ALIRAN DAYA
ALIRAN DAYA
&
&
RUGI-RUGI DAYA
RUGI-RUGI DAYA
Aliran Daya Reaktif
Aliran Daya Reaktif
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aliran daya reaktif
Faktor daya daerah Faktor daya daerah perumahan (Rabu) perumahan (Rabu)
Faktor daya daerah Faktor daya daerah perumahan (Minggu) perumahan (Minggu)
0
0,,7
7 –
– 0
0,,9
9
0
0,,7
7 –
– 0
0,,9
9
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 “typical” berdasarkan jenis
industri dan jenis proses
Faktor Daya motor induksi sangat
tergantung pada beban
Peralatan/Beban yang menyerap
Daya Reaktif
1. Motor Induksi
Power Ml M2 M3 M4 M5 M7 HP 1 5 25 50 100 200 kW - 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 Efficiency [%] 73 83 89 90.5 91.5 93Power 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 Rugi-Rugi Magnetik
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)
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
4. Transformator
Rugi-rugi trafo tergantung pada besar arus beban dan tahanan belitan primer & sekunder trafo
Bila mengalir arus nominal, rugi-rugi trafo :
Atau,
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 Total Losses [W] 2194.5 2690 4462 7415 11550 Losses [%] 95.6 78 47 28 18
Transformer load and no-load losses as a function of load
Rugi-rugi per kVA
Beban optimal (ekonomis) trafo :
Secon : pembebanan ekonomis trafo
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
Daya reaktif beban penuh juga dapat ditentukan sbb.:
Untuk trafo besar, S n > 1 MVA
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 per kVA dari daya VA :
Contoh 2 :
Diketahui trafo dng data 1000 kVA, u% = 5%, io = 4.5%, ΔPo = 4000kW , ΔPn = 14000W
Rugi-rugi tanpa beban :
Untuk kq = 0.15 , beban ekonomis trafo
:
When total losses appearing in both transformers and distribution lines are taken into account, the optimal
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 (ΔP
n), dan beban nol
(ΔP
0)
• Hitung rugi-rugi total (ΔP
t) dari trafo.
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
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
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
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
Reactive power may be generated by rotating compensators or capacitors
mutual interaction of inductive and capacitive currents, by their arithmetic summation, leads to high values for a power factor,
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
Daya reaktif yang dibangkitkan oleh kapasitor :
Arus kapasitor :
(sin φ = 1.0) HUBUNGAN DELTA
HUBUNGAN STAR
Tegangan pada kapasitor :
V : Tegangan antar fasa
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.
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
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)
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
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)
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 It Ir φ P (kW) S (kVA) Q (kvar) φ
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
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
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
-Sebagai contoh perhitungan praktis, perhatikan saluran ke 4 dari sistem dibawah ini (lihat tabel)
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 Classical Approach Substation QC[kVAr] Losses QC[kVAr] Losses
[kW] [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.
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
2.2 x 500 + 1.5 x 1000 + 0.5 x 2000 2.2
Lcenter = = 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
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
Untuk menghindari masalah ini,
Untuk menghindari masalah ini, individual compensationindividual compensation
should never
should never have an outphave an output higher thaut higher than the outputn the output
correspond
corresponding to the ing to the no-load currenno-load current of the mot of the motortor , ,
where
where Io Io = = no-load no-load currentcurrent
Bila data Io tidak ada, Io dapat ditentukan sbb.: Bila data Io tidak ada, Io dapat ditentukan sbb.:
When a capacitor is to be connected to
When a capacitor is to be connected to a motor with aa motor with a star/delta
star/delta switchswitch, it is important to check that there is , it is important to check that there is nono switching position in which
switching position in which the capacito the capacitor is either r is either directlydirectly short-circuited
Capacitors reduce motor supply
Capacitors reduce motor supply
currents, so the setting of the m
currents, so the setting of the m
otor
otor
protection switch shou
protection switch shou
ld be adjusted
ld be adjusted
to
to
give the same protection to the mot
give the same protection to the motor
or
as it was before compensation
Case Study
Case Study
Reactive Power Compensation in Industrial
Reactive Power Compensation in Industrial
Networks
Networks
An industrial customer intends to increase production by An industrial customer intends to increase production by thethe installation of new machines. The new
installation of new machines. The new installation comprisesinstallation comprises seven induction motors of 22 kW. A new supply line of 185 seven induction motors of 22 kW. A new supply line of 185 mm2 (copper) is proposed.
mm2 (copper) is proposed. An initial An initial calculation calculation showed showed
that the n
that the new installatew installation would ion would cause the tracause the transformer to nsformer to bebe
overloaded
overloaded .. Consider compensation of reactive power toConsider 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 Total 429 - 433.5 610 847 596 602.4
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 Total 429 - 125 308.6 446.8 620 596 173.8
Parameters Uncompensated OPTION 1 Compensated OPTION 1
S [kV A] 610 447
Transformer load [%] maximum 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
