Mixing Problems 1

Problem 1.

A tank contains $10$ L of water in which initially $y_0$ kg of salt is dissolved. Brine runs in $2$ L per minutes containing $30$% of salt per liter and runs out $2$ L per minutes. The mixture in the tank is kept uniform by stirring.
(a) Determine the amount of salt $y(t)$ in the tank at all times $t>0$.
     Show that $\displaystyle{\lim_{t\to\infty}}y(t)=y_{*}$ independently of $y_0$ where $y_{*}$ is the     equilibrium solution of the problem.
(b) If initially there is no salt in the tank, i.e., $y(0)=y_0=0$, determine $y(5)$.

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(a) Since
$$y'(t)=\text{salt inflow rate} -\text{salt outflow rate},$$
we get
$$y'(t)=0.6-0.2 y(t).$$
The equilibrium solution is
$$y_{*}=3.$$
From the differential equation
$$\frac{1}{0.6-0.2 y}\,dy=1\,dt.$$
Integrating both sides we obtain

$$\begin{align*} &\Step{1A}{-5\ln\left| 0.6-0.2 y\right|=t+c,}\\ &\Step{2A}{\ln\left| 0.6-0.2 y\right|=-\frac{t}{5}+c,} \\ &\Step{3A}{0.6-0.2 y=c e^{-t/5},} \\ &\Step{4A}{y(t)=3+c e^{-t/5}.}\\ \end{align*}$$

From the initial condition
$$c=y_0-3.$$
Using this value of $c$ we obtain
$$y(t)=3+(y_0-3) e^{-t/5}.$$
It yields
$$\lim_{t\to\infty}y(t)=3=y_{*}.$$
(b)
$$y(5)=3-3e^{-1}=1.89636.\qquad\blacksquare$$

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