High 3rd harmonic on the neutral

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!

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!