View the thread, titled "3 X single phase showers" which is posted in UK Electrical Forum on Electricians Forums.

I've been asked to price 3 X 7kw showers. There is a three phase board about 40 meters away. Can anyone see an issue with 40 amp three phase MCB with 10mm four core swa to a sub board then 3 X 32 amp rcbos for each individual shower.

Cheers
 
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.
 
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 :) .
 
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
 
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
 
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 ?
 
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.
 
Last edited:
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?
 

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