RPM of variable speed DC Motors

After some web surfing, with HP being the same on a permanent magnet DC motor, is a lower RPM for the given HP better than a higher RPM for the same HP?

Motor A: 1HP, 90V, 2500 RPM, 10.7A 4.28A per 1,000 RPM Motor B: 1HP, 90V, 1750 RPM, 9.2A 5.26A per 1,000 RPM

Since I'm a recovering college graduate, does this mean more kick per RPM by the lower speed drive?

Reply to
Louis Ohland
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"Louis Ohland" wrote: (clip) does this mean more kick per RPM by the lower speed drive? ^^^^^^^^^^^^^^^^ I've never heard of anyone calculating amps per RPM, and I think that is because it is meaningless. First question: How do you know that both motors are developing 1 HP? Are the figures you listed measured or read off the nameplates? Of course, you get more torque at lower speed for a given horsepower. If the figures are accurate, then motor A is slightly more efficient than motor B.

Reply to
Leo Lichtman

Well, some manufacturers do not make finding their torque graphs readily available. The amps were off a manufacturer's website, but they are just summaries. No tables or graphs at all.

Efficiency aside, neglecting heat loss, wouldn't using a motor rated to deliver 1HP at a lower RPM be better?

Reply to
Louis Ohland

Definitely means more torque. Less reduction required if you need less than 2500 (0r 1750) RPM. More torque means mor amops, and vise versa.

Reply to
clare at snyder.on.ca

A theoretically perfect DC motor has a 1:1 relationship between armature current and shaft torque, and a 1:1 relationship between armature voltage and speed. It is, in fact, the same relationship if you use the right units (Newtons, meters, volts, amps, radians and seconds) to do your calculation.

Real DC motors, unless they're horribly inefficient, don't stray too far from the theoretical -- so you can pretty much calculate the torque vs. current characteristic from the speed vs. voltage characteristic.

Define "better". The motor rated to deliver 1HP at the lower RPM is clearly capable of delivering more torque without overheating. Assuming that it's bearings and armature construction will stand up to the stress, you'll be able to get nearly that torque out of it at higher speeds (and voltages), but it'll probably be bigger.

If you need more than 1750 RPM and you don't want to mess with giving the motor higher than rated voltage, then the second motor is clearly worse. If you need more torque than the first motor can give and you'll always be operating at less than 1750 RPM, the second motor is clearly better. If neither of these two conditions hold, then you haven't given enough information.

If you're going to be gearing either of them down no matter what, and you haven't selected a gear ratio yet, then the speed and torque don't matter, so you need to look at efficiency (and heat generation), weight, and other factors that you haven't mentioned.

Reply to
Tim Wescott

Certainly you must mean Volts per 1000 RPM. A common specification. At 52.6 volts per 1000 RPM, 92 volts will drive it to 1.75 K RPM.

Which motor is better? That is like asking which gear is better for a bicycle. Won't it depend on what you intend to use it for? One motor appears to be designed for more RPM, and less torque, (if they are really both 1 HSP motors) It appears that the specifications listed are somewhat ambiguous/ questionable. What are you planing to do? How many RPM do you really want?

Reply to
Half-Nutz

Motor B has more torque at rated current. The A/RPM is meaningless, derived by dividing the rated full-load current by the rated RPM. One interesting observation, though, is that motor B is 16% more efficient than motor A, since they both deliver 1 HP at the same voltage, but B draws a lot less current.

Jon

Reply to
Jon Elson

maybe it will help to understand that horsepower is a measure of power, not force - torque is "force", RPM is "speed", power is the product, so to speak. one horsepower is 3300 ft-lbs per minute - e.g. a one horsepower motor can lift a 3300 pound weight (not mass) one foot in one minute, or it can lift a 1 pound weight about 2/3 of a mile in the same time.

Reply to
William Noble

Ya dropped a zero. 1 HP = 33,000 ft-lbf/min.

Reply to
Don Foreman

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This is only true of permanent magnet motors or shunt field motors operated with a fixed field voltage.

In the case of a perfect lossless SERIES field motor the shaft speed is directly proportionate to supply voltage AND inversely proportionate to supply current..

Because increase of supply current increases BOTH the field and the armature excitation, the shaft torque is proportional to current squared.

With regard to the original query, because the manufacturer quotes RPM/supply current, these are clearly series wound motors possibly intended for wheeled transport.

There is little to choose between them. The higher speed motor is likely to be slightly smaller and lighter than the lower speed machine. The lower speed machine is a bit more efficient and the lower speed and higher torque may be an advantage.

Neither machine is suitable if the main aim is constant speed with variable load.

Jim

Reply to
pentagrid

Theoretical concerns aside, then how can one tell if a DC motor is appropriate for use on a lathe?

snipped-for-privacy@yahoo.com wrote:

Reply to
Louis Ohland

Ya added an f in lbf ? never heard of that one :-)

Mart> >

Reply to
Martin H. Eastburn

That is to distinguish it as pound (force) as opposed to pound (mass)

lbf .... lbm

Here is a link that discusses the relationship with Newton's second law.

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Mart> Ya added an f in lbf ? never heard of that one :-)

Reply to
JohnB

If it is shunt or PM feild and enclosed frame of adequate horsepower at the correct rated speed, it's useable. The motor with the lower native speed for the same horsepower will be the stronger motor when used variable speed, as it will provide more torque at lower speeds than the highspeed motor will.

Reply to
clare at snyder.on.ca

Reply to
Martin H. Eastburn

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