidak lepas sinkron (stabil)
b. Sama dng (a), tetapi generator dlm keadaan beban
nol (tegangan internal generator disumsikan tetap,
sama dengan yang telah ditentukan di (a))
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0 20 40 60 80 100 120 140 160 1800
0.5
1
1.5
2
Equal-area criterion applied to the sudden change in power
Power angle, degree
P o w e r , p
e r u n i t
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Initial Power = 0.600 pu
Initial Power Angle = 16.791 degree
Sudden Initial Power = 1.084 pu
Total Power for Critical Stability = 1.684 pu
Maximum Angle Swing =125.840 degree
New Operating Angle = 54.160 degree
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0 20 40 60 80 100 120 140 160 1800
0.5
1
1.5
2
Equal-area criterion applied to the sudden change in power
Power angle, degree
P o w e r , p
e r u n i t
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Initial Power = 0.000 pu
Initial Power Angle = 0.000 degree
Sudden Initial Power = 1.505 pu
Total Power for Critical Stability = 1.505 pu
Maximum Angle Swing =133.563 degree
New Operating Angle = 46.437 degree
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F
E’g E~
A1 A2
A3 A4 A5
A6
Psh
d6
P
d1 d2 drcl dm d
NORMAL
SELAMA
GANGGUAN
A1 + A2 = A3 STABIL
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Exercise 1
• Sketch (dng Matlab) utk gangguan
yang bersifat sementara, tapi TAK
STABIL dan STABIL• Sketch (dng Matlab) utk gangguan
permanen. Jelaskan apakah sistem
STABIL atau TAK STABIL.
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F
E’g E~
A1
A5
Psh
P
d1 d2drcl
dm d
2 SALURAN
SELAMA
GANGGUAN
2 SALURAN
dlimit
A2 A3
A4
di
A1 + A3 = A2 + A4 STABIL
1 SALURAN
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Exercise 2
• Sketch (dng Mat lab) utk gangguan
yang bersifat sementara, tapi TAK
STABIL (dan STABIL)• Sketch (dng Matlab) utk gangguan
permanen, tapi STABIL.
• Sketch (dng Matlab) utk gangguanpermanen, tapi TAK STABIL
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Psh
d1 dc dm d
Pm sin d
r 1 Pm sin d
r 2 Pm sin d
SUDUT KRITIS (δC)
d d d d
d
d
d P r P A
c
mcch sin1
111
111 coscos d d d d cmc sh P r P
)(sin22 cm sh
m
c
m P d P r A d d d d
d
d
cm shmcm P P r d d d d coscos2
A1 = A2
A2
A1
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12
11211 coscos/
cos r r
r r P P mmm sh
c
d d d d
d
m
sh
m
m
sh
mm sh
m sh
P R
P
P
P
P r P
P P
2
1
1
1
2
1
sin
sin
sin
sin
d
d
d
d
(Satuan RADIAN)
(Satuan RADIAN)
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X12
=0.3
X12
=0.3
Xt=0.2
X’d=0.3
E’d
1 2
Inf.
V=1.0
F
Generator sinkron dng konstanta inertia H=5
MJ/MVA. Data pada diagram segaris mempunyai
base sama. Daya yang dibangkitkan generator pada
Infinite bus : P = 0.8 pu., Q = 0.074 pu. Tegangan
infinite bus 1.0 pu
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a. Hubung singkat 3 fasa temporer terjadi di F. Bila
gangguan diamankan langsung dengan kedua CB
dari dua saluran yang terhubung pada bus 1
terbuka/trip (tidak ada kurva P vs δ utk selama
gangguan), tentukan sudut kritisnya (δc) – saatkedua CB reclose.
b. Hubung singkat 3 fasa permanen terjadi pada
pertengahan salah satu saluran. Gangguan
diamankan dengan membuka CB-CB pada saluranyang mengalami gangguan. Tentukan sudut
kritisnya (δc) – saat CB trip/membuka.
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0 20 40 60 80 100 120 140 160 1800
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Application of equal area criterion to a critically cleared system
Power angle, degree
P o w e r , p
e r u n i t
Pm
Critical clearing angle = 84.7745
CB trip CB reclose
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0 20 40 60 80 100 120 140 160 1800
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Application of equal area criterion to a critically cleared system
Power angle, degree
P o w e r
, p e r u n i t
Pm
Critical clearing angle = 98.8335
P = 1.80 sin d
P = 1.46 sin d
r 2
= 0.8
P = 0.65 sin d
r 1=0.36
d1= 26.390 dm= 146.580dc= 98.800
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CONTOH SOAL
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Jika CB terbuka, tentukan :
a. Sudut 1, 2 dan m ( gambar P vs di bawah ).
b. Apakah sistem stabil atau tidak ? Jika stabilhitung 3 (gambar P vs di bawah ).
Contoh 1 :
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Kurva P vs d
P
P sh
A 1 A2
dd d d d1 2 3 m
sebelum gangguan
selama gangguan ( CB terbuka )
P1 = P 1m sin d
P2 = P2m sin = r 1.P 1m sind d
gangguan terjadi (CB terbuka)
sebelum gangguan
d sin11 m P P
d d sinsin 1222 mm P r P P
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Sebelum gangguan terjadi :
4,0)2,02,0(4,0
)2,02,0.(4,02,01 j j j j
j j j j X eq
d d d sin0,3sin4,0
1.2,1sin
.
1
1 eq
g
X
V E P
d sin11 m P P jika P1 = Psh, d = d1, maka :
1sin0,35,1 d
5,0sin 1
d
o1
1 30)5,0(sin d
G (C )
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Gangguan terjadi (CB terbuka) :
6,04,02,02 j j j X eq
d d d sin0,2sin6,0
1.2,1sin
.
2
1 eq
g
X
V E P
d sin22 m P P jika P2 = Psh, maka :
d sin0,25,1 75,0sin d oo1 4,1316,48)75,0(sin atau d
oo
2
4,1316,48 m
dan d d
d d d sin0,2sin6,0
0,1.2,1sin
.
2
2 eq
g
X
V E P
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SISTEM STABIL, KARENA DAPAT
DIPEROLEH LUASAN YANG SAMA DIATAS
Psh DENGAN LUASAN DIBAWAH Psh.
Contoh 2 : (lihat diagram contoh 1)
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Jika terjadi Hubung Singkat 3 Fasa di F (permanen)
Sebelum gangguan terjadi : d d sin0,3sin11 m P P
Selama gangguan terjadi (CB belum trip):
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d d d sin2,1sin1
1.2,1sin
.
3
3 eq
g
X
V E P
d d sin2,1sin33 m P P 2,13 m P
Setelah gangguan terjadi (CB trip) :
d d sin2sin22 m P P 22 m P
Bila CB trip pada δ=60o :
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P
P s h
A 1A 2
1
dd d
d
Sebelum gangguan :
d d sin0,3sin11 m P P
Selama gangguan :
d d d sin2,1sin.sin 1233 mm P r P P
3
2,1
sin.
sin.
1
3
2 d
d
m
m
P
P r
Setelah gangguan :
d d sin0,2sin22 m
P P
3
2
sin.
sin.
1
21
d
d
m
m
P
P r
sebelum gangguan
setelah gangguan
selama gangguan
A1 A2
Bila CB trip pada δ 60 :
δ=600
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0,3462
4392,07854,0
)30cos60(cos33
2,1)3060(5,1)cos.(cos.)(
sin..)(
oooo
121212
2
112121
d d d d
d d d d d
d
m sh
m sh
P r P
d P r P A
0,4534
8692,13226,2
)304,131(5,1)4,131cos60(cos33
2)()cos.(cos.
)()sin..(oooo
2211
22 112
d d d d
d d d d
d
d
m shmm
m sh
m
m
P P r
P d P r A
Karena luas A2 ( 0,4534) lebih besar dari
A1 ( 0,3462 ) maka sistem stabil.
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P
P s h
A 1A 2
1
dd d
d
d
Menghitung δkritis
sebelum gangguan
setelah gangguan
selama gangguan
A1 A2
δ1 δmδc
δc = δkritis
P
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32,1
32
30cos3
2,1)4,131cos(
3
2)304,131(
3
5,1
cos
coscos)(
cos
1
21
1211
11
oooo
c
mm
m
sh
c
r r
r r P
P
d
d d d d
d
1,1962rad68,54o
c
c
d
d )3659,0(cos
8,0
)866,0(2,1)-0,6613(2)0,52362,2934(5,1cos 11
Diubah radian
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Bila CB terbuka pada δ=90o
sebelum gangguan
setelah gangguan
selama gangguan
G T Si Tid k S bil
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Psh
o
A 1A2
A3
sebelum gangguan
setelah gangguan
selama gangguan
d do
od
ddf
Gangguan Temporer, Sistem Tidak Stabil
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Keterangan :
Pada saat 1=30o terjadi gangguanhubung singkat ke tanah.
Pada saat 2=105o CB-CB terbuka ( trip )
dan gangguan masih berlangsung,beban hanya disuplai melalui satu
saluran,
Pada saat 3=120
o
CB-CB menutup(recloser bekerja) dan gangguan hilang
G T Si t St bil
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Psh
= o
sebelum gangguan
setelah gangguan
selama gangguan
A1
A2
A3
90d
d= o30d
= o70d ds
Gangguan Temporer, Sistem Stabil
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Exercise 3
• Kerjakan lagi contoh 1 dan 2, sketch
(dng Matlab) P vs δ -nya
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CONTOH SOAL
Charles Gross prob 12 5 dan 12 7
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b c
0,20
0,30
1 ,0 0o
a d
e
B 1 B 2
Generator 0,050,01 0,05
d=Inf in ite busX' = X = 0 .15
d q
0,1
Charles Gross, prob. 12.5 dan 12.7
Tentukan :
a. Eg’
b. Sudut kritis, bila gangguan hubung singkat 3 fasa
terjadi pada bus e, dan gangguan diisolasi dengan
terbuka/trip nya CB : B1 dan B2
Eg’=?
S=1+j0.2 pu
P=Psh Q
PREFAULT
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b c
j0,15 j0,05
j0,30
E'q 1,0 0o
a d
e
j0,05
j0,20 j0,10
j X j0,15 j0,05 j0.30 j0,20 j0,10 j0.30 j0,20 j0,10
j0,05 j0,15 j0,05 j0,15 j0,05 j0,4
I (conjugate)S
V
1,0 j0,2
1 j0 1,0 j0,2 I 1.0 j0,2 1,0198 11,31
E V I j x 1 1,0198 11,31 0,4 90 1 j0 0,0799 j0,4
1,0799 j0,4 1,1516 20,32
E
0
q' 0 0
0
0
q'
Jadi E 1,1516 pu dan 20,32
Pe 2,88 sin
0d
d d d V
X
sin,
,
sin11516 1
0 4
FAULTED
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b c
j0,15 j0,05
j0,30
E'q 1,0 0o
a d j0,05
j0,20 j0,10
1
3
2
Transformasi
Z j j
j j j j j
Z
j j
j j j j j
Z j j
j j j j j
;
, ,
, , ,
,
,,
, ,
, , ,
,
, ,
, ,
, , ,
,
,,
1
2
3
0 3 01
0 3 0 2 01
0 03
0 60 05
0 3 0 2
0 3 0 2 01
0 06
0 6 01
01 0 2
0 3 0 2 01
0 02
0 60 0333
j0,2 j0,05 j0,05 j0,1
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E'q 1,0 0o
j0,0333
j0,2
E'q 1,0 0o
j0,05 j0,05 j0,1
j0,0333
jX
Transformasi
j X j j j j j j
j
j j j j
j j
;
, , , , , ,
,
, ,, ,
,,
0 25 0 0333 0 25 015 015 0 0333
0 0333
0 25 0150 25 015
0 033315261
j1 5261
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E'q 1,0 0o
j1,5261
jX
Pe 0,7549
E V
jX
q'sin
,
,sin sind d d
1152 1
15261
POSTFAULT
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POSTFAULT
b c
j0 ,1 5 j0 ,0 5
j0 ,3 0
E'q 1 , 0 0o
a d j0 ,0 5
j X j0,15 j0,05+ j0,30 j0,05 j0,55
Eq'
Pe 2,0945 sin
28 52 151 480 0
V
X
Pe Pm Sin
dan
sin
,
, sin
, sin,
sin ,
d d d
d d d
d d
11516 1
0 55
1 2 09451
209450 4774
180 28 52
1
0 0