Power factor correcting 14

[sparkodean]

I am looking to calculte the required KVAR power factor unit that would be required to get my power fator above 85%. My current is approx 800 amps at 480 volt and is basically all inductive (motors). Also, can I calculate the current per KVAR this unit will draw.

 

[Skogsgurra]

sparkodean,

The calculation is simple and forward, but one must have the actual P.F. to start with. It cannot be guessed from your data (800 A @ 480 V) since the P.F. is heavily dependent on motor output i.e. shaft loading. Do you have the actual Power Factor available?

 

[sparkodean]

Thank-you for your response. No, I do not have the PF available. We currently have a 250kvar unit on this plant, and according to our utility company, out power factor runs between 75 and 80 percent.

 

[busbar]

On the second question, current for given kVAR is the same as for a resistive or inductive load. Per-phase current in amperes = {capacitor reactive} voltamperes/480V/sqrt3. A rule of thumb for 480V 3ø loads is 1.2 amperes per kVA {or kVAR/”kVAC”.}

Assuming an ANSI installation, NEC Article 460 would likely apply, with capacitor-circuit conductors sized 135% of capacitor rated current.

Watch out for harmonic sources, higher-yet line voltage and high ambient temperatures.

http://www.geindustrial.com/products/manuals/GEH-2738D.pdf

http://www.geindustrial.com/products/applications/GEP-974F.pdf

http://www.aerovox.com/catalogs/power_factor_correction/toc.htm

http://www.aerovox.com/technical/index.html

http://www.aerovox.com/catalogs/power_factor_correction/pg16.htm

http://www.epcos.com/inf/20/50/db/power_99/02970311.pdf

 

[wareagle]

The utility should be able to give your the kw and kva demand for your company. If you know the kw and pf you can calulate the kva if it is not available. You said that you want to improve your pf to 0.85. Why not improve it to 0.95 or 0.99. If you are being billed on a kva demand it is cost effective to improve the pf to a hige value. Can you say what your utility is billing you, kw or kva? Cost per kva or kw? Usually the demand cost is about $5.00/kva.

 

[cuky2000]

Check if the enclose calculator could help you.

http://www.monachos.gr/eng/calculators/kvar_calculator.htm

 

[ctpt]

In other words, the required kVAR = kW(tan[ø]1-tan[ø]2),
where, [ø]1= initial power factor
and [ø]2= required power factor

 

[ctpt]

correction to above,
"where, cosø1= initial power factor
and cosø2= required power factor"

 

[nostrayvoltage]

1). Assume single phase for illustration. Then 800amps at 480volts is equal to 384 kva (800*480/1000).

2). If power factor is 0.8 this means cos=0.8 and then the sine = 0.6. KVAR=KVA times sine = 384*0.6 = 230.4kvar.

3). So the circuit is has 230.4 kvar of inductive reactance, so if you install an additional 230.4 kvar of capacitive reactance you will pf=1.

4). Follow same procedure if three phase.

I know this is not the answer to your question, but it is simple way to get to a pf of 1. Similiar procedure works for 0.9 pf if that is what you really want.

no_stray_voltage@yahoo.com

 

[vtakev]

Hi Sparkodean

Everything said from other guys is correct. Just pay attention to high level harmonics. If you have frequency drives or softstarters for your motors they could create 5, 7 and 11 harmonics. In this case you will need reactors on the capacitor banks or harmonic filter. NB: LV capacitors, which will work with reactors must ne able to withstand at their terminals 1000V!

I'm trying to attach somehow...obviously I'm stupid... simple kVAr calculation xls sheet. Anyway, you can find a lot of this stuff in Internet.

I made simple calculation for you and it gave appr. 165 kVAr in order to achieve 0.95 cosphi from 0.85. Of course, the characteristics of the installation (even daily load diagram) have to be taken in consideration.

 

[powercam]

What is Power Factor Correction?

Most loads on an electrical distribution system fall into one of three categories; resistive, inductive or capacitive. In your own plant, the most common is likely to be inductive. Typical examples of this include transformers, fluorescent lighting and AC induction motors. Most inductive loads use a conductive coil winding to produce an electromagnetic field, allowing the motor to function.

All inductive loads require two kinds of power to operate:

Active power (kwatts) - to produce the motive force
Reactive power (kvar) - to energise the magnetic field
The operating power from the distribution system is composed of both active (working) and reactive (non-working) elements. The active power does useful work in driving the motor whereas the reactive power only provides the magnetic field.
The bad news is that you are charged for both!


Active Power
Reactive Power
Available Active Power
As the power factor drops the system becomes less efficient. A drop from 1.0 to 0.9 results in 15% more current being required for the same load.

A power factor of 0.7 requires approximately 43% more current; and a power factor of 0.5 requires approximately 100% (twice as much) to handle the same load.


The objective, therefore, should be to reduce the reactive power drawn from the supply by improving the power factor.

If an AC motor were 100% efficient it would consume only active power but, since most motors are only 75% to 80% efficient, they operate at a low power factor. This means poor energy and cost efficiency because the Regional Electricity Companies charge you at penalty rates for a poor power factor.

By installing capacitors to improve your power factor you could SAVE MONEY on your electricity bill.

Additional potential benefits include:

Reduction of heating losses in transformers and distribution equipment
Longer plant life
Stabilised voltage levels
Increase in capacity of your existing system and equipment
Improved profitability

 

[harmonicoscillation]

Someone asked why he doesn't want to correct higher than 0.85. I don't know for sure in his case, but many utilities, instead of direct kVA metering or PF billing, only penalize you if you are below 0.85 PF, and don't reward you above that. Other PFs may apply, but .85 is a common threshold.

I'd also like to mention that it is most efficient to correct PF at each inductive load. Often, plant managers will perceive it as cheaper to just correct above .85 at the service and be done with it. A more thought out presentation must be done on why it is beneficial to the plant to correct close to the inductive load. The concerns presented by powercam (above) give a well thought out summary of those issues.

 

[sparkodean]

This is true. In our case, the utility company only penalizes for pf below 85%. I am not sure of the benefits of have separate correction units close to the actual load. In our case however, most of our larger loads (Motors) are on soft starts, and the manufacturer of the soft start claims that it could be detramental to the soft start to do pf correction on the load side. To do it on the line side, I am at the same place as to where my service is coming in and can do the correction on the whole service. Am I looking at this correctly, or am I way off.

 

[tinfoil]

Most utilities only examine the power factor (kW / kVA) at the main service supply point, and then only for the values at peak for any given billing period.

In much of Canada, the magnitude used for the demand charge is expressed as "90% of the peak kVA, or 100% of the peak kW, whichever is greater".

These means that the demand portion of the bill will be the same whether the pf is 90.001% or 100%, but will increase if the pf (at peak) becomes lower than 90%. Interestly, older 'inductive-wheel' kW meters cannot sense the difference between 'leading' and 'lagging' pf's.

I am aware of a case where an industry had a leading pf, much less than 90%, and paid a 'power factor penalty' for the priviledge of furnishing some excess kVARs back into the grid, which actually HELPED the utility with its current flow and voltage support! The industry had shut down most of its process, but had neglected to review the size of the fixed capacitors installed at their main.

So, many industries can predict what their peak load and pf will be at peak, and if this is expected to be more-or-less constant in the long run, then they install a suitible FIXED capacitor at their main to improve their bills.

This saves them money from the utility, but they lose the opportunity to reduce current flows and correct phase angles within their facilities. Correcting a pf from 70% to 90% at the point of consumption (such as a motor) reduces incoming currents by over 20% all the way back to the main, improving voltages, extending the life of plant electrical infrastructure and returning extra capacity to the system for other new loads.

BTW, the use of reactors is very important if drive controllers are present.

 

[peebee]

Two thoughts on harmonics:

1. As mentioned above, if your low PF is due to harmonics, capacitors won't help. You'll need filters. Caps only help with displacement power factor, they don't help with harmonic power factor.

2. Your capacitor can form an LC circuit with the inductance in the rest of the system. You should check that the resonant frequency of the LC circuit is not near any of the harmonic frequencies. If so, you might have problems ranging from voltage distortion to frequent failure of capacitor fuses to violent, dazzling failures of capacitor contacts and capacitors. Check the harmonics.

Also - I don't believe anyone mentioned that there are a few ways that capacitors can be installed:

1. One big old cap at the mains entrance.
2. One big bank of small capactiors at the mains entrance, with a PLC adding and removing capacitance as required.
3. Lots of small caps scattered around the plant, usually one at each motor starter or such that switches on and off with the motor by the motor contactor.

#2 & #3 will prevent you from overcorrecting (going past 100% and into leading PF) which can easily happen with #1 during times of low loads. Overcorrection can be bad.

You should consider hiring a PE for this. Or maybe getting some engineering assistance from a cap vendor.

Just curious -- you have by far the lowest PF I've heard of in a long long time. What kind of plant is this? Are you melting steel or something?

 

[mc5w]

There are some things to also watch with capacitors:

1. Motor reactive power at full load is about twice as much as at no load. If the capacitor is greater than the motor no load KVAR and is connected between the controller contacts and the motor you are likely to get an overvoltage when the motor shuts off. This is because the motor turns into an overexcited induction generator.

If you are installing capacitors at each motor you should put the capacitor on its own contactor and "motor" overload relay with the contactor connected to the supply side of the motor controller - this works fine with soft starts. Both the motor and capacitor overload relays should be connected in series with the stop button so that both devices shut off if there is an overload. This also avoids the overvoltage problem when shutting off the motor and allows for a capacitor more appropriate for the actual motor load.

Putting the capacitors close to motors has a number of benefits such as wiring running cooler which will save on the power bill and possibly save on upgrades. Wiring resistance will also help damp out resonances but that is NOT foolproof. Wiring impedance also helps with inrush current but you should consider using Telemecanique's capacitor switching contactor that cuts in some high resistance air core inductors before closing the main contacts on a case by case basis.

2. Some industrial and commercial time of use rates will reward you for having a capacitive power factor during peak periods and penalize you for power factor that is too good during off peak periods. This is because transmission lines and substations have to be maintained at 90% inductive to 95% capacitive for good voltage regulation during off peak periods and more like 90% inductive to 85% capacitive during peak periods. If say half your capacitance is at the motors and the other half at your service or industrial substations you could benefit from time of use reactive compensation.

There is a very good chapter on reactive compensation for transmission lines, both telephone and power, in Engineering and Science in the Bell System.

An extreme example is that Consolidated Edison has too much system capacitance because of their thousands of miles of underground cable. They just about have to bar the use of power factor correction capacitors.

3. The usual reasons for installing power factor correction capacitors are not to save on the reactive power charges but rather to free up system capacity or to make the system operable off of a standby generator. Not having to upgrade a motor control center or panelboard from 1200 amps to 1600 amps can save you a lot of dough. OPerating a generator at 85% to 95% inductive power factor will make the voltage regulator's life easier and make the rotor run cooler and live longer.

Caution: Most generators cannot tolerate operation past 95% capacitive power factor. Going past 95% power factor will make the rotor incapable of sustaining full torque from the prime mover.

Mike Cole, mc5w@earthlink.net

 

[sparkodean]

Peebee,
We are a Stone quarry with approx a total of 700Hp on this service.

Thank-you all

We recently replaced bad cap's in our 250kvar pf unit. The pf is now around 87%, according to the last utility bill we received. I appreciate all the responses, and in the future probably do things a little different due to the responses received and knowledge received. At this point the pf unit is not energized until the last motor in the plant is started, however it is possible to start this motor by itself, which would would give me a leading pf. I am planning to change this, in that the pf unit will not get energized until all the motors are running via the PLC.