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.

Presentasi berjudul: "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."— Transcript presentasi:

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 (  ) = 31,7 o

13 Perhitungan hubungan faktor daya 0,85 (Pinalti PLN) dengan biaya kVArh adalah sebagai berikut : P (kW) S (kVA) Q (kVAr) Cos  = 0,85 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 (I 2 R). 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 a) Lamp 100 Watt, PF = 0,5 b) 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

21 Example

22 cont’d If PF need to become 0,95

23