Motor question

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Feb 6, 2010
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I read somewhere that a 3 phase 2hp two pole 3450 rpm motor only delivers about 1 hp at half speed when run through a VFD. Is this true? Thanks for any information. Larry Lehman
 
Yes.

The motor will run at constant torque up to the rated speed. If you set the drive up to go above the rated frequency (usually 60 Hz), it will run at constant power between rated frequency and maximum frequency.

Here's a graph showing the difference between the outputs of a 2HP 2-pole (3500 RPM) and a 2HP 4-pole (1750 RPM) motor.

 
Tim, I've seen that graph before, and it shows HP as staying constant. Help me understand here, I know you've got LOTS of experience and knowledge on VFD drives 'n motors. Since the HP stays the same as the RPM increases, doesn't the torque go down? When you're "hogging" on profiling a blade, isn't torque what's important rather than pure HP? The graph below is one I found a while back showing hp staying close to the same, but also showing how the torque drops. Tim, would you comment on this please?

HP%20vs%20RPM%20curve_zpsy8y8sdom.jpg
 
The short answer to the original question is yes. Horsepower is the product of torque times speed, divided by a constant. Half the speed at the same torque is half the horsepower. Torque is a function of the amperage flowing through the motor and the magnetic force it produces, the VFD can send the rated amperage of the wiring in the motor to the motor at any frequency up to the frequency the motor is rated for (60 hz usually). There is some "slippage" at the lower rpms as shown in the graph but the motor can produce almost constant torque up to the rated speed. As speed rises above the motor's rated speed, available torque must decrease or the VFD or motor will have a problem with overheating (i.e. the VFD limits the power to the motor). Reduced torque output is not an issue unless you are using a VFD to "over speed" a motor.
 
The short answer to the original question is yes. Horsepower is the product of torque times speed, divided by a constant. Half the speed at the same torque is half the horsepower. Torque is a function of the amperage flowing through the motor and the magnetic force it produces, the VFD can send the rated amperage of the wiring in the motor to the motor at any frequency up to the frequency the motor is rated for (60 hz usually). There is some "slippage" at the lower rpms as shown in the graph but the motor can produce almost constant torque up to the rated speed. As speed rises above the motor's rated speed, available torque must decrease or the VFD or motor will have a problem with overheating (i.e. the VFD limits the power to the motor). Reduced torque output is not an issue unless you are using a VFD to "over speed" a motor.

Larryz10, i hope you don't mind me piggybacking on this thread.

So when people are using the vfd to double a 1750 motor, on say the kbac vfd with the 2x jumper, are they doubling the hz? Is it the same for a teco fm50?
Are they programming the vfd to run at 120 hz in order to maintain the torque on the 4 pole motor?
 
I've been trying to post to this thread for several hours, but it doesn't allow a long post. Let's see if this short post will make it.

edit: yep, short post ok, but wouldn't even save a long post. Keep getting following msg
[h=4]Website is offline[/h] No cached version of this page is available.

[h=1]Error 524 Ray ID: 220c333453e309fa • 2015-09-04 19:48:02 UTC[/h] [h=2]A timeout occurred[/h]
Ken H>
 
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Ken, yes the torque drops, but it at 120 Hz and 3500 RPM on the 1750 RPM motor, it only drops to the same level as you'd get if you used a 3500 RPM motor at 60 Hz.

I'll concede that your "better" graph, based on measurements of an actual motor/VFD combination, shows power dropping to about 5% less than rated at 3500 RPM (5.25HP vs 5.5HP), but it's a pretty small effect.

Both the 2HP 2-pole and the 2HP 4-pole will give 2 HP (give or take 5%) at 3500 RPM and therefore the same torque (give or take 5%).

If you need more torque at 3500 RPM than a 2HP motor can give you, you need a bigger motor.

Whatever size motor you use, the 4-pole will give you the extra area in the triangle over the 2-pole.
 
This question came about because of direct drive belt grinders where a 3450 rpm motor would give enough speed for knife making versus a 1725 rpm motor and jackshaft to double the speed for grinding. Larry
 
Tim, as Mahoney says, isn't hp a product of rpm times torque divided by a constant? If so, then if the torque stays the same from 100% to 200%, wouldn't that double the hp from 2ph (in our example) to something greater than 2hp depending on the constant of course?

From the graph I posted, doesn't it show the torque dropping considerable at 200% rpm? There is the peak torque line on graph, then there is the gray area showing total torque (or work?).

torque is a measure of how many pounds can be moved 1 foot. For grinding this is how hard you can press against the belt without stalling motor. The more pounds of pressure against belt, the more torque is required before bogging motor.

hp is how many pounds can be moved 1 foot in 1 sec. 1 hp = 550 pounds moved 1 ft in 1 second.

So, in our world of grinding a blade, the max hp is normally used when profiling at max speed - let's say 3600 SFPM. Let's assume both motors are geared so both run the belt at 3600 SFPM at 100% rpm. The 2-pole motor will have about a 4” drive pulley and 4-pole motor will have about a 8” drive pulley.

I've no idea what the actual forces used, but let's "assume" that a 2 hp motor running at 100% RPM will handle a big chuck of metal pressed against a 36 grit belt with 50 lb of pressure before the motor starts bogging down. (I have no idea if this is a valid number, but let's just assume it is). This will be true for both 2 and 4 pole motors. Remember, the 4 pole motor has a 9” drive wheel, not often seen. For a real world setup, let's say both motors are setup with 4” drive wheels. Then, the 2 pole motor will be running at 100% rpm which is max torque, but the 4 pole motor will require 200% rpm to get 3600 rpm, which would have around half torque. At 100% rpm we've “assumed” 50 lb pressure against grinding belt, but with 200% rpm only 25 lb or so could be pressed against grinding belt.

Let me say again, the above is my understanding how torque and hp work – anyone have corrections? I'd be happy to hear them.

Tim, Mahoney - others please do comment on the above, I'm always open to learning.

Ken H>
 
Yes, when running above rated speed the torque drops to half at double the speed.

Generalizing a bit, most motors tend to be designed with the rotating parts good to 3600 RPM, as it's the highest speed they'll ever need to run on mains power (a 2-pole motor at 60 Hz under no load). Once they've designed the motor frame, the manufacturer can put in 2-pole, 4-pole or more-pole windings and get lower speeds as the pole-count increases, safe in the knowledge that the rotating parts will take it. The windings are in the static part of the motor.

When we use a VFD, we can run over 60 Hz electrically and this lets a 4-pole motor reach the 3600 RPM design speed of the motor frame (3500 RPM under load at 120 Hz).

Because our 2-pole, 2HP motor and our 4-pole, 2HP motor both reach 3500 RPM and both produce 2 HP at that speed, both motors produce the same torque at 3500 RPM and we fit them both with the drive wheel that performs best at that speed.

With the VFD, we don't have to put a bigger wheel on the 4-pole motor because we are no longer constrained to run at 1750 RPM.

The difference now is that the 2-pole motor produces only 1 HP at 1750 RPM, where the 4-pole produces 2 HP at 1750 RPM, so at 1750 RPM, the 4-pole motor produces twice the torque of the 2-pole.

Using your pressure example above, both motors would take 50 lb of pressure at 3500 RPM.

The 2-pole would also take 50 lb of pressure at 1750 RPM.

The 4-pole would take 100 lb of pressure at 1750 RPM.

The 4-pole also has an advantage at low speed.

Most motor/VFD combinations will run smoothly down to around 10 Hz. For a 2-pole motor, that's around 600 RPM. For a 4-pole, it's around 300 RPM. If you sharpen with the grinder, having the lower minimum speed available can make a real difference.
 
Tim, I needed to think on that a bit, you are correct. Since hp = (torque * rpm)/K, and both produce 2 hp (or very close) at 3600, then both MUST be producing the same torque.

you said: "the 2-pole motor produces only 1 HP at 1750 RPM" which is correct, but since at slower rpm/lower belt speed I'm not hogging, but doing sensitive grinding, the fact it's only 1hp is no problem.... and I saved $50 with the 2 pole motor.

Tim - thanks for hanging in there with me.

Ken H>
 
Excuse me bumping this thread, but I've been trying to decide which type of motor to use as well and found this thread informative - but I noticed some "anomalies" to the conclusions:

1) By my account 2-pole motors are MORE expensive than their equivalent 4-pole motors (compare prices of Baldor vm3550/vm3554 or Leeson 110125/110192).

2) When reviewing the KBAC-27D manual (page 18) it clearly shows torque dropping down to 50% above 60hz when using the 2x speed multiplier (50% torque was at the max of 120hz).

Tim, you said that the 4-pole motor produces 2 hp at both 1750 and 3500 - and that is reflected in your chart from post 1. But you also said "both motors produce the same torque at 3500 RPM". I can't however figure out how this could be (for 2 otherwise equivalent 2hp motors).

The calculation for torque is:

Torque = (HP * constant) / rpm.

So if the power is the same then the torque must go DOWN if the rpm goes UP.


Having said that, and assuming I am correct - which is better for our purposes? The 4-pole will have better lower/mid torque while the 2-pole will have better higher end torque (50% more at 3450 rpm). Which speeds typically need more torque for knife making? Assume a 5" drive wheel...so ~2257 fpm at 1725 motor rpm and ~4514 fpm at 3450 motor rpm.
 
Ah nvm, I think I have found the error in my logic. The torque DOES go down from 1750-3500. But it will be the same as that of the 2-pole motor at that speed like you say. And the 4-pole will always have higher torque than the 2-pole at lower speeds because it reaches max HP at a lower speed.

tl;dr - buy 4-pole, use 2x VFD multiplier and profit.
 
Java - if there is a true "Profit" to using a 1725 rpm motor vs 3400 rpm motor is open to discussion. While the 3400 rpm motor might have less power at lower speeds, you seldom (if ever) use much power at those low speeds. The place you will use full power is in the upper RPM ranges where the 2 pole and 4 pole motors both will have lots of power..... and saving money on purchase was my deciding factor.

You might check here for 3 ph motors for less price: http://tinyurl.com/q4d85bn $127 shipped will get you a 4 pole 2hp and $166 will get the 2 pole motor.

Ken H>
 
Having seen those import prices I take back my "profit" comment. It's interesting that those import motors have the 2-poles cheaper than the 4-poles like you would expect while Baldor and Leeson are pricing higher by rpm instead.

I guess they have so much markup in them they are market pricing them instead because "3500 is faster than 1750 therefore it should cost more". Unless maybe there is an actual difference in the build quality of their 3500 and 1750 motors (which I doubt)?
 
Seems it should actually cost more to manuf a 4 pole motor than a 2 pole motor because of the extra copper. Note the extra weight of the 4 pole motor due to the extra copper. Does that make it better? Heck if I know - if you're using your grinder 8 hr/day, 5 or 6 days/wk, it might be worth while spending the extra money for a Baldor industrial rated motor. Myself, I'd rather have a spare sitting on the shelf if I could not handle a couple days of down time due to bad motor. Not that I think the Iron Horse motors will go bad. I'm using a Leeson on lathe and the Ironhorse on grinder. Had I known about Ironhorse when buying the Leeson for lathe that's what I would have went with then.
 
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