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Non-linear ODE: Use change of variable to solve the equation for initial conditions

  1. Jul 12, 2007 #1
    1. The problem statement, all variables and given/known data

    When an insoluble bubble rises in a deep pool of liquid, its volume increases according to the ideal gas law. However, when a soluble bubble rises from deep submersion, there is a competing action of dissolution that tends to reduce size. Under practical conditions, it has been proven that the mass transfer coefficient, [itex]k_c[/itex], for spherical bubbles in free-fall (or free-rise) is constant. Thus, for sparingly soluble bubbles released from rest, the applicable material balance is

    [tex]\frac{d}{dt}\,\frac{4\,C\,\pi}{3}\,R^3\,=\,-k_c\,C^*\,4\pi\,R^2[/tex]

    where the molar density of the gas

    [tex]C\,=\,\frac{P}{R_g\,T}[/tex]

    with [itex]R_g[/itex] and T as the ideal gas constant and temperature, respectively. [itex]C^*[/itex] is the molar solubility of the gas in the liquid, and R(t) is the bubble radius, which changes over time. The pressure P at a distance z from the top of the liquid surface is

    [tex]P\,=\,P_A\,+\,\rho_L\,g\,z[/tex]

    where [itex]\rho_L[/itex]is the liquid density and g is the gravitational acceleration. The rise velocity, [itex]\frac{dz}{dt}[/itex], follows a linear relation between speed and size

    [tex]\frac{dz}{dt}\,=\,\beta\,R(t)[/tex]

    where [itex]\beta[/itex] is a constant that depends on the liquid viscosity.


    SHOW that a change of variables allows the material balance equation to be written as

    [tex]R\,\frac{dR}{dP}\,+\,\left(\frac{1}{3}\right)\,\frac{R^2}{P}\,=\,-\frac{\lambda}{P}[/tex]

    and

    [tex]\lambda\,=\,\frac{k_c\,R_g\,T\,C^*}{\rho_L\,g\,\beta}[/tex]



    2. Relevant equations

    Algebra and the Calculus.



    3. The attempt at a solution

    [tex]\frac{dz}{dt}\,=\,\beta\,R(t)[/tex]

    [tex]\frac{d}{dt}\,=\,\beta\,R(t)\,\frac{d}{dz}[/tex]

    Substituting into the original material balance equation:

    [tex]\beta\,R(t)\,\frac{d}{dz}\,\left(\frac{P}{R_g\,T}\,\frac{4\pi\,R^3}{3}\right)\,=\,-k_c\,C^*\,4\pi\,R^2[/tex]

    Here I am stuck, how do I "show" that the two versions of the material balance equations are equivalent?
     
    Last edited: Jul 12, 2007
  2. jcsd
  3. Apr 30, 2011 #2
    Re: Non-linear ODE: Use change of variable to solve the equation for initial conditio

    Hey I am doing this question at the moment, I found this thread while doing a google search. Do you remember how you solved it?

    I've reached about just as far, but I've found lambda by using dP/dz = rhoL*g

    (Don't really know how to use html code tbh.)

    Any help is appreciated, thanks.
     
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