Auto transformer secondary available fault current

[PE (always learning)]

I have a step-up auto transformer that feeds an elevator disconnect on it's secondary. The auto transformer goes from 208V, 3 phase to 480V, 3 phase.

The available fault current on the primary side of the autotransformer is 6,700 amps.

The transformer is 32 kVA and I'm assuming 1.5% impedance. The available fault current at the elevator disconnect needs to be below 5,000 amps.

I would assume that the high voltage side of the transformer would reduce my available fault current and I could calculate it the same way that I would calculate fault current for a step down transformer. My secondary (480V) side is showing 1,359 amps. In my eyes, this would allow me to meet the short circuit current rating of the 5,000 amps listed on the elevator equipment.

I thought I was reading somewhere that autotransformers are different when it comes to fault current. Please let me know

 

[PE (always learning)]

Attached is a one line and a spec of the autotransformer for more information

 

[wwhitney]

Looks like the installation doesn't comply with 215.11 / 210.9.

Cheers, Wayne

 

[PE (always learning)]

Looks like the installation doesn't comply with 215.11 / 210.9.

Cheers, Wayne

I'm not the designer on this project, I'm just hired to do the study..

the autotransformer is provided by the elevator manufacturer in this scenario and I'm not showing you the full specifications which may provide more insight

I'm a little confused with your statement as grounding is shown for the auto transformer

 

[wwhitney]

I'm a little confused with your statement as grounding is shown for the auto transformer

215.11 / 210.9 is not about grounding, it's about whether the circuit contains a grounded conductor, i.e. neutral. So if I understand them correctly, they prohibit delta-delta autotransformers, except as indicated in the exceptions.

But as you say the autotransformer is provided by the elevator manufacturer, maybe it's consider part of the utilization equipment and not subject to those limits? Not sure.

Sorry not to be able to help with your actual question.

Cheers, Wayne

 

[David Castor]

It's very odd to see an autotransformer used here, but it's elevator equipment so expect the unusual. That said, the autotransformer impedance will be (generally) extremely low compared with a two-winding transformer.

 

[GoldDigger]

I have a step-up auto transformer that feeds an elevator disconnect on it's secondary. The auto transformer goes from 208V, 3 phase to 480V, 3 phase.

The available fault current on the primary side of the autotransformer is 6,700 amps.

The transformer is 32 kVA and I'm assuming 1.5% impedance. The available fault current at the elevator disconnect needs to be below 5,000 amps.

I would assume that the high voltage side of the transformer would reduce my available fault current and I could calculate it the same way that I would calculate fault current for a step down transformer. My secondary (480V) side is showing 1,359 amps. In my eyes, this would allow me to meet the short circuit current rating of the 5,000 amps listed on the elevator equipment.

I thought I was reading somewhere that autotransformers are different when it comes to fault current. Please let me know

With an isolation transformer the fault current on the secondary is basically limited by the available fault current reduced by the impedance of the primary and secondary windings and any impedance associated with core losses.
With an autotransformer the point where the input tap is located on the transformer can deliver the full primary circuit fault current without any limitation due to the transformer windings, while the ouput fault current will be reduced by the impedance of only that portion of the winding which is between the input and the output taps, considered as the impedance of that secondary winding given that the primary winding is effectively short circuited.
The result is that the reduction in output fault current will be less than the reduction caused by an isolation transformer of the same general characteristics. I do not, however, have the exact calculations.

 

[Strathead]

With an isolation transformer the fault current on the secondary is basically limited by the available fault current reduced by the impedance of the primary and secondary windings and any impedance associated with core losses.
With an autotransformer the point where the input tap is located on the transformer can deliver the full primary circuit fault current without any limitation due to the transformer windings, while the ouput fault current will be reduced by the impedance of only that portion of the winding which is between the input and the output taps, considered as the impedance of that secondary winding given that the primary winding is effectively short circuited.
The result is that the reduction in output fault current will be less than the reduction caused by an isolation transformer of the same general characteristics. I do not, however, have the exact calculations.

That makes some sense. An Autotransformer would seem to be nothing but an impedance in series with the load when stepping down voltage, but does that also mean that it would actually be a negative impedance and the fault current would go up proportionally when used as a step up? It seems that it would. Also wouldn't the formulas be 2*pi*f*l for the portion of the coil in question?

 

[wwhitney]

The result is that the reduction in output fault current will be less than the reduction caused by an isolation transformer of the same general characteristics. I do not, however, have the exact calculations.

That makes sense for the single phase 2-wire case.

But in the OP there's a 3-wire delta autotransformer. If it is an "open" arrangement of just two autotransformers, then the behavior will differ on the 3 lines. One of the line conductors will just be connected straight through, with no reduction in fault current from the autotransformer.

Cheers, Wayne

 

[MyCleveland]

Jim Dungar:

Can you provide some insight on this thread please.

 

[synchro]

I have a step-up auto transformer that feeds an elevator disconnect on it's secondary. The auto transformer goes from 208V, 3 phase to 480V, 3 phase.

The available fault current on the primary side of the autotransformer is 6,700 amps.

The transformer is 32 kVA and I'm assuming 1.5% impedance. The available fault current at the elevator disconnect needs to be below 5,000 amps.

Is the step-up autotransformer similar to the one below that was posted by "coachmike" at https://forums.mikeholt.com/threads/elevator-boost-transformer.2569261/post-2772116 ? This particular one only has a 0.85% impedance.

The following article describes how you can calculate the effect of the impedance on the SCCA:
https://iaeimagazine.org/2018/july-...tions-using-transformer-and-source-impedance/

 

[jim dungar]

Jim Dungar:

Can you provide some insight on this thread please.

There will be two effects from the autotransformer.
1. The (1) three-phase fault current (L-L-L) will be affected by the transformer impedance.
2. The (3) single-phase fault currents will be affected differently. The common line to ground fault current will be unchanged while the other two will be impacted by the transformer, if the transformer is an open delta.

The open-delta transformers calculations are usually best left to software rather than hand calculations if you are trying to get the single line to ground currents.

While you can use the standard infinite source method of FLA/%IZ, you really should consider the source impedance. Note how the nameplate, in synchro's post #11, limits the output fault current to 118X from an infinite bus using a closed delta.