Small generator for house - inverter or non-inverter?

[news.valornet.com]

Hi,

I am planning on installing a transfer switch and generator to power my
house when utility fails. My main items for power are the furnace blower
(natural gas), refrigerator, and a chest freezer. It would be nice if I
could run some lights as well.

I've measured the furnace blower while running (5.5A @ 120v) but I don't
know what its startup is yet. I have a fluke clamp meter and using the
inrush feature to measure startup current it said my refrigerator (21cf)
pulled 15.3A @ 120. Normal running is 1A although I did see it go to 3.4A
when not running, so perhaps this was the defrost mode. I have no idea
about the chest freezer yet.

I am looking at two Yamaha models, the ef2800i (120V: 20.8 A continuous,
23.3 A max) and the ef4000de (120V: 29 A continuous, 33 A max). The
ef4000de is $350 more, but does NOT provide inverted power like the ef2800i
does.

My question is, how important is the inverter? I could probably live with
the smaller output of the ef2800i if I was more careful about not using
things at the same time.

What I really don't want is damage to my home electrical items from a
non-inverted generator. Is this type of damage a common issue?

Thanks,

Alan

 

[John Gilmer]

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.

 

[TimPerry]

Hi,

I am planning on installing a transfer switch and generator to power
my house when utility fails. My main items for power are the furnace
blower (natural gas), refrigerator, and a chest freezer. It would be
nice if I could run some lights as well.

I've measured the furnace blower while running (5.5A @ 120v) but I
don't know what its startup is yet. I have a fluke clamp meter and
using the inrush feature to measure startup current it said my
refrigerator (21cf) pulled 15.3A @ 120. Normal running is 1A
although I did see it go to 3.4A when not running, so perhaps this
was the defrost mode. I have no idea about the chest freezer yet.

I am looking at two Yamaha models, the ef2800i (120V: 20.8 A
continuous,
23.3 A max) and the ef4000de (120V: 29 A continuous, 33 A max). The
ef4000de is $350 more, but does NOT provide inverted power like the
ef2800i does.

My question is, how important is the inverter?

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

with the smaller output of the ef2800i if I was more careful about
not using things at the same time.

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?

What I really don't want is damage to my home electrical items from a
non-inverted generator. Is this type of damage a common issue?

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

 

[Roby]

Hi,

I am planning on installing a transfer switch and generator to power my
house when utility fails. My main items for power are the furnace blower
(natural gas), refrigerator, and a chest freezer. It would be nice if I
could run some lights as well.

I've measured the furnace blower while running (5.5A @ 120v) but I don't
know what its startup is yet. I have a fluke clamp meter and using the
inrush feature to measure startup current it said my refrigerator (21cf)
pulled 15.3A @ 120. Normal running is 1A although I did see it go to 3.4A
when not running, so perhaps this was the defrost mode. I have no idea
about the chest freezer yet.

I am looking at two Yamaha models, the ef2800i (120V: 20.8 A continuous,
23.3 A max) and the ef4000de (120V: 29 A continuous, 33 A max). The
ef4000de is $350 more, but does NOT provide inverted power like the
ef2800i does.

My question is, how important is the inverter? I could probably live with
the smaller output of the ef2800i if I was more careful about not using
things at the same time.

What I really don't want is damage to my home electrical items from a
non-inverted generator. Is this type of damage a common issue?

Thanks,

Alan

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

 

[Spokesman]

Hi,

I am planning on installing a transfer switch and generator to power my
house when utility fails. My main items for power are the furnace blower
(natural gas), refrigerator, and a chest freezer. It would be nice if I
could run some lights as well.

I've measured the furnace blower while running (5.5A @ 120v) but I don't
know what its startup is yet. I have a fluke clamp meter and using the
inrush feature to measure startup current it said my refrigerator (21cf)
pulled 15.3A @ 120. Normal running is 1A although I did see it go to 3.4A
when not running, so perhaps this was the defrost mode. I have no idea
about the chest freezer yet.

I am looking at two Yamaha models, the ef2800i (120V: 20.8 A continuous,
23.3 A max) and the ef4000de (120V: 29 A continuous, 33 A max). The
ef4000de is $350 more, but does NOT provide inverted power like the ef2800i
does.

My question is, how important is the inverter? I could probably live with
the smaller output of the ef2800i if I was more careful about not using
things at the same time.

What I really don't want is damage to my home electrical items from a
non-inverted generator. Is this type of damage a common issue?

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.

 

[news.valornet.com]

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

 

[news.valornet.com]

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).

Thanks!!!

Alan

 

[[email protected]]

| 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?

 

[daestrom]

On Fri, 20 Oct 2006 09:25:26 -0500 news.valornet.com
| 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

 

[[email protected]]

|> 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).