Ch 9.8 Diffusion

• Particle diffusion shares characteristics with heat conduction.
• Similar to Fourier's Law for heat conduction, Fick's Law of Diffusion will apply to mass transport.
• This will lead to equations that model particle diffusion.

What do you call a fake noodle?

An impasta.

• Diffusion is a mechanism for mass transport.
• Particles flow from high concentration to low concentration.
• Movement towards uniform concentration in available space.

Concentration is the key variable in mass transport, and can be defined in a number of ways, including the following:

• The mass of particles divided by volume containing the particles, with units kg/liter or mg/liter.
• The number of particles in a volume divided by the volume.
• The volume of all the particles (the solute) as a fraction of the total volume containing the particles (the solution).

Formulate differential equation for equilibrium concentration of fluid particles in pipe as a function of distance from end of pipe.

• \( \small{x =} \) distance into pipe from left end, in meters.
• \( \small{C(x) =} \) equilibrium concentration at \( x \), in \( \small{kg/(m^3)} \).
• \( \small{J(x) =} \) mass flux, with units \( \small{kg/(m^2 sec)} \).
  

Relates mass flux to the rate of change of concentration:

\[ J(x) = -D \frac{dC}{dx} \]

Similar to Fourier's law of heat conduction.

• \( D \) is the diffusion coefficient, or diffusivity, with units \( m^{2}/s \)
  
• Diffusivity depends on particles, fluid, and temperature.
  
• Particles diffuse more easily at higher temperatures.
• Diffusion is a slow process (see values in table).

\[ J(x,t) = -D \frac {\partial C (x,t)}{\partial x} \]

• Rate of mass flow is expressed in terms of kg/sec.
• Units of \( \small{J(x)A(x)} \) are \( \small{\left(\frac{kg}{m^2 sec}\right)(m^2) = \frac{kg}{sec}} \).
• Starting with the word equation, we proceed as follows:

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{rate \, of \, change \, of } \\ \mathrm{mass \, within \, region} \end{Bmatrix} &= \begin{Bmatrix} \mathrm{rate \,mass \, flows} \\ \mathrm{into \, region} \\ \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate \,mass \, flows} \\ \mathrm{ out \, of \, region} \end{Bmatrix} \\ 0 & = J(x)A - J(x+ \Delta x)A \end{aligned} } \]

• Let \( \small{h = \Delta x} \), divide both sides by \( \small{-hA} \), and let \( \small{h \rightarrow 0} \):

\[ \small{ 0 = \lim_{h \rightarrow 0} \frac{J(x+ h) - J(r)}{h} = \frac{dJ}{dx}} \]

Our ODE for mass flux \( J(x) \) is:

\[ \frac{dJ}{dx}=0 \]

To get an ODE in terms of \( C(x) \), recall Fick's Law:

\[ J(x) = -D\frac{dC}{dx} \]

Thus the ODE for the concentration \( C(x) \) at equilibrium is

\[ \frac{d^2C}{dx^2}=0 \]

For linear diffusion inside cylinder, the ODE we derived is

\[ \frac{d^2C}{dx^2}=0 \]

For radial diffusion inside the cylinder, the ODE is

\[ \frac{d}{dr} \left( r \frac{dC}{dr} \right) = 0 \]

This follows since the area through which diffusion occurs changes with distance \( r \) from the center.

For spherical geometry, we have

\[ \frac{d}{dr} \left( r^2 \frac{dC}{dr} \right) = 0 \]

This follows since the area through which diffusion occurs changes with square of the distance \( r^2 \) from the center.

If we add an additional source/sink of particles into the system at distance \( r \) from the center of the sphere, then the ODE gets more complicated; see next slide.

\[ D \frac{1}{r^2} \frac{d}{dr} \left( r \frac{dC}{dr} \right) + M(r) = 0 \]

Assuming volumetric mass source at distance \( r \) from center:

\[ D \frac{1}{r^2} \frac{d}{dr} \left( r \frac{dC}{dr} \right) + M(r) = 0 \]

• \( D \) is diffusion coefficient and \( M(r) \) is rate of “production” of mass, in \( \frac{kg} {m^3 s} \).
• Note that \( M(r) \gt 0 \) and \( M(r) \lt 0 \) with the addition and removal of mass, respectively.