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Evaporation differential problem

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data
    ltQxz.jpg


    2. Relevant equations
    Given in problem


    3. The attempt at a solution
    3uXPE.jpg

    Hey guys

    So I have this problem I'm working on and I just can't seem to get too far with it. I've gotten half of the first part...I'm just having trouble integrating the equation I got...and I don't even know if the equation I got is right in the first place!

    I set:

    Rin = pi
    Rout = pi(20h-h^2)

    dh/dt = Rin - Rout


    I'm trying to solve the DE right now but don't know how to proceed...

    Any help would be great,
    Thanks.
     
  2. jcsd
  3. Oct 16, 2012 #2
    Anyone? Sorry for bumping this so early!
     
  4. Oct 16, 2012 #3
    I looked at it and it looks very messy to include the evaporation term. But maybe I'm missing something. So instead, I worked an analogous problem by filling an inverted right triangle (like the cross-section of a trough made with two pieces of wood) with area in the same way: We can fill the area at a rate of [itex]k[/itex] and let the evaporation rate be proportional to the length of the base (on top of it) with proportionality constant [itex]c[/itex]. So, I would write the change in area as:

    [tex]\frac{dA}{dt}=f(A,t)=k-cL[/tex]

    but I can get the length of the base as a function of A. For a right triangle, that's [itex]L(A)=\sqrt{4 A}[/itex] so then we have including evaporation:

    [tex]\frac{dA}{dt}=k-c\sqrt{4A}[/tex]

    Integrating:

    [tex]\int_0^A \frac{dy}{k-c\sqrt{4y}}=\int_0^t dt[/tex]

    And even that simple case is still very messy and the function A(t) is in terms of a Lambert function.

    Now, if you used the same principle, in your case, expressing area of the pond surface as a function of volume beneath it, the DE I think would be:

    [tex]\frac{dV}{dt}=\pi-0.01 A(V)[/tex]

    but the function A(V) looks way too messy to do it symbolically.

    Maybe someone else can help.
     
    Last edited: Oct 16, 2012
  5. Oct 16, 2012 #4
    Thanks I greatly appreciated your time and effort.

    Hopefully someone else will chime in soon.
     
  6. Oct 17, 2012 #5
    How about we try this:
    We know from part (a) that [itex]A(h)=\pi(20 h-h^2)[/itex] and:

    We can write:

    [tex]\frac{dh}{dt}=\frac{dh}{dV} \frac{dV}{dt}[/tex]

    and first without evaporation we have:

    [tex]\frac{dh}{dt}=\frac{dh}{dV} \frac{dV}{dt}=\frac{1}{2\pi Rh-\pi h^2} \pi[/tex]

    we can integrate that:

    [tex]\int_0^h (2Rh-h^2)dh=\int_0^t dt[/tex]
    and get:

    [tex]Rh^2-1/3h^3=t[/tex]
    and use that to figure out how long it takes to fill without evaporation.

    Now with evaporation, we have:

    [tex]\frac{dV}{dt}=\pi-kA(h)[/tex]

    so substituting, we obtain:

    [tex]\frac{dh}{dt}=\frac{1}{2\pi Rh-\pi h^2}\left[\pi -k(20\pi h-\pi h^2)\right][/tex]

    Now the question is, can we integrate that and is it correct? Gotta' try things in math you know, and learn to tolerate making wrong turns. :)
     
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