| Installing an outdoor construction power transformer...4160 Delta/600 Delta.
|
| Need phase ground lamps on the 600V side.
My non-engineering guess:
Get 4 347-volt bulbs. Wire 3 of them in a WYE arrangement to the three
phase lines. Wire the 4th bulb between the star point and ground. Label
the first three A and B and C. Label the 4th as G. You should see dim
light in A and B and C, and no light in G, in the normal case. In the
case of one phase faulting to ground, you should see them deviate towards
no light in the faulted phase, and light in G, varying depending on the
degree of fault impedance.
This setup is subject to false negatives. If bulb G has burned out, it
won't show a fault. Multiple bulbs in parallel helps reduce the risk,
but 347-volt bulbs are not cheap, and do poorly at low wattage. These
would need to be incandescent bulbs.
Three test switches that connect a heavy duty impedance to ground (be
sure to choose values that won't cause a problem if 2 or 3 switches
are closed at the same time) can be used to verify if the bulbs are
working.
| Anyone know of a circuit diagram...specs for a simple phase loss lamp box?
Loss of phase on the primary would give you mixed voltages at single
phase on the secondary. If phase B goes out, you would see 300 volts
on A-B and B-C. The simple circuit would be 3 600 volt bulbs in delta,
and watch for 2 of them going very dim; their common phase being the
lost one.
http://ka9wgn.ham.org/ |
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Your idea of a Y set of lamps is good. However, there is no need for a
neutral to ground lamp and use of such may lead to questionable situations.
Suppose we have what you describe. Normally, the 3 lamps would see 347V and
the G lamp would be dark as there would be little or no neutral current to
ground. This is as you say. Now consider a solid fault on phase C. There are
two parallel impedances (lamps) between neutral and ground. Effectively C
would be at ground potential. The result would be something like 150V
between C and the neutral point while the A to N and B to N voltages would
be about 370V. The ground and C lamps would be of equal brightness. If the
ground impedance was infinite, then we have the normal situation. The
voltage of the C lamp under fault would be between 347 and 150V while the
voltage of the A and B lamps would be between 370 and 347V corresponding- I
haven't worked out the relative brightness balance between the ground and C
lamps in intermediate cases.
If phase C was open, then the AC and BC voltages would be 300V as you have
indicated but the C to neutral voltage would be 0 . As there is no ground
loop, there will be no neutral to ground current so the neutral will be at
ground potential (and A, B will be at 300V to ground).
Of course, any other grounded loads on the system will have a considerable
effect.
Suppose that the neutral is tied directly to ground with no lamp. This has
the effect of holding phase to ground voltages at 347V. In normal operation,
the 3 lamps will see 347V. If phase C is solidly grounded, then the C lamp
will be out and the A and B lamps will see 600V. As long as fault impedance
is low compared to the lamp impedance, this would be the approximate case.
Again with infinite ground impedance, the normal situation results.
In the case of an open on phase C, the situation would be the same as with a
lamp in the neutral to ground path.
Saving is 1 lamp and probably a more sure indication of a fault. Problem is
that the lamps must take 600V. Neons with resistors might work better than,
say, 5 -120V bulbs in series.
Anyhow- check this out- no guarantees are given for the analysis which was
done fairly crudely and may be affected by wine
.
