Question

Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains 90 L (liters) of water and 290 g of salt, while tank 2 initially contains 60 L of water and 245 g of salt. Water containing 50 g/L of salt is poured into tank1 at a rate of 3.5 L/min while the mixture flowing into tank 2 contains a salt concentration of 40 g/L of salt and is flowing at the rate of 4 L/min. The two connecting tubes have a flow rate of 6 L/min from tank 1 to tank 2; and of 2.5 L/min from tank 2 back to tank 1. Tank 2 is drained at the rate of 7.5 L/min. You may assume that the solutions in each tank are thoroughly mixed so that the concentration of the mixture leaving any tank along any of the tubes has the same concentration of salt as the tank as a whole. (This is not completely realistic, but as in real physics, we are going to work with the approximate, rather than exact description. The 'real' equations of physics are often too complicated to even write down precisely, much less solve.) How does the water in each tank change over time

Answer 1

Let $A_1$ be the amount of salt in tank 1 at time $t$, and $A_2$ the amount of salt in the tank 2.

The volume of solution in either tank stays constant. In tank 1, at time $t$ (min) we have

$90\,\mathrm L+\underbrace{\left(3.5\dfrac{\rm L}{\rm min}-6\dfrac{\rm L}{\rm min}+2.5\dfrac{\rm L}{\rm min}\right)}_0t=90\,\mathrm L$

In tank 2,

$60\,\mathrm L+\underbrace{\left(4\dfrac{\rm L}{\rm min}+6\dfrac{\rm L}{\rm min}-2.5\frac{\rm L}{\rm min}-7.5\frac{\rm L}{\rm min}\right)}_0t=60\,\mathrm L$

Then the concentration of salt in tanks 1 and 2 at any given time is $\dfrac{A_1\,\rm g}{90\,\rm L}$ and $\dfrac{A_2\,\rm g}{60\,\rm L}$.

The net rate of change of the amount of salt in tanks 1 and 2 follows a simple rule:

$\dfrac{\mathrm dA_i}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

Each rate is in units of g/min. Each L coming in or going out contributes or removes some salt depending on the flow rate (L/min) and concentration (g/L) of the solution in either tank. For tank 1, we have

$\text{rate in}=\left(50\dfrac{\rm g}{\rm L}\right)\left(3.5\dfrac{\rm L}{\rm min}\right)+\left(\dfrac{A_2}{60}\dfrac{\rm g}{\rm L}\right)\left(2.5\dfrac{\rm L}{\rm min}\right)$

$\text{rate out}=\left(\dfrac{A_1}{90}\dfrac{\rm g}{\rm L}\right)\left(6\dfrac{\rm L}{\rm min}\right)$

Then the amount of salt in tank 1 has rate of change (ignoring units)

$\dfrac{\mathrm dA_1}{\mathrm dt}=-\dfrac{A_1}{15}+\dfrac{A_2}{24}+175$

A similar breakdown for tank 2 shows a rate of change of

$\dfrac{\mathrm dA_2}{\mathrm dt}=\dfrac{A_1}{15}-\dfrac{A_2}6+160$

In matrix form, the system is described by

$\begin{pmatrix}A_1\\A_2\end{pmatrix}'=\begin{pmatrix}-\frac1{15}&\frac1{24}\\\frac1{15}&-\frac16\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}+\begin{pmatrix}175\\160\end{pmatrix}$

You can solve this with the usual eigenvalue method and method of undetermined coefficients. You should get a general solution of

$\begin{pmatrix}A_1\\A_2\end{pmatrix}=C_1\begin{pmatrix}5\\-6-2\sqrt{19}\end{pmatrix}e^{\frac{-7+\sqrt{19}}{60}t}+C_2\begin{pmatrix}5\\-6+2\sqrt{19}\end{pmatrix}e^{\frac{-7-\sqrt{19}}{60}t}+\begin{pmatrix}4300\\2680\end{pmatrix}$

Then use the initial values $A_1(0)=290$ and $A_2(0)=245$ to solve for $C_1,C_2$ and find the particular solution.