Laplace Transforms on PDEs: Non-dimensionalization

[VinnyCee]

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 \left(\frac{D_{AB}\,t}{L^2}\;>>\;1\right). We wish to reconsider this analysis in order to derive a suitable relationship for short time conditions \left(\frac{D_{AB}\,t}{L^2}\;<<\;1\right). For short time conditions, the effect of concentration-dependent diffusivity is minimized. In the analysis to follow, use dimensionless independent variables

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

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

\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]

from the "fraction solute A remaining" equation

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

Homework Equations

PDEs, Non-dimensionalization, Laplace Transforms.

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

\frac{{\partial x_A }}{{\partial t}}\; = \;D_{AB} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}

with initial and boundary conditions

t\,=\,0,\;\;x_A\,=\,x_0

z\,=\,0,\;\;x_A\,=\,0

z\,=\,L,\;\;D_{AB}\,\frac{\partial\,x_A}{\patial\,z}\,=\,0

The Attempt at a Solution

D_{AB}\,=\,\frac{\theta\,L^2}{t}

z\,=\,\zeta\,L

Using the first equation to substitute into the PDE above

\frac{{\partial x_A }}{{\partial t}}\; = \;\frac{\theta\,L^2}{t} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}

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!

 

[J77]

Well, how do \partial t and \partial z nondimensionalise?

 

[VinnyCee]

I don't know, how do I figure that out?

 

[J77]

What's \partial z/\partial \zeta?

 

[VinnyCee]

Let's focus on \partial z first!

How would I get that knowing that z\,=\,\zeta\,L?

Is \partial z\;=\;0?

 

[J77]

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

This is how you get \partial z...

 

[VinnyCee]

\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

\partial\,t\;=\;\frac{L^2}{D_{AB}}\,\partial\,\theta

And for the other "dimensionless independent variable"

\zeta\;=\;\frac{z}{L}\;\;\longrightarrow\;\;z\;=\;\zeta\,L

\frac{\partial\,z}{\partial\,\zeta}\;=\;L

Now get an expression to change the variable from z to \zeta

\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

\frac{1}{L^2}\,\frac{\partial^2\,x}{\partial\,\zeta^2}\,=\,\frac{\partial^2\,x}{\partial\,z^2}

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

\frac{{\partial\,x_A}}{{\partial\,t}}\;=\;D_{AB}\,\frac{{\partial^2\,x_A}}{{\partial\,z^2}}

\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}

\frac{\partial\,x_A}{\partial\,\theta}\;=\;\frac{\partial^2\,x_A}{\partial\,\zeta^2}

now take a Laplace transform

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

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

s\,X\,-\,x_0\;=\;\frac{d^2\,X}{d\zeta^2}

X''\,-\,s\,X\,=\,-x_0

 

[VinnyCee]

Solving the ODE

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

Now use the R equation?

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

 

[adebam]

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