What is the particle residence time for given flow rates of gas mixture components?

Question

For two different chemical substances there are two open valves with flow rates $$Q_1=a\frac{m^3}{h}\ \text{ and }\ \ Q_2=b\frac{m^3}{h},$$ leading into seperate cables.

Next, the cables join, the substances mix perfectly and they flow along together.

Then during the route, there is an observational volume $V$ (e.g. a cylinder or length $d$ and cross section $A$).

How long is the residence time $\Delta t$ of a particle of the mixture inside the volume?

Answer 1

First, I assume no chemical reactions take place, as I think you implied. Second, absence of a single velocity for a "perfectly mixed" fluid would meant diffusion. Let's rule it out for the diffusion velocities being small. Anyway, to separate once mixed fluid would require something, like temperature gradient or gravity.

Now to the question itself, when we call a speed of the particle a mean speed defined as:

$$\boldsymbol v = \frac{\rho_1 \boldsymbol v_1 + \rho_2 \boldsymbol v_2}{\rho_1 + \rho_2}$$

As far as I see, all comes to the mass conservation law

$$\rho_1 Q_1 + \rho_2 Q_2 = \rho_o Q_o$$

$o$ for the observational volume.

The only thing that hinders from calculating the resulting flow rate is that you don't know the resulting density $\rho_o$. One can do it if the flow is incompressible and one knows the density of the mix. Assuming the incompressible flow, the simplest mix model I guess will yield:

$$\rho_o = \frac{\rho_1 Q_1 + \rho_2 Q_2}{Q_1 + Q_2}$$

The residence time itself:

$$\Delta t = V / Q_o$$

With the simplest model it is

$$\Delta t = \frac{V}{Q_1 + Q_2}$$

So the question is all about compressibility of the flow and the density of the mix.