Residual Grounding

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mull982

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I was reading an article on Ground fault sensing using a residual grounding detection method and had a question in regards to 3-Phase 4-Wire Residual grounding. I understand that in 3-Phase 3-Wire residual grounding under normal operation the vector sum of the phase currents is zero but do not understand why in a 3-Phase 4-Wire system the vector sum of the phase currents equals the nuetrual current. Is this because all of the phase currents return to their source via the nuetral. If this is the case, then how do the phase currents in a 3-Wire system return to their source?

I was hoping someone could shed some light on the situation and explain why in a 3-wire system the vector currents equal zero, and in a 4-wire system they equal the nuetral current?

Thanks

Mull982
 
mull982 said:
I was hoping someone could shed some light on the situation and explain why in a 3-wire system the vector currents equal zero, and in a 4-wire system they equal the nuetral current?
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It is just an optical illusion.
Simply put, in a wye system the vector sum of the phase currents always equals the neutral current. It is just that in a 3-wire system there is no neutral conductor therefore the neutral current is zero.
 
I'm assuming that in a 3-wire delta system that the vector currents just cancel each other (add to zero) after the load and do not return to the source is this correct?

In a four 4-wire system you mentioned that the vector sum of the currents equals the nuetral current. Does this mean that they add to zero and the neutral current is also zero, or do the add to a current value which actually flows through the neutral back to the source? In a 4-wire system does current flow through the nuetral back to the source?
 
mull982 said:
I'm assuming that in a 3-wire delta system that the vector currents just cancel each other (add to zero) after the load and do not return to the source is this correct?

In a four 4-wire system you mentioned that the vector sum of the currents equals the nuetral current. Does this mean that they add to zero and the neutral current is also zero, or do the add to a current value which actually flows through the neutral back to the source? In a 4-wire system does current flow through the nuetral back to the source?
Click to expand...

All current flows in a "circular" path. So even in a delta arrangement when some current flows to the load there is an equal amount returning to the source.

In a wye system, the neutral carries the unbalanced load from the three phases. For example, 10A on phase A to neutral and 0A on B and C, would result in 10A on the neutral. But, 10A on each of phase A, B, and C would result in 0A on the neutral. Remember that in the real world it is extremely hard to have exactly balanced loading so there is almost always some neutral current (things like power factor come into play).
 
You bump into a terminology issue here, and so you may be correct, if your understanding of the real physics matches the terminology that you are using :)

Current _always_ returns to its source (at least in the approximation of ignoring capacitance, and that is another terminology issue!).

Consider for a moment a single phase, two wire system. The current going out on one wire returns on the other wire, forming a closed circuit. There is no 'cancelation' going on; the current in the two conductors is exactly the same except for direction of flow.

In a three phase three wire system, the same thing is happening; but now the current is dividing amongst several paths. The current that goes _out_ on the A leg returns on some combination of the B and C legs. Current going _in_ on the A leg must be supplied by some combination of the B and C legs. All current flowing on each leg is _balanced_ by the other legs.

This is where the terminology issue comes in. If you represent the current flowing _out_ on leg A as a vector, with magnitude set by current and direction set by phase angle, and similarly represent leg B and leg C, then you will find that the sum of all three vectors is zero. This might be described as 'canceling' out; but it really just says that the current flowing on one leg is balanced by the current on the other legs.

The exact same thing remains true when you have a 4 wire system with a neutral. If you represent the current on all _4_ conductors using vectors, then the sum of all 4 vectors will be zero (this of course assuming no ground faults), and the sum of the three phase conductors must therefore balance the remainder on the neutral conductor.

-Jon
 
The question that is still confusing me, is in a 3-Wire system, how does the current return to the source? What path do each of the phase currents take back to return to the source?
 
3-wire Currents

3-wire Currents

The 3-wire discussion above is correct. At any moment in time (remember 60Hz alternating current!), the current flowing in A-phase must equal the sum of the current in B and C-phases. There is no place else for the current to flow. But, because this s Tesla's wonderful 60Hz AC with 120 degree phase shift between voltages, in the next instant, current may flow "into" B and out A and C.

With 4-wires (neutral), the system is just more complex, but the same rules apply, the total current into a "black box" (A+B+C+N) must be zero. Therefore the residual A+B+C=-N. Ignore the signs and you see that what goes in must come out.
 
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