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Sugar water tank problem (again)

  1. Oct 21, 2009 #1
    Thank you for any help that you can give me. I am very bad at differential equations and this one is really stumping me.
    1. The problem statement, all variables and given/known data
    A tank contains 2740 L of pure water. A solution that contains 0.05 kg of sugar per liter enters a tank at the rate 3 L/min The solution is mixed and drains from the tank at the same rate.


    2. Relevant equations
    I guess I need to solve for ds/dt, which I think would be something along the lines of
    ds/dt=.15-3s/2740, or the rate at which the sugar enters minus the rate at which it leaves.

    3. The attempt at a solution

    So far I've found the equation above and I think that the solution is something like e^[(-3*t)/2740], but my online homework thing keeps saying that it is wrong. Also, that solutino doesn't work for s(0)=0 and subtracting one from the above equation does not work. Does anybody have any suggestions? I think I might be solving the differential equation wrong, but I'm sure that my integrating factor is correct. Thank you very much.
     
  2. jcsd
  3. Oct 21, 2009 #2

    lanedance

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    Homework Helper

    the rate of change will be the difference between the sugar in (conc of 0.05kg/l at 3L/M per min) & the sugar out (conc of tank)

    I am guessing you have used s for the sugar in the tank?

    one would expect it would start at zero and increase assymotically to the concentration of the incoming stream

    inlet stream (conc c_0=0.05kg/L) sugar in
    (ds/dt)_in = c_0*3L/min = 3*(0.05)
    (ds/dt)_out = c_tank*3L/min = (s/2740)*3

    so the differential equation looks correct
    ds/dt=3*(0.05-s/2740) with initial condition s(0) = 0

    (2740/3)*ds/dt = 2740*0.05-s

    maybe show how you solve this & we can have a look at it

    hint you could try the subtitution & variable change
    u = s-2740*0.05
     
  4. Oct 21, 2009 #3
    Yeah, I chose s for the amount of sugar in the tank.
    Anyway, here is how I would try to solve this (probably wrong)
    (2740/3)*ds/dt=2740*0.05-s
    ds/dt=.15-3s/2740
    ds/dt+3s/2740=.15 I(t)=e^(3t/2740)
    ds/dt*e^(3t/2740)+(3s/2740)e^(3t/2740)=.15*e^(3t/2740)
    integrate both sides:
    e^(3t/2740)s=.15e^(3t/2740)*(2740/3)
    simplify:
    e^(3t/2740)s=137(e^3t/2740)
    And this is where I get stuck. Because then I get that s(t)=137, which obviously is wrong. Sorry about the slow post.
     
  5. Oct 21, 2009 #4

    lanedance

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    i haven't used integrating factors for ages, but how about the substitution?

    so if you have
    (2740/3)*ds/dt = 2740*0.05-s

    let u = s-2740*0.05
    then
    du/dt = ds/dt

    do the equation becomes
    (2740/3)*du/dt = -u

    should be able to separate, integrate & substitute back in for s
     
    Last edited: Oct 21, 2009
  6. Oct 21, 2009 #5
    I ended up with -e^(3t/2740)+137=s but my homework thing still said that the answer was wrong.

    My steps were:
    3dt/2740=du/-u integrate both sides:
    3t/2740=ln|-u|
    e^(3t/2740)=137-s
    -e^(3t/2740)+137=s
    The problem is that this just doesn't fit the initial condition of 0 kg sugar. And when I changed the constant so that it was (+/-)e^3t/2740) (either way) -1, it still said that I was wrong, even though the initial condition was met.
     
  7. Oct 21, 2009 #6

    lanedance

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    you're right that its wrong, that answer makes no sense, s will tend toward negative infinity and the initial condition is not met -> s(0) =-e^(0) + 137, which is not 0

    i think the problem is with the negative, move it to the t side of the equation before integrating, then keep a constant when you integrate as well, then choose it to satisfy your initial condition
     
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