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POWER ANDPOWER FACTOR

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Impedance (Ohm)

Impedance is a ratio between voltage and

current

Unit of impendance is Ohm and simbolized by Z

CjZ

LjZ

RZ

C

L

R

1

Series Impedance

Pararel ImpedanceThe Operation is same with Resistant

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Resistance

The angle between current and voltage are in

phase

R

VI

R

00

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Inductance

0

Inductive Reactance (Ohm)

Current flow in pure inductance :

0 0 0 9090

L

L

L

X j L

V V V V IX j L L L

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Capacitance

0

1Capacitive Reactance (Ohm)

Current flow in pure capacitance :

0 0 901/

C

L

C

Xj C

V VI CVX j C

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Phasor Diagram for Impedance

and Current-Voltage Relationship

R

XL

Xc

Z = R + Xtot

IR

IL

IC

V

I = IR + (IC+IL)

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Resitive-Inductive Load

LoadVs

i

R

XL

Xc

Z = R + Xl

0 ( )

( )eq L

Voltage V assume

impedance Z R X Z

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Current Flow ..

IR

IL

IC

V

I = IR+ IL

Z

V

Z

VI

0

EXAMPLE !!!

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PowerThere 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

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Phasor Diagram for Three Component of

Power Relationship :

R

XL

Xc

Z = R + Xtot

P (watt)

QL(VAr)

QC(VAr)

(S = P + Q)

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Resitive-Inductive Load

BebanVs

i

dari data diatas diketahui :

Tegangan = V

Impedansi = Z = =XLR

=

Z

XLR

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Phasor Diagram

P(Watt)

QL (VAr)

Qc(VAr)

S = P + QL

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

Complex Power (S)

Pinalty PLN (Electrical Company)

Cos = 0,85

The angle () = 31,7o

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Perhitungan hubungan faktor daya 0,85 (Pinalti PLN)

dengan biaya kVArh adalah sebagai berikut :

P (kW)

S (kVA)

Q (kVAr)

Cos = 0,85

22Q

PS

PcosP

Jika cos = 0,85Maka 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

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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 = 619,7kVArh

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Impact of Power Factor

Lower Power Factor cause negative

impact, there are:

Increase Line Losses (I2

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

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Example of Impact lower PF

Decrease Maximum Capacity Load

ContohPower Contract (VA) = 1000 VA

a) Lamp 100 Watt, PF = 0,5

b) Lamp 100 Watt, PF= 1

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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

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Equipment to Increase Power Factor is

Capasitor Bank/Power Factor Correction

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Capacitor Instalation Circuit

BebanVs

i

Kapasitor

Bank

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P(Watt)

Qc(VAr)

S (VA) last

lastlast

QL(VAr)

S new

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Example

P = 3 kW

PF = 0,6Vs

i

kVArkVASQ

kVAkWP

S

arc

PF

kWP

48,0.553sin

56,0

3

cos

536,0cos.

6,0cos

3

4 kVAr

3 kW

5 kVA

53

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contd

If PF need to become 0,95

3 kW

3,16 kVA

2,18

0,98 kVAr

kVArxSQ

kVAkWP

S

arc

PF

kWP

b

b

b

b

98,0312,016,32,18sin

16,395,0

3

cos

2,1895,0cos.

95,0cos

3

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4 kVAr

3 kW

3,16 kVA

2,18

0,98 kVAr

5 kVA

QL

QCkVArQc

QQQcBL

02,398,04