VFD Conversion again

[alexmin]

My PM1236 has 2hp 1 phase motor.
I was contemplating getting a new 10hp VFD (derated 5hp) and new
5hp 3 phase motor so I can run my lathe without changing gears.

Unfortunately this setup costs around $600 or more.
Due to certain budget constrains imposed by my lovely spouse all I can afford in near future is a 2HP 3phase motor that I can use with the VFD I already have(it can drive a motor up to 2hp)

A new motor will give me a soft start, smoother motor and really low min RPM. I still have to keep changing gears though.

What are your thoughts about it?

 

[65535]

Save up your budget constrained allowances for the 5HP motor?

No point in going with the same HP motor IMO you'll just regret not having the cash to buy the 5HP setup.

A 5HP motor will produce more power than the 2HP stock motor until you dip below 40% RPM, at that point you'll want to change gears no matter what unless you're working on a small part.

Anyways that's what I would do.

 

[liveforphysics]

My VFD's let me over-drive the motors up to about 300% of rated capacity.

I run 1hp TEFC 3P motors at 3hp for about 20 minutes continously, and I've never had any trouble. The motor gets warmer than usual, but motor heat isn't a big deal for brushless induction motors.

I would imagine you could run that 2hp motor as a 5hp motor just fine as long as you're not planning on doing very extended time runs. :)

 

[alexmin]

My VFD's let me over-drive the motors up to about 300% of rated capacity.

What is this "over-drive" option? Do you just set your VFD like if it is driving 3hp motor but connect it to 1 hp motor?
My VFD (KBAC-27D) has jumpers to select different horse power modes but the max is 2 hp.

 

[65535]

Running a motor above 100% RPM (or above 60Hz) will start to lose torque very quickly. At 300% rpm, if the motor hand not grenaded, it would produce poor torque and no where near 300% HP.

I'd like to know what brand VFD and motor are capable of that, and what machinery they are on.

 

[brickbat]

Running a motor above 100% RPM (or above 60Hz) will start to lose torque very quickly. At 300% rpm, if the motor hand not grenaded, it would produce poor torque and no where near 300% HP.

Right. When a VFD is used to operate a motor above its rated speed, torque drops off. Generally, the motor enters a constant horsepower region above 60 Hz, where, for example a doubling of speed will mean a halving of torque.

That said, fitting your lathe with a 2HP VFD and motor will still be a significant improvement, IMO. It will reduce the number of gear shifts. BTW, check prices on a 3HP drive - the big step in pricing seems to be at 5HP, and the 3HP drives aren't much more expensive than the 2HP models.

I put a VFD on my 1 HP Tree milling machine to reduce the number of times I need to change the belt drive ratio (It has multiple sheaves), and it's been great. Sure, some times, I need lots of torque at the spindle, and have to change the belt ratio, but lots of times, I don't and just leave the belt ratio at a mid-value and adjust the VFD.

My lathe is variable speed, and it's really nice to see the way you can dial in a sweet spot for the best surface finish during a cut - kinda hard to do that if you have to change gears!

And, while you might NOT want to attempt to run your motor at 300% of its nameplate speed, you can certainly hit 150% with no issues...

 

[wquiles]

What are your thoughts about it?

I would wait until you can afford what you "really" want. Second choice, get at least a 3HP 1750 rpm quality motor - that will be a significant improvement over the stock motor.

Will

 

[liveforphysics]

I do not triple the RPM of the motor. I go from 60hz to maybe 90-100hz max, but I don't require any over speed to double-triple the motor power.

I use a Yaskawa VFD that takes in 220 single phase, and outputs 440v 3p with user selectable current weakening and current phase limits. The drive was ~$320 used on Ebay if I remember correctly.

I pair the motor windings to be configured for 220v 3P (to lower winding resistance to enable higher current draw ability), then wire it to the 440v 3P VFD output, and simply use phase current controls to limit the amount of power the motor is getting. I can make it some the motor is drawing less than 100w, all the way up to 4kw on the same motor by setting the phase current limits. The phase currents determine the torque of the motor, the frequency determines the RPM, and the RPM*torque determines the power.

Best Wishes,
-Luke

 

[brickbat]

The phase currents determine the torque of the motor, the frequency determines the RPM, and the RPM*torque determines the power.

Exactly.

So, help us understand what's happening when you get triple the motor's rated power. If you're getting that at the motor's rated speed, as you say, then wouldn't the motor be delivering triple the torque? And in doing so, draw triple the current? And it does that for 20 minutes without emitting smoke? WOW!! That's one amazing motor, or you have one helluva cooling system for it...

 

[liveforphysics]

Exactly, triple the current, roughly triple the torque (not a perfect triple due to saturation losses, increased eddy currents, etc.), and triple the power for a given RPM.

Tripleing the current has the same effect on torque as adding a 3:1 gear stage, but with out the 1/3 drop in RPM, but at the expense of drawing 3 times the power (and outputting 3 times the power).

Of course that's all +-10-15%. Efficiency curves are funny with induction motors. :)


The motor can handle things thermaly because the rate it can transfer heat into the air around it depends on the difference in temp between the air exposed surfaces, and the rate air moves past it. When they rate a quality motor (I use baldors), they rate it for essentially a worst case scenerio situation. I give the motors a situation with active cool airflow, in return, they increase to a higher temperature with higher power levels, and find the new thermal equilibrium temp (still in a safe zone) and hold just fine. :)

 

[brickbat]

A motor drawing three times its nameplated current is dissipating roughly 9 times its rated power in resistive losses. (Most of the power lost to heat in an induction motor is lost in the resistance of the stator and rotor conductors) And of course, tripling the current through a resistor causes 9 times the power to be lost in it.

Now, as I understand heat transfer, in order for the motor to dissipate that much heat, that motor would need to be at (again roughly) 9 times the temperature RISE for which it was rated. In other words, if the motor rating was at, say 40C rise over ambient at its nameplate current, and given that you'r not really changing the cooling system (Its TEFC, right), wouldn't the motor now rise to roughly 360C above ambient? In steady state, anyway.

Any chance your VFD is computing input power incorrectly, since its wired for 480V? Or have you actually confirmed you really are putting a line current of 3 times the nameplate into the motor? - triple 3.5 amps is about 10.5 amps, right? A decent motor could do that for short periods of time, but after 20 minutes, it'd be a TESG motor (totally enclosed smoke generator)

I'm sure you've heard of the magic smoke theory of electrical devices.... :) Just because it's a "totally enclosed" motor, doesn't mean the magic smoke won't leak out :). Trust me on this - I have some recent experience with a 6" belt sander in this area. There's a lot of nasty smelling smoke contained in a TEFC motor...

 

[precisionworks]

My VFD's let me over-drive the motors up to about 300% of rated capacity.

I run 1hp TEFC 3P motors at 3hp for about 20 minutes ...

I'm not an electrical engineer, nor a VFD expert, but I've set up a few dozen drives. Never have seen one that allows what you describe unless you have a 3hp drive controlling a 1hp motor. Configure the drive parameters so that the drive is set up for the full load amps of the 3hp motor & the drive will happily pump out all the current demanded by a 3hp motor. It will work until the motor gets tired of that abuse & smokes out, but 3ph motors are cheap enough that you can keep a few on hand :oops:

Speaking of which, it makes a lot more sense to buy a 3hp motor. With 3X the torque and 3X the hp, the larger motor will do more work more easily than a little 1hp motor that's operated well beyond design limits.

FWIW, a number of current CNC lathes, mills & machining centers are dual hp rated. Something like 10hp continuous and 15hp for 15 minutes. Again, this is done by sizing the drive for the largest hp (15hp in this example), running a constant speed cooling fan, limiting the over drive to 150%, and limiting the time to 15 minutes of full load use.

300%, no matter how you get there, is a recipe for failure. Even the most heavily constructed Vector Drive motor or TENV motor will have a short life under those conditions.

 

[precisionworks]

A new motor will give me a soft start, smoother motor and really low min RPM. I still have to keep changing gears though.

The VFD will do all of the above, plus provide instant reverse for power tapping, plus instant stop, plus allow for an increased top speed (run the max freq to 90 Hz and your top speed will be 150% of current max, as long as the spindle bearings will tolerate that).

No reason not to do that since you already have the drive & a 3ph 2hp motor is pretty cheap on eBay :D

 

[alexmin]

The VFD will do all of the above, plus provide instant reverse for power tapping, plus instant stop, plus allow for an increased top speed (run the max freq to 90 Hz and your top speed will be 150% of current max, as long as the spindle bearings will tolerate that).

No reason not to do that since you already have the drive & a 3ph 2hp motor is pretty cheap on eBay :D

Barry,

you made my day! :hitit:

I am ordering a new 3ph 2hp motor.
Eh, what the hell, when I get a 5hp setup I can always use extra 2hp motor to make a VFD conversion for my drill drill press or band saw!


BTW Where can I get a pulley for PM1236? It seems to have unusual double belt configuration.

 

[precisionworks]

Where can I get a pulley for PM1236?

As someone will surely say ... you have a lathe, make one :rolleyes: But I wouldn't (been there, done that, costs way too much in time).

You may want to look at stacking a pair of single groove cast iron sheaves, like those in the Browning BK series:

http://www.emotorstore.com/productD..._A_subCatID_E_325_A_catID_E_17_A_brandID_E_23





Or look at a Baldor/Maska sheave like the MAS42 or MAS53.

http://www.maskapulleys.com/images/produit/sheaves.pdf

Look on page 20 for specifics.

 

[liveforphysics]

Never have seen one that allows what you describe unless you have a 3hp drive controlling a 1hp motor.

Just a 3hp drive on a 1hp motor alone wont do it, unless you're loading it to have a crazy slip percentage, but then inductive heating would be awful.

The trick is to run a 3hp drive at 440v on a 1hp motor terminated for 220v. The winding resistance is halved, allowing the drive to easily control phase currents to be 2-3x higher (just like if it had a real 2-3hp 440v motor for a load).

Now, as I understand heat transfer, in order for the motor to dissipate that much heat, that motor would need to be at (again roughly) 9 times the temperature RISE for which it was rated. In other words, if the motor rating was at, say 40C rise over ambient at its nameplate current, and given that you'r not really changing the cooling system (Its TEFC, right), wouldn't the motor now rise to roughly 360C above ambient? In steady state, anyway.

If the motor winding was only a resistor, this would be the case. Fortunately it's an inductor/resistor. :)

Motor heat is best modeled at Pin-Pout = Heat. Otherwise in the example above, you've got some over-unity power creation happening in the motor, and I wana get in on that gig! ;)


The difference between Pin to Pout is determined by the efficiency of the motor. It also is roughly a constant value up until you get saturation. This is why it's important to use a quality motor with thin lams.

Likewise, efficiency of the motor is of course critical.
A motor making 3hp at 70% efficiency is making over double the heat of a motor making 3hp at 85% efficiency.

A relatively small improvement in motor efficiency has huge impact on heat.


Lastly, looking at the heat transfer abilities, what is different in the cooling abilities of motors in the same NEMA package size (at least for TEFC)? Same rotor, same lams, same little fan, same RPM, same case material, same surface area. The difference between the 1hp and the 3hp for motors that have the same package is just the winding resistance. You can copy that lower resistance by terminating the motor voltage tap windings in parallel rather than series, or if you wish to go one step further (or an in-between step) you can terminate in delta rather than wye, and use the the drive frequency to compensate for the RPM change.

300%, no matter how you get there, is a recipe for failure. Even the most heavily constructed Vector Drive motor or TENV motor will have a short life under those conditions.

I agree a motor sure can cook inside, and it's so cheap it's not a big deal to replace.

However, with the VFD's automatic field weakening when it's not loaded, and my heavy load cutting duty cycle on the lathe being low, generally always sub 30%, the average thermal loading on the motor isn't too bad. I'm on the same motor, and it's been a champ for me for a few years now. It may be cooked and just about to let go inside, or maybe my low duty cycle never gets it over-heated, but as long as it keeps chugging along for me, I will keep being happy with it. :)




I always chuckle at how big induction motors are relative to the power they produce.

I build PMBLDC3P (permanent magnet brushless dc 3-phase) motors for light electric vehicles and large scale RC toys. I've got 9hp motors that are the size of a softball and 4lbs each, and affordable off the shelf items! Too bad the RPM range makes them require complex reduction stages for most applications.

Here is a pic of a pair of these monsters on my 65mph DIY E-bike. :) The same power with induction motors would be 5 times the weight of the bike and batteries. lol

http://www.hobbycity.com/hobbyking/...80-100-B_130Kv_Brushless_Outrunner_(eq:_70-55)




The record right now for a PMBLDC motor is 1.4lbs for 7hp continous. I'm shooting to break the record, so far I'm not even half way there though. lol. Dream big. ;)
http://www.launchpnt.com/about-us/n...owerful-new-halbach-array-electric-motor.html

 

[brickbat]

… and my heavy load cutting duty cycle on the lathe being low, generally always sub 30%, the average thermal loading on the motor isn't too bad.

OK, wait. Earlier you said you got 3HP for 20 minutes continuous, and you got that at 100% of rated speed, right?

The trick is to run a 3hp drive at 440v on a 1hp motor terminated for 220v. The winding resistance is halved, allowing the drive to easily control phase currents to be 2-3x higher (just like if it had a real 2-3hp 440v motor for a load).

I understand, but don't see how it's a trick. I can see how, with a V/Hz drive, it would allow you to get, say 2HP at twice the rated speed out of a 1 HP motor, but it has nothing to do with your claimed ability to get 300% torque out of a motor. It’s a 480/240V machine, right? And rated line current is what, 3.5A @ 240V? And you get 300% torque, right?

At what line current? Wouldn’t it be in the neighborhood of 10.5 A?

OK, let’s agree on a simple motor model. (If you please, I know a motor is not a resistor, and I’m not making any extraordinary “Free Energy” claims.)

Your machine is rated at 1HP, that’s represents about 750 W of mechanical output power. And let’s just say that your motor is 80% efficient at full load, which is typical for common 1 HP 3-phase induction motors. Therefore the input power is about 940 W. So, the motor is dissipating about 190W of heat, right?

I propose we model this loss electrically as a resistor, as is commonly done in the texts I’ve seen on AC machines. Physically, it arises from many sources, but the copper wire resistance of the stator, and the resistance of the conductors in the rotor dominate. If we were to increase the current through the wire and rotor conductors (as we’d have to do to get more torque) the current through these conductors also has to increase proportionally. Sure, we get more torque and thus more mechanical power out of the motor, but, if we double the current, the loss increases by the square, (P=I^2 R in a resistive component) so 4 times the resistive loss. Our original 190W of loss becomes 760W. And if we triple the current, the resistive loss again increases by the square, so 9 times 190W = 1710W of loss.

Without a mod to the cooling system, being able to dissipate 9 times the power means roughly 9 times the rise over ambient (assuming here that radiative loss is not significant for a motor's cooling).

Now, if you say you're running a short duty cycle, we can drop all this. Because at some small duty cycle, you can get 300% torque out of a motor without smoke, but not for 20 min at a time...

 

[precisionworks]

Why does this remind me of the cheapo air compressors that claim 10hp from a 3hp motor :crackup:

 

[liveforphysics]

I propose we model this loss electrically as a resistor, as is commonly done in the texts I’ve seen on AC machines. Physically, it arises from many sources, but the copper wire resistance of the stator, and the resistance of the conductors in the rotor dominate. If we were to increase the current through the wire and rotor conductors (as we’d have to do to get more torque) the current through these conductors also has to increase proportionally. Sure, we get more torque and thus more mechanical power out of the motor, but, if we double the current, the loss increases by the square, (P=I^2 R in a resistive component) so 4 times the resistive loss. Our original 190W of loss becomes 760W. And if we triple the current, the resistive loss again increases by the square, so 9 times 190W = 1710W of loss.

Without a mod to the cooling system, being able to dissipate 9 times the power means roughly 9 times the rise over ambient (assuming here that radiative loss is not significant for a motor's cooling).

This is simply not the way a motor works my friend. :)

The efficiency of the motor includes the resistive loss, and the efficiency is roughly fixed assuming the saturation points aren't reached.

Lets use the 80% efficiency example.
At 1hp, it's 932w in and 746w (1hp) out, and 186w of motor heating.

At 3hp, it's 2796w in, and 2238w (3hp) out, and 558w of motor heating.

Lets say the motor efficiency tapers off to 75% when it's over driven. This would mean:
2984w in, 2238(3hp) out, and 746w of motor heating.

What does it take to handle and increase of 4x the heating on a motor?
Lots of easy options. 4x the airflow over the motor surface, 2x the airflow and 2x the delta-T, a motor mount that sinks to the large metal frame of the machine (as my lathe motor does), or lots of other options.


Now think about this. When they make the 3hp version of the motor in the same NEMA frame/package, if the efficiency is similar, and the winding resistance is setup to enable it to draw the current needed to make the power, is there any difference? ;)


I play with the motor speed all the time for an easy adjustment, I think it's likely the biggest advantage of a VFD. If I need to chug through some harsh rough cuts or whatever, the current of course naturally ramps up as the slip% increases, and keeps ramping till it hits the VFD's programmed phase current limits (giving roughly 3hp). I just do hobby work rather than production work, so my loading is different than factory machine running constantly of course. I make a cut, measure, changing tooling, check run-out, make another cut, measure, work the math to figure out how much to trim off next, etc. Lots of time where the motor is just sitting spinning the lathe unloaded, and with the field weakening, it pulls a little under an amp at 220v (measuring power into the VFD) just free running. It's definitely getting cooling time and a friendly duty cycle. Most lathe jobs I do are finished in maybe ~20mins if I had to take a wild guess, it's not like I time myself when I'm just tinkering in the garage. In a situation with my Lathe where I needed extended high torque, I would gear down and freq up the motor to 90hz or so, which would of course let the phase currents decrease and still maintain the same torque at the lathe.


Lots and lots of ways to skin a cat. :) Lots of motor options, drive options, and setup of motors and drives.

If you guys run your machines making cuts that put maximum motor loading on them long periods continuously, then by all means, get the toughest setup you can find, maybe over-size the motor.

I may have badly perceived the posters intent for the machines use, but it seemed to me like he was more of a hobbiest like myself looking to add some function and ease of use to a machine, and do it on a tight budget.

I offered a suggestion that has been working great for me. Didn't mean to stir up any trouble. :(

 

[brickbat]

Lets say the motor efficiency tapers off to 75% when it's over driven.

That's a fascinating theory of yours - that an induction motor efficiency drops to only 75% at THREE times its rated torque output. Got a link to Baldor's specs showing that ?

Efficiency curves for induction motors shown in my texts don't go to 300% of rated torque, but at 150% of rated torque, efficiency is dropping off FAST. The fact is that losses in an induction motor are predominately (Not totally, granted) resistive at high loads, and resistive losses increase as the square of the current. No way around Ohm's law that I've yet found...

No offense intended, but I'm concluding that your 20 minute/300% claim was, erm, optimistic. :)

Maybe all we can agree on is that a 2HP VFD will be a nice setup for the OP's machine...

I suppose you have a physics background. Do you, by chance, have a text on AC machines handy? Mine are all pretty old, and I'd like to get something newer...

On edit:

This is simply not the way a motor works my friend. :)

The folks at Mathworks seem to think it is. Note particularly R1 and R2 in the model - they are resistors used to model the stator and rotor losses.

http://www.mathworks.com/access/helpdesk/help/toolbox/physmod/elec/ref/inductionmotor.html