Answer to Question #226915 in Chemical Engineering for Lokika

Question

What should be the diameter (mm) of set rolls which accepts feed equivalent to spheres of 50 mm in diameter and crush them to spheres having in diameter of 15 mm? Here the co-efficient friction is 0.30

A) 250

B) 500

C) 600

D) 750

E) None of these

Answer 1

The following formula relates the coefficient of friction ($\mu$), radius of rolls (r), radius of product (d), and radius of feed (R):

$\cos a = \dfrac{(r + d)}{ (r + R)}$

where a is related to the coefficient of friction by the relation,

$\mu= \tan a$

$\mu = 0.30 \\a = \tan^{-1}(0.30) = 16.7°$

And we have, d = 7.5 cm; R = 25cm

Substituting for the known quantities in the first equation

$\cos (16.7°) = \dfrac{(r + 7.5)}{(r + 25)}$

$0.96 = \dfrac{(r + 7.5)}{(r + 25)}$

$r + 7.5 = 0.96 (r +25)\\ r - 0.96 r = 24-7.5\\ 0.04r= 16.5\\ r = 412.5\text{ cm}$

Radius of rolls = 412.5 cm

The closest of which is b.