A cyclone is operated in a closed circuit with a ball mill. The cyclone is feed from a rod mill with a slurry that has a density
of 1250 kg/m3 while the solid bulk density is 2900 kg/m3. Cyclone overflow goes to flotation circuit consists of a rougher, a scavenger and one cleaner cell. The slury feed to the rougher has 15% solids (overflow) while the feed to ball mill (underflow) from the cyclone has 50% solids with a mass flow rate of 40 t/h.
• Draw a possible flow sheet of the process
• Calculate the mass flow rate of the feed from rod mill to cyclone (hint: you need to calculate % solids in feed and dilution ratios)
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• Draw a possible flow sheet of the process
• Calculate the mass flow rate of the feed from rod mill to cyclone (hint: you need to calculate % solids in feed and dilution ratios)
Get more help from Chegg
1 answer:
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Answer:
4m/s
Explanation:
We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted
Mathematically
\
We also know that Power = force x velocity ..................(i)
The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load
=> power = mg x Velocity........(ii)
While hoisting the load at at constant speed only the potential energy of the mass increases
Thus Potential energy = Mass x g x H...................(iii)
where
g = accleration due to gravity (9.81m/s2)
H = Height to which the load is hoisted
Equating equations (ii) and (iii) we get
m x g x v =
thus we get v = H/t
Applying values we get
v = 6/1.5 = 4m/s
Given:
diameter of sphere, d = 6 inches
radius of sphere, r = = 3 inches
density, = 493 lbm/
S.G = 1.0027
g = 9.8 m/ = 386.22 inch/
Solution:
Using the formula for terminal velocity,
=
(1)
where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A =
Volume of sphere, V =
Using the above formulae in eqn (1):
=
=
=
Therefore, terminal velcity is given by:
=
inch/sec