# Vortex tube should have a supersonic nozzle for efficiency; a simple drilled hole will not allow supersonic flow, right?

Scientific American had an article on supersonic nozzles in Amateur Scientist.  The nozzle needs to first narrow and then widen, so that maximum velocity is reached by the expanding air.
The Hilsch tube seems to be very primitive because of details like the primitive nozzle.  Perhaps, the Hilsch tube was designed to have too much friction.  Perhaps, the tube was built to be sure it was inefficient and not very useful.  The tube can only violate the Second Law if it is built with a minimum of friction losses.

sort by: active | newest | oldest
Mar 6, 2010. 10:13 PMkelseymh says:
Why do you need supersonic exhaust?  What exactly is preventing you from putting a supersonic nozzle on the "hot" end of the vortex tube?

And why do you think there's any violation of the normal laws of physics here?
Mar 8, 2010. 8:45 PMkelseymh says:
Ah, so you want to feed supersonic flow into the vortex tube.  Thanks for clarifying that!

If you're going to do a thermodynamic analysis, you'd better do it right before you try to claim you're violating well-known, and well-understood, physical laws.  The cold outlet is not a closed, nor even the entire, system.  The airflow is segregated by temperature.  You need to take account of both the cold and hot outputs, and calculate the total change in entropy at both ends.

You also need to take accout of how air is being introduced into the system.  If you are getting hgh flow rates out of the system, then you'd better have something pushing (or sucking) air into the system at the same rate.  Otherwise you just cavitate and it stops.

So now you have a pump or some such thing, doing work on the air coming in.  Now you'd better be taking account of that input work (energy) in your calculations before you try to claim you're violating physical law.
Mar 9, 2010. 11:11 PMkelseymh says:
Except that a volume of gas has a density which varies inversely with pressure, leading to a net buoyant force upward, and a thermal gradient in opposition to what you describe.

Regardless, you are still failing to consider the full system.  The gravitational field itself, is doing work on the falling object.  You seem so bent on discovering "violations of physical law" that you ignore the fact that you're dealing an open system.  Such systems can very well demonstrate either increases or decreases in entropy, depending on the external inputs, with no violation of any physical law.

If you don't understand the simple and obvious difference between open and closed systems in thermodynamics, then how can I take any of your claims seriously?
Mar 10, 2010. 9:28 PMkelseymh says:
No, it doesn't.  Temperature measures the variance of a distribution of velocities of a system.  A single particle does not (cannot) have a well-defined temperature.

Since you don't understand this definition, nor do you understand the difference between an open vs. closed systems as your repeated use of gravitational examples shows, my confidence that you understand thermodynamics at all is quite low.
Mar 6, 2010. 9:07 PMorksecurity says:
It can't violate the Second Law if it *is* built with minimum friction losses, either.

(Localized reversals of entropy are possible if driven with additional energy from outside. Closed systems don't.)
Mar 9, 2010. 11:12 PMkelseymh says:
It's pretty clear, now, that this fellow doesn't understand what you mean by "closed system."  Never mind.
Mar 10, 2010. 8:41 PMorksecurity says:
Sorry, no. Energy is being pumped into the system to drive the heat transfer. (An air conditioner or refrigerator can also move heat against the natural gradient, but requires an external energy source.). No violation of Second Law.