Mixing Problems (Tank Problems)

Why mixing problems are the canonical applied first-order ODE

A tank of liquid with something dissolved in it — salt, pollutant, drug, dye — receives inflow of one concentration and loses outflow of current concentration. The "current concentration" part is what makes it a first-order linear ODE instead of plain algebra: the rate of change depends on how much is in the tank right now.

The template is $\frac{dS}{dt} \;=\; (\text{rate in}) \;-\; (\text{rate out})$ $= \bigl(c_{\text{in}} \cdot r_{\text{in}}\bigr) \;-\; \bigl(\tfrac{S(t)}{V(t)} \cdot r_{\text{out}}\bigr)$

where $S$ is the amount of salt (grams), $V$ is the volume of liquid, $c_{\text{in}}$ is inflow concentration (g/L), and $r_{\text{in}}, r_{\text{out}}$ are flow rates (L/min).

When inflow and outflow rates are equal, $V$ stays constant and you get a clean first-order linear ODE in $S$.

The given problem

• Tank holds $V = 100$ L of pure water → $S(0) = 0$ g.
• Inflow: $r_{\text{in}} = 2$ L/min of brine at $c_{\text{in}} = 5$ g/L.
• Outflow: $r_{\text{out}} = 2$ L/min of well-stirred mixture.

Because flows match, $V(t) \equiv 100$ forever. Perfect.

Step-by-step

Step 1 — Set up the ODE.

Rate in: $c_{\text{in}} \cdot r_{\text{in}} = 5 \cdot 2 = 10$ g/min.

Rate out: $\dfrac{S(t)}{V} \cdot r_{\text{out}} = \dfrac{S(t)}{100} \cdot 2 = \dfrac{S(t)}{50}$ g/min.

$\frac{dS}{dt} = 10 - \frac{S}{50}$

Rearrange into standard first-order linear form (#174): $\frac{dS}{dt} + \frac{1}{50} S = 10$

Step 2 — Solve via equilibrium + transient.

For $S' + P S = Q$ with constants, the equilibrium is $S_{\text{eq}} = Q/P = 10 / (1/50) = 500$ g.

General solution: $S(t) = S_{\text{eq}} + (S_0 - S_{\text{eq}})\, e^{-P t} = 500 + (0 - 500)\, e^{-t/50}$ $\boxed{\, S(t) = 500 \bigl(1 - e^{-t/50}\bigr) \text{ g} \,}$

Step 3 — Interpret the two pieces.

• $S_{\text{eq}} = 500$ g is the equilibrium. At that level, in-rate exactly cancels out-rate. Reachable only asymptotically.
• Time constant $\tau = 1/P = 50$ minutes. After one $\tau$, the gap to equilibrium has shrunk to $1/e \approx 37\%$. After $5\tau = 250$ min, the gap is under 1%.

Equilibrium concentration: $500 \text{ g} / 100 \text{ L} = 5 \text{ g/L}$ — exactly the inflow concentration. Makes physical sense: given infinite time, the tank becomes indistinguishable from the inflow.

Verification

$S(0) = 500 (1 - 1) = 0$ ✓. $S'(t) = 500 \cdot \dfrac{1}{50} e^{-t/50} = 10 e^{-t/50}$. Plug in: $S' + S/50 = 10 e^{-t/50} + 10(1 - e^{-t/50})/1 = 10$. ✓

When inflow ≠ outflow — changing tank volume

If $r_{\text{in}} \ne r_{\text{out}}$, the volume becomes $V(t) = V_0 + (r_{\text{in}} - r_{\text{out}}) \, t$ and the ODE gets messier — still linear in $S$, but with time-varying coefficient: $\frac{dS}{dt} + \frac{r_{\text{out}}}{V(t)} S = c_{\text{in}} r_{\text{in}}.$ Solve using the integrating factor $\mu(t) = \exp\!\left(\int r_{\text{out}} / V(t) \, dt\right)$.

Also, watch for the tank overflowing (filling case) or emptying (draining case); those impose natural stopping times.

Where mixing ODEs actually show up

• Pharmacokinetics: drug concentration in the bloodstream under continuous IV infusion + renal clearance.
• Pollution in lakes/rivers: pollutant mass in a well-mixed body of water with inflow and outflow streams.
• HVAC / indoor air: contaminant concentration with ventilation.
• Chemical reactors: CSTR (continuous stirred-tank reactor) — the bread and butter of chemical engineering.

Same ODE, different units. The story of "well-mixed" is everywhere in process engineering.

Common mistakes

• Wrong concentration in the out-rate. It's the current concentration $S(t) / V(t)$, not the inflow concentration $c_{\text{in}}$.
• Forgetting the initial condition. The ODE alone has infinitely many solutions; $S(0)$ picks one.
• Units. If you mix L and gallons or minutes and hours, your time constant will be off. Pick one system.
• Assuming equal flow rates without checking. Keep the $V(t)$ tracker alive until you know the tank stays filled.

Try it in the visualization

Animate the tank filling with coloured fluid proportional to $S(t)$. Slide $c_{\text{in}}$ to change the equilibrium concentration; slide the flow rate to see the time constant $\tau = V / r$ shrink or grow. Overlay the $S(t)$ curve next to the tank so you can read off how close you are to equilibrium.