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 (μ), radius of rolls (r), radius of product (d), and radius of feed (R):
cos α = (r + d) / (r + R) → 1
where α is related to the coefficient of friction by the relation,
μ = tan α
Angle of nip = 2 α
We have, μ = 0.29
Therefore, α = tan-1(0.29) = 16.17o
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 cmLast Modified on: 30-Apr-2024
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