A cyclone separator is being used to remove particulates from a gas stream. The

Question

A cyclone separator is being used to remove particulates from a gas stream. The current removal efficiency is 96% for an air flow rate through the separator equal to 18 m3/s. Due to a process change, the air flow rate is expected to increase to 22 m3/s.

What percent removal efficiency may be expected under the new condition?

Removal efficiency (R, in %) has been expressed as an empirical correlation in terms of the ratio (r) of particle diameter (d) to %u201Ccut diameter%u201D (dC), defined as the diameter of particles for which removal efficiency is 50%.

R   =   1.5 x %u2013 17.92 x2 + 72.37 x3 %u2013 6

The cut diameter is usually assumed to be inversely proportional to the square root of the air flow rate, i.e., dC = k Q%u20121/2.

Explanation / Answer

amount of air flow inversly proportional to the efficiency

so v=k/efficiency

18=k/96

k=1728

v=k/efficiency

22=1728/efficiency

efficiency=78.54%

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