Variable frequency drive 10

[asifraza0]

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

 

[Skogsgurra]

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

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[waross]

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

 

[GroovyGuy]

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.

 

[asifraza0]

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

 

[jraef]

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

 

[itsmoked]

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

 

[waross]

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

 

[LionelHutz]

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.

 

[asifraza0]

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

 

[GroovyGuy]

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.

 

[asifraza0]

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

 

[GroovyGuy]

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

 

[asifraza0]

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

 

[jraef]

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

 

[dogzzz]

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
[highlight #EF2929]Yes you can run the motor from 0 to rated RPM.
[/highlight]
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

 

[LionelHutz]

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.

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.

 

[waross]

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

 

[asifraza0]

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

 

[Skogsgurra]

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

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.