Answer to Question #226917 in Chemical Engineering for Lokika

Question

Mesh no 28, 35 and 48 are arranged in Taylors standard sieve series, in Taylors series mesh no 35 is the next smaller screen to the mesh no 28 & mesh no 48 is the next smaller screen to the mesh no 35. Here screen opening size of mesh no 28 is 0.589 mm. then what is the screen opening size of mesh no 48....?

Answer 1

From the given data and mass balance equation, we can determine the mass flow rate of the undersize and overflow streams.

F = 0(Oversize)+U(Undersize)

0+U = 100

(0 x 0.163) + (U x 0) = (100 x 0.106)

0 = 65t/h

11 = 100-65 = 35 t/h 

from the graph observe the aperature size at 50% partition coefficient and we will get = 3.2 mm

d₅₀ = 3.2 mm