Suppose we have a thermally isolated container of a homogeneous gas at temperature $T$. If, for example, the container is filled with xenon at STP, then we know that the RMS speed of each xenon particle, $S_p$, is about 240 m/s.

Now we insert a solid fan blade on a shaft into the center of the container and begin to rotate it. As it rotates it alters the velocity of each particle with which it interacts via simple, inelastic collisions.

If the fan blade speed $S_b$ is small relative to the RMS speed of the particles (i.e., $S_b \ll S_p$), then it seems like it can't increase the average RMS particle speed because on average it is just as likely to collide with a particle with a velocity component opposite the motion of the blade (reducing that particle's velocity component by $S_b$) as it is to collide with a particle to which $S_b$ is additive. Also, particle collisions are almost as likely to occur with the receding face of the blade as with the advancing face. (But that almost seems like a hint....)

However, as the fan speed becomes large relative to the RMS speed of the particles – i.e., $S_b \gg S_p$:

  1. The blade is much more likely to increase any particle's RMS speed, because on average even if a particle is moving opposite the direction of the advancing blade, $|S_b - S_p| > |S_p|$.
  2. It is much less likely for a particle to collide with the receding face of the blade.

So point #1 seems to suggest that the fan will in fact increase RMS speed of the gas by something like $|S_b - S_p|$, but point #2 seems to suggest this won't happen for long, because eventually the average velocity of the particles will match the speed of the blade.

(If necessary, we can assume that the container is cylindrical or spherical, and the blade is in the limit a point solid rotating at a fixed radius from the center. Let us further assume that external energy is added to the blade as needed to maintain its speed at $S_b$ ... although we will note that if we run the experiment over time such energy decreases rapidly as the fan "sets the gas spinning.")

What really happens in this simple model? Does the fan increase the container's temperature by a fixed amount related to $S_b$? Does the conversion of energy to heat depend on the relative velocities of the blade and the average particle, and if so what is the nature of that relationship?

  • $\begingroup$ +1 and thanks for the headache......what about turbulence outside the sweep of the blades, can that be ignored in this experiment? Collisions between particles are allowed. I just wonder how "pure" you want to work this? $\endgroup$
    – user163104
    Jul 24, 2017 at 20:26
  • $\begingroup$ @Countto10 - I'm curious about the full "mechanical" model, so all inelastic collisions (with particles and container walls) and turbulence should be included (if they are factors). Explicitly excluded would be any radiation or energy conducted through the container walls or out of the shaft that would have to drive the fan. (Bonus points for answers that address the complications introduced by removing the "ideal gas" simplifications, and that address the same question for liquids.) $\endgroup$
    – feetwet
    Jul 24, 2017 at 21:15
  • $\begingroup$ Is the fan transferring energy to the gas to move it around. If so, the work done by the fan on the gas is increasing the internal energy of the gas (temperature). You don't need to consider non-ideal gas effects if the pressure within the gas is less than about 10 bars. $\endgroup$ Jul 24, 2017 at 22:27
  • $\begingroup$ @ChesterMiller: I don't think it's that easy. At least some significant amount of the fan's energy "sets the gas spinning" in a uniform direction. That is largely mechanically reversible (we can use the same blade as a "brake" to remove that spinning velocity component as mechanical energy from the container) and therefore (I think) means that component of the energy has not been converted into heat (i.e., it has not increased the gas's temperature). $\endgroup$
    – feetwet
    Jul 25, 2017 at 0:33
  • $\begingroup$ Your question seems to boil down to: how does mechanical energy input through fan blade get converted into internal energy of gas inside an adiabatic container? If there was no such conversion, the fan blade would rotate forever with no external power input being necessary (after being set into motion initially). Therefore I think what you are seeking is an explanation of viscous action of a fluid at the molecular level. Here's a link: physics.stackexchange.com/questions/129676/… $\endgroup$
    – Deep
    Jul 25, 2017 at 4:11

2 Answers 2


The answer to your first question in the comment "Can I increase the temperature of a gas by mechanically whacking the particles with a macroscopic solid?", the answer is "usually Yes", and in regard to the particular setup you have mentioned the answer is "definitely Yes". Comparison of fan blade speed ($S_b$) and rms speed of gas particles ($S_p$) is irrelevant to this particular question, by which I mean that so far as $S_b\neq 0$ it is possible to heat up the gas by stirring it. In fact in the particular flow setup you have suggested, it is inevitable that the gas is heated up. This is most easily understood by looking at the flow at the macroscopic level. Wherever there are velocity gradients in the flow, there the viscous action converts part or all of mechanical energy in the flow into its internal energy. In your setup, if nowhere else, there will velocity gradients at the wall of the container.

Therefore to answer your first question succintly, you can increase the temperature of a gas by mechanically whacking the particles with a macroscopic solid, provided that you whack them in such a way that you create regions of higher average motion and regions of lower average motion in the domain of the flow. Molecular collisions then guarantee dissipation of kinetic energy contained in the average motion. Exactly how this happens requires an explanation of viscosity at the molecular level and is explained in this post that I mentioned earlier. If you whack the gas particles in such a way that all of them have the same average motion (for e.g. solid body rotation, where the container wall does the "whacking") then of course there would be no viscous dissipation and consequently no rise in gas temperature.

Your second question in the comment " If so, how much [can the temperature be increased]?" is a bit vague, since temperature can be increased indefinitely so far as you keep injecting energy into the flow by means of the fan (of course several constraints, practical and theoretical, shall put a brake to this indefinite rise, but that is another question). A better question would be "How fast can the temperature rise?", and the answer to this question does depend on $S_b$ and $S_p$.

From the continuum viewpoint, the rate of dissipation at a point per unit mass of fluid is given by $\epsilon=\nu s_{ij}s_{ij}$, where $\nu$ is the kinematic viscosity of the fluid and $s_{ij}$ is the strain rate at the given point. The rate of dissipation over the entire flow is then given by $Q=\int_VdV\rho\epsilon$, where $V$ is volume of the fluid and $\rho$ is the fluid density. The kinematic viscosity $\nu\sim S_pl_{mean}$ in which $l_{mean}$ is the mean free path of gas molecules. An average magnitude of strain rate may be taken to be $S_b/L$ in which $L$ is the blade length (perhaps a better length scale may be derived from more careful considerations). Therefore $Q\sim \rho V S_p l_{mean}(S_b/L)^2$. You see that for a given fluid, small blade velocities results in small strain rates which results in small rates of dissipation which results in slow increase in temperature of the gas. But the rate of increase is never zero for non-zero $S_b$ no matter how small it is.


Putting a spinning blade into an ideal gas, must increase the total energy of the system.

No matter what the speed of the fan is, you are adding, even for a short while, energy to the system, which will increase the temperature of the gas.

I am ignoring the Maxwell velocity distribution, as it just complicates a problem that is really based around the mechanics of the particles colliding with the spinning blade.

then it seems like it can't increase the average RMS particle speed no matter how long it runs because on average it is just as likely to collide with a particle with a velocity component opposite the motion of the blade (reducing that particle's velocity component by Sb) as it is to collide with a particle to which Sb is additive.

For simplicity, imagine two gas particles with velocity vectors normal to the surface of the oncoming blade, but one is heading towards the blade, the other directly away.

The particle heading directly away, will receive a velocity boost from the moving blade.

The particle heading towards the blade will have 1 of three speeds relative to the blade.

  1. Its velocity exceeds that of the blade, so that it slows the blade slightly.

  2. It's velocity matches the blade, it which case, its velocity momentarily drops to zero, for an infinitesimal moment, then it acquires the velocity of the blade.

  3. It's velocity is less than that of the blade, so when it collides with the blade, it acquires extra velocity.

Behind the blade, particles moving away from it, will have no effect.

Particles moving towards the blade, will only serve to increase the blades velocity.

When you average the effect of all of the these possible situations, you have a net increase in velocity of the gas in the system, and an associated increase in energy and temperature.

  • $\begingroup$ "The fan does work, so it must add heat," seems almost like a restatement of a thermodynamic law, as does "if heat is added, it must be fully reflected in the temperature and therefore the RMS speed of the gas." But I think we're cheating without explaining or confirming the mechanism behind both of those statements. (NB: In this thought exercise I am allowing that energy is added as necessary to hold the fan's rotational velocity $S_b$ constant, although once we play this out over time we will notice that increasingly less energy is needed to do so....) $\endgroup$
    – feetwet
    Jul 25, 2017 at 0:02

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