Screw Conveyor Speed for a Target Capacity
The auger speed needed to hit a target volumetric capacity.
Example
You enter
- Target capacity (ft³/hr) 220.157
- Screw diameter (in) 9
- Shaft / pipe diameter (in) 2.5
- Pitch (in) 9
- Trough loading fraction (CEMA class) 0.3
You get
- Required screw speed 40.0 RPM
Details, formula, and sources
rpm = target_ft3_hr / (flight_area x (pitch/12) x 60 x loading) (CEMA Book No. 350). A 9 in screw with a 2.5 in shaft, 9 in pitch, at 30% loading needs 40 RPM for 220 ft^3/hr; double the target and it needs 80 RPM (capacity is linear in speed). Divide a mass rate by the bulk density for the volumetric target first. A flagged high RPM means step up a screw size. CEMA and the manufacturer govern.
rpm = target_ft3_hr / ((pi/4)(D^2 - d^2) x (pitch/12) x 60 x loading), all lengths in feet. The inverse of Q = (pi/4)(D^2-d^2) x pitch x RPM x loading x 60.
CEMA Screw Conveyor standard (Book No. 350) capacity method solved for speed, by name.
CEMA-published; the swept-volume capacity relation is public. CEMA and the manufacturer govern.
Estimate. AHJ and licensed professional govern.
Field names used by the API: target_ft3_hr, screw_diameter_in, shaft_diameter_in, pitch_in, loading_fraction, rpm
- Linear in speed capacity is proportional to RPM; CEMA caps the speed by screw diameterCEMA Book No. 350
- Loading fraction per the CEMA material class (light ~30-45%, heavy/abrasive lower); user-suppliedCEMA Book No. 350