Two phases to house - loss of neutral

Yep, that is how 415VAC is obtained.

With the exception of some pump motors, nothing in the house runs on

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.

Ummmm ... The O/P is in AUSTRALIA, where the supply is nominal 230V single phase, NOT in Merka like you presumably are.

Do it the old fashioned way: make a test! Can't be that difficult, can it?

WHere are you located that this is true? Surely not in the US/North America.

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

Sylvia.

Different system, there is no 110 here, only 240v and 415v

As for Sylvia, yes you would be charged for the consumption. this situation would work similarly to a 3 phase delta type load

The load, even unbalanced as it is, would be a certain VA at 415v - across the 2 phases

How the meter responds to power factor of the load in question would be the only thing that may or may not register less KWH than actually used, but this should be to the same degree as the same load in a single phase install.

IF this wasn't the case, we would have big problems regarding billing on installs having both single and 3 phase loads on the same meters in

Two large lamps in series.

I don't have any. I'd have to buy them.

Anyway, all a test would do is show that the answer is probably correct. It wouldn't make it any more obvious.

Sylvia.

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

Sylvia Else Inscribed thus:

I have three meters, one for each phase.

Sylvia,

Draw yourself a vector diagram. Then with some simple trigonometry it

Assume a resistive load between two of the three phases, with a load of

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

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.

Yep. A pair of 100W/240V lightbulbs should do the trick.