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Dilution and Accretion Problem - Diff. Eq.

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

    [tex]\Delta[/tex]x = change in salt content in time interval [tex]\Delta[/tex]t

    3. The attempt at a solution

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

    [tex]\Delta[/tex]x = 6[tex]\Delta[/tex]t - (2x[tex]\Delta[/tex]t)/(100 + t)

    6[tex]\Delta[/tex]t because salt content is increasing this much every interval.

    - (2x[tex]\Delta[/tex]t)/(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
     
  2. jcsd
  3. Nov 13, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi ZachN! Welcome to PF! :smile:

    (have a delta: ∆ :wink:)

    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 :wink:)

    ok, solve the thingy equation, dx/du = -2s/u, and then look for a particular solution (there's a really easy one :biggrin:)
     
  4. Nov 13, 2008 #3
    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...
     
  5. Nov 13, 2008 #4

    tiny-tim

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    oops! :redface:

    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. :redface:
    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! :rolleyes:), and it should give the same result. :smile:
     
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