5
$\begingroup$

Summary

From dimensional analysis I find that the dynamic viscosity of an ideal gas must depend on its pressure $p$, density $\rho$ and mean molecular free path $l$ in this way:

$$ \mu = C \sqrt{\rho p} l.\quad $$

Here, $C\geq0$ is a non-dimensional constant.

However, I find it counter intuitive that the dynamic viscosity, the 'internal friction', of the fluid increases with an increasing mean free path. My intuition tells me that the internal friction is low if the molecules are widely separated.

  • Have I missed some quantity that should enter the expression?
  • Has my derivation failed in some other way?
  • Is my intuitive picture wrong?

The derivation

In an ideal gas, molecules are interacting only through ellastic collisions. The equation of state is:

$$ p = \rho R T. \quad (1) $$

The variables and their units are:

  • $p$: Pressure [kg/(m s$^2$)]
  • $\rho$: Density [kg/(m$^3$)]
  • $R$: Specific gas constant [m$^2$/(s$^2$ K)]
  • $T$: Temperature [K]

In general, these are field variables, so $p = p(\mathbf{x},t)$, $\rho = \rho(\mathbf{x},t)$ and $T = T(\mathbf{x},t)$. In fluid dynamics, a common assumption is that each infinitesimally small volume is in thermodynamic equilibrium, so that (1) holds at every point in the fluid. I make this assumption. I also assume that the fluid is 'Newtonian', so that the viscous stress tensor is proportional to the rate of strain. The constant of proportionality is the dynamic viscosity, $\mu$, whose unit is [kg/(m s)].

The dynamic viscosity is a 'material property'; it is independent of the motion of the fluid. In general, it is varying over space, so that $\mu = \mu(\mathbf{x},t)$. It's value is a property of the material and depends on its thermodynamic state.

It seems impossible to find how $\mu$ depends on the thermodynamic state from (1). Pressure has 'almost' the correct units, but I need to multiply the pressure by some time scale $\tau$ [s]. This time scale must depend on the microscopic properties of the material, and the only way I find it possible to construct it is by using the $l$ [m] the mean free path of the molecules in the fluid. The time scale contructed is:

$$ \tau = \sqrt{\frac{\rho}{p}} l.\quad (2) $$

Using (2) I find that the dynamic viscosity must depend on $p$, $\rho$ and $l$ in this way:

$$ \mu = C \sqrt{\rho p} l,\quad (3) $$

where $C\geq0$ is a non-dimensional constant.

Improve this question
$\endgroup$
3
  • $\begingroup$ Actually, the viscosity of an ideal gas (i.e., a real gas in the limit of low pressures) does not depend on p. See Bird, Stewart, and Lightfoot, Transport Phenomena for the derivation you are seeking. $\endgroup$
    – Chet Miller
    Feb 9, 2017 at 12:39
  • $\begingroup$ Is that really true? In such case, for a fixed $R$, I am quite sure that I need a quantity additional to $\rho$, $p$, $T$, and $l$ to construct the viscosity. I can not think of any such quantity. Which one would that be? $\endgroup$
    – Frysen
    Feb 9, 2017 at 12:53
  • $\begingroup$ See my answer below. $\endgroup$
    – Chet Miller
    Feb 9, 2017 at 13:08

2 Answers 2

Reset to default
3
$\begingroup$

Your intuition is wrong on this. Consider one-dimensional steady flow, say in the $x$-direction, with a velocity gradient in the $y$-direction. Thus the particles at a given level have average velocity $$\bar{\mathbf u}=(u(y), 0, 0)^T,$$ and fluctuating velocities $${\mathbf u}'=(u', v', w')^T.$$

Let's consider particles that at time $t_0$ are located at $(x,y_0)^T$, which have velocities ${\mathbf u}=(u_0+u', v', w')^T.$ These particles will, on average, travel a distance of the mean free path length $l$ at that velocity, before hitting other particles. The particles will thus have migrated to a different $y$ position, where the average particle velocity will be $$\bar{\mathbf u(y)}=(u(y_0)+(y-y_0)\frac{\partial u}{\partial y}, 0, 0)^T.$$ Notice that the average difference in the $x$-component of the velocity of such particles will therefore be proportional to the mean free path $l$ times an integral $I$ over the distribution of $v'$ and $w'$ velocities which does not matter here: We have $y-y_0=I\,l$. The mean velocity difference for such particles is therefore just $\bar{\Delta u}=I\,l\,(\partial u/\partial y)$.

Since the mean velocity is assumed to stay constant, such particles will have their velocity adjusted to the one at their new $y$-position. Viscous forces correspond to the work required to achieve this. These forces must therefore be proportional to the velocity gradient times the mean free path length.

P.S.: Also see the derivation in the Wikipedia article on viscosity.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for a very good argument for why the visosity increases with $l$. This has developed my intuitive picture of viscosity. Do you agree to this picture of your argument: "If the mean free path is long, the fluid is more 'entangled'. If momentum exchange between different fluid parts are illustrated by springs, this will to some extent resemble a web of springs that are pushing/pulling the other over some distance. Making each spring longer (increasing the mean free path) makes different parts of the fluid more 'connected' (actually more viscous)." $\endgroup$
    – Frysen
    Feb 9, 2017 at 17:25
  • $\begingroup$ Yes, the notion of "increased entanglement" seems to be a meaningful intuitive metaphor for what is happening. The increased "entanglement" means there will be higher resistance towards shearing deformation. $\endgroup$
    – Pirx
    Feb 9, 2017 at 17:27
2
$\begingroup$

According to Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 1, Maxwell(1860) obtained the following expression for the viscosity of an ideal gas comprised of rigid spheres: $$\mu=\frac{2}{3\pi}\frac{\sqrt{\pi mkT}}{\pi d^2}$$where m is the mass of each sphere and d is its diameter. The derivation is presented in the book. Note that this expression is indeed independent of pressure. They also present a graph of reduced viscosity as a function of reduced temperature and reduced pressure.

Improve this answer
$\endgroup$
4
  • $\begingroup$ What is $k$? Boltzmann's constant? $\endgroup$
    – Frysen
    Feb 9, 2017 at 13:34
  • $\begingroup$ If that is the case, then my expression is consistent with this. Consider that the mass $m$ of each sphere equals the density $\rho$ divided by the number density $\eta$: $$ m = \frac{\rho}{\eta}. \quad (1) $$ The equation of state is: $$ p = \eta k T. \quad (2) $$ (1) and (2) gives that: $$ m k T = \frac{p \rho}{\eta^2}. $$ Thus: $$ \frac{\sqrt{m k T}}{d^2} = \frac{\sqrt{p \rho}}{\eta d^2}. $$ $\eta d^2$ should be proportional to the inverse of the mean free path (inverse of cross section area times number density), so: $$ \mu \propto \sqrt{\rho p}l. $$ $\endgroup$
    – Frysen
    Feb 9, 2017 at 14:19
  • $\begingroup$ But you have an important point in that, for the rigid sphere case and if the derivation in Transport Phenomena is correct, the viscosity for a given molecular composition depends only on the temperature. The reson that $\rho$ and $p$ enters my equation must be that they in turn depend on $m$ and $d$. $\endgroup$
    – Frysen
    Feb 9, 2017 at 14:25
  • $\begingroup$ Check out the derivation in Transport Phenomena. This is a book that has stood the test of time. It has been in use for almost 70 years now, with an updated version coming out in around 2002. I have used this book throughout my long career more than all the other books combined. $\endgroup$
    – Chet Miller
    Feb 9, 2017 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.