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Re: Two phases to house - loss of neutral



"Trevor Wilson"

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** TW has misread the Q.

 It is only about how domestic watt hour meters respond.


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** Indefinitely, actually.



....   Phil



Re: Two phases to house - loss of neutral




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Yep, that is how 415VAC is obtained.

Re: Two phases to house - loss of neutral


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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


On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else

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

Re: Two phases to house - loss of neutral


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

Sylvia.

Re: Two phases to house - loss of neutral


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Two large lamps in series.


Re: Two phases to house - loss of neutral


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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.

Re: Two phases to house - loss of neutral


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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


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<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.

Re: Two phases to house - loss of neutral


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

--
    W
  . | ,. w ,   "Some people are alive only because
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Re: Two phases to house - loss of neutral


On Fri, 27 Nov 2009 13:21:08 +1100, Sylvia Else

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

Re: Two phases to house - loss of neutral


Sylvia Else Inscribed thus:

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I have three meters, one for each phase.

--
Best Regards:
                Baron.

Re: Two phases to house - loss of neutral


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With the exception of some pump motors, nothing in the house runs on
220. Even the stove splits 110 to the burners and 110 to the oven. No
neutral and no 110's to anything.

Re: Two phases to house - loss of neutral
On Thu, 26 Nov 2009 16:46:51 -0500, Van Chocstraw

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

Re: Two phases to house - loss of neutral
On Thu, 26 Nov 2009 16:46:51 -0500, Van Chocstraw

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

Re: Two phases to house - loss of neutral
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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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