Mixing-Tank Models
Mixing-Tank Models
Assume the liquid is thoroughly mixed, so the solute concentration is uniform throughout the tank.
Quantities and units
| Symbol | Meaning | Units |
|---|---|---|
| \(V\) | constant liquid volume | volume |
| \(x(t)\) | amount of solute at time \(t\) | mass |
| \(a\) | equal volumetric inflow and outflow rate | volume per unit time |
| \(m\) | solute inflow rate | mass per unit time |
The subscripts “in” and “out” denote non-negative stream rates; “net” denotes the change inside the tank. Here \(a>0\) and \(V>0\).
\(\begin{aligned}\left(\dfrac{\mathrm dV}{\mathrm dt}\right)_{\!\mathrm{net}}&=\left(\dfrac{\mathrm dV}{\mathrm dt}\right)_{\!\mathrm{in}}-\left(\dfrac{\mathrm dV}{\mathrm dt}\right)_{\!\mathrm{out}}\\ \dfrac{\mathrm dV}{\mathrm dt}&=a-a=0\end{aligned}\)
Concentration and solute rates
| Quantity | Expression | Units |
|---|---|---|
| Tank concentration | \(\displaystyle\frac{x(t)}{V}\) | mass per volume |
| Solute inflow | \(m\) | mass per unit time |
| Solute outflow | \(\displaystyle\frac{x(t)}{V}a\) | mass per unit time |
For each stream, \(\left(\dfrac{\mathrm dx}{\mathrm dV}\right)_{\!\mathrm{stream}}\) means solute mass per unit volume carried by that stream (its concentration).
\(\begin{aligned}\left(\dfrac{\mathrm dx}{\mathrm dt}\right)_{\!\mathrm{in}}&=\left(\dfrac{\mathrm dx}{\mathrm dV}\right)_{\!\mathrm{in}}\left(\dfrac{\mathrm dV}{\mathrm dt}\right)_{\!\mathrm{in}}=\dfrac{m}{a}\,a=m\\ \left(\dfrac{\mathrm dx}{\mathrm dt}\right)_{\!\mathrm{out}}&=\left(\dfrac{\mathrm dx}{\mathrm dV}\right)_{\!\mathrm{out}}\left(\dfrac{\mathrm dV}{\mathrm dt}\right)_{\!\mathrm{out}}=\dfrac{x}{V}\,a\end{aligned}\)
Apply “rate in minus rate out” separately to liquid volume and solute mass. Equal liquid flow rates keep \(V\) constant; they do not generally keep \(x\) constant.
\(\begin{aligned}\left(\dfrac{\mathrm dx}{\mathrm dt}\right)_{\!\mathrm{net}}&=\left(\dfrac{\mathrm dx}{\mathrm dt}\right)_{\!\mathrm{in}}-\left(\dfrac{\mathrm dx}{\mathrm dt}\right)_{\!\mathrm{out}}\\ \dfrac{\mathrm dx}{\mathrm dt}&=m-\dfrac{x}{V}\,a\\ &=m-\dfrac{ax}{V}\end{aligned}\)
The coefficient \(\dfrac{a}{V}\) has units of inverse time, so \(\dfrac{a}{V}x\) has units of mass per unit time, matching \(m\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\).
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