Subject
- Posted on
Material flow rate to power required.
- 08-03-2006
August 3, 2006, 6:10 am
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.
Re: Material flow rate to power required.
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
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
Re: Material flow rate to power required.
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
Re: Material flow rate to power required.
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
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
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