Maker Pro
Maker Pro

Two phases to house - loss of neutral

P

PeterD

Jan 1, 1970
0
Of course not. I'll just get out the 415VAC resistive load I happen to
have lying around, and see what registers.

Sylvia.

If you can't find it, you can borrow mine... Oh, wait, too far away!
<bg> Couple of (identical) resistance electric heaters, in series?
 
B

baron

Jan 1, 1970
0
Sylvia Else Inscribed thus:
But is it how it's metered?

I haven't gone through the math, and I'd overlooked the fact that each
meter sees a power factor of less than one, so I can't say now whether
I think the meters would read correctly. But if there's a way of
looking at the problem that makes the answer obvious, I've yet to see
it.

Sylvia.

I have three meters, one for each phase.
 
D

David

Jan 1, 1970
0
Sylvia said:
Anyway, all a test would do is show that the answer is probably correct.
It wouldn't make it any more obvious.

Sylvia.

Sylvia,

Draw yourself a vector diagram. Then with some simple trigonometry it
*should* be more obvious.

Assume a resistive load between two of the three phases, with a load of
1 unit current and 1 unit voltage. The load will thus be 1 unit power.

Each single phase meter will see the in phase voltage as 1 / sqrt(3)
(240/415).

Each single phase meter will see an in phase current of 1 x cos(30).
Remember that cos(30) = 1/2 sqrt(3).

This gives the power measured by each meter as
V*I = 1/sqrt(3) * 1/2 sqrt(3)

The two square roots of three cancel, which, unsurprisingly leaves 1/2.
Thus each meter records 1/2 the power in the load, and you will thus get
billed correctly.

David
 
S

Sylvia Else

Jan 1, 1970
0
David said:
Sylvia,

Draw yourself a vector diagram. Then with some simple trigonometry it
*should* be more obvious.

<sigh>

The test wouldn't make it more obvious.

I said I hadn't done the math. The math would give the result, not make
it obvious.

Sylvia.
 
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