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Power Factor CorrectionPower Factor Correction

Save Money and Energy

Presented by:Alan L. OQuinn, P.E.

Team Power Solutions, LLC

aoquinn@teampowersolutions.com

(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