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Laplace Transforms on Partial Differential Equations - Non-dimensionalization too

  1. Jul 26, 2007 #1
    Laplace Transforms on Partial Differential Equations - Non-dimensionalization too!!!

    1. The problem statement, all variables and given/known data

    The experiment described in the previous problem was analyzed from the point of view of long time [itex]\left(\frac{D_{AB}\,t}{L^2}\;>>\;1\right)[/itex]. We wish to reconsider this analysis in order to derive a suitable relationship for short time conditions [itex]\left(\frac{D_{AB}\,t}{L^2}\;<<\;1\right)[/itex]. For short time conditions, the effect of concentration-dependent diffusivity is minimized. In the analysis to follow, use dimensionless independent variables

    [tex]\theta\;=\;\frac{D_{AB}\,t}{L^2},\;\;\;\zeta\,=\,\frac{z}{L}[/tex]

    Apply Laplace transforms to the non-dimensional transport equation to show that

    [tex]\bar R\left( s \right)\; = \;\frac{1}{s}\; - \;\frac{1}{{s^{\frac{3}{2}} }}\;\left[ {\frac{{1\; - \;e^{ - 2\,\sqrt s } }}{{1\; + \;e^{ - 2\,\sqrt s } }}} \right][/tex]

    from the "fraction solute A remaining" equation

    [tex]R\;=\;\frac{1}{L}\,\int^L_0\,\frac{x_A\left(z,\,t\right)}{x_0}\,dz[/tex]



    2. Relevant equations

    PDEs, Non-dimensionalization, Laplace Transforms.

    A hint is given that we must obtain an expression for the Laplace transform of the composition [itex]x_A[/itex] that appears in the PDE below

    [tex]\frac{{\partial x_A }}{{\partial t}}\; = \;D_{AB} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}[/tex]

    with initial and boundary conditions

    [tex]t\,=\,0,\;\;x_A\,=\,x_0[/tex]

    [tex]z\,=\,0,\;\;x_A\,=\,0[/tex]

    [tex]z\,=\,L,\;\;D_{AB}\,\frac{\partial\,x_A}{\patial\,z}\,=\,0[/tex]



    3. The attempt at a solution

    [tex]D_{AB}\,=\,\frac{\theta\,L^2}{t}[/tex]

    [tex]z\,=\,\zeta\,L[/tex]

    Using the first equation to substitute into the PDE above

    [tex]\frac{{\partial x_A }}{{\partial t}}\; = \;\frac{\theta\,L^2}{t} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}[/tex]

    But what do I do about the squared partial derivative of z in the last term?

    I have been thinking about this problem for hours, I just don't know where to start in order to derive the equation that is requested. Any help would be greatly appreciated!
     
    Last edited: Jul 26, 2007
  2. jcsd
  3. Jul 26, 2007 #2

    J77

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    Well, how do [tex]\partial t[/tex] and [tex]\partial z[/tex] nondimensionalise?
     
  4. Jul 26, 2007 #3
    I don't know, how do I figure that out?
     
    Last edited: Jul 26, 2007
  5. Jul 26, 2007 #4

    J77

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    What's [tex]\partial z/\partial \zeta[/tex]?
     
    Last edited: Jul 26, 2007
  6. Jul 26, 2007 #5
    Let's focus on [tex]\partial z[/tex] first!

    How would I get that knowing that [tex]z\,=\,\zeta\,L[/tex]?

    Is [tex]\partial z\;=\;0[/tex]?
     
  7. Jul 26, 2007 #6

    J77

    User Avatar

    Sorry, the earlier post should've been [tex]\partial z/\partial \zeta[/tex].

    This is how you get [tex]\partial z[/tex]...
     
  8. Jul 26, 2007 #7
    [tex]\theta\;=\;\frac{D_{AB}\,t}{L^2}\;\;\longrightarrow\;\;\frac{\partial\,\theta}{\partial\,t}\;=\;\frac{D_{AB}}{L^2}\;\;\longrightarrow\;\;\partial\,\theta\;=\;\frac{D_{AB}}{L^2}\,\partial\,t[/tex]

    [tex]\partial\,t\;=\;\frac{L^2}{D_{AB}}\,\partial\,\theta[/tex]

    And for the other "dimensionless independent variable"

    [tex]\zeta\;=\;\frac{z}{L}\;\;\longrightarrow\;\;z\;=\;\zeta\,L[/tex]

    [tex]\frac{\partial\,z}{\partial\,\zeta}\;=\;L[/tex]

    Now get an expression to change the variable from z to [itex]\zeta[/itex]

    [tex]\frac{\partial\,x}{\partial\,\zeta}\;=\;\frac{\partial\,x}{\partial\,z}\,\frac{\partial\,z}{\partial\,\zeta}\,=\,\frac{\partial\,x}{\partial\,z}\,L\;\;\longrightarrow\;\;\frac{\partial^2\,x}{\partial\,\zeta^2}\,=\,\frac{\partial^2\,x}{\partial\,z^2}\,L^2[/tex]

    [tex]\frac{1}{L^2}\,\frac{\partial^2\,x}{\partial\,\zeta^2}\,=\,\frac{\partial^2\,x}{\partial\,z^2}[/tex]

    Now substitute the two change of variable expressions above into the hint equation from the top

    [tex]\frac{{\partial\,x_A}}{{\partial\,t}}\;=\;D_{AB}\,\frac{{\partial^2\,x_A}}{{\partial\,z^2}}[/tex]

    [tex]\frac{D_{AB}}{L^2}\,\frac{\partial\,x_A}{\partial\,\theta}\;=\;D_{AB}\,\frac{1}{L^2}\,\frac{\partial^2\,x_A}{\partial\,\zeta^2}[/tex]

    [tex]\frac{\partial\,x_A}{\partial\,\theta}\;=\;\frac{\partial^2\,x_A}{\partial\,\zeta^2}[/tex]

    now take a Laplace transform

    [tex]s\,X\,-\,x\left(\theta\,=\,0\right)\;=\;\frac{d^2\,X}{d\zeta^2}[/tex]

    and set [itex]x(\theta\,=\,0)\;=\;x_0[/itex] to get a second order ODE

    [tex]s\,X\,-\,x_0\;=\;\frac{d^2\,X}{d\zeta^2}[/tex]

    [tex]X''\,-\,s\,X\,=\,-x_0[/tex]
     
    Last edited: Jul 27, 2007
  9. Jul 26, 2007 #8
    Solving the ODE

    [tex]x_A\;=\;A\,e^{\sqrt{s}\,\zeta}\,+\,B\,e^{-\sqrt{s}\,\zeta}\,-\,\frac{C}{s}[/tex]

    Now use the R equation?

    [tex]R\;=\;\frac{1}{L}\,\int^L_0\,\frac{x_A\left(z,\,t\right)}{x_0}\,dz[/tex]
     
    Last edited: Jul 26, 2007
  10. Dec 4, 2008 #9
    VinnyCee,
    Just wondering whether we have to dimensionless everything you solve PDE. Is it possible to solve the PDE with the DAB in it.
    Regards
    Daivd
     
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