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In article , (Jeff Wisnia) writes:
| Dan Lanciani wrote:
| In article .com, (kevin) writes:
|
| | So can anyone answer the op's original question? Does a mechanical
| | meter overestimate when the load is imbalanced?
|
| Yes, in the sense that the typical 4-terminal (1.5 element) meter most
| often used in residential split-phase service charges the customer for
| 150% of the losses in the neutral on the utility's side of the meter.
| (That's the simple case of one service drop from transformer to customer.
| For multiple connections the analysis gets more complicated.) This seems
| to surprise many people. Google for Blondel's theorem before trying to
| come up with an argument that the meter does not make this error.

|
|
| I believe you,

That's comforting.

I've had some incredibly long threads (last in the
alt.electrical.engineering group I think) trying to convince people that
the meter could be wrong.

| and If I understood you, when speaking of mechanical
| meters the 1.5 element meter you describe has one voltage field across
| the 240 volt line and two current fields, one in series with each "hot"
| conductor.

Yes, and I suspect most electronic meters are the same. After all,
they plug into the same meter base as the mechanical meter they replace
and there is usually no way for them to connect to ground/neutral
reliably.

| I think I see what you're referring to.
|
| Reducing it to absurdity, if you only load one side of the line, all the
| return current flows through the neutral, but the voltage field still
| "sees" the full 240 volts at the meter terminals even though the power
| you are using isn't the product of the current drawn time half of the
| 240 that voltage, it's the current times half of that voltage MINUS the
| voltage drop in the neutral.

Yes, though there will be some measured drop in the hot line as well so
if hot-to-hot started out as 240V at the transformer it won't be 240V
at the meter. It works out to your paying for 150% of the loss in the
neutral: the sum of all the loss in the neutral itself plus half the
loss in the loaded hot leg.

Now on a multi-drop setup you could in theory make this work to your
advantage. Take note of which hot leg appears higher with respect to
the neutral (on account of your neighbors' unbalanced loads) and shift
your own loads to this leg (but no more than will tend to pull the
neutral to the other side of center). Now you are getting free power
to the same extent of the metering error and at the expense of your
neighbors. The power company always wins (or at worst breaks even if
everything is perfectly balanced), though, which may be part of the
reason they like undersized neutrals.

| N.B. This has nothing to do with the meter's being mechanical. You can
| build a 5-terminal (2 element) mechanical meter that does not have the
| error.
|
| I'm guessing that kind of meter must have TWO voltage fields?

Yes, the terminology seems to be (1 element) = (1 voltage field plus 1
current field). I don't really like this since it seems to me that
"1.5 elements" could be ambiguous. Is the extra element a voltage or
a current winding?

Dan Lanciani
ddl@danlan.*com