Mixing Problems

A typical mixing problem deals with the amount of salt in a mixing tank. Salt and water enter the tank at a certain rate, are mixed with what is already in the tank, and the mixture leaves at a certain rate. We want to write a differential equation to model the situation, and then solve it.

The independent variable will be the time, t, in some appropriate unit (seconds, minutes, etc). It may not be so obvious what to use for a dependent variable: the concentration of salt in the liquid, or the amount of salt. It is usually simpler to use the amount, Q (again, in appropriate units such as kilograms). Then the basic principle determining the differential equation is

$$\frac{dQ}{dt} = \begin{array} {c}\mbox{rate of salt} \\ \mb... ...in{array} {c}\mbox{rate of salt} \\ \mbox{leaving}\end{array}$$

Since salt is not created or destroyed in this process, the change in the amount of salt in the tank comes from the balance between what enters and what leaves.

Example:

Initially a tank contains 10000 litres of brine with a salt concentration of 1 kg salt per 100 litres. Brine with 2 kg salt per 100 litres enters the tank at a rate of 20 litres per second. The well-stirred mixture leaves at the same rate. Find the concentration of salt as a function of time.

Solution: Let Q be the amount (in kg) of salt in the tank, and t the time in seconds, with t=0 initially. We have Q(0) = 1/100 (10000) = 100 kg.

The rate at which salt enters is 20 (2/100) = 0.4 kg/sec. The rate at which salt leaves is 20 C(t), where C(t) is the concentration of salt in the tank at time t. Now C(t) = Q(t)/V(t) where V(t) is the volume of liquid in the tank. Since liquid leaves at the same rate as it enters, the volume is a constant (10000 litres). So the differential equation is

$$\frac{dQ}{dt} = 0.4 - \frac{Q}{500}$$

This is a first-order linear equation. The integrating factor is $\mu(t) = \exp(t/500)$, and the general solution is

$$Q(t) = 200 + C \exp(-t/500)$$

Substituting the initial condition t=0, Q = 100, the answer is

$$Q(t) = 200 - 100 \exp(-t/500)$$

Note what happens as $t \to \infty$: $Q(t) \to 200$. Q = 200 is an equilibrium solution: at this value of Q, dQ/dt = 0, so Q would stay constant at this value.

Example:

Take the same setup as in the previous example, but the mixture leaves the tank at only 10 litres per second. Of course the tank will eventually fill up, but we want to know the amount of salt at any time before this.

Solution:

This time instead of being constant, the volume V(t) is a function of time: V(t) = 10000 + 10 t. The differential equation is

$$\frac{dQ}{dt} = 0.4 - \frac{Q}{1000 + t}$$

The integrating factor is $\mu(t) = 1000 + t$. The general solution is

$$Q(t) = \frac{400 t + 0.2 t^2 + C}{1000 + t}$$

and using the initial condition we get

$$Q(t) = \frac{400 t + 0.2 t^2 + 100000}{1000 + t}$$