3 sp circuits off a tp isolator

[netblindpaul]

OK, typo, 20A TP mcb, sorry, same answer though, each phase of the 20A TP mcb will carry 20A without "flagging" a fault, as each phase will be independent and all connected loads will be 1ph, all of the current will return down the N, resulting on 60A of N current, why is this so hard?

 

[netblindpaul]

Where are the phase currents going to cancel themselves out in 3 single phase final circuits using a common N conductor?

 

[Tidy Max]

Because each phase is still being generated 120degrees out? If you were to run a 4 core cable for a three phase piece of equipment the neutral csa isn't 3 times the size of the phase conductor is it? Is this not the same principle?

 

[netblindpaul]

NO it is NOT!
FFS, you have 3 independent single phase circuits.
Think 3 radial socket circuits in a domestic, would you use the SAME, N conductor for all of them, each circuit could take 20A.
Let me turn this around a bit.
3x1ph mcb in a DB fed in 2.5 singles to a JB, lines only running 3 off 20A radial socket circuits, line conductors always separate.
Single 2.5 N from the DB to the JB.
From the JB, you feed each 1ph socket circuit with a separate Line & N conductor.
What current could you see in the N from the DB to the JB?

Now swap the 3x1ph MCB's for a 1x3ph MCB, the difference is?
You STILL have 3 off 1ph loads.
There is no balanced 3ph load here, where does that bit come from?
If you have 3x20A even through a 3ph TP mcb, you can draw 20A per phase through this & do this for each phase, there will be NO balancing as they ARE NOT LINKED.
FFS how hard is this.
The N will have to take the sum of the single phase currents.

 

[Tidy Max]

No because 3 radials in a domestic would all be off the same phase? Where as with a 20A tp MCB, it is three different phases?

If this is the case why do three phase installations not have a seperate neutral for each phase?

Please calm down, its not even my work.

 

[netblindpaul]

I can't calm down this is basic stuff!

What is the difference, where are the phases going to cancel?
They can't until they become a balanced 3ph circuit.
Once they are a balanced/linked 3ph circuit, then they can have the same size N.

You WILL NOT get phase balancing and cancellation on 3 single phase circuits fed from a 3ph mcb.

The 3ph circuit feeding the DB that contains the 3ph MCB feeding the 3phs mcb feeding the 3x1ph circuits will have its N conductor protected by the CPD at the origin of that circuit.

Come, I can't believe that this is not a wind up now.

 

[snowhead]

What is the difference, where are the phases going to cancel?
They can't until they become a balanced 3ph circuit.
Once they are a balanced/linked 3ph circuit, then they can have the same size N.

You WILL NOT get phase balancing and cancellation on 3 single phase circuits fed from a 3ph mcb.

I take it you don't work with 3 phase.


The Neutrals WILL balance at the point they join , an extended Star point, in the 3ph isolator.

If you can't get your head around it you'll have to either trust what others say or get a clamp meter and go and measure neutral currents in 3 ph DB's.

 

[Deleted member 72666]

NO it is NOT!
FFS, you have 3 independent single phase circuits.
Think 3 radial socket circuits in a domestic, would you use the SAME, N conductor for all of them, each circuit could take 20A.
Let me turn this around a bit.
3x1ph mcb in a DB fed in 2.5 singles to a JB, lines only running 3 off 20A radial socket circuits, line conductors always separate.
Single 2.5 N from the DB to the JB.
From the JB, you feed each 1ph socket circuit with a separate Line & N conductor.
What current could you see in the N from the DB to the JB?

Now swap the 3x1ph MCB's for a 1x3ph MCB, the difference is?
You STILL have 3 off 1ph loads.
There is no balanced 3ph load here, where does that bit come from?
If you have 3x20A even through a 3ph TP mcb, you can draw 20A per phase through this & do this for each phase, there will be NO balancing as they ARE NOT LINKED.
FFS how hard is this.
The N will have to take the sum of the single phase currents.

Paul you are wrong here chap!
The reason the Neutral can cope is because of the rotating phases (120 degrees), Or how else would you size the neutral in say a 100 amp tp & n db? Or would it need to be 3 times larger than the phases (lines)

 

[netblindpaul]

FFS snowhead, that is EXACTLY WHAT I HAVE JUST SAID!

The OP has 3 SEPARATE SINGLE PHASE CIRCUITS FROM A 3phMCB.


Oh & YES I DO work with 3ph and stuff MUCH more complex that 3ph install ****.

 

[netblindpaul]

stan,
I am NOT wrong, you have 3x2.5 that is 1x2.5 for EACH independent ph of the 3ph circuit, which ARE NOT LINKED, how can they balance FFS.
Then you have 1x2.5 N conductor carrying ALL the return current.
That is unless the OP is describing things wrongly.

If you have a 3ph & N 100A TP db then you will have the N protected by a 100A upstream over current protective device.

 

[Deleted member 72666]

That's where the square root of 3 comes into play! I think the confusion lies with the fact that they're single phase loads but in reality they are split over 3 phases and if you measure the voltages between them you can see the clearer picture. They may not be balanced but the neutral current is the sum of the 3 circuits / 1.732.
Hope that makes it a little clearer.

 

[snowhead]

FFS snowhead, that is EXACTLY WHAT I HAVE JUST SAID!


Oh & YES I DO work with 3ph and stuff MUCH more complex that 3ph install ****.

You should be able to understand this very basic situation then.

The 3 neutrals carry full current for their respective circuits up to the point they join together, in the 3 phase switch.
As the 3 neutral currents are all out of phase they then balance and the main neutral from the switch would never carry more than full load current of any 1 circuit.

 

[netblindpaul]

stan,
They are NOT balanced.

Draw the circuit the OP has described.
FFS
You have a 5 core 2.5, three phases, N & cpc.

Each phase can draw what it likes as it is independent up to 20A.
ALL return current has to flow down the N conductor as there is ONLY 1 feeding ALL 3 circuits.
You have no control over how they are loaded etc.
Yes the currents are spread over 120 degrees, but you have no control over this.
You can not say that they will balance.
In a 3ph load then yes, but in 3 x 1ph loads how do you know that the return current will be in phase, you don't know what it is feeding, all return currents could end up being in phase due to the load characteristics.

IF you have a 3ph load, then it's fine, if you have a 3ph board with say a 5c 25mm sq supply cable then that would be fine as it would be protected by an overcurrent protective device that would protect the N, plus, it would be a balanced load.
Let the MCB feeding your 3ph circuit be a 100A, to a 3ph DB, so you have 3x100A/ph with 25mm sq line & 25mm sq N, now you have no idea what the outgoing loads can be, this is OK up to 100A single phase or 300A balanced 3ph, however, it is ONLY acceptable up to the N carrying a current, including harmonics of the current carrying capacity of the N conductor for a 25mm sq 5C cable of whatever type it is in its installed configuration.
Hence why 1/2 size N conductors are no more.

You cannot guarantee that the return current flow from the loads will cancel out.

 

[netblindpaul]

You should be able to understand this very basic situation then.

The 3 neutrals carry full current for their respective circuits up to the point they join together, in the 3 phase switch.
As the 3 neutral currents are all out of phase they then balance and the main neutral from the switch would never carry more than full load current of any 1 circuit.

Where did you get 3 N conductors from the OP said there is ONLY one hence my issue, there is ONE N conductor for all 3 off single phase loads, when are you lot going to get this.
The OP stated 5C x 2.5 one core for each line, one shared N conductor, & 1 cpc, thus you have 3 single phase circuits sharing a common N conductor, each phase is protected @ 20A, the N is 2.5, you are still saying this is OK, when you have NO IDEA of the load?

 

[yellowvanman]

stan,
I am NOT wrong, you have 3x2.5 that is 1x2.5 for EACH independent ph of the 3ph circuit, which ARE NOT LINKED, how can they balance FFS.
Then you have 1x2.5 N conductor carrying ALL the return current.
That is unless the OP is describing things wrongly.

If you have a 3ph & N 100A TP db then you will have the N protected by a 100A upstream over current protective device.

You need to think about this a bit more.

First of all take a 3 phase motor - no need for a neutral because all the line currents cancel each other out.

If you draw 3 sine waves 120deg apart, when L1 is at 1.0 then L2 and L3 will cross at -0.5 (2 x -0.5= -1.0). When L2 is @ 1.0, L1 and L3 at at -0.5 and so on - hence no need for a neutral.

Now take a 3 phase distribution baord feeding various 1ph circuits. because of the diversity of the circuits there is no overall balancing of the phase currents and any inbalance has to flow down the neutral - kirkchoff law - some of currents at a node = 0.

Now take the example given for simplicity lets say each phase feeds 1 socket and the neutrals are all linked. If L1 is used current goes up L1 and returns by N. if L2 is now used same current goes up L1 get to the N, where it meets the returning L2 current, as its 120 deg out of phase then it cancels out part of the original L1 current.

So taking my example above if L1 current at an instantaneous point is 10A, L2 will be 120 deg out of phase so at that point L2 will be rising (or falling) on the sine wave @ 5A, so the net N current at that point in time will be 5A.

So in effect the addition of the phase current cancels each other because of the phase difference.

If you can't get a grasp of this basic 3 phase theory you could get yourself in bother!