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Differential equations - mixing problem (more complicated)

  1. Feb 12, 2007 #1
    A 500 gallon tank originally contains 100 gallons of fresh water. Beginning at time t=0, water containing 50 percent pollutants flows into the tank at the rate of 2 gal/min and the well stirred solution leaves at the rate of 1 gal/min. Find the concentration of pollutants in the tank at the moment it overflows.

    The answer is 48%

    my attempt, something is wrong as i did not get that answer or anywhere near it.

    rate in: 2 x 0.50
    rate out: 1 x s(t)/(100+t)

    s'(t) = 1- s(t)/(100+t)
    s'(t) + s(t)/(100+t) = 1

    a(t) = 1/(100+t) , b(t) = 1

    using u(t) = exp(integ(a(t)dt)):
    u(t) = exp(integ(a(t)dt))
    u(t) = exp (ln (100+t))
    u(t) = 100+t

    using d/dt u(t)s(t) = u(t)b(t):
    d/dt [(100+t)s(t)]= (100+t)(1)
    (100+t)s(t) = 100t + 1/2t^2 + C
    s(t) = [100t + 1/2t^2 +C ]/(100+t)

    sub s(0) = 0 into s(t) [is this even correct?]
    0 = [0+0+c]/[100+0]
    c= 0
    s(t) = [100t + 1/2t^2]/(100+t)

    c(t) = s(t)/100
    c(t) = [100t + 1/2t^2]/100(100+t)

    then i found when the tank overflow:
    500 = 100+t
    t= 400
    then found c(t)

    c(400) = [100(400)+1/2(400)^2]/100(100+400)
    c(400) = 2.4

    which is wrong.

    can anyone help me?
  2. jcsd
  3. Feb 12, 2007 #2


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    Staff Emeritus
    Science Advisor

    Okay, good so far.

    Yes, that's correct- initially the tank contained "100 gallons of fresh water"

    Here's your error!! The tank will contain 500 gallons when it overflows, not 100. The concentration is s/500, not s/100.

    Obviously, a concentration can't be larger than 1! Again, your error is in dividing by 100 gallons, the initial amount of water in the tank, when, at the time of overflow, there are 500 gallons of water in the tank. Divide your answer by 5.
  4. Feb 12, 2007 #3
    thanks very much
    that was a stupid mistake on my part!
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