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Variable frequency drive
10

Variable frequency drive

Variable frequency drive

(OP)
Hello Experts:
Need your advise in designing an electrical system

I'm designing a screw Auger system for conveying solid material from ground to inside of a tank. Based on the solid loading and the solid flowrate the calculated rated power is about 1-2 hP. I want to use a 4-5 hP motor with VFD to drive the Auger; I'm using 4-5 hP motor based on two reasons 1) The calculated power is approximate and I want to be conservative 2) Future expansion
My questions are:
1)) Is there any issue with using a 4-5 hP motor with a VFD even though the calculated Auger power is much lower
2) My calculated Auger speed is 5 rpm; can I turn down the motor to 5 rpm using the VFD
3) Shall I use Induction or synchronuous motor?
4) Shall I use the motor with 3600 rpm or with 1750 rpm
5) Any issue or suggestion with the overall design

Thanks in advance for your help
Araza

Asif Raza

RE: Variable frequency drive

There are two fundamental issues:
1. There is no mentioning of a gear to reduce motor speed from (as you say) 3600 or 1750 RPM down to screw speed, which in your case is 5 RPM.
2. The margin between calculated (from mass/second and elevation) power and power needed by the auger system seems to be very small. Most of the power is used to overcome friction. In comparison the actual transportation work is very low.

If you try to reduce speed with the aid of a VFD only, you will need a much larger motor (more than 400 times larger) and VFD than the ones you mention. The reason is that the motor's torque is by no means increased when you reduce speed with the VFD. Rather the opposite. If you have a suitable transmission in place, it may work with your numbers. But the margin is still very narrow.

The other questions cannot be answered before you have got the fundamental numbers right.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: Variable frequency drive

3 You won't easily find a synchronous motor at that small power. (Except possibly a very expensive special design.)
Use an induction motor. (asynchronous)
4 Use 1760 RPM. Plentiful and cheap. If you use 3560 RPM you will have to use double the gear reduction ratio.
5 What Gunnar said.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Variable frequency drive

Hi Azif,
Per the replies above, what is the turn-down ratio of the drive system, (ie belt sheave diameters and gear box ratio)?
Also what is the min & max speeds that you want the auger to turn at? I suspect that you would like the min auger speed to be 5rpm, correct?
Since this is a constant torque application, you must be careful wrt the motor's min speed, else them motor will overheat and burn out. Most motors can typically only operate down to 1/2 speed with a constant torque load, without forced cooling (ie 600prm for a 1200rpm, or 6-pole, motor). The good news is that you can overspeed the motor beyond it's synchronous speed of 1200 rpm, to say 1800rpm (at 90hz). This effectively increases the overall turn-down ratio. Therefore, the largest turndown that you can expect from a constant torque system, without forced cooling, is (say) 3 or 4 to 1, perhaps a little higher if the motor is oversized.

You are correct that the motor and VFD need to be slightly oversized. It is a good thing to have some additional torque in your back pocket for a sticky start. Just be sure that the mechanical components can handle this additional torque without breaking.

Of course there are vector drives, that can provide NP torque down to 0 rpm, but I don't think you need to go down that path for this application.

RE: Variable frequency drive

(OP)
Hello All:

Thanks for all the advice
The minimum auger rpm is 5-10 rpm
The Auger is belt driven
Yes I realize now that it would not be possible to turn the motor from 1750 rpm down to 5 rpm with VFD only
Till now I was not even planning to have a gearbox
I've talked to one electrical shop and they said that use a VFD to turn the motor down to 450 rpm and then use a gear box with a gear reducer of 1:90 to turn it down to 5 rpm
Now I'm planning to use 5 hP motor with a 5 hP VFD and a gear reducer
Araza

Asif Raza

RE: Variable frequency drive

Out of curiosity, how did you determine the motor HP requirement without knowing the motor speed? The two are integral. If you determined it based on the torque requirement at 5RPM, 5HP at 1750RPM is going to be a major over kill! That's not to say it will not work, but you also don't want so much excess torque that you twist off the auger shaft if it jams.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

RE: Variable frequency drive

It may not be possible but if it is this is by far the best way to determine your needed torque. Load the auger with whatever it will be angering and then turn it with a bending bar torque wrench. You simply cannot beat actual measurement on something as squirrelly as an auger. You only alternative is lots of guessing and then putting in a way-too-big motor and more expensive VFD to run it.

BTW This would be a constant torque (higher capacity) VFD application.

Keith Cress
kcress - http://www.flaminsystems.com

RE: Variable frequency drive

2
I suggest selecting a gear ratio that will let you run the motor at full rated speed. If you run the motor at half speed you will need double the HP to develop the same torque at the same final speed.
I live in the land of augers. The rural population of Alberta may be exceeded by the number of augers used for moving grain.
An additional ratio reduction is often achieved by driving the gear box with a belt and pulley reduction of 10:1 or even 15:1
You may not be able to find the exact ratio gear box that you want. Use a high ratio belt reduction to fine tune the final reduction ratio.
Most augers here run the motor at full speed. I have never seen a VFD on a grain auger. Gas engine augers are usually run at full speed.
Consider running the auger faster than 5 RPM (quite a bit faster) and using some sort of gate to restrict the feed rate.
I hope that you are going to drop the material into the top of the tank and not try to force it in the bottom against the weight of existing material.
If this is a stoker such as is used for a coal burner please tell us. That may change some suggestions.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Variable frequency drive

2
You do realize that with a VFD the HP is more or less the ratio of the speed reduction? A 5hp, 1750rpm motor run at 450rpm via a VFD is capable of approximately 1.3hp. This falls well outside

jraef - Can you explain that comment? The required motor HP is the required HP at the load shaft + connection losses. The connection can be a gear reducer or belt drive or simply a direct coupling.

RE: Variable frequency drive

(OP)
All:

The calculated horsepower required is only 0.3 hP, it is based on Auger loading and the solid flow-rate. Assuming an Auger efficiency of 20% and a motor efficiency of 90%. I'm getting a motor power of 1.7 hP. Hence I want to use a 3 hP motor
The auger is feeding the material inside the tank from the top, it is not being forced against the existing material

Any feedback on using the 3 hP motor with a gear reducer of 1:90 would be highly appreciated
Araza

Asif Raza

RE: Variable frequency drive

Hi Azif,
Please find below, a typical Speed-Torque capabilities curve for a 3hp induction motor. This curve is based on a GE 4-pole (ie 1800rpm) motor under PWM type control. This curve is 25 years old, but your motor OEM should be able to provide a similar curve. With this motor your can achieve a (almost) 10:1 turn-down, with a derate of 25% motor torque.

This implies that the mechanical turndown required is now 50:1 (ie 250/5). This can easily be accomplished via a belt drive and gear-box combination.

I am curios however in that you stated above that 'The minimum auger rpm is 5-10 rpm'. What is the max rpm that you require? The scheme below can provide a maximum auger speed of approx 50rpm (assuming a mechanical turndown of 50:1).

I have successfully used this scheme many times on constant-torque applications much-much larger than 3hp.

RE: Variable frequency drive

(OP)
Hi Groovyguy: Thanks

Are you turning down the motor speed from 1800 rpm to 250 by a belt drive?
Based on the solid loading and the feed flow-rate the auger needs to run at 5 rpm min at 20 kg/hr, 12 rpm at 50 kg/hr and 20 rpm at 100 kg/hr. So the auger speed is between 5 - 20 rpm

Araza


Asif Raza

RE: Variable frequency drive

Hi Asif,
No, the above graph illustrates that, with this particular motor, you can achieve a 10:1 turndown with the motor alone.( ie this motor can provide 75% NP torque from 250rpm thru 2430rpm.)
Now that I know that your auger speed is 5 to 20 rpm (ie a 4:1 turndown), please find the curves below for both a 4-pole (1800rpm) motor and a 6-pole (1200rpm) motor.
Note,
1) Either of these options will work fine, although I prefer the 6-pole option as it has a lower bottom end speed. (ie less onerous requirements for the mechanical speed reduction hardware).
2) There is no motor derate for either the 4-pole or 6-pole option (in the case of these particular GE motors).
3) You should obtain similar derate curves from your motor OEM. I wouldn't think that they will be much different from the curves below, but I would still confirm.
4) I suspect that the mechanical turndown hardware will consist of a belt-drive and gear box, with an overall turndown of:
- 90:1 for the 4-pole option, or
- 60:1 for the 6-pole option



RE: Variable frequency drive

(OP)
Thanks Groovyguy:

Can you please tell me how to read these curves?
what significance it has on the motor selection? I plan to use 3 hP motor. Would this be sufficient using your de-rating curves
Araza

Asif Raza

RE: Variable frequency drive

Quote:

jraef - Can you explain that comment? The required motor HP is the required HP at the load shaft + connection losses. The connection can be a gear reducer or belt drive or simply a direct coupling.

He didn't say the required HP at the SHAFT was 5HP, he said he wanted to use a 5HP MOTOR, but had not yet picked the speed. I was pointing out that he put the cart before the horse.

Bottom line, HP is a function of torque at a speed. How much torque in a 5HP 4 pole motor vs a 5HP 2 pole motor? TWICE as much right? So if he had not yet decided on the speed of the MOTOR, how did he determine 5HP was right, or even the 1-2HP he said at first, or even the 0.3HP he says now? And now we know there IS a belt drive already, but we don't know if it's 1:1 or a ratio?

Asif,
If you have determined 0.3HP at 5RPM we can extrapolate the required torque to be .3 x 5250/5 = 315 ft-lb. If you buy a 5HP 4 pole (@1750RPM) motor, that's 15 ft-lb at the motor shaft. So to take it to 5RPM, a 350:1 gear reduction gets you 5250 ft-lbs of torque at the auger! Even using a 90:1 reduction gives you 1350 lb-ft of torque at the shaft at roughly 20RPM, and then if turned down to 5RPM with a VFD (constant torque) at 15Hz, is still 1350 ft-lb at the shaft, roughly 4X what the auger needs, and likely way more than the shaft can handle if jammed.

2HP 4 pole motor and a 90:1 reduction nets out to 540 ft-lb at the shaft, plenty of fudge (>40%) to account for losses* I would think. But I like waross' idea better, 3HP at a lower ratio so that the motor runs faster and can keep itself cooler, then IF you need to vary the speed turn it down with the VFD. If you do NOT need to vary the speed, don't use a VFD. Go with a larger reduction.

* By the way "Motor at 90%" would have to do with the ELECTRICAL efficiency of the motor. The HP rating IS ALREADY the mechanical rating, you don't need to factor efficiency in mechanically again. And 20% efficiency of the auger??? What does that mean? 80% of the torque input to the shaft will be lost in friction? Wow...


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

RE: Variable frequency drive

My questions are:
1)) Is there any issue with using a 4-5 hP motor with a VFD even though the calculated Auger power is much lower
There will no issues . I hope the VFD is rated for 5HP or 3.67 Kw.

2) My calculated Auger speed is 5 rpm; can I turn down the motor to 5 rpm using the VFD
Yes you can run the motor from 0 to rated RPM.

3) Shall I use Induction or synchronuous motor?
Induction motor is ok , less expensive , and your torque requirement is less and is not more then 300 N-m.
Also you will need DC source for providing field to the Synchronous motor


4) Shall I use the motor with 3600 rpm or with 1750 rpm
Any one is OK , anyway you will be using gear box to reduce the speed to 5RPM , other thing is if you use 3600 RPM torque on the shaft will be less and with 1750 it will be more .
5) Any issue or suggestion with the overall design
I think no problem , I just want to know the inverter rating of VFD
Thanks in advance for your help
Araza

RE: Variable frequency drive

If you want to best utilize the motor then size the gear reduction so the auger runs at the highest required speed when the motor is at rated speed. In other words, the auger is at 20rpm when a 4-pole motor is at 1750rpm. Otherwise, you have to oversize the motor to get the HP you want at the auger.

As examples.

Setup a 5hp motor to run the auger at 20rpm with the motor at rated speed and you get an available shaft torque of about 1313 ft-lbs. In other words, you will get 5hp at the auger shaft, minus the reducer losses. To run the auger at 5rpm will require the motor to run at 25% of its rated speed. This is called a 4:1 turn down ratio, which should be a relatively easy motor to source. You will have the same torque available across the 5-20rpm range if you use a motor capable of a 4:1 turn down ratio.

Setup a 5hp motor to run the auger at 20rpm with the motor at 26% rated speed and you get an available shaft torque of about 338 ft-lbs. In other words, you will get 1.28hp at the auger shaft, minus the reducer losses. Further, to run the auger at 5rpm will require the motor to run at 6.5% of it's rated speed. This is a 15:1 turn down ratio and will most likely require a specialized motor. If your motor isn't capable of a 15:1 turn down ratio then the available shaft torque will be lower when running at 5rpm.



Quote:

He didn't say the required HP at the SHAFT was 5HP, he said he wanted to use a 5HP MOTOR, but had not yet picked the speed. I was pointing out that he put the cart before the horse.

Bottom line, HP is a function of torque at a speed. How much torque in a 5HP 4 pole motor vs a 5HP 2 pole motor? TWICE as much right? So if he had not yet decided on the speed of the MOTOR, how did he determine 5HP was right, or even the 1-2HP he said at first, or even the 0.3HP he says now? And now we know there IS a belt drive already, but we don't know if it's 1:1 or a ratio?


HP measures how much work the motor can do, regardless of the shaft speed. Sure, the 4-pole motor has twice the torque, but it can still only do the same amount of work as the 2-pole motor. If you mechanically gear reduce both motors to 5rpm then you get the same torque at 5 rpm from both motors.

If the auger requires 0.3hp then a 0.5hp motor can work no matter what rpm the motor is rated for, unless the gear/belt reduction is below 60% efficiency.

When you decide to electrically reduce the motor speed via a VFD then you need to know that the motor HP reduces by the speed ratio as I pointed out in my last post.

RE: Variable frequency drive

Quote (OP)

Any feedback on using the 3 hP motor with a gear reducer of 1:90 would be highly appreciated
1760 RPM/90 = 19.6 RPM
Close enough to 20 RPM.
Use a VFD to turn down to 450 RPM for an output speed of 5 RPM.
If you need the full 20 RPM, set the VFD for 1840 RPM (61.3 Hz)

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Variable frequency drive

(OP)
Hi All:
Thanks for the excellent feedback
One last question:
What is the minimum turndown achievable on the motor using a VFD
If I'm using a 1800 RPM inverted duty motor; can I achieve turndown lower than 1:4 i.e lower than 450 rpm. How much lower can I go?
Thanks very much
Araza

Asif Raza

RE: Variable frequency drive

With en external fan or water cooling, plus a vector drive and an encoder - you can have 100% turn down. There are also sensorless vector drives that can do it without any encoder.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: Variable frequency drive

just to clarify some numbers:

you said that the calculated power at auger shaft is 1/3 HP @ 5 rpm. This corresponds to 477,5 N.m at auger`s shaft

you want to use a 3HP @ 1750 rpm. This corresponds to 12,27 N.m at motor shaft (rated torque)

total speed reduction i=1750 rpm / 5 rpm = 350 (mechanical reduction)

total torque a 3HP @ 1750 rpm motor can give at auger shaft will be: 12,27 x 350 = 4294,5 N.m

3HP is 9 times bigger than what you need (4294,5/477,5 = 8,99)

1HP @ 1750 rpm .... T=4,1 N.m

477,5/350 = 1,36 N.m at motor shaft is what you need.

4,1 / 1,36 = 3,01

I would use a 1 HP @ 1750 rpm which has a 3 times safety marging and cost less than a 3 HP. Also the reducers will cost less

RE: Variable frequency drive

If buying an "inverter duty" motor, most of the responsible suppliers will provide you with a maximum turn down ratio. The highest I have seen is 1000:1 for a motor that has an integral separately powered cooling fan (blown motor). That's not to say it cannot operate at zero speed for braking operations on things like hoists, but that is never something one would do continuously, so it would be ludicrous to state that. 1000:1 however would be 0.06Hz continuously, so slow it may as well be stopped. You will find that the "inverter duty" moniker will come with turn down ratios all over the map, but I generally ignore those that say less than 100:1 if I really need it to turn that slow, and if it says 100:1 without a separate blower, I am suspiciously cautious.

But the reason I put "inverter duty" in quotes is that this term has multiple definitions, many of them driven by marketing, not engineering. The onus is upon you to investigate the specifications and source of the specifications to the best of your abilities. As a general rule dating back to the days of the Roman Empire, remember "Caveat emptor" which means "Let the buyer beware", telling us that even that long ago, if the price looks too good to be true, it probably is. I have seen motors burn out lately when used on inverter drives and when I've seen classic signs of standing wave insulation damage, was told "but these are inverter duty motors, that should not have happened!" Pulled the spec sheets and all they meant was that the motor was TEFC derated by one size to allow for added heating at low speeds, defined as 6:1 turn down, total BS as far as being suitable for being run on an inverter.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

RE: Variable frequency drive

The minimum depends on how cheap you go with a motor.

The maximum can be much more than 4:1. Look at the Marathon inverter rated motors.
microMAX - 20:1 turndown ratio for the fan cooled version.
Black Max - 1000:1 turndown ratio.

RE: Variable frequency drive

Hi Azif,
I apologize for taking so long to get back to you, but occasionally I need to do some work. Bosses are funny that way.bigsmile

The intent of the curves, that I attached above, was to illustrate the turn-down ratio that an typical induction motor can provide without over-heating.

From the above curves, most motors driving a constant-torque load, can provide a 4:1 turndown without having to derate the motor or having to supplying an external blower fan. With a derate of 25% on a (6-pole) motor, a turndown of 10:1 can be achieved, again without an external blower fan. A turndown of > 10:1 will require either further derating or a blower fan.

BTW, the value of 0.3hp that you provided above, please confirm that this is the maximum power requirement for a fully-loaded auger at 20rpm. Do you have a (full-load) torque value for the auger? Just curious ....

It is possible to go as low as zero rpm with a VFD controlled motor, but you don't want to go there unless it's absolutely necessary. I can't imagine that you would require a turndown > 10:1, but perhaps I should not make that assumption.

jraef is correct in his statements wrt marketing hype wrt 'inverter rated' motors. I usually will ask for a IEEE-841 inverter rated motor, they cost a tad more. If cost is a prime concern, you could ask for a 'IEEE-840' (or '841 light') motor, they cost less due to less onerous testing, but will have all of the 841 features.

I would further recommend a dv/dT filter on the VFD output. These filters are inexpensive and help to protect the motor windings from stress.

Regards,
GG

RE: Variable frequency drive

(OP)
Here is what I'm doing now:
I buying a inverted duty, 3 hP, 4 pole motor with 1750 rpm. I'm turning it down to approximately to 450 rpm using a VFD
I also have a gear reducer 1:20 to turn down the rpm further
There is a pulley of 8" on the auger and a 3" pulley on the motor which gives further rpm reduction

Lukin1977 I liked your calculation. Can you use the above values to generate a new set of calculation for the torque requirement and to check whether there is enough torque available to move the auger
The auger needs 0.1 hP at 5 rpm and 0.3 hP at 17 rpm

Thanks Araza

Asif Raza

RE: Variable frequency drive

It looks like at your top auger speed, 20 RPM, the motor will be turning at 1067 RPM.
It is wasteful to design a 1750 RPM motor to run at 1067 RPM.
The 3 HP motor will develop (3x1067/1750=2) 2 HP at 1067 RPM.
You have had several suggestions to gear the motor to run at full speed at your highest desired output speed.
This will allow the motor to develop full HP, and run cooler. Less load due to higher reduction ratio and better cooling due to a higher motor speed when the auger is turning at slower speeds.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Variable frequency drive

(OP)
Waross:

You are absolutely right; this is the correct way to doing it.
This is a trial project and the budget is tight. I'm using whatever equipment I can find in my store to make it happen
I just happen to have a 3 hp motor mounted on a 1:20 gear box
Araza

Asif Raza

RE: Variable frequency drive

Asifraza0: Use the following formula:

P = T x S / 9550

P = Power in kW
T = Torque in N.m
S = Speed in rpm

1 HP = 750 W (aprox)

RE: Variable frequency drive

If that is what is available, then by all means use it.
Yours
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Variable frequency drive

Should work, but you really only have a 1.8hp motor. This means the maximum torque would be 472ft-lbs on the auger shaft. Operating in constant torque mode means you could get this torque at all speeds from 5-20rpm.

The motor will have to vary from 267 to 1067 rpm. Why would you turn the motor down to 450rpm???

If you were buying the motor like you first indicated, you'd likely be better off with a 1200rpm, 2hp motor.

RE: Variable frequency drive

2
(OP)
All:

I replying to close the loop here
I've designed the system using a 3 hP, 1800 rpm motor with a gear reducer of 1:30 and a Auger pulley with a belt ratio of 3/8
The Auger is moving very smoothly and doing the desired work and I'm quite happy with its performance
Thanks all for all your help
Araza

Asif Raza

RE: Variable frequency drive

Great to hear asifraza0!

Thanks for coming back and telling us.

Keith Cress
kcress - http://www.flaminsystems.com

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