What method do you use to size the neutral conductor to prevent excessive neutral shift with line to neutral loads starting? At what point would you consider over voltage to be unacceptable? Feeders and branch circuits.
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Inverter supplied induction motor it is not my specialty [in our power station even BFPump of 10 MW starts D.O.L.!]
What I know about neutral shift it is what NEMA MG1 and IEC 6003417 [up to 1000 V] recommend.
Since the voltage to ground reaches 1.86 times the rated, a motor up to 500 V rating will not be in danger. For higher than 500 mainly for 4.16 kV an isolation transformer is required. Seefor instanceToshiba Application Guide:
https://toshont.com/index.php?gfdow...TB_iframe=true

Originally posted by Julius Right View PostInverter supplied induction motor it is not my specialty [in our power station even BFPump of 10 MW starts D.O.L.!]
What I know about neutral shift it is what NEMA MG1 and IEC 6003417 [up to 1000 V] recommend.
Since the voltage to ground reaches 1.86 times the rated, a motor up to 500 V rating will not be in danger. For higher than 500 mainly for 4.16 kV an isolation transformer is required. Seefor instanceToshiba Application Guide:
https://toshont.com/index.php?gfdow...TB_iframe=true
Thanks but I had feeders and branch circuits in mind. The neutral conductor has impedance, and it goes up as the circuit run increases.
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You either need to calculate the greatest possible neutral current (presume someone is intentionally trying to create the greatest load on only one phase) or match the line conductors.Master Electrician
Electrical Contractor
Richmond, VA
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Good point. Code only address nonlinear loads with Yconnected transformers. Delta xfmrs are not considered.
However, when sizing neutrals for continuous loading, or shortcircuit current, you might be interested in the last paragraph of example D3(a) in Annex D.Roger Ramjet NoFixNoPay
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Originally posted by ramsy View PostGood point. Code only address nonlinear loads with Yconnected transformers. Delta xfmrs are not considered.
However, when sizing neutrals for continuous loading, or shortcircuit current, you might be interested in the last paragraph of example D3(a) in Annex D.
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Many or most of you probably already know this but here goes:
On a Yconnected system with balanced linear loads, at every point in time there is at least one opposing current for every nonzero phase current. So the phase currents (ideally) will cancel at the neutral connection
Nonlinear loads like rectifiers draw current only near the voltage peaks, and therefore the phase currents have short durations that don't overlap. And so there is little or no cancelation of currents at the common neutral point. Therefore the mean square current in the neutral is the sum of the three mean square currents on each phase. So for current pulses with equal value, the total neutral RMS current is sqrt (3) = 1.73... times the RMS current on one phase. If there's any overlap of the phase currents then the neutral current will be less.
I don't think the same worst case condition can exist on the line currents of a delta because there's no single conductor where all of the peak currents flow through. However, there can be circulating harmonic currents in a delta transformer that can cause other problems.Last edited by synchro; 091619, 02:22 PM.
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