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.