A tank with 5000 kg capacity initially contains 1000 kg of water. A 50% by weight solution in ethanol is added at a rate of 500 kg / h. The solution comes out at a rate of 200 kg / h. What is the percentage of ethanol in the tank just when it is completely full?
Total mass at any time in tank = 1000 + (500 - 200)t = 1000 + 300t
Let, E be concentratio of ethanol in the tank after mixing, and it be the exit concentration
(1000+300t)dE/dt = 500*0.5 - 200*E
When the tank is completely full, 1000 + 300t = 5000 thus, t= 13.333 hrs
Thus, integrating, we get
dE/(250 - 200E) = dt/(1000 + 300t)
We get at t=13.333 hrs, E = 0.8225
Thus, the tank has 82.225% Ethanol.
This result was arrived at by using Polymath. The concentration variation with time is as follows:
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