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Suppose the concentration $c(x,y,t)$ inside a reactor evolves the following $$\frac{\partial c}{\partial t}=D\frac{\partial^2 c}{\partial x^2}-kc , $$ where $D$ is the diffusion coefficient and $k$ is the reaction rate. If the characteristic length of the reactor is $L$, how does one make the equation dimensionless and identify important dimensionless parameters?

I imagine that the "new" unit of length must be $L$, however, I don't know how to continue (by the way, I don't fully understand what exactly this implies).

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The problem seems to contain three dimension parameters, $D$, $k$ and $L$. The first thing would be to determine what the typical scale parameters are. You need a length scale $x_0$ and a time scale $t_0$. The options for the length scale are $x_0=L$, $x_0=D/k/L$ or $x_0=\sqrt{D/k}$ and those for the time scale are $t_0=1/k$, $t_0=L^2/D$ or $t_0=L/\sqrt{kD}$. To determine which one to choose, one needs to think a little about the physics.

Next, one convert the function (concentration) into a function of dimensionless variable $$ c(x,t) \to c_0(u,v) = c_0\left(\frac{x}{x_0},\frac{t}{t_0}\right) . $$ Then we plug this back into the equation. The result is $$ \frac{1}{t_0} \partial_v c_0(u,v) = \frac{D}{x_0^2}\partial_u^2 c_0(u,v) - kc_0(u,v) . $$ Based on this result, it seems that the appropriate choice for the scale parameters would be $x_0=\sqrt{D/k}$ and $t_0=1/k$, because then the dimensionless of the equation becomes completely independent of any parameters: $$ \partial_v c_0(u,v) = \partial_u^2 c_0(u,v) - c_0(u,v) . $$

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    $\begingroup$ Note, that by rendering equation independent of any parameter, you need to fix boundary conditions at variable values of $u$, so the implicit dimensionless parameter would be $L/x_0$. Another possibility would be fixing boundary conditions (say, at $u=0$ and $u=1$) by setting $x_0=L$ and retaining dimensionless parameter in the equation (for example $\alpha=D/(L^2 k)$ as a coefficient in front of $\partial^2_u$). $\endgroup$ – A.V.S. Jan 13 '18 at 20:52
  • $\begingroup$ @A.V.S. - +1 Something both answers forget to mention is the role of the boundary conditions in determining the scales. $\endgroup$ – nluigi Jan 16 '18 at 9:52
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First divide all the equation by some reference concentration $c_0$, so that $c/c_0$ is dimensionless.

Now, $D$ has units of $[\text{space}]^2/[\text{time}]$, and $k$ has the units of $1/[\text{time}]$, so multiplying all the equation by a reference time $t_0$ is the same as writing the dimensionless $t/t_0$ on the left and $t_0 k$ on the right. Moreover dividing and multiplying by some reference position $x_0$ squared in the first term of the right gives: $$\dfrac{\partial (c/c_0)}{\partial(t/t_0)} = \dfrac{D}{D_0} \dfrac{\partial^2(c/c_0)}{\partial(x/x_0)^2} - (t_0k)(c/c_0)$$ where $D_0 = \dfrac{x_0^2}{t_0}$ has units of diffusion coefficient.

This is the same as adopting a system to measure things. For instance, you can take $x_0 = 1 \text{meter}$ , $t_0 = 1 \text{sec}$ and $c_0$ some reference concentration you like. This fixes the reference diffusion coefficient to be $1\text{meter}^2\text{sec}^{-1}$.

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  • $\begingroup$ A word of caution. If any of the parameters (in this case $D, k, c$) are zero you can mess yourself up by mindlessly setting the scaled values to $1$. And be careful about the signs used - the mathematics for some problems becomes ugly if you use the wrong choice of sign, most typically you can end up with $i=\sqrt{-1}$ popping up all over the place. $\endgroup$ – StephenG Jan 11 '18 at 23:11

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