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Math Help - Solving Mixture Problem using 1st Order Lin Dif. Eq.

  1. #1
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    Solving Mixture Problem using 1st Order Lin Dif. Eq.

    word problems. awesome

    Water flows through a purifying machine at the rate 50 gal/min. The machine removes 10% of the impurities as the water flows through. Suppose I have 1000 gallons of water containing 5 gallons of impurities. How long do I have to run it through the machine to reduce the amount of impurity to half a gallon?

    So I tried setting this up in the form dx/dt + r_out*(x/V) = r_in*c_in, where r_in and r_out are 50 gal/min, V is 1000 and my initial condition is x(0) = 5 gal/V. The problem is, I don't know what to use for c_in (concentration of the inflow).

    I'm pretty sure the problem wants me to use the integrating factor method and I'm comfortable doing so, I just can't figure out how to set up the problem in its entirety. Anybody want to help me figure out how to find c_in?
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    Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

    Quote Originally Posted by rhobere View Post
    word problems. awesome

    Water flows through a purifying machine at the rate 50 gal/min. The machine removes 10% of the impurities as the water flows through. Suppose I have 1000 gallons of water containing 5 gallons of impurities. How long do I have to run it through the machine to reduce the amount of impurity to half a gallon?

    So I tried setting this up in the form dx/dt + r_out*(x/V) = r_in*c_in, where r_in and r_out are 50 gal/min, V is 1000 and my initial condition is x(0) = 5 gal/V. The problem is, I don't know what to use for c_in (concentration of the inflow).

    I'm pretty sure the problem wants me to use the integrating factor method and I'm comfortable doing so, I just can't figure out how to set up the problem in its entirety. Anybody want to help me figure out how to find c_in?
    I would say that c_in = 0. If you're flowing pure water into the purifier, then c_in = 0.
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  3. #3
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    Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

    I think the idea is more that the water is pumped from the container through the pump, where 10% of the impurities are removed before being pumped back into the tank. otherwise the fact that the pump is removing 10% of the impurities would be irrelevant. if the water isn't being pumped back into the tank, the problem would at least have to make a mention of there being a new source of water I would think.

    the best I could come up with is that c_in is a continuous growth rate. as in A_0e^rt, where A_0 is equal to 5 gallons of impurities and r is equal to .9 (-10% impurities per cycle = 1-.1 = .9). when I use it try to solve it in that form I get a negative number for t.
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    Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

    Quote Originally Posted by rhobere View Post
    I think the idea is more that the water is pumped from the container through the pump, where 10% of the impurities are removed before being pumped back into the tank. otherwise the fact that the pump is removing 10% of the impurities would be irrelevant. if the water isn't being pumped back into the tank, the problem would at least have to make a mention of there being a new source of water I would think.

    the best I could come up with is that c_in is a continuous growth rate. as in A_0e^rt, where A_0 is equal to 5 gallons of impurities and r is equal to .9 (-10% impurities per cycle = 1-.1 = .9). when I use it try to solve it in that form I get a negative number for t.
    Yeah, I was totally off in my response. Why would you put pure water into a purifier?

    So, I'm thinking your P&ID would look something like this: one tank, line out to a pump, which leads to the purifier, which comes back into the tank. Would that be correct?

    If so, then I would agree that r_in and r_out are equal. (The pressures on either side of the pump would not be equal, but the flows would.) Wouldn't the c_out be 0.9 * c_in? And if the concentration in is just x/V, then you've got the DE

    \frac{dx}{dt}=r_{\text{in}}c_{\text{in}}-r_{\text{out}}c_{\text{out}}=50\,(0.9)\frac{x}{V}-50\,\frac{x}{V}=-50(0.1)\frac{x}{V}.

    How does that strike you?
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  5. #5
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    Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

    Well, I went and talked to my professor about it yesterday and you're spot on although he got to the answer a little differently. I, however, was WAY over-analyzing it.

    c_out was the unknown x/V = x/1000
    c_in was simply c_out - the fraction of c_out that was removed per cycle. so (x/1000 - .1x/1000)

    ends up being:

    dx/dt = 50(x/1000-.1x/1000)-50(x/1000) which simplifies to

    dx/dt = -.005x and so on from there. (I'm writing this based off memory right now so there may be some subtle errors. lol)
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    Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

    So the solution to the DE is... what?
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