In inclined screw conveyor, as we increase the angle of conveyor from ground, its capacity decreases. Here inclination factor comes in play. Say, if angle of inclination is 15%, inclination factor is 0.75. It means, conveyor capacity will be 0.75times of horizontal conveyor capacity. Now, my question is how this inclination factor affects motor power & torque. Where does this factor fit in motor power, toque and flow rate calculation. Is it something like, I should divide the designed flow rate (calculated assuming horizontal conveyor) by inclination factor? and use this value for subsequent calculation? Am I on right track?

Hi Rutujab, I recommend you to check the techincal features of the conveyor you are using. Generally the technical features have different kind of informations such as: - the basic formulas for their functioning - operating conditions (...and also if they can work with an inclination angle) Maybe it could be also a problem of leakage between the screw and the body of the conveyor when the gravity increases due to inclination.

Good day Rutujab, I would refer you to the CEMA 350 standard Ch. 7. It details the calculations required to determine motor horsepower, and has information regarding inclined and vertical screw conveyors. It and the CEMA 300 standard are available as a bundle at the CEMA webstore.

It would appear that the horsepower required to power the conveyor is dependant on the mass of the material to be moved and the length of the conveyor. I have a slide rule published by FortWorth Steel and Machinery Co about their BeeLine screw conveyors that has no mention of the incline angle in calculations of HP, only the amount to material to be moved. As you mentioned, the mass of material to be moved decreases with the increase of the inclination of the conveyor so the HP requirements would also decrease all other factors being the same.

Actually, since an inclined screw conveyor is used to move material up vertically as well as horizontally, the amount of energy (and therefore over time, horsepower) needed to move a given amount of material would increase based on the overall height which needs to be traversed, in addition to the energy needed to move the material horizontally along the conveyor. Your statement would only be correct if the screw conveyor were moving material down, in which case a horizontal auger would be much more efficient.

Hello, PierArg - I am designing conveyor, not using one. So, in this case, I will have to give all these details Randall Wink: - you are right that mass of material to be moved decreases with the increase of the inclination of the conveyor. But it means, if you have to lift material to higher position (i.e. if you are conveying matt. from bottom to top position), HP requirement will increase, as conveyor's capacity is decreased to carry material. Right jthutcheson?

Corrected Posting Based on MY interpretation of what my little slide rule indicates that if you're moving 10,000 lbs of portland cement per hour in an 20 inch diameter screw that is 45% (max recommended) loaded, the screw speed should be about 100 RPM. If the conveyor is 100 feet long the HP required is about 1.4 HP to move the material. To run the basic empty conveyor, babbet bearing take about 2.9 HP and with ball bearings it takes 1.45 HP. So with ball bearings my conveyor would take about 2.85 HP (1.4 material + 1.45 conveyor). Now, let's say that the inclined angle is 45 degrees and the loading goes from 45% to 30%, the screw speed goes to 150 RPM to maintain the material flow, the material HP stays the same at 1.4 HP but now is divided by the cosine of the angle of 45 or 1.4/.70711 or about 1.98 HP. Additionally, the screw turning at a higher RPM to maintain the flow means that the empty conveyor HP increases to about 2.15 HP. So with ball bearings my conveyor would now take about 4.13 HP (1.98 material + 2.15 conveyor). Hope this helps in your quest.

I found this reference. http://www.krk.com.br/html/produtos/phoenix/Design_Fundamentals.pdf In the first pages there are the basic formulas to desing a conveyor belt. As you can see, it reflects what Randall wrote.

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