Finding Salt Concentration in a Tank with Limited Capacity

[alane1994]

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1lb of salt per gallon is entering at a rat of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity.

How on earth...

This is so similar I feel like I should be able to do it, yet I am drawing a big blank.

 

[Chris L T521]

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1lb of salt per gallon is entering at a rat of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity.

How on earth...

This is so similar I feel like I should be able to do it, yet I am drawing a big blank.

Again, we need to make a list of the things we know and then end up using the differential equation $\dfrac{dx}{dt} = r_ic_i - r_oc_o$ from the previous thread.

So, we know that $x(0)=100\text{ lb}$, $c_i = 1\text{ lb/gal}$, $V(0) = 200\text{ gal}$, $r_i = 3\text{ gal/min}$ and $r_o=2\text{ gal/min}$.

The fun part is figuring out the amount of concentration leaving the tank. Just as before, it's a quantity that depends on $t$: $c_o(t) = \dfrac{x(t)}{V(t)}$. However, the rate of change of the volume of solution in the tank is the difference of the entry/exiting rates. Thus, we see that
\[\frac{dV}{dt} = r_i-r_o\implies V(t) = (r_i-r_o)t+C\]
Since $V(0)=V_0$, we have that $C=V_0$ and thus $V(t) = (r_i-r_o)t+V_0$. In the context of this problem, that means $V(t)= t+200$ and thus $c_o(t) = \dfrac{x(t)}{t+200}$.

Therefore, the differential equation this time is
\[\frac{dx}{dt} = 3(1) - 2\cdot\frac{x}{t+200}\implies \frac{dx}{dt} = 3-\frac{2x}{t+200}\implies \frac{dx}{dt} +\frac{2}{t+200}x = 3\]
which is a linear first order equation.

To solve it, you'll need to use the integrating factor. Once you find $x(t)$, you need to compute the amount of concentration of salt in the tank right when it's about to overflow (i.e. when $V(t)=500\implies t=t_0$) by computing $c(t_0)=\dfrac{x(t_0)}{500}\text{ lb/gal}$.

The limiting concentration amount is then $\displaystyle\lim_{t\to\infty}\frac{x(t)}{t+200}$.

I hope this makes sense!

 

[MarkFL]

You've got more going on in this problem than the last. You have brine coming in and brine leaving, and at different rates. You want to determine the amount $A(t)$ of salt present in the tank at time $t$. Amount equals concentration per volume, and here the volume is a function of $t$, let's call it $V(t)$. So, let's begin with:

\(\displaystyle \text{time rate of change of salt = time rate of salt coming in minus time rate of salt going out}\)

\(\displaystyle \text{time rate of salt coming in=concentration of brine coming in times the time rate of volume coming in}\)

\(\displaystyle \text{time rate of salt going out=concentration of brine going out times the time rate of volume going out}\)

Hence, stated mathematically, we may write:

\(\displaystyle \frac{dA}{dt}=1\frac{\text{lb}}{\text{gal}}\cdot3 \frac{\text{gal}}{\text{min}}-\frac{A(t)}{V(t)}\frac{\text{lb}}{\text{gal}}\cdot2\frac{\text{gal}}{\text{min}}\)

And so, in \(\displaystyle \frac{\text{lb}}{\text{min}}\), we have:

\(\displaystyle \frac{dA}{dt}=3-\frac{2A(t)}{V(t)}\)

Now, how can we determine $V(t)$? What kind of ODE do we have, and how should we solve it?

Hahaha...this time I got ninja'd...:D

 

[Chris L T521]

... Hahaha...this time I got ninja'd...:D

Touché. (Rofl)

 

[alane1994]

Again, we need to make a list of the things we know and then end up using the differential equation $\dfrac{dx}{dt} = r_ic_i - r_oc_o$ from the previous thread.

So, we know that $\color{red}{x(0)=200\text{ lb}}$, $\color{red}{c_i = 1\text{ lb/gal}}$, $V(0) = 200\text{ gal}$, $r_i = 3\text{ gal/min}$ and $r_o=2\text{ gal/min}$.
\(\color{red}{\text{I believe this is supposed to be something else}}\)
The fun part is figuring out the amount of concentration leaving the tank. Just as before, it's a quantity that depends on $t$: $c_o(t) = \dfrac{x(t)}{V(t)}$. However, the rate of change of the volume of solution in the tank is the difference of the entry/exiting rates. Thus, we see that
\[\frac{dV}{dt} = r_i-r_o\implies V(t) = (r_i-r_o)t+C\]
Since $V(0)=V_0$, we have that $C=V_0$ and thus $V(t) = (r_i-r_o)t+V_0$. In the context of this problem, that means $V(t)= t+200$ and thus $c_o(t) = \dfrac{x(t)}{t+200}$.

Therefore, the differential equation this time is
\[\frac{dx}{dt} = 3(1) - 2\cdot\frac{x}{t+200}\implies \frac{dx}{dt} = 3-\frac{2x}{t+200}\implies \frac{dx}{dt} +\frac{2}{t+200}x = 3\]
which is a linear first order equation.

To solve it, you'll need to use the integrating factor. Once you find $x(t)$, you need to compute the amount of salt in the tank right when it's about to overflow (i.e. when $V(t)=500\implies t=\ldots$).

Couple questions mentioned above. If they are in fact true, doesn't that alter all of what you have said? Just input different values in?

 

[Chris L T521]

Couple questions mentioned above. If they are in fact true, doesn't that alter all of what you have said? Just input different values in?

I noticed I had the wrong x(0) originally (read the problem a little too fast) and edited my original post. It should be x(0)=100, not x(0)=200.

Sorry about that!