Two phases to house - loss of neutral
in [Repair]
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From: PeterD on 27 Nov 2009 09:13 On Fri, 27 Nov 2009 13:21:08 +1100, Sylvia Else <sylvia(a)not.at.this.address> wrote: >PeterD wrote: >> On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else >> <sylvia(a)not.at.this.address> wrote: >> >>> terryc wrote: >>>> On Thu, 26 Nov 2009 22:12:24 +1100, Phil Allison wrote: >>>> >>>> >>>>>>> Would the two utility meters correctly reflect the energy I consumed? >>>>>> **Briefly, yes. >>>>> ** Indefinitely, actually. >>>> Yep, that is how 415VAC is obtained. >>> 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. >> >> >> Do it the old fashioned way: make a test! Can't be that difficult, can >> it? > >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?
From: baron on 27 Nov 2009 10:57 Sylvia Else Inscribed thus: > terryc wrote: >> On Thu, 26 Nov 2009 22:12:24 +1100, Phil Allison wrote: >> >> >>>>> Would the two utility meters correctly reflect the energy I >>>>> consumed? >>>> **Briefly, yes. >>> ** Indefinitely, actually. >> >> Yep, that is how 415VAC is obtained. > > 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. -- Best Regards: Baron.
From: PeterD on 27 Nov 2009 18:58 On Mon, 23 Nov 2009 18:38:47 -0700, D Yuniskis <not.going.to.be(a)seen.com> wrote: >This is how GFCI breakers work -- they watch for current "leaking" >off to ground someplace other than in the "return" conductor. No they don't. They look for imbalanced current flow between the two conductors.
From: David on 27 Nov 2009 19:28 Sylvia Else wrote: > > 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
From: D Yuniskis on 27 Nov 2009 19:40
PeterD wrote: > On Mon, 23 Nov 2009 18:38:47 -0700, D Yuniskis > <not.going.to.be(a)seen.com> wrote: > > >> This is how GFCI breakers work -- they watch for current "leaking" >> off to ground someplace other than in the "return" conductor. > > No they don't. They look for imbalanced current flow between the two > conductors. Gee, isn't that what I *said*? How do you get an imbalance if current isn't *leaking* off to ground someplace other than in the "return" conductor? |