CL: Molecular Origins of Viscosity

Returning to our molecular picture of shear stresses, we can think of the momentum flux (shear stress) as being related to the movement (swapping) of molecules across a plane. We might write this flux (for Z molecules that moved) as:

$\displaystyle{\tau = - \sum_{n=1}^Zm_n\left (v_x|_{y-}-v_x|_{y+}\right)}$

where the |y- refers to the value at the location just below our plane. If we relate this value (and the + variety) to our slope of the velocity (shear rate or rate of strain) by:

$\displaystyle{v_x|_{y-} = v_x|_{y_0}-\left (\frac{dv_x}{dy} \right)|_{y_o}\delta}$

$\displaystyle{v_x|_{y+} = v_x|_{y_0}+\left (\frac{dv_x}{dy} \right)|_{y_o}\delta}$

we can then combine these equations to give

$\displaystyle{\tau = 2 \sum_{n=1}^Zm_n\frac{dv_x}{dy}|_{y_o}\delta}$

Here, we can think of $\delta$ as the average distance between molecular collisions. This can be related to a statistical mechanics quantity, the mean free path as $\delta=(2/3)\lambda$ so that our sum becomes (for molecules of the same mass):

$\displaystyle{\tau = \frac{4}{3}\lambda Z m\frac{dv_x}{dy}|_{y_o}}$

which we can show gives us

$\displaystyle{\mu = \frac{4}{3}\lambda Z m}$

Z can be related to the molecular concentration N and the average random molecular velocity C as Z = NC/4. Using this and the statistical mechanics expressions for the hard-sphere mean free path and average velocity:

$\displaystyle{\lambda = \frac{1}{\sqrt{2}\pi N d^2}}$

$\displaystyle{C = \sqrt{\frac{8\kappa T}{\pi m}}}$

where d is the molecular diameter. Combining these expressions gives the hard sphere viscosity:

$\displaystyle{\mu = \frac{2}{3\pi^{3/2}}\frac{\sqrt{m\kappa T}}{d^2}}$


Explain the molecular origins of fluid viscosity and shear stresses