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Mixture Problem (differential equations)

  1. Jul 13, 2005 #1
    Q: A large tank initially contains 100 gal of brine in which 10lb of salt is sissolved. Starting at t=0, pure water flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simulaneously flows out at the slower rate of 2 gal/min.

    a) How much salt is in the tank at the end of 15 min and what is the concentration at that time?

    A: Initial condition x = 10, t=0

    [tex]IN = (0 lb/gal)(5 gal/min) = 0 lb/min[/tex]
    [tex]OUT =(\frac{x}{100+3t} lb/gal)(2 gal/min) = \frac {2x}{100 + 3t}lb/min [/tex]


    [tex]\frac{dx}{dt} = 0 - \frac{2x}{100+3t}[/tex]
    [tex]\frac{dx}{dt} + \frac{2x}{100+3t} = 0[/tex]

    [tex]IF = (3t + 100) ^\frac{2}{3}[/tex]

    so [tex]\frac{d(3t+100)^\frac{2}{3}x}{dt} = dt[/tex]
    [tex](3t + 100)^\frac{2}{3}x = c[/tex]
    [tex]x = \frac{c}{(3t + 100)^\frac{2}{3}}[/tex]

    Substituting initial condition x = 10, t=0,

    [tex]10 =\frac{C}{(3t +100)^\frac{2}{3}}[/tex]
    [tex]10(100)^\frac{2}{3}= C[/tex]
    C = 215.443

    Hence the amount of salt in time t is

    [tex]x = \frac{215.443}{(3t + 100)^\frac{2}{3}}[/tex]

    Solving problem a) using this equation, I got

    t = 15

    [tex]x =\frac{215.44}{(3*15 + 100)^\frac{2}{3}}= 7.8lb.[/tex]

    But the textbook answer was 0.054lb. I understand that since there was only 10lb in 100 gal to start with, it doesn't make sence it still contains 7.8lb. of salt after 15min., but I don't know where I made a mistake.
    Will somebody tell me where I should revise?
    Last edited: Jul 13, 2005
  2. jcsd
  3. Jul 13, 2005 #2


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  4. Jul 13, 2005 #3
    Saltydog, I understood the problem you suggested me to take a look.
    Only the difference between mine and that problem is that instead of conc water, flesh water is added to the tank. That is why I thought
    [tex]IN = 0 [/tex]
    Am I wrong?
  5. Jul 13, 2005 #4
    You actually have the right answer, but to the wrong problem. Your answer is in pounds, but the question is asking for concentration.
  6. Jul 13, 2005 #5


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    No, the problem asked two things: the amount of salt in the tank and the concentration.

    Beez, I did the problem solving the d.e. slightly differently:
    [tex]\frac{dx}{dt}= -\frac{2x}{100+3t}[/tex]
    is separable:
    [tex]\frac{dx}{x}= -\frac{2dt}{100+3t}[/tex]
    but that gives exactly the same answer.

    Why do you think 7.8 lb doesn't make sense? Initially there is 0.1 pound of salt in each gallon so 0.2 pounds of salt going out each minute. At that rate, after 15 minutes, 3.0 pounds of salt would have left, leaving 7 pounds. Since the concentration is declining we would expect that actually less salt would leave so that there would be more than 7 lb of salt left in the tank. Your answer is much more sensible than "0.054 lbs".

    Oh, wait- now Jelfish's response makes sense- in part! The correct answer is:
    After 15 minutes there will be 7.8 pounds of salt in the tank and the concentration will be 7.8/145= 0.054 pounds of salt per gallon!
  7. Jul 13, 2005 #6
    you guys this may look like a simple mixing problem but it isnt. it is a exponential decay problem. the concentration at time t is dependent only on the rate in which it is decreasing not on concentration, therefore you need to use exponetial decay. you may even be able to use the equation of half-life
  8. Jul 13, 2005 #7
    Yes, but the rate at which it is decreasing is also dependant on the concentration since it is constantly being mixed with pure water. Therefore the change in concentration decreases as well.

    If you think you know an easier (or the correct) way of solving this problem, please show everyone, because I'm not certain of what you mean.
  9. Jul 13, 2005 #8


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    Beez, you straight with this or what? Your answer above, 7.8 lbs is the amount of salt in the tank after 15 minutes. Since the amount of water is increasing by 3 gal/min, the tank has 145 gal. of water after 15 minutes. So the concentration of salt is thus 7.8/145=0.054 lb/gal. Ok?
  10. Jul 13, 2005 #9
    Thank you everyone.

    Well, at last it was not a math but English problem with me (my excuse is English is not my first language :smile: )
    Yes, I understood what I missed. I also see that 7.8lb was the right amount of salt left in the tank as well.
    Sorry that I did not respond right away since I live in Japan, it was past midnight while you guys were helping me.

    Thanks again all!

  11. Jul 13, 2005 #10
    I did it

    Everyone, you should be proud of me. With your help, I think I completely understood the mixing problem. The second part of the question was as follows
    b) If the capacityof the tank is 250 gal, what is the concentration at the instant the tank overflows?

    This is how I solved
    Since the water in the tank increases 3 gal/min, it will take 50 min (150/3) for the tank to get full.
    Next obtain the amount of the salt in the water in 50 min.
    so substitute 50 to t in the equation
    [tex]x = \frac{215.44}{(3t + 100)^\frac{2}{3}}[/tex]

    [tex]x = \frac{215.44}{(3*50 + 100)^\frac{2}{3}}[/tex]
    and I got x = 5.428 lb.

    Since it is asking the concentration level, I do the following to obtain the answer.


    And the answer is 0.022lb. which was the answer in the textbook!
    Thanks again guys! :rofl:
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