3-phase fault 1

[KVX10]

I'd appreciate it for some feedback/explanation of the following occurance...

Presently there is a 3 phase 4 wire Y feeding a new subdivision at the end of the line with phase "a" and "c". On the way to the subdivision, there is a pump station with a 120/208 service that is radial fed. Before this there are 3 single phase reclosers that feed all of the above.

The issue: "a" phase recloser opens by a fault. The subdivision that is fed by this phase is not occupied, there are 4 empty houses with temp services and a worker complains that his tools are not working. 55v is measured at the transformer.

We investigated and found that the "a" phase was being backfed through a heating element in the pump station and continuing to the subdivision which virtually has no load besides the transformer it backfed through at the pump station and the transformer at the subdivision.

My explanation: is that because there is "no-load" at the subdivision, the breaker for the heating element will not trip and backfeed will continue until the heater is turned off or the other recloser is opened that is feeding the other phase of the pump station transformer.

Correct me if I'm wrong.
Thanks.

 

[waross]

The heater is back feeding the "A" phase transformer at the pumping station. The recloser is open so the sub does not see anything on "A" phase.
Now you have several impedances in series and possibly some in parallel.
The heater is supplied from a healthy phase and is the first impedance. The "A" phase transformer windings at the pump house are the second and third impedances. The fourth and fifth impedances are the transformer windings feeding the workers tool. The tool is the sixth impedance.
The primaries of any other transformers on this line will be in parallel with the fourth fifth, and sixth impedances.
Let's ignore other transformers for the sake of explanation.
A current flows from "B" phase through the heater and and through the transformer secondary to the neutral. You have two impedances in series and the voltage across each will equal the current times the impedance. There will probably be about 55 volts across the transformer winding and possibly almost 100 volts across the heater. Don't expect the voltages to add to 120 volts, the phase angles are almost 90 degrees apart.
The impedance of the transformer secondary will change to reflect any load on the transformer. The voltage would probably have been less than 55 volts if the tool was operated as the voltage reading was taken.
The primary voltage on "A" phase would have been about 3400 volts on a 13,000 volts system. That is why linemen never assume that anything is "dead".
The heater breaker, the heater is seeing about 100 volts instead of 208 volts. Current will be about 1/2 normal. The breaker won't trip on that current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

Bill's explanation is good if the pump station is served with a grd wye - grd wye transformer. If it is served by a delta-wye transformer, then another explanation of the voltage on "A" phase is:

With "A" phase open, there is voltage on the pump station transformer "A" phase because it is connected between two windings that are energized on "B" and "C" phases. The voltage on "A" phase will be 1/2 of normal. This will energize the rest of the opened "A" phase to the subdivision.

 

[KVX10]

Thanks Bill and jghrist,
Bill's explanation is along the lines of what I believed what is going on...with extra detail! Thank you for your comments. Could be awhile before a decent load will be picked up as the housing market is pretty bad.