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8/3/2019 PF Correction
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POWER FACTOR IMPROVEMENT
EPM333 APRIL 2011Prof. Hossam Talaat
Electrical Power Eng.
Ain Shams Univ
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POWER FACTOR DEFINITION
S = V I*
PF = Cos()
Q = V I sin()+V
-
Ia Ir
I
R L+V
-
Ia Ir
I
R C
Ir
Ia
I
V
Ir
Ia
I
V
Inductive Load Capacitive Load
Leading Power Factor
Lagging Power Factor
Q
P
S
V
Lagging Q
Q
P
S
V
Lagging Q
Q = S sin() = +vePower factor = P / S (lagging)
Q = S sin() = -vePower factor = P / S (leading)
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PRACTICALVALUESOFPOWERFACTORFORDIFFERENTAPPLICATIONS
Load Type Plant / Appliance PF
Lighting
Incandescent Lamps
Fluorescent Lamps (Uncompensated)
Fluorescent Lamps (Compensated)
1.0
0.5
0.93
Heating
Ovens Using Resistance Elements
Induction Heating Ovens (Compensated)
1.0
0.85
Welding
Fixed 1-Phase Arc-Welding Set
Arc-Welding Transformer-Rectifier Set
0.5
0.7 to 0.8
Induction
Motors
Loaded At 0%
Loaded At 50%
Loaded At 100%
0.17
0.73
0.85
Arc Furn. Arc Furnace 0.8
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RELATIONSHIPBETWEENREACTIVEPOWER
ANDPOWERFACTOR
P = 1000 kW
S = P/PF
= 1000/PF
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RELATIONSHIPBETWEENSYSTEMLOSSESAND
POWERFACTOR
Ploss = |I|2 r = [Ia /PF]
2 r
Ploss-normalized = Ploss / Ploss-min = 1 / PF2
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CONCEPTOF PF IMPROVEMENT
PL
QL
SL Qc
QTSTT
L
PL
QL
SL
Qc
QT PL,ST
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ADVANTAGESOF PF IMPROVEMENT
1- Reduce Electricity Billl (customer)
2- Reduction of cable size (Customer)
Multiplying factor for the cross-sectional
area of the cable core
1 1.25 1.67 2.5
Power Factor 1 0.8 0.6 0.4
3- Reduction of Losses (Utility + Customer)
4- Reduction of Voltage Drop (Utility)
5- Increase Equipment Capacity (utility)
Transformers, Feeders, T.L.,
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TYPICAL EXAMPLE
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POWERWORLD SIMULATOR MODEL(WWW.POWERWORLD.COM)
slack
Line Losses =
Line Loading Transformer Loading
12
3
11.31 kV 11.31 kV 0.391 kV
0%
A
Amps
0%
A
MVA
0.00 AMP
0.000 MW
0.000 Mvar
0.000 MW0.000 MW
0.000 Mvar
0.000 Mvar
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UNCOMPENSTED LOAD
slack
Line Losses =
Line Loading Transformer Loading
12
3
11.31 kV 10.71 kV 0.349 kV
39%
A
Amps
40.45 AMP
0.500 MW
0.500 Mvar
0.030 MW0.530 MW
0.589 Mvar
0.000 Mvar
94%
A
MVA
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COMPENSATED LOAD
slack
Line Losses =
Line Loading Transformer Loading
12
3
11.31 kV 11.02 kV 0.380 kV
25%
A
Amps
63%
A
MVA
26.23 AMP
0.500 MW
0.500 Mvar
0.012 MW0.512 MW
0.037 Mvar
0.500 Mvar
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COMPARISON
CaseVL (V) ILine (A)
Ploss
(kW)
Trans
Load %
Feeder
Load %
Pgenerated
MW
No-Load 391 0 0 0 0 0
Uncomp.
Load 349 40.45 30 94 39 0.530Compens.
Load 380 26.23 12 63 25 0.512
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PRACTICAL MEASUREMENTOF PF
Average tan()= Annual kVArh / Annual kWh
Cos() = ..
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CREATINGCUSTOMERMOTIVATION
(PENALTYINEGYPT)
Penalty = (0.9-PF) * kWh charges(0.7 < PF < 0.9)
When the average annual PFis less than 0.7, thepenalty is:
Penalty= [ (0.9-0.7) + 3(0.7-PF)/2 ] * kWh charges(PF < 0.7)
If the consumer does not correct the PF, the penaltyincreases after three months to:
Penalty= 2 * (0.9-PF) * kWh charges
The utility also has the right to discontinue service ifthe consumer does not correct the PFwithin anothersix months.
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FIXED VS VARIABLE CAPACITOR BANKS
Fixed capacitors This arrangement employs one or more capacitor(s) to
form a constant level of compensation.
This configuration is usually applied in cases where the
level of load is almost constant. Automatic capacitor banks
This kind of equipment provides a fast adaptation ofcompensation to match the level of load such that aselected level of power factor is maintained within a
tolerated limit. Such equipment is applied at points in an installation
where the active-power and/or reactive-power variationsare relatively large, e.g., at the busbars of a generalpower distribution board, at the terminals of aheavily-loaded feeder cable,
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AUTOMATIC CAPACITORBANK
A bank of capacitors is divided into a number ofsections, each of which is controlled by a contactor.
The size of the bank increased or decreased in steps,
A regulator monitors the power factor of the controlledcircuit(s) and is arranged to close and openappropriate contactors to maintain a reasonablyconstant system power factor.
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AUTOMATIC CAPACITORBANK (CONT.)
Regulator
CT
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AUTOMATIC CAPACITORBANK (CONT.)
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