ALIRAN DAYA & RUGI-RUGI DAYA
Aliran Daya Reaktif :
- mengak mengakiba ibatka tkan n rugirugi-rug rugii pada pada saluran dan transformator - menuru menurunka nkan n kapasi kapasitas tas jari jaringa ngan n distribusi - faktor faktor daya daya menun menunjuk jukkan kan besar besar aliran daya reaktif
Faktor daya daerah perumahan (Rabu)
0,7 – 0,9
Faktor daya daerah perumahan (Minggu)
0,7 – 0,9
Faktor Daya “typical” berdasarkan jenis industri dan jenis proses (motor industri menyerap energi listrik > 50%) No
Industry
Power Factor
Process
Power Factor
1
Auto parts
0.75÷0.8
Air Compressing
0.75÷0.8
2
Brewery
0.76÷0.8
Welding
0.35÷0.6
3
Clothing
0.35÷0.6
Machining
0.4÷0.65
4
Hospital
0.75÷0.8
Stamping
0.6÷0.7
5
Commercial Building
0.8÷0.9
Spraying
0.6÷65
Faktor Daya motor induksi sangat tergantung pada beban
Peralatan/Beban yang menyerap Daya Reaktif 1. Motor Induksi Distribusi Rugi-Rugi “typical” Power
Ml
M2
M3
M4
M5
M7
1
5
25
50
100
200
0.746
3.7
18.65
37.3
74.6
149.2
Output[W]
746
3,730
18,560
37,300
74,600
149,200
Input [W]
1,020
4,491
20,946
41,217
81,530
160,432
73
83
89
90.5
91.5
93
HP kW
Efficiency [%]
-
Rugi-Rugi Magnetik Power
Ml
M2
M3
M4
M5
M7
HP
1
5
25
50
100
200
kW
0.746
3.7
18.6
37.3
74.6
149.2
MagnetiC Core Loss [W]
76
225
351
765
906
1,650
Total Loss [W]
274
761
2,296
3,917
6,930
11,232
. Magnetic Loss [%]
27
29
15
19
13
15
Magnetic Loss current [A]
0.1
0.31
0.5
1.06
1.2
2.3
Komposisi Rugi-Rugi Motor Component Loss
Loss [%]
Standard power loss
37
Rotor power loss
18
Magnetic core loss
20
Friction and windings
9
Stray load loss
10
Efisiensi motor induksi tergantung pada besar/size dari motor (makin besar motor makin tinggi efisiensinya)
Rata-rata rugi magnetik 20 % dari rugi-rugi total (cukup significant)
2. Alat Pengatur Kecepatan (Variable Speed Drive)
Untuk
aplikasi industri, motor induksi membutuhkan
pengaturan kecepatan Digunakan
variable speed drive system yang
menghasilkan variable frequency dan variable voltage
Faktor daya dari three phase diode bridge rectifiers sangat tinggi (teoritis : 0.955)
Bila digunakan thyristor bridge rectifiers , faktor daya menjadi fungsi dari firing angle dan overlap angle yang akan meningkatkan konsumsi daya reaktif
3. Discharge Lamps Rangkaian lampu yang menggunakan choke/leakage transformer ballast mempunyai faktor daya lagging yang rendah Faktor daya dikoreksi dengan kapasitor menjadi 0,85 atau lebih (rangkaian < 30 watt biasanya tidak dikoreksi)
Koreksi faktor daya dari 0,5 menjadi 0,85 akan menghasilkan penurunan arus sebesar 40 % Keuntungan Electronic Ballast :
Improved circuit efficiency i.e. reduced ballast loss Reduction in weight, particularly for larger lamp sizes. Improved luminous efficacy for many lamp types Absence of flicker. Elimination of audible ballast noise. Elimination of supply current harmonics and provision of unity power factor without the use of a correction capacitor. Facility for accurate control of lamp power or current. Reduced run-up time and restart time for high-pressure lamps. Controlled starting and operating conditions leading to improved lamp life.
4. Transformator Rugi-rugi trafo tergantung pada besar arus beban dan tahanan belitan primer & sekunder trafo
Bila mengalir arus nominal, rugi-rugi trafo :
Atau,
Rugi-rugi total :
Contoh 1 : Diketahui transformer dng data Sn : 500 kVA, V:11/0.4 kV, ΔPo : 2100 W and ΔPn = 9 450 W, hitung dan plot rugi-rugi sebagai fungsi beban. Load [%]
10
25
50
75
100
Load [kVA]
50
125
250
375
500
No-load Losses [W]
2100
2100
2100
2100
2100
Load Losses [W]
94.5
590
2362
5315
9450
2194.5 2690
4462
7415
11550
47
28
18
Total Losses [W] Losses [%]
95.6
78
Transformer load and no-load losses as a function of load
Rugi-rugi no-load constant tidak tergantung pada beban
No load losses (in % of total losses) as a function of transformer load
Rugi-rugi per kVA
Beban optimal (ekonomis) trafo :
Secon : pembebanan ekonomis trafo
Transformer losses per KVA as a function of transformer load
CATATAN : 1. Minimum losses per kVA terjadi pada beban trafo kira-kira 50% rated capacity
2. Hanya rugi-rugi trafo yang diperhitungkan (tidak termasuk rugi-rugi saluran/supply lines)
Daya reaktif trafo tanpa beban : i0 = arus tanpa beban (%)
Daya reactif yang diserap trafo :
Daya reactif beban penuh atau
Daya reaktif beban penuh juga dapat ditentukan sbb.:
Untuk trafo besar, S n > 1 MVA
Daya reaktif total trafo :
Aliran daya reaktif menghasilkan rugi-rugi pada jaringan distribusi :
kq : 0.1 ÷ 0.2
Rugi-rugi total : Rugi-rugi trafo dan rugi-rugi jaringan distribusi
Rugi-rugi tanpa beban Rugi-rugi berbeban
Rugi-rugi per kVA dari daya VA :
Beban optimal (ekonomis) trafo :
Contoh 2 : Diketahui trafo dng data 1000 kVA, u% = 5%, io = 4.5%, ΔPo = 4000kW , ΔPn = 14000W Rugi-rugi tanpa beban :
Rugi-rugi beban nominal (rated load) :
Untuk kq = 0.15 , beban ekonomis trafo
:
When total losses appearing in both transformers and distribution lines are taken into account, the optimal transformer load is about 70% of full load.
TUGAS-2 • • • •
Siapkan sistem jaringan distribusi (1 feeder) Run Load Flow Check rugi-rugi trafo-nya Cari data typical dari rugi-rugi trafo untuk pembebanan nominal (ΔPn), dan beban nol (ΔP0) • Hitung rugi-rugi total (ΔPt) dari trafo. • Note: S ditentukan dari hasil run Load Flow.
Perhitungan Rugi-Rugi akibat Aliran Daya Reaktif ΔP : rugi-rugi akibat aliran daya reactif φ : sudut fasa antar tegangan dan arus supply R : tahanan saluran supply
Rugi-rugi transmission dan distribusi (dikompensasi oleh capacitor banks) Q : P tan φ Qc : kapasitas dari compensating devices
Losses in distribution lines depend on the location of customers, and they should be calculated for each customer individually To obtain losses of electrical energy, power losses should be multiplied by the number of hours of demand. This is a relatively easy task when demand is constant. Unfortunately, in practice, demand varies during the day , so there is a need for the introduction of a measure allowing the determination of energy losses for varying demand.
TUGAS : RUMUS UTK MENENTUKAN RUGI-RUGI ENERGI DNG MENGGUNAKAN LOSS FACTOR
Kompensasi Daya Reaktif Pembangkitan Daya Reaktif Kapasitif Daya reaktif induktif yang dibutuhkan peralatan listrik dapat dng mudah diperoleh secara lokal dari kapasitor yang terhubung paralel (shunt capacitors). Dengan demikian aliran daya reaktif dari sumber/pembangkit yang jauh bisa dihindari, sehingga dapat mengurangi rugi-rugi akibat aliran daya reaktif.
Flow of active and reactive power without compensation
Flow of active and reactive power with compensation
Reactive power may be generated by rotating compensators or capacitors Synchronous generators at power stations that produce and supply reactive power. Such generators can be used to supply reactive power to local customers. Transmission of reactive power to distant customers is associated with network losses and is not cost effective. Synchronous generators are designed in such a way that the optimal operating point requires some reactive power generation, so a very high power factor is not feasible. Synchronous condensers that consist at unloaded generators connected in various places within the supply network. Their primary role is to supply only reactive power. Due to high initial cost and significant losses, synchronous compensators are only used in applications where their voltage regulating and stabilizing effects are necessary. Synchronous motors can produce reactive power when overexcited. Since small synchronous motors are expensive, this method is rarely used.
Capacitors are the best solution to producing reactive power, due to their low initial cost and inexpensive maintenance
mutual interaction of inductive and capacitive currents, by their arithmetic summation, leads to high values for a power factor, calculated as cos φ, and reduction of supply current magnitude
Kapasitor
pada sistem 3 fasa dapat dihubungkan
delta atau star. Hubungan delta
memberikan daya reaktif lebih besar
dari hubungan star dengan harga kapasitor (μF) yang
sama. Ini
diakibatkan oleh tegangan antar fasa yang lebih
besar pada capacitor bank dengan hubungan delta
HUBUNGAN DELTA
Daya reaktif yang dibangkitkan oleh kapasitor :
(sin φ = 1.0)
Arus kapasitor :
HUBUNGAN STAR
Tegangan pada kapasitor : V : Tegangan antar fasa Daya reaktif yang dibangkitkan :
Although a delta configuration provides three times more reactive power than a star arrangement, capacitors connected in delta are subjected to higher voltages ; therefore, this arrangement is not recommended for HV installations.
Lokasi Kapasitor Central compensation capacitor bank terhubung pada HV incoming feeder. Group compensation capacitor bank terhubung pada MV/LV buses. Individual compensation unit capacitor terhubung pada motor2.
Central compensation mengakibatkan penurunan rugi2 pada sisi supply, rugi2 pada jaringan industri tidak terkompensasi. Kompensasi ini hanya untuk memenuhi persyaratan perusahaan listrik agar mempunyai faktor daya diatas yang ditentukan (PLN menentukan faktor daya minimal 0.85 lag)
Keuntungan Utilization
of reactive power compensating banks, since all motors do not operate at the same time. Low maintenance cost.
Kerugian Switching
the protection equipment may cause explosions Transients caused by energizing of a large capacitor group Space requirements Provides only upstream compensation
Group compensation Keuntungan : Low
installation cost Ability to utilize installed capacitance Low maintenance cost
Aspek negatif : Necessity
to install capacitor banks on each MV/LV bus Only upstream compensation Space requirements
Individual capacitor units Keuntungan : increased
capacity of the supply lines provide direct voltage support capacitor and load are switched ON/OFF together, which does not require expensive switching equipment easy selection and installation of capacitor units
Aspek negatif : high
installation cost, since price per kVAr is higher for small units requires lengthy calculations installation is not fully utilized over-excitation of a motor large transients generated in frequently switching the installation (ON/OFF) longer running off periods for some loads
The best solution is usually a combination of individual and group compensation. Although this solution involves lengthy calculations, it can be cost effective in many installations.
1. Release Line Capacity
The flow of reactive power causes not only energy losses due to resistance of distribution lines, but it also reduces the capacity of transmission lines, in particular in peak demand periods.
Contoh 3: Diketahui beban yang menyerap daya aktif yang dinyatakan dengan arus aktif sebesar 50A. Harga awal cos φ nya adalah 0.5 lagging, yang menghasilkan arus total sebesar 100A. Hitung peningkatan kapasitas saluran, bila faktor daya diperbaiki dengan compensating devices. Hasil perhitungan diberikan pada tabel2 berikut. Ia
P (kW)
φ
φ It
Ir
Q (kvar) S (kVA)
cos φ
0.5
0.6
0.7
0.8
0.9
0.98
sin φ
0.866
0.8
0.71
0.6
0.43
0.2
Reactive current [A]
86
66.4
50.7
37.5
23.8
10.2
Total current [A]
100
83
71.4
62.5
55.5
51
Active current [A]
50
50
50
50
50
50
Line Capacity Increase (%)
0
17
28,6
37.5
44.5
49
Increase of supply line capacity due to power factor improvement
It is very difficult to evaluate the exact increase of line capacity, in particular when many distribution lines are not fully loaded, even in peak periods
Increase of supply line capacity due to power factor improvement
2. Jaringan Industri Untuk menekan biaya peralatan kompensasi daya reaktif adalah dengan menentukan lokasi yang tepat untuk pemasangan kapasitor, yang tidak hanya mengurangi kebutuhan daya reaktif tetapi juga biaya peralatan dan pemasangan yang minimal. Jaringan industri umumnya mempunyai konfigurasi radial . Kompensasi daya reaktif di industri bertujuan untuk mencapai faktor daya tertinggi di gardu induk yang menghubungkan sistem kelistrikan industri tsb dengan jaringan distribusi/transmisi, dan mengurangi rugi-rugi pada sistem kelistrikan industri
Metode paling efektif adalah individual compensation karena memberikan kompensasi daya reaktif langsung ke beban. Pada kondisi-kondisi tertentu metode ini sangat mahal, dan bisa mengakibatkan kenaikan tegangan yang cukup besar pada belitan/kumparan motor induksi. Central compensation adalah metode kompensasi yang paling sederhana, karena kompensasi dipasang pada main substation. Tetapi metode ini tidak mengurangi rugi-rugi didalam sistem kelistrikan. Pada Group compensation , peralatan kompensasi dipasang pada substation yang mensupply kelompok2 beban. Kompensasi ini bertujuan untuk mengurangi rugirugi energi pada semua saluran supply.
Fungsi objective untuk memperoleh total rugi2 minimal pada semua saluran supply (group compensation) diformulasikan sbb.:
Qi = reactive power consumed in each substation, Qci = reactive capacitance to be installed in each substation Ri = resistance of supply lines V = supply voltage
Constraint dari permasalahan ini adalah harga dari faktor daya pada main bus (yang telah ditentukan), dan kapasitas kapasitor harus sama dengan besar daya reaktif yang harus dikompensasi untuk faktor daya tertentu.
Qctotal = the total value of capacitance to be installed. Permasalahan ini dapat diselesaikan dengan menggunakan 2 metode : by application of Lagrange multiplier by application of Dynamic Programming
Sebagai contoh perhitungan praktis, perhatikan saluran ke 4 dari sistem dibawah ini (lihat tabel) Line
P [kW]
Q[kV A]
R [Ω]
1
150
60
4.0
2
110
60
2.0
3
100
130
0.5
4
150
250
0.2
Total
510
500
-
Faktor daya pada main bus ditentukan, yaitu 0.96, maka kapasitas kapasitor yang dibutuhkan adalah 350 kVAr. Hasil optimasi diberikan pada tabel dibawah ini. Untuk memberikan gambaran dari metode optimasi ini, lokasi kapasitor dan rugi-rugi ditentukan dengan menggunakan "classical approach" dimana kompensasi faktor daya dilakukan pada setiap substation untuk mencapai 0.96.
Optimal Calculation Substation
QC[kVAr]
Losses
Classical Approach QC[kVAr]
[kW]
Losses [kW]
1
50
0.77
20
12.3
2
50
0.43.
30
3.5
3
100
0.87
100
0.87
4
150
3.87
200
0.967
Total
350
5.94
350
17.64
TUGAS • Dengan metode “trial and error”, tentukan nilai kapasitor yang menghasilkan total rugi2 antara 5,94 kW dan 17,64 kW • Buat 2 (dua) scenario
3. Penyulang (Feeders) Distribusi Pada jaringan supply distribusi, optimisasi aliran daya reaktif bertujuan untuk menentukan lokasi kapasitor pada suatu feeder sehingga rugi-rugi yang disebabkan oleh aliran daya reaktif minimal, untuk mempertahankan tegangan (maximum) sepanjang feeder tsb, dan untuk meminimisasi biaya pemasangan.
check
Untuk menghindari over kompensasi pada saluran supply , capacitor banks should be split into two parts: fixed capacitors and switched capacitors.
When demand is not known, it is assumed, as a rule of thumb, that 1/3 of the units are fixed on line year around, while 2/3 of the units are switched due to the variable demand -
Bila capacitor banks akan dipasang pada satu titik di saluran supply, maka circuit load centre ditentukan sbb.: where Pi : loads in a particular section of the supply line li : distances between nodes
Contoh 4:
Lcenter = 2.2 x 500 + 1.5 x 1000 + 0.5 x 2000 2.2
= 1636 m
Load centre mendekati node 2, sehingga capacitor bank dipasang pada node 2 . This simple rule-of-thumb does not take into account either varying demand or differences in line resistance between feeder nodes. It is estimated that this method provides the optimal solution in less than 40% of cases.
Side Effects 1. Overcompensation
Bila terjadi overcompensation, tegangan saluran supply menjadi lebih tinggi dari nominalnya, yang akan mempengaruhi beban-beban lain yang terhubung pada saluran tsb. Hal ini mengakibatkan berkurangnya umur isolasi, dan mempunyai dampak negatif pada beban-beban sensitif, misalnya lampu pijar.
Automatic capacitor control In order to avoid overcompensation, equipment for group or central compensation is often provided with automatic regulation, switching capacitors in and out in step with the load. When large load fluctuations exist, it is recommended to use an automatic bank with several steps. Switching of the capacitors is regulated by a power factor relay, keeping the power factor at the setting value
2. Induction Motor Excitation Kompensasi individual dari motor-motor induksi dengan ratings sampai dengan 8 kW bisa digunakan besaran standard dari kapasitor tegangan rendah. However, capacitors used for individual compensation should not be too large . This limitation results from possible motor excitation. When an induction motor is disconnected from the supply and continues to rotate, the capacitor feeds excitation current to the motor that starts operating as an induction generator. If the capacitor is too large, the self-excitation voltage is higher than the rated voltage. This can damage both the motor and capacitor.
Untuk menghindari masalah ini, individual compensation should never have an output output higher than than the output corresponding correspond ing to the no-load current current of the motor motor , where Io = no-load current
Bila data Io tidak ada, Io dapat ditentukan sbb.:
When a capacitor is to be connected to a motor with a star/delta switch, switch, it is important to check that there is no switching position in which the which the capacitor capacitor is either directly short-circuited or in series with the motor motor windings windings
Capacitors reduce motor supply currents, so the setting of the motor motor protection switch should should be adjusted to to give the same protection to the motor motor as it was before compensation
Case Study Reactive Power Compensation in Industrial Networks An industrial customer intends to increase production by the installation of new machines. The new installation comprises seven induction motors of 22 kW. A new supply line of 185 mm2 (copper) is proposed. An proposed. An initial calculation showed that the new new installation installation would cause the transformer transformer to be overloaded . Consider compensation of reactive power to reduce energy demand.
No
Load
P kW
cosφ
Q kVAr
S kVA
I A
Iactive A
Ireactive A
1
Metal Halide Lamp 220*400W
88
0.85
54.5
103.5
144
122
76.4
2
Induction Motors 5.5 kW*16
88
0.7
90
125
174
122
124
3
Induction Motors 11kW*9
99
0.6
132
165
230
138
184
4
Induction Motors 22kW*7
154
0.7
157
220
306
214
218
429
-
433.5
610
847
596
602.4
Total
No
Load
P kW
cosφ
Qnew kVAr
QC kVAr
S kVA
I A
Iactive A
Ireactive A
1
Metal Halide Lamp 220*400W
88
0.96
25.6
28.9
91.6
127
122
35.6
2
Induction Motors 5.5 kW*16
88
0.96
25.6
64.4
91.6
127
122
35.6
3
Induction Motors 11kW*9
99
0.96
28.6
103.2
103.1
143
138
40.25
4
Induction Motors 22kW*7
154
0.96
44.9
112.1
160.4
223
214
62.4
429
-
125
308.6 446.8
620
596
173.8
Total
OPTION 1 Uncompensated
OPTION 1 Compensated
610
447
122(*)
89
Total current [A]
847
620
Total active current [A]
596
596
Total reactive current [A]
602
174
Average cos φ
0.7 .
0.96
Active power [kW]
429
429
Reactive power [kVAr] .
433
125
-
308.6
Parameters S [kV A] Transformer load [%] maximum
Compensation devices [kVAr]
Tugas UTS SIAPKAN SISTEM KELISTRIKAN PLN (JTM)
1. RUN LOAD FLOW (BASE CASE) PROFILTEGANGAN, DAYA INPUT, FAKTOR DAYA, LOSSES JARINGAN, LOSSES TRAFO, PEMBEBANAN TRAFO 2. PEMBEBANAN DIUBAH SHG TERJADI MASALAH POWER QUALITY PROFILTEGANGAN, DAYA INPUT, FAKTOR DAYA, LOSSES JARINGAN, LOSSES TRAFO, PEMBEBANAN TRAFO
Tugas UTS 3. BERIKAN KOMPENSASI DAYA REAKTIF YANG MENGHASILKAN PENGURANGAN LOSSES (C DI PASANG PADA 2/3 PANJANG FEEDER, PADA LOAD CENTER, DLL) PROFILTEGANGAN, DAYA INPUT, FAKTOR DAYA, LOSSES JTM, LOSSES TRAFO, PEMBEBANAN TRAFO (GI & DISTRIBUSI) 4. UTK NO 3, HITUNG PENINGKATAN AVAILABLE POWER, RELEASE CAPACITY & PENGURANGAN LOSSES, FAKTOR DAYA
Tambahan Kuliah • Tgl 03 – 04 – 08 , Kamis, jam 18.30, di Lab SSTL (reg) • Tgl 03 – 04 – 08 , Kamis jam 16.00, ruang C103 (ext)
Tugas UTS SIAPKAN SISTEM KELISTRIKAN INDUSTRI
1. RUN LOAD FLOW (BASE CASE) PROFILTEGANGAN, DAYA INPUT, FAKTOR DAYA, LOSSES JARINGAN, LOSSES TRAFO, PEMBEBANAN TRAFO 2. PEMBEBANAN DIUBAH SHG TERJADI MASALAH POWER QUALITY PROFILTEGANGAN, DAYA INPUT, FAKTOR DAYA, LOSSES JARINGAN, LOSSES TRAFO, PEMBEBANAN TRAFO