3 Phase 4 wire, Line Harmonics, Common N question

[whproject]

Theoretically speaking If I have a 3 phase 4 wire supply L1 L2 L3 and N
[FONT=verdana, arial, helvetica]And the 3 hot phases are 120 degrees offset, bar any faults and line harmonics etc, the max load on the common (shared) N, would be that of the heaviest load on a single phase right , EG[/FONT]
[FONT=verdana, arial, helvetica]L1 draws 63a L2 draws 69a and L3 draws 72a (at the same time), the load on the common N, would be 72a right, provided of coarse my 3 live phases are offset by 120 degrees.[/FONT]

[FONT=verdana, arial, helvetica]Take this hypothetical scenario, I am using 100a rated wire for the live phases, I want to use the same 100a cable for the common N, problem is, I understand [/FONT]HID[FONT=verdana, arial, helvetica] light ballast can create harmonics.
Does this mean the phases degree offset can slightly fluctuate , if so, would it be a dangerous problem when using the amperages mentioned above, to overload a common N, I hear harmonics can cause 150% of the hot load load to return through the N, would the theory be right to think that this would still only mean 150% of 72 amps, (the heaviest load on any hot phase), thus deeming the 100 amp rated wire for the common N not ideal but theoretically safe?

Thanks for your time, [/FONT]

 

[Lucien Nunes]

Without harmonics, the neutral current will be the vector sum of the line currents, so in your example would only be a few amps. Harmonics are multiples of the supply frequency, so e.g. 3rd harmonic is 150Hz, at which freq the phase angle is zero so the currents add. The total neutral current would then be the algebraic sum of the harmonic currents, plus the fundamental as before. You should do the calcs based on the ballasts you are using

 

[Knobhead]

In = √((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic))) For no harmonics.

 

[Lucien Nunes]

Thanks Tony, I can't type those symbols on my mobile, sitting here in the lorry in a quarry.

 

[Knobhead]

I have a Word document I copy and past from. Took a while to build up, I'll send it to you.

What are you doing in a quarry?

 

[ackbarthestar]

In = √((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic))) For no harmonics.

That will give you the magnitude but you then you are relying on intuition to know its displacement angle from the line conductors and the sense or direction of flow

From memory
I make the neutral about 8A leading L3 by about 40 deg

or Sqrt((63.cos90+69cos.210+72cos.330)^2 +(63.sin90+69.sin210+72.sin330)^2)

I too am unable to place extended ASCII characters onto the screen without changing the keyboard

 

[Spoon]

I have a Word document I copy and past from. Took a while to build up, I'll send it to you.

Please can you send it to me as well.... You can never have too much info...

 

[Lucien Nunes]

As for the harmonic content of the load current, you need to check it on the specific model of ballast in use. The ultimate example of currents summing in the neutral was Strand Electric's first attempt at a thyratron dimmer. To save valves they used 3-pulse rectification so the entire theatre's lighting load was returned as DC in the neutral equal to all three lines added. The cable didn't like it very much, prompting a hasty re-think.

Btw Tony I am drying out wet electronics

 

[Knobhead]

Btw Tony I am drying out wet electronics

Have fun!

Thats one ting I don't miss.

 

[DPG]

Sorry, I'm in a lorry in a quarry - sounds like some sort of Allo Allo sketch Daz

 

[Lucien Nunes]

I dont care about the candle with the handle but I could demolish the gateau from the chateau