# What is the particle residence time for given flow rates of gas mixture components? [closed]

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?

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## closed as off-topic by JamalS, Kyle Kanos, Danu, Bernhard, JimMar 30 at 14:23

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do you want a resulting flow rate or do you mean by particle a molecule of either of the substances? –  Yrogirg Aug 15 '12 at 16:56
@Yrogirg: I was expecting to get a single particle velocity. In that case, if I can compute a single flow rate, I guess the residence time for a particle of the the mixture can be computed. If there is not a single velocity, then a velocity for each of the substances is good too, of course. –  NikolajK Aug 16 '12 at 8:10
What kind of answer did you expect? My answer is really elementary, what do you want above that (if any)? –  Yrogirg Aug 16 '12 at 9:01
@Yrogirg: No, the answer is good, I just don't want to miss something. –  NikolajK Aug 16 '12 at 9:03

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.

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Okay, so your answer comes down to "The flow rate of the mixed substance is the addition of the two flow rates before (independed on anything, especially independend of geometry)", am I right? –  NikolajK Aug 16 '12 at 9:01
No, no. First, it is true if the flow is incompressible, just like water or slow gas (< 0.3 Mach) without heating. Second it is true if you take m^3 of the first substance, m^3 of the second, mix and you get 2 m^3 as the result. It is not true for example for water and ammonia (?). I can't remember that gas, water just sucks it almost without changing its volume. –  Yrogirg Aug 16 '12 at 9:06
oh it might be true if these two effects compensate each other. For example during the mixing volume increased due to tricky interaction between molecules, but then it decreased as the result of (undersonic) gas being compressed in a tighter pipe. –  Yrogirg Aug 16 '12 at 9:22
At least mass flow rates add up :) –  Bernhard Aug 16 '12 at 10:51