3 phase question

calstar

calstar

Active User
H-M Supporter Gold Member
Joined
Sep 26, 2013
Messages
408
Look at the spec on this motor, it shows 1hp at 60hz and 3/4 at 50. Does this mean that when using a VFD that much hp is proportionally lost as you dial down the hz to slow the motor speed? Seems like you lose a lot of power if you slowed the rpm by dropping the hz to 20% or so.

thanks, Brian

dfd35961-dac0-4b09-8948-5d374d8afd77_zpsb16098b9.png
 
You do, basically because you are limited at the amount of current you can put in. Since current is (more or less) proportional to torque, lower speed at a fixed torque is less power.
 
Also depends on how your drive is configured. They come as constant torque, variable horsepower or constant horsepower, variable torque. It depends on the type load you are driving. I am usually ask "Is this for pump/fan duty or conveyor duty."
 
HP is a function of RPM x Torque. HP= (Torque x RPM)/5250. So the short answer is yes, as you dial down the speed the HP drops off. A 3450 RPM motor has half of the torque of a 1725 RPM motor of the same rated HP. The motor is rated for 1 HP at 60Hz and 230V, (USA standard) and 3/4 HP at 50Hz and 220V (European standard). So you can see that both the voltage as well as the frequency are lower.

Most VFD's have a torque boost (sometimes labeled as Slip Compensation) function that increases the voltage at lower RPM, this also increases the current draw. You have to be a bit careful with this feature because the motor may overheat if adding too much boost. It is normally used for starting large loads and thus is a temporary condition. The VFD will also current limit to protect it's self and the motor to some degree.
 
Just a tidbit, but a lot of motors (especially those rated for VFDs) have temperature sensors embedded in them to deal with overheating.

I also noticed that everyone on the thread so far is from the US West Coast. YAY WEST COAST!
 
CURSES! Ah well, everyone's welcome, I just thought it was interesting ;)
 
JimDawson said:
HP is a function of RPM x Torque. HP= (Torque x RPM)/5250. So the short answer is yes, as you dial down the speed the HP drops off. A 3450 RPM motor has half of the torque of a 1725 RPM motor of the same rated HP.
Click to expand...

I like where you're going with that and I'd like to expound a bit for/to Brian..

Let's say we hooked up that 3450 RPM motor to a lathe and fed it 230V 3phase from mains. No work piece in the chuck, just free spinning at rated RPM. Are we getting 1HP out of it? nope. We might be getting 1/10th HP from it. It's not until we chuck up a big bit of metal and start mercilessly gouging into it with a pointy object that current starts to rise and torque starts to rise, and thus HP ((torque x RPM)/5250) starts to rise. When we gouge into the workpiece hard enough to make the motor draw 3.2A, that's when we know we've hit 1HP.

So, let's say we empirically discover how aggressive of a cut it takes to draw the rated 3.2A @ 230V (1HP) and then we take that same depth of cut repeatedly at slower and slower speeds... would you expect the "diminishing HP" to eventually cause the lathe to bog down and come to a stop? I wouldn't. I'd expect to be able to maintain the same cut almost to zero speed, because my torque isn't changing, just my speed is; and that's the reason for the "diminishing HP" ... diminishing speed ((Torque x RPM)/5250). (example is not perfect; real world observations subject to change in behavior of cutting action per feeds & speeds and also effects of electrical characteristics in the motor which I don't trust myself to explain accurately)

So yes, you "lose HP," but no, you don't really "lose" anything that you would miss. Kinda like how a car can be rated 245HP but only require 25HP to maintain freeway speed - set your cruise control and you'd never know that those other 220 ponies were asleep in the stable.
 
You must log in or register to reply here.
Back
Top