winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Again, trying to keep it relevant to the OP's question:
The calculation requires that you determine the net 'triplen' harmonic content of the current drawn by your lamps, and then add up this content for all 3 phases. As Besoeker keeps stating, the Triple harmonic currents add on the neutral.
However as everyone else has been saying, you can approach this from the other side of the calculation to determine if there is likely a problem.
1) You look at the power factor of the lamp, or the specification for the total current drawn. The idea is to figure out the maximum 'not utilized' current going to the lamp in addition to that actually powering the lamp.
You have a 22W lamp operating at 277V. If it had unity power factor it would require a current of 0.079A. Say the datasheet tells you that the lamp actually draws 0.125A; this means that 0.051A of current is somehow 'circulating'. This could be standard 'displacement power factor', or it could be caused by harmonics. If you know the power factor; take 1/PF * <ideal current> to get the total current drawn.
2) Calculate the total current on each of your supply legs. In this case you have 108 lamps on each leg, for a current of 13.5A (this with my example of 0.125A per lamp; you need the actual datasheet value).
3) Calculate the total _unutilized_ current on each supply leg 0.051*108= 5.5A
4) Now you make the _assumption_ that _all_ of the 'not utilized' current is triplen harmonic content. This is _not_ a valid assumption, because there are almost certainly non-triplen harmonics such as 5th or 7th. The point is to do a _fast_ calculation that will tell you if you need to look closer or not. Again: triplen harmonics add, so in the worst case we get 3 * 5.5A = 16.5A on the neutral.
If you have 12AWG conductors this is likely _not_ a problem.
Factors which could push things in either direction:
The standard for small LED lamps permit 3rd harmonic content of up to 86% of fundamental!!!. Run the same calculation above, but with this 86% value and you've overloaded your neutral.
The specs for a lamp like the one you are considering state a PF of > 0.9 which suggests a lamp current of 0.088A; run the same calculation above and you will find very little current on the neutral. The 9 mA of 'unused' current is certainly no all triplen harmonic content, but even if it is there would not be a problem.
-Jon
The calculation requires that you determine the net 'triplen' harmonic content of the current drawn by your lamps, and then add up this content for all 3 phases. As Besoeker keeps stating, the Triple harmonic currents add on the neutral.
However as everyone else has been saying, you can approach this from the other side of the calculation to determine if there is likely a problem.
1) You look at the power factor of the lamp, or the specification for the total current drawn. The idea is to figure out the maximum 'not utilized' current going to the lamp in addition to that actually powering the lamp.
You have a 22W lamp operating at 277V. If it had unity power factor it would require a current of 0.079A. Say the datasheet tells you that the lamp actually draws 0.125A; this means that 0.051A of current is somehow 'circulating'. This could be standard 'displacement power factor', or it could be caused by harmonics. If you know the power factor; take 1/PF * <ideal current> to get the total current drawn.
2) Calculate the total current on each of your supply legs. In this case you have 108 lamps on each leg, for a current of 13.5A (this with my example of 0.125A per lamp; you need the actual datasheet value).
3) Calculate the total _unutilized_ current on each supply leg 0.051*108= 5.5A
4) Now you make the _assumption_ that _all_ of the 'not utilized' current is triplen harmonic content. This is _not_ a valid assumption, because there are almost certainly non-triplen harmonics such as 5th or 7th. The point is to do a _fast_ calculation that will tell you if you need to look closer or not. Again: triplen harmonics add, so in the worst case we get 3 * 5.5A = 16.5A on the neutral.
If you have 12AWG conductors this is likely _not_ a problem.
Factors which could push things in either direction:
The standard for small LED lamps permit 3rd harmonic content of up to 86% of fundamental!!!. Run the same calculation above, but with this 86% value and you've overloaded your neutral.
The specs for a lamp like the one you are considering state a PF of > 0.9 which suggests a lamp current of 0.088A; run the same calculation above and you will find very little current on the neutral. The 9 mA of 'unused' current is certainly no all triplen harmonic content, but even if it is there would not be a problem.
-Jon