# Material flow rate to power required.

• posted

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.

• posted

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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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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-- 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

Michael

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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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