# Dilution Problems

Consider a tank which initially holds ${V}_{{0}}$ gal. of brine that contains ${a}$ lb. of salt. Another brine solution containing ${b}$ lb. of salt per gallon is poured into the tank at a rate of ${e}$ gal./min., while, simultaneously, the well-stirred solution leaves the tank at a rate of ${f{}}$ gal./min. The problem is to find the amount of salt in the tank at any time ${t}$.
Let ${Q}$ denote the amount (in pounds) of salt in the tank at any time ${t}$. The time rate of change of ${Q}$, $\frac{{{d}{Q}}}{{{d}{t}}}$, equals the rate at which the salt enters the tank minus the rate at which the salt leaves the tank. The salt enters the tank at a rate of ${b}\cdot{e}$ lb./min. To determine the rate at which the salt leaves the tank, we first calculate the volume of brine in the tank at any time ${t}$, which is the initial volume ${V}_{{0}}$ plus the volume of brine added ${e}\cdot{t}$ minus the volume of brine removed ${f{\cdot}}{t}$. Thus, the volume of brine at any time is ${V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}$. The concentration of salt in the tank at any time is $\frac{{Q}}{{{V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}}}$, from which it follows that the salt leaves the tank at a rate of ${f{\ }}\frac{{Q}}{{{V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}}}$ gal./min.

Thus, $\frac{{{d}{Q}}}{{{d}{t}}}={b}\cdot{e}-{f{\ }}\frac{{Q}}{{{V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}}}$, or $\frac{{{d}{Q}}}{{{d}{t}}}+\frac{{f}}{{{V}_{{0}}+{\left({e}-{f}\right)}\cdot{t}}}{Q}={b}{e}$.

Example 1. A tank initially holds 100 gal. of a brine solution containing 20 lb. of salt. At ${t}={0}$, fresh water is poured into the tank at a rate of 5 gal./min., while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time ${t}$.

Here, ${V}_{{0}}={100}$, ${a}={20}$, ${b}={0}$, and ${e}={f{=}}{5}$. So, the equation is

$\frac{{{d}{Q}}}{{{d}{t}}}+\frac{{1}}{{20}}{Q}={0}$. The solution to this linear equation is ${Q}{\left({t}\right)}={C}{{e}}^{{-\frac{{t}}{{20}}}}$.

At ${t}={0}$, we are given that ${Q}={a}={20}$; so, ${20}={C}{{e}}^{{-\frac{{0}}{{20}}}}$, or ${C}={20}$.

Thus, ${Q}{\left({t}\right)}={20}{{e}}^{{-\frac{{t}}{{20}}}}$.

Note that ${Q}\to{0}$ when ${t}\to\infty$, as it should, since fresh water is being added.

Now, let's consider one more example.

Example 2. A tank contains 40 l of solution with 2 g of substance per liter. Salt water containing 3 g of this substance per liter runs in at a rate of 4 l/min., and the well-stirred mixture runs out at the same rate. Find the amount of substance in the tank after 15 minutes.

Note that here we have a problem in terms of liters and grams, not in terms of pounds and gallons. It doesn't matter. The main thing is that all variables should be presented in the same units.

Since initially there are 40 l of solution and each liter contains 2 grams of substance, we have that initially there are ${40}\cdot{2}={80}$ grams of substance.

So, ${V}_{{0}}={40}$, ${a}={80}$, ${b}={3}$, ${e}={f{=}}{4}$.

And the equation is written as $\frac{{{d}{Q}}}{{{d}{t}}}+\frac{{1}}{{10}}{Q}={12}$. This is a linear equation. Its soltuion is ${Q}{\left({t}\right)}={120}+{C}{{e}}^{{-\frac{{t}}{{10}}}}$.

At ${t}={0}$, we are given that ${Q}={a}={80}$; so, ${80}={120}+{C}{{e}}^{{-\frac{{0}}{{10}}}}$, or ${C}=-{40}$.

Thus, ${Q}{\left({t}\right)}={120}-{40}{{e}}^{{-\frac{{t}}{{10}}}}$.

We need to find ${Q}{\left({15}\right)}$: ${Q}{\left({15}\right)}={120}-{40}{{e}}^{{-\frac{{15}}{{10}}}}\approx{111.075}$ grams.