A pair of rolls is to take a feed equivalent to spheres of 3 cm in diameter and crush them to spheres having 1 cm diameter. If the coefficient of friction is 0.29, what would be the diameter of rolls?

Calculations:

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

cos a = (r + d) / (r + R) à 1

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

m = tan a

Angle of nip = 2 a

We have, m = 0.29

Therefore, a = tan^{-1}(0.29) = 16.17^{o}

And we have, d = 0.5 cm; R = 1.5 cm

Substituting for the known quantities in equn.1,

cos (16.17) = (r + 0.5)/(r + 1.5)

0.9604 = (r + 0.5)/(r + 1.5)

r + 0.5 = 0.9604 (r + 1.5)

r - 0.9604 r = 1.4406 - 0.5

r = 23.753 cm

Radius of rolls = 23.753 cm

Dia of rolls = 2 x 23.753 = 47.5 cm