3 X single phase showers

[David Rann]

Picking up the big hint from @westward10 - I had a go at a phasor diagram. If phase 1 shower is on, I get 30A (say) on the distribution circuit N. If phase 1,2 and 3 showers are all on then I get 0A on N, as they all cancel out. No big surprise as it's a balanced 3 phase load at that point. So what about when 2 showers are on? Hint - I was wrong before in #14 - it's my brain that's been flogged .

I give up Wilko. Do tell

 

[David Rann]

Okay just been speaking to some that said even in the worst case scenario on an unbalanced system the neutral current will not exceed the highest of the phase. So if it's a 30 amp circuit then the maxmima neutral current is 30amp.

So many different responses and its clear I'm not the only one confused on this.

Cheers

 

[Andy78]

Okay just been speaking to some that said even in the worst case scenario on an unbalanced system the neutral current will not exceed the highest of the phase. So if it's a 30 amp circuit then the maxmima neutral current is 30amp.

So many different responses and its clear I'm not the only one confused on this.

Cheers

This rings a bell, but honestly it's not a calculation I have performed many times since originally learning it. Makes more sense to me than the neutral current being the pure sum of the line currents.

I'd honestly have to dig out the books to draw the diagram properly but it shouldn't be too complex as the resistive loads should be considered in phase, i.e no lag or lead angles to adjust for ? Am I right ?

 

[marconi]

A shower passes a current of Iphase.

There are 3 phase currents I1, I2 and I3

Taking I1 as reference at 0deg, the I2 is =120 phase angle and I3 is 240deg (or -120deg)

Assume two showers on using I2 and I3, then In is I2 + I3 =

Iphase(cos120 +jsin120) + Iphase (cos-120 +jsin-120) =

Iphase (-0.5 + j0.58) + Iphase(-0.5 -j0.58)=

Ip (-0.5 + j0.58 -0.5 -j0.58) = -Iphase = -7000/240.

If you now add in a switched on I1 fed shower In = I1 + I2 + I3 =

Iphase-Iphase = 0

So, one shower magnitude of In = 7000/240A

Two showers magnitude of In = 7000/240A (but in anti-phase to the current the third off shower would pass).

Three showers In = 0A

Thus In is 7000/240 A for one or two showers or 0A fro three showers.

No need for N csa to be larger than line csa.

 

[David Rann]

A shower passes a current of Iphase.

There are 3 phase currents I1, I2 and I3

Taking I1 as reference at 0deg, the I2 is =120 phase angle and I3 is 240deg (or -120deg)

Assume two showers on using I2 and I3, then In is I2 + I3 =

Iphase(cos120 +jsin120) + Iphase (cos-120 +jsin-120) =

Iphase (-0.5 + j0.58) + Iphase(-0.5 -j0.58)=

Ip (-0.5 + j0.58 -0.5 -j0.58) = -Iphase = -7000/240.

If you now add in a switched on I1 fed shower In = I1 + I2 + I3 =

Iphase-Iphase = 0

Huh?

 

[westward10]

If your selected cable is suitable for the 40A triple-pole device then the neutral conductor will be adequate.

 

[marconi]

 

[marconi]

View attachment 44713

 

[westward10]

The question is Marconi what will the neutral current be with two 30A loads across two phases but no load in the third phase. People are stating double the line current at 60A, this is not correct.

 

[marconi]

The question is Marconi what will the neutral current be with two 30A loads across two phases but no load in the third phase. People are stating double the line current at 60A, this is not correct.

The magnitude of In will be 30A for one shower on, 30A for two showers on and 0A for three showers on.

 

[David Rann]

Thanks Marconi! Thanks evreyone

 

[westward10]

Delete, total miscalculation

 

[davesparks]

Got it - in the case where the 4 conductors are the same ccc but fully loaded and all the loads are single phase only then N will be flogged surely (?).

No, with each phase loaded equally the neutral current will be zero.

 

[Andy78]

After a bit of digging I'm thinking this will be appropriate for unbalanced loads of a resistive nature ?

Edit: image of formula won't show

This would give the following
1 shower on - N current = line current
2 showers on - N current = just slightly less than line current, but near enough.

This is turning into a bit of a revision session for me so please correct me if I'm wrong.

 

[davesparks]

If, you’ve stopped posting for a bit, I’ll try to explain.
25mm2 is over sized for 100A to allow for imbalances.
The outgoing circuits from the 3 phase board may all be single phase, 3phase or a mixture of both.
The conductors for each outgoing circuit should be sized appropriately.
When the neutral current from phase L1 gets back to the board, it can now use the L2 neutral conductor to flow to the L2 phase, and the neutral current from the L2 phase can use the L3 neutral conductor to flow to the L3 phase, etc.

So once the neutral currents get back to a single point, they can then start to balance themselves out.

In the OP’s situation, there is a very good chance that the 3 showers will not be used at the same time.
However there is a chance that 2 will be used at the same time.
If only 1 or 2 are used, the neutral currents will not be able to balance out.
If all 3 are used at the same time, they will balance out.
As such you need to size the neutral conductor to handle up to 2 lots of current.

Utter nonsense, where on earth did you get this from?

With simplevresistive loads like these showers the neutral current will be at its highest when only one phase is loaded where it will be equal to the loaded phase. any load on one or both of the other two phases will reduce the neutral current.

 

[snowhead]

Can anyone who thinks that the Neutral cable from a 3 phase DB, whether the DB has all 3ph loads or a mixture or all single phase, has any possibility of having to carry more than the load on any 1 phase then please show the rest of us non believers where 4 core S.W.A with a larger neutral cable can be found or identify the reg prohibiting the use of 4 (equal) core SWA to a 3 ph D.B.

 

[westward10]

Yes we would have triple pole boards with neutral supply conductors ten times the size of the lines.

 

[Wilko]

Hi - I now think 30A for 1, 30A for 2 and 0A for 3. From the vector diagram of the currents, have all 3 running, then turn one off (the dashed line), what is the result? A 30A vector at 60 degrees (the dotted line).

 

[Andy78]

Thanks to the OP for this thread.
Always good to be reminded of what I've forgotten through not using it and get the bits clicked back into place.

 

[Lucien Nunes]

Crikey, it's a long time since I saw so much gibberish condensed into one thread. Reminds me of a nursery rhyme;
Up and down the single phase
In and out of the neutral
That's the way the current goes
Pop! goes the cable.

The neutral current will never exceed the highest line current provided the loads have either unity power factor (as in this case, with resistive heating in the showers), or a non-unity power factor caused by phase displacement only, similar on all phases. In practice, with a mixed load it is very unlikely that the neutral current will ever reach the circuit rated current.

The neutral current can easily exceed the actual line current with phase-angle controlled loads such as dimmed tungsten lighting. The waveform distortion introduces a third-harmonic component which adds in the neutral rather than cancelling. Also any assymmetry, such as half-wave rectification, can produce additive harmonics. If the harmonic distortion is severe, and can take place when the line currents are at maximum, then the neutral conductor may need to be larger than the lines.