Calculate Torque Needed for 6" Auger on 17% Incline

[SevenToFive]

I am trying to figure out the torque needed for an auger, but I seem to run into a wall. The auger is 6.00 inches in diameter and 44 feet long and on a 17% incline. The weight of the product going through the auger is 29lbs per cubic foot. So I calculated the cubic feet of the auger, 3.14 * r^2*H = 3.14*0.5^2*44=34.57 cubic feet. Then 34.57ft^3*29lbs*ft^3 = 1002lbs.

I am having a difficult time figuring out how to calculate the torque needed, even if it is a conservative value to know how large of a gearbox we would need with a 40:1 gear ratio and if we can use a 1HP motor on it. The only other information that I have is that the flow rate is 1200 bushels per hour.

Any help is greatly appreciated.

 

[haruspex]

From the throughput, the slope and the length you can calculate the power needed. Through what vertical height is the feedstock lifted, what work is done in the process, and at what rate?
For the torque there is not enough info. You need to know the pitch of the auger, i.e. in each turn how far does the feedstock move.

 

[Asymptotic]

While looking around to see how larger conveyor augers are dimensioned (it appears 1-1/4" ID and 6" pitch is a common size for 6" OD) found an engineering resource http://www.kwsmfg.com/services/screw-conveyor-engineering-guide/kws-calculators.htm# that may be useful. The industry magazine Powder and Bulk Engineering has a variety of articles on the nitty-gritty of conveying, and this one in particular is a good read http://www.powderbulk.com/enews/2013/editorial/story_pdf/pbe_112013RIHF.pdf

 

[SevenToFive]

Haruspex, thanks for reconfirming that I am missing something to complete the calculations.

Asymptotic, Thanks for the links.