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POWER AND POWER FACTOR

POWER AND POWER FACTOR. Impedance (Ohm). Impedance is a ratio between voltage and current Unit of impendance is Ohm and simbolized by Z. Series Impedance Pararel Impedance. The Operation is same with Resistant. Resistance. The angle between current and voltage are in phase. Inductance.

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POWER AND POWER FACTOR

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  1. POWER AND POWER FACTOR

  2. Impedance (Ohm) • Impedance is a ratio between voltage and current • Unit of impendance is Ohm and simbolized by Z Series Impedance Pararel Impedance The Operation is same with Resistant

  3. Resistance • The angle between current and voltage are in phase

  4. Inductance

  5. Capacitance

  6. Phasor Diagram for Impedance and Current-Voltage Relationship

  7. Resitive-Inductive Load

  8. Current Flow ….. EXAMPLE !!!

  9. Power There are three component of Power : 1. S = Complex Power (VA) S = V.I* 2. P = Real/Active Power (Watt) P = V x I* x PF (cos phi) 3. Q = Reactive Power (Var) Q = V x I* x sin phi

  10. Phasor Diagram for Three Component of Power Relationship :

  11. Resitive-Inductive Load = dari data diatas diketahui : Tegangan = V Impedansi = Z = =

  12. Phasor Diagram Power Factor (PF) is ratio between Real Power (P) to Complex Power (S) Pinalty PLN (Electrical Company) Cos  = 0,85 The angle () = 31,7o

  13. Q (kVAr) S (kVA) Cos  = 0,85 P (kW) Perhitungan hubungan faktor daya 0,85 (Pinalti PLN) dengan biaya kVArh adalah sebagai berikut : • Jika cos  = 0,85 • Maka Q = 0,6197 P • Artinya Jumlah maksimum kVArh adalah 0,6197 besar kWh • Jika Jumlah kVArh lebih dari 0,6197 kWh, • Maka kelebihan kVArh harus dibayar oleh konsumen

  14. Example --- Electrical Bill : If sum of our total energy (kWh) consume (LWBP + WBP) are 1000 kWh, so the totals kVArh permitted : 0,6197 x 1000 = 619,7 kVArh

  15. Impact of Power Factor Lower Power Factor cause negative impact, there are : • Increase Line Losses (I2R). • Decrease system efficiency. • Increase abondement cost • Increase Electrical Bill (cost) --- if get pinalty • Need to increase the capacity of equipment (Trafo) --- increase investation cost

  16. Example of Impact lower PF Decrease Maximum Capacity Load Contoh Power Contract (VA) = 1000 VA • Lamp 100 Watt, PF = 0,5 • Lamp 100 Watt, PF= 1

  17. Number of lamps a) can be install is : • S (VA) lamp a) = 100 W/ 0,5 = 200 VA • Number of lamps = Langganan VA / S • Number of lamps = 1000 VA / 200 VA = 5 lamps • Number of lamps b) can be install is • S (VA) lamp b) = 100 W/ 1 = 100 VA • Number of lamps = Langganan VA / S • Number of lamps = 1000 VA / 100 VA = 10 lamps

  18. Equipment to Increase Power Factor is Capasitor Bank/Power Factor Correction

  19. Capacitor Instalation Circuit

  20. Example

  21. cont’d • If PF need to become 0,95

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