Dilution and Accretion Problem - Diff. Eq.

[ZachN]

Homework Statement

A tank initially contains 100 gal of brine whose salt concentration is 1/2 lb/gal. Bine whose salt concentration is 2 lb/gal flows into the tank at the rate of 3 gal/min. The mixture flows out at the rate of 2 gal/min. Find the salt content of the brine and its concentration at the end of 30 min. Hint. After 30 min, the tank contains 130 gal of brine.

Homework Equations

\Deltax = change in salt content in time interval \Deltat

The Attempt at a Solution

This is the model I have come up with but can't get the correct answer:

\Deltax = 6\Deltat - (2x\Deltat)/(100 + t)

6\Deltat because salt content is increasing this much every interval.

- (2x\Deltat)/(100 + t) because the salt content is decreasing this much every interval. (100 + t) represents that the total amount of gallons of brine are increasing this much every minute because 3 gals are entering and 2 gallons are leaving - hence a net of 1 gal/min increase.

If this is the correct equation, I am having a problem solving the diff. eq.

Thanks

 

[tiny-tim]

Welcome to PF!

This is the model I have come up with but can't get the correct answer:

\Deltax = 6\Deltat - (2x\Deltat)/(100 + t)

6\Deltat because salt content is increasing this much every interval.
…
If this is the correct equation, I am having a problem solving the diff. eq.

Hi ZachN! Welcome to PF!

(have a delta: ∆ )

Yes, I get the same equation:

dx/du = 6 - 2s/u, where u = t + 100.

(i'm using u 'cos it's easier to write )

ok, solve the thingy equation, dx/du = -2s/u, and then look for a particular solution (there's a really easy one )

 

[ZachN]

Thanks for the response, yeah I'm new here and I may be here a long time since I am teaching myself some maths.

Where is the "s" coming from and why do you only solve dx/du = - 2s/u? I tried to solve dx/du = 6 - 2x/u using an integrating factor but...

 

[tiny-tim]

Where is the "s" coming from

oops!

I was using s for salt, and then I noticed you were using x, and I meant to change them all but I forgot …

so s = x.

and why do you only solve dx/du = - 2s/u? I tried to solve dx/du = 6 - 2x/u using an integrating factor but...

It's another way of doing it …

I think it's called the "general solution and particular solution method" …

you solve it with any awkward stuff = 0, and then put the awkward stuff back in and more-or-less guess a "particular solution" to add on …

I prefer it (when it works! ), and it should give the same result.