1 Mixing Problem

Background

Mixing problems occur quite frequently in chemical industry. We explain here how to solve the basic model involving a single tank (see the figure on the right). The tank contains $1000\;\text{gal}$ of water in which initially $100\;\text{lb}$ of salt is dissolved. Brine – i.e. salt water – runs in at a rate of $10\;\text{gal/min}$ , and each gallon contains $5\;\text{lb}$ of dissolved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at $10\;\text{gal/min}$ .

Problem

Find the amount of salt in the tank at any time $t$

Solution Step 1: Setting up a model

Let $y(t)$ denote the amount of salt in the tank at time $t$ . Its time rate of change is

y^{\prime}=\text{salt inflow rate}-\text{salt outflow rate}

(1)

$5\;\text{lb}$ times $10\;\text{gal}$ gives an inflow of $50\;\text{lb}$ of salt. Now, the outflow is $10\;\text{gal}$ of brine. This is $10/1000=0.01$ (=1%) of the total brine content in the tank, hence 0.01 of the salt content $y(t)$ , that is, $0.01y(t)$ . Thus, from (1) we obtain the following ODE as a model:

y^{\prime}=50-0.01y=-0.01(y-5000).

(2)

Solution Step 2: Solution of the Model

The ODE (2) is separable. Separation, integration, and taking exponents on both sides gives

\frac{dy}{y-5000}=-0.01dt,\qquad\ln|y-5000|=-0.01t+c^{\ast},\qquad y-5000=ce^{% -0.01t}.

Initially, the tank contains $100\;\text{lb}$ of salt. Hence $y(0)=100$ is the initial condition that will give the unique solution. Substituting $y=100$ and $t=0$ in the last equation gives $100-5000=ce^{0}=c$ . Hence $c=4900$ . Hence the amount of salt in the tank at time $t$ is

y(t)=5000-4900e^{-0.01t}.

(3)

This function (see the graph on the right) shows an exponential approach to the limit $5000\;\text{lb}$ . Can you explain physically that $y(t)$ should increase with time? That its limit is $5000\;\text{lb}$ ? Can you see the limit directly from the ODE?