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Power Factor CorrectionPower Factor Correction
Save Money and Energy
Presented by:Alan L. OQuinn, P.E.
Team Power Solutions, LLC
(404) 213-8368
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What is Power Factor
Correction?
Usually adding capacitance to PowerSystem.
Ideal power factor is 1.0
Typical industrial complex is 0.65 0.8 pf.
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How Do You Get Low PF?
Motor loads are inductive and need VARs.
R
M => L
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Whats Wrong With Low PF?
Wastes Power Overloads Equipment
Costs Money Uses Up Natural Resources
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Complex Power
Apparent Power = S = P + j Q VA
Real Power = P = | V | | I | cos () W
Reactive Power = Q = | V | | I | sin () VAR
Power Factor = PF = cos ()
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Single Phase Power
I = kVA / kV = kW / (kV PF)
kVA = kV I
kW = kVA PF
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Three Phase Power
I = kVA / ( kV 1.732 ) = kW / (kV 1.732 PF)
kVA = kV 1.732 I
kW = kVA PF
Note: kV is the phase-to-phase voltage
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Power Triangle
KVA
KVAR
KW
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Classic PF Example
Facility has peak demand of 1000 kW Load Power Factor is 0.707 lagging
Calculate kVAR. Calculate amount of capacitance in kVAR
needed to correct to a 0.9 PF
How much is capacitance in microfarads?
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Draw Power Triangle
KVA = 1000/0.707 = 1414
= acos(0.707) = 45o
KVAR =
sqrt (1414^2 1000^2)
= 1000
KW = 1000
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Graphical Representation of PowerPF = 0.707
-400
-200
0
200
400
600
800
1000
0 0.005 0.01 0.015 0.02
Time
Po
wer
Real Reactive Apparent
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PF Example Continued Slide 1 of 4
KVAR =sqrt (1111^2 1000^2)
= 484
KVA = 1000/0.9 = 1111
= acos(0.9) = 25.8o
KW = 1000
Added kVAR = 1000 - 484 = 516
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Comparison
KVAR = 1000
KVAR = 484
KW = 1000
Added kVAR = 1000 - 484 = 516
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Graphical Representation of PowerPF = 0.9
-300
-200-100
0
100
200
300
400
500
600
700
800
0 0.005 0.01 0.015 0.02
Time
Po
wer
Real Reactive Apparent
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PF Example Continued Slide 2 of 4
C = ( kVAR 103
) / (2 f ) (kV)2
= ( 500 103 ) / (2 60 ) (0.480)2
= 5,759 microfarads
Note: 480 volts was the assumed voltage.
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PF Example Continued Slide 3 of 4
Load
1000 kW
0.707 pf
A
B
M
UtilitySource
C
Meter reads 0.9 pf
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PF Example Continued Slide 4 of 4
What is the reduction in amps supplied by the utility?
kVA before was 1414
kVA after pf correction is 1111
Reduced kVA is 303 kVA
Reduced amps is 303 / ( 0.480 1.732 ) = 364 amps
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Benefits
Reduced I2
R losses Less voltage drop. VLN = I Zeq
Reduced conductor size. Possible reduce substation transformer size.
ProvidesVoltage support
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Harmonics
Multiple of the fundamental frequencydistortions. 5th = 300 Hz, 7th = 420 Hz
Created by non-linear devices such as UPS
systems, rectifiers, VFDs, welders,
fluorescent ballasts, personal computers
IEEE 519-1992
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5th Harmonic Distortion
-1.5
-1
-0.5
0
0.5
1
1.5
0 2 4 6 8 10 12
Fundamental 5th Harmonic Total
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5th, 7th, 11th 13th Harmonics
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6
Total Fundamental
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Harmonic Resonance
Occurs when capacitance reactance andsystem reactance are equal.
Large currents circulate between
transformer and capacitor. Higher voltages may occur across
equipment.
Usually 5th, 7th, 11th, and 13th are thefrequencies of most concern
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Systems Harmonic ResonancePrevious Example: 1500 kVA Transformer, %Z = 5.75
500 kVAR of Capacitance
Harmonic Order = h = sqrt( kVASC/ kVAR )
kVASC = kVA / (%Z/100) = 1500 = 1500/0.0575 = 26,087
h = sqrt (26,087 / 500 ) = 7.22
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Load has 5th, 7th, 11th and 13th Harmonics each with 10%Current Distortion
Easy Power Model
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1 . 0 2 . 2 3 . 4 4 . 6 5 . 8 7 . 0 8 . 2 9 . 4 1 0 . 6 1 1 . 8 1 3 . 0
0 . 0 0
1 5 . 0 0
3 0 . 0 0
4 5 . 0 0
6 0 . 0 0
7 5 . 0 0
9 0 . 0 0
1 0 5 . 0 0
1 2 0 . 0 0
1 3 5 . 0 0
1 5 0 . 0 0
F re q ue n c y S c a n - In j e c t io n o n B U S - 2
H a r m o n i c
ZpuM
agnitude
B U S - 2
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Documented Sources GE Corporation, GE Low Voltage Power Factor Correction Capacitors GEP-974-G,
2003 Su, Kendall L. Fundamentals of Circuit Analysis, Waveland Press, 1993
Matthes, John H. Introduction to the Design and Analysis of Building Electrical
Systems. Van Nostrand Reinhold, 1993