Question

How do you find the amount of sugar in the tank after t minutes if a tank contains 1640 liters of pure water and a solution that contains 0.09 kg of sugar per liter enters a tank at the rate 5 l/min the solution is mixed and drains from the tank at the same rate?

Answer 1

This needs a bit of calculus and some thought.

Let's start by defining a few useful quantities:
`\sigma` the rate at which solution enters and drains (l/min);
`\rho` the sugar solution concentration entering the tank (kg/l);
`m(t)` the total mass of sugar in the tank at time `t` (kg);
`V` the volume of the tank (l)

The rate of sugar mass gained from the solution entering is simply `\sigma \rho`. And since the sugar solution concentration at time `t` inside the tank is `m(t)`/`V`, the mass-loss rate by draining is `\sigma m(t)`/`V`

The overall gain rate is just the difference between above gain and loss rates:

`{dm(t)}/{dt} = \sigma \rho - \sigma {m(t)}/V`.

Rearranging this to integrate, we have

`\int_0^{m(t)}{dm'}/{\sigma \rho - \sigma m'\mbox{/}V} = \int_0^t dt'`,

which leads to

`V/\sigma\ln({V\rho}/{V\rho-m}) = t`.

And solving for `m` we finally obtain

`m(t) = V\rho(1-e^{-\sigma t\mbox{/}V})`.

To check this makes sense, note that at the start `(t=0)` the mass of sugar is zero and then slowly rises, reaching an aymptotic value (i.e. after an infinite time) corresponding to the concentration of the solution entering: `m(t\mbox{=}\infty)\mbox{/}V = \rho`.