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Mixing Problem Help.

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A brine solution of salt flows at a constant rate of 7 L/min into a tank that holds 30 L of brine solution in which was dissolved 1.7 kg. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine entering the tank is 0.33kg/L, find the mass of the salt in the tank after "t" minutes.


    2. Relevant equations

    [tex]\frac{dy}{dt}[/tex]= (in rate) - (out rate)

    3. The attempt at a solution

    We know that the in rate will equal [tex]\frac{7L}{min}[/tex]*[tex]\frac{1kg}{3L}[/tex] = [tex]\frac{7kg}{3min}[/tex]. The out rate will equal [tex]\frac{7 L}{min}[/tex]*[tex]\frac{y(t) kg}{30L}[/tex] = [tex]\frac{7y(t)kg}{30min}[/tex]

    Now here is where I set up the d.e. mentioned above.

    [tex]\frac{dy}{dt}[/tex] = [tex]\frac{7}{3}[/tex] - [tex]\frac{7 y(t)}{30}[/tex]

    which equals

    [tex]\frac{dy}{dt}[/tex] = [tex]\frac{210-21y(t)}{90}[/tex]

    Now we can separate and integrate to find y(t), which i will just call y.

    so we have:

    [tex]\int[/tex][tex]\frac{dy}{210-21y}[/tex] = [tex]\int[/tex][tex]\frac{dt}{90}[/tex]

    which all boils down to:

    -21*ln|210-21y|=[tex]\frac{t}{90}[/tex]+C

    Now we can plug in our initial condition of y(0)=1.7 to solve for C.

    -21*ln|210-21(1.7)|=0+C so -21*ln|210-35.7|=C

    And from here I am stuck because the value i am getting for C doesnt seem correct.

    Should I solve the d.e. outright for y, then plug in the initial conditions.. or what??

    Thanks for any help or comments!!
     
  2. jcsd
  3. Mar 10, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Don't you want (-1/21)*ln|210-21y|??? Take the derivative.
     
  4. Mar 10, 2008 #3
    Yeah, thats quite embarrassing. Thanks for pointing it out though!
     
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