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Differential Equations Water Evaporation

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    A pond contains 1,000,000 gal of water and an unknown amount of chemical. Water containing .01 gram of this chemical per gallon flows into the pond at a rate of 300 gal/hour. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume the chemical is uniformly distributed throughout the pond. The differential equation for the amount of chemical in the pond at any time is:

    [itex]\frac{d amount(t)}{dt} = 300(\frac{1}{100} - \frac{amount(t)}{1000000} )[/itex] (Finding this equation was the first part of the problem).

    Suppose that in addition to the flow into and out of the pond described above, that pure water is evaporating from the pond at the rate of one gallon per hour. Write down a differential equation for the amount of chemical in the pond at any given time.

    3. The attempt at a solution
    I'm tempted just to write the following:

    [itex]\frac{d amount(t)}{dt} = 300(\frac{1}{100} - \frac{amount(t)}{(1000000 - 1*t)}) [/itex]

    Does this not take into account the change in concentration of the chemical in the water over time? I'm not sure if there's something more sophisticated going on here.
     
  2. jcsd
  3. Mar 26, 2012 #2
    Maybe I can get my work checked for the first part of the problem, too. I said:

    Define a(t) to be the amount of chemical at any given time, t. The inflow of the chemical into the pond does so with concentration 1 gram/100 gallons. Chemical outflow has concentration a(t) / 106 gallons. Thus, I reasoned that the change in concentration will be given by the difference between inflow and outflow, multiplied by the flow rate, or 300. I wrote,

    [itex]\frac{da(t)}{dt} = 300 (\frac{1}{100} - \frac{a(t)}{10^6})[/itex], which gives me my equation.

    Does that seem right?
     
  4. Mar 26, 2012 #3
    That looks fine, just remember that t is in hours. These are called "mixture" problems, and the general form of them is usually:
    [tex]\frac{dA}{dt} = (rate in) - (rate out)[/tex]
    Where A(t) is some function representing amount of stuff in the container.
     
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