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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 $c$an do is to keep separate equations for matter and volume balan$c$e a$n$d o$n$ly deduce concentration equations in the very end.2011-04-20

1 Answers 1

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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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    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.2011-04-22