OT... How do I Emulate power factor?

[hoof]

Hello folks, I am allways amazed of how well rounded this forum is. I spend most of my time pretty evenly dividing my hobby time between manchine fun and electrical fun. I have no problem trying to get both of these hobbies to offest some of the expenses incured either. That being said I am trying to develope an electrical load device that is capable of emulating/simulating a high power factor load. .5 to be precise. I realize what PF is, and how it's corrected. And there is alot of information on the web for just about most stuff related to this subject. But nothing on how one would create a bad power factor. Any thoughts from the electrical guru's here? Thanks in advance for your assistance.

Ray

[dp]

Use a synchronous motor for a load? http://wiki.answers.com/Q/How_synchr...tor_correction

[becksmachine]

Wouldn't .5 actually be a LOW (poor) power factor?

Dave

[Paul Alciatore]

Power factors other than unity (1) are caused by a load that is either inductive or capacitive. So you can simulate any power factor with the proper combination of resistance, capacitance, and inductance. The actual circuit can take several forms, but the resistive element is probably needed in all of them so it is the base, starting place. The capacitive element would probably be connected in parallel with the resistance and the inductive element in series with it.

In order to figure the values for each of these elements for a particular power level and power factor, you have to do the math. For that, you should read some electric theory books.

[J Tiers]

If you know the VA you want, at what voltage, then you can determine the impedance needed.

What the power factor is, amounts to the proportion of the apparent power (volts * amps, or "VA")that is actual watts.

The power factor is defined as the cosine of the phase angle between volts and current, or; Volts * amps * cosine(phase angle). In a resistor they are in phase, and in a pure inductance or capacitor, they are exactly 90 degrees out of phase.

the PF alternately defined is: PF=watts/VA It comes out the same either way.

if the load were resistive, the VA would equal the watts, phase angle would be zero, and PF would be 1 (cosine of 0). If the load were all reactive, the VA would naturally be all reactive, no watts, phase angle would be 90, so PF would be 0 (cosine of 90). That is not possible in actual normal physical "stuff", since there will be resistance of at least some amount.

You want the power to be half the VA, or 0.5. The arc-cosine of 0.5 (angle having cosine of 0.5) is 60 degrees. So you want a load with a phase angle of 60 degrees, and an impedance corresponding to the VA you figured.

This is best done using vectors.... so your "unit circle" has a radius corresponding to the impedance you want. With horizontal as resistance, vertical as reactance, draw a line at 60 deg to horizontal from the center to the circle. That represents the total impedance.

Dropping a line from the end of that radius to the horizontal gives a right triangle, and the cosine is equal to the adjacent side / hypotenuse. Hypotenuse is the impedance, and at 60 deg the cosine is known to be 0.5, so the adjacent side, representing resistance, is 0.5 x the radius, which is what is wanted.

Now for the reactance.... there is another right triangle formed, if you put a line from the end of the radius to the vertical (reactive) axis. In that case, the angle is 30 degrees, and the cosine of 30 is 0.866. Therefore the reactive impedance should be 0.866 x the total impedance.

You already know the total impedance, since you know the total VA you wanted, so it comes down to calculating the resistance and the inductance or capacitance based on the needed impedance figured out per the above, the choice of component being decided by whether you want leading or lagging current.

This is all apparent if you use vector analysis.

Don't forget to measure the resistance of the inductor if you want lagging phase, and subtract that from the resistance you figured. In fact, ideally you would find the resistance of the inductor when it is as hot as it gets at the current you will have, and subtract that.

Capacitors are more trouble to measure resistance in, and usually are of fairly low resistance at power frequencies, so you can either ignore it, or estimate it, or simply adjust the resistance to get the perfect 0.5 PF.

It looks as if the reactive plus resistive impedances would add up to be too much, but don't worry. It all comes out OK in the math.

[Old Hat]

You should put that in wiKo-pediA! It's the closest I've ever come to having a clue
what the heck power factor even is. Electro-friends talk of it just to agrivate me.

[hoof]

Thank you for your replies, Most helpful indeed. Dave yes, .5 is a poor power factor. Thats actualy what I'm after here. It is for testing some switching apperatus under worst case senerio's. J Tiers, (Wish I knew your real name) That was a very eloquent explanation I find most helpful. I will post when the sparks start to fly. Again thank you all.

Ray

[garyhlucas]

You know, when a motor is under loaded the power factor drops as the voltage and current are not in phase. So simply running a motor with no load should give you the low power factor you are looking for.

[J Tiers]

You know, when a motor is under loaded the power factor drops as the voltage and current are not in phase. So simply running a motor with no load should give you the low power factor you are looking for.

Not quite right, actually. I assume "under loaded" means a light load... right?

An unloaded motor is pulling largely the magnetizing current for the core material, and not much power, just enough to overcome losses in bearings and in "windage" for the cooling fan, etc.. The power factor may be as low as 0.1, or 0.2.

When a LOAD is put on the motor, the same magnetizing current still flows, but the resistive power goes UP, because shaft power is resistive, it is real watts output. The ratio of watts to reactive power is bigger.

Therefore, a loaded motor has a better power factor, closer to 1, most likely between 0.6 and 0.8. the phase angle is less between volts and amps.

My real name?

I guess it is a mystery........

[dp]

My real name?

I guess it is a mystery........

I thought you sed it was Jerrold.

I have a recollection from long ago that you could use a synchronous motor with a variac across the shunt to emulate a wide range PF loads. I don't think you can take it negative but if you put a variable speed pony motor on it you should be able to. I've stashed away my Electrical Engineer's Reference Manual which is where I think I found that factoid.

Rotary converters are expensive to run because they have a very low PF number.

[J Tiers]

I thought you sed it was Jerrold.

Yah think?

I have a recollection from long ago that you could use a synchronous motor with a variac across the shunt to emulate a wide range PF loads. I don't think you can take it negative but if you put a variable speed pony motor on it you should be able to. I've stashed away my Electrical Engineer's Reference Manual which is where I think I found that factoid.

Rotary converters are expensive to run because they have a very low PF number.

Unloaded synchronous motors with wound field can be made to have various inductive or capacitive characteristics by adjusting the field. strength. They can therefore act to correct PF.

RPCs can have the PF corrected very easily by capacitors added to the input line.

But they are not expensive.... single phase residential service is not (yet?) charged for power factor, , and in fact, the old mechanical meters actually have an adjustment to balance out and discard reactive power, charging only for "real" power in watts.

So a low PF does nothing for your bill.

If you have 3 phase power, and so are charged for PF, you do not need an RPC....

[dp]

But they are not expensive.... single phase residential service is not (yet?) charged for power factor, , and in fact, the old mechanical meters actually have an adjustment to balance out and discard reactive power, charging only for "real" power in watts.

So a low PF does nothing for your bill.

If you have 3 phase power, and so are charged for PF, you do not need an RPC....

The OP was looking for a PF load emulator, not 3-phase power.

Can I Save Money on My Utility Bill?

In the case of utility bills, power factor correction capacitors are usually justified from a cost-effectiveness standpoint when the utility rate structure includes a power factor penalty. When power factor penalties are applied, correcting power factor will save energy costs by reducing the power factor penalty portion of their bill, but it will save very little energy. Sometimes the power factor penalty is explicit, and sometimes it is built in. That is, some power companies charge an effective power factor penalty by charging for demand using kVA instead of kW. kVA is calculated simply by multiplying the Volts x Amps, which gives you the apparent power (uncorrected). In the example above, the kVA would be 240 V x 100 A /1000 = 24 kVA. With the power factor correction, the new demand is 240 V x 70 A/1000 = 16.8 kVA. If the demand charge is high, this can be significant.


Some power companies (notably Seattle City Light and Puget Sound Energy) actually measure kVARh (kilovolt-amps-reactive-hours), and bill customers a power factor charge based on that. Rarely, if ever, are power factor penalties charged to residential customers. They usually only apply to the largest accounts.

http://energyexperts.org/EnergySolutionsDatabase/ResourceDetail.aspx?id=4158

I wonder if it is rarely or if ever.

[J Tiers]

The OP was looking for a PF load emulator, not 3-phase power.



I wonder if it is rarely or if ever.

hey, you mentioned rotary converters..... don't blame me......

I do not know of any utility that charges for PF to residential. They would have to get the new charges through the Public Service Commission, first, in many states of the US. That might not be easy when it is realized that everyone's bills would suddenly jump up as much as 50%.

[dp]

That was another example of a PF load emulator.

Puget Power allows up to 7.5HP single-phase motors on residential lines. I would bet a home shop could easily exceed that with AC heat pump, a lathe turning, air compressor running, and then the wife turns on the washing machine and dryer.

[J Tiers]

Puget Power allows up to 7.5HP single-phase motors on residential lines. I would bet a home shop could easily exceed that with AC heat pump, a lathe turning, air compressor running, and then the wife turns on the washing machine and dryer.

is that aggregate?

Most of those restrictions are for any single motor. It's based on the starting surge, what that does as far as a line disturbance, as well as the capacity of the transformer, or "pole pig" that steps down the voltage to 240, and not on the PF.

As mentioned, a motor operating at nearly full load has a high PF, easily over 0.6 and probably in the area of 0.7. A/C is often loaded heavily. If that were as bad as it gets, things would be far better than they are.

Also, a run-capacitor motor can have a very high PF.... A PSC motor can have a PF of essentially 1.0 at full load, by selecting the capacitor.