Modeling, Units, and Nondimensionalization

Example A tank holds $V = 10\ \mathrm{L}$ of salt water. Brine with concentration $c_{\text{in}} = 3\ \mathrm{g/L}$ flows in at $q_{\text{in}} = 2\ \mathrm{L/min}$, and the mixture flows out at $q_{\text{out}} = 2\ \mathrm{L/min}$, so the volume stays $10\ \mathrm{L}$. At the start the tank holds $M(0) = 10\ \mathrm{g}$ of salt. A natural state is the salt amount $M(t)$ in grams, with $t \ge 0$ in minutes. The numbers $V$, $q_{\text{in}}$, $q_{\text{out}}$ and $c_{\text{in}}$ are parameters, and the concentration in the tank is the derived quantity $c(t) = M(t)/V$.

Example For the tank, salt enters at $q_{\text{in}}c_{\text{in}} = 6\ \mathrm{g/min}$. If the tank is well mixed, the outflow has the tank’s concentration $M/V$, so salt leaves at $q_{\text{out}}M/V = M/5$ grams per minute, and no salt is produced or destroyed. With $M$ in grams and $t$ in minutes, the model is

with $M(0) = 10$. The rate is zero at the equilibrium amount $M = 30$. The difference $D = M - 30$ satisfies $dD/dt = -D/5$, so $D(t) = D(0)e^{-t/5} = -20e^{-t/5}$ and

Its derivative $4e^{-t/5}$ equals $6 - M(t)/5$, and $M(0) = 10$. The amount rises toward $30\ \mathrm{g}$, and the concentration $M/V = 3 - 2e^{-t/5}$ in grams per liter approaches the incoming $3\ \mathrm{g/L}$. The model assumes perfect mixing, a constant incoming concentration and equal flow rates. If the flow rates differed, the volume $V(t)$ would change and the outflow term would become $q_{\text{out}}M/V(t)$.

Example In the inflow rate the liters cancel,

The outflow coefficient $q_{\text{out}}/V = \tfrac{1}{5}\ \mathrm{min}^{-1}$ has units of inverse minutes, so $M/5$ is in grams per minute. Measured in seconds, with $t_{\text{sec}} = 60t$, the same rates are $6\ \mathrm{g/min} = 0.1\ \mathrm{g/s}$ and $\tfrac{1}{5}\ \mathrm{min}^{-1} = \tfrac{1}{300}\ \mathrm{s}^{-1}$, so the model reads $dM/dt_{\text{sec}} = 0.1 - M/300$. Its solution $M = 30 - 20e^{-t_{\text{sec}}/300}$ is the same curve, because $t_{\text{sec}}/300 = t/5$.

Example Every term of

is in grams per minute. In the solution $30 - 20e^{-t/5}$ with $t$ in minutes, the $5$ stands for $5\ \mathrm{min}$, so $t/5$ is a pure number and the exponential makes sense. Writing the outflow as the amount $M$ itself, without a rate coefficient, would subtract grams from grams per minute, and the check fails at once. The equation $dM/dt = 6 + M/5$ passes the check, but its sign says that the outflow adds salt, so it is still wrong.

Example For the tank, $M_{*} = 3\ \mathrm{g/L} \times 10\ \mathrm{L} = 30\ \mathrm{g}$ and $T = (10\ \mathrm{L})/(2\ \mathrm{L/min}) = 5\ \mathrm{min}$. The time scale is a relaxation time. After $t = T$ the gap to equilibrium has shrunk from $20\ \mathrm{g}$ to $20e^{-1}\ \mathrm{g} \approx 7.36\ \mathrm{g}$, about $37\%$ of its starting size. It is not the time at which the tank reaches equilibrium, which never happens at a finite time, because $20e^{-t/5} > 0$ for every $t$.

Example For the tank, $u = M/(30\ \mathrm{g})$ and $\tau = t/(5\ \mathrm{min})$, so $dM/dt = 6\, du/d\tau$ in grams per minute. Substituting into $dM/dt = 6 - M/5$ gives $6\, du/d\tau = 6 - 6u$, and dividing by the common rate $6$ leaves

Its solution $u = 1 - \tfrac{2}{3}e^{-\tau}$ is the amount $M(t)$ found above divided by $30$, with $t = 5\tau$. The same steps for any constant-volume, well-mixed tank with the scales $M_{*} = c_{\text{in}}V$ and $T = V/q$ give $du/d\tau = 1 - u$ again, so only the starting value $u(0) = M(0)/(c_{\text{in}}V)$ tells such tanks apart.