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

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.

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## 11 Replies

kelseymh (author)2010-03-06

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?

R.Blakely (author)2010-03-08

A supersonic nozzle should be placed at the compressed air input side of the Vortex tube.  A plain drilled hole does not permit supersonic flow into the vortex.  A vortex separates molecules by temperature, which could violate the Second Law in a very efficient Vortex tube.  Pressure drop  occurs, for example, in a free expansion, but temperature does not drop in such an expansion.  In a Vortex tube, both temperature and pressure drop for the air exiting the cold outlet port.  Air can be extraordinarily cold at the cold outlet thus violating the Second Law.

kelseymh (author)2010-03-08

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.

R.Blakely (author)2010-03-09

A vortex of air conducts heat outward by molecular collisions.  For example, using a gravity analog, a molecule falling hits molecules below with more violence than normal, and so heat is conducted downward. A gas in a gravity field thus acquires a thermal gradient naturally, which violates the Second Law! To understand this concept we can imagine a box in a gravity field. If the box contains one molecule of air then the molecule will hit the bottom of the box with more violence than when it hits the top of the box. Thus, the box will become hotter at the bottom than at the top.

kelseymh (author)2010-03-09

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?

R.Blakely (author)2010-03-10

Your response is off the topic. A single molecule in a box demonstrates that the Second Law is not always true. In a gravity field the molecule must hit the bottom of the box with more violence than it hits the the top of the box. Violence equates with temperature.

kelseymh (author)2010-03-10

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.

orksecurity (author)2010-03-06

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.)

R.Blakely (author)2010-03-09

A perfect vortex, such as can be created in a rotor filled with air (rotor spinning in a vacuum) can conduct heat outwards. A turblent vortex in a Vortex Tube can also conduct heat outwards, but the vortex can conduct at a higher rate. The heat conduction outwards in a vortex violates the Second Law! This is because the heat is conducted from cold air towards hot air.
An example of Second Law violation occurs in the atmosphere; heat conducts downward continuously until a tornado occurs due to the generated thermal inversion, at least in theory because air conducts heat poorly unless the air is turbulent.

orksecurity (author)2010-03-10

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.

kelseymh (author)2010-03-09

It's pretty clear, now, that this fellow doesn't understand what you mean by "closed system."  Never mind.