Download POWER AND POWER FACTOR Impedance

POWER AND
POWER FACTOR
Impedance (Ohm)


Impedance is a ratio between voltage and
current
Unit of impendance is Ohm and simbolized by Z
ZR  R
Z L  jL
1
ZC 
j C
Series Impedance
Pararel Impedance
The Operation is same with Resistant
Resistance
V0 0
IR 
R

The angle between current and voltage are in
phase
Inductance
Inductive Reactance (Ohm)  X L  j L
Current flow in pure inductance :
V 0 V 00
V 0
V
IL 



  90
XL
j L  L90  L
Capacitance
1
Capacitive Reactance (Ohm)  X C 
jC
Current flow in pure capacitance :
V 0 V 00
IL 

 CV 90
XC
1/ jC
Phasor Diagram for Impedance
and Current-Voltage Relationship
IC
XL
I = IR + (IC+IL)
Z = R + Xtot
IR
R
Xc
IL
V
Resitive-Inductive Load
Voltage  V 0(assume)
impedance(Zeq )  R  X L  Z  
XL
i
Z = R + Xl

Xc
R
Vs
Load
Current Flow …..
V0 V
I
   
Z  Z
IC
IR

I = IR + IL
EXAMPLE !!!
IL
V
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
Phasor Diagram for Three Component of
Power Relationship :
QL(VAr)
XL
Z = R + Xtot
(S = P + Q)
R
P (watt)
QC(VAr)
Xc
Resitive-Inductive Load
i
Vs
dari data diatas diketahui :
Tegangan = V
Impedansi = Z = R  XL= Z 
Beban =
R  XL
Phasor Diagram
QL (VAr)
S = P + QL

P(Watt)
Power Factor (PF) is ratio between Real Power (P) to
Complex Power (S)
Qc(VAr)
Pinalty PLN (Electrical Company)
Cos  = 0,85
The angle () = 31,7o
Perhitungan hubungan faktor daya 0,85 (Pinalti PLN)
dengan biaya kVArh adalah sebagai berikut :
Q (kVAr)
S (kVA)
cos   P 
S
P
Q2  P 2
Cos  = 0,85
P (kW)
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 :
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
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
Decrease Maximum Capacity Load
Contoh
Power Contract (VA) = 1000 VA
a) Lamp 100 Watt, PF = 0,5
b) Lamp 100 Watt, PF= 1

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
Equipment to Increase Power Factor is
Capasitor Bank/Power Factor Correction
Capacitor Instalation Circuit
i
Vs
Kapasitor
Bank
Beban
QL(VAr)
S (VA) last
S new
 last
 last
P(Watt)
Qc(VAr)
Example
i
P = 3 kW
PF = 0,6
Vs
4 kVAr
5 kVA
53
3 kW
P  3kW
PF  cos   0,6
  arc. cos 0,6  53
P
3kW
S

 5kVA
cos 
0,6
Q  S sin 53  5kVA.0,8  4kVAr
cont’d

If PF need to become 0,95
0,98 kVAr
P  3kW
PFb  cos   0,95
3,16 kVA
18,2
b  arc. cos 0,95  18,2
P
3kW
Sb 

 3,16kVA
cos  0,95
Qb  S sin 18,2  3,16 x0,312  0,98kVAr
3 kW
QL
5 kVA
4 kVAr
Qc  QL  QB
QC
0,98 kVAr
3,16 kVA
18,2
3 kW
Qc  4  0,98  3,02kVAr