Two phases to house - loss of neutral

"Trevor Wilson"

"Sillier than Anyone Else Alive "

Leaving the neutral failure issue aside, what would happen if I disabled
the Earth leakage detectors and drew power across the two phases.

**Depends on the equipment you connected. The damage might range from mild
to catastrophic.

** TW has misread the Q.

It is only about how domestic watt hour meters respond.


Would the two utility meters correctly reflect the energy I consumed?

**Briefly, yes.
** Indefinitely, actually.



..... Phil
 
On Wed, 25 Nov 2009 18:32:10 -0700, D Yuniskis
<not.going.to.be@seen.com>wrote:

PeterD wrote:
On Wed, 25 Nov 2009 17:28:50 -0700, D Yuniskis
not.going.to.be@seen.com> wrote:

Hi Sylvia,

Sylvia Else wrote:
Are you really thick or just pretending.
I'll opt for neither. So far you've said nothing relevant to the thread,
which concerns a house with two connected phases.
Have we decided that you really have two different *phases*
vs. two different *legs* (but really "single phase" -- think
center tapped transformer).

E.g., 3 phase "220" (US) is three conductors plus a ground.
If you look at any two of those conductors (ground referenced)
you would see a 120 degree PHASE lag between them.

A US residential service is "single phase, 220" in which
you have two 110V circuits ("legs") 180 degrees out of phase
with each other (i.e., the center tapped xformer concept).

So, here, half of your lighting/living circuits are on
one *leg* (so those appliances are nominally 110VAC)
and the other half on the other "leg". Some bigger
loads tend to straddle both legs (220) -- e.g., the
blower motor in the evaporative cooler. Other
appliances *use* both legs but actually as two 110V
*circuits* (i.e., higher ampacity available on two
conductors).


US standard is 240 volts, not 220.

If we want to be *pedantic*, there are LOTS of "little
errors" in this discussion! ;-)
My Yasnac/Yaskawa Acroloc CNC machine is supplied 3 phase 208 volts.
 
Sylvia Else Inscribed thus:

Baron wrote:
Sylvia Else wrote:

baron wrote:
Sylvia Else Inscribed thus:

Baron wrote:
**********************************
Don't know about the UK.
The lights would go out ! Domestic properties usually have just
a 240v single phase feed. Usually you would get a single phase
running down one side of the street and another phase running
down the other.
**********************************

I said I have two phases.

Sylvia.
Yes I know ! I answered the question above !

You think I'm mistaken about having two phases?

Sylvia.

Are you really thick or just pretending.


I'll opt for neither. So far you've said nothing relevant to the
thread, which concerns a house with two connected phases.

If you think otherwise, try explaining yourself better.

Sylvia.
I wasn't answering about two phase systems. Someone said he didn't know
about the UK. That was the bit I was replying to.

--
Best Regards:
Baron.
 
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.
 
Sylvia Else wrote:
I have to phases of power supply to my house - so three power lines, two
phases plus neutral.

I've on occasion wondered what would happen if we lost the neutral line.
It seems to me that we'd then have the voltage between the two phases
across two sets of appliances, one set attached to one phase, and the
other set attached to the other phase, with the two sets in series as a
result of their common connection to the neutral wire. Since the two
sets are unlikely to represent equal loads, the net result would be a
large overvoltage on one set of appliances.

My electrician says it's not an issue, but I can't see why.

Any thoughts?

Sylvia.
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.
 
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.
 
On Thu, 26 Nov 2009 16:46:51 -0500, Van Chocstraw
<boobooililililil@roadrunner.com> wrote:

Sylvia Else wrote:
I have to phases of power supply to my house - so three power lines, two
phases plus neutral.

I've on occasion wondered what would happen if we lost the neutral line.
It seems to me that we'd then have the voltage between the two phases
across two sets of appliances, one set attached to one phase, and the
other set attached to the other phase, with the two sets in series as a
result of their common connection to the neutral wire. Since the two
sets are unlikely to represent equal loads, the net result would be a
large overvoltage on one set of appliances.

My electrician says it's not an issue, but I can't see why.

Any thoughts?

Sylvia.

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.
Ummmm ... The O/P is in AUSTRALIA, where the supply is nominal 230V
single phase, NOT in Merka like you presumably are.
 
On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else
<sylvia@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?
 
On Thu, 26 Nov 2009 16:46:51 -0500, Van Chocstraw
<boobooililililil@roadrunner.com> wrote:

Sylvia Else wrote:
I have to phases of power supply to my house - so three power lines, two
phases plus neutral.

I've on occasion wondered what would happen if we lost the neutral line.
It seems to me that we'd then have the voltage between the two phases
across two sets of appliances, one set attached to one phase, and the
other set attached to the other phase, with the two sets in series as a
result of their common connection to the neutral wire. Since the two
sets are unlikely to represent equal loads, the net result would be a
large overvoltage on one set of appliances.

My electrician says it's not an issue, but I can't see why.

Any thoughts?

Sylvia.

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.
WHere are you located that this is true? Surely not in the US/North
America.
 
PeterD wrote:
On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else
sylvia@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.
 
On Nov 27, 7:46 am, Van Chocstraw <boobooililili...@roadrunner.com>
wrote:
Sylvia Else wrote:
I have to phases of power supply to my house - so three power lines, two
phases plus neutral.

I've on occasion wondered what would happen if we lost the neutral line.
It seems to me that we'd then have the voltage between the two phases
across two sets of appliances, one set attached to one phase, and the
other set attached to the other phase, with the two sets in series as a
result of their common connection to the neutral wire. Since the two
sets are unlikely to represent equal loads, the net result would be a
large overvoltage on one set of appliances.

My electrician says it's not an issue, but I can't see why.

Any thoughts?

Sylvia.

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.

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.
 
Sylvia Else wrote:
PeterD wrote:
On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else
sylvia@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.
Two large lamps in series.
 
F Murtz wrote:
Sylvia Else wrote:
PeterD wrote:
On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else
sylvia@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.
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.
 
On Fri, 27 Nov 2009 13:21:08 +1100, Sylvia Else
<sylvia@not.at.this.address> wrote:

PeterD wrote:
On Fri, 27 Nov 2009 10:33:40 +1100, Sylvia Else
sylvia@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?
 
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.
 
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
 
David wrote:
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.
<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.
 
PeterD wrote:
On Mon, 23 Nov 2009 18:38:47 -0700, D Yuniskis
not.going.to.be@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?
 
On Mon, 23 Nov 2009 18:38:47 -0700, D Yuniskis
<not.going.to.be@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.
 
PeterD wrote:

On Mon, 23 Nov 2009 18:38:47 -0700, D Yuniskis
not.going.to.be@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.
And how do you think imbalance is going to take place?

Think, Think, before hitting the send button.
 

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