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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?


Question.

(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 http://concise.britannica.com/ebc/article-9030421/diffusion [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.

1 Answers 1

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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...2011-12-02
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    @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.2011-12-02
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    Great explanation so far, just seeking a further clarification though - could you please explain what "steady state" means in this context? What would be the situation described by non-steady state flux and concentration? How do I know when $q$ and $c(x)$ are in a steady state?2011-12-03
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    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.]2011-12-03
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    Hey Srivatsan, I found this [webpage](http://www.sv.vt.edu/classes/MSE2034_NoteBook/MSE2034_kriz_NoteBook/diffusion/examples/examples.html) which refers to Fick's first law as the stead-state scenario and Fick's second law as the non-steady state scenario. Does this make sense to you? Also see [this](http://www.physicsforums.com/showthread.php?t=132588#post_message_1093616).2011-12-03
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    Er, not really. I don't quite follow it, sorry.2011-12-03
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    According to mybsaccownt at [PF](http://www.physicsforums.com/showthread.php?t=132588#post_message_1093616) "in the 1st law, we're looking at steady state diffusion, and how the concentrations vary with respect to position" and that "in the 2nd law, we are looking at concertration with respect to time and it may be increasing or decreasing". "the 1st law is used for steady state diffusion (concentration does not change with respect to time), that's why D is a constant" and "2nd law is used for non-steady state diffusion, where the concentration does depend on time".2011-12-03
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    Srivatsan, when you integrated both sides of $\frac{dc}{dx} = -\frac{q}{D}$, are you treating $q(x,t)$ as a constant? Shouldn't $q(x,t)$ be a function, as the notation implies?2011-12-04
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    @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.2011-12-05
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    But $q$ still depends $x$, and since you are integrating $-\frac{q(x)}{D}$ with respect to $x$, how is it that $q$ is being treated as a constant in your solution...? (I'm not suggesting it's wrong; just trying to follow. :) )2011-12-05
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    @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.2011-12-05
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    That is kind of you to say. Whenever you are ready; I won't be going anywhere.2011-12-06