# What is the concentration as a function of time for an open system

I know how to find the concentration as a function of time for a closed (or constant volume) system.

if x is the amount of solute;

X'=(rate inflow)(concentration in)-(rate outflow)(concentration outflow)

However I have trouble with systems that have different inflows and outflows (changing volume).

The (concentration outflow) term is the one I am haveing difficulty writing. I know it is (amount of solute out/volume out), but what is the amount of solute out?

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The best thing you can do is to keep separate equations for matter and volume balance and only deduce concentration equations in the very end. –  Raskolnikov Apr 20 '11 at 20:50
The bounty is to reward the current answer. –  JoeHobbit Oct 30 '11 at 1:19

Suppose we have a system with a constant inflow of a solute and a constant volume outflow, although not necessarily constant concentration. The change in volume per unit of time of the system is then

$$\frac{dV_S}{dt}=RV_{in}-RV_{out} \; .$$

This one can be solved easily giving

$$V_S(t)=V_S(0)+(RV_{in}-RV_{out})t \; .$$

For the mass of solvant in the system, we have the equation

$$\frac{dM_S}{dt}=RM_{in}-\frac{dM_{out}}{dt} \; .$$

Of course, the rate of outflow of mass of solvant will depend on the concentration of the solute in the system and the rate of outflow so that

$$\frac{dM_{out}}{dt}=RV_{out} \cdot C_S \; ,$$

leading to the mass balance equation

$$\frac{dM_S}{dt}=RM_{in}-RV_{out} \cdot C_S \; .$$

The concentration of solute in the system is then $C_S=M_S/V_S$ which changes in time as

$$\frac{dC_S}{dt}=\frac{d}{dt}\left(\frac{M_S}{V_S}\right)= \frac{1}{V_S}\frac{dM_S}{dt}-\frac{M_S}{V_S^2}\frac{dV_S}{dt} \; .$$

Combining our equations, we get

$$\frac{dC_S}{dt} = \frac{1}{V_S}\left(\frac{dM_S}{dt} -C_S\frac{dV_S}{dt}\right)$$

$$\frac{dC_S}{dt} = \frac{1}{V_S}\left(RM_{in}-RV_{out} \cdot C_S-C_S(RV_{in}-RV_{out})\right)$$

$$\frac{dC_S}{dt} = \frac{1}{V_S}\left(RM_{in}-C_S\cdot RV_{in}\right)$$

Remembering our solution for the volume

$$\frac{dC_S}{dt} = \frac{RM_{in}-C_S\cdot RV_{in}}{V_S(0)+(RV_{in}-RV_{out})t}$$

which can be solved by separation of variables

$$\frac{dC_S}{RM_{in}-C_S\cdot RV_{in}} = \frac{dt}{V_S(0)+(RV_{in}-RV_{out})t}$$

Integrating gives

$$\frac{-1}{RV_{in}}\log(RM_{in}-C_S\cdot RV_{in}) = \frac{1}{(RV_{in}-RV_{out})}\log(V_S(0)+(RV_{in}-RV_{out})t) + K$$

where we introduced some integration constant $K$ which we'll specify later. Working further out

$$RM_{in}-C_S\cdot RV_{in} = A (V_S(0)+(RV_{in}-RV_{out})t)^{\frac{RV_{in}}{(RV_{out}-RV_{in})}}$$

in which $\exp(K)=A$. The constant $A$ should be chosen in such a way that

$$RM_{in}-C_S(0)\cdot RV_{in} = A (V_S(0))^{\frac{RV_{in}}{(RV_{out}-RV_{in})}} \; .$$

Finally,

$$C_S = \frac{RM_{in} - A (V_S(0)+(RV_{in}-RV_{out})t)^{\frac{RV_{in}}{(RV_{out}-RV_{in})}}}{RV_{in}} \; ,$$

or with the formula for $A$ substituted in

$$C_S(t) = \frac{RM_{in} - (RM_{in}-C_S(0)\cdot RV_{in}) (1+(\frac{RV_{in}-RV_{out}}{V_S(0)})t)^{\frac{RV_{in}}{(RV_{out}-RV_{in})}}}{RV_{in}} \; .$$

Hope this helps. Sorry for the longwinded derivation with little text and too many formulas. Feel free to ask questions if something is unclear.

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I like this! Thankyou! I followed the derivation until d/dt(Ms/Vs). I tried the Product Rule and got something a little different. Did you do more than the product rule here? The d/dt(Vs) that is multiplied by the Ms/(Vs)^2 seems in my estimation to appear out of thin air. Also, did you derive this yourself? (picks jaw up off floor) –  JoeHobbit Apr 21 '11 at 3:16
Hello, I indeed used the product rule as well as the chain rule. If you have a quotient of functions depending on time, say $a(t)/b(t)$, then deriving with respect to time gives $a'(t)/b(t) + a(t) (1/b(t))'$. For the derivative of the inverse function $1/x$ is $-1/x^2$, so with the chain rule, this becomes $a'(t)/b(t) - a(t) (1/b(t)^2) b'(t)$. I did derive this myself, but it is an "extreme" coincidence that I explained the very same problem to a student last year. It was in the analysis text of Prof. Quaghebeur of the KULeuven. You don't know the guy? –  Raskolnikov Apr 21 '11 at 7:36
I haven't heard of the student or the Prof. Quaghebeur. But I have Googled him now. What book was the question in? I found an Analysis book or two written by him on Amazon.com, but they were in French. My current Differential Equations textbook by Edwards and Penney (Prentice Hall) only superficially touches on constant in-and-out flow systems. I have a strong background in Chemistry and naturally was curious. –  JoeHobbit Apr 21 '11 at 16:11
No, it's not the same guy. Besides, the book is a university textbook tailored for the specific analysis course he is giving. So I don't think you can buy it through Amazon. –  Raskolnikov Apr 22 '11 at 9:31