Capacitors,%20Line%20or%20Load%20Side%20of%20Motor%20Disconnect?%20(01/25/2K)
  Question:

Where do you place a capacitor (line side, load side of motor disconnect, short-circuit protection device, overload) for a 400-hp motor 3-phase power factor correction capacitor? Also, why do capacitors lower the current?

Steve Dodd

Response No. 1:

In regard to the PF correction question, place the capacitors on the LOAD side of the motor starter, as near as possible to the motor(s). Give them overcurrent protection, also [fuses]. Also, place a "draining resistor" in parallel with them, with a value of something like 22K. This will assure that there is no stored charge left in them after the circuit is opened. The motor will drain them, but if the motor is disconnected from the circuit and the circuit is energized, this will charge the capacitors and leave them charged after the circuit is re-opened.

The capacitors will offer Capacitive Reactance [Xc] to the circuit, while the typical loaded induction motor will offer Inductive Reactance [Xl] to the circuit. The [Xl] causes a lagging power factor and the CURRENT will lag behind the voltage. [Xc] causes a leading power factor and the VOLTAGE will lag behind the current. Low [50%] lagging power factor will cause the circuit [and load] to draw twice the current, which is the total kVA of the load. The kVA equals the TRUE POWER in watts, plus the REACTIVE POWER, in Volt-Amps Reactive [or kVAr's]. The kVA is called the APPARENT POWER. An example circuit below:

1-hp motor Ė 100-volt AC Squirrel Cage - nominal power factor [50% PF]. When motor is developing one full horsepower at the shaft, the branch circuit will be delivering 1,492 volt-amps [1.49kVA], which is 14.92-amps at 100-volts. Half of this kVA load is true power - in watts, which is 746-watts. The other half is reactive power - in VARs, which is 746-VAR. The total is 1,492kVA. By adding capacitors [Xc] in with the capacitance - in microfarads Ė it will equal somewhere near 746-VARs, the circuit's total kVA and power factor changes. At the transformer, up to where the capacitors are connected, the power factor is near unity [close to 100%]. This causes the total current on the branch circuit to be almost 7.46-amps. The reason why the capacitors will cause a lower current flow is because they "tune" the circuit and correct the power factor. Therefore, there is a lower current flowing because the overall change results in a lower kVA load on the circuit.

I tried to keep the answers concise and much shorter than the typical reply I give, so I sure hope this makes some sense plus explains what you were asking. It's a simple thing to understand once you grasp the basic concept behind power factor [which is not something that can be explained on the back of a box of Cocoa Puffs.]. So, once again - here is my favorite slogan: "I suggest that you get at least two good books that cover basic AC theory" in order to fully grasp the reasons why this phenomena occurs. The books will also answer other pending questions that will come up.

Scott E. Thompson, adst@SoCA.com

Response No. 2:

Please review this and pass it on to Steve. I am not a college professor when it comes to power factor correction but I believe I can answer Steve's question in a way that does not use the heavy math. Please also forward other answers you have received to me so I can share that with my students.

First, place the capacitor on the line side if you wish to extend the life of the disconnect contacts. If, however, you desire better "power factor correction," connect capacitors to the load side of the disconnecting means as close to the motor as possible. Next, capacitors do not lower the current; they shift the current or allow it to lead to compensate for the lagging current caused by any inductive load. They do raise the current.

I teach my students as was taught to me in Tech School. Just remember the phrase "ELI the ICE man. "E = Electromotive force or voltage measured in Volts. I = Current or induced current measured in Amps. The L stands for inductance, which occurs with coils or motors and is measured in Henrys. The C stands for capacitance, which occurs with capacitors and is measured in Farads. The ELI means that in an inductive circuit the voltage leads the current (amps.) The ICE means that in a capacitive circuit the current (amps) leads the voltage.

Think of it in terms of DC instead of AC. When you hook a battery up to a coil the current does not flow at the moment of connection but must build up the flux or magnetic field and as the flux field builds, the current grows to maximum. The voltage, on the other hand, is applied directly across the coil and battery voltage is measured on the coil from the moment of connection.

On the other hand, when you hook a battery up to a capacitor the current flows immediately but the voltage requires time to build to the maximum (battery) level. The current flows full strength inhibited only by the resistance of the source, wire and reactance of the capacitor. While the voltage may be applied, it does not appear across the capacitor until full capacitor charge occurs.

In a pure capacitive -- non-resistive or inductive -- AC circuit, the phase shift from the leading current to the lagging voltage is equal to 90-degrees. While in a pure inductive -- non-resistive or capacitive -- AC circuit, the phase shift from the leading voltage to the lagging current is equal to 90-degrees. If you need to see the oscilloscope graph for that, I could show it on an MS word document but I recommend you look in any basic electronics or industrial electricity book.

Richard A. Melcher

Response No. 3:

Your E-mail of 11/1 regarding sizing of a capacitor to a 400-hp 3-phase induction motor does not specify either the rated voltage or the rpm of the motor. Based on a typical installation (1800-rpm synchronous speed/probably 1750 full-load speed), I would suggest a 60kVAr capacitor would provide correction to approximately 95% and a reduction of about 6% in line current.

If the capacitor is installed after the OL relays, the short circuit protection for the motor provides protection also for the capacitor conductors and no additional disconnect device is needed. Conductor size can be calculated but is also available from charts: based on a 480-volt installation, 60kVAr draws about 72-amps, will require a #2 conductor, probably THWN [don't forget to apply 110-14(c)(1)]. This is not only the most economical method but also provides the best P.F. correction without the problem of voltage rise when the motor is not running.

If you want to install the capacitor ahead of the OL's (and possibly ahead of the motor short circuit protection if a combination starter is used), a separate disconnect must be used to provide this protection. Again, based on the above, a 200-amp device with a 125-amp fuse is suggested. This does not provide the best protection and can lead to serious overvoltages during curtailed plant operations (weekends).

Article 460 of the NEC contains all of the relevant details and the handbook (fig 460.1) gives a picture of both of these installations. The OL's must be reduced in value since the line current is reduced (only when the cap is on the load side of the OL's). Article 460-9 discusses this issue.

Regarding reduction in line current: induction motors are essentially inductors and, as such, have a lagging power factor (the current in the circuit lags the voltage) and therefore, the input is measured in volt-amperes, not watts. For an example, using nameplate data, convert the horsepower to watts (746 w/hp) and compare to the current times the rated voltage. In an AC circuit, volts times amps are not watts as in a DC circuit. The motor requires lagging vars to create the magnetic energy in the motor. The capacitor has the opposite electrical effect; the current leads in the circuit and provides leading vars. Properly selected values of capacitance (a capacitor) will, in effect, offset the lagging requirements of the motor by providing leading vars. It is important to size the capacitor on the basis of the motor's rpm since when the motor is turned off, the inertia of the load causes the motor to become an induction generator for a few seconds (capacitor providing the excitation) and the induced voltage only has the impedance of capacitor to discharge in (bad situation). There is also the possibility that should the motor be restarted while it is coasting (and functioning as a generator) the phase angle of the motor's voltage and the reconnected line voltage probably are not in phase and could cause serious mechanical and electrical damage to the motor.

Most capacitor manufacturers will provide all of this information and calculate the amount needed for the entire plant. DO NOT forget, if the electrical system contains nonlinear loads (VSD's, etc.) the electrical system will contain various harmonics, probably 3rd and 5th. The impedance of a capacitor inverts to the apply frequency; i.e., as the frequency goes up, the impedance (or better, reactance) goes down, meaning the current will increase often well beyond the rated value. Capacitor suppliers will evaluate your system for these harmonics.

Bill Ramage, CatsRus64@cs.com

Response No. 4:

In response to some of your teacher questions:

1.     I know of a couple of companies that sell electrical labs and support materials. The best one that I have seen for electricians (have not personally used them) is sold by Hampden Engineering Corp.-P.O. Box 563, East Longmeadow, MA. 01028ó800- 253-2133. They have several different models depending on the subject. They also have dissectible motor labs for single and 3-phase motors. They also have transformers, power supplies and other equipment (meters etc.).

2.     You might also try Mid West Tech--206 West 2nd Street Madrid, IA. 50156ó800- 358-5702 www.midwestech.com They sell Lab-Volt lab equipment (which I have used). Lab-Volt's address is P.O. Box 686, Farmingdale, New Jersey. 07727---800- 522-8658. Not sure if you can buy direct or not!

Terry Crowe, tlcrowe@dmacc.cc.ia.us

Response No. 5:

In response to Steve Dodd's questions about capacitors: Article 460 states that capacitors may be installed on either the line or load side of the disconnecting means provided that the capacitor(s) can also be disconnected from the energy source for maintenance, etc. I have installed them at the motor side of the motor starter so that they would be out of the circuit whenever the motor is not running. The whole idea is to provide phase correction for poor power factor situations and when the motor is off, there is no need for that. Also, to leave them energized would cause the power factor to be over-corrected and most Power Companies don't want that either. I don't know if the contacts last longer or shorter on the starter because of inrush current to the capacitor. In theory it should help!

Another way is to disconnect the capacitor(s) with a contactor (relay) at the same time the motor is turned off. No overload protection is needed for just the capacitor but overcurrent protection is required along with some kind of discharge device as capacitors can (and often do) store electricity. This can be quite lethal. Always check for AC & DC voltage before working on them! NEC requires an automatic discharge within one minute. Never assume that it is working! Do NOT short the terminals with a screwdriver or similar device. Use a resister or maybe a light bulb if voltage isn't too high. It is typical to use a contactor with a capacitor bank (several in parallels) to correct several motors at the same time. Works especially well when they all start/stop at the same time (a lot easier to install).

The reason the circuit uses less amps with the capacitor in the circuit resides in the phase shift between voltage and current in the circuit. An inductive circuit causes a 90-degree shift with voltage leading current. The net effect makes the circuit appear to use more energy than what the load requires (we call this apparent power). A capacitive circuit does a phase shift in the opposite direction (current leads voltage). If the right balance of capacitance is added to an inductive load (motor) then the phase shifts cancel and the current and voltage peak and zero at the same time (we call this true power). This causes the supply to only see the load and maybe a little circuit resistance (wiring, contacts, terminations, etc.). Net result reduces the amount of current needed and allows the generating part of the system to run cooler. If too much capacitance is in the circuit (not disconnected with motor, wrong value, etc.), then the circuit will appear to use more power again.

Terry L. Crowe, tlcrowe@dmacc.cc.ia.us

Response No. 6:

Article 460-8(b) and (c) exceptions allow connection to the load side of the motor controller or disconnect. That's where I would put it; otherwise you will need a disconnecting means and separate overcurrent protection for the capacitor, which could lead to other problems. The reason they lower RMS current is that they increase RMS voltage. The motor and capacitor are basically in a dance together. The capacitor stores electrons on the upward travel of the AC sine wave. When the sine wave falls is when the capacitor discharges and puts voltage into the motor. The motorsí part of the dance is that it "borrows" energy to build the magnetic field on the upward travel and puts it back into the circuit as the sine wave falls (inductance), just like a transformer does when its field collapses. This is known as reactance in a circuit. This portion of borrowed energy that does not get consumed and the amount of energy (watts) that's consumed makes up your power factor. Power Companies know that industrial users have a lot of inductive loads and charge a power factor penalty because they have to have big enough wires to supply the demand. The capacitor closes the distance between the borrowed and consumed loads.
If you need more, let me know.

Rick Hart, rick.hart@med.va.gov

Response No. 7:

Without looking it up, I would say on the line side (to provide a means of isolating the motor from the capacitor for motor servicing). Capacitors would lower the current cause after they discharge their storage and they must recharge for the next cycle, thus creating a drop in current.

DunnElectr@aol.com

Response No. 8:

In answer to the question asked by Steve Dodd, you would place the capacitor on the load side of the motor. Capacitors lower the current with magnetizing. This magnetized current will result in considerable savings in the amount of current used.

Claridge Wilson

Response No. 9:

Electrical Motor Repair by Robert Rosenberg and August Hand (Third Edition) is full of these types of motors and others.

Dennis Carlson, carlsond@ticus.com

Response No. 10:

The phase angle of the capacitor in an AC circuit is opposite to the phase angle of the inductive motor load in the AC circuit (when the correct size capacitor is selected for the motor). For example, in a parallel circuit, if there were 15-amps in the inductive branch and 15-amps through the capacitor branch they would be 180 out of phase with each other. As a result, the currents cancel each other out. If plotted as vectors, they will cancel each other out creating a resonance circuit. (Resonance Circuit? when the line currents and line voltages are in phase).

Placement of the capacitor should be in parallel with the 400-hp motor. If at all possible, place the capacitor as close to the motor as possible to avoid other electrical issues. See Article 460 for Code compliance.

llarue@nwtc.tec.wi.us

Response No. 11:

Capacitors for power factor correction of individual motors should be placed on the load side of the motor controller. Once the motor is online, the effects of the capacitor would be in place. When the motor is off-line, the capacitor would also be disconnected. The overloads may need to be adjusted for the improved power factor and lower operating current of the motor. This also exempts you from having a separate disconnecting means for the capacitor bank. Persons working on this system should be aware of the possible stored energy contained in capacitors. Almost all correction capacitors that I have seen do have a method for discharging when switched off. No.2 capacitors will provide the stored electrical energy needed to magnetize the windings during its discharge cycle and allow the incoming line currents to provide the actual energy needed to produce the work required by the motor. The phase shift allowed by the insertion of a capacitor more closely aligns the current and voltage waveforms to allow a greater portion of input volt-amps to be used as work (watts).

Compare it to the timing of your cars' engine. If the timing is off enough, that spark is provided prior to proper fuel mixture delivery and we have low efficiency. If we can time the spark at the exact moment the mixture is provided to the cylinder, we have a greater combustion and complete use of our available resource. The same is true if the current and voltage waveforms are timed correctly, or corrected by use of capacitors. The closer the timing, the more improved use of our resource, in this case, volt--amp--input. Line currents won't have to be so high to account for the out-of-phase losses we often experience with large inductive loads, such as motors.

Phil Burke, pburke@nekatech.net

Response No. 12:

To answer the questions raised, I will briefly and simply describe the actions and reactions. The capacitor should be connected on the load side of the motor disconnect, which will activate the capacitor only when the motor is running. Capacitor protection, wiring and discharge requirements must be in compliance with NEC Article 460.

Motor current to the motor consists of two component (vectors) of currents added vectorially to form a "total" current to the motor which can be read by a current meter. One component (vector) current is "reactive" and for an induction motor it is "inductive.Ē The second component (vector) of this current is "real," the current which is useful current that produces torque, work and other motor requirements.

The capacitor connected in line with the motor produces "capacitive" reactance, which cancels part of the "inductive" reactance of the motor. Since the "total" current to the motor equals the sum of the "reactive" and "real" currents, the "total" current will be less due to "capacitive" reactance canceling part of the "inductive" reactance of the motor.

Alexander I.Orloff, http://members.aol.com/ocsoft/

Response No. 13:

"Where do you place a capacitor?" Common Wealth Sprague (CWS) Capacitors, Inc. provides a publication with a good description of where and how to install capacitors. You can call them at 413-664-4466 and ask for "Power Factor Correction - A guide for the Plant Engineer". In the meantime, here are some general guidelines. It would be best to call CWS with the exact system description before proceeding though.

If the motor has an across-the-line starter and will not be reversed or jogged and is not multi-speed, then the capacitor can be installed between the starter and the overloads. If the motor is reversed or jogged or is multi-speed, then the cap should be installed between the disconnecting means and the starter. If the motor starter is a Reduced Voltage Autotransformer (RVAT), you may also need an additional contactor to keep the capacitor off-line until the motor is up to full speed. The CWS publication explains why.

Is the 400-hp motor started with an adjustable speed drive or a "soft start"? If so, then someone should look at the harmonics generated by the drive and ensure that the harmonic frequencies don't match the system resonance frequency. There is a section in the guide that talks about this also. It is possible that an additional contactor might be needed to keep the capacitor off-line until the motor is started and/or a filter might be required for the harmonics.

In addition, the CWS guide gives the recommended sizing for capacitors so that the PF is not corrected beyond .95, which could result in damage to the motor.

"Why do capacitors lower current?" This is a hard topic to explain without all the mathematical analysis that goes along with it. There are many explanations out there and the CWS guide does a very good job and includes all the pictures and formulas. I hope this make sense.

When you increase the power factor of a motor load by installing power factor correction capacitors, you reduce the kVA requirements between the utility and the point at which the capacitor is connected to the system. The motor still requires the same kW and kVArís but now the capacitor is supplying the kVArís instead of the utility. The current between the motor and the capacitor hasn't changed. But, the current between the utility and the capacitor is reduced.

There are a lot of things to think about regarding PF capacitors. Hopefully, I've answered your immediate questions. If you need clarification or have any further questions don't hesitate to call or E-mail.

H. (Scott) Gibson, hsgibson@snopud.com

Response No. 14:

In my limited experience a kVAr bank generally is installed depending on what it serves. If more than one motor, then on the line side of each motor and with a disconnect to allow the kVAr bank to be isolated from the system. When serving just one motor it may be installed load side but again with a disconnect to allow for isolation of the kVAr bank. I have also seen kVAr banks installed so that they can be taken out of a system but allow the motors to continue to operate so a plant or system shutdown is not necessary. One big thing to remember when working on these systems is to be sure to bleed off the charge the capacitor has before doing any work on these systems.

Capacitors change the voltage-current relationship by changing the phasing of this relationship. If the capacitor is effective in correcting the power factor situation the system becomes more efficient and the current demand is lower because there is less waste (more efficient). The capacitor offsets/counteracts the inductive signature of the motor or system and helps make the complete system more efficient. I donít know if this helps or just confuses this more.

Brian Cox, brianc@trib.com

Response No. 15:

The basic answer to the question is, capacitors (sized for the specific motor characteristics) should be installed at the motor, where the kVAr is required. Installing capacitors will improve the power factor relationship, thus reducing the current flow to the motor. kVA is required to operate an induction motor, which includes the reactive requirements, which is not read by a standard amp meter. By improving the power factor to 95%, the total current flow is reduced through all equipment components between the serving transformer and the motor terminals (if the capacitors are installed at the terminals of the motor). Therefore, it is important to locate the capacitor at a point so when the motor is operated the capacitor is in operation with it. If the capacitor is connected to the motor circuit before the motor starter, then the capacitor will need to be turned on and off manually with the operation of the motor.

Capacitors lower the reactive current needed from the transformers to operate inductive loads. However, it is important to remember the capacitor only corrects that situation between the power source, transformer(s), and the capacitor. That's why itís important to locate the capacitance as close to the motor terminals as possible. Usually one will record the operating characteristics of the motor operation, then size and install the capacitors needed. The motor installation should be designed for the installation of capacitors even if they will be added later.

Another option is synchronous motors generate reactive power as they operate so they can actually help the entire plant power factor problem. Since I work at a Utility, power factor is usually part of the electric rate structure, so it is also important to operate the plant at the power factor level, which does not penalize the customer on the electric bill.

Ron Nelson, ronn@wrecc.com

Response No. 16:

The VARs that will be provided by the capacitor does not care what side of the line you place them; however, you need to make sure you protect the capacitor with short-circuit protection. The current is only lower on the upstream side of the capacitor because the VAR current component of the total current is now being provided by the capacitor and not the line. Try to place the capacitor as close to the load as possible to take advantage of line loss characteristics.

Karl Hauer, Karl.Hauer@austinenergy.com

Response No. 17:

We usually start motors at full-line voltage regardless of how large the motor is. I would think the capacitor would be needed as near as possible to the motor terminal because that is where you really need it. Capacitors at the motor terminal serve two objectives:

1.     It will try to maintain the voltage level during motor startup. Remember that it is charged before the motor is started.

2.     It supplies the motor with reactive power; that is, it will feed the motor with some of its magnetizing current. Thatís why the current supplied to the motor from the utility will be decreased accordingly.

Mohammad S. Qahtani

Response No. 18:

The capacitor shall be connected on the motor side of the overload relay and the relay shall be set for the reduced line current, which will be lower than the motor current at all loads. Also, care shall be taken while sizing the capacitor to prevent overvoltages due to self-excitation and transient torques. Since the magnetizing reactive current that is required by the motor is supplied by the capacitor, the line current drawn from the source reduces.

T.Prasada Reddy, preddy@polysindo.co.id

Response No. 19:

See NEC 460-9. Capacitors are most generally used on the load side of the overload relays. They affect the sizing of the overload relays because the relay sees less total "imaginary" current. Normal motor load is somewhat inductive. That means that there is some imaginary current and some "real" current. A capacitor also uses imaginary current, but it is of opposite polarity, so it can cancel out the imaginary current drawn by the motor. Overload relays must be sized for all the current, both the real current and the imaginary current. The addition of a capacitor cancels out some of the imaginary current, so the overload relay needs to be sized smaller so as to provide the same level of protection for the motor.

Vince Saporita, vsaporita@buss.com

Response No. 20:

The power factor capacitors must be connected ahead of the overloads in order for the overloads to function properly. They lower the current by bringing the power factor to unity, which eliminates excess current draw.

Paul E. Keller, Pek@tlc-engineers.com

Response No. 21:

It is beneficial to install capacitors associated with large motors in a manner that they turn on and off with the motor. This will make sure the capacitor is off when the motor is off and keep the capacitor from causing a problem with leading power factor when the motor is off.

A capacitor reduces amperage by offsetting the inductive VARs necessary to magnetize the motor. Example, if a motor with no capacitor correction has a 70% power factor, it uses 400kW, 400kVAr and 565kVA. The kW is the useful work, the kVAr is the energy needed to magnetize the windings, and kVA, apparent power, is the vector sum of these two. If a capacitor is installed that is about 270kVAr, it will provide that much of the magnetizing current. Leaving 400kW, 130kVAr (which is 400kVAr minus the 270kVAr of the capacitor) and 420kVA (the square root of 400kW squared plus 130kVAr squared), and the new power factor is 95%. The reduction in kVA is proportional to the reduction in amperage. So, in this example, the amperage is reduced by 26%. Note: the amperage is only reduced from the point at which the capacitor is installed, back towards the source. So, if you put the capacitor at the main service, only the main breaker and the electric utility see the benefit. If you put the capacitor at the motor, you will reduce the amperage in the feeder to the motor also.

Always have a person that is trained in sizing and installing capacitors look at the design for the installation. Variable frequency drives and other equipment may have problems when capacitors are installed.

John McComb, john_mccomb@fpl.com

Response No. 22:

You would place the capacitor on the load side of the motor disconnect, short-circuit protection device, overload.

Gary Gilbert, gilbertg@globalserve.ne

Response No. 23:

Most electrical loads are inductive in nature and have a lagging power factor (i.e., motors, light ballasts, etc.). This means their current draw is made up of two mathematical components: resistance and inductance. Most electricians know these terms as kWs and kVArs, the combination of the two being kVA. One must keep in mind that the two components are actually two-dimensional vectors and cannot be directly added to each other. Nor can the kVA of one load be directly added to the kVA of another without first breaking the value into its individual components.

But without getting into the math side of this issue, just understand that a motor load is comprised of two components: a resistance (due to the copper conductors in the windings) and an inductance (due to the magnetic fields set up in the motor which causes it to rotate in the first place) which, when combined, form what is known as impedance. Therefore, for any given supply voltage the motor is going to draw a current from the power distribution system that is comprised of both a resistive component and an inductive component.

To reduce the current the power distribution system must supply to the motor, we need to supply the motor requirements by other means. To reduce the resistive current the motor draws from the system, we could locate a real-power generator right at the motor to supply its load. However, this is generally expensive and impractical as the generator must be synchronized with the rest of the power distribution system, and it has on-going operational costs. Or, we could reduce the inductive component the motor draws from the power distribution system by locating a reactive-power generator at the motor to supply its inductive load. A capacitor just happens to be a type of reactive-power generator. Capacitors are compact and relatively inexpensive, and can be located just about anywhere. By connecting a capacitor near the motor terminals, the inductive requirements of the motor are supplied by the capacitor and the power distribution system only needs to supply the resistive component of the load. The net result is a reduction in the current flowing in the power distribution system, a lowering of the ratio of the real current to the inductive current flowing in the system (i.e., an increase in the overall power factor of the system), a reduction in the heat loss a power distribution system generates, a reduction in the HVAC required to cool the area, and a reduction in the size of the power distribution system required to supply the load. All good things!

However, one needs to be careful when applying capacitors for power factor correction or for any other purpose. Power distribution systems are not the simple things they used to be. There has been a proliferation of electronic loads, DC drives, and other components out there known as non-linear loads. Their complex impedances, when coupled with the capacitance of power factor correction capacitors can cause a system resonance that is both dangerous and destructive. The addition of power factor correction capacitors should be reviewed by a professional engineer competent in the evaluation of harmonic impedance studies to assure a resonance condition is not created while trying to solve a simple problem.

As for the location of the power factor correction capacitor, they should be located as electrically close to the load as possible (i.e., load-side of the starter). One should review the requirements of ANSI/NFPA.70, Article 460 for details. If the starter is located in a MCC, the capacitor should be mounted nearby and the associated protective equipment should be located in the MCC. This makes connection of the capacitor terminals to the load side of the starter simple. If the starter is self-contained and stands alone, locate the capacitor and associated protective equipment adjacent to the starter and add an extra set of load-side lugs in the starter for connection of the capacitor. In either case, the components of the combination starter will operate properly and protect the motor for both overload and overcurrent regardless of the status of the capacitor.

In all cases the motor feeder, branch circuit, and starter need to be sized for worst-case operation (i.e., assuming the capacitor is out of operation). The benefits of multiple power factor correction capacitors are reaped upstream in the reduction in size of the larger supply equipment where the loss of one capacitor does not significantly impact the system.

Marc Goslow

  Go to top of page
Newsletter Registration   |   Stay Connected:
 

888.NEC.CODE (632.2633) 3604 PARKWAY BLVD, STE 3, LEESBURG FL 34748  

Tell a Friend About This Site

  NEC® and National Electrical Code® are registered trade marks
of the National Fire Protection Association (NFPA).
  ©Copyright 2011 Mike Holt Enterprises, Inc.