3 phase Wye system with broken/missing neutral

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shespuzzling

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I have a theoretical question. Wondering what would happen in the following scenario:

3 phase wye connected grounded generator providing power to a distribution board. The distribution board has 4 wire loads (i.e. lighting) with neutral conductors. There is no neutral conductor from the board back to the generator. I'm assuming there would be ground fault tripping issues but wondering what other takes might be.

Thanks.
 
Imbalanced voltages that depend on imbalanced currents that depend on imbalanced load impedances, just like single-phase open-neutral conditions.
 
jumper said:
Maybe I am misreading. If there no neutral back to the genny, then no current flow unless you have a short circuit and current flows on the EGC.
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No, there will still be line-to-line currents because the individual phase loads are connected through the floating neutral bus in the distribution panel.
 
No current will seek the ground because there is only one earthed connection: the source neutral. If you understand what happens with an open neutral on 120/240 1ph, you should be able to figure out what would happen here.

Unless all three phases happen to be equally loaded, the voltage on the common neutral will be skewed away from zero toward the most heavily loaded phase and, to a lesser degree, toward the second most heavily loaded phase.
 
LarryFine said:
No, there will still be line-to-line currents because the individual phase loads are connected through the floating neutral bus in the distribution panel.
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Yes, you are correct. I thought I was misreading the question and I was.:ashamed1:

I was mixing up voltage imbalance, current flow, GFCI, and circuit complete path.

Oops....brain fart.... it happens.
 
Is the gen set neutral grounded?
Is the board neutral grounded?

If no to both, voltage will divide between the phases and let the magic smoke out unless loads are equal.
 
LarryFine said:
No current will seek the ground because there is only one earthed connection: the source neutral. If you understand what happens with an open neutral on 120/240 1ph, you should be able to figure out what would happen here.

Unless all three phases happen to be equally loaded, the voltage on the common neutral will be skewed away from zero toward the most heavily loaded phase and, to a lesser degree, toward the second most heavily loaded phase.
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And the result of this neutral shift, of course, is to cause the resulting neutral currents to cancel, thus satisfying K's Law of Currents at the open neutral node.
 
shespuzzling said:
Thanks all. Anybody have a good resource for what happens with an open neutral on a 120/240 system?
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mwbc_open1.gif
 

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shespuzzling said:
Thanks all. Anybody have a good resource for what happens with an open neutral on a 120/240 system?
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Even simpler: imagine a 100w 120v bulb in series with a 10w 120v bulb, and wire the two bulbs across 240v. If the center point of the two bulbs is connected to the supply neutral, each bulb receives 120v and burns as its rated power and brightness. (This is the description of the power normally coming into your home.)

Now, break the neutral connection between the supply neutral and the center point between the two bulbs, so now they are actually in series across 240v. The 100w bulb, having a much lower resistance than the 10w bulb, will force the 10w bulb to see most of the 240v and burn out rapidly from the gross over-voltage.

Now, do you have any more questions?
 
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