Who do I get to change my meter tails?

[AndyPRK]

remember yours in a house.

You could have your heaters between A & N
domestic load between B & N

That would require a thicker N

 

[JohnW2]

remember yours in a house. You could have your heaters between A & N domestic load between B & N. That would require a thicker N

As I said, certainly if the loads are all resistive, that is not true. With resistive loads, the maths is such that the highest possible neutral current is the highest of the currents in any of the phase conductors - and even that extreme case only happens if only one phase is loaded.

Kind Regards, John

 

[AndyPRK]

Arh, because of the phase shift on the sine wave

 

[SUNRAY]

With resistive loads it is not possible for the neutral current to exceed any of the phase currents. a single phase loaded will result in a similar In. With 2 phases equally loaded In will be the same current as it would for a single phase but the angle will be half way between the two phases, or to put it another way - opposite the third phase.
Try adding the currents using something like matchsticks. The first time it was explained to me saw with screwdrivers.

 

[SUNRAY]

remember yours in a house.

You could have your heaters between A & N
domestic load between B & N

That would require a thicker N

No this is not so.

 

[JohnW2]

With resistive loads it is not possible for the neutral current to exceed any of the phase currents. a single phase loaded will result in a similar In.

Yes, that's what I said.

With 2 phases equally loaded In will be the same current as it would for a single phase but the angle will be half way between the two phases, or to put it another way - opposite the third phase.

I'm not totally sure what you mean by 'the same current as it would be for single phase' - but if you mean that, in that situation (with no current in third phase), the neutral current would be the same as the current in either of the phase conductors, then I agree.

With resistive loads, those are the two 'worst cases' (only one phase loaded or two phases equally loaded), in which cases the neutral current will be equal to the current in one of (or the only one of) the loaded phases. When there are loads on all three phases then (with all resistive loads), the neutral current will always be less than that in any of the phase conductors.

Kind Regards, John

 

[SUNRAY]

I was on a refurb site with a 300A 3p feed to a DB. Over night the neutral connexion failed (which I was being blamed for initially as I was the last person to leave that evening), damaging most of the AV kit I had commissioned and nearly one hundred PC's.
There were 150A fuses in the DB for the roof chiller, one of the supplies crimps had failed and the wire touched the adjacent crimp, blowing the other phases fuse and the connexion onto the main neutral bar.

The following morning there were dozens of people there. Almost in unison the ensemble cringed when the MD of the E&M company commented that as it was expected to run at 900A the metal link looked very small.

 

[333rocky333]

That reminds me of a conference room at a Holiday inn, they said that some of the clients had plugged there laptops in and the chargers were smoking, turned out to be a failed neutral, that 2 phased the socket outlets.

As for 3 phase neutral currents, I recall they can be higher than the phase current, but i dont recall how.
We fit 3 phase gantries that have 3 outlets per phase, the 9 outlets allow a maximum 32a per phase, these gantries plug in with a 6mm 4 core to a 32 amp socket.
We had a run of the Neutral pin burning up, so i looked into the matter to no avail, due to the many combinations of being unbalanced, i think a clamp meter would have been easier.
something to do with working out the Vectors Victor

 

[SUNRAY]

Yes, that's what I said.
I'm not totally sure what you mean by 'the same current as it would be for single phase' - but if you mean that, in that situation (with no current in third phase), the neutral current would be the same as the current in either of the phase conductors, then I agree.

With resistive loads, those are the two 'worst cases' (only one phase loaded or two phases equally loaded), in which cases the neutral current will be equal to the current in one of (or the only one of) the loaded phases. When there are loads on all three phases then (with all resistive loads), the neutral current will always be less than that in any of the phase conductors.

Kind Regards, John

Yeah, I didn't explain very well did I

Let's try an example. with phase 3 (240°)unused and
10A on P1 (0°), In is 10A @ 0°
Adding 10A on P2 (120°), In is 10A @ 60° (which happens to be opposite P3)
Any reduction in P2 current will reduce In to a minimum of 8.66A @ 30° when Ip2 is 5A.
It's been many years since I did any 3P calculations and I'm working from gut feeling here so apologies if there are errors in the numbers.

Any current in P3, up to the 10A will reduce In.

I'm trying to be simplistic so ignoring those who may describe the current as being antiphase to the voltage.

 

[JohnW2]

As for 3 phase neutral currents, I recall they can be higher than the phase current, but i dont recall how.

If the loads on the phases are markedly imbalanced not only in terms of magnitude but also PF, then it can happen. I doubt that such a situation is (to any significant extent) all that common, and, I would think, most unlikely to be a significant issue in a domestic or similar installation.

Kind Regards, John

 

[JohnW2]

Yeah, I didn't explain very well did I Let's try an example. with phase 3 (240°)unused and
10A on P1 (0°), In is 10A @ 0° ... Adding 10A on P2 (120°), In is 10A @ 60° (which happens to be opposite P3)

Well, it's no co-incidence - it obviously 'happens' because I1=I2 (and I3=0).

Any reduction in P2 current will reduce In to a minimum of 8.66A @ 30° when Ip2 is 5A. It's been many years since I did any 3P calculations and I'm working from gut feeling here so apologies if there are errors in the numbers.

No errors - that is correct!

Any current, up to the 10A will reduce In.

I presume you meant 'any current in P3', in which case I agree (provided, I think, that the current does not exceed 20A, not 10A). Indeed, in your simple situation, with I1=I2, In will reduce by the same amount as I3, falling to zero when I3 = 10A (i.e. perfectly balanced loads), If I3 increases above 10A, In will then start increasing, getting back to 10A (with the opposite phase) when I3 = 20A (and I1=I2=10A).

However, all our discussion has been about resistive loads. If they are reactive (and not balanced with respect to PF), then everything changes (and gets more complicated).

Kind Regards, John

 

[SUNRAY]

That reminds me of a conference room at a Holiday inn, they said that some of the clients had plugged there laptops in and the chargers were smoking, turned out to be a failed neutral, that 2 phased the socket outlets.

As for 3 phase neutral currents, I recall they can be higher than the phase current, but i dont recall how.
We fit 3 phase gantries that have 3 outlets per phase, the 9 outlets allow a maximum 32a per phase, these gantries plug in with a 6mm 4 core to a 32 amp socket.
We had a run of the Neutral pin burning up, so i looked into the matter to no avail, due to the many combinations of being unbalanced, i think a clamp meter would have been easier.
something to do with working out the Vectors Victor

Well we have only mentioned resistive loads until now, in which case no In cannot exceed the highest phase current.

If a single phase has an inductive load then the current will be at a different angle to the voltage, it will lag the voltage by a maximum of 90° (for pure inductance) so adding that to P2 of a 3 phase (120°) will put the current @ 30°, let's stick with the 10A of the previous example added to P1 will result in In of well over 19A @ 15°.
If we added a pure capacitive load of 10A on P3 the current will be at 330° with a resulting In of 29A or so @ 0°.

Again Gut feelings so numbers may be inaccurate.

BUT I stress the chances of attaining such extreme opposite loads is almost incomprehensible. Most 3Ph devices are by their nature usually balanced even if they are very inductive or capacitive.

 

[JohnW2]

BUT I stress the chances of attaining such extreme opposite loads is almost incomprehensible. Most 3Ph devices are by their nature usually balanced even if they are very inductive or capacitive.

Quite - that's much the same as I said a couple of posts ago. Even without resorting to extreme scenarios, I very much doubt that, in practice, In will hardly ever be appreciably greater than the highest of the phase currents - so I remain doubtful as to whether it is appropriate (particularly in my domestic installation) to have a neutral conductor with a higher CSA than the phase ones....

... and, in case you don't believe me, I think you can see that the ca ble in the bottom of the neutral Henley is larger than anything else ...


... and (as commented before) note the red visible insulation on one of the neutrals (now covered with grey silicone by the most recent meter changer!) - doubly noteworthy because (a) it is visible and (b) it is red!

Kind Regards, John

 

[JohnW2]

What happened too my piccie? Maybe too large ... try again ....

Kind Regards, John

 

[SUNRAY]

Quite - that's much the same as I said a couple of posts ago. Even without resorting to extreme scenarios, I very much doubt that, in practice, In will hardly ever be appreciably greater than the highest of the phase currents -

I think you are correct, and it used to be quite common to have a smaller neutral wire.

... and, in case you don't believe me, I think you can see that the ca ble in the bottom of the neutral Henley is larger than anything else ...

Kind Regards, John

I promise I have seen far too many dalliances from the norm or regs to disbelieve people.