wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
OK, I've been trying to understand exactly what you mean by "random" loading conditions, so I checked the following version of what the above might mean:I will add that, a decade or three ago, I did a detailed analysis of the heat generated by all three phases and the neutral under random loading (balanced and not balanced) loading conditions. I proved that if the three phase loads are not equal (i.e., some neutral current), the total heat generated by the four wires will be lower than the heat generated if the three phases alone (i.e., no neutral current).
For a wye secondary, the currents IA, IB, IC, and IN will sum to zero as phasors for an appropriate sign convention. So take IA and IB to be of random magnitude uniformly distributed between 0 and some fixed maximum current, which can be taken to be 1 without loss of generality, and of zero phase shift (resistive loads). Then we have two cases:
(1) IN = 0, so IC = - IA - IB. Here IC will have a non-zero phase shift unless | IA | = | IB |. Then the expected value of |IA|2 + |IB|2 + |IC|2 = 13/12. (Compared to the maximum possible value of 3 when IA, IB and IC are all maximum).
(2) Instead IC is also of random magnitude uniformly distributed between 0 and 1, and of zero phase shift. Then IN = - IA - IB - IC. The expected value of |IA|2 + |IB|2 + |IC|2 + |IN|2 = 23/16.
Now 23/16 > 13/12, so for this formulation of "random loading", the heating is less when there is no neutral current. Did you mean something else? I could try allowing random phase shifts for the line currents, between some chosen bounds, although then the integration for the expected values would be at least twice as complex
Cheers, Wayne
P.S. If the "no neutral current" case is supposed to also have a resistive load for IC, i.e. there is only one degree of freedom as all the line currents are of equal magnitude, then the expectation value of the sum of square currents is just 1, so still less than case (2) above.