Q3. 0.4 kg\'s of dry sea-shore sand, containing I per cent by mass of salt, is t

Question

Q3. 0.4 kg's of dry sea-shore sand, containing I per cent by mass of salt, is to be washed with 0.4 kg's of fresh water running countercurrently to the sand through two classifiers in series. It may be assumed that perfect mixing of the sand and water occurs in each classifier and that the sand discharged from each classifier contains one part of water for every two of sand by mass. If the washed sand is dried in a kiln dryer, what percentage of salt will it retain? What wash rate would be required in a single classifier in order to wash the sand to the same extent?

Explanation / Answer

SOLUTION:

Let underflow salt flow rate from first stage = x kg/s

Salt inlet to second stage = 0.4 kg/s * 1%

= 4 x 10^-3 kg/s
sand outlet in the underflow from first stage
= 0.4 kg/s

For constant underflow

Mass flow rate of water = 0.4 kg/s / 2 = 0.2 kg/s

Water inlet to first stage in underflow = 0.2 kg/s

Water inlet to first stage in overflow = 0.4 kg/s

Water balance around first stage

water flow rate in overflow = 0.4 kg/s

Salt rate in overflow = (x * 0.4)/0.2

= 2x kg/s

At constant molal overflow, the concentration is same in overflow and under flow.

Water inlet Flow rate in second stage overflow = 0.4 kg/s

Water outlet Flow rate in second stage underflow = 0.2 kg/s

Overall Water balance in second stage
water flow rate in overflow from second stage

= 0.4 / 0.2

= 0.2 kg/s

Salt inlet rate in underflow = 4 * 10^-3 kg/s

in overflow = 2x kg/s

Total salt = (4*10^-3 + 2x) kg/s

salt flow rate with water stream

= (4*10^-3 + 2x)/2

= (2*10^-3 + x) kg/s

overall salt balance
4*10^-3 = x + (2*10^-3 + x)

x = 1*10^-3 kg/s

salt percentage in dried sand = (1*10^-3 × 100)/(0.4 + 2*10^-3)

= 0.249 %

Part b
For a single classifier

Water flow rate in overflow = z kg/s

Water outlet in underflow = 0.2 kg/s

Water balance

Water in outlet overflow = (z - 0.2) kg/s

Salt rate in feed = 4*10^-3 kg/s

Salt flow rate in underflow outlet = 1*10^-3 kg/s

Salt flow rate in overflow outlet = 3*10^-3 kg/s

For constant molal overflow

(1*10^-3)/(0.20 + 1*10^-3) = 3*10^-3/(3*10^-3 + z - 0.2)

z = 0.8 kg/s

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