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Homework Help: Can anyone see what's wrong with my solution

  1. Oct 17, 2004 #1
    Ok, first order linear mixing problem.

    "A 400 gal. tank initially contains 200 gal. of water containing 2 ppb by weight of dioxin, an extremely potent carcinogen. Suppose water containing 5 ppb of dioxin flows into the top of the tank at a rate of 4 gal/min. The water in the tank is kept well mixed, and 2 gal/min are removed from the bottom of the tank. How much dioxin is in the tank when the tank is full?

    Ok. Vo = 200 gal., a = 2 ppb, b = 5 ppb, e = 4 gal/min, f = 2 gal/min
    (Vo = initial volume of water, a = initial amount of dioxin, b = amount of dioxin flowing in, e = rate of b, f = rate going out)
    So, if Q is the amount of dioxin in the mixture at any time t,

    dQ/dt + 2/(200 + 2t) = 20 (from dQ/dt + f/(Vo + (e-f)t) = be)

    So, my a(t) = 2/(200+2t), r(t) = 20. u(t) = e^((int)a(t)dt),

    u(t) = e^(int)(2/(200 + 2t))dt

    u(t) = e^(ln|200 + 2t|) = 200 + 2t
    //(int) -integral (too lazy for Latex)

    So, Q(t) = 1/u(t)[(int)u(t)r(t)dt] + c/u(t)

    Q(t) = 1/(200 + 2t)[(int)(200 + 2t)(20)dt] + c/(200 + 2t)

    Q(t) = [1/(200 + 2t)](4000t + 20t^2) + c/(200 + 2t)

    Q(t) = (20t^2 + 4000t + c)/(200 + 2t)

    Ok, since Q(0) = a = 2ppb then

    Q(0) = a = 2 = (c/200), so c = 400, then

    Q(t) = (20t^2 + 4000t + 400)/(200 + 2t)

    Now, when the tank is full (400 gal, I need the amount of dioxin, so
    200 + 2t = 400, when t = 100min)

    Q(100) = (20(100^2) + 4000(100) + 400)/(400)

    Q(100) = 1501 ppb.

    Which is a little absurd(from 2ppb initially). Also, this is wrong because the answer is 4.25ppb.

    Can someone go through my math and see if I made a mistake? Because I can't seem to find what went wrong.

  2. jcsd
  3. Oct 18, 2004 #2


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    I think your starting equation is incorrect. The rate at which dioxin leaves the container is proportional to the amount of dioxin in the tank.

    Incidentally, 4.25 ppb is the correct answer!
    Last edited: Oct 18, 2004
  4. Oct 18, 2004 #3
    Oops, I did mean dQ/dt + (2/200 + 2t)Q = 20, but that's how I did work it out on paper and got that answer. What do you mean exactly?
    Last edited: Oct 18, 2004
  5. Oct 18, 2004 #4


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    Your notation is confusing and it would help a LOT if you would take the time to LaTeX your equations!

    In any case, I find it easier to work with concentrations rather than the actual amount of contaminant. You should be able to recast your equations into this form:

    [tex]\frac {dC}{dt} = \frac {R_{in}}{V}\left(C_i - C\right)[/tex]

    where C(t) is the concentration of contaminant in the tank, [itex]C_i[/itex] is the concentration of the inflow, V is the volume of water in the tank which can be written as

    [tex]V = V_0 + (R_{in} - R_{out})t[/tex]

    The R's are the rates (constant) of flow into and out of the tank. The integration is straightforward.
  6. Oct 18, 2004 #5
    Thanks Tide, I got it. Sorry about the confusing notation. I'm not too savvy in LaTeX yet, and I was sort of in a rush. Been studying all night for my exam.
  7. Feb 13, 2011 #6
    I'm having an issue with this same exact problem. This is the DE that I got from the initial problem.

    [tex]\frac{\mathrm{d} C}{\mathrm{d} t} = \frac{4gal}{min}(5ppb)-\left( \frac{2ppb}{200gal+(2gal)t}\right )C[/tex]

    Is this anywhere close to the equation that I should be finding? I've solved the differential and obtained an enormous number for the final ppb (parts per billion) when the tank is full. I am very uncertain about how to do the math where the ppb is concerned. I am counting it as a whole number and am unsure whether I should treat it as [tex]5\times10e^{-9}[/tex] instead. Thanks for the help.
    Last edited: Feb 13, 2011
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