Imagine a steady state, one-dimensional, compressible flow in a horizontal pipe
ID: 1373988 • Letter: I
Question
Imagine a steady state, one-dimensional, compressible flow in a horizontal pipe of constant cross sectional area. This flow can be isothermal, adiabatic (Fanno), or diabatic (Rayleigh). As an example, the relevant macroscopic energy balance for Fanno flow is:
?h=??(KE)
In other words, Fanno flow converts enthalpy into kinetic energy. Below are Fanno lines for a real gas (steam) for different flow rates at the same pipe ID and inlet conditions. enter image description here
The lines clearly indicate the "efficiency" with which enthalpy is converted to kinetic energy. At low velocity (left upper portion of lines), almost no enthalpy is converted while the entropy generation is large (the line is nearly horizontal). Near Mach 1 (the rightmost maximum entropy point for each curve) the enthalpy conversion is high and entropy generation is nearly zero (the line is nearly vertical). This happens whether you are originally subsonic (the upper branch) or supersonic (the lower branch). Another way of saying no entropy generation occurs is that the viscous losses (friction) are zero. This would imply that the molecules no longer impart (as much?) force upon each other near the choke point. This seems absurd.
In isothermal flow, you also find that as you near the choke point (interestingly, below Mach 1!), the flow approaches frictionless.
As I understand it, these examples are physically realizable flow regimes (if obviously very short in length scale). Why is compressible flow so efficient near the choke point? Is something at the molecular scale really going on or is it simply a poor mathematical model at these velocities?
UPDATE: Obviously, in the supersonic regime (lower branch), the reverse process takes place, i.e. kinetic energy is converted into enthalpy. Still, my question holds since this conversion is nearly perfect as you approach Mach 1.
Explanation / Answer
Instead of thinking about the efficiency of the acceleration, think about the flow itself; the flow is in its maximum entropy state at the throat. Any additional dissipative force would shift the choking location, but it would still occur at Mach 1.
so why is Mach 1 the maximum entropy state. I think it's that the thermal boundary layer is fully developed at the throat in adiabatic flow; that is, while there is still momentum transfer, there is no heat transfer (thermal equilibrium == maximum entropy)
At the microscopic level, entropy is defined by Boltzmann's equation S=k?ln?. Boltzmann's constant (k) is, well, constant; for S to increase, ? must increase. I've only worked with microstates for stationary gas, so I don't know offhand why the number of microstates consistent with the macrostate is maximum at Mach 1