Can Cold Air Be 100% Humid in a Vortex Tube?

[pranj5]

I hope most here are familiar with vortex tube i.e. Ranque-Hilsch tube. I just want to guess if humid air used. We can get some idea about functionality of vortex tube from this http://www.airtx.com/how-does-a-vortex-tube-work.htm. Now take the example for the lowest pressure level mentioned in the page. As per this page, when the cold: hot ratio is 4:!, then the temperature of the hot flow is 107°C above the inlet temperature. That means if the inlet temperature is at around 30°C, then the temperature at the hotter end is 137°C and the colder end will be at 2.5°C. Suppose 100% humid air compressed isothermally to 20 psiG (the pressure level mentioned) and then used in the vortex tube.
Now, at 30°C, the moisture content of 100% humid air is 30.32 g/m³ while at 2.5°C, it's 5.74 g/m³. That simple means that the moisture content of the hot flow is 24.58 g/m³. All the information given about moisture content of humid air has been collected from this page.
What made above is just an assumption because I am not sure whether the cold air can be 100% humid or not. It's highly possible that all the moisture will be concentrated at the hotter end as water vapour molecules are lighter than both Nitrogen and Oxygen. I am requesting those who have done some work on vortex tube can show me light in this regards,

 

[RonL]

I hope to see some input from others, as the tube function has always interested me. Long ago I benefited from the use of a unit to provide cold air inside of a sandblast hood, when building a metal push boat, years later I found the explanation of their operation in an air conditioning service manual, the biggest surprise was the mention of a 5,000,000 rpm spin speed of the air inside?
We had very good dryers on our Sulair Compressor, so moisture did not come into play, I suspect water inside a tube would present a few difficulties and because of the motion of air in the tube, the calculator might not be usable.

My impression from the drawings in both links, is that the tube is little understood even by those manufacturing them...maybe we are close to a time when some computer software program can give more cross section numbers?

Again I hope to see some input here...

 

[256bits]

That simple means that the moisture content of the hot flow is 24.58 g/m³.

does it simply mean that?
Why did the moisture content drop, and where did the difference go?

 

[Baluncore]

Compressed air is dried before use. If wet compressed air is used, injection nozzles may ice-up due to adiabatic cooling.

What made above is just an assumption because I am not sure whether the cold air can be 100% humid or not.

If the cold fraction reaches 100% RH, denser liquid condensate will move out to dissolve in the hot fraction, while water vapour, being lighter than air will continue to move to the centre. If sufficient water was present in the compressed air, then I would expect the cold fraction to reach 100% RH before exit from the tube.

 

[pranj5]

We had very good dryers on our Sulair Compressor, so moisture did not come into play, I suspect water inside a tube would present a few difficulties and because of the motion of air in the tube, the calculator might not be usable.

The air already contains the moisture inside even before being compressed.

Why did the moisture content drop, and where did the difference go?

I want to mean that the difference will be concentrated at the hotter end. As it's hotter, therefore it can contain far more amount of vapour than the amount of mentioned. And at that point I am not sure whether all the vapour will be concentrated at the hotter end or there will be some left in the cooler flow.

Compressed air is dried before use. If wet compressed air is used, injection nozzles may ice-up due to adiabatic cooling.

Not possible! This will happen when this compressed air will either pass through a Joule-Thompson nozzle or rotate a turbine.

If the cold fraction reaches 100% RH, denser liquid condensate will move out to dissolve in the hot fraction, while water vapour, being lighter than air will continue to move to the centre.

My knowledge of physics and common sense tells the opposite. As far as I know (I am sure many here will agree with me) that lighter particles will go to the periphery and heavier particles concentrate at the centre if the rotation is happening in uniform angular momentum.

 

[pranj5]

Actually, I am thinking about one point. Vortex tubes has been considered to be less efficient than compressors based coolers. But that can be altered. Nobody so far has given attention to a fact that both the hot and cold flow coming out of a vortex tube has velocities. If the cold flow has been used to rotate a turbo expander, than more of its internal energy will be converted into power and the flow will become colder.

 

[RonL]

Actually, I am thinking about one point. Vortex tubes has been considered to be less efficient than compressors based coolers. But that can be altered. Nobody so far has given attention to a fact that both the hot and cold flow coming out of a vortex tube has velocities. If the cold flow has been used to rotate a turbo expander, than more of its internal energy will be converted into power and the flow will become colder.

It's been too many years ago, but I remember a pleasant stream of cool air into my sandblast hood and a muffled, shielded hot air exhaust that had little velocity, not anything that would make any useful power. About 30% of the air's energy is lost to internal friction, while the air volume is split into roughly 75% cold and 25% hot discharges. (that split is based on a 170 degree hot and a 40 degree cold discharge)
I am curious about the comments of where particles would go would a sand particle and water do the same thing? I have my doubts .
Extending the vortex tube upward in scale (size and volume flow) I believe would lead you to a Tesla Turbine or the real principles of jet engine technology in order to convert airflow and heat exchange into mechanical working power.

 

[pranj5]

What's the velocity of the cold air flow with the mentioned i.e. 75:25 ratio?

 

[pranj5]

About 30% of the air's energy is lost to internal friction, while the air volume is split into roughly 75% cold and 25% hot discharges. (that split is based on a 170 degree hot and a 40 degree cold discharge)

Do you mean that with this level of separation, the temperature of the hot exhaust is at 170°C above the inlet temperature and the colder is -40°C below the inlet?

 

[RonL]

Do you mean that with this level of separation, the temperature of the hot exhaust is at 170°C above the inlet temperature and the colder is -40°C below the inlet?

Sorry, I should have made it clear the numbers are Fahrenheit temperatures. (40 F cold and 170 F hot) I don't know how to get a degree symbol to appear in my typing.

As for velocities at discharge points, I never have had thoughts or reason to quantify them.

From the wiki link in your first post, the vortex tube is said to have been discovered in 1933...I was using one in about 1975 and I think their application as cooling devices were just coming into practice.
You might find it interesting that Tesla patented his turbine in 1916 and some of his talking points in his application are about the thermodynamic conversion to energy that takes place as air flows through his design.

 

[pranj5]

Well! In that case kindly convert those into °C and tell me the hot and cold outlet temperature. By the way, what's the pressure of the inlet?

 

[Kevin Halagan]

I am currently using a exair vortex system and am having problems with the condensation on the generator. If anyone knows how to get the right fraction so I can get colder air that would be great. As far as what you are saying I have been working on keeping the pressure down and decreasing diameter and length to get more laminar flow... I doubt that helps I haven't seen a lot of vortex discussions and am new to using them.

 

[pranj5]

At which part the condensation occurs? What's the input pressure and how much is the exit velocity of both hot and cold?

 

[Kevin Halagan]

Condensation occurs around 80 PSI. We use an input pressure of 40 psi because we cannot exceed a certain decibel level. That gives us an cold air exit of 28 ft/s and a hot air exit of 33 ft/s.There is three junctions and about a foot of .25" diameter tubing with the cold air exit. I did a rough calculation and we are having a heat loss of about 500 joules due to this tubing and junctions.

 

[pranj5]

In that case, I want to know what's the cold and hot exit temperature in °C.

 

[Bruce Zerr]

Now, we're getting somewhere! I finally found a refrigeration diagram on the Web, that I could cut-and-paste over to Physics Forums.
That's a start:

Greetings pranj5,

Are you still interested in exploring your question on running a Vortex Tube with moist air?
If so, I can probably help some. I've never done a phase-change, vortex tube calculation before, so
it should be interesting. We will both learn something.

Meanwhile, I can show and explain how these devices work. I'll walk you thru the Theory, the math,
the calculations, too.

Can you tell me something about your background? Are you and "Engineer"? What kind? How much
do you know about "Thermodynamics", and the "refrigeration cycle". Are you familiar with the "Enthalpy",
and "Entropy"?

I'm not looking for much detail here - just hoping that you're somewhat familiar with these subjects.
- - - - - - -
Does anyone else, on this forum, also have question(s)?

-Bruce Zerr
Senior Process Design Engineer - Gas Centrifuges

 

[RonL]

@Bruce Zerr , welcome to PF
Short on time at the moment, but a little later I might have a couple of questions for you.
Might be that a new thread should be started, not sure of pranj5's status ?

Later
RonL

 

[Bruce Zerr]

Greetings RonL,

I have some good news/bad news. The Head Member of this site sent me an
e-mail - stating that this thread is to be shut down now. One of the reasons given, was
that this site only allows material coming from either textbooks on the subject being discussed, or
refereed material in published journals. The good news is that we can still 'talk' and discuss
this material between ourselves - if you are interested. Would love to show you what I've got - the Engineering
Theory and the math.

So - here's my email address
<< e-Mail address deleted by Mentors >>

 

[berkeman]

@Bruce Zerr -- Please do not post your e-mail address in the open forums. It attracts spam bots and other bad things. If someone wants to contact you by starting a Private Conversation with you, that it up to them. This thread is closed.