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

 

[lukin1977]

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

 

[jraef]

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

 

[LionelHutz]

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.

 

[GroovyGuy]

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.

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

 

[asifraza0]

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

 

[waross]

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

 

[asifraza0]

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

 

[lukin1977]

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)

 

[waross]

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

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

 

[LionelHutz]

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.

 

[asifraza0]

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

 

[itsmoked]

Great to hear asifraza0!

Thanks for coming back and telling us.

Keith Cress
kcress -

http://www.flaminsystems.com