nawao
Member
- Location
- Abuja - Nigeria
If one phase of the high legs on a delta is missing, what will be the 120/208 relationship on a wye bank
Phase missing
- Thread starter nawao
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boboelectric
Senior Member
- Location
- Eighty Four,Pa.15330
It would be missing. You might have motors and lights in trouble.Some undervoltage,making them over current,(single phasing).
Volta
Senior Member
- Location
- Columbus, Ohio
What system are you speaking of? A high-leg delta, or a wye?
If a phase is out on a wye, you still have two 120 volt and one 208 volt connections.
If high-leg delta, which phase?
Is the delta a transformer primary to the wye?
- Location
- Wisconsin
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- PE (Retired) - Power Systems
Are you asking what the wye secondary voltages would be if the delta primary lost 1 phase? Or, are you asking what they would be if you lost one of the delta primary windings, but keep all 3-phases?
nawao
Member
- Location
- Abuja - Nigeria
Yes, this is exactly what I'm asking for
ptonsparky
Tom
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- EC - retired
You loose voltage on one of the secondary phases, for three phase motor loads on the secondary, the load "single phases".
I suspect the exact voltage ph to ph and ph to neut will depend entirely on the type of loads you have connected.
I suspect the exact voltage ph to ph and ph to neut will depend entirely on the type of loads you have connected.
1 line or 1 winding?
If you lost 1 line on the primary you will lose 2 lines on the secondary.
If you lost 1 winding on the primary, you will lose 1 line on the secondary
If one of the primary lines opens, you are left with a single phase system. If we lost Vb for example,
Van = 120V @ phi
Vbn = Vcn = 104V @ phi +/- 180
This occurs because two of the primaries are now in series, driven by Vac.
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My earlier statement would be more accurate if it were stated, "If you lost 1 of the primary lines, 2 lines of the secondary will have reduced voltage and no longer have a 120? out-of-phase relationship."
I agree with your phasing assessment. However, I believe it would be less than 104V, because the two primary windings would be operating at half voltage. Say on a 480 Delta to 208Y/120 xfmr, the primary windings typically fed by the lost line would normally have 480V across each. With that line missing, they would have 480V across both in series, thus 240V across each. That figures out to 60V to Neutral output on the affected secondary windings.
Yer Right!
Yer Right!
Oh I hate to be wrong! Worse yet, I hate to admit being wrong!
And, Vbc = 0
Yer Right!
Oh I hate to be wrong! Worse yet, I hate to admit being wrong!
And, Vbc = 0
I know the feeling :grin:
Hmmm... Hate to say it, but wouldn't line-to-line secondary voltages be 180, 120, and 60? That is, Van = 120V, then + and – 60V for the two other affect windings.
Vab = Van + Vnb = 120 + -60 = 60V
Vbc = Vbn + Vnc = 60 + 60 = 120V
Vac = Van + Vnc = 120 + 60 = 180V
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nawao
Member
- Location
- Abuja - Nigeria
How did you calculate Vbn = Vcn = 104V @ phi +/- 180?
Not quite:
Not quite:
Look carefully at the polarity dots. Both Vbn and Vcn are 180 out from Van.
Not quite:
Look carefully at the polarity dots. Both Vbn and Vcn are 180 out from Van.
Mistake:
Mistake:
As Smart pointed out, these voltages should be 60V because their primaries are in series with a single phase voltage applied. The phase angles are determined by inspection of the primary and secondary polarity dots.
Mistake:
As Smart pointed out, these voltages should be 60V because their primaries are in series with a single phase voltage applied. The phase angles are determined by inspection of the primary and secondary polarity dots.
OK... my turn to say I was wrong
Vab = Van + Vnb = 120 + 60 = 180V
Vbc = Vbn + Vnc = -60 + 60 = 0V
Vac = Van + Vnc = 120 + 60 = 180V
That's OK:
That's OK:
That's OK Smart. I made a mistake once--thought I was wrong!
That's OK:
OK... my turn to say I was wrong
Vab = Van + Vnb = 120 + 60 = 180V
Vbc = Vbn + Vnc = -60 + 60 = 0V
Vac = Van + Vnc = 120 + 60 = 180V
That's OK Smart. I made a mistake once--thought I was wrong!
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