480 Delta-Wye

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infinity

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I have a 480 volt, 10 kva Delta-Wye transformer to create a neutral for an AC unit. Primary OCPD is 15 amps. Does this require secondary protection?
 
The secondary conductors are tap conductors and require overcurrent protection.
 
infinity said:
I have a 480 volt, 10 kva Delta-Wye transformer to create a neutral for an AC unit. Primary OCPD is 15 amps. Does this require secondary protection?
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It's rather convoluted but see Section 240.4(F):
(F) Transformer Secondary Conductors.

Single-phase (other than 2-wire) and multiphase (other than delta-delta, 3-wire) transformer secondary conductors shall not be considered to be protected by the primary overcurrent protective device...
Click to expand...
 
I said it was required but I'm being asked why? It's a 15 amp OCPD on the primary feeding a 15 amp load so the question is would the primary trip with no secondary OCPD.
 
Only two-wire single-phase or three-wire, three-phase (delta-delta) truly reflect "proportional" secondary to primary currents. Otherwise, there's a phase-shift between them. Personally, I wouldn't worry too much about overloads, but faults could be a nightmare.
 
rbalex is correct. In a delta-wye, there is a one-to-one correspondence between the currents of the primary and secondary windings on the same leg of the core. In a 480/480V delta-wye, the current through each primary winding will be 1/1.73 = 0.577 times the current drawn through each secondary winding, If there's a 3-phase resistive load that draws balanced load currents of 15A on the secondary, the current through each primary winding will be 0.577 x 15A = 8.66A.

But say that the fuse on the B phase to the primary has blown, so that it no longer contributes current to the two windings of the delta primary that it's connected to. One of the three secondary L-N windings will be on the same leg of the core as the primary winding that is still connected and powered by phases A and C. Now if we keep raising the load current on that L-N secondary winding, it would take approximately 15A x 1.73 = 26A of load current before we get up to15A on phases A and C to the primary. And so in this situation, both the primary and secondary windings can be overloaded by a factor of 1.73 when there are only primary fuses for protection.
 
synchro said:
But say that the fuse on the B phase to the primary has blown, so that it no longer contributes current to the two windings of the delta primary that it's connected to.
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OK, but say you had a delta-delta transformer. Then 240.4(F) allows you to consider the secondary conductors to be protected by the primary side OCPD (at the proper ratio). Now suppose the B-phase to the primary is lost. As I understand it, that effectively gives you a single phase 2-wire to split phase 3-wire transformer (where the B secondary terminal is now the midpoint of the AB-BC combined coil), for which 240.4(F) would not consider the secondary conductors to be protected by the primary side.

Therefore it seems that the reason that delta-wye transformers are not permitted to have their secondary conductors protected by the primary OCPD under 240.4(F) would not be related to loss of a primary leg, as that logic would also apply to delta-delta. Or have I overlooked something?

Cheers, Wayne
 
wwhitney said:
OK, but say you had a delta-delta transformer. Then 240.4(F) allows you to consider the secondary conductors to be protected by the primary side OCPD (at the proper ratio). Now suppose the B-phase to the primary is lost. As I understand it, that effectively gives you a single phase 2-wire to split phase 3-wire transformer (where the B secondary terminal is now the midpoint of the AB-BC combined coil), for which 240.4(F) would not consider the secondary conductors to be protected by the primary side.

Therefore it seems that the reason that delta-wye transformers are not permitted to have their secondary conductors protected by the primary OCPD under 240.4(F) would not be related to loss of a primary leg, as that logic would also apply to delta-delta. Or have I overlooked something?

Cheers, Wayne
Click to expand...
The B terminal is not electrically equal to the midpoint of A-C.
 
jim dungar said:
The B terminal is not electrically equal to the midpoint of A-C.
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I think I see what you mean now, thank you.

With a delta-delta transformer, and the B primary supply disconnected, only the AC primary coil is energized directly by the supply, which will induce the usual L-L voltage on the AC secondary coil. So an AC connected 2-wire secondary load will see the usual voltage and current. But the AB and BC connected secondary loads will ends up in series across that AC L-L secondary voltage, and with the B secondary terminal floating, it will end up as a voltage divider depending on the impedances connected AB vs BC.

Also, since the secondary B terminal is connected to the interior point on that voltage divider, that means the secondary AB and BC coils will have applied voltages per that voltage divider, and there will be an induced voltage back on the primary AB and BC coils proportionately? I.e. in the diagram below, for an idealized transformer, if the primary on the left is a 240V delta, the measured voltages at the primary terminals will be 80V AB and 160V BC?

Cheers,
Wayne

1678158232_connecttion.png
 
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wwhitney said:
... Therefore it seems that the reason that delta-wye transformers are not permitted to have their secondary conductors protected by the primary OCPD under 240.4(F) would not be related to loss of a primary leg, as that logic would also apply to delta-delta. Or have I overlooked something?
Click to expand...

The only reason I mentioned the case where a delta-wye had a blown fuse is that I thought it would be simpler to understand in terms of the line currents drawn by a delta when there's a single L-N load current on one of the the wye secondary windings. This is because with a blown fuse on the B phase, obviously there will only be line currents on the A and C phases. But a blown fuse is not really necessary to explain the inadequacy of having primary overcurrent protection but no secondary protection on a delta-wye transformer. And I did not intend to imply that the reason behind 240.4(F) was dependent on having an open phase on the primary. But thanks for bringing this up.

The concept I was intending to convey is that in a delta-wye, the ratio of a secondary phase current to the line current feeding the delta can be significantly higher with unbalanced loads than with balanced loads. This is because with balanced loads, two L-N winding currents are contributing to the line current drawn at each input of the delta. But if there's only one such L-N winding current, then that L-N current will have to be significantly higher to produce the same input line current as with equal L-N loads. Therefore with a delta-wye, overcurrent protection on the primary selected for the transformer's rating when driving balanced loads will be insufficient to protect the transformer secondary in all cases.
 
synchro said:
The concept I was intending to convey is that in a delta-wye, the ratio of a secondary phase current to the line current feeding the delta can be significantly higher with unbalanced loads than with balanced loads.
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By significantly, do you mean a factor of sqrt(3), or worse? I have been pursing the math in this thread: https://forums.mikeholt.com/threads...elationships-split-single-phase-case.2574979/

So far I've gotten to the formula (IA, IB, IC) = (Ia - Ib, Ib - Ic, Ic - Ia), where the LHS are the primary delta line currents (as complex numbers), the RHS side is the differences of the secondary wye line currents (the neutral current has dropped out), and each coil has a turns ratio of 1 (e.g. 240V delta to 416Y/240V wye).

For the balanced case (the secondary line currents differ only by 120 degree phase rotation), this gives |IA| = sqrt(3) * |Ia|. For the case of only one secondary line current Ia nonzero (and In therefore its negative), this gives |IA| = |IC| = |Ia|. Are there any worse cases to consider?

Thanks, Wayne
 
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