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Differential Equations Mixture problem

  1. Feb 26, 2013 #1
    at t=o an evenly mixed 40 liters solution with 100grams of salt is poured into a container. solution comes in at 2 grams per liter at 3 liters per minute and flows out at the same rate. At what time is the salt solution 120 grams?

    I got -9.something minutes for my answer :/

    This question was on a test I just took. It's from memory. The linear equation I got was
    dA/dt+3/40*A=6

    What should the answer be?
     
    Last edited: Feb 26, 2013
  2. jcsd
  3. Feb 26, 2013 #2

    haruspex

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    Your equation looks right.
    What is the initial concentration? What is the target concentration? What is the concentration coming in? Any surprise here?
     
  4. Feb 26, 2013 #3

    rude man

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    There is 6g of salt going into the vessel and 3x/40 g leaving, so the salt contents in the vessel is changing at a rate of
    dx/dt = 6 - 3x/40
    which is not your equation. Your equation is adding the stuff that is supposed to be leaving!

    (Jeez, I hope I got that right ... will trade answers when appropriate).
     
  5. Feb 26, 2013 #4

    haruspex

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    Good catch - I missed that. But I suspect it's just a transcription error in the post. The D.E as posted would not have led to a negative answer.
     
  6. Feb 26, 2013 #5
    yea. I mistyped it. should've been dA/dt+3/40*A=6

    The problem still remains though. He states that the mixture was just poured in at t=0 implying that a negative time is even more likely to be a wrong answer :/
     
  7. Feb 26, 2013 #6

    haruspex

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    Pls try to answer my questions in post #2.
     
  8. Feb 26, 2013 #7
    100 grams per 40 liters initial
    120 grams target
    2 grams per liter at 3 liters per minute is coming in

    None of that helps me in any way :/

    Is my answer correct?
     
  9. Feb 26, 2013 #8

    haruspex

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    Yes, but let's put that in consistent units... 2.5g/l initially; 2g/l coming in; 3g/l target. See a problem with that?
     
  10. Feb 26, 2013 #9
    there's only a problem because the same amount that is coming in is also leaving correct? that's why it never gets to 120? to add to that, the resulting function showed the amount gradually decreasing over time.
     
  11. Feb 26, 2013 #10

    haruspex

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    Having expressed it in terms of concentrations, it doesn't matter whether the volume in the tank is increasing or decreasing. If you start with a solution more dilute than the target, and you add solution more dilute than the target, how are you ever going to reach the target concentration?
     
  12. Feb 26, 2013 #11
    ok. i see
     
  13. Feb 27, 2013 #12

    rude man

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    It does get to 120. In 63.83 sec.
     
  14. Feb 27, 2013 #13
    what function did you get?


    jsRdctc.jpg
     
    Last edited: Feb 27, 2013
  15. Feb 27, 2013 #14

    rude man

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    EDIT: right answer after all. Sorry to confuse.

    OK, we all agree on the ODE by now I take it, so have you tried to solve it? Show your effort.

    Sorry again, I see you did, but I can't make out the top part.
     
  16. Feb 27, 2013 #15
    I edited the post above to show most of the work. You can't have negative time though especially since he says "at t=0 the well mixed solution was poured in" meaning that before that, the tank was empty
     
  17. Feb 27, 2013 #16

    rude man

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    Hold everything, I have to reexamine what I did.

    EDIT: the problem has no solution and haruspex is right.
     
    Last edited: Feb 27, 2013
  18. Feb 27, 2013 #17

    haruspex

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    Looking back on this problem, you might consider that the problem setter made an error. OTOH, it's quite a good exercise in recognising the significance of an apparently silly answer.
     
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