# Thread: Diluition problem

1. ## Diluition problem

Dear all,
I have a practical problem to submit to you:

I have a reservoir of "R" cubic meters of water.
The water flow in/out from this reservoir is 80 liters/seconds.

To the inlet stream, I add a solution which contain 20% Aluminium, with a rate of "A" mg of Al/minute.

The question is:
which is the maximum concentration, measured in the outlet stream, that I can reach ?

Which is the equation that can describe the increasing in concentration of Al in the outlet tube, expressed as mg Al/minute (starting from the start of addition) ?

Thank you a lot for your help !

Giova

2. You have to make an assumption about how well the reservoir is "stirred", and the only one I can think of to make is that it is very well stirred, so the Al at the input is immediately at the output. Physically, though, this may not be likely.

Anyway, if you let x be the concentration of aluminum in the reservoir, then dx/dt will consist of two components - it will increase due to adding the aluminum, and decrease due to replacing the water at the current concentration with fresh water.

To solve equations like these, you need to separate the variables, so you'll have dx/(cx+d)=dt. Then integrate both sides and solve for x.

Post again in this thread if you're still having trouble.

3. Hi Hollywood.

Ok, I can accept your assumption.
But how I can practically integrate both sides and solve ?
Can you do a formula that I can to applicate (or a little Excel sheet) ?
Or can you solve for me for a time = 0.5, 1, 2,3 hrs, A=0.4 mg Al/min and volume = 1 cubic meter ?

Thank you,

Giova

4. Ok. Let's first convert the units of all your data.

Let r be the size of the reservoir in liters, so that will be R times 1000.
Let f be the flow rate in and out of the reservoir, 80 liters/second.
Let a be the rate of addition of aluminum, A mg/minute times (1 minute/60 seconds) times (1g/1000mg) = A/60000 grams/second.
So x is the concentration in grams per liter.

The aluminum in the reservoir changes due to two processes: addition of aluminum at rate a, and removal of reservoir fluid at rate f, so removal of aluminum at rate f times x. So we have:

$\displaystyle \frac{dx}{dt}=\frac{1}{r}(a-fx)$,

$\displaystyle dx=(\frac{a}{r}-\frac{f}{r}x)dt=-\frac{f}{r}(x-\frac{a}{f})dt$

$\displaystyle \frac{dx}{x-(a/f)}= -\frac{f}{r} dt$. This is called the method of separation of variables, by the way.

$\displaystyle \int{\frac{dx}{x-(a/f)}}=\int-\frac{f}{r}dt$

$\displaystyle \ln{(x-\frac{a}{f})}=-\frac{f}{r}t+C$, where C is the constant of integration.

$\displaystyle x-\frac{a}{f}=Ce^{-ft/r}$. I'll transform the constant of integration without changing its name (here I replaced e^C with C).

$\displaystyle x=\frac{a}{f}+Ce^{-ft/r}$

We can evaluate C by using the initial condition x=0 when t=0, so C=-a/f.

$\displaystyle x=\frac{a}{f}(1-e^{-ft/r})$

And just as a sanity check, we can see x=0 when t=0, and x goes to a/f at infinity, where a/f is the concentration of the water coming in.

I think that answers all your questions. Post again in this thread if you're still having trouble.

- Hollywood