MYRON ZUCKERCALMANUAL

POWER FACTOR CORRECTIONAPPLICATION GUIDE

INC

.

®

PAGE - 2

HOW TO APPLY CAPACITORS TOLOW VOLTAGE POWER SYSTEMS.

SECTION INDEX SECTION I POWER FACTOR UNDERSTANDING POWER FACTOR........................ PG 3 IMPROVING POWER FACTOR...................................PG 4

SECTION II ADVANTAGES OF MAINTAINING A HIGH POWER FACTOR ELIMINATION OF PENALTY DOLLARS......................PG 5 ADDITIONAL CAPACITY IN ELECTRICAL SYSTEM ......................................................................PG 5 REDUCTION OF I2R LOSSES..................................... PG 5

SECTION III HOW TO DETERMINE AMOUNT OF KVAR REQUIRED ANALYSIS OF UTILITY BILLS.....................................PG 6-7

SECTION IV LOCATION OF REQUIRED CAPACITORS

SECTION V HARMONIC DISTORTION PROBLEMS ......................................................................................PG 11

SECTION VI ENGINEERING DATA DEFINITIONS ..............................................................PG 12

BASIC RELATIONS .....................................................PG 12

315 East Parent St.Royal Oak, Michigan 48067Tel. (248) 543-2277 (800) 245-0583 Fax (248) 543-1529www.myronzuckerinc.com

CALMANUAL

METHOD #1 CAPACITOR AT LOAD (CALMOUNT ® brand capacitor).................. PG 8-9 METHOD #2 FIXED CAPACITOR BANK (CAPACIBANK ® brand capacitor)................PG 10 METHOD #3 AUTOMATIC CAPACITOR BANK (AUTOCAPACIBANK ™ brand capacitor)...PG 10 METHOD #4 COMBINATION OF METHODS .............PG 10

INC

.

®

MYRON ZUCKER

PAGE - 3

POWER FACTOR

SECTION IUNDERSTANDING POWER FACTOR

In most modern electrical distribution systems, the predominantloads are resistive and inductive. Resistive loads are incandescentlighting and resistance heating. Inductive loads are A.C. Motors,induction furnaces, transformers and ballast-type lighting. Inductiveloads require two kinds of power: (1) active (or working) power toperform the work (motion) and (2) reactive power to create andmaintain electro-magnetic fields. The vector sum of the activepower and reactive power make up the total (or apparent) powerused. This is the power generated by the utility for the user toperform a given amount of work.

* Active power is measured in KW (1000 Watts) * Reactive power is measured in KVAR (1000 Volt-Amperes

Reactive) * Total Power is measured in KVA (1000 Volts-Amperes)

Power factor then is the ratio of active power to total power. Wecan illustrate these relationships by means of a right triangle.(See Figure 1.)

KVA or KVA-HRS(Total Power )

PF = KW = COS O KVA

O

Figure 1:

KVAR orKVAR-HRS( Reactive Power )

KW or KW-HRS(Active Power)

Note that a low power factor requires a larger amount of KVAto accomplish a fixed amount of work (KW), whereas a high powerfactor would require a lesser amount of KVA to accomplish thesame amount of work. Utilities provide the KVA to the user, andby means of continuous metering, they bill the user each month,and provide actual values of the components of power shown inFigure 1. If the values shown on the bill indicate a low powerfactor, many utilities will add a penalty to the bill. In like manner, ahigh power factor may result in a reduction in the over-all cost oftotal power consumed.

PAGE - 4

IMPROVING POWER FACTOR

The solution is to add power factor correction capacitors tothe plant power distribution system. They act as reactive powergenerators, and provide the needed reactive power to accomplishKW of work. This then reduces the amount of reactive power, andthus total power, generated by the utility. Let’s look at an actualcase of power factor improvement to an industrial plant, and thesavings that resulted. (See Figure 2)

Figure 2:

Because the utility applied a penalty formula when the powerfactor fell below 85%, this user had a penalty of $650.00 addedto the bill. To accomplish 812KW of work the 1500KVAtransformer was almost 78% loaded. (1160÷1500 = 77.3%)

The solution in this case was to add capacitors to the systemby installing them at each of thirteen large motors. The totalKVAR added was 410. This improved the power factor to 89%,and reduced the required KVA to 913, which is the vector sumof KW and KVAR. (See Figure 3)

The user, doing the same amount of work, but now withcapacitors installed, has eliminated the $650.00 monthly penalty.This would be an annual savings of $7,800.00. The capacitorsand the labor to install them cost $7,351.00, a payback of lessthen 12 months.

The utility has to generate 247 less KVA (1160 - 913 = 247),and the user has the 1500 KVA transformer now loaded onlyto 60% of capacity. This will allow the addition of more loadin the future to be supplied by the transformer.

COS O = .70

KVAR = 828KVA = 1160

KVAR =828 - 410 = 418

Figure 3:

COS O = .89

KW = 812

KW = 812

KVA = 913

PF = 812

x 100 = 70%

PF = 812 x 100 = 89%

1160

913

PAGE - 5

ELIMINATION OF PENALTY DOLLARS

A high power factor eliminates penalty dollars imposed whenoperating with a low power factor. For many years, most utilitiesdemanded a minimum of 85% power factor as an average foreach monthly billing. Now many of these same utilities aredemanding 95%...or else pay a penalty!

The actual wording or formula in the utility rate contractmight spell out the required power factor, or it might refer to KVAbilling, or it might refer to KW demand billing with power factoradjustment multipliers. Have your utility representative explain theparticular rate contract used in your monthly bill. This will insureyou are taking the proper steps to obtain maximum dollar savingsby maintaining a proper power factor.

ADDITIONAL CAPACITY IN ELECTRICAL SYSTEM

A high power factor can help you utilize the full capacity ofyour electrical system. To refresh our memory, let’s look again atthe power triangle story, shown on Pages 3 & 4, Figures 1, 2 ,and 3. Remember that KVA is a measure of the total powergenerated by the utility for you to accomplish your KW of work.Remember that the KVA figure is the amount of power passingthrough your plant transformer, and limited by its rated size: e.g.750 KVA, 1500 KVA, 2500 KVA, etc. In the previous example,we reduced your transformer loading from 1160 to 913 KVA, thusallowing for more load to be added in the future.

THE ADVANTAGESOF MAINTAININGA HIGH POWERFACTOR

A potential savings in billed KW-Hrs can be realized dependingupon where the capacitors are located in your electrical system.When capacitors are energized they reduce the total power usage(KVA) from their location in the system up to the utility source.In other words, capacitors reduce the current in amperes that hadbeen flowing from the utility to the capacitor location. This amperereduction might be as high as 20%. Since watt loss generated bycurrent passing through a conductor is expressed by the formula ...watt loss = (Ampere) 2 x Conductor Resistance (W=I2R)... it isobvious that locating the capacitors at the extremities of thefeeders and branch circuits (where the loads are) can result ina sizeable reduction in total KW-Hrs usage every month.

REDUCTION OF I 2R LOSSES

SECTION II

PAGE - 6

SECTION III

HOW TODETERMINEAMOUNT OFKVAR REQUIRED

ANALYSIS OF UTILITY BILLS

Monthly utility bills should be studied and analyzed todetermine this requirement. Since loads vary from month tomonth, or season to season, it is well to cover the last twelvemonths of bills. Almost all utilities print out the average powerfactor for the month, and the total KW-Hrs consumed duringthat billing period. If this period happens to cover 30 days,then we have 30 x 24, or 720 hours. Divide the billed KW-Hrsby 720 and you will obtain the average KW for the billing period.

With this information, we can once again draw our powertriangles to determine how much KVAR would be required toimprove the power factor to some new desired level. Or, wecan proceed to use Table 1 which simplifies the calculations.For example see below.

INSTRUCTIONS FOR USING TABLE 1:

1. Find the billing (original) power factor in column (1).2. Read across for desired power factor.3. Multiply number shown by average KW obtained above.

EXAMPLE:

The utility bill shows an average power factor of .72 with anaverage KW of 627. How much KVAR is required to improvethe power factor to .95 ?

STEPS:

1. Locate .72 (original power factor) in column (1).2. Read across desired power factor to .95 column. We find .635 multiplier3. Multiply 627 (average KW) by .635 = 398 KVAR.4. Install 400 KVAR to improve power factor to 95%.

Now that we have determined that capacitors totaling 400KVAR must be installed, we must decide where to locate them.

PAGE - 7

Desired Power FactorOrig-inalPowerFactor

TABLE 1: MULTIPLIERS TO DETERMINE CAPACITOR KVAR REQUIRED FOR POWER FACTOR CORRECTION

0.60

PAGE - 8

SECTION IV

LOCATION OFREQUIREDCAPACITORS

HERE ARE 4 METHODS USED IN LOCATING CAPACITORSWITHIN AN ELECTRICAL SYSTEM.

Method #1: CAPACITOR AT LOAD (CALMOUNT ® brand capacitor)

Install a single capacitor at each sizeable motor and energizeit whenever the motor is in operation. We refer to this asCalmount ® brand capacitor (Capacitor At Load). Tables 2 and3 show suggested KVAR ratings to be selected.

This method usually offers the greatest advantages of all,and the capacitors could be connected either in location (A) or(B) in Figure 4 below:

MOTOR

SWorCB

overload relays

CapacitorA

Starter

Figure 4:

CapacitorB

Location A - Normally used for most motor applications.

Location B - Used when motors are jogged, plugged, reversed;for multi-speed motors, or reduced-voltage start motors.

The advantages of method #1 are many:

(A) Corrects PF, unloads the transformer, reduces losses in conductors (KW-Hrs) from source to motor location.

(B) Voltage drop to motor is reduced - thus optimizing motor performance.

(C) Installation simple - no new switches or circuit breakers required.

PAGE - 9

*For use with 3-phase, 60 hertz NEMA Classification B Motors to raise full load power factor to approximately 95%

TABLE 2: SUGGESTED MAXIMUM CAPACITOR RATINGS USED FOR HIGHEFFICIENCY MOTORS AND OLDER DESIGN (PRE "T-FRAMES") MOTORS*

*For use with 3-phase, 60 hertz NEMA Classification B Motors to raise full load power factor to approximately 95%

TABLE 3: SUGGESTED MAXIMUM CAPACITOR RATINGS "T-FRAME" NEMA "DESIGN B" MOTORS*

PAGE - 10

Method #2: FIXED CAPACITOR BANK (CAPACIBANK ® brand capacitor)

Install a fixed quantity of KVAR electrically connected at one ormore locations in the plant’s electrical distribution system, andenergized at all times.

This method is often used when the facility has few motors ofany sizeable horsepower to which capacitors can economicallybe added. A fixed amount of KVAR can easily be added to anexisting run of plug-in bus by installing a Busmount™ brandcapacitor. A fixed amount can be added to the main buses in amotor control center. In most cases, however, the fixed bank(Capacibank™ brand capacitor) is usually located near theservice entrance switchboard. In all cases, a separate fusedswitch, or circuit breaker, must be provided ahead of thecapacitor bank.

There is one most important fact to remember whenever youinstall a fixed bank. When the system is lightly loaded (perhapson Sundays or holidays), and you have too large a bank of KVARenergized, the voltage can be so great that motors, lamps, andcontrols can burn out. Unbalanced load or other similar conditionscan aggravate the trouble with harmonics. Our research indicatesthat KVAR equal to 20% of the transformer KVA is the maximumsize of a fixed KVAR bank that should be installed. Values largerthen this can result in a large resonant current, potentially harmfulto the system.

Remember, that while the fixed bank can unload thetransformer, and show an improved power factor on your monthlybill, it does nothing to reduce the conductor watt loss (and thusbilled KW-Hrs).

Method #3: AUTOMATIC CAPACITOR BANK (AUTOCAPACIBANK ™ brand capacitor)

Install an automatically controlled capacitor bank(Autocapacibank™ brand capacitor) that will closely maintain apre-selected value of power factor. This is accomplished by havinga controller switch steps of KVAR on, or off, as needed. This typeof bank eliminates the concern of having too much KVARenergized at light load periods.

This method would seem to have much appeal, but it also hasa real disadvantage. Since it is usually located near the incomingservice entrance switchboard, we find that like the fixed bank thisautomatic bank does nothing to reduce the conductor losses (andthus billed KW-Hrs). Remember that the reduction in conductorlosses using Calmount ® brand capacitor (method #1) canbe sizeable.

Method #4: COMBINATION OF METHODS

Since no two electrical distribution systems are identical, eachmust be carefully analyzed to arrive at the most cost-effectivesolution, using one or more of the methods.

PAGE - 11

SECTION V

Starting in the late 1970's commercial, institutional, andindustrial plants have experienced a tremendous growth in theuse of equipment that can generate “harmonic” distortion in powersystems. Some examples of such equipment will include DCdrives, AC variable frequency drives, rectifiers, induction furnaces,and UPS systems. This harmonic distortion develops a currentwave shape which results in higher than normal RMSamperes (and heat) which will result in nuisance fuse-blowing,circuit-breaker tripping, over-heated transformers, and prematurecapacitor failure.

If a facility has but a few pieces of the above-mentionedequipment in use, Myron Zucker, Inc. can pinpoint the harmonicnumber and amplitude present in the system. If the facility is alarge one with many sources of harmonic distortion, then acomplete audit of the total electrical system with a harmonicanalysis must be made. We can provide such services.

The solution to all of the above is the installation of harmonicfilters that not only correct or improve the power factor, butalso prevent harmonics from damaging existing equipment online. We have developed the Caltrap™ brand harmonic filters forapplication to the actual harmonic source equipment. We alsohave the larger Capacitrap™ brand harmonic filters (large filterbanks) to provide overall system correction when many types ofharmonic-producing equipment exist. This has become quite aspecialized field, and we consider ourselves as leaders in low-voltage filter application. We are ready to help you eliminate your“Dirty Power” problems! Let us furnish you our new applicationguide for solving harmonic distortion problems.

HARMONICDISTORTIONPROBLEMS

* RATED CURRENT BASED ON OPERATION AT RATED VOLTAGE, FREQUENCY, AND KVAR† CONSULT NATIONAL ELECTRICAL CODE FOR OTHER WIRE TYPES. ABOVE SIZE BASED ON 350C AMBIENT OPERATION. (REFER TO NEC TABLE 310-16.)NOTE: FUSES FURNISHED WITHIN CAPACITOR ASSEMBLY MAY BE RATED AT HIGHER VALUE THAN SHOWN IN THIS TABLE. THE TABLE IS CORRECT FOR FIELD INSTALLATIONSAND REFLECTS THE MANUFACTURER'S SUGGESTED RATING FOR OVERCURRENT PROTECTION AND DISCONNECT MEANS IN COMPLIANCE WITH THE NATIONAL ELECTRICAL CODE.

FOR 3-Phase 60 Hz CAPACITORS(These wire sizes are based on 135% of rated current in accordance with the 1999 National Electrical Code. Article 460)

RECOMMENDED WIRE SIZES, SWITCHES AND FUSES

SECTION VI CAPACITOR DEFINITION & APPLICATION DATA DEFINITIONS BASIC RELATIONS

C: Capacitance (farads)

KW: Kilowatts, measure of active power

KVA: Kilovolt-amperes, measure

of apparent power

KVAR: Kilovolt-amperes reactive

µF: Microfarads, measure of capacitance

(farads x 10 - 6 )

f: Frequency of voltage or current in Hz

Ic: Capacitor current in amperes

W: Dissipated power, in watts

V: Voltage (Volts)

I or A: Current (Amperes)

R: Resistance (ohms)

1 2.4 14 5 30 1.2 14 3 30 1 14 3 30 11.5 3.6 14 6 30 1.8 14 3 30 1.4 14 3 30 1.52 4.8 14 10 30 2.4 14 5 30 1.9 14 3 30 2

2.5 6 14 10 30 3.0 14 6 30 2.4 14 5 30 2.53 7.2 14 I5 30 3.6 14 6 30 2.9 14 5 30 34 9.6 12 20 30 4.8 14 10 30 3.8 14 6 30 45 12 12 20 30 6 14 10 30 4.8 14 10 30 56 14 10 25 30 7.2 14 15 30 5.8 14 10 30 6

7.5 18 10 30 30 9 14 15 30 7.2 14 15 30 7.510 24 8 40 60 12 12 20 30 9.6 12 20 30 10

12.5 30 8 50 60 15 10 25 30 12 12 20 30 12.515 36 6 60 60 18 10 30 30 14 10 25 30 15

17.5 42 6 70 100 21 8 35 60 16 10 30 30 17.520 48 4 80 100 24 8 40 60 19 8 35 60 20

22.5 54 4 90 100 27 8 50 60 22 8 35 60 22.525 60 2 100 100 30 8 50 60 24 8 40 60 25

27.5 66 2 125 200 33 6 60 60 26 8 45 60 27.530 72 2 125 200 36 6 60 60 29 8 50 60 30

32.5 78 1/0 150 200 39 6 65 100 31 8 50 60 32.535 84 1/0 150 200 42 6 70 100 34 6 60 60 35

37.5 90 1/0 150 200 45 6 75 100 36 6 60 60 37.540 96 2/0 175 200 48 4 80 100 38 6 65 100 40

42.5 102 2/0 175 200 51 4 90 100 41 6 70 100 42.545 108 3/0 200 200 54 4 90 100 43 6 75 100 4550 120 3/0 200 200 60 2 100 100 48 4 80 100 50

52.5 126 3/0 200 200 63 2 110 200 50 4 80 100 52.555 132 4/0 250 400 66 2 125 200 53 4 90 100 5560 144 4/0 250 400 72 2 125 200 58 2 100 100 6065 156 4/0 250 400 78 1/0 150 200 62 2 110 200 6570 168 300M 300 400 84 1/0 150 200 67 2 125 200 7075 180 300M 300 400 90 1/0 150 200 72 2 125 200 7580 192 350M 350 400 96 2/0 175 200 77 1/0 150 200 8090 216 500M 400 400 108 3/0 200 200 86 1/0 150 200 90

100 240 500M 400 400 120 3/0 200 200 96 2/0 175 200 100125 300 (2)4/0 500 600 150 4/0 250 400 120 3/0 200 200 125150 360 (2)300M 600 600 180 300M 300 400 144 4/0 250 400 150200 480 (2)500M 800 800 240 500M 400 400 192 350M 350 400 200225 540 (3)300M 900 1200 270 (2)4/0 500 600 216 500M 400 400 225250 600 (3)350M 1000 1200 300 (2)4/0 500 600 240 500M 400 400 250300 720 (3)500M 1200 1200 360 (2)300M 600 600 288 (2)4/0 500 600 300350 420 (2)350M 700 800 336 (2)300M 600 600 350400 480 (2)500M 800 800 384 (2)350M 700 800 400450 540 (3)300M 900 1200 432 (2)400M 750 800 450500 600 (3)350M 1000 1200 480 (2)500M 800 800 500550 660 (3)500M 1100 1200 528 (3)300M 900 1200 550600 720 (3)500M 1200 1200 576 (3)350M 1000 1200 600

600 VOLTS240 VOLTS 480 VOLTS

Wire Size Wire Size Wire Size 9O°C-Type C.B. 90°C-Type C.B. 90°C-Type C.B.

THHN or THHN or THHN orCurrent* XHHW* Fuse Switch Current* XHHW* Fuse Switch Current* XHHW* Fuse Switch

KVAR (Amps) or Equiv.† (Amps) (Amps) (Amps) or Equiv.† (Amps) (Amps) (Amps) or Equiv.† (Amps) (Amps) KVAR

208V 240V 480V 600V 2.78 2.41 1.20 0.96

Applied Voltage Amps / KVAR

KW PF = KVA = cos O

KVAR = KVA 2—KW 2

KVAR x lO 3

C in µF = (2 π f) x (KV) 2

W = I2 R

KVA = 3 x V x A (3-phase) 10 3

IC = (3-phase)KVAR x 10 3

3 x V

PAGE - 12