Applications - Mixing Problems

You will learn how to set up and solve differential equations for mixing problems involving tanks with inflow and outflow of different concentrations. This is a classic applied problem that appears throughout engineering and chemistry.

Imagine a tank containing a fixed volume of liquid. A solution with a certain concentration flows in, while the well-mixed solution flows out at the same rate (so the volume stays constant).

Let $Q(t)$ be the amount of solute (in grams, pounds, etc.) in the tank at time $t$. Then the rate of change is:

$$\frac{dQ}{dt} = \text{rate in} - \text{rate out}$$

If the inflow has concentration $c_{\text{in}}$ and flow rate $r$, and the outflow has concentration $\frac{Q}{V}$ (where $V$ is the constant volume), the equation becomes:

$$\frac{dQ}{dt} = r \cdot c_{\text{in}} - r \cdot \frac{Q}{V}$$

This is a first-order linear differential equation, which we already know how to solve using an integrating factor.

The diagram above shows a typical mixing tank setup. Solution flows in from the top with concentration $c_{\text{in}}$ at rate $r$, mixes thoroughly, and flows out the bottom at the same rate.

Example 1: Basic Mixing

A 100-liter tank initially contains 10 kg of salt. Pure water flows in at 5 L/min while the well-mixed solution flows out at the same rate. How much salt remains after 20 minutes?

Solution: $V = 100$, $r = 5$, $c_{\text{in}} = 0$.

$$\frac{dQ}{dt} = 0 - \frac{5Q}{100} = -\frac{Q}{20}$$

Solution: $Q(t) = 10e^{-t/20}$

At $t = 20$: $Q(20) \approx 3.68$ kg.

Example 2: With Nonzero Inflow

A 200-gallon tank starts with 40 pounds of salt. Brine containing 2 lb/gal flows in at 3 gal/min while the mixture flows out at 3 gal/min. Find the amount of salt after 30 minutes.

Solution:

$$\frac{dQ}{dt} = 6 - \frac{3Q}{200}$$

Integrating factor: $e^{3t/200}$. Solving gives:

$$Q(t) = 400 - 360e^{-3t/200}$$

At $t = 30$: $Q(30) \approx 184.6$ pounds.

Example 3: Changing Volume

(Advanced) A tank starts with 50 liters of pure water. Brine at 0.5 kg/L flows in at 2 L/min while mixture flows out at 3 L/min. Set up the equation (note volume is decreasing).

Solution: Volume at time $t$: $V(t) = 50 - t$. The equation becomes more complex but is still solvable with similar techniques.

Mixing problems model real industrial processes: chemical plants blending solutions, water treatment facilities adding chlorine, pharmaceutical companies mixing medications, and even home brewing or cooking when ingredients are continuously added and drained. Environmental engineers use them to track pollutant levels in lakes or rivers when there is inflow from contaminated streams and outflow to the ocean.