In most situations in physics, the effect of kinetic friction is to reduce the macroscopic kinetic energy of a system and convert it into heat, thereby increasing its temperature. but in the case of subsonic Fanno flow, the opposite happens: temperature decreases and velocity increases. I know the mathematical theory behind this, but can someone explain how friction is playing its dissipative role over here? There is the familiar effect of entropy increase but I'm also looking for a answer to how reduction of temperature and increase in velocity can have the above effect which seems quite counter intuitive to me.


1 Answer 1


Friction still dissipates kinetic energy from the fluid. However, since we are assuming that the flow is adiabatic, heat is not transferred out of the system and is instead fed back into the fluid. To conserve total energy, this thermal energy can either increase the temperature and decrease velocity or decrease velocity and increase temperature. The choice is governed by the second law of thermodynamics: total entropy must increase.

As a thermodynamic state, entropy can written in terms of two other states. While it may be intuitive that a process that increases temperature increases the number of microstates (and therefore increases entropy), we also need to consider how the other states are affected.

For an ideal gas, the speed of sound $a$ is related to temperature by $$ a^2 = \gamma R T. $$ Using the definition of the Mach number $M = u/a$, we can get the following relationship between the flow velocity and temperature $$ \frac{\mathrm{d}T}{T} + \frac{\mathrm{d} M^2}{M^2} = \frac{\mathrm{d}u^2}{u^2}. $$

Since we are also assuming that the flow rate is constant, the continuity equation reduces to $\mathrm{d}(u \rho) = 0$. This assumption fixes the relationship between flow velocity and density $$ \frac{\mathrm{d}\rho}{\rho} + \frac{1}{2} \frac{\mathrm{d}u^2}{u^2} = 0. $$ So as the velocity increases, the density must decrease in order to maintain flow rate.

The change in entropy is then determined by the rate of temperature change and the rate of density change, both of behave differently at different Mach numbers. For subsonic flows, the combination that will result in increasing entropy is to drop temperature and density.

Let me know if this is still too hand-wavy, I can dig up some old notes and go over it step by step.

  • $\begingroup$ The KE can be divided into two parts: microscopic and macroscopic, corresponding to the temperature and velocity respectively.Since dissipating macroscopic KE and putting it back there isn't dissipation at all,I assume you are referring to micro energy. While in full agreement that "dissipating" micro KE into macro KE is exactly what is happening, I still can't seem to find the answer to how trading the disorderly micro temperature related energy to the much more orderly macro velocity related energy increases the entropy, as mentioned in the 2nd part of my question. $\endgroup$
    – alex
    Commented Dec 30, 2015 at 17:27
  • $\begingroup$ Also please don't hesitate to go into the maths if you feel it is required to provide an answer to my question. Iv'e edited my question to remove the "conceptual" $\endgroup$
    – alex
    Commented Dec 30, 2015 at 17:35
  • $\begingroup$ So pressure (i.e internal energy density) is converted into kinetic energy. Could you expand upon: "For subsonic flows, the combination that will result in increasing entropy is to drop temperature and density"? This doesn't appear quite clear. $\endgroup$
    – Mathews24
    Commented Oct 21, 2018 at 14:33

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.