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

In summary, Mixing salt with a common drain causes a change in the total mass of salt in the two tanks.
  • #1
mathcoral
5
0
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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  • #2
I would begin with:

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

Hence:

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

Then:

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

I would put the ODE into standard linear form:

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

Our integrating factor is:

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

And we get:

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

Or:

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

Can you proceed?
 
  • #3
After integrating I got
x2(t)=\(\displaystyle \frac{3-a}{10a}\)(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.
 
  • #4
mathcoral said:
After integrating I got
x2(t)=\(\displaystyle \frac{3-a}{10a}\)(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:

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

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

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

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 $x_2$:

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

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

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

Solve this for $t=t_{\max}$ and then evaluate $x_2\left(t_{\max}\right)$.
 
  • #5
Setting the derivative to 0 I got t=\(\displaystyle \frac{ln(\frac{3}{3-a})}{a}\)

Plugging in for t in x2(t) I got
x2(t)=.1(\(\displaystyle \frac{3-a}{3}\))3/a kg
 
  • #6
mathcoral said:
Setting the derivative to 0 I got t=\(\displaystyle \frac{ln(\frac{3}{3-a})}{a}\)

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

Yes, I get equivalent results. (Yes)
 
  • #7
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)
 

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

1. How does the mass of salt in two tanks affect the mixing process in a common drain system?

The mass of salt in two tanks plays a crucial role in determining the rate and efficiency of mixing in a common drain system. The greater the mass of salt present in the tanks, the slower the mixing process will be, as it takes longer for the salt to disperse evenly in the solution.

2. What factors can affect the mass of salt in two tanks in a common drain system?

The mass of salt in two tanks can be influenced by several factors, such as the initial concentration of salt in each tank, the flow rate of the solution, and the size and shape of the tanks. Additionally, any changes in these factors during the mixing process can also impact the mass of salt present in the tanks.

3. Is the mass of salt in two tanks the only factor that affects the mixing process in a common drain system?

No, the mass of salt in two tanks is not the only factor that influences the mixing process. Other factors, such as the flow rate, tank size, and initial concentration, also play important roles in determining the overall mixing efficiency.

4. How can the mass of salt in two tanks be controlled in a common drain system?

The mass of salt in two tanks can be controlled by adjusting the initial concentration of salt in each tank or by changing the flow rate of the solution. By maintaining a consistent flow rate and carefully measuring the initial salt concentration, the mass of salt in two tanks can be controlled and optimized for efficient mixing.

5. What are some potential applications of studying mixing with common drain in relation to the mass of salt in two tanks?

Studying mixing with common drain and the mass of salt in two tanks has many practical applications, such as in industrial processes where mixing is necessary to achieve a homogeneous solution. This can include applications in chemical and food processing, water treatment, and pharmaceutical manufacturing, among others.

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