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Determining Concentration of a Tank (Using Linear Differential EQ)

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem states that we have a tank with a capacity of 400L filled with a mixture that has a concentration of 0.05g Chlorine per Liter. The concentration is reduced by pumping in pure water (zero concentration of chlorine), at a rate of 4L/s, and pumped out at 10L sec.


    2. Relevant equations I know that the rate of change in concentration is equal to the rate in minus the rate out.


    3. The attempt at a solution The rate of concentration input is zero since it's pure water, and I set the concentration of mixture out at x/(400-6t) since the volume is decreasing at some concentration at 6L per second due to the larger amount of mixture being pumped out. So the equation is then dx/dt=0.05-x/(400-6t), at least from what I understand.

    Then, using the Linear Differential EQ method, I found I(x)=(400-6t)^-6. Multiplying both sides of the equation, I get (400-6t)^2(dx/dt)+(400-6t)^-7x=0.05(400-6t). The left side is suppose to resemble the product rule, but I'm thinking I really messed something up at some point.

    Could someone point out my mistake so I can do a retake? Sorry if this is hard to read, I'm not good at inputting equations. Thanks for any help.
     
    Last edited by a moderator: Mar 13, 2013
  2. jcsd
  3. Mar 13, 2013 #2

    vela

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    You're saying the concentration of chlorine the mixture is equal to
    $$\frac{x}{400-6t}.$$ Since V=400-6t is the volume of the liquid in the tank, ##x## must represent the total amount of chlorine in the tank. In that case, the units on the lefthand side would be g/s, but on the righthand side, the units would be g/L. The two sides of your equation can't be equal.

    At what rate is chlorine leaving the tank when the concentration is x/(400-6t)? That's what you want to set equal to dx/dt.

    I couldn't follow what you did here, but I think you're trying to use an integrating factor. The correct differential equation is separable, so you shouldn't need to use this technique.
     
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