Screw Loads
My first point is that to handle 1 Kg/hr of fruit with a 5 cm diameter screw at 88 rpm seems to be an extremely low cross sectional loading that is grossly inefficient for a crushing operation. There are many factors to take account of when calculating the forces on a screw. The torque resulting from a force at the end of a helicals crew depends on the axial load applied at this point, the inclination of the flight face in contact and the friction coefficient of the flight surface with the product. Whereas the blade helix angle varies from the flight tip to its root at juntion with the shaft, the dominant area is near the periphery. This can by crudely treated as the mechanics of a rotating wedge on the bald at a geometric mean diameter for a reasonable approximation. To this must be added the torque to convey the material, deal with any shear if the inlet is flooded and overcome tip drag and possible trapping in the clearance layer between the screw flight and the casing at the rate of feed involved. This is a much more detailed analysis and highly dependent on the nature of them material being handled and the circumstances of the equipment geometry. For such a modest duty, a one off appliciation would be best undertaken by a trial, if the drive specification were sensitive, or accept the cost of supply by a specialist manufacturer. ■
Auger Design Calculations
Hi
I'm designing an auger inside a tube for crushing fruit and I'm trying to work out the forces acting on it.
The design is 16cm long and 5 cm diameter and it turns at 88rpm. I think it needs a torque of 6 NM. Its throughput needs to be 1kg a hour.
I was trying to calculate the horziontal and shear forces if possible ??
Thanks. ■