High 3rd harmonic on the neutral

Is it correct that the effect of the harmonic is to shift part of the neutral current out of sync so it’s returning with the neutral current of another phase adding to the overall neutral current from where they are combined in a three phase distribution board busbar?


Andy Betteridge

Is the question behind the original post due to a reduced size neutral conductor having been installed and it is now running too hot?


Andy Betteridge

"So how can you cancel single phase return currents down to zero in the neutral conductor?


Can someone sketch out the layout of a building installation such as this with a X to mark the point where the neutral currents cancel each other out?"


You can't. But then 3rd harmonic current isn't an issue in a single phase installation. It is only when you get out into the street that the 50Hz neutral currents start to cancel but the 150HZ neutral currents are additive.

Consider a 3-phase DB with 3 equally-loaded single-phase resistive circuits currently active (say a heater in each of 3 rooms). The individual SP circuit neutrals will of course take the full current of one heater. Where the 3 neutrals join and become a single neutral at the DB's N bar, it becomes more interesting. At the moment in time where L1 is at the peak of the cycle, there is a current I1 say flowing back from the L1 heater. Consider what is happening to L2 at this exact moment. The voltage (and hence current, for a resistive load) are delayed by 120 degrees compared to L1. So when L1 is at its positive peak, L2 is at -0.5 of peak, so I2 = -0.5I1. Similarly, I3 = -0.5I1 at this moment in time. So there is a big current flowing back into the neutral bar from the L1 heater, and two half-sized currents flowing from the neutral bar out to L2 and L3 heaters. So there's a high current on the neutral bar, but zero current flowing in or out of the DB's supply N.


Over time, the voltages and hence currents on L1,L2,L3 vary in a sine wave, but at all times the sum of the 3 neutral currents is zero.

This isn't something which I would normally think about, but I think that the penny has dropped.


As WW says above, at the fundamental frequency of 50 Hz, when L1 is at its peak, the vector is pointing straight upwards. 120 degrees behind, L2's vector is pointing to the left and down by half as much. L3's is pointing to the right and also down by half as much. The two halves of L2 and L3 cancel out L1's vertical component; and the left and right components of L2 and L3 cancel each other. So if L1 = L2 = L3, there is no current at the fundamental frequency in the neutral.


Now introduce a harmonic at 150 Hz i.e. 3 times the fundamental. Because they are now going three times as fast, the vectors are not 120 degrees out of phase with each other, but in phase so they add up.


ETA: unless I have forgotten all my mathematics, the neutral current can never be more than 300% of the lines (assuming that they are equally loaded) and even then, the whole current would be a third (or other triplen) harmonic. Even if a single piece of equipment could cause the current drawn to be entirely at the third harmonic, it would be very odd indeed if all of the many occupiers' equipment did the same thing.


However, if L1 = L2 = L3 = 100 A and N = 50 A, only about 17% of the load is producing the 3rd harmonic. 83 A in each phase cancel each other out; the remaining 17 A are additive and 17 x 3 = 50 (approximately).


HTH!

The thing that is going around in my head is trying to balance three single phase supplies connected to a three phase supply is chasing unicorns.


A large care home I did work at had a three phase supply, but three single phase meters and installations around the building. You had to be careful what you connected to what, but any neutral current issues are in the DNO network, not in the installations.


With three single phase installations connected to a three phase distribution board within the installation then presumably from what has been said when all the single phase installations are in use the neutral bar in the distribution board becomes a star point and because it’s AC with a positive and minus cycle current neutral flows back through the phases.


We all assume when we open up a single phase consumer unit the neutral bar is tied to earth and has negligible voltage on it although it could a lot of current flowing through it, you can touch it but try removing a conductor under load and sparks will fly with the end of the conductor immediately coming up to 240 volts.


With the three phase board as well as being tied down to earth the voltages should cancel as they are on different parts of the cycle? In addition the current should cancel in the neutral conductor by feeding back into one or more phases? But it will only cancel completely if the single phase loads are perfectly matched?


That’s three questions ?


Then the harmonics throw it all out of balance.


Andy B.

In the TP DB with SP circuits example (ignoring harmonics), what you might have at a particular snaphot in time for three nearly but not quite equal loads is: 50A flowing into the N bar from circuit 1's N, 15A flowing out from the N bar to circuit 2's N, 25A to circuit 3, and 10A flowing from the N bar back up the TP supply neutral. The N bar is nearly at earth potential, drifting by whatever voltage drop there is along the supply N to the earthing point due to the 10A current flowing along it.


Over the course of the cycle, the 50/-15/-25 vary, but always add up to +10 (which represents the load imbalance), so there's a constant 10A along the supply N. That's the beauty of TP; compared with 3 SP supplies -each with a full-size N - by combining the N conductors the currents mostly cancel each other out and the combined N can be a lot smaller, or indeed removed completely when you know the load will be fully balanced.

I don't think anyone tries to balance three single phase loads.


The important issue is that, without harmonics , the neutral current cannot be larger than the largest single phase current. Any additional loads on a different phase reduce the neutral current so the neutral does not need to be larger than the phase conductors.


Now start to consider harmonics and this logic goes out the window as, mathematically, the neutral current can be twice the largest phase current.

It’s over twenty years since I did three phase theory, so feel free to add a few sketches and notes to jog my memory, as I’m not getting a picture in my head of where the neutral current is supposed to cancel in this installation.

If it helps any, and all the sine wave stuff isn't exactly obvious, try thinking of a simpler situation as a means of introduction...


Say we had a 3-wire d.c. system - +12V - 0V - 12V say with two sets of 12V loads - some connected between +12V and 0V and the rest connected between 0V and -12V.  Say the first group of loads were drawing a total of 10A and the second group 8A. Hopefully it should be 'obvious' that the +12V supply line needs to deliver 10A, the -12V line 8A. If you then think about the point where the -v side of the first group of loads and the +ve side of the second group of loads meet (at the load end of the 0V supply wire) then we have 10A 'in' and 8A 'out' to/from the loads - and as we know that the total of currents into and out of a point must sum to zero, you should be able to deduce that the 0V wire must carry 2A back to the supply. (Probably helps to draw a diagram). So that's a simple example of currents cancelling in a common conductor. If both sets of loads drew 10A then you'd have 0A flowing in the 0V conductor.


I struggled with sine waves cancelling until it eventually dawned on me that the result of adding two pure sine waves of the same frequency together was always a pure sign wave - it doesn't matter if they're out of sync with each other or of different amplitudes you still get a pure sign wave. Nothing I've read seemed to clearly state that, but without that observation much of the rest of the explanations didn't make much sense to me. Shifting a sine wave by 180 degrees is the same as negating it - so adding two sine waves 180 degrees out of phase with each other is the same as subjecting one from the other. So subtracting one pure sine wave from another (of the same frequency) also always produces anther pure sign wave. Those with mathematical training probably find such things obvious, but it took me a long while to figure it out.


Add two sine waves together - one offset by 120 degrees (L2) and the other offset by 240 degree (L3) and you end up with a sine wave offset by 180 degrees - i.e. the exact opposite of one with a 0 degree offset (L1) so adding in L1 brings to total to a pure zero-amplitude sine wave (or nothing at all, if you prefer).


   - Andy.