A solute added to a container with a small hole

[Adel Makram]

Homework Statement

Suppose there is solute $s$ in a bowel containing fluid. There is a tiny hole near the bottom which leaks a small fixed volume of solute $\lambda$ per unit time $dt$. In addition, there is a small added solute to the fluid in a constant rate $\alpha$ so as the volume of the fluid is constant.
What is the equation of $s$ at any $t$?

Homework Equations

The Attempt at a Solution

If there is no added solute, the equation of $s$ should be $s=s_0 e^{-\lambda t}$ where $s_0$ is the initial amount of the solute at $t_0$.
Now, I tried to include the added solute as linear function of $t$.
$ds=(-\lambda s+\alpha)dt$. If $\alpha=0$, we get the equation above. But I do not know how to get $s$ if $\alpha$ is not zero. Or may be, $s$ can be obtained from different approach!

 

[LCKurtz]

Homework Statement

Suppose there is solute $s$ in a bowel containing fluid. There is a tiny hole near the bottom which leaks a small fixed volume of solute $\lambda$ per unit time $dt$.

Do you really mean the hole just leaks solute and no fluid? Wouldn't it leak solute + fluid and concentration would matter?

 

[Adel Makram]

Do you really mean the hole just leaks solute and no fluid? Wouldn't it leak solute + fluid and concentration would matter?

Yes, it leaks fluid and solute. I also assume that the added solute is instantaneously solved in the fluid with no sediments.

 

[LCKurtz]

I have been busy today so haven't had time to reply. I should have commented earlier, but if I understand the problem correctly, you are adding pure solvent to keep the volume steady. I think you need the volume of the bowl in your equations because it affects the concentration. Also I would expect that as time passes the volume of the solute to increase towards the volume of the bowl. Are we thinking of the same problem here?

 

[Ray Vickson]

Homework Statement

Suppose there is solute $s$ in a bowel containing fluid. There is a tiny hole near the bottom which leaks a small fixed volume of solute $\lambda$ per unit time $dt$. In addition, there is a small added solute to the fluid in a constant rate $\alpha$ so as the volume of the fluid is constant.
What is the equation of $s$ at any $t$?

Homework Equations

The Attempt at a Solution

If there is no added solute, the equation of $s$ should be $s=s_0 e^{-\lambda t}$ where $s_0$ is the initial amount of the solute at $t_0$.
Now, I tried to include the added solute as linear function of $t$.
$ds=(-\lambda s+\alpha)dt$. If $\alpha=0$, we get the equation above. But I do not know how to get $s$ if $\alpha$ is not zero. Or may be, $s$ can be obtained from different approach!

Your problem statement is very confusing, and what you write is possibly not what you mean. You say that solute leaks out at rate $\lambda$, and that solute is added so that the fluid volume remains constant. So, for example, if the solute is salt and the fluid is water, the container leaks salt water at rate $\lambda$. To keep the volume constant, something needs to be added at the same volumetric rate $\lambda$. Is the added material just salt, or just water, or is it also salt-water? The wording seems to imply that you add salt alone, and in that case the concentration of salt in the solution must increase over time.

Your analysis seems to be incorrect: without adding salt to the vessel, the fluid (salty water) leaks out at a constant rate $\lambda$, so the salt concentration would remain constant but the volume of fluid would be $v(t) = v_0 - \lambda t$, where $v_0$ is the initial volume. The actual amount of salt present in the vessel at time $t$ would be the $s(t) = \rho v(t)= s_0 - (\lambda s_0/v_0) t$, where $s_0$ is the original volume of salt present in the solution and $\rho = s_0/v_0$ is the initial salt concentration (which remains constant over time). Of course, all this changes considerably if you start adding salt.

Please tell us if the intended problem is as I have described it.

 

[Adel Makram]

Sorry that I was not clear in the description of the problem. I like the analogy of salt and water, so I will consider it here.
Let`s consider a tank has water and salt. The hole near the bottom leaks water and salt. To keep the volume of the fluid constant, we add from the top a volume of water and salt in a same rate of the lost volume from the hole. The added solution has fixed concentration of salt.
I am now wondering what will be the concentration of the salt as a function of time inside the container.
Intuitively, the concentration will depend, somehow, on the ratio between the concentration of the salt in the added volume relative to the rate of lost salt and water. This is because, had we only added pure water, the concentration of the salt would have decreased exponentially with time. And the question, what will be the equation of the salt concentration.

 

[Ray Vickson]

Sorry that I was not clear in the description of the problem. I like the analogy of salt and water, so I will consider it here.
Let`s consider a tank has water and salt. The hole near the bottom leaks water and salt. To keep the volume of the fluid constant, we add from the top a volume of water and salt in a same rate of the lost volume from the hole. The added solution has fixed concentration of salt.
I am now wondering what will be the concentration of the salt as a function of time inside the container.
Intuitively, the concentration will depend, somehow, on the ratio between the concentration of the salt in the added volume relative to the rate of lost salt and water. This is because, had we only added pure water, the concentration of the salt would have decreased exponentially with time. And the question, what will be the equation of the salt concentration.

OK. so the added solution may have a different salt concentration from the original brine in the vessel.

If the incoming salt concentration were the same as the initial in-vessel salt concentration, the in-vessel salt concentration would not change---it would be equivalent to a problem with no hole in the vessel and no added fluid. The case of unequal concentrations leads to a more interesting problem, and I assume that is the one you are asked to address.

 

[Adel Makram]

OK. so the added solution may have a different salt concentration from the original brine in the vessel.

If the incoming salt concentration were the same as the initial in-vessel salt concentration, the in-vessel salt concentration would not change---it would be equivalent to a problem with no hole in the vessel and no added fluid. The case of unequal concentrations leads to a more interesting problem, and I assume that is the one you are asked to address.

Correct, this is the one I am seeking to solve.

 

[LCKurtz]

OK, I think the problem is now clearer. Thinking of salt dissolving in water, we would assume the volume of water is unchanging as the salt dissolves. I would suggest setting it up this way. Let V = volume of bowl in liters (L), r = rate of water entering and leaving the bowl in L/min, $\alpha$ = concentration of salt entering in kg/L. Then r$\alpha$ is rate of salt entering in kg/min. Then if s(t) is amount of salt in the bowl at time t in kg, the concentration of salt in the bowl is $\frac{s(t)} V$ in kg/L and the rate of salt leaving is $\frac {rs(t)} V$ in kg/min. This should enable you to find the DE for $\frac{ds}{dt}$ and solve it. You should find that as $t\to \infty$ the amount of salt in the bowl approaches $V\alpha$.

 

[Adel Makram]

Then if s(t) is amount of salt in the bowl at time t in kg, the concentration of salt in the bowl is $\frac{s(t)} V$ in kg and the rate of salt leaving is $\frac {rs(t)} V$ in kg/min.

Thank you, I think this should take me to my proposed solution under the assumption that the rate of the salt leaving the bowel is proportionated to the total salt in the bowel. (The proportion constant should be negative sign because the leaving salt reduces the total amount of salt in the bowel).
So:
$ds_{out}=-\lambda s dt$
$s_{in}=\alpha dt$.
Therefore, the total change of the salt is $ds=ds_{in}+ds_{out}=\alpha dt-\lambda sdt$.
This gives $ds/dt+\lambda s=\alpha$ which solves $s=\frac{\alpha}{\lambda}+C e^{-\lambda t}$ where $C$ is a constant.
Still, I am not sure about the competency of this solution.

 

[LCKurtz]

Still, I am not sure about the competency of this solution.

Then you might want to actually try my suggestion. If you do, notice that I have corrected the units on $\frac{s(t)} V$. Incidentally, the word you want is "bowl", not "bowel".

 

[Adel Makram]

Then you might want to actually try my suggestion. If you do, notice that I have corrected the units on $\frac{s(t)} V$. Incidentally, the word you want is "bowl", not "bowel".

But I do not see where to start from your post #9. You gave the variable in units and a boundary condition when t approaches infinity.

 

[Ray Vickson]

But I do not see where to start from your post #9. You gave the variable in units and a boundary condition when t approaches infinity.

I think you are supposed to set up a differential equation, then solve it and show that the solution approaches a certain limiting value as $t \to \infty.$