An \(800 \mathrm{gal}\) tank is filled with \(500 \mathrm{gal}\) containing \(80 \mathrm{lb}\) of salt. Brine containing \(5 \dfrac{\mathrm{lb}}{\mathrm{gal}}\) is pumped into the tank at a rate of \(6 \dfrac{\mathrm{gal}}{\mathrm{min}}\text{.}\) The well-mixed solution is then pumped out at a rate of \(8 \dfrac{\mathrm{gal}}{\mathrm{min}}\text{.}\) Find a differential equation for the amount \(A(t)\) of salt in the tank at time \(t\) (in minutes).
Section 1.6 Mixtures
The rate of change in the amount of a substance in a well-mixed solution is equal to the input rate in of that substance minus its output rate. Let \(A(t)\) denote the amount of a substance at time \(t\text{.}\) Further, let \(f_i\) be the flow in, let \(c_i\) be the concentration in, and let \(f_0\) be the flow out. Notice that the rate \(R=(\text{flow})(\text{concentration})\text{.}\) So the volume in the tank is given by Table 1.6.1. Volume in the tank at integer time steps
Then the rate of change of \(A\) is given by
| Time | Volume |
|---|---|
| \(0\) | \(V_0=\text{initial volume}\) |
| \(1\) | \(V_0+(f_i-f_0)\) |
| \(2\) | \([V_0+(f_i-f_0)]+(f_i-f_0)=V_0+2(f_i-f_0)\) |
| \(\vdots\) | \(\vdots\) |
| \(t\) | \(V_0+t(f_i-f_0)\) |
\begin{align*}
\dfrac{\mathrm{d}A}{\mathrm{d}t} = R_i-R_0 \amp = f_i c_i - f_0 c_0\\
\amp = f_ic_i - f_0 \left( \dfrac{\text{mass}}{\text{volume}} \right)\\
\amp = f_ic_i - f_0 \left( \dfrac{A}{V_0+(f_i-f_0)t} \right)
\end{align*}
