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