I have a customer that needs a well wired up. the well head is 410 feet from the house and a 30 amp circuit. using the normal volt drop formula to stay within 3-5% for the motor i would use #2AL use and run a direct burial ground with it. after talking to the well company they said i can use normal 10-3 uf to feed the pump motor in fact the spec sheet they faxed me says i could go 1020 feet. why is this? is the volt drop not a factor because the motor is in water vitually liguid cooled? i was curious to other ideas or if anyone else has run across this scenerio too.
You said a "30 amp circuit", but what is the nameplate amps of the pump motor? A 2 HP pump would be a huge well pump, and would have a nameplate value of about 12 amps. Code permits motor breakers to be upsized quite a bit, so its hard to know what amps to calculate when all you see is the breaker size. I would expect your pump to be 1 HP or less (about 8 amps at 240V).
I would be surprised to see a 30A breaker on a motor that had a nameplate value of 10A or less, and I'd also be surprised to see 10-2 allowed for 1020 feet on a motor with a nameplate amp value over 8 amps or so.
If you want to use the breaker amp rating for "I" you can, and that would give you an extremely conservative circuit that will have acceptable voltage drop even during starting. But I think this is overkill, and its going to cost way more than what I think you really need.
I don't believe there are any code rules with respect to voltage drop, just some guidelines. I would use the nameplate * 1.25 in the voltage drop calculation and try to keep it under 3% (7.2V).
I think that 30A breaker requirement must be wrong. Perhaps if it was wired as a 120V motor, but that still seems too large. For a 5 amp 240V motor, I would think a 15A breaker would be just fine.
My V/D calculator comes up with the same thing so I think you will be OK on #10 and I agree on the 15a breaker. I have a 1/2hp well pump that is doing fine on a 15. Them pushing this out to 1020' may be a stretch but it still only works out to about 5% V/D (12v from 240). I am guessing that is where the number came from.