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Can anyone tell me where to begin?

How do I find the expression for steady state flux and steady state concentration for example?

What assumed knowledge is implicit in the question?

What common mathematical facts are relevant?


(4.) Consider a substance diffusing in one dimension with diffusivity $D$ from $x=0$ where the concentration is maintained at $c(0,t)=c_0$ to $x=L$ where the concentration is maintained at $c(L,t) = 0$ (i.e., the substance is removed as soon as it gets to $x = L$).

(a) Find an expression for the steady state flux and the steady state concentration.

(b) Find an expression for the total amount of substance $m$ in the region $(0,L)$ in steady state.

(c) The average transit time $\tau$ from $x=0$ to $x=L$ can be estimated as the time for the total amount $m$ to leave the region given the flux $q$ for the amount that leaves per unit time, i.e., $\tau = m/q$. Show how this estimate of $\tau$ relates to the mean square displacement.

(d) A typical neurotransmitter has a diffusivity $\approx 10^{-6} \mathrm{cm}^2 \mathrm{s}^{-1}$. How long does it take the neurotransmitter to diffuse across a synaptic junction that is about $0.02$ micron. How does this synaptic time delay compare with the typical speed of a neutron pulse ($\approx 27 \mathrm{m}\mathrm{s}^{-1}$).

(e) NEW UNANSWERED QUESTION: The concise edition of the Encyclopedia Britannica [sic] defines diffusion as the "process resulting from random motion of molecules by which there is a net flow of matter from a region of high concentration to a region of low concentration. A familiar example is the perfume of a molecule that quickly permeates the still air of a room. A typical perfume molecule has a diffusivity of $\approx 10^{-5} m^2 s^{-1}$. How long would it take a typical perfume molecule to diffuse across the still air of a room that is $\approx 10m$ across?

To be helpful, please explain the solution thoroughly in a way that a beginner can follow.

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This is an exercise in Fick's law, that postulates the following formula for the diffusion flux: $$ q = - D \frac{\partial c}{\partial x}, \tag{$\ast$} $$ where

  • $q$ is the diffusion flux, i.e., the amount of substance that flows through the cross-section at a given position. To prevent accumulation or draining of material at any $x \in (0, L)$, we assume that this flux is the same, say $q$, at all points.

  • $D$ is the diffusivity (again given in the question).

  • $c(x)$ is the concentration at time $t$. I am showing only the dependence on $x$, not on $t$.

It just remains to solve $(\ast)$. In steady state, $c$ depends on $x$, not on $t$. Therefore, $$ \frac{dc}{dx} = -\frac{q}{D}, $$ which gives $c = C - \frac{q}{D} x$ for some constant $C$. We are yet to use the two boundary conditions:

  • Setting $c=c_0$ at $x=0$ gives $C = c_0$, which implies that $c(x) = c_0 - \frac{qx}{D}$.
  • Setting $c = 0$ at $x=L$ gives $c_0 = \frac{qL}{D}$. You are supposed to solve for the flux $q$ in terms of the other constants $c_0$, $L$ and $D$.

Plugging this back in $(\ast)$, we get $$c = c_0 \left( 1 - \frac{x}{L} \right) .$$

(a) The steady state flux is $q = \cdots$, and the state state concentration is $c(x) = \cdots$.

(b) The total amount of substance at steady state is $$ m = \int_0^L c(x) ~dx = \int_{0}^L c_0 \left( 1 - \frac{x}{L} \right) ~dx = \cdots. $$

(c) You can find the average transit time $\tau$ from $\tau = m / q = \cdots$.

Finally, I believe the mean square displacement is given by $$ \frac{\int_{0}^{L} c(x) x^2 ~ dx}{\int_0^L c(x) ~ dx} = \cdots. $$

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What does $x$ actually represent? - conceptually it is been defined as length (or position, or even direction in some sources)? But length of what? I Googled for 2 hours and couldn't find any resource which explains this properly... – ptrcao Dec 2 '11 at 15:14
@ptrcao It's the position, not length or direction. [The direction is perhaps the X-axis.] Imagine a substance flowing in a tube (or something like it) where the tube extends from $x=0$ to $x=L$. Then $x$ is a general position on the tube. – Srivatsan Dec 2 '11 at 17:35
Steady state means the concentration, flux etc. do not change with time. In non-steady state, things change with time, so the problem gets more complicated. However, as far as I know, Fick's law does not directly cover this, so I cannot comment on how to find non-steady state flux and concentration. [And in most physical systems, steady state is attained as $t \to infty$, i.e., "at infinite time", and is never actually reached. However, after sufficiently long time, you might get a good enough approximation to steady state.] – Srivatsan Dec 3 '11 at 4:02
@ptrcao For some strange reason, I was not notified of your last comment. At steady state, $q$ will be independent of $t$. I will expand my answer soon. – Srivatsan Dec 5 '11 at 3:36
@ptrcao [I have been meaning to tell this for sometime: I admire your persistence and clarity of your queries.] That needs a bit of an explanation and it's getting late; I will expand my answer tomorrow. – Srivatsan Dec 5 '11 at 5:00

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