"Karl Townsend" wrote in message
anews.com...
My apple water bin dump uses a .5 horse 3 phase 220 VAC motor
attached by
pulleys and belt to an outboard motor prop in a tube to pump water.
Maybe 1"
head at 100+ GPM. I just measured the current draw on this motor at
1.5 amp,
2.0 amp, and 2.2 amp with a clamp A-meter.

Na, I'm not totally re doing something that has run 20 years. But I'd
like to try a different prop - it might be better. If not, I'll go
back. My only question is if this motor will have enough power at 1/2
speed. Do you know how to figure that?

First, the mass of water moved and the head determine the hp, ignoring
frictional losses, so gearing down and changing the prop pitch to move
the same amount of water should require the same hp so I'd start with
the motor you have now if you gear down with pulleys. If a vfd gives
constant torque but now the motor is running slower, then it will
develop less hp so you need a bigger motor. hp=torque*rpm/5252 in sae
units, so right now at .5 hp at the motor you are getting 1.52 ft-lbs at
1725 rpm. If you wanted .5 hp at 863 rpm that means you need 3.04
ft-lbs of torque, and if you want to get that by running a 1725 rpm
motor at half speed with a vfd that would be a motor that makes 3.04
ft-lbs at 1725 rpm or a 1 hp motor. Clear as murky mud? :-)

--
Regards,
Carl Ijames carl.ijames at verizon.net