Small generator for house - inverter or non-inverter?

Since you have a transfer switch, why not gut get a 5 kW plant for about $500.

You can live almost normally with that kind of power if your heat comes from gas. You can even run a window air conditioner.

If you have any problem with a 5 kW generator it will come from the "feature" than causes them to shift to idle when there is low demand. This causes a frequency change.

Some folks would use a generator 24/7 while the power is out. Most, however, just run it during "comfort hours" (7-11) and the morning (6-8). This is usually enough to keep food cold or the house from freezing (if you have gas heat).

To address your question" I would go for the "inverter" model that produces "clean" power.

Be sure and contact your local utility and get an OK for what you are doing. Some utilities have strict rules for transfer switches and alternate sources of power.

this is the type of thing the guy in the other thread with the office full of computers should be looking at.

I could probably live

lets see, you want to power your house from a 2.8 kW generator. after the trouble and expense of installing a transfer switch would it not seem reasonable to get something in the 5 to 10 kW range?

you should see what happens when a governor fails and the generator runs wild.... better yet hope the a breaker trips.

I've never seen anything that I would consider a definitive answer as to whether non-inverted generators will damage "sensitive electronics". My bet is this falls into the category of urban legends and old wive's tales, but the bottom line is I don't know.

A big question is what happens when a generator malfunctions and the voltage goes to high? What I need to do and what you should probably do is look for some good surge suppressors. There's an article on surge suppressors at:

If you have a computer, it might crash when the furnace, etc., starts up. I think people usually solve this problem with a UPS.

If noise is a big issue, I would probably go with the Honda EU3000is. It has a noise rating of 49 to 58 dBA vs. 60 to 67 for the Yamaha EF2800i. The EF4000DE is even louder at 69 dBA. In other areas, though, the Yamaha EF2800i seems to have the advantage. The cost is only about $1300 vs. $2000 and the weight is only 64lbs vs 134lbs.

An inverter allows engine speed to vary as total load changes. That should reduce fuel consumption ... and fuel might be hard to get during a widespread power outage. Seems like the average noise level would also be lower, which your neighbors would like.

A possible downside of an inverter is very special electronic replacement parts which might be both expensive and hard to find. Power electronics can be very robust, but that depends on design.

Roby

Why do you think a non-inverted sytem would damage anything? They are more reliable with less parts to go wrong and no more likely to damage your household equipment than an invertor unit. In fact many items in your house will run much better on a non-invertor generator.

That's precisely what makes them so appealing in my opinion. The Honda EU3000, for instance, has a noise rating of only 49 dBA at 1/4 load, or

700W. Depending on your situation, that means there is a very good chance you could run the generator all night long if it only powers your furnace blower fan, for instance and you and your neighbors wouldn't hear it. And it would only sip a small amount of gasoline while it's doing it.

Hi,

Thanks for all the posts guys. I have one more question which is more of a general electricity question.

In a 240V connectionm say a 14-30R, you have a ground, common, hot 1 (x), hot 2(y). Lets say you use 10 gauge wire between the inlet and the transfer switch which should be able to handle 30 amps on each line. So, the hot 1 could be 30A, and the hot 2 could be 30A, we are talking about 240*30 or

7200 watts. Lets say all the circuits in the transfer panel are 120V, so no 240V circuits. Half of them are on hot 1 (X), and the other half on hot 2 (Y).

If hot 1 is running 30A @ 120V and hot 2 is running 30A @ 120V, my question is, does the common handle 60A in this case?

Does this question make sense?

Thanks,

Alan

I'm not an electrician, but here's my unprofessional opinion on it. If you have a 240V generator, hot 1 & hot 2 will both have a potential of

120V relative to neutral and ground. However, there will be a potential of 240V between hot 1 and hot 2 since they are 180 degrees out of phase. If you connect hot 1 and hot 2 to a 240V load with a neutral, there will be very little current flowing in the neutral wire since the current will mostly cancel out.

Similarly, if you connect hot1 and hot2 to a circuit breaker panel that has only 120V loads for instance, the current will still cancel out to the extent that you have perfect sine waves that are exactly 180 degrees out of phase.

However, if you buy a 60A, 120V generator (I don't think they make them, BTW) and connect hot 1 (the only hot) to both the X and Y terminals of the hypothetical NEMA 14-30R and then plug it into your circuit breaker box, you will have a total of 60A flowing through you circuit breaker box and into a 30A common wire and the currents will all be in the same phase and there will be no cancellation and you will have an unsafe, dangerous situation.

The bottom line is if you connect 120V generator to both branches of your circuit breaker box make sure that the common wire can handle the total current that the generator is capable of producing. In addition, you should be sure to switch the breakers to the off position on any

240V loads.

Again, I'm not an electrician, so don't take all this as gospel.

Hi,

Thanks, I was thinking about this more after I posted the question and your explanation sounds just right. For example, if the circuits on one side of panel are running 30A and the other side also 30A, then because they are 180 deg out of the phase, then common wire will have almost no current. But, if one side was running 20A and the other side was running 30A, then the difference would be running on the common wire (10A).

Thanks!!!

Alan

I think that's correct. Another way to look at it is that the worst-case scenario is when all of the current is on one side. After that, they start subtracting.

In the worst case scenario, though, where you were running a lot of motors or power supplies, etc., that exhibit reactance and scramble the sine wave, you could get over 30-Amps on the 30-amp wire. This obviously isn't something people worry about, though, because when you go to Home Depot and buy 240-V wire, the neutral is always the same size as the 2 hot wires--at least I think they are.

That brings up an interesting subject that I hadn't thought about before. If you use a non-inverter generator where the sine waves are supposedly pretty ugly, are they ugly in exactly the same way on both poles? If not, would there be less cancellation with a non-inverted generator than an inverted generator? That's an interesting question, I think, but in practice I doubt if the problem would ever come up unless the wire was under-spec'd to begin with.

On Fri, 20 Oct 2006 09:25:26 -0500 news.valornet.com wrote: | Hi, |> I'm not an electrician, but here's my unprofessional opinion on it. If |> you have a 240V generator, hot 1 & hot 2 will both have a potential of |> 120V relative to neutral and ground. However, there will be a potential |> of 240V between hot 1 and hot 2 since they are 180 degrees out of |> phase. If you connect hot 1 and hot 2 to a 240V load with a neutral, |> there will be very little current flowing in the neutral wire since the |> current will mostly cancel out. |>

|> Similarly, if you connect hot1 and hot2 to a circuit breaker panel that |> has only 120V loads for instance, the current will still cancel out to |> the extent that you have perfect sine waves that are exactly 180 |> degrees out of phase. | | Thanks, I was thinking about this more after I posted the question and your | explanation sounds just right. For example, if the circuits on one side of | panel are running 30A and the other side also 30A, then because they are 180 | deg out of the phase, then common wire will have almost no current. But, if | one side was running 20A and the other side was running 30A, then the | difference would be running on the common wire (10A).

And now for the tough question. What if one side has a very inductive load of 20A (power factor 0.25) and the other side has a very capacitive load of 20A (power factor 0.25) ... now what is the current on the common wire?

Well..... Since you asked....

For the inductive load, 20A @ 0.25 means there is 5A of 'real' current and sqrt(20^2-5^2)=19.4 A of lagging reactive current. Similarly, for the capacitive load you would have 5A of 'real' current and sqrt(20^2-5^2)=19.4A of leading reactive current.

But the two supplies are '180-out' from each other. So the 5A real current in one load will exactly cancel the 5A real current in the other load. Unfortunately, because of the 180 shift, the lagging reactive current will not cancel the leading reactive current, instead they will actually add. So you'll end up with about 38.8 A of imaginary current in the neutral, possibly overloading it.

While this does overload the neutral, it's such a remote/rare possibility, that it is seldom worried about. Another 'neutral danger' is when running high harmonic loads on each phase of a wye-connected three-phase system. The harmonics from each single phase load will not cancel perfectly at the neutral and can lead to more current in the neutral than in any of the three phase conductors.

daestrom

|> And now for the tough question. What if one side has a very inductive |> load of 20A (power factor 0.25) and the other side has a very capacitive |> load of 20A (power factor 0.25) ... now what is the current on the common |> wire? |>

| | Well..... Since you asked.... | | For the inductive load, 20A @ 0.25 means there is 5A of 'real' current and | sqrt(20^2-5^2)=19.4 A of lagging reactive current. Similarly, for the | capacitive load you would have 5A of 'real' current and sqrt(20^2-5^2)=19.4A | of leading reactive current. | | But the two supplies are '180-out' from each other. So the 5A real current | in one load will exactly cancel the 5A real current in the other load. | Unfortunately, because of the 180 shift, the lagging reactive current will | not cancel the leading reactive current, instead they will actually add. So | you'll end up with about 38.8 A of imaginary current in the neutral, | possibly overloading it.

Bingo.

| While this does overload the neutral, it's such a remote/rare possibility,

Of course. Anyone stupid enough to hook up "reatively unbalanced" loads like that deserves what they get :-)

| that it is seldom worried about. Another 'neutral danger' is when running | high harmonic loads on each phase of a wye-connected three-phase system. | The harmonics from each single phase load will not cancel perfectly at the | neutral and can lead to more current in the neutral than in any of the three | phase conductors.

Delta can even potentially see a 15% current increase from such loads. That's not something you'd likely worry too much about, though.

And even single phase is not completely immune. If the loads on each side of the neutral are designed enough differently that they conduct at different times, such as one conducting before peak and the other conducting after peak, you can see this kind of overload even for single phase.

And then there is a potential issue of current peaking causing extra heating on conductors due to the I^2*R losses. Drawing 400 amps at 50% duty cycle heats the wiring more than drawing 200 amps at 100% duty cycle. Where harmonics raise the current peaks beyond what they would normally be in a sinusoidal current, there will be more heating in all the conductors. I would not expect it to be that big a deal, but it could crop up in an unexpected place some day (like somone thinking that

48 amps of lights at 25% on time for a big sign averages out to work on a 15 amp circuit and wonders why the breaker trips eventually).