Reflected Power

I think the private customers wouldn't understand, and therefore not accept it. Industrial customers can understand why they'd have to pay more if they draw more current, even if it isn't all active.

joe

On Wednesday, December 2, 2015 at 5:01:57 AM UTC-5, snipped-for-privacy@gmail.com wrote :

OMG.. Don't give them any ideas!

Thats actually a good question, the standard residential watthour meter mea sures true watts so it is fair.

I know the industrial watthour meters also measure PF and I think there is a penelty fee if the PF us below some limit. Easy to do with the new elect ronic meters. How did the old mechanical meters for industrial customers take PF into account, how did the mechanical meter actually work and measur e PF?

It might be easier to think in terms of energy.

Power is a rate of flow of energy.

So called real power is energy actually consumed.

So called Reative power is energy circulating back and forth and not actually consumed (except in the wire losses).

Mark

The strongly time-varying character of the drive impedance of a Class C or switching amplifier does make the question a bit less obvious. One would need to analyze the final + tank + tuner almost like a SMPS to get it really right, I expect.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Because you're not using volts*amps. Do you really want to pay for imaginary power? Besides, real power is usually less than apparent power, so you're getting a bargain.

The problem is determining who's problem a bad power factor should be. Originally, the power company was responsible for delivering power. If the load wasn't quite right, the power company would plant PF correcting capacitors on top of the power poles. When the PF would change, such as at night or weekends, someone would drive around town and flip big knife switches to connect and disconnect these caps. In southern California, during the 1980's and 90's, this was automated with a 900 MHz SCADA system (where I designed some of the radios that were NOT used).

That worked acceptably for residential power, but not industrial scale power. So, the power companies dumped the responsibility for correcting the PF to the industrial customers, which is the current (pun intended) situation. The utility companies could also, and probably will, transfer the cost of power factor correction to residential customers, but only after smart meters are set to produce a PF reading, and it appears on your power bill. I would guess about

5 years from now. Of course, it will be justified as part of saving the environment: More: I dunno if my smartmeter will display PF, or both VA and Watts. I'll check later.

Yep. The regulated industries tend to control the regulating authorities charged with regulate them. In other words, the tail does wag the dog.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

"The classic example of a non-linear load is a rectifier with a capacitor input filter, where the rectifier diode only allows current to pass to the load during the time that the applied voltage exceeds the voltage stored in the capacitor, which might be a relatively small portion of the incoming voltage cycle."

"Other examples of nonlinear loads are battery chargers, electronic ballasts, variable frequency drives, and switching mode power supplies."

It's more like the other way around. Power line harmonics, due to non-linear loads, makes power factor correction difficult.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

Please cease using the impedance matching analogy (reflections), because it doesn't work for utility power.

Reflected power or VSWR as in an RF system does not apply to electric power systems. In power systems, the source, transmission lines, and load are NOT impedance matched. If the were matched, then the source would dissipate as much power as the load, and the generator would instantly burn up. There are reflections, but at 60 Hz, with a wavelength of 5,000 km, you're not going to see any voltage nodes (nulls) or anti-nodes (peaks) necessary to measure VSWR unless you wrap your extension cord around the planet.

In utility power, all that PF correction does is adjust the voltage to current phase angle to zero. In effect, the addition of caps across the power line resonates with the load inductance so that the generator sees a purely resistive load. However, for RF, that's not the case, as the antenna tuner simply matches the source to 50 ohms. The 50 ohms can be any combination of R, L, and C as long as the vector sum is equal to 50 ohms. You can have a 50 ohm matched load, and still have a rotten and uncorrected phase angle.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

However, for RF, that's not

No.... not any combination... THe antenna tuner matches the source to 50 Ohms __RESISTIVE__. i.e. to the center point of the Smith chart. Not just any 50 Ohms will do.

A "50 Ohm matched load" __means__ 50 Ohms __resistive.__

A 50 Ohm capacitor is not a "matched load" and will create a large VSWR.

A tuner will (try to) cancel the reactive part of the load AND transform the resistive part to 50 Ohms.

Analogously, the power company adds capacitors near the load to cancel the reactive part of the load.

So in RF land we want to get to THE CENTER POINT on the Smith chart which is 50 Ohms resistive.

In power land they want to get to the center Horizontal LINE which is simply resistive (no reactance) of whatever value.

Mark

With the exception of my initial post, I wasn't aware that I was! Apologies if it appeared that way, obviously my error (presumably).

what? What do you think reactive loads do the phase ?

My god, I am not even going to get into this one..

Some of you that should know better, do not, what a big let down.

Jamie

Can you read? No, I didn't think so.

They certainly don't distort.

It really isn't a letdown that you can't read (or think). It's been obvious for a long time.

mako

** For the energy company to know if a customer's PF is poor, or not, they have to monitor the VA consumption over time and compare that with energy consumption the same period.

If the PF is 1 the readings will be the same, but if the PF is poor the kVA hours readings will be higher than the kW hours.

A meter that shows the instantaneous PF is more interesting to the customer as it allows them to know if they have enough PF correction in place.

.... Phil

Depends on whether the customer has linear loads or not. If the loads are linear, then PF=cos(phi) (with phi the phase angle between voltage and current). If the loads are non-linear, then PF=(I1,rms/Irms)*cos(phi) in which I1, rms is the RMS value of the base harmonic current component (50 Hz or 60 Hz depending on your location) and Irms the RMS value of the total current.

If the PF is poor, then the kVA reading could still be the same as the kW reading, but be poor because of a high Irms.

joe

There is more too power factor correction than just shifting the phase back and forth. Try googling "PFC IC" if you don't see what I mean.

Jeroen Belleman

** No it does not.

** Utter drivel.

The basic DEFINITION of PF is the ratio of real power to VA. PF = real power / VA

Please go away.

.... Phil

For *any* load, the power is defined as the average of the integral of V(t)*I(t)*dt over the period of interest. Then the power factor will be that average divided by the product of Vrms*Irms.

PF=cos(phi) is for the special case of sine waves only.

Not so. A poor power factor is 0. The kW reading is 0. The VA reading is the product of Vrms*Irms. Consider an inductor with no internal resistance. Let's say you have an inductor that draws 1Arms at 1Vrms. That's one VA. But, the power is 0. There fore the power is 0/1 or zero.

Therefore the power factor is

0/1 or zero.

My apologies. You're correct and I was wrong. For an "L" network of one inductor and capacitor, an antenna tuner will produce a purely resistive match at a given frequency. "L" networks can't match every possible load, but do a good enough job for many real antenna.

Agreed. The further away one gets from the center of the Smith Chart, the higher the VSWR (i.e. constant VSWR circle about the 50 ohm center point). One can't do that with PF correction because there's no characteristic impedance value for a power line (where everything is mismatched) with which to normalize the Smith Chart. Therefore, matching for minimum VSWR (or minimum reflected power) in a power line system makes little sense and is quite different from an RF system, which was the point I was attempting to make.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

You *can* look at this as a matching problem. Fortunately the mismatch between your little main-fed gadget and the big mains generating plant is so great that all but a very small amount of power is reflected, and the superposition of the incident and reflected waves yields a small current. It's fortunate also that the transmission line length is very short compared to the wavelength, so that this current is reasonable everywhere on the line. ;^)

Jeroen Belleman