Chemical Engineering - Mechanical Operations - Discussion
Discussion Forum : Mechanical Operations - Section 4 (Q.No. 10)
10.
For a cyclone of diameter 0.2 m with a tangential velocity of 15 m/s at the wall, the separation factor is
Discussion:
3 comments Page 1 of 1.
Mahesh said:
9 years ago
Separation factor = 2 * v^2/gD.
= 2 * 15 * 15/9.81 * .2.
= 230.
= 2 * 15 * 15/9.81 * .2.
= 230.
Sai said:
8 years ago
Separation factor for cyclone Fc/Fg = u^2 / r.g.
So, the answer is A.
So, the answer is A.
(1)
Manoj said:
5 years ago
Separation factor =v^2/rg.
v=15,(r=d/2 (0.2/2) = 0.1), g=9.81.
Ans =229.9.
So the Answer is D.
v=15,(r=d/2 (0.2/2) = 0.1), g=9.81.
Ans =229.9.
So the Answer is D.
(3)
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