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Homework Help: Diff. Eqs. Mixing Problem

  1. Jan 29, 2007 #1
    I haven't found any examples anywhere on how to do a mixing problem where you don't know the rate of volume change, could someone give me a hand just setting up the equation? Here is the problem. . .

    1. A tank contains 500 gal of a salt-water solution containing 0.05 lb of salt per gallon of water. Pure water is poured into the tank and a drain at the bottom of the tank is adjusted so as to keep the volume of solution in the tank constant. At what rate (gal/min) should the water be poured into the tank to lower the salt concentration to 0.01 lb/gal of water in under one hour?
  2. jcsd
  3. Jan 29, 2007 #2
    Sorry I didn't post what work I've tried:
    if x(t) is amount of salt in that tank then
    dx/dt=Volume in*Salt Concentration in-Volume out*Salt Contentration out
    dx/dt=Volume in*(0)-Volume Out*(x/500)
    dx/dt=-(x/500)*(Volume Out)
  4. Jan 29, 2007 #3


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    Hey there, in the beginning before pumping out the salt (x(0)=25 lb), you know the tank will have 25 pounds of salt dissolved, then you want to diminish that to 5 pounds of salt (x(t) = 5 lb). You are given the time too, so t = 60, if the time is in minutes.
    Last edited: Jan 29, 2007
  5. Jan 30, 2007 #4
    Ok, maybe I'm missing some obvious thing but does the information you gave let me calculate dx/dt? I'm still confused. I was pretty sure I'd use that information after I had calculated the differential equation.
  6. Jan 30, 2007 #5
    You're looking for volume/time at time t where salt concentration =.01, essentially what it's asking you to do is to rearrange the equation so that the unknown is = to the variables you are given, have a play around untill you find an equation that relates pounds and time to volume

    Once you have that you should be able to integrate to find your answer. at f'(59.59) t=59.59 or just under an hour where salt concentration=.01.

    ie with these values of the variables in your equation with the unknowns (x) etc.

    EDIT: Sorry I meant integrate obviously.
    Last edited: Jan 30, 2007
  7. Jan 30, 2007 #6


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    And "Volume Out" is the constant you want to find: call it v.

    dx/dt= -(v/500)x

    Can you integrate that? What must v be so that x(1)< 0.01?
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