Laplace Transforms on PDEs: Non-dimensionalization

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VinnyCee
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Laplace Transforms on Partial Differential Equations - Non-dimensionalization too!

Homework Statement



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]

Homework 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]

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!
 
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Well, how do [tex]\partial t[/tex] and [tex]\partial z[/tex] nondimensionalise?
 
I don't know, how do I figure that out?
 
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What's [tex]\partial z/\partial \zeta[/tex]?
 
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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]?
 
Sorry, the earlier post should've been [tex]\partial z/\partial \zeta[/tex].

This is how you get [tex]\partial z[/tex]...
 
[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]
 
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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]
 
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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