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Thread: Mixture problem

  1. #1

    Mixture problem

    I've been racking my brain and Can't figure part of this question out...

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    A tank initially contains 100 gallons of water with 20 lbs of salt dissolved in int. A solution containing 2 lbs of salt per gallon of water is pumped into the tank at a rate of 1 gallon per minute and the well-mixed solution is pumped out of the tank at a rate of 2 gallons per minute.


    A) How much salt will be in the tank after 25 mins?
    B) When will there be 55 lbs of salt in the tank? [Note: There are two answers]
    C) How much salt will be in the tank after 100 minutes?
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    I started out with my initial equation
    dS/dt = rate in - rate out
    dS/dt = (2)(1) - (s/100)(2)

    This was rewritten into a linear equation form
    S' + (1/50)S = 2

    and plugged in as
    S = (∫e∫1/50 dt * 2 dt) / (e∫1/50 dt)

    Some more magic gets me to
    S = 100+Ce-t/50

    and solving for C using information given in the problem gets me
    S=100-80e-t/50


    for A its just plug and chug - t=25 and I get S=51.47 lbs.
    for B its more plug and chug - S=55 so I get T=28.76 Mins. However I cant find the second answer.
    for C its pretty easy - the tank empties at T=100 so it's 0.

    I am really stuck on part B. I know there are two answers, because the salt in the tank fills up to a maximum, then the flow of water out begins to overcome the saltwater solution going in, so my salt level in the tank decreases - the whole thing looks like an inverted parabola shape. However my equation doesn't give me this.

  2. #2
    Elite Member
    Join Date
    Apr 2005
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    USA
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    8,907
    The water volume isn't fixed. Work on that.
    Make sure your rates have the same scale.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Yeah I think I figured it out

    my initial equation is
    dS/dt = 2 - (S/100-t)(2)

    so I get
    S' + (2/100-t)S = 2

    plugging that in I get
    S = (∫e2∫1/100-t dt * 2 dt) / (e2∫1/100-t dt)

    and integrating I end up with
    S = 2(100-t) + C(100-t)2

    Plugging in S(0)=20, I get
    C= -0.018

    so my equation ends up as
    S = 2(100-t) - 0.018(100-t)2

    This gives me the answers
    A) 48.75 lbs
    B) 38.9 mins and 50 mins
    C) 0 lbs


    I think I figured it out. Double checking with a graphing calculator, the curve looks true and my numbers line up with my T and S at those points.

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