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Finding concentration of an incoming solution

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A tank holds 10 L of pure water. Brine (unknown constant concentration) is flowing into the tank at 1 L/min. The water is mixed well and drained at 1 L/min. After 20 min, there are 15 g of salt left in the tank. What is the concentration of the salt in the incoming brine?

    2. Relevant equations

    3. The attempt at a solution

    Let A(t) = amount of salt (g) at time t (min).
    I can see that A(20) = 15, but I'm confused on how I should set up the diff eq.

    I know that A'(t) = (concentration of incoming * rate of incoming) - (concentration of outgoing * rate of outgoing)

    I let
    r = concentration of incoming
    s = concentration of outgoing​
    so I think that gives me
    ##\dfrac{dA}{dt} = (r \frac{g}{L})(1 \frac{L}{min}) - (\frac{1}{10} s \frac{g}{L})(1 \frac{L}{min})##

    ##\dfrac{dA}{dt} = r - \frac{1}{10} s##

    But then I've no idea where to go from here.
     
  2. jcsd
  3. Jan 26, 2013 #2

    haruspex

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    Why are you dividing the concentration of outgoing by 10?
    How is the quantity of salt in the tank, A, related to the outgoing concentration, given that it's well mixed?
     
  4. Jan 26, 2013 #3
    Well, the concentration of the solution in the tank at any point would be (amount of salt in tank)/(volume of tank), so I figure the concentration of the outgoing solution ##s## at any given time will be ##\frac{A}{10}##. Looks like I'd mistaken ##s## for ##A##. Is that it? Thanks for the reply
     
  5. Jan 26, 2013 #4

    haruspex

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    Yes, that's what it looked like.
     
  6. Jan 26, 2013 #5
    Ah, okay, thanks for that.

    So I've got the equation
    ##\dfrac{dA}{dt} + \frac{1}{10}A = r##, which gives me the general solution
    ##A = 10x + Ce^{-\frac{1}{10}t}##.

    Using A(0) = 0,
    ##0 = 10x + C##,
    ##C = -10x##.

    Using A(20) = 15,
    ##15 = 10x - 10x e^{-2}##
    ##x = \dfrac{15}{10(1-e^{-2})}##
    ##x \approx 1.735 \frac{g}{L}##.

    Can someone check my work on that? I'd hate to make another mistake. Thanks!
     
  7. Jan 26, 2013 #6

    haruspex

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    Looks right to me.
     
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