A tank initially contains $200$ gallons of brine (solution of water and salt) whose salt concentration is $3 \text {lb/gal}$. Brine with salt concentration $2\text {lb/gal}$ flows into the tank at $4\text {lb/min}$. The solution in the tank is constantly thoroughly mixed and flows out at the same rate of $4\text {lb/min}$.

Find the salt content of the brine at the end of 20 minutes.

I got the differential equation $\frac{\text{dx}}{\text{dt}}=8-\frac{4x}{200}$ where $x$ is the amount of salt in the solution at time t

this leads to the following $$\int_{600}^x\frac{\text{dx}}{0.02x-8}=\int_{0}^t -t\text{dt}$$ where the limits of integration represent the initial and final conditions, for example, $x=600$ initially because we start off with $600$ lbs of salt in the tank

solving yields:

and after solving got $$x=200(2+e^{\frac{-t}{50}})$$

Is this correct?

Also, when will the salt concentration be reduced to $2.5\text {lb/gal}$?

My intuitive approach is: $$\frac{\text{dx}}{\text{dt}}=2.5$$ $$2.5=8-\frac{4x}{200}$$ $$x=275$$

And then substitute $x$ into $x=200(2+e^{\frac{-t}{50}})$

However I got a non-real answer, so something must be off.

  • $\begingroup$ Hi, what do you mean by non-real answer? Normally in math if a value is not real it is complex. $\endgroup$
    – Taozi
    Mar 31 '16 at 21:40
  • $\begingroup$ it was my calculator's output: non real ans $\endgroup$ Apr 1 '16 at 16:18

I see, my original intuitive approach was wrong. The concentration is just $\frac{4x}{200}$, and since net inflow, the difference between the inflow and outflow rate, is $0$ gallons per minute, does not depend directly on time, only on amount of salt content in the brine at time t.

So $$\frac{4x}{200}=2.5$$ $$x=500$$ There must be $500$ pounds of salt at this time.

Equating $$t=-50\log(\frac{0.02*500-8}{4})$$ will yield the time where there are $500$ pounds of salt in the concentration, which implies the concentration at that time must be $2.5$ lbs per gallon.


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