How much kg of salt is in the tank after 30 minutes?

Question

A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.5kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min. How much kg of salt is in the tank after 30 minutes?

Answer 1

Accumulation = in - out for mass of salt and for Volume of solution

1) dm/dt = .5 kg/L *(6 L/min) - (m/V)*(4 L/min) = 30 - 4 m/V m(t=0) = 0

2) dV/dt = 6 L/min - 4 L/min = 2 L/min V(0) = 100 L

From (2) V = 100 + 2t

Subbing into (1) : dm/dt = 30 - 4m/(100+2t)

1st order equation in the form dm/dt + 4/(100+2t) * m = 30 is classic m' + p(t) m = q(t) problem which is covered in your textbook. (IC is m(0) = 0). Once you have solution, plug in t = 30 minutes. You can check answer on Wolfram alpha.

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