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Re: Two phases to house - loss of neutral

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.

Re: Two phases to house - loss of neutral

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

Re: Two phases to house - loss of neutral

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

3 phase premises.

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