alane1994 said:
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200L of dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Any assistance would be very appreciated.
These types of problems need to be set up carefully.
Let $x(t)$ denote the amount of solute in the tank, $c(t)$ denote the concentration of the solute in the solution, and let $V(t)$ denote the volume of the solution at time $t$. Thus, we have that $x(t)= c(t)V(t)$. In this problem we're told that water is flowing in (denote this rate by $r_i$) and being drained (denote this rate as $r_o$) at the same rate (thus $r_i=r_o=r=2\text{ L/min}$); hence the volume $V(t)$ of the solution remains constant (i.e. $V=200\text{ L}$). Now, the amount of concentration $c_i$ flowing into the tank is $0\text{ g/L}$ since fresh water is being added. The amount of concentration leaving the tank is $c_o(t) = \dfrac{x(t)}{V}=\dfrac{x(t)}{200}\text{ g/L}$. With this, we have enough information to set up the differential equation.
If $\Delta x$ denotes the change in solute in the solution, then we have that
\[\Delta x = \{\text{grams in}\} - \{\text{grams out}\} = r_i c_i \Delta t - r_oc_o\Delta t \implies \frac{\Delta x}{\Delta t} = r_i c_i - r_oc_o\]
and thus as $\Delta t\to 0$, we get the differential equation
\[\frac{dx}{dt} = r_i c_i - r_oc_o\]
which in our case is
\[\frac{dx}{dt} = -\frac{2x}{200}\implies \frac{dx}{dt} = -\frac{x}{100}\]
Once you solve this simple ODE, you're then left with finding $t$ such that $x(t)= .01x(0)$.
I hope this makes sense!
EDIT: Was ninja'd by MarkFL; I should be quicker when it comes to these things... (feel free to delete if you think it's necessary, Mark; didn't mean to sideswipe you in any way).
