MHB Mixing with Common Drain: Mass of Salt in Two Tanks #46 Nagle

mathcoral
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Mixing with a Common Drain. Two tanks, each holding 1 L of liquid, are connected by a pipe through which liquid flows from tank A into tank B at a rate of 3-a L/min (0<a<3). The liquid inside each tank is kept well stirred. Pure water flows into tank A at a rate of 3 L/min. Solution flows out of tank A at a L/min and out of tank B at 3-a L/min. If, initially, tank B contains no salt (only water) and tank A contains 0.1 kg of salt, determine the mass of salt in each tank at time T>=0. How does the mass of salt in tank A depend on the choice of a? What is the maximum mass of salt in tank B?Here is picture of the tank View attachment 8094

I was going to make a normal equation x'=Ax to find the eigenvectors.

I got stuck on the set up
x1=rate in - rate out
x1(t)=0 - (3-a)x1(t) - ax1(t)
x1(t)= -3x1(t)
The rate in is 0 because the liquid is pure water.

x2(t) =rate in - rate out
x2(t)= (3-a)x1(t) - (3-a)x2(t)

I am unsure how to put the 3 L/min coming out of from both tanks (bottom pipe).
 

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I would begin with:

$$\d{x_1}{t}=-3x_1$$ where $$x_1(0)=\frac{1}{10}$$

Hence:

$$x_1(t)=\frac{e^{-3t}}{10}$$

Then:

$$\d{x_2}{t}=(3-\alpha)\left(\frac{e^{-3t}}{10}-x_2\right)$$ where $$x_2(0)=0$$

I would put the ODE into standard linear form:

$$\d{x_2}{t}+(3-\alpha)x_2=(3-\alpha)\frac{e^{-3t}}{10}$$

Our integrating factor is:

$$\mu(t)=e^{(3-\alpha)t}$$

And we get:

$$e^{(3-\alpha)t}\d{x_2}{t}+(3-\alpha)e^{(3-\alpha)t}x_2=(3-\alpha)\frac{e^{-3t}}{10}e^{(3-\alpha)t}$$

Or:

$$\frac{d}{dt}\left(e^{(3-\alpha)t}x_2\right)=(3-\alpha)\frac{e^{-\alpha t}}{10}$$

Can you proceed?
 
After integrating I got
x2(t)=$$\frac{3-a}{10a}$$(eat-1)e-3t

The mass of salt in tank A does not depend on the choice of a.
Tank B depends on a.

I am unsure how to find the maximum mass for tank B.
 
mathcoral said:
After integrating I got
x2(t)=$$\frac{3-a}{10a}$$(eat-1)e-3t

The mass of salt in tank A does not depend on the choice of a.
Tank B depends on a.

I am unsure how to find the maximum mass for tank B.

I got the equivalent:

$$x_2(t)=\frac{\alpha-3}{10\alpha}e^{-3t}\left(1-e^{\alpha t}\right)$$

Now, for the optimization of this function, recall we had:

$$\d{x_2}{t}=(3-\alpha)\left(\frac{e^{-3t}}{10}-x_2\right)$$

Note: We could also differentiate the function we found, but I think this will be less work. ;)

We can equate this derivative to zero to find the turning point, and substitute for $x_2$:

$$(3-\alpha)\left(\frac{e^{-3t}}{10}-\frac{\alpha-3}{10\alpha}e^{-3t}\left(1-e^{\alpha t}\right)\right)=0$$

Since we are given $0<\alpha<3$, we may divide through by $$\frac{3-\alpha}{10\alpha}$$ to obtain:

$$\alpha e^{-3t}-(\alpha-3)e^{-3t}\left(1-e^{\alpha t}\right)=0$$

Solve this for $t=t_{\max}$ and then evaluate $x_2\left(t_{\max}\right)$.
 
Setting the derivative to 0 I got t=$$\frac{ln(\frac{3}{3-a})}{a}$$

Plugging in for t in x2(t) I got
x2(t)=.1($$\frac{3-a}{3}$$)3/a kg
 
mathcoral said:
Setting the derivative to 0 I got t=$$\frac{ln(\frac{3}{3-a})}{a}$$

Plugging in for t in x2(t) I got
x2(t)=.1($$\frac{3-a}{3}$$)3/a kg

Yes, I get equivalent results. (Yes)
 
Hi mathcoral, welcome to MHB!

It seem to me that we still don't have the actual maximum mass of salt in tank B.
For that we need to maximize for $\alpha$ as well.
We can make it ourselves a little easier, since to achieve that maximum all liquid from tank A should go through tank B, shouldn't it?
So $\alpha = 0$.
How much salt would that make at its maximum in tank B? (Wondering)
 
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