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For a 3 phase open delta transformers:

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When you are computing the short circuit current between the 208v high leg (point B) and neutral. Do you only use one 75kVA or do you add the other transformer to produce 75kVAx1.5 (or other value) = 112.5kVA?

Im thinking if you need to add because between point B and neutral, there are more than 1 winding (or 1 transformer) involved. What do you think?

The following is the nameplate of the transformers and the lever was set to 2nd:

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    \$\begingroup\$ I suspect it may be a bit more complicated. You may need to look at the equivalent circuits of the individual single-phase transformers that make up the specific configuration. You may need to vectorially add two components of short circuit current. \$\endgroup\$ – Charles Cowie Mar 11 at 0:09
  • \$\begingroup\$ How do you do that? any idea what kVA would be equivalent of it? \$\endgroup\$ – Jtl Mar 13 at 2:22
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    \$\begingroup\$ I think it is more like two impedances in series rather than increased kVA. It would be two voltage sources, 120V at zero degrees in series with 240V at 120 degrees. The 120V source would have half of the single phase transformer impedance in series and the 240V source would have the full transformer impedance in series. The voltages add up to 208V at 90 degrees as you know. The shout circuit impedance may be just 1.5 X the impedance of a single transformer. I think that, because each impedance has the same angular relationship with its own source voltage, they just add together. \$\endgroup\$ – Charles Cowie Mar 13 at 2:44
  • \$\begingroup\$ If the impedances increase, would this produce a greater short circuit current, how? Thanks. \$\endgroup\$ – Jtl Mar 13 at 14:41
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    \$\begingroup\$ If the impedance increases, the short-circuit current is lower. \$\endgroup\$ – Charles Cowie Mar 13 at 14:55
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To calculate the current in a short circuit between point B and the center tap of the A to C winding I would first calculate the winding impedance. The full 240 volt winding is rated 75 kVA. The impedance of a load that would draw 75 kVA is 240^2/75000 = 0.768 ohms. The transformer impedance is 2.59% or 0.0259 x 0.768 = 0.0199 ohms. The total impedance of a 120 V winding in series with a 240 V winding is 1.5 x 0.0199 = 0.0298 ohms. The short circuit current would then be 208/.0298 = 6971 amps.

I used the percent impedance to calculate impedance in ohms because the percent impedance is based on 75 kVA at 240 volts and the combination of windings results in 208 volts. It seems easier and more clear to used a fixed ohmic value when dealing with one single-phase transformer and part of another connected to three-phase power.

I believe the equivalent circuit is a shown below. Each transformer or part of a transformer is represented as an ideal voltage source and an impedance. The impedance of the 3-phase source is assumed to be zero.

enter image description here

Transformer resistive and reactive components

The transformer nameplate states the impedance as a percentage. Knowing only the percentage does not allow the resistive and reactive components, R and X, to be determined. Therefore the result of the above calculation is the combined resistance and reactance, Z. Often nameplates state X/R as well as percentage. That allows Z to be calculated when necessary. Another alternative is to find a listing of typical X/R for various transformer sizes. At this kVA level, X/R is likely more than 5 and therefore the impedance can be considered to be mostly inductive reactance.

If any added external impedance is mostly inductive, it can simply be added. If it is mostly resistive, it can be added using the square root of the sum of squares. In the case of adding the impedance of half of an additional winding, X/R is assumed to be the same for both components and can be simply added.

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    \$\begingroup\$ For wiring individual circuits on the load side of the distribution panel for a 75 kVA transformer, the fault current at the end of a 20 foot 8 AWG, 50 amp feeder for an individual load circuit would be 20% less. \$\endgroup\$ – Charles Cowie Mar 17 at 13:15
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    \$\begingroup\$ In the last line I suppose you mean 25% less not 25 times. I suspect that arc fault calculations may be based of the worst case being maximum power transfer to the arc. This is all about arc fault calculation and not the subject of the original question. You need to find a tutorial on arc flash calculation. I am not familiar with that. \$\endgroup\$ – Charles Cowie Mar 17 at 14:09
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    \$\begingroup\$ Certainly 20 feet of wire has some inductance, but what little I found about that seemed to indicate that it would not be significant. If you assumed the wire had inductive impedance rather than resistance, the effect would still not be significant. The issue is not worth pursuing. \$\endgroup\$ – Charles Cowie Mar 19 at 3:19
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    \$\begingroup\$ Adding Z1 + Z2 is based on the assumption that X/R is the same for both. I am surprised that that would be the case. However I believe that I had come to the conclusion that the full range of possible assumptions about X/R for wire would result in about 1% to 5% reduction of short circuit current. I think I said something to that effect in a comment to another question. \$\endgroup\$ – Charles Cowie Mar 27 at 12:59
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    \$\begingroup\$ I was under the impression that X/R is 5 or more for most power distribution transformers, but I have recently seen some tables with X/R below 5. The variation is not as consistently related to kVA as I believe I saw some time ago. It may not be possible to reasonably estimate transformer X/R. Wire X/R probably varies with standing specifications, type of conduit, conduit size relative to wire size and other factors. Information about that seems to be difficult to find. \$\endgroup\$ – Charles Cowie Mar 28 at 14:00

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