0.4 kg/s of dry sea-shore sand, containing 1 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?
Question : 0.4 kg/s of dry sea-shore sand, containing 1 per cent by mas...
The problem involves a mass balance around the two stages. If x kg/s salt is in the underflow discharge from stage 1, then:
salt in feed to stage 2 = (0.4 × 1)/100 = 0.004 kg/s.
The sand passes through each stage and hence the sand in the underflow from stage 1 = 0.4 kg/s, which, assuming constant underflow, is associated with (0.4/2) = 0.2 kg/s water. Similarly, 0.2 kg/s water enters stage 1 in the underflow and 0.4 kg/s enters in the overflow. Making a water balance around stage 1, the water in the overflow discharge = 0.4 kg/s.
In the underflow discharge from stage 1, x kg/s salt is associated with 0.2 kg/s water, and hence the salt associated with the 0.4 kg/s water in the overflow discharge = (x × 0.4)/0.2 = 2x kg/s. This assumes that the overflow and underflow solutions have the same concentration.
In stage 2, 0.4 kg/s water enters in the overflow and 0.2 kg/s leaves in the underflow.
Thus: water in overflow from stage 2 = (0.4 – 0.2) = 0.2 kg/s.
The salt entering is 0.004 kg/s in the underflow and 2x in the overflow — a total of (0.004 + 2x) kg/s. The exit underflow and overflow concentrations must be the same, and hence the salt associated with 0.2 kg/s water in each stream is:
(0.004 + 2x)/2 = (0.002 + x) kg/s
Making an overall salt balance:
0.004 = x + (0.002 + x) and x = 0.001 kg/s
This is associated with 0.4 kg/s sand and hence:
salt in dried sand = (0.001 × 100)/(0.4 + 0.001) = 0.249 per cent
The same result may be obtained by applying equation 10.16 over the washing stage:
S_{n+1}/S_{1}=(R-1)/(R^{n+1}-1) (equation 10.16)
In this case: R = (0.4/0.2) = 2, n = 1, S_{2}=x, S_{1}=(0.002+x) and :
x/(0.002+x)=(2-1)/(2^{2}-1)=0.33x = (0.000667/0.667) = 0.001 kg/s
and the salt in the sand = 0.249 per cent as before.
Considering a single stage:
If y kg/s is the overflow feed of water then, since 0.2 kg/s water leaves in the underflow, the water in the overflow discharge = (y – 0.2) kg/s. With a feed of 0.004 kg/s salt and 0.001 kg/s salt in the underflow discharge, the salt in the overflow discharge = 0.003 kg/s.
The ratio (salt/solution) must be the same in both discharge streams or:
(0.001)/(0.20 + 0.001) = 0.003/(0.003 + y – 0.2) and y = 0.8 kg/s