Design of a Screw Feeder

[bjhershberger]

Does anyone have any information on how to calculate starting torque for a screw feeder?

The auger will be used to empty a hopper sized to hold 25 tons of Portland. Currently we use a drag chain to pull the material out but the customer wants us to switch to an auger design. Since it is a mobile application the hopper has to be in the approximate shape of a rectangular box and the material will pack down on the auger.

Any suggestions would be greatly appreciated.

-Bernard

 

[PeterCharles]

You need to give better details of your requirements. Don't expect too much 'book learning' in materials handling, most knowledge resides with specialist equipment manufacturers accumulated over years (and bitter, costly experiences)
Approach reputable suppliers with your requirement and let them supply a screw feeder suitable for the application.

 

[ccw]

Hi BJHershberger,
While I agree with PeterCharles about seeking help agressively from material handling engineers in cases like this, you do have an experience base at your hands that should be utilized. The draw chain has to shear and propel the compacted portland similar to what the auger will have to do, so the scale of reaction forces, mechanical power, etc., for the same delivery, should be about the same. Of course, this assumes that the draw chain feed design was successful. Other important parameters will be the pitch, diameter, and length of the auger. It seems that a half pipe, slightly larger than the diameter of the auger, would be more efficient than a rectangular box hopper bottom

--CCW

 

[XELR8]

Since it's New Years and I want you to keep your job, I will perform a basic approach that you will need to verify.

Assume: M = 4 ton/hr, Rho = 150 lbs/ft^3, 10" Auger Conveyor, 30'-0 long, the motor will be direct coupled with 1800 RPM.

Conservatively calculate the mass of material to be displaced per second. Assume the material will fill a 10" diameter cylinder 30 feet long.

Vol of material =(Pie*D^2)/4) * L = (Pie*(10/12)^2)/4) * 30' = 16.4 ft^3

Mass of material = Rho*Vol = 150 lbs/ft^3 * 16.4 ft^3 = 2460 lbs.

Calculate the required 10" shaft speed in RPM for 4 tph: 1- 10" screw revolution is 7.57 ft^3/hr

Volumetric flow rate = (4 ton/hr) * 2000 lbs/ton / 150 lbs/ft^3 = 53.33 ft^3/hr

RPM = Volumetric flow rate / displaced material in one rev = (53.33 ft^3/hr) / (7.57 ft^/hr) = 7 RPM

Calculate the acceleration to move the entire mass. (This will be conservative because the entire mass is not moved instantaneously at start up).

7 RPM = 0.117 rev/sec, thus 8.57 sec/rev

V2 = V1 + at
a = 0 + V2/t
a = 0 + (0.117 in/sec)/8.57 sec = 0.014 in/ sec/sec

So the force on the screw shaft is:

F = ma
F = 2460 * 0.014 = 33.59 force lbs

The starting Torque to turn the 10" screw = F * r = 33.59 * ( 10/2) = 167.9 in lbs or 14 ft lbs

The motor shaft HP without reduction is:

HP = 14 * 1800 / 5252 = 4.79 HP. I would use a 10 HP motor.

Rule of thumb:
1) 3 phase motor at 460 V - 1.25 Amp/HP

Therefore, you should be pulling about = 4.79 HP * 1.25 AMP/HP = 5.99 AMP, without efficiency or friction loss accounted.

I hope this helps your situation.