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Help with Tank problem- NOT TYPICAL

  1. Jun 24, 2011 #1
    Help with Tank problem-- NOT TYPICAL

    It's been a few years since my last experience with differential equations so this is giving me some issues... Any and all help is appreciated.

    Problem:

    A 65 gallon tank has 20 pounds (or ~1 gallon) of sand in it with 35 gallons of water (completely mixed). We pump/circulate the solution at 5 gallons/minute and filter it, then put it back in the tank. Clean water goes in at the same rate that sandy water is pulled out, so there is no overflow or volume change.

    How many minutes must we filter the mixture before we have a tank that only contains 0.2 pounds (0.01 lbs, or 99% clean) of sand?
     
  2. jcsd
  3. Jun 24, 2011 #2
    Re: Help with Tank problem-- NOT TYPICAL

    Subscribed.
     
  4. Jun 24, 2011 #3
    Re: Help with Tank problem-- NOT TYPICAL

    What is the rate that sand would be removed as a function of the amount of sand in the tank?

    By the way, the problem you described isn't realistic; there would be a volume change because the sand would take up volume.
     
  5. Jun 24, 2011 #4
    Re: Help with Tank problem-- NOT TYPICAL


    The pump circulates on the tank through the filter, and the filter catches all the sand. The concentration of the water at that time is pulled out, and clean (filtered) water is put back in.


    What I mean by "no volume change" is that in the entire system, there is no change. Sure, there is water in the hoses/pump, etc. but the tank isn't going to overflow or drain completely.



    ...this is currently happening where I work right now, BTW. I can post pictures if you'd like. :D
     
  6. Jun 24, 2011 #5
    Re: Help with Tank problem-- NOT TYPICAL

    dx/dt = IN - OUT

    In: 0 (Because it's clean water)

    Out = (xlbs of sand/35 gallons of water)(5gallons/min)

    dx/dt = 0 - 5x/35

    x(0) = 20 (at 0 minutes, there is exactly 20 pounds of sand in the tank)

    you need to find when x(t) = .2 this will tell you at what time t the tank has .2 pounds in it
     
  7. Jun 25, 2011 #6
    Re: Help with Tank problem-- NOT TYPICAL


    Bear with me here, it's been a while since I've done this...

    Integrate both sides with respect to X then plug in 0.2 for X and the resulting number is my time in minutes?
     
  8. Jun 25, 2011 #7
    Re: Help with Tank problem-- NOT TYPICAL

    You need to separate dx and dt.
     
  9. Jun 26, 2011 #8
    Re: Help with Tank problem-- NOT TYPICAL

    Ohhh, that's right!

    So I'd have 7*dx/x = dt then integrate each side w.r.t. the variable, then plug in 0.2 for values of x. Correct?
     
  10. Jun 26, 2011 #9
    Re: Help with Tank problem-- NOT TYPICAL

    dx/dt = 0 - 5x/35

    dx/dt = -x/7
    7/x dx/dt = -1
    7/x dx = -dt

    yep, sounds good
     
  11. Jun 27, 2011 #10
    Re: Help with Tank problem-- NOT TYPICAL

    approximately 11.5 minutes. Thanks a LOT folks!
     
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