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Dilution problem

  1. Nov 24, 2009 #1
    Ok, I'm working through some dilution/accretion problems in my ODEs book (not homework), all of them centered around draining brine from a tank. The author illustrates the method for solving the problem when you're adding more brine to the tank, and the total water level stays the same, you simply separate the variables. The author also illustrates the method for solving the problem when the water level is changing, but you're not adding more brine to the tank. Again you simply separate the variables.

    And that's it, that's all my book explains. It doesn't explain at all what to do when you're both changing the water level and adding more brine to the tank. The author doesn't even hint at a method for solving it:


    Adding brine, but not changing the water level:

    100 gallon tank, 3 gals/min brine flow in at 2 lbs of salt per gallon, 3 gals/min of the mix flow out.
    dx = 6dt (brine) - (x/100)3dt.
    dx/(x-200) = -.03dt.
    Separable, hooray.


    Not adding brine, changing the water level:

    100 gallon tank, 2 gals/min fresh water flow in, 3 gals/min of the mix flow out.
    dx = (x/(100-t))3dt.
    dx/x = (3/100-t)dt
    Again, separable.


    Adding brine, changing the water level:

    100 gallon tank, 3 gals/min brine flow in at 2lbs of salt per gallon, 2 gals/min of the mix flow out.
    dx = 6dt - (x/(100+t))2dt
    That doesn't look very separable to me....
     
  2. jcsd
  3. Nov 25, 2009 #2
    You are right, it isn't. However, it can be solved. First define [tex]\tau=t+100[/tex]. Then:
    [tex]x'=6-2x/\tau[/tex]
    which is homogeneous (invariant under [tex]x\rightarrow ax, \tau \rightarrow a\au [/tex]). For that kind, you have the standart change [tex] z=x/\tau [/tex], which transforms it into the new equation
    [tex] z'\tau=6-3z [/tex]
    which is now separable. The full solution is then
    [tex] x=2(t+100)-\frac{C}{(t+100)^{2}} [/tex]
     
    Last edited: Nov 25, 2009
  4. Nov 25, 2009 #3
    I see how that works. Thanks a lot for the help
     
  5. Nov 25, 2009 #4
    Actually on second thought I'm not quite sure what you're trying to specify with z'. What exactly are you differentiating Z with respect to?
     
  6. Nov 26, 2009 #5

    HallsofIvy

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    He is differentiating the new dependent variable, z, with respect to the new independent variable, [itex]\tau[/itex].
     
  7. Nov 26, 2009 #6
    that is [tex]x=z(\tau)\tau\rightarrow x'(\tau)=z'\tau+z[/tex]
     
  8. Dec 28, 2009 #7
    at what speed should a clock be moved so that it may appear to lose 1 minute in each hour?
     
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