But wouldn't apparent power (as measured via true RMS devices) = average power even if the device is only drawing current in various parts of the sine wave? Voltage and current are still in phase as you say- just stopping and starting at various times. I personally would not call this power factor, rather average power consumed over a period of time. Much like a diode in series with a bulb or a triac dimmer. Less work is done per second. To lower power factor (shift current in relation to voltage) you need either capacitance or inductance, devices which briefly "store" electricity and then "send" it back to the source after having "taken" it from the source. For example a capacitor charging (taking) and then discharging (sending back) electrons creates added current on the line in addition to the current already present actually going to do work. Because of I2R losses and amps being amps in terms of thermal limits, we restrict equipment to VA instead of watts. The more VA is doing work relative to "bouncing" back and forth, the more we consider assets to be well utilized.

Cheesy explanation, but that is how I was taught it- I'm sure you can word it better and by the looks of it you did.   

Correct me if I am wrong.

You were taut of only small part of reasons and causes why a heavily loaded transmission transports not much power, although this part is occurring as problem in majority of real life power distribution systems...
But in recent years, when the active power electronic become wide spread, the nonlinear problems start to become in the focus as well.


For the terminology: The "work done for a second" is exactly how the term "Real power" is defined. And what the "average power" mathematically is. Because it describes the ong term average energy transfers, it makes sense to be used in any circuit, either linear harmonic (mains), nonlinear AC or even pulsed DC.

The term "apparent power" means a value, how the circuit value "appear" (so hence the product "Vrms * Arms"). And because it does not mean any energy transfer, it can not bear "W" as units, so it has just the "V*A" aka VA as unit. Because it means how the wiring is loaded and what losses are to be expected there, it is useful again in any circuits, mainly for wiring sizing.

The "reactive power" is what is causing energy bouncing there and back, so e.g. the pure capacitor and inductor. Again there is no long term energy transfer, so the units are VA. Normally this value makes sense to use only with linear circuits with harmonic single frequency feed (so inductors, capacitors, resistors on an AC mains), it is useful as base for what compensation is needed to get unity power factor. It could be expressed for nonharmonic or pulsed circuits as well, but it does not have any practical use there, because the power factor deterioration factors are so broad in nature, they need dedicated correction for each anyway and for most cases nothing like a compensation element exist at all (and sometimes "the higher the better" is what works - like a DC blocking tank capacitor with a DC supply and a pulsed load combination).


The phase shift cause you were taut is what is called "linear distortion". So you have distortion in the signals (the voltage and currents are different), but it is of a linear nature (scaling and superposition is working,...). In an environment with sinewave source of a fixed frequency the only way how the distortion may pop up is just a phase shift, nothing else.
There indeed PF = cos(Phi), because all the rest is ideal.

But when dealing with nonlinear circuits and more generic systems, the non-unity power factor problems appear and they can not anymore be described just a plain phase shift.
In fact even when you may define a phase shift (e.g. a phase shift between sine voltage vs 1'st harmonic of the current) as a "Phi",
PF is not anymore equal to cos(Phi). In fact PF < cos(Phi), because the phase shift is only one component of the PF problem and there are more of them.

Another cases of a non unity PF could be:
- Sinewave current feeding a load that act as a voltage clamp, so makes the voltage rectangular. There is no shift, but PF=2/Pi/sqrt(2).
- A DC voltage source (e.g. a battery) feeds a pulsed load with 50% duty (PWM dimmer). There the PF=sqrt(1-DutyRatio) = sqrt(1/2).
Note: Even the DC circuits may have below unity power factor. There the correction uses to be simple: Large blocking tank capacitor smoothing out the pulses. That means from the capacitor downstream the PF is still 0.707, but upstream the current is filtered, so the PF becomes closer to 1. And the larger the capacitor, the better the filtering, so the closer the PF goes towards 1, so the "the larger the better" case, something unheard of in text book phase compensations...
- An input of a single phase full wave rectifier loaded by a constant current (an inductor filter) means sinewave voltage, but rectangular current. PF=2/pi/sqrt(2).
- A phase cut dimmer set to 90deg conduction angle (= half power). Voltage is sinewave, current is only half cut sine pulses. PF=sqrt(1/2)
- A 2kW heating plate controlled by a thermostat for 10 seconds ON and 10 seconds OFF, fed from an AC mains. Power factor here is again sqrt(1/2).
Note: We are speaking about a resistor load (text books say the resistor is unity, right?). But it is not there alone, it is there with the cycling thermostat. And that makes the total load power factor only 71%. Actually the same apply as for the PWM with DC supply.
- Two 2kW heating plates, both controlled by a thermostat 10s ON/10s OFF (each plate has its own), but the hermostats are synced so they alternate. Because all the time is one plate connected and both are the same resistance, the overall power factor is unity.
Note: Here we have two loads, each the same 0.7 power factor, both because of the same mechanism (the 20s period PWM at 50%), the overall PF becomes unity again (simply the thermostats cancel each other). Again you won't see that in linear phase shift area (two capacitors won't cancel each other).

So the problematic becomes way broader. Yet still the "apparent power" and "real power", so the "power factor" terms make their practical use the way how they were defined originally for just the linear case.

   
Ahhh, thank you- almost did not consider that. But, I have to ask- at what point is a Mercury arc tube the most efficient (lumens per watt)? If we are dealing with a ballasted roadway or tunnel fixture longevity would be the primary goal, but because in this case we are using filaments (for the sake of the discussion we will assume they are an integral part of the bulb inside the envelope) I am less concerned about arc tube longevity as the filament will probably break before the tube fails even if over driven. At least it is my thin belief that over driving does not shorten life that much.

The higher arc load, the higher efficiency. But it tends to saturate at extreme high loads (see the efficacy vs arc loading graph ; it is valid enev when varying power to a given lamp)

Also deluxe phosphor- how would that play with the filament's light? 

- A phase cut dimmer set to 90deg conduction angle (= half power). Voltage is sinewave, current is only half cut sine pulses. PF=sqrt(1/2)
- A 2kW heating plate controlled by a thermostat for 10 seconds ON and 10 seconds OFF, fed from an AC mains. Power factor here is again sqrt(1/2).
Note: We are speaking about a resistor load (text books say the resistor is unity, right?). But it is not there alone, it is there with the cycling thermostat. And that makes the total load power factor only 71%. Actually the same apply as for the PWM with DC supply.
- Two 2kW heating plates, both controlled by a thermostat 10s ON/10s OFF (each plate has its own), but the hermostats are synced so they alternate. Because all the time is one plate connected and both are the same resistance, the overall power factor is unity.
Note: Here we have two loads, each the same 0.7 power factor, both because of the same mechanism (the 20s period PWM at 50%), the overall PF becomes unity again (simply the thermostats cancel each other). Again you won't see that in linear phase shift area (two capacitors won't cancel each other).

So the problematic becomes way broader. Yet still the "apparent power" and "real power", so the "power factor" terms make their practical use the way how they were defined originally for just the linear case.

Excellent explanation! But if I may, I'd call the above, especially the heaters, duty cycle rather than power factor. At least that is how the NEC views, and terminology used in designing power systems. But the rest is spot on and I thank you for that putting it mildly.

The higher arc load, the higher efficiency. But it tends to saturate at extreme high loads (see the efficacy vs arc loading graph ; it is valid enev when varying power to a given lamp)

Over driving the arc would be in my benefit then.


You know lighting history a lot more than I do- what is the highest lumen per watt commercially produced self ballasted mercury vapor lamp that you know of? The 16,000 hours you mention earlier is very lucrative, I'm wondering how efficiency does with it.     

Excellent explanation! But if I may, I'd call the above, especially the heaters, duty cycle rather than power factor. At least that is how the NEC views, and terminology used in designing power systems. But the rest is spot on and I thank you for that putting it mildly.

Yes, the codes tend to keep separating the Duty Ratios (switching slower than mains frequency) vs Power Factor (switching at mains frequency - e.g. phase cut). But when you are about to calculate e.g. the wire loading/selfheating/losses, you get the same results when you are strictly using the Apparent power way for the calculation regardless of the frequency.
By the way "Duty Cycle" is technically a nonsense term for that, however it is still widely used; it is speaking about ratio of time sections and not time alone.

Over driving the arc would be in my benefit then.

You know lighting history a lot more than I do- what is the highest lumen per watt commercially produced self ballasted mercury vapor lamp that you know of? The 16,000 hours you mention earlier is very lucrative, I'm wondering how efficiency does with it.     

I do not have any overview what was ever produced.
Regarding the lifetime: The 16khour extrapolation I've made assumed the only aging is the filament evaporation (i.e. the principal unavoidable incandescent aging mechanism). But the real products are way more than just that and with such lifetime expectations there may be another mechanisms killing the lamp at the end (cement disintegration, stresses/cracks in the glass,...), which is not possible to extrapolate the same way  and what is a result of the general quality. The difference towards the filament aging is, these could be made long lasting without impacting the lamp performance (efficacy,...; making the filament alone longer lasting means you get lower efficacy, because that is plain physics).

Regarding efficacy vs longevity, yup, its a compromise. I wish there were more self ballasted mercury lamps used in the past. I mean there is something very elegant about the idea, however, efficiency at 24,000 hours would equal an incandescent roughly interpreting your graph and equations. 


Does anyone know where self ballasted lamps are being used in China? Researching Chinese makers of SBMV, there seem to have many and of a wide variety.   

Does anyone know where self ballasted lamps are being used in China? Researching Chinese makers of SBMV, there seem to have many and of a wide variety.   

In south east Asia these are popular as cheap high CCT light sources, while the intense high CCT is there treated as a luxury light.
Therefore they are mainly in clear versions - the lowering CCT by a phosphor is not desirable there, the halogen ballast provides just perfect color correction there...
So it is mainly for their domestic market there...
Only recently the high CCT LEDs gradually take over, but the high intensities are still way too expensive for that market, so these SBMV's are still popular...

Dumb question- but what is the color temperature and color rendering? Would phosphorus make these lamps more efficient? Or increase rendering?

Dumb question- but what is the color temperature and color rendering? Would phosphorus make these lamps more efficient? Or increase rendering?

More efficient? Not much, because there is not that much UV power to utilize, yet any coating absorbs some light. So the gain in utilizing some of the UV is not much more than what becomes lost.
It is to a big extend linked to the color: Unlike the western market, the east market wants high CCT, what would mandate the use of green and blue phosphor emission. But in that area the phosphors are way less efficient than the orange emitting western mixes. And even the orange emitting mixes boost the efficacy by barely 20% and only on lower wattage lamps (with lower arc loading). The gain in higher wattages (with inherently more efficient arc due to higher loading) way smaller, or even nearly no efficacy gain at all, the phosphor is there just for the collor correction.
With the halogen ballast the rendering is achieved by the ballast incandescent radiation, it can not get much better with the sharp green and blue peaks of the MV burner.
Plus the market wants sparkling light, so the diffusion is not welcome there.

Makes sense, but as you say the US market wants diffusion. I remember years back (late 90s before the fluorescent retrofits) supermarket chains around me used coated metal halide, as did Lowes/HomeDepot/Menards until the late 2000s.