Ok I may have missed some thing here, say you have a tp&n supply connected to a switch fuse with say a 4 core swa feeding a tp&n db and all loads on that db are single phase, would that be incorrect? Would you oversize your neutral?
 
No it's not a tp+n supply & it's not to a db, it 3 off final circuits fed from the same 3p mcb, with a common N to all circuits, and the cct's are socket outlet circuits.
 
What local distribution network, all of these things can be found on a small domestic install these days, I'm not going to back down and no one has yet given proof that the N cannot be overloaded in the scenario given by the OP.
So until this is done I stand by my arguments, I DESIGN, I DO NOT GUESS values, so the N can be overloaded, & if I was being paid for my consultancy then I would be bothered to calculate things, as I am not, it;s down to you lot to prove otherwise, which, as yet, you have failed to do.

I never said never, I just said given the OP, it was highly unlikely.

On another note, PF is not cumulative in the sense that if a system is made up of loads at a PF of .9, the overall PF is still .9. The average domestic install is expected to average out at roughly .9 - .95, three of those domestic installs across three phases and the slightly negative effects caused by the non-linear loads within each dwelling become meaningless. The exact reason for this can be found within our discussion regarding our three 20A socket circuits.

We all suffer from a lack of understanding at times (myself included! :D) you have since seen sense but still seem to be fixated on being theoretically correct. It seems to me that you're just not looking at the bigger picture.
 
I am not talking about PF, I am mainly talking about distortion of the supply, I have never said anything about PF.
Supply distortion these days has a much bigger affect on installs than PF as this is normally taken care of, but, distortion IS cumulative.
 
Fantastic thread guys.............well done to all.

Great stuff:)

Hahaha...must have a proper read tomorrow!
The original Q just appears to be an unbalanced Star with a Neutral, what's the problem?!?! :sifone::smilielol5:
 
Hahaha...must have a proper read tomorrow!
The original Q just appears to be an unbalanced Star with a Neutral, what's the problem?!?! :sifone::smilielol5:

If you sort it out let me know mate.................I've agreed with at least 10 differing versions up till now lol
 
TO be honest, whist there seems to be a lack of belief from posters here with regard to what I am saying, there seems to be a significant lack of understanding of what I am trying to put across as well, I don;t know how else to describe things.

Thing is see, I'm not a spark, I operate mostly as an Engineer, & consultant thus I am paid for my opinions, & I don't start throwing arguments about unless I can back them up, because that is how I make my living, and I'm doing OK, thank you and have been for a few years now, based solely on my own personal opinion, which is hen backed up by fact once the money starts flowing, why should I do otherwise when it is my living.
I'm not proposing anything as stupid as payment here, I'm here for the fun of it. but, I will never post anything that I cannot back up, and no one has managed to prove with science that my arguments are incorrect yet, once they do I will give an apology, until then I stand by them.

In the OP, the N conductor CAN be overloaded.
The circuit is not acceptable.
If anyone can PROVE otherwise in all scenarios then I will bow down, until then I await proof.
 
Well I thought I got pretty ****ty at one point so I WILL apologise for that, but, I stand by my engineering arguments even if my attitude stunk!
If it did, sorry.
 
In the OP, the N conductor CAN be overloaded.
The circuit is not acceptable.
If anyone can PROVE otherwise in all scenarios then I will bow down, until then I await proof.

And pigs CAN fly, albeit thrown out of a hot air baloon strapped to a hang glider!

You are theroetically correct, just not realistic.
 
I don't doubt your conviction Paul, but in my mind, I can't see the theoretical difference between the OPs scenario, and the scenario of a feeder feeding houses on different phases in the street.
 
I think it depends on how it's wired (to state the obvious!)
Trying some sketches as we speak.
Can't see the Neutral carrying 60A though...
 
Set the circuit up and clamp it mate then tell us what the neutral was taking...........I'll trust you.

3 single phase circuits feeding 3 heaters(all on different phases.....or should that be lines !) through same mcb with one neutral for all 3 circuits.
 
Ahh, now johnny, there you have a good point, they don't comply with BS7671 as a start! ;)#
They don't have to!
Also, the scenario is again different, because they will approximate phase balance across their feeder, and, the N conductor at each house will be protected load end by the fuse in the head, & the N conductor in the street will be either earthed "PME" or protected supply end by a suitable OCPD, which will prevent unbalanced currents overloading the N.
In the OP this is not the situation.
 
OK maybe not 60A in all scenario's but it could take that depending on the loads...

Also no one has come back with a calculation that can prove that the N cannot be overloaded regardless of the connected loads yet.
 
The Sqrt(((A^2)+(B^2)+(C^2))-((AB)+(AC)+(BC))) should prove that. I'm fully aware that not all loads are linear, and socket outlet circuits will always be prone to this, whatever configuration they are installed in.

You are absolutely correct in saying that we cannot guarantee the nature of the load, but this is true for most installs nowadays.


This is an interesting debate though.
 
Sorry johnny, not convinced! ;)

Also the DNO argument is not along the same lines.

You have to look at the origin over current devices, and the circuit conductor sizes in relation to the prospective loads...
 
Draw it out, two ways of doing it, it's an unbalanced star!!!
OP best go back and make sure all his Neutral connections are secure, floating Star point is another box of frogs!
 
I'm aware that the DNO abide by the ESQCR regs (or should do) and not BS7671.
however I don't see how a standard cutout with three BS88 fuses is any different in its theory to what we have in the OP.
i can see where I think you are coming from.
are you along the lines of the neutral current from say the far sockets will sum with the neutral currents of the middle sockets and then the near sockets.
Current always sums in series, (the way the neutrals are wired) but only if they are in phase. I understand that they will not always be 120 degrees exactly out of phase because of load characteristics, but the regs have a built in safety margin to allow for the unknowns. It's not always wise to rely on them, but we are talking about assessing an existing install, not installing from new.
 
At last Archy!
Also there are other issues.
I was not going to say this, but, I did suggest this when I said that the currents could not balance without affecting the other 1ph circuits.
I can't believe how difficult it can be to get people to understand?
Also from the star point to the protective device there could be other issues...
Are there not?...
 
Paul mate, no one is arguing the point that the circuit is badly designed! We all agree, just not on the neutral overload point.
 
I still maintain that the circuit can be overloaded on the N conductor, as no-one has proven otherwise with 3 off socket outlet circuits as the point of use.
I'm not talking about PF, here either, we are at least talking harmonics & distorted N currents, which you cannot ignore.
Whilst the simplification I posted earlier I agree may have been a little "too" simple at 60A due to the 3 x 20A phase currents, no one has proven as yet that this cannot happen, & I don't believe that anyone can, when the final loads are unknown!
 
Point 1: I still maintain that the circuit can be overloaded on the N conductor, as no-one has proven otherwise with 3 off socket outlet circuits as the point of use.
Point 2: I'm not talking about PF, here either, we are at least talking harmonics & distorted N currents, which you cannot ignore.
Whilst the simplification I posted earlier I agree may have been a little "too" simple at 60A due to the 3 x 20A phase currents, no one has proven as yet that this cannot happen, & I don't believe that anyone can, when the final loads are unknown!

Point 1:
Without posting my sketches, I'm convinced that the max current that Neutral will carry is one phase loaded and the other two unloaded, as many have already said.

Point 2:
On a few 13A SOs, would that really come into play?
 
Well Archy, I'm not and no one has convinced me that the N cannot carry more than 20A without affecting the other circuits.
Also no one seems to be prepared to consider that the N current & voltage may not be in phase, nor that there may be no harmonics in the N current.
On a few 13A socket outlets yes I do believe it could come into play.

I'm not convinced & I still stand by the fact that the N currents can sum, and no one yet has proven otherwise, it is actually easy to prove they can, but I'm not here to do that.
 
Harmonics have got all the DNO's very worried these days, and for very good reason too!! Those 3 1/2 core cables they installed just about everywhere, are coming back to bite them in the arse!! Not quite the money saving solution they once thought them to be!! lol!!
 
Ok Paul, three phases (120deg out), three sockets (one on each phase), one common neutral. A 2.99kW light bulb (13A) is plugged into each socket. Will the neutral current sum to 39A?

Edit: LAMPS! I mean lamps, before you pedants jump on board! Lol.
 
Last edited:
So you are drawing 52A per phase so it will sum to more than 39A.
Remembering that you have 2x2x13A socket outlets per phase & you are now drawing 13A per outlet, that is 26A per pair, thus the circuit is non compliant, however, you would hope that the circuit protective device would operate, to protect the conductors.

However, at 26A per phase how long would the MCB take to trip, if you were to only connect a single lamp to each twin socket outlet, and would the 2.5 be rated at this for a single phase, in its installed configuration, you would hope it was, and would there be no instances where it could be possible for the N to be overloaded, possibly not in a perfectly balanced loading scenario.

However, you cannot guarantee that this is all that will be connected.

I'm not going to do the maths at this time of night, sorry, I can see where you are coming from and it is possible for the N, not to be overloaded as the circuits could be perfectly balanced. though I doubt it.

Also, I'm now getting very tired and I'm getting to the end of the computer repair I'm doing tonight so it's time for bed as I'm up @ 06:00 in the morning for work.

This will have to continue the next time I am free to debate things.

However, I'm still maintain that the N CAN be overloaded.
 
No, three 13A lamps in total. A 13A load per phase in simple.

- - - Updated - - -

No, three 13A lamps in total. A 13A load per phase simply put.
 
OK, sorry missed the one per phase bit, read it as one per socket!
So currents will be reduced as per load reduction, the loads are now perfectly balanced (within reason) so the current drawn per phase will be say for arguments sake 13A
 
I'm not convinced & I still stand by the fact that the N currents can sum, and no one yet has proven otherwise, it is actually easy to prove they can, but I'm not here to do that.

You're pulling our chains now, bog standard 3 phase theory.
 
Fancy talk like non linear loads, reactance, harmonics and fancy things like that...I'm off down the pub.
 
OK, sorry missed the one per phase bit, read it as one per socket!
So currents will be reduced as per load reduction, the loads are now perfectly balanced (within reason) so the current drawn per phase will be say for arguments sake 13A

Forgive me if I'm wrong, but it does appear to me that you're being a tad evasive with your answers now. I know what the current per phase will be, what I want to know is what you believe the current in the common neutral will be at any given moment?

Would you be so kind as to answer the question with a simple A, B or C please:

A. The current in the neutral will sum to 39A.
B. The current in the neutral will be exactly the same as the current per phase (13A).
C. The current will be somewhere between 13A and 39A.


Secondly, what I would like to know is what you think the current in the common neutral would be if ONE of the 13A loads was then disconnected from one phase?

A. The current in the neutral will sum to 26A.
B. The current in the neutral will be exactly the same as the current per phase (13A).
C. The current will be somewhere between 13A and 26A.
 
Harmonics have got all the DNO's very worried these days, and for very good reason too!! Those 3 1/2 core cables they installed just about everywhere, are coming back to bite them in the arse!! Not quite the money saving solution they once thought them to be!! lol!!

Well it serves them right in not investing in the distribution system when they saw how the power requirements were changing, which will be borne, no doubt, by the tax payer. Every installation now relies on SMPs and almost all industrial sites uses VSD control.

I believe they call it CPD for their designers and workers.
 
Muesli, btw.
 
which I believe to be a linear load. Had it been eggs, well those currents can hang around for days.
 
Again just going to point out that this isn't actually my work! Didn't expect to see 8 pages of posts!
 

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3 sp circuits off a tp isolator
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