sa02607001e eng guide pwr fact correciton

8/2/2019 SA02607001E Eng Guide Pwr Fact Correciton

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POWER FACTOR CORRECTION:

A GUIDE FOR THE PLANT ENGINEER

CAPACITOR BANKS &

PASSIVE HARMONIC FILTERS

See labels on product for additional information.


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SA02607001E

Power Factor Correction:A Guide for the Plant Engineer

Description

PART ONE: POWER FACTOR

What is Power Factor? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Should I Be Concerned About Low Power Factor? . . . . . . . . . . . . . . . . . . . . .

What Can I Do to Improve Power Factor? . . . . . . . . . . . . . . . . . . . . . . . . . . . .

How Much Can I Save by Installing Power Capacitors?. . . . . . . . . . . . . . . . . . .

How Can I Select the Right Capacitors for My SpecificApplication Needs? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

How Much kvar Do I Need? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Where Should I Install Capacitors in My Plant Distribution System? . . . . . . 1

Can Capacitors Be Used in Non-Linear, Non-Sinusoidal Environments? . . . 1

What About Maintenance? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Code Requirements for Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Useful Capacitor Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

PART TWO: HARMONICS

Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1What are Harmonics?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

What are the Consequences of High Harmonic Distortion Levels? . . . . . . . . 2

IEEE 519. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

How are Harmonics Generated? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

What Do Power Factor Correction Capacitors Have to Dowith Harmonics? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

How Do I Diagnose a Potential Harmonics Related Problem? . . . . . . . . . . . . 2

How Can Harmonics Problems Be Eliminated? . . . . . . . . . . . . . . . . . . . . . . . . 2

What is a Passive Harmonic Filter?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Do I Need to Perform a System Analysis to Correctly Apply

Harmonic Filters?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

What is Eatons Experience in Harmonic Filtering?. . . . . . . . . . . . . . . . . . . . . 2


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PART ONE: POWER FACTOR

What is Power Factor?

Special Electrical Requirementof Inductive LoadsMost loads in modern electrical distribution systemsare inductive. Examples include motors, transform-ers, gaseous tube lighting ballasts and inductionfurnaces. Inductive loads need a magnetic fieldto operate.

Inductive loads require two kinds of current:

Working power (kW) to perform the actual work ofcreating heat, light, motion, machine output, etc.

Reactive power (kvar) to sustain the magnetic field

Working power consumes watts and can be read ona wattmeter. It is measured in kilowatts (kW). Reac-tive power doesnt perform useful work, but circu-lates between the generator and the load. It places aheavier drain on the power source, as well as on thepower sources distribution system. Reactive poweris measured in kilovolt-amperes-reactive (kvar).

Working power and reactive power together makeup apparent power. Apparent power is measured inkilovolt-amperes (kVA).

Note: For a discussion on power factor in non-linear,non-sinusoidal systems, turn to Page 17.

Figure 1. kW Power

Fundamentals of PowerPower factor is the ratio of workinapparent power. It measures howelectrical power is being used. A signals efficient utilization of elecwhile a low power factor indicateof electrical power.

To determine power factor (PF), dpower (kW) by apparent power (k

or sinusoidal system, the result isto as the cosine .

For example, if you had a boring operating at 100 kW and the appaconsumed was 125 kVA, you wou125 and come up with a power fa

Figure 3. kVA Power

Hot Plate

Light

ResistiveLoadG

G MMotorFi ld

PF kWkVA------------- cosine = =

kW( )100kVA( )125------------------------------- PF( ) 0.80=

G

k

kVA

COS kWkVA----------- PF= =


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Should I Be ConcernedAbout Low Power Factor?Low power factor means youre not fullyutilizing the electrical power youre paying for.

As the triangle relationships in Figure 5 demon-strate, kVA decreases as power factor increases.At 70% power factor, it requires 142 kVA to produce100 kW. At 95% power factor, it requires only105 kVA to produce 100 kW. Another way to lookat it is that at 70% power factor, it takes 35% morecurrent to do the same work.

Figure 5. Typical Power Triangles

100 kW

33kvar

100kvar

100 kW

142kVA

105kVA

PF100142-------- 70%= =

= =PF100

105

-------- 95%


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What Can I Do to ImprovePower Factor?You can improve power factor by adding power factorcorrection capacitors to your plant distribution system.

When apparent power (kVA) is greater than workingpower (kW), the utility must supply the excessreactive current plus the working current. Powercapacitors act as reactive current generators.(See Figure 6.) By providing the reactive current,they reduce the total amount of current your systemmust draw from the utility.

95% Power FactorProvides Maximum BenefitTheoretically, capacitors could provide 100% ofneeded reactive power. In practical usage, however,power factor correction to approximately 95%provides maximum benefit.

The power triangle in Figure 7 shows apparentpower demands on a system before and afteradding capacitors. By installing power capacitorsand increasing power factor to 95%, apparent power

is reduced from 142 kVA to 105 kVA a reductionof 35%.

Figure 6. Capacitors as kvar Generators

Figure 7. Required Apparent Power BefAdding Capacitors

18

Amperes

16Amperes

10 hp, 480 V Motorat 84% Power Facto

3.6 Amperes

3 kvar

Capacitor

Power Factor Improved to 95%Line Current Reduced 11%

105kVA

After

95% PFAfter

70% PFBefore

142kV

ABefore

1

2

COS1

100142---------- 70% PF= =

COS2

100105---------- 95% PF= =


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SA02607001E

How Much Can I Save byInstalling Power Capacitors?Power capacitors provide many benefits:

Reduced electric utility bills

Increased system capacity

Improved voltage

Reduced losses

Reduced Utility BillsYour electric utility provides working (kW) andreactive power (kvar) to your plant in the form of

apparent power (kVA). While reactive power (kvar)doesnt register on kW demand or kW hour meters,the utilitys transmission and distribution systemmust be large enough to provide the total power.Utilities have various ways of passing along theexpense of larger generators, transformers, cables,switches, etc., to you.

As shown in the following case histories, capacitorscan save you money no matter how your utility billsyou for power.

kVA BillingThe utility measures and bills every ampere ofcurrent, including reactive current.

Case 1

Case 2

Assume an uncorrected 460 kVA demand, 480 volt,three-phase at 0.87 power factor (normally good).

Billing:

$4.75/kVA demand Correct to 0.97 Power Factor

Solution:

From Table 6 kW Multipliers, to raise the Power Factor from0.87 to 0.97 requires capacitor:

Multiplier of .316 x kW.316 x 400 = 126 kvar (use 140 kvar)

kWPF---------- kVA=

4000.97-------------- 412 Billing Demand=

Corrected

kVA Power Factor kW=460 .87 400 kW Actual Demand=

460 kVA $4.75 $2185/mo.=

412 kVA $4.75 $1957/mo.=

$1957

$ 228/mo. Saving 12 =

Uncorrected Original Billing:

Corrected New Billing:

$2736 Annual Saving

Assume the same conditions except that:400 kW @ 87% = 460 kVA

400 kW @ 97% = 412 kVA Corrected Billing

kVA Demand Charge: $1.91/kVA/month (112,400 kWh/month energy consumed).

Energy Charge:

$0.0286/kWh (first 200 kWh/kVA of demand)$0.0243/kWh (next 300 kWh/kVA of demand)$0.021/kWh (all over 500 kWh/kVA of demand)

Energy:kWh = 112,400

460 x 200 = 92,000 kWh@ 0.0286 = $2631.20

460 x 300 = 138,000but balance only = 20,400

@ $0.0243 = $ 495.72

kWh = 112,40412 x 200 = 82,400

@ $0.0286 = $2356.

412 x 300 = 123but balance only = 30,0

@ $0.0243 = $ 7

(9600 kWh in first step reduced by $0.0043)

Note: This is not a reduction in energy consumed,but in billing only.

A 130 kvar capacitor can be paid for in less than 14 mo

460 kVA $1.91 $878.60=412 kVA $1.91 $786=

$786.92

Uncorrected: Corrected:

Demand:

$ 91.68 Savings in Demand Charg

$2631.20$ 495.72$3126.92--------------------------------

+

UncorrectedEnergy Charge

$2356.64$ 729.00$3085.64--------------------------------

+

CorrectedEnergy Char

Saving in Energy Charge Due to Rate Cha$3085.64$ 41.28------------------------------

$ 41.28$ 91.68$ 132.96

12$1595.52------------------------------

--------------------------------

+EnergyDemand

Monthly Total Saving

Saving:


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kW Demand Billing withPower Factor AdjustmentThe utility charges according to the kW demand and

adds a surcharge or adjustment for power factor.The adjustment may be a multiplier applied to kWdemand. The following formula shows a billingbased on 90% power factor:

If power factor was 0.84, the utility would require7% increase in billing, as shown in this formula:

Some utilities charge for low power factor but givea credit or bonus for power above a certain level.

Case 1

Case 2

kW Demand 0.90Actual Power Factor--------------------------------------------------------------------

kW 0.90

0.84--------------------------------

107 (Multiplier)=

Assume a 400 kW load, 87% power factor with thefollowing Utility tariff.

A. Demand Charges:

First 40 kW @ $10.00/kW monthly billing demand

Next 160 kW @ $ 9 .50/kWNext 800 kW @ $ 9 .00/kWAll over 1000 kW @ $ 8.50/kW

B. Power Factor Clause:

Rates based on power factor of 90% or higher.When power factor is less than 85%, the demand willbe increased 1% for each 1% that the power factor isbelow 90%. If the power factor is higher than 95%, thedemand will be decreased 1% for each 1% that thepower factor is above 90%.

Note: There would be no penalty for 87% powerfactor. However, a bonus could be credited if the power

factor were raised to 96%.

To raise an 87% power factor to 96%, refer to Table 6.Find 0.275 x 400 kW = 110 kvar (Select 120 kvar to ensurethe maintenance of the 96% level).

To Calculate savings:

Normal 400 kW billing demandFirst 40 kW @ $10.00= $ 400.00Next 160 kW @ $ 9.50 = $1520.00Bal. 200 kW @ $ 9.00 = $1800.00Total 400 kW $3720.00 Normal Monthly Billing

First 40 kW @ $10.00= $ 400.00Next 160 kW @ $ 9.50 = $1520.00Bal. 175 kW @t $ 9.00= $1575.00

$3495.00 Power FactorAdj d Billi

kW 0.90New Power Factor-------------------------------------------------------------- =

400 0.900.96

--------------------------------- 375 kW Demand=

New Billing:

With the same 400 kW load, the powerIn this example, the customer will pay

(From Case 1. When the Power Factor demand is 375 kW = $3495.00 per mon

First 40 kW @ $10.00= $ 400.00Next 160 kW @ $ 9.50 = $ 1520.00Next 244 kW @ $ 9.00 = $ 2196.00Total 444 kW $ 4116.00 $3

12 = $7452.00

Yearly savings if corrected to 96%.

To raise 81% power factor to 96%, selecTable 6. 0.432 x 400 kW = 173 kvar. Usea 96% power factor. The cost of a 180 k$1900.00, and the payoff is less than fo

55 kvar would eliminate the penaltyfactor to 85%.

400 0.900.81

--------------------------------- 444 Billing kW Demand=

$4116.00$3720.00$ 395.00--------------------------------

Charge at 81%Normal kW Demand ChargPower Fac17tor Adjustmen

81% Power Factor


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SA02607001E

kvar Reactive Demand Charge

The utility imposes a direct charge for the useof magnetizing power, usually a waiver of some

percentage of kW demand. For example, if thischarge was 60 cents per kvar for everything over50% of kW, and a 400 kW load existed at the time,the utility would provide 200 kvar free.

Case 1

Increased System CapacityPower factor correction capacitors increase sycurrent-carrying capacity. Raising the power faon a kW load reduces kVA. Therefore, by addin

capacitors, you can add additional kW load to system without altering the kVA.

Figure 8. Correcting Power Factor IncreasesTransformer Output

The same principle holds true for reducing cur

on overloaded facilities. Increasing power factfrom 75% to 95% on the same kW load results21% lower current flow. Put another way, it tak26.7% more current for a load to operate at 75and 46.2% more current to operate at 65%.

Assume a 400 kW load demand at 81% power factor.

Tariff Structure:

Demand Charge is:$635.00 for the first 200 kW demand

$ 2.80 per kW for all additionReactive demand charge is:

$ 0.60 per kvar in excess of 50% of kW demand

In this example, kW demand = 400 kW, therefore 50% =200 kvar which will be furnished at no cost.

This ratio is the basis for the Table of Multipliers (Table 6).

Cos PF kWkVA-------------= =

Tan kvarkW--------------=

OppAdj--------------or

AdjHyp-------------or

200kvar

289.6kvar

12

400 kW

With 200 kvar allowed at no cost, then

From 1.0 or unity power factor column. Table 6, note that0.500 falls between 89% and 90% power factor. The billing

excess kvar is above that level 81% power factor.

Since 200 kvar is allowed, the excess kvar is 89.6 (round to90) x $0.60 = $54.00 per month billing for reactive demand.

Solution:

To correct 400 kW from 81% to 90% requires 400 x 0.240 (fromTable 6) = 96 kvar. (Use 100 kvar.) The approximate cost forthis capacitor is $1250.00. The payoff is about 23 months.

2200400------------ 0.5 or 50% of kW= =

289.6 kvar=Tan1 0.724 kvar= kW Tan1= 400 0.724 =

360 kW

75% PFOriginal

Condition

480 kVA578 Amperes

450 kW

95% PFCorrected

474 kVA570 Amperes

A plant has a 500 kVA transformer operating near capaIt draws 480 kVA or 578 amperes at 480 volts. The prespower factor is 75%, so the actual working power avaiis 360 kW.

It is desired to increase production by 25%, which meathat about 450 kW output must be obtained. How is thaccomplished? A new transformer would certainly beone solution. For 450 kW output, the transformer woube rated at 600 kVA to handle 75% power factor load.More likely, the next size standard rating would be nee

(750 kVA).

Perhaps a better solution would be to improve the powfactor and release enough capacity to accommodate tincreased load.

To correct 450 kW from 75% to 95%, power factor requ450 x 0.553 (from Table 6) = 248.8 kvar use 250 kvar at $2800.00.


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Industries with Low Power FactorBenefit Most from CapacitorsLow power factor results when inactive motors

are operated at less than full load. This often occursin cycle processes such as those using circularsaws, ball mills, conveyors, compressors, grinders,punch presses, etc. where motors are sized forthe heaviest load. Examples of situations wherelow power factor (from 30% to 50%) occur includea surface grinder performing a light cut, anunloaded air compressor and a circular sawspinning without cutting.

The following industries typically exhibit lowpower factors:

Table 1. Typical Low Power Factor Industries

Include Power CapacitoNew Construction/ExpaIncluding power capacitors in you

tion and expansion plans can redtransformers, bus, switches, etc., project in at lower cost.

Figure 9 shows how much systemreleased by improving power factpower factor from 70% to 90% relper kW. On a 400 kW load, 128 kV

Improved Voltage CondiLow voltage, resulting from exces

draw, causes motors to be sluggisAs power factor decreases, total lincreases, causing further voltagecapacitors to your system and imyou get more efficient motor perflonger motor life.

Reduced LossesLosses caused by poor power facreactive current flowing in the syswatt-related charges and can be e

power factor correction.

Power loss (watts) in a distributiocalculated by squaring the currenit by the circuit resistance (12R). Tloss reduction:

Industry UncorrectedPower Factor

Saw MillsPlastic (Esp. Extruders)Machine Tools, StampingPlating, Textiles, Chemicals, BreweriesHospitals, Granaries, Foundries

45% 60%55% 70%60% 70%65% 75%70% 80%

% Reduction Losses 100 100Orig

Ne---------------

=

Orig

ina

lPower

Fac

tor

1.00

.90

.80

.70

.60

0.90

0.95

1.00

0.80

0.70

Corrected Power Factor


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SA02607001E

How Can I Select the Right Capacitorsfor My Specific Application Needs?

Once youve decided that your facility can benefitfrom power factor correction, youll need to choosethe optimum type, size and number of capacitorsfor your plant.

There are two basic types of capacitor installations:individual capacitors on linear or sinusoidal loads,and banks of fixed or automatically switchedcapacitors at the feeder or substation.

Individual vs. Banked InstallationsAdvantages of individual capacitors at the load:

Complete control; Capacitors cannot causeproblems on the line during light load conditions

No need for separate switching; Motor alwaysoperates with capacitor

Improved motor performance due to more effi-cient power utilization and reduced voltage drops

Motors and capacitors can be easily relocatedtogether

Easier to select the right capacitor for the load

Reduced line losses

Increased system capacity

Advantages of bank installations at the feederor substation:

Lower cost per kvar

Total plant power factor improved reducesor eliminates all forms of kvar charges

Automatic switching ensures exact amountof power factor correction, eliminates over-

capacitance and resulting overvoltages

Table 2. Summary of Advantages/Disadvantages ofIndividual, Fixed Banks, Automatic Banks, Combination

Consider the Particular Needsof Your PlantWhen deciding which type of capacitor installabest meets your needs, youll have to weigh thadvantages and disadvantages of each and conseveral plant variables, including load type, losize, load constancy, load capacity, motor starmethods and manner of utility billing.

Load Type

If your plant has many large motors, 50 hp and ait is usually economical to install one capacitor pmotor and switch the capacitor and motor togetIf your plant consists of many small motors, 1/2 hp, you can group the motors and install one capat a central point in the distribution system. Oftebest solution for plants with large and small mois to use both types of capacitor installations.

Load SizeFacilities with large loads benefit from a combtion of individual load, group load and banks o

fixed and automatically-switched capacitor unA small facility, on the other hand, may requironly one capacitor at the control board.

Sometimes, only an isolated trouble spot requirpower factor correction. This may be the case if plant has welding machines, induction heaters odc drives. If a particular feeder serving a low powfactor load is corrected, it may raise overall planpower factor enough that additional capacitorsare unnecessary.

Load ConstancyIf your facility operates around-the-clock and hasconstant load demand, fixed capacitors offer theest economy. If load is determined by eight-hourfive days a week, youll want more switched unidecrease capacitance during times of reduced lo

Load CapacityIf your feeders or transformers are overloadedif you wish to add additional load to already lolines, correction must be applied at the load. If

facility has surplus amperage, you can install citor banks at main feeders. If load varies a greadeal, automatic switching is probably the answ

Utility BillingThe severity of the local electric utility tariff for pfactor will affect your payback and ROI. In many an optimally designed power factor correction swill pay for itself in less than two years

Method Advantages Disadvantages

IndividualCapacitors

Most technicallyefficient, most flexible

Higher installationand maintenance cost

Fixed Bank Most economical,fewer installations

Less flexible,requires switchesand/or circuit breakers

Automatic

Bank

Best for variable loads,prevents overvoltages,low installation cost

Higher equipment cost

Combination Most practical for largernumbers of motors

Least flexible


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How Much kvar Do I Need?The unit for rating power factor capacitors is akvar, equal to 1000 volt-amperes of reactive power.

The kvar rating signifies how much reactive powerthe capacitor will provide.

Sizing Capacitors forIndividual Motor LoadsTo size capacitors for individual motor loads, useTable 3 below. Simply look up the type of motorframe, RPM and horsepower. The charts indicate

the kvar rating you need to bring 95%. The charts also indicate how

is reduced when capacitors are in

Sizing Capacitors for EntiIf you know the total kW consumpplant, its present power factor anyoure aiming for, you can use Tato select capacitors.

Table 3. Suggested Maximum Capacitors Ratings

Induc

tion

Mo

tor

hp

Ra

ting Number of Poles and Nominal Motor Speed in RPM

2 3600 RPM 4 1800 RPM 6 1200 RPM 8 900 RPM 10 720 RPM

Capac

itor

kvar

Curren

t

Re

duc

tion

% Capac

itor

kvar

Curren

t

Re

duc

tion

% Capac

itor

kvar

Curren

t

Re

duc

tion

% Capac

itor

kvar

Curren

t

Re

duc

tion

% Capac

itor

kvar

Curren

t

Re

duc

tion

%

Used for High Efficiency Motors and Older Design (Pre T-Frame) Motors

357.5

1.522.5

141211

1.522.5

151312

1.523

201715

234

272522

2.545

353230

101520

345

1099

345

111010

356

141312

567.5

211816

689

272321

253040

679

988

679

1099

7.59

10

111110

91012.5

151413

1012.515

201816

506075

12.51517.5

888

101517.5

988

12.51517.5

101010

1517.520

121110

2022.525

151514

100125150

22.527.530

888

202530

888

253035

999

27.53037.5

101010

354050

131312

200250300

405060

888

37.54550

877

405060

988

506060

1099

607080

121111

350400450500

60757575

8888

60607575

7666

75758085

8888

758590

100

9999

9095

100100

101099

T-Frame NEMA Design B Motors

235

11.52

141414

11.52.5

242322

1.523

302826

234

423831

234

404040

7.5

1015

2.5

45

14

1412

3

45

20

1818

4

56

21

2120

5

67.5

28

2724

5

7.58

38

3632

202530

67.58

121211

67.58

171716

7.58

10

191919

91015

232322

1012.515

292524

405060

12.51517.5

121212

1517.520

161515

152022.5

191917

17.522.525

212120

2022.530

242422

75 20 12 25 14 25 15 30 17 35 21


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SA02607001E

For use with three-phase, 60 Hz NEMA Classification B Motors to raise full load power factor to approximately 95%.

Table 4. Suggested Capacitor Ratings, in kvars, for NEMA Design C, D and Wound-Rotor Motors

Note: Applies to three-phase, 60 Hz motors when switched with capacitors as single unit.

Note: Use motor manufacturers recommended kvar as published in the performance data sheets for specific motor tdrip-proof, TEFC, severe duty, high efficiency and NEMA design.

InductionMotor Rating (hp)

Design C Motor Design D Motor1200 r/Minimum

Wound-RotorMotor

1800 and 1200 r/Minimum 900 r/Minimum

152025

556

566

566

5.577

304050

7.51012

91215

101215

111317.5

6075

100

17.51927

1822.527

1822.530

202533

125150

200

3537.5

45

37.545

60

37.545

60

4050

65

250300

5465

7090

7075

7585


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Table 5. Suggested Capacitor Ratings for Medium Voltage Motors

Above sizes are intended to provide a corrected power factor of approximately 95% at full the limited number of capacitor ratings available, it is not possible to raise every motor PF

To calculate kvar required to correct power factor to a specific target value, use the followin

Wherehp: Motor Nameplate Horse Power%EFF: Motor Nameplate Efficiency (Enter the value in decimal)PFa: Motor Nameplate Actual Power FactorPFt: Target Power Factor

Note: Consult the motor manufacturers data sheet to verify the maximum kvar of capacitors that can bconnected at motor terminals. To avoid self-excitation, do not exceed the maximum kvar rating that the motor manufacturer.

Induction

Mo

tor

hp

Ra

ting Number of Poles and Nominal Motor Speed in RPM

2 3600 RPM 4 1800 RPM 6 1200 RPM 8 900 RPM 10 720 RPM

Capacit

or

kvar

Curren

t

Re

duct

ion

% Capacit

or

kvar

Curren

t

Re

duct

ion

% Capacit

or

kvar

Curren

t

Re

duct

ion

% Capacit

or

kvar

Curren

t

Re

duct

ion

% Capacit

or

kvar

Curren

t

Re

duct

ion

%

2400 and 4160 Volts Open

100125150200250

2525255050

87777

2525255050

109887

2525255050

1110998

2525255075

1110999

2525255075

1211111010

300350400

450500

505075

7575

766

65

505075

7575

766

66

757575

75100

887

66

7575

100

100125

999

99

7575

100

100125

999

99

600700800900

10001250

75100100125150200

555555

100100150150200200

666665

100125150200250250

666655

125150150200250300

887766

150150200250250300

988876

2400 and 4160 Volts Totally Enclosed Fan Cooled

100125150

200250

252525

5050

776

66

252525

5050

877

77

252525

5050

988

88

252550

5075

111111

1111

252550

5075

111111

1111

300350400450500

50757575

100

66665

507575

100125

77777

75100100100125

88887

75100100125150

1010101010

100100100125150

1111111111

kvar(required)hp 0.746

% EFF----------------------------

1 PFa2PFa

---------------------------1 PFt2

PFt--------------------------

=


15/28

SA02607001E

Instructions:1. Find the present power factor in column 1.

2. Read across to optimum power factor column.

3. Multiply that number by kW demand.

Example:If your plant consumed 410 kW, was currently oper-ating at 73% power factor and you wanted to correctpower factor to 95%, you would:

1. Find 0.73 in column 1.

2. Read across to 0.95 column.

3. Multiply 0.607 by 410 = 249 (round to 250)

4. You need 250 kvar to bring your plant to 9

power factor.

If you dont know the existing power factor levof your plant, you will have to calculate it befousing Table 6 below. To calculate existing powfactor: kW divided by kVA = Power Factor.

Table 6. Multipliers to Determine Capacitor Kilovars Required for Power Factor Correction

OriginalPowerFactor

Corrected Power Factor

0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.9

0.500.51

0.52

0.53

0.54

0.9820.937

0.893

0.850

0.809

1.0080.962

0.919

0.876

0.835

1.0340.989

0.945

0.902

0.861

1.0601.015

0.971

0.928

0.887

1.0861.041

0.997

0.954

0.913

1.1121.067

1.023

0.980

0.939

1.1391.094

1.050

1.007

0.966

1.1651.120

1.076

1.033

0.992

1.1921.147

1.103

1.060

1.019

1.2201.175

1.131

1.088

1.047

1.2481.203

1.159

1.116

1.075

1.2761.231

1.187

1.144

1.103

1.3061.261

1.217

1.174

1.133

1.3371.292

1.248

1.205

1.164

1.3691.324

1.280

1.237

1.196

1.4031.358

1.314

1.271

1.230

1.4401.395

1.351

1.308

1.267

1.4811.436

1.392

1.349

1.308

1.5291.484

1.440

1.397

1.356

1.581.54

1.50

1.45

1.41

0.55

0.56

0.57

0.58

0.59

0.769

0.730

0.692

0.655

0.619

0.795

0.756

0.718

0.681

0.645

0.821

0.782

0.744

0.707

0.671

0.847

0.808

0.770

0.733

0.697

0.873

0.834

0.796

0.759

0.723

0.899

0.860

0.822

0.785

0.749

0.926

0.887

0.849

0.812

0.776

0.952

0.913

0.875

0.838

0.802

0.979

0.940

0.902

0.865

0.829

1.007

0.968

0.930

0.893

0.857

1.035

0.996

0.958

0.921

0.885

1.063

1.024

0.986

0.949

0.913

1.093

1.054

1.016

0.979

0.943

1.124

1.085

1.047

1.010

0.974

1.156

1.117

1.079

1.042

1.006

1.190

1.151

1.113

1.076

1.040

1.227

1.188

1.150

1.113

1.077

1.268

1.229

1.191

1.154

1.118

1.316

1.277

1.239

1.202

1.166

1.37

1.33

1.29

1.26

1.22

0.60

0.61

0.62

0.630.64

0.583

0.549

0.516

0.4830.451

0.609

0.575

0.542

0.5090.474

0.635

0.601

0.568

0.5350.503

0.661

0.627

0.594

0.5610.529

0.687

0.653

0.620

0.5870.555

0.713

0.679

0.646

0.6130.581

0.740

0.706

0.673

0.6400.608

0.766

0.732

0.699

0.6660.634

0.793

0.759

0.726

0.6930.661

0.821

0.787

0.754

0.7210.689

0.849

0.815

0.782

0.7490.717

0.877

0.843

0.810

0.7770.745

0.907

0.873

0.840

0.8070.775

0.938

0.904

0.871

0.8380.806

0.970

0.936

0.903

0.8700.838

1.004

0.970

0.937

0.9040.872

1.041

1.007

0.974

0.9410.909

1.082

1.048

1.015

0.9820.950

1.130

1.096

1.063

1.0300.998

1.19

1.15

1.12

1.091.06

0.65

0.66

0.67

0.68

0.69

0.419

0.388

0.358

0.328

0.299

0.445

0.414

0.384

0.354

0.325

0.471

0.440

0.410

0.380

0.351

0.497

0.466

0.436

0.406

0.377

0.523

0.492

0.462

0.432

0.403

0.549

0.518

0.488

0.458

0.429

0.576

0.545

0.515

0.485

0.456

0.602

0.571

0.541

0.511

0.482

0.629

0.598

0.568

0.538

0.509

0.657

0.626

0.596

0.566

0.537

0.685

0.654

0.624

0.594

0.565

0.713

0.682

0.652

0.622

0.593

0.743

0.712

0.682

0.652

0.623

0.774

0.743

0.713

0.683

0.654

0.806

0.775

0.745

0.715

0.686

0.840

0.809

0.779

0.749

0.720

0.877

0.846

0.816

0.786

0.757

0.918

0.887

0.857

0.827

0.798

0.966

0.935

0.905

0.875

0.846

1.02

0.99

0.96

0.93

0.90

0.70

0.71

0.72

0.73

0.74

0.270

0.242

0.214

0.186

0.159

0.296

0.268

0.240

0.212

0.185

0.322

0.294

0.266

0.238

0.211

0.348

0.320

0.292

0.264

0.237

0.374

0.346

0.318

0.290

0.263

0.400

0.372

0.344

0.316

0.289

0.427

0.399

0.371

0.343

0.316

0.453

0.425

0.397

0.369

0.342

0.480

0.452

0.424

0.396

0.369

0.508

0.480

0.452

0.424

0.397

0.536

0.508

0.480

0.452

0.425

0.564

0.536

0.508

0.480

0.453

0.594

0.566

0.538

0.510

0.483

0.625

0.597

0.569

0.541

0.514

0.657

0.629

0.601

0.573

0.546

0.691

0.663

0.635

0.607

0.580

0.728

0.700

0.672

0.644

0.617

0.769

0.741

0.713

0.685

0.658

0.817

0.789

0.761

0.733

0.706

0.87

0.84

0.82

0.79

0.76

0.750.76

0.77

0.78

0.79

0.1320.105

0.079

0.052

0.026

0.1580.131

0.105

0.078

0.052

0.1840.157

0.131

0.104

0.078

0.2100.183

0.157

0.130

0.104

0.2360.209

0.183

0.156

0.130

0.2620.235

0.209

0.182

0.156

0.2890.262

0.236

0.209

0.183

0.3150.288

0.262

0.235

0.209

0.3420.315

0.289

0.262

0.236

0.3700.343

0.317

0.290

0.264

0.3980.371

0.345

0.318

0.292

0.4260.399

0.373

0.346

0.320

0.4560.429

0.403

0.376

0.350

0.4870.460

0.434

0.407

0.381

0.5190.492

0.466

0.439

0.413

0.5530.526

0.500

0.473

0.447

0.5900.563

0.537

0.510

0.484

0.6310.604

0.578

0.551

0.525

0.6790.652

0.626

0.599

0.573

0.730.71

0.68

0.65

0.63

0.80

0.81

0.82

0.83

0.84

0.000 0.026

0.000

0.052

0.026

0.000

0.078

0.052

0.026

0.000

0.104

0.078

0.052

0.026

0.000

0.130

0.104

0.078

0.052

0.026

0.157

0.131

0.105

0.079

0.053

0.183

0.157

0.131

0.105

0.079

0.210

0.184

0.158

0.132

0.106

0.238

0.212

0.186

0.160

0.134

0.266

0.240

0.214

0.188

0.162

0.294

0.268

0.242

0.216

0.190

0.324

0.298

0.272

0.246

0.220

0.355

0.329

0.303

0.277

0.251

0.387

0.361

0.335

0.309

0.283

0.421

0.395

0.369

0.343

0.317

0.458

0.432

0.406

0.380

0.354

0.499

0.473

0.447

0.421

0.395

0.547

0.521

0.495

0.469

0.443

0.60

0.58

0.55

0.52

0.50

0.85

0.86

0.87

0.880.89

0.000 0.027

0.000

0.053

0.026

0.000

0.080

0.053

0.027

0.000

0.108

0.081

0.055

0.0280.000

0.136

0.109

0.083

0.0560.028

0.164

0.137

0.111

0.0840.056

0.194

0.167

0.141

0.1140.086

0.225

0.198

0.172

0.1450.117

0.257

0.230

0.204

0.1770.149

0.291

0.264

0.238

0.2110.183

0.328

0.301

0.275

0.2480.220

0.369

0.342

0.316

0.2890.261

0.417

0.390

0.364

0.3370.309

0.47

0.45

0.42

0.390.36

0.90

0.91

0.92

0.93

0.94

0.000 0.028

0.000

0.058

0.030

0.000

0.089

0.061

0.031

0.000

0.121

0.093

0.063

0.032

0.000

0.155

0.127

0.097

0.066

0.034

0.192

0.164

0.134

0.103

0.071

0.233

0.205

0.175

0.144

0.112

0.281

0.253

0.223

0.192

0.160

0.34

0.31

0.28

0.25

0.22

0.95

0 96

0.000 0.037

0 000

0.079

0 041

0.126

0 089

0.18

0 14


16/28werFa

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Table 7. Recommended Wire Sizes, Switches and Fuses for Three-Phase, 60 Hz CapacitorsNote: These wire sizes are based on 135% of rated current in accordance with the National Electrical C

90C Copper Type THHN, XHHW or equivalent, applied at 75C ampacity. Rate current based on operation afrequency and kvar. Consult National Electrical Code for other wire types. Above size based on 30C Ambie(Refer to NEC table 310-16.)

Note: Fuses furnished within Capacitor Assembly may be rated at higher value than shown in this tablefor field installations and reflects the manufacturers suggested rating for overcurrent protection and dcompliance with the National Electrical Code.

kvar 240 Volts 480 Volts 600 Volts

Current

(Amps)

Wire

Size

Fuse

(Amps)

Switch

(Amps)

Current

(Amps)

Wire

Size

Fuse

(Amps)

Switch

(Amps)

Current

(Amps)

Wire

Size

0.511.5

1.22.43.6

141414

366

303030

1.21.8

1414

33

3030

1.01.4

1414

22.53

4.86.07.2

141414

101015

303030

2.43.03.6

141414

666

303030

1.92.42.9

141414

456

9.61214

141414

202025

303030

4.86.07.2

141414

101015

303030

3.84.85.8

141414

7.58

10

1819

24

1210

10

3035

40

3060

60

9.09.6

12

1414

14

1520

20

3030

30

7.27.7

9.6

1414

14

12.51517.5

303642

886

506080

6060

100

151821

141210

253040

303060

121417

141412

2022.525

485460

644

80100100

100100100

242730

10108

405050

606060

192224

101010

303540

728496

321

125150175

200200200

364248

866

608080

60100100

293438

886

4550

60

108120

144

1/02/0

3/0

200200

250

200200

400

5460

72

44

2

100100

125

100100

200

4348

58

66

4

758090

180192216

250M300M350M

300350400

400400400

9096

108

1/01/01/0

150175200

200200200

727786

331

100120125

241289300

400M(2)3/0(2)3/0

400500500

400600600

120144150

2/03/03/0

200200250

200200400

96115120

12/02/0

150180200

361432481

(2)250M(2)350M(2)400M

600750800

600800800

180216241

250M350M400M

300400400

400400400

144173192

3/0250M300M

240250

300

289300

361

(2)3/0(2)4/0

(2)250M

500500

600

600600

600

231241

289

400M400M

(2)3/0

360400

432480

(2)350M(2)500M

750800

800800

346384

(2)250(2)300


17/28

SA02607001E

Where Should I Install Capacitorsin My Plant Distribution System?

At the LoadSince capacitors act as kvar generators, the mostefficient place to install them is directly at the motor,where kvar is consumed. Three options exist forinstalling capacitors at the motor. Use Figures 10 16, and the information below to determine whichoption is best for each motor.

Location A Motor Side of Overload Relay New motor installations in which overloads can

be sized in accordance with reduced current draw

Existing motors when no overload change isrequired

Location B Line Side of Starter Existing motors when overload rating surpasses

code (see Appendix for NEC code requirements)

Location C Line Side of Starter Motors that are jogged, plugged, reversed

Multi-speed motors

Starters with open transition and starters that

disconnect/reconnect capacitor during cycle Motors that start frequently

Motor loads with high inertia, where disconnect-ing the motor with the capacitor can turn themotor into a self-excited generator

At the Service FeederWhen correcting entire plant loads, capacitor bcan be installed at the service entrance, if loadditions and transformer size permits. If the amof correction is too large, some capacitors caninstalled at individual motors or branch circuit

When capacitors are connected to the bus, feemotor control center or switchboard, a disconand overcurrent protection must be provided.

Figure 11. Installing Capacitors Online

Locating Capacitors on Motor Circuits

Figure 10. Locating Capacitors on Motor Circuits

Main Busor Feeder

CapacitorBank

Fused SwitchCircuit Breake

Motor

Motor

Feed

MotorStarter

FusedSafety

Switch orBreaker

Installat Location:

CapacitorC

CapacitorB

CapacitorA

Thermal

OverloadB AC


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Locating Capacitors on Reduced Voltage and Multi-Speed Motor

Figure 12. Autotransformer Closed Transition

Note: Connect capacitor on motor side of starting contacts(2, 3, 4) at points A B C.

Figure 13. Series Resistance Starting

Note: Connect capacitor on motor side of starting contactor(1, 2, 3) at points A B C.

Figure 14. Part-Winding Starting

Note: Connect capacitor on motor side of starting contacts(1, 2, 3) at points A B C.

Figure 15. Wye-Delta Starting

Note: Connect capacitor on motor sid(1, 2, 3) at points A B C.

Figure 16. Reactor Starting

Note: Connect capacitor on motor side(1, 2, 3) at points A B C.

MotorStator

5

4

3

2

1

C

B

A

6

7Line

Start:Close 6-7-2-3-4

Transfer:Open 6-7

Run:Close 1-5

MotorStator

1

2

3 A

B

6 9

5 8

4 7

Line

Start:Close 1-2-3

Second Step:Open 4-5-6

Third Step:Close 7-8-9

C

Motor

StatorStart: Close 1-2-3Run: Close 4-5-6

A

B

C2

1

3Line

6

5

4

1

2

Line

Wye Start:Close 1-2-3-7-8

Delta Run:Close 1-2-3-4-5-6

4

B 5

A

7

8

6

3

C

A

Line

3

B2

C1

6

5

4

Start:Close 1-2-3

Run:

Close 4-5-6


19/28

SA02607001E

Can Capacitors Be Used inNon-Linear, Non-Sinusoidal

Environments?Until recently, almost all loads were linear, withthe current waveform closely matching sinusoidalvoltage waveform and changing in proportion to theload. Lately, non-linear loads which draw currentat frequencies other than 60 Hz have increaseddramatically.

Examples of linear and non-linear devices areas follows:

Linear Devices Motors

Incandescent lighting

Heating loads

Non-Linear Devices dc drives

Variable frequency drives

Programmable controllers

Induction furnaces

Arc-type lighting

Personal computers

Uninterruptible power supplies (UPS)

The increase in non-linear loads has led to harmonicdistortion in electrical distribution systems. Althoughcapacitors do not cause harmonics, they can aggra-vate existing conditions.

Because harmonic voltages and currents are

affected by all of the equipment in a facility, theyare sometimes difficult to predict and model.

Capacitor Banks and TransformersCan Cause ResonanceCapacitors and transformers can create dangerousresonance conditions when capacitor banks areinstalled at the service entrance. Under theseconditions, harmonics produced by non-lineardevices can be amplified many fold.

Problematic amplification of harmonics becomesmore likely as more kvar is added to a system whichcontains a significant amount of non-linear load.

You can estimate the resonant harmonic by usingthe following formula:

For example, if the plant has a 1,500 kVA transformer with 5-1/2% impedance and the short-crating of the utility is 48,000 kVA, then kVAsys wequal 17,391 kVA.

If 350 kvar of capacitors were used to improvepower factor, h would be:

Because h falls right on the 7th harmonic,these capacitors could create a harmful resonacondition if non-linear devices were present infactory. In this case the capacitors should be aponly as harmonic filtering assemblies. For furtinformation, see Harmonic Filtering on Page 1

of this document.

See Page 19 (Part 2) for an additional discussion harmonics.

What About MaintenanceCapacitors have no moving parts to wear out require very little maintenance. Check fuses onregular basis. If high voltages, harmonics, swit

surges or vibration exists, fuses should be chemore frequently.

Capacitors from Eaton operate warm to the tIf the case is cold, check for blown fuses, openswitches or other power losses. Also check forbulging cases and puffed up covers, which sigoperation of the capacitor interrupter.

Code Requirements

for CapacitorsNameplate kvar: Tolerance +15, -0%.

Discharge Resistors: Capacitors rated at 600 vand less must reduce the charge to less than 5volts within 1 minute of de-energization. Caparated above 600 volts must reduce the chargewithin 5 minutes.

Continuous Operation: Up to 135% rated (nam

plate) kvar, including the effects of 110% ratedvoltage (121% kvar), 15% capacitance tolerancand harmonic voltages over the fundamentalfrequency (60 Hz).

Dielectric Strength Test: Twice the rated acvoltage (or a dc voltage 4.3 times the ac ratingfor non-metalized systems).

hkVAsys

k-----------------------=

h 17,391350

---------------------- 49.7 7.0= = =


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Useful Capacitor Formulas

Nomenclature: C = Capacitance inFV = Voltage

A = CurrentK = 1000

A. Additional Data

B. Miscellaneous

C. Standard Data

Note: Above is at nominal voltage @ 6kvar F and current.

1.

% V.R.kvar Cap.( ) % Transformer Reactance

kVA Transformer( )-------------------------------------------------------------------------------------------------------------=

Simplified Voltage Rise:

2. Losses Reduction:

% L.R. 100 100Original PF

Improved PF-----------------------------------

2=

3. Operation at other than rated voltage and frequency:

Note: Use of voltages and frequencies abovetherated values can be dangerous. Consult the factoryfor any unusual operating conditions.

a. Reduced Voltage:

Actual kvar Output( ) Rated kvar Actual VoltageRated Voltage

-----------------------------------------

2=

b. Reduced Frequency:

Actual kvar Rated kvarActual Freq.

Rated Freq.---------------------------------

=

c. Examples:

kvar 208( ) kvar 240( ) 208

240----------

20.75= =

Voltage Reduction:

kvar 120( ) kvar 240( ) 120

240----------

20.25= =

10 kvar @ 240 V 7.5 kvar @ 208 V=( )

10 kvar @ 240 V 2.5 kvar @ 120 V=( )

Frequency Reduction:

60 kvar @ 480 V 60 Hz 50 kvar, 480 V, 50 Hz=( )

kvar 50 Hz( ) kvar 60 Hz( )5060------

0.83= =

(a)

(b)

Voltage F/kvarTotal

Amperes/kvar

Single-Phase

208240480

61.246.011.5

4.814.172.08

60024004160

7.370.460.153

1.67

kVA =

Power Factor Cos kWkVA-------------= =

Tan kvarkW--------------=

1.

2.V A PF

103---------------------------kW =

3.

4. Line Current

(Amperes)

kVA 103

V

---------------------------=

V A

103--------------

Single-Phase Thre

3 -----------

3

1------------

kVA

3

------------

(See Table 6)

5. Capacitor Current(Amperes)

2f( )CV 106

=

kvar 103

V----------------------------

kvar 103

3 V----------------------------also:

6. kVAkWPF

--------=(kW Motor Input)

kW (Motor Input)hp 0.746efficiency

----------------------------=7.

8. Approx. Motor kVA Motor hp (at=


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SA02607001E

PART TWO: HARMONICS

IntroductionThere has been much discussion and interestin recent years on the subject of power quality.Whereas in the past, power received from theelectric utility and utilized by an industrial plant wasgenerally a pure sinusoidal waveform i.e., cleanpower more frequently today industrial plants arefinding they have to deal with the problem of dirtypower. Dirty power is a slang expression used todescribe a variety of voltage and current contamina-tions on the pure sinusoidal waveform. Dirty powercan come in the form of short-term transients or

steady-state, continuous distortions. In addition,the sources of dirty power can be external to aplant (as might be the case if a neighboring plantis contaminating the utilitys distribution system)or the source can reside within the plant itself.

Harmonic distortion is a specific type of dirty powerthat is usually associated with an industrial plantsincreased use of adjustable speed drives, powersupplies and other devices using solid-state switch-ing. However, harmonic distortion can be generatedby any of a variety of non-linear electrical devicesexisting within a manufacturing plant or withinnearby plants. Because harmonic distortion cancause serious operating problems in certain plantenvironments, it is important that the plant engineeror facilities personnel understand the fundamentalsof harmonic distortion, know how to recognize thesymptoms of this problem, and know what can bedone to solve the problems once they are identified.

What are Harmonics?A harmonic is a component of a periodic wavehaving a frequency that is an integral multiple ofthe fundamental power line frequency of 60 Hz. Forexample, 300 Hz (5 x 60 Hz) is a fifth order harmonicof the fundamental frequency Figure 17. Figure 18shows the resultant wave when the fundamentaland fifth harmonic are combined. The result isharmonic distortion of the power waveform.

Harmonics which are typically seen on a power

system can be subdivided into two distinct cate-gories by the nature of problems they create andthe remedies they usually require.

1. Those harmonic currents which are thedominant harmonic orders created by three-phase non-linear loads 5th, 7th, 11th, 13th,and higher order odd harmonics which are not

lti l f 3

Figure 17. Fundamental and 5th Harmonic

Figure 18. Fundamental and 5th Harmonic Combined

Harmonics are a steady-state phenomenon anshould not be confused with short-term phenomena that last less than a few cycles. Transienelectrical disturbances, overvoltage surges anundervoltage sags in the supplied voltage are harmonics. Some of these short-term disturbain voltage or current can be mitigated by transvoltage surge suppressors, line reactors, or isotransformers. However, these devices usually little, if any, effect on harmonic currents or vo

The level of voltage or current harmonic distoexisting at any one point on a power system cexpressed in terms of the total harmonic disto(THD) of the current or voltage waveform. The

Volts

T

Fundamental

5th Harmonic

Volts

T


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What are theConsequences of High

Harmonic Distortion Levels?Just as high blood pressure can create stress andserious problems in the human body, high levels ofharmonic distortion can create stress and resultantproblems for the utilitys distribution system, theplants distribution system, as well as all of theequipment that is serviced by that distributionsystem. The result may be the plant engineersworst fear the shutting down of important plantequipment ranging from a single machine to anentire line or process.

Equipment shutdown can be caused by a numberof events. As an example, the higher voltagepeaks that are created by harmonic distortion putextra stress on motor and wire insulation whichultimately can result in insulation breakdown andfailure. In addition, harmonics increase rms currentresulting in increased operating temperatures formany pieces of equipment resulting in greatlyreduced equipment life.

Table 8 summarizes some of the negativeconsequences that harmonics can have on typicalequipment found in the plant environment. Whilethese effects are categorized by problems createdby current and voltage harmonics, current andvoltage harmonic distortion usually exist together(current harmonic distortion causesvoltageharmonic distortion).

Harmonic distortion disrupts plants. Of greatestimportance is the loss of productivity, throughputand possibly sales. These occur because of process

shutdowns due to the unexpected failure of motors,drives, power supplies or just the spurious trippingof breakers. Plant engineers realize how costlydowntime can be and pride themselves in main-taining low levels of plant downtime. In addition,maintenance and repair budgets can be severelystretched. For example, every 10C rise in theoperating temperatures of motors or capacitorscan cut equipment life by 50%.

Table 8. Negative Consequences of Harmonics onPlant Equipment

IEEE 519IEEE Standard 519-1992, IEEE RPractices and Requirements for Ha

in Electrical Power Systems reprerecent effort to establish a standardharmonic distortion levels on a powfollowing two Tables 9 and 10 sumand current harmonic distortion lim

Note: The current distortion limits are size of the customers load relative to tcircuit capacity of the utility (stiffness). Iwhose loads potentially have more effetem and neighboring customers are he

Table 9. End User Limits Current DistGeneral Distribution Systems End-User LiMaximum Harmonic Current Distortion Individual Harmonic Order (Odd Harmo

All power generation equipment is limof current distortion, regardless of act

Note: Even harmonics are limited to 2harmonic limits above.

Note: Current distortions that result inoffset, e.g., half wave converters are n

Note: Where /SC = Maximum Short-Ciand /L = Maximum Demand Load Currfrequency component) at PCC.

Table 10. Util ity Limits Voltage Distort

Note: High voltage systems can have where the cause is an HVDC terminal by the time it is tapped for a user.

Two very important points must reference to the above limitation

1. The customer is responsible forent distortion to within accepta

utility is responsible for limiting2. The limits are only applicable

Common Coupling (PPC) betwand the customer. The PCC, wdefined, is usually regarded asthe utility equipment ownershtomers, or the metering pointabove limits cannot be meani

Equipment Consequences

Current Harmonic Distortion Problems

Capacitors Blown fuses, reduced capacitor life

Motors Reduced motor life,inability to fully load motor

Fuses/Breakers False/spurious operation,damaged components

/SC/IL


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SA02607001E

of IEEE 519, may be forthcoming.

How are Harmonics Generated?Harmonics are generated by non-linear loads.A non-linear load is a circuit element that drawscurrent in a non-sinusoidal manner. Until recently,most factory loads were primarily linear, withcurrent waveform closely matching the sinusoidalvoltage waveform and changing in proportion tothe load. More recently, however, factory loadswith major non-linear components have increaseddramatically. Table 11 gives typical examples oflinear and non-linear devices.

Table 11. Examples of Linear and Non-Linear Devices

Figure 19. Harmonic Magnitude as Percentage of theFundamental Current

Note: Harmonic currents typically injected by six-pulse VFDs.

Non-linear devices that cause the most problemscan generally be grouped into two categories electronic power converters and arcing devices.

Electronic Power ConvertersElectronic power converters e.g., adjustablespeed drives, power supplies are by far thelargest contributors to harmonic distortion intodays plant environment. An electronic poweconverter changes electrical energy from one to another, typically by rectifying the ac voltagdc and utilizing the dc voltage directly or synthing a new ac voltage. This change is accompliby using solid-state devices silicon control

ifiers (SCRs), diodes, transistors to periodicswitch in the conducting circuits of the converFigure 19 shows a typical harmonic currentspectrum for six-pulse electronic power conveBelow are some common names that are typicassociated with electronic power converters.

Common Names Given to ElectronicPower Converters Adjustable Speed Drives

Variable Frequency Drives

SCR Drives ac Motor Drives (ac/dc/ac)

dc Motor Drives (ac/dc)

Three-phase Full Wave Rectifiers

Three-phase Full Wave Converters

Six-pulse Converters

As most plant engineers appreciate, there isan increased use of electronic drives due to thability to more efficiently or precisely drive a moor process.

Arcing DevicesArc furnaces and welders are the two types of adevices that cause the most harmonic distortioalthough arc lighting (fluorescent, mercury vawill also cause small degrees of harmonic disto

Other EquipmentMotors, generators, transformers and arc lightalso have small non-linear components, althothe contribution of these devices to total harmdistortion in a plant tends to be relatively sma

PrimarilyLinear Devices

Devices withMajor Non-Linear Components

Some MotorsIncandescent LightingHeating Loads

dc DrivesVariable Frequency DrivesProgrammable ControllersInduction FurnacesSolid-State Uninterruptible Power

Supplies (UPS)Arc FurnacesArc Welders

HarmonicSix-PulsePower Converter

3

5

7

9

11

13

15

17

0.0%

17.5%

11.1%

0.0%

4.5%

2.9%

1.0%

0.8%


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What Do Power FactorCorrection Capacitors Have

to Do with Harmonics?A discussion of power system harmonics is incom-plete without discussing the effects of power factorcorrection capacitors. In an industrial plant contain-ing power factor correction capacitors, harmoniccurrents and voltages can be magnified consider-ably due to the interaction of the capacitors with theservice transformer. This is referred to as harmonicresonance or parallel resonance. For a typical plantcontaining power factor correction capacitors, theresonant frequency (frequency at which amplifica-tion occurs) normally falls in the vicinity of the 5th tothe 13th harmonic. Since non-linear loads typicallyinject currents at the 5th, 7th, 11th and 13thharmonics, a resonant or near-resonant conditionwill often result if drives and capacitors are installedon the same system, producing the symptomsand problems outlined in the previous section.

Note: Capacitors themselves do not cause harmonics,but only aggravate potential harmonic problems. Often,harmonic-related problems do not show up untilcapacitors are applied for power factor correction.

It is a common misconception that the problem ofapplying capacitors in harmonic environments islimited to problems caused for the capacitor itself that the capacitors lower impedance at higherfrequencies causes a current overload into thecapacitor and, therefore, must be removed.However, the capacitor/harmonics problem mustbe viewed from a power system standpoint. Thecapacitor-induced increase of harmonic voltagesand currents on a plants system may be causingproblems while the capacitor itself remains within

its acceptable current rating.

How Do I DiagnosePotential Harmonic

Problem?If a plant engineer suspects that hharmonics problem, the followingbe performed as an initial investigtial problems:

1. Look for symptoms of harmoin Table 8. If one or more of toccurs with regularity, then thsteps should be taken.

2. If the plant contains power facapacitors, the current into thshould be measured using a meter. If this value is higher trated current at the system voor so), the presence of harmodistortion is likely.

3. Conduct a paper audit of the producing loads and system This analysis starts with the gor horsepower data on all the

devices in the plant, all capacinformation on service entranEaton has specific analysis fotions to guide the plant enginthis information, and engineesentatives can provide assistaThis data is analyzed by Eatodetermine whether the condito create unfavorable levels o

4. If the electrical distribution sy e.g., multiple service entra

capacitors or if the paper aor considered to be too burdedefinitive way to determine ware causing a problem is throplant audit. This audit involveof the electrical system layouloads, as well as harmonic mat strategic locations. This daassembled and analyzed to oconcise understanding of theEaton provides an engineerin

conduct these on-site plant au

SA02607001E


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SA02607001E

How Can Harmonics ProblemsBe Eliminated?

When power factor correction is required in thepresence of non-linear loads, or the amount ofharmonic distortion must be reduced to solvepower quality problems or avoid penalties, themost reliable, lowest cost solution is often realizedwith the use of harmonic filters.

What is a PassiveHarmonic Filter?A shunt harmonic filter (see Figure 20) is,essentially, a power factor correction capacitorcombined with a series iron core reactor. A filterprovides power factor correction at the fundamentalfrequency and becomes an inductance (like a motor)at frequencies higher than its turning point.Cutler-Hammer harmonic filters are almost alwaystuned below the 5th harmonic. Therefore, the filterprovides an inductive impedance path to thosecurrents at harmonic frequencies created by nearlyall three-phase non-linear loads (5th, 7th, 11th,13th, etc.). Since the filter is not capacitive at thesefrequencies, the plant electrical system can nolonger resonate at these frequencies and cannotmagnify the harmonic voltages and currents.A shunt harmonic filter therefore accomplishesthree things:

1. Provides power factor correction.

2. Prevents harmonic overvoltages due toparallel resonance.

3. Reduces voltage harmonic distortion andtransformer harmonic loading at frequenciesabove its turning point.

In some circumstances, a harmonic resonancecondition may accrue gradually over time ascapacitors and non-linear loads are installed ina plant. In those instances, replacement of suchcapacitors with harmonic filters is in order tocorrect the problem.

Figure 20. Shunt Harmonic Filter

Phase

A

B

C

Reacto

CapacBank


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Do I Need to Perform aSystem Analysis to Correctly

Apply Harmonic Filters?The proper application of harmonic filters can begreatly simplified by recognizing that there are onlya few pieces of pertinent information and analysissteps which need to be taken for most systems inorder to correctly deal with the problem.

Eatons approach to power system harmonicdiagnosis and solutions has evolved, via years ofaccumulated knowledge and experience, to anapproach which eliminates needless analytical

complexity in the majority of circumstances. Rather,it relies on the relatively few pieces of informationwhich are required to make the correct applicationsdecision. If this information indicates that sometype of metering and measurement is required, thenonly those select measurements which yield usefulinformation will be performed, keeping the com-plexity and cost to a minimum without sacrificingsolution correctness.

Our abilities in the area of harmonic analysis, how-ever, extend to the state-of-the-art in computerized

analysis tools should the customer require thethoroughness afforded by such tools.

One of the most basic and useful pieces of informa-tion which mustbe known before attempting to cor-rect power factor in the presence of non-linear loadsis the ratio of the total non-linear kVA to the servicetransformer kVA rating. This ratio alone can oftenbe used to determine whether harmonic filters arenecessary to correct power factor or whether plaincapacitors can be added without experiencing

problems as follows:1. If the plants total three-phase non-linear load

(in kVA, 1 hp = 1 kVA) is more than 25% of themain transformer capacity, harmonic filterswill almost always be required for powerfactor correction.

2. If the plants total three-phase non-linearload is less than 15% of the main transformercapacity, capacitors can usually be appliedwithout problems.

3. If the plants total non-linear load is between15 and 25%, other factors should be considered.

Starting with this most basic information, yourEaton sales representative will work with you todetermine what additional information or measure-ment, if any, is required in order to recommend thecorrect solution to your problem.

What is Eatons Exin Harmonic Filteri

Eaton has been in the power capa70 years, manufacturing power facapacitors used in low voltage anapplications. In the 1980s, we bega number of customers to help rerelated to harmonic distortion. Duperiod, we developed a number oincorporate harmonic filtering in automatic capacitor banks. The suinstallations has made Eaton the of harmonic filters.

With a fully integrated manufactuwe maintain the strictest quality cthe industry. All power capacitorsat various stages in their manufacshipment to ensure a long serviceEaton provides on-site supervisioassistance to ensure that all capacfiltering assemblies are properly aplant environment.

Please call 1-800-809-2772 for assstanding and solving any probleminvolving capacitors or harmonics


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Eaton Corporation is a diversified industrial manufacturer ranked amongthe largest Fortune 500 companies. The electrical group is Eatons largestdivision and is a global leader in electrical control, power distribution,power quality, automation, and monitoring products and services.Eatons electrical products include brands such as Cutler-Hammer,MGE Office Protection Systems, Powerware, Holec and MEM.Eaton provides PowerChain Management solutions to serve the needsof the industrial, institutional, IT, data center, mission critical, utility,residential and OEM markets worldwide.

PowerChain Management solutions help enterprises achieve sustainableand competitive advantages through proactive management of thepower system as a strategic, integrated asset throughout its life cycle.With Eatons distribution, generation and power quality equipment;full-scale engineering services; and information management systems,

the power system is positioned to deliver powerful results: greaterreliability, operating cost efficiencies, effective use of capital, enhancedsafety and risk mitigation.

Eaton Corporation

Electrical Group

1000 Cherrington Parkway

Moon Township, PA 15108

United States

877-ETN-CARE (877-386-2273)

Eaton.com

2008 Eaton CorporationAll Rights ReservedPrinted in USASA02607001E / Z6962April 2008

Eaton, Cutler-Hammer, Powerware,PowerChain Management, Holecand MEM are trademarks of EatonCorporation. All other trademarks are

the property of their respective owners.

sa02607001e eng guide pwr fact correciton

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