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# POWER AND POWER FACTOR - PowerPoint PPT Presentation

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

• 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

• The angle between current and voltage are in phase

Current Flow ….. Relationship

EXAMPLE !!!

Power Relationship

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

=

dari data diatas diketahui :

Tegangan = V

Impedansi = Z = =

Phasor Diagram Relationship

Power Factor (PF) is ratio between Real Power (P) to Complex Power (S)

Pinalty PLN (Electrical Company)

Cos  = 0,85

The angle () = 31,7o

Q (kVAr) Relationship

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

Example --- Electrical Bill : Relationship

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

Impact of Power Factor Relationship

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

Example of Impact lower PF Relationship

Contoh

Power Contract (VA) = 1000 VA

• Lamp 100 Watt, PF = 0,5

• Lamp 100 Watt, PF= 1

• Number of lamps a) can be install is : Relationship

• 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

Equipment to Increase Power Factor is Relationship

Capasitor Bank/Power Factor Correction

Example Relationship

cont’d Relationship

• If PF need to become 0,95