Transformer Tap rule

[jim dungar]

Yes, but the latter case won't give you 100A in the primary. You only get 100A through the primary if you have 100A through all of the secondary pieces.

The amp-turns on the primary and the secondary are equal. So if, say, the full primary and secondary are 10 turns, and we have 100 amps through, say, 3 turns on the secondary, and 0 amps through the other 7 turns on the secondary,

This is so far off.

Assume you have a transformer with 100A flowing through a 4 turn primary, on the secondary there are also 4 turns so there would be a corresponding 100A. This works out to be 25A per turn of flux.

Now take that same transformer secondary and cut it into two pieces/windings of 2 turns each, with no other changes passing 100A across 4 windings with 25A per turn of flux will result in 100A across the 4 turns of the primary.

Now with a center tap between the two secondary windings. Have 100A flow through one winding and 0A flow through the other. This results in 50A per turn, which is 2X the original amount of flux

This gives us the same flux density as the original condition so we would get the same 100A across the 4 turns on the primary.

 

[LarryFine]

This is so far off.

Assume you have a transformer with 100A flowing through a 4 turn primary, on the secondary there are also 4 turns so there would be a corresponding 100A. This works out to be 25A per turn of flux.

Now take that same transformer secondary and cut it into two pieces/windings of 2 turns each, with no other changes passing 100A across 4 windings with 25A per turn of flux will result in 100A across the 4 turns of the primary.

Now with a center tap between the two secondary windings. Have 100A flow through one winding and 0A flow through the other. This results in 50A per turn, which is 2X the original amount of flux

This gives us the same flux density as the original condition so we would get the same 100A across the 4 turns on the primary.

Are you saying that the primary current would be the same 100a with either:

1. 100a at 120v across one 120v secondary

Or:

2. 100a at 240v across both 120v secondaries

 

[wwhitney]

This is so far off.

Sorry, you're still mistaken.

In transformers the conserved quantities (same on the primary and secondary) are volts/turn and amps * turns (not amps/turn). E.g. from this source: https://eepower.com/technical-articles/power-transformer-basics-the-magnetic-circuit-part/ we have "the product of current and number of turns in each winding N·i must balance. This balance is called ampere-turns,"

So in your example, with 4 turns on the primary and secondary, and 100A through the full primary and secondary coils, we have 400 amp-turns on both sides, which is balanced. Change the secondary to 100A through just 2 of the turns, and 0A through the other 2 turns, and we have 200 amp-turns. The primary current changes to 50A to maintain the balance of 200 amp-turns on each side.

Cheers, Wayne

 

[jim dungar]

My apologies for my post #61. I should no better than to try create and edit posts on my phone.

I do know the basis of transformers is that the primary current times the number of primary turns is equal to the secondary current times the number of secondary turns. This is taught very early on any transformer course.

 

[wwhitney]

I do know the basis of transformers is that the primary current times the number of primary turns is equal to the secondary current times the number of secondary turns.

OK, now that is clarified, would you reconsider the issue in the OP?

Namely, for a 480D : 208Y/120V transformer, 100A on the secondary L-N will only cause 25A on the primary L-L. So when applying 240.21(C)(6), if the goal is to have parity with 240.21(B)(2), the appropriate ratio to use is 4:1.

If you agree, do you think the language in 240.21(C)(6) adequately reflects that, and if not, do you have a suggestion on an improvement?

Thanks,
Wayne

 

[Strathead]

OK, now that is clarified, would you reconsider the issue in the OP?

Namely, for a 480D : 208Y/120V transformer, 100A on the secondary L-N will only cause 25A on the primary L-L. So when applying 240.21(C)(6), if the goal is to have parity with 240.21(B)(2), the appropriate ratio to use is 4:1.

If you agree, do you think the language in 240.21(C)(6) adequately reflects that, and if not, do you have a suggestion on an improvement?

Thanks,
Wayne

I know you guys think on a level above me, but I feel like the 1/3 is arbitrary in the first place. On top of this, we are using 480 transformers for all of our discussion, but this section doesn't limit the voltage differential. A 4160 to 480 transformer would have have quite different ratios. But I also feel the wording of 240.21(C)(6) is lacking. As I stated, the wording certainly justifies using L-N voltage to arrive at the wire size, but I also feel that many inspectors would balk at this since using L-L would result in a larger minimum wire.

 

[wwhitney]

As I stated, the wording certainly justifies using L-N voltage to arrive at the wire size, but I also feel that many inspectors would balk at this since using L-L would result in a larger minimum wire.

For the secondary side, the lower voltage (L-N) gives the larger ratio and the larger minimum wire size. Which is what I am arguing for (except in the case of the wye-wye).

Cheers, Wayne

 

[Strathead]

For the secondary side, the lower voltage (L-N) gives the larger ratio and the larger minimum wire size. Which is what I am arguing for (except in the case of the wye-wye).

Cheers, Wayne

Yeah, I had it backwards. My head still spins every time I try to do this math.