Material flow rate to power required.

How to convert bulk material flow rate available in kg per second (kg/sec) to power required at its driving shaft (of some big drum like
pulley)? Consider that the angle while conveying is an average of theta degrees.
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SandeepSubrati wrote:

You need to know the distance that the material is raised. If it's falling out of a hopper, no power is needed.
Jerry
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Engineering is the art of making what you want from things you can get.

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Jerry Avins wrote:

CONSIDER A DISTANCE OF L METRES FOR THE MATERIAL TO BE RAISED AT ANGLE THETA. BELT SPEED IS UNIFORM AT V METRES/SECOND.
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Jerry Avins wrote:

-- CONSIDER A DISTANCE OF L METRES FOR THE MATERIAL TO BE RAISED AT ---ANGLE THETA. BELT SPEED IS UNIFORM AT V METRES/SECOND.
What is with the shouting? See that key called caps lock? Tap it once and see if perhaps you get lower case. Those who actually worry about communication know that words with lower case can be read much more easily than those in all upper case. The rest of us just call it "polite".
Look at the increase in potential energy of the material (mgh, or mg L sin Theta) per second. Add the energy needed to accelerate the material from zero to the belt speed (mv^2), since you are dropping things onto a belt at an even rate, it comes out to mv^2/sec Next, look at the amount of friction in the system. Loading per roller * friciton coefficient * number of rollers. Add a little bit for the cantankerous roller which wants to sit still, probably about 10 percent. Divide by the efficiency of your motor, probably around 0.9
Then to be safe, double the answer and add 10 percent.
Michael
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Herman Family wrote:

Thank you, Michael. I was preparing a detailed response, but you beat me to it. My opening comment was "Really now, there's no need to shout."
We should add for the naive that engineers and physicists (and this is an exercise in elementary physics) consider that a kilogram is a unit mass. To use formulas that include so-called weights which are given in Kg, one must multiply by g, the acceleration due to gravity. A kilogram weighs 9.8 newtons; forgetting that will yield a result that is an order of magnitude too small. Remember: w = mg.
Jerry
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