How can I derive Fick's second law for a fluid in a narrow pipe?

[jason177]

Homework Statement

Imagine a narrow pipe, filled with fluid, in which the concentration of a certain type of molecule varies only along the length of the pipe (in the x direction). By considering the flux of these particles from both directions into a short segment \Deltax, derive Fick's second law, dn/dt = D d²n/dx² (those should be partial derivatives not normal ones) where n is the particle concentration and D is the diffusion coefficient.

Homework Equations

J_x = -D dn/dx
where J is the particle flux

The Attempt at a Solution

I don't even know where to start

 

[vanesch]

The question is: can you take Fick's first law for granted ?

 

[jason177]

It doesn't say whether we can or not so I assume we can.

 

[vanesch]

Well, then this is not so hard. What you have to do is to consider a piece of medium with thickness \Delta X and "do the bookkeeping" of what goes in, what goes out, and hence how things change locally (also called "mass conservation") during a time \Delta t.

 

[jason177]

Alright well after playing around with it for a while I still have no idea what to do. How would I do it if we couldn't take Fick's first law for granted?

 

[vanesch]

Consider a position x0, and a position a bit further, at x0 + \Delta x.

Consider a time t0 and a time t0 + \Delta t.

Consider a density n(x,t) that is function of x.

Now consider how much is "in" the box \Delta x at time t.

Consider how much "comes in" at the "x" wall and how much "goes out" at the "x + \Delta x side during the time \Delta t.

That should be more than enough to get you going...