240 Delta to 208Y No XO

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LarryFine said:
A neutral is still required. We don't see one.

The incoming and outgoing neutrals are individually capped.
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winnie said:
It isn't clear to me that this is a wye connected transformer. Despite the wye in the label, it might actually be delta connected.

If the L-N voltages are stable, and the connected load working properly, and the 240V supply high leg delta, I would leave everything alone.

If something isn't working, then the next steps would be to find out what the 240V supply grounding is, and get a circuit diagram for the transformer

Jon
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What I meant was to show what a typical wye autotransformer is - I didn't mean to say that is what the OP has. I agree it is ambiguous and a wiring diagram would greatly help, or at the very least a better picture where we can make out the entire label.
 
electrofelon said:
A picture is worth 1000 words. Here is a diagram of a typical wye connected autotransformer. But I agree with Augie, I dont like referring to the "output" or "input" of an autotransformer. It should just be called a wye connected autotransformer and nothing else.
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In this case the input is fused.

Just because the coils are connected in a Wye configuration, there is no requirement to have a neutral point connection.

The OP should put a label on the unit identifying it as equal to a 208V high leg, for troubleshooting purposes, and then move onto another project
 
winnie said:
It isn't clear to me that this is a wye connected transformer.
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Taking the source voltage system to be a 240V 4-wire delta (midpoint of A-C is grounded), we can conclude the autotransformer is wye connected based on the L-G voltages in the OP.

For a delta connected autotransformer, the transformed voltages A'-G and C'-G would end up different, while for the wye connected case they will be equal as in the OP. Also, it's not too hard to calculate the B'-G voltage for the wye connected case, and it matches the OP (see below).

Cheers,
Wayne

For a wye connected autotransformer, the voltages relative to the center point M (derived 3 phase neutral) will all scale by 208/240 = sqrt(3)/2. The voltage point M is 1/3 of the way from G towards B, i.e. M-G = 208/3 and B-M = (2/3)*208. So after transformation the new voltage B'-M = (sqrt(3)/2)*(2/3)*208 = 208/sqrt(3), and as B', M, and G are colinear, the voltage B'-G will be 208/3 + 208/sqrt(3) = 189V.
 
late to the party but i'll just add I have never run into an elevator motor that must be fed by 208.
And I have more than a few times had the elevator company we work for provide me with one of those autotransformers.
When I cheeked the nameplate of the motor (not as easy as you would think) it was actually a 240v motor that said ok for 208.
 
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