Crushing, Screening & Conveying

Crushing, Screening & Conveying 2017-04-04T06:57:13+00:00
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Screen efficiency calculation (3 replies)

10 months ago
firebird 10 months ago

Recently part of our linear screen is damaged. Most of the Coarse ones are going to the undersize chute. The formula for screen efficiency is 100-(% of undersize in the coarse product). However since the oversize is going to the undersize, could I determine my screen efficiency using this formula?

100-(% of oversize in the undersize).

10 months ago
raghdmuhi 10 months ago
1 like by David

Screen efficiency is obtained using different equations, depending on whether your product is the oversize or undersize fraction from the screen. The following information is required to calculate the two screen efficiencies;-

  • Qms(f)  = Mass flow rate of solid feed.
  • Qms(o) = Mass flow rate of solid in the screen overflow.
  • Qms(u) = Mass flow rate of solid in the screen underflow.
  • Mu(f)    = Mass fraction of undersize in the feed.
  • Mu(o)    = Mass fraction of undersize in the overflow.
  • Mu(u)    = Mass fraction of undersize in the underflow.


The screen efficiency based on the oversize (Eo) is then given by the equation;-


            Qms(o) * [1 - Mu(o)]
Eo =     -----------------------------                           ( Equation 1)
             Qms(f) * [1 - Mu(f)] 


and the screen efficiency based on the undersize (Eu) is then given by;


            Qms(u) *  Mu(u)
Eu =  -------------------------------                        ( Equation 2)
             Qms(f) *  Mu(f)


Overall screen efficiency is then calculated by multiplying these two efficiencies together ie;-

E = Eo * Eu                                               ( Equation 3)

Paul Morrow
10 months ago
Paul Morrow 10 months ago

Efficiency is the ratio of the undersize obtained in screening to the amount of undersize available in the feed. It is found by the formula:

And more

10 months ago
firebird 10 months ago

Really appreciate that guys pawl and raghdumhi. Helps a lot.thanks


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