What's the efficiency rating for a magnetic MH ballast?
I was thinking 50 percent, and this is for powering a 1650 Watt MH connected to a 120 Volt 20 Amp outlet.
And what's the efficiency for electronic?

50% is not realistic. 2 wire choke ballasts are around 90..92% in the higher wattages, i'd expect less for HX/CWA but no less than say 80%

You should not mix the efficiency with a power factor.
20A draw on 120V does not mean 2400W power input, it just means the apparent power (= a hypothetical maximum power you may get when loading the wires in the same way) is 2400VA (it is not "W", because it does not mean any real power, but really "VA", as it is just multiplication of some two circuit parameters)
The efficiency of a ballast at this power level I would expect somewhere around 90..95%, so the real input power would be about 1730 to 1830W (assume the output power is really the 165W).
The ratio of these two numbers is called Power Factor (PF = 1800/2400 = 0.75). The 25% difference does not mean losses, but power which is "bouncing" between the mains and ballast without doing any work and by that only loading the wiring (and transformers).

You should not mix the efficiency with a power factor.
20A draw on 120V does not mean 2400W power input, it just means the apparent power (= a hypothetical maximum power you may get when loading the wires in the same way) is 2400VA (it is not "W", because it does not mean any real power, but really "VA", as it is just multiplication of some two circuit parameters)
The efficiency of a ballast at this power level I would expect somewhere around 90..95%, so the real input power would be about 1730 to 1830W (assume the output power is really the 165W).
The ratio of these two numbers is called Power Factor (PF = 1800/2400 = 0.75). The 25% difference does not mean losses, but power which is "bouncing" between the mains and ballast without doing any work and by that only loading the wiring (and transformers).

So I can power a 1650 Watt MH with ballast safely in this situation?
What about a 2000 Watt MH with ballast?

With all above I assumed that ballast is RATED for THAT 1650W lamp, so then the only explanation for the 2.4kVA input would be the low power factor.
If that ballast is not rated specifically for this lamp type, then most likely the ballast feeds the lamp by about 2kW and so the lamp is overdriven: Most ballasts of such power levels are designed with power factor around 0.9 (when they are complete and run the rated lamp), that means with the 93% efficiency the lamp is fed by the 2.4kVA*0.9*0.93 = 2kW.
Most, but not all.
The typical cases which tend to have lower power factor are the low power ballasts (that isn't the case here; these use to have power factor of around 0.4 till 0.6) and
ballasts designed to run multiple lamp wattages (if these lamp types have the same rated current and only differ in the arc voltage). These ballasts are then optimized to have the maximum current with any lamp type as low as possible (so usually the higher power lamp yields higher power factor).

I do not know the exact rating of the 1650 vs 2kW MH lamps, so I have no idea, if they have so similar rated current  so a single ballast type could serve both lamp types or not.
Ballasts are quite close to a constant current source, so they feed the lamp by current they are designed for, regardless what is the real power going to the lamp and what is the lamp rating. The lamp then dictates the voltage there. Of course, the arc voltage the ballast supports is limited, but it will always cover the rated lamp types.

So you may have one single ballast for small fluorescents from F4T5 till F8T5 (in 230V areas this includes even two of these operated in series pair and the F13T5), just because all these lamps have their rated current around 0.17A and the ballast feeds all lamps by the 0.17A. Well, in this case the lamp ratings were intentionally designed so they may share the same ballast type, so to made the ballast cheaper. And for the same reason the PLS5W till PLS9W (again in the 230V areas it includes their series pairs plus the PLS11W) were designed for exactly the same ballast as well.

With all above I assumed that ballast is RATED for THAT 1650W lamp, so then the only explanation for the 2.4kVA input would be the low power factor.
If that ballast is not rated specifically for this lamp type, then most likely the ballast feeds the lamp by about 2kW and so the lamp is overdriven: Most ballasts of such power levels are designed with power factor around 0.9 (when they are complete and run the rated lamp), that means with the 93% efficiency the lamp is fed by the 2.4kVA*0.9*0.93 = 2kW.
Most, but not all.
The typical cases which tend to have lower power factor are the low power ballasts (that isn't the case here; these use to have power factor of around 0.4 till 0.6) and
ballasts designed to run multiple lamp wattages (if these lamp types have the same rated current and only differ in the arc voltage). These ballasts are then optimized to have the maximum current with any lamp type as low as possible (so usually the higher power lamp yields higher power factor).

I do not know the exact rating of the 1650 vs 2kW MH lamps, so I have no idea, if they have so similar rated current  so a single ballast type could serve both lamp types or not.
Ballasts are quite close to a constant current source, so they feed the lamp by current they are designed for, regardless what is the real power going to the lamp and what is the lamp rating. The lamp then dictates the voltage there. Of course, the arc voltage the ballast supports is limited, but it will always cover the rated lamp types.

So you may have one single ballast for small fluorescents from F4T5 till F8T5 (in 230V areas this includes even two of these operated in series pair and the F13T5), just because all these lamps have their rated current around 0.17A and the ballast feeds all lamps by the 0.17A. Well, in this case the lamp ratings were intentionally designed so they may share the same ballast type, so to made the ballast cheaper. And for the same reason the PLS5W till PLS9W (again in the 230V areas it includes their series pairs plus the PLS11W) were designed for exactly the same ballast as well.

I'm asking if a 1650 Watt lamp + appropriate ballast will be safe to run on a 20 Amp outlet, and then the same but with a 2000 Watt lamp + appropriate ballast.

I see, now I got the question... Before I thought you have measured the ballast current and it read the 20A...

There is no reason why it would be unsafe, assuming the installation corresponds to the circuit breaker rating.
If it will be reliable? In other words will not the breaker trip when everything is OK? Well, that is a question.
Definitely the expected line current should be printed on the ballast label, for sure for normal operation. For abnormal conditions like startup and eventually open circuit, you should consult the ballast documentation.
If it is not printed, then given the power rating, I would expect the PF to be above 0.9 and the efficiency as well. So I would expect about 16.5A for the 1650W one and 20A for the 2kW one.
So the 2kW has no margin, so I would expect it may trip the breaker, mainly when some abnormal conditions happen (mains brief small overvoltage,...).
For the 1650W there is some margin, but not that big.
Assume there is nothing else on that circuit, so when the breaker trips, only the lamp will fail to work.
It depends, how the ballast behaves mainly during runup (that will happen regularly, so it should be below trip point; but it take rather limited time, so you should look into more detailed breaker charts).
When the lamp extinguish (a mains dip or so), it will take quite long time before it reignites. If the ballast tend to draw way higher current than normally (pretty common for power factor compensated HX or so) and that will likely trip the breaker. If the ballast current is lower than rated, it should be OK even there.
Then there is condition, when the lamp is going to fail. Usually that means increasing the arc voltage, what means increasing of the power transferred from the ballast to the lamp. That of course means the input current goes up as well, so it may trigger the breaker. In this case it may be quite a useful feature - as it will very likely reduce the probability of lamp explosion. But it won't eliminate it, so the fixture should be still properly robust.

I see, now I got the question... Before I thought you have measured the ballast current and it read the 20A...

There is no reason why it would be unsafe, assuming the installation corresponds to the circuit breaker rating.
If it will be reliable? In other words will not the breaker trip when everything is OK? Well, that is a question.
Definitely the expected line current should be printed on the ballast label, for sure for normal operation. For abnormal conditions like startup and eventually open circuit, you should consult the ballast documentation.
If it is not printed, then given the power rating, I would expect the PF to be above 0.9 and the efficiency as well. So I would expect about 16.5A for the 1650W one and 20A for the 2kW one.
So the 2kW has no margin, so I would expect it may trip the breaker, mainly when some abnormal conditions happen (mains brief small overvoltage,...).
For the 1650W there is some margin, but not that big.
Assume there is nothing else on that circuit, so when the breaker trips, only the lamp will fail to work.
It depends, how the ballast behaves mainly during runup (that will happen regularly, so it should be below trip point; but it take rather limited time, so you should look into more detailed breaker charts).
When the lamp extinguish (a mains dip or so), it will take quite long time before it reignites. If the ballast tend to draw way higher current than normally (pretty common for power factor compensated HX or so) and that will likely trip the breaker. If the ballast current is lower than rated, it should be OK even there.
Then there is condition, when the lamp is going to fail. Usually that means increasing the arc voltage, what means increasing of the power transferred from the ballast to the lamp. That of course means the input current goes up as well, so it may trigger the breaker. In this case it may be quite a useful feature - as it will very likely reduce the probability of lamp explosion. But it won't eliminate it, so the fixture should be still properly robust.

One more note, I plan to run this setup eventually on a portable generator that is only rated for 5000 constant Watts, so if the lamp fails and draws more power, it can't because the generator will not supply enough power. And I think you could likely pull about 20.5-21 amps, because it can handle 5000 Watts, and if it is 120 Volts, that means about 21 Amps, but only if it is right on 120 Volts.
The other thing is I plan to use both of the 20 amp outlets, which have their own "button" type breaker for each of them, if I remember.
Any possible surge currents would be an issue because it can't handle very much surge power.

5kW at 120V means 40A. So I guess you meant 240V (two 120V outlets), didn't you?

The generators use to be shaft horsepower limited, so I think it will have enough juice to power the 2kW lamp. But only if the breaker would not be complaining.
Regarding an inrush current: Thhe inrush current may be limited, so the voltage just sags. But I don't think that should be any problem with a single lamp.
However it may become a problem, if you want to power two lamps from that generator. Because when one of them causes some severe transient (e.g. restrike after cool down), it will very likely extinguish the other one. So I guess you may power two lamps, but switch them ON before you start the engine and once one get extinguished (it starts to cycle), disconnect it and let it disconnected. Or if that was some glitch in connections (someone accidentally clicked the switch, the engine runs out of fuel so starts "coughing",...), kill the engine, if required refill the fuel tank, let both lamps cool down longer than they usually need for restrike and only then restart the engine again. The transient from one lamp igniting will very likely extinguish the other one, so you really have to shut down and then restart them both at once...
If you do not plan to use these lamps elsewhere, just wire them without any switch, as that will only make problems on the generator.
And of course, there should be nothing else connected to the generator, mainly nothing should be switched ON or OFF, as even seemingly small transients will extinguish the lamps and then cause problems with restart.

5kW at 120V means 40A. So I guess you meant 240V (two 120V outlets), didn't you?

The generators use to be shaft horsepower limited, so I think it will have enough juice to power the 2kW lamp. But only if the breaker would not be complaining.
Regarding an inrush current: Thhe inrush current may be limited, so the voltage just sags. But I don't think that should be any problem with a single lamp.
However it may become a problem, if you want to power two lamps from that generator. Because when one of them causes some severe transient (e.g. restrike after cool down), it will very likely extinguish the other one. So I guess you may power two lamps, but switch them ON before you start the engine and once one get extinguished (it starts to cycle), disconnect it and let it disconnected. Or if that was some glitch in connections (someone accidentally clicked the switch, the engine runs out of fuel so starts "coughing",...), kill the engine, if required refill the fuel tank, let both lamps cool down longer than they usually need for restrike and only then restart the engine again. The transient from one lamp igniting will very likely extinguish the other one, so you really have to shut down and then restart them both at once...
If you do not plan to use these lamps elsewhere, just wire them without any switch, as that will only make problems on the generator.
And of course, there should be nothing else connected to the generator, mainly nothing should be switched ON or OFF, as even seemingly small transients will extinguish the lamps and then cause problems with restart.

The generator has 2 options:
2 120 Volt 20 Amp outlets, I will be using this.
1 240 Volt 20 Amp outlet, I will not be using this because of the 240 Volts.
And how much power can I expect to be drawn if there is a transient or surge?
A note about the generator is that the spin speed stays the same, even if you draw 1 Watt, or 5000.

And how much power can I expect to be drawn if there is a transient or surge?

When the ballast core saturates, the peaks could be easily about 5x the normal current (ironically more efficient ballast => higher these peaks). I doubt the generator will be able to hold it's voltage with these, most likely it will cause undervoltage dips. The exact extend depends on the generator impedance itself, but they are usually quite saggy. Don't be fooled by practically static measurements with different loads - there the voltage regulator corrects the excitatikon so, the voltage is exactly where it should be. But this regulation has very slow response (100's ms till seconds) to really cover the inrush peaks, so those nearly always lead to temporarily sagging voltage.
And it is this brief voltage sag, what inject consequent transient into the other ballast circuit and very frequently leads to arc extinction.

The fact there are two 120V branches does not mean they would be separated or so. They are both on the same core, so very tight inductively coupled to each other. And that means whatever disturbance on one branch is directly transferred to the other one as well.

A note about the generator is that the spin speed stays the same, even if you draw 1 Watt, or 5000.

That is the function of the engine governor and it would be very bad, if it does not keep it constant in the rated power range (the rpm means directly the output frequency). Of course, it is again a feedback system with a nonzero response time, so when you change the load, the rpm temporarily changes and it takes some time for the governor to settle it back. Usually a heavy flywheel helps to minimize this, but it makes the whole machine heavier as well.
And of course, if the load power (plus the generator losses) exceeds the engine power, the governor output gets stuck on full throttle and then the rpm drops down.
And what I wanted to say, the 5kW maximum load is usually imposed by the engine power, so it is the maximum, where the rpm could be still maintained under control of the governor.

Would having 2 parallel ballasts on the 240 Volt outlet help with the voltage sagging?

Would having 2 parallel ballasts on the 240 Volt outlet help with the voltage sagging?

Nope.
For the generator it does not matter at all, which way the overload spikes come from, it will always sag in the same way for the same surge power.
And the ballast will create the same surge power regardless on which voltage tap it gets its input.

So whether to use 2x120V and connect each ballast separately or 1x240V and then wire both ballasts (of course switched to 240V inputs) parallel is up to your wiring preference for the installation.
Maybe the ballast losses may differ a bit, but that you have to check.
But you can not connect one as 240V and other as 120V, that will overload one of the 120V generator branch.

Quote from: wattMaster on August 07, 2016, 10:17:10 AM

Would having 2 parallel ballasts on the 240 Volt outlet help with the voltage sagging?

Nope.
For the generator it does not matter at all, which way the overload spikes come from, it will always sag in the same way for the same surge power.
And the ballast will create the same surge power regardless on which voltage tap it gets its input.

So whether to use 2x120V and connect each ballast separately or 1x240V and then wire both ballasts (of course switched to 240V inputs) parallel is up to your wiring preference for the installation.
Maybe the ballast losses may differ a bit, but that you have to check.
But you can not connect one as 240V and other as 120V, that will overload one of the 120V generator branch.

I would use to two 120 Volt outlets, but then how to we turn them on without cycling from the voltage surges?

I would use to two 120 Volt outlets, but then how to we turn them on without cycling from the voltage surges?

When the lamps are still cold, they have pretty low arc voltage, so are inherently way more robust against these surges. In that state the lamps are capable of immediate restrike, even before the electrodes cool down again, so although it may flicker few seconds, it will settle and start after that.
The problem is when the lamps warm up, then there is not much margin. Plus once it get extinguished, it is not able to reignite immediately, so it means another cool down/ignition cycle.
So how to power them?

So how to use the thing:
Because even an apparently low load (even a 100W notebook power adapter) could have quite high inrush currents, it is not any good idea to power anything else from that generator than the two lights.
Then just wire them without any switch or so, connect them when the engine is still stopped and then power them ON/OFF by just starting/killing the engine. That will ensure all the transients happen when the lamps are still cold and so capable to restrike immediately. Since the lamps start, nothing will change abruptly in the system, so unless it becomes really unstable, it will light steadily till the gas runs out.
And that is the reason, why I said the fixtures should have no switches: My experience with such outdoor mobile installations is, sooner or later someone will accidentally touch some of the switches and that will extinguish the lamps. Having no switches means this problem becomes eliminated...