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Differential equation mixing problem

  1. Jan 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A vat with 2000L of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 20L/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

    2. Relevant equations


    3. The attempt at a solution
    At ##t = 0## the amount of alcohol is 80L, let the amount at time ##t## be ##y##.
    ##\frac{dy}{dt} = rate in - rate out##
    ##rate in = (20L/min)(0.06) = 1.2L/min##
    ##rate out = \frac{y}{2000}(20L/min) = \frac{y}{100}L/min##

    Which leads to:
    ##\frac{dy}{dt} = \frac{1200 - y}{t}##

    ##\int\frac{dy}{1200 - y} = \int\frac{dt}{100}##

    ##-\ln|1200 - y| = \frac{t}{100} + k##

    ##1200 - y = Ae^{-t/100}##, yielding
    ##y(t) = 1120e^{-t/100}##

    Substituting ##t = 1## gives ##y \approx 91.14## and converting to a percentage gives about ##4.6%## while the actual answer is ##4.9##. The general solution works fine for ##t = 0## which implies that the value of ##A## is correct. Where lies the error here?
     
  2. jcsd
  3. Jan 14, 2016 #2

    Samy_A

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    You made a computation error, as ##\frac{dy}{dt} = \frac{1200 - y}{t}## is wrong.

    There a further error(s) down the line, so after fixing the formula for ##\frac{dy}{dt}##, carefully redo your calculations.
     
  4. Jan 14, 2016 #3
    I made a typo there, it's supposed to be

    ##\frac{dy}{dt} = \frac{1200 - y}{100}##
     
  5. Jan 14, 2016 #4

    Samy_A

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    Still wrong.
     
  6. Jan 14, 2016 #5
    Multiplication error...

    ##\frac{dy}{dt} = \frac{120 - y}{100}##
     
  7. Jan 14, 2016 #6
    Still a balls-up. I just don't see it.
     
  8. Jan 14, 2016 #7

    Samy_A

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    Yes, that's correct.

    Now just continue from this.
    This will lead to ##120 - y = Ae^{-t/100}## (exactly as in your first computation, with 120 instead of 1200).

    Now use y(0)=80 to get A.
     
  9. Jan 14, 2016 #8
    More than one typo, actually. Anyway, now I get

    ##y = 120 - 40e^{-t/100}## with ##t = 1## giving ##y = 80.398...## which is not correct. That's what I meant by the persistent balls-up.
     
  10. Jan 14, 2016 #9

    Samy_A

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    How many minutes are there in 1 hour?
     
  11. Jan 14, 2016 #10
    Problem solved and the need for a break confirmed. I can't believe I missed that. Much appreciated.
     
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