Saturated steam inside a vortex tube

[pranj5]

We all are more or less familiar with the performance of vortex tubes regarding how they can make hot and cold flow by pressurised and/or air/gas at high speed. What I am wondering is the performance of saturated steam inside vortex tube.

As for example, suppose we are using saturated steam at 100°C temperature and at 1 barA pressure in a vortex tube. The speed has been created by a high speed blower instead of compressing the saturated steam to get the forced vortex. Now, if the temperature of the colder flow will be at 80°C, then how much steam will be condensed?

For vortex tubes, the density and pressure of both the input and the output flows remains the same. That means, the flow coming out of the colder section at 80°C temperature and at the same pressure i.e. 1 barA; that will mean that around 54% of steam in the colder stream has lost their LHV. Just imagine such huge amount of enthalpy has been transferred to the hot flow and where its temperature will rise if all the enthalpy lost from the cold flow will be added to the hot flow.

That's why I want to know the behaviour of saturated steam inside a vortex tube, especially about the enthalpy distribution between the hot and the cold flow. I have searched net but haven't found even a paper on this matter and it seems that this subject need research and that may lead to some good heat pump mechanism in future.

 

[RonL]

We all are more or less familiar with the performance of vortex tubes regarding how they can make hot and cold flow by pressurised and/or air/gas at high speed. What I am wondering is the performance of saturated steam inside vortex tube.

As for example, suppose we are using saturated steam at 100°C temperature and at 1 barA pressure in a vortex tube. The speed has been created by a high speed blower instead of compressing the saturated steam to get the forced vortex. Now, if the temperature of the colder flow will be at 80°C, then how much steam will be condensed?

For vortex tubes, the density and pressure of both the input and the output flows remains the same. That means, the flow coming out of the colder section at 80°C temperature and at the same pressure i.e. 1 barA; that will mean that around 54% of steam in the colder stream has lost their LHV. Just imagine such huge amount of enthalpy has been transferred to the hot flow and where its temperature will rise if all the enthalpy lost from the cold flow will be added to the hot flow.

That's why I want to know the behaviour of saturated steam inside a vortex tube, especially about the enthalpy distribution between the hot and the cold flow. I have searched net but haven't found even a paper on this matter and it seems that this subject need research and that may lead to some good heat pump mechanism in future.

I know there are some here that are knowledgeable enough to answer your questions, but my only experience with a vortex tube had a very good moisture extraction device to eliminate as much water as possible, in order to prevent ice inside the unit. I will be surprised if any information is going to turn up, that involves water in a vortex tube ?
I'm interested in your question because I have a (I think) similar thought of a process that involves, a higher temperature change of state, water to steam cycle.

 

[pranj5]

I specifically want to know how saturated steam at high speed will perform inside vortex tube along with specific details like the temperatures of both the outputs for a specific input temperature. I don't know whether anywhere such experiment has been done or not because I have searched net and haven't been able to found out any data in this regards.
As far as I know, vortex tubes are used only with air and it's a question how saturated steam will perform inside vortex tube.

 

[russ_watters]

As for example, suppose we are using saturated steam at 100°C temperature and at 1 barA pressure in a vortex tube. The speed has been created by a high speed blower instead of compressing the saturated steam to get the forced vortex. Now, if the temperature of the colder flow will be at 80°C, then how much steam will be condensed?

How much air and how much steam? At what temperature and humidity is the air? Does it lose any additional heat through the casing of the tube? As always, your problem needs to be more specifically defined in order to be answerable.

For vortex tubes, the density and pressure of both the input and the output flows remains the same.

If some is condensing, the density is not the same.

That means, the flow coming out of the colder section at 80°C temperature and at the same pressure i.e. 1 barA; that will mean that around 54% of steam in the colder stream has...

Where did you get 80C for the output temperature?

...lost their LHV.

LHV? Lower Heating Value? Are we burning something here too?

Just imagine such huge amount of enthalpy has been transferred to the hot flow and where its temperature will rise if all the enthalpy lost from the cold flow will be added to the hot flow.

What's a "cold flow"? What's a "hot flow"?

... that may lead to some good heat pump mechanism in future.

Heat pump? What does any of this have to do with a heat pump? Are you using that term the way it is normally defined?

 

[pranj5]

How much air and how much steam? At what temperature and humidity is the air? Does it lose any additional heat through the casing of the tube? As always, your problem needs to be more specifically defined in order to be answerable.

The amount isn't a factor for vortex tube. For such case, we can consider that the loss through casing can be negligible. My question is very much clear. I just want to know if saturated steam at high speed is used in the vortex tube, then what would be the possible temperature at the hot flow and at the cold flow. Kindly have some basic knowledge about how vortex tube works.

If some is condensing, the density is not the same.

Here you can use the term being liquefied. The steam looses it Latent Heat of Vaporisation and become water.

Where did you get 80C for the output temperature?

I have said "suppose" in this case just to show that huge amount of Latent Heat has been released by such small drop in temperature. With air as working fluid, 20C temperature drop is very common and easy to achieve.

LHV? Lower Heating Value? Are we burning something here too?

LHV means Latent Heat of Vaporisation.

What's a "cold flow"? What's a "hot flow"?

Any vortex produces two outflows. One hot at higher temperature than the input and one cold lower than the input.

Heat pump? What does any of this have to do with a heat pump? Are you using that term the way it is normally defined?

Yes. The gross enthalpy of the input fluid has been divided unequally among the two flows and like a refrigerator, a vortex tube too acts like a heat pump.

 

[russ_watters]

The amount isn't a factor for vortex tube.

It must be.

Yes. The gross enthalpy of the input fluid has been divided unequally among the two flows and like a refrigerator, a vortex tube too acts like a heat pump.

Please provide a reference demonstrating that what you describe is true, both about the way a vortex tube works and that it constitutes a "heat pump".

According to the wiki on vortex tubes, they use compressed air, harnessing the expansion of the air in one direction to create a cold stream. If you intend to use atmospheric pressure air and steam, the vortex tube won't do anything useful.

 

[pranj5]

It must be.

For a vortex, it isn't a factor that how much fluid flows through it. The main factor is the speed with which the fluid will enter the tube. Because on that factor, it will depend what would be the temperatures at the hot and cold flow coming out of the vortex tube.

Please provide a reference demonstrating that what you describe is true, both about the way a vortex tube works and that it constitutes a "heat pump".

The term COP is often used for vortex tubes. Just google with vortex and you will see the results. The term COP is only associated with heat pumps. You can also see that how performance of vortex tube has been compared with compressor based refrigeration.

 

[russ_watters]

For a vortex, it isn't a factor that how much fluid flows through it. The main factor is the speed with which the fluid will enter the tube.

Those two statements contradict each other. The flow rate is the area times the speed. In addition, if you are mixing two streams, then the properties of each will determine the properties of the mixture.

I'll repeat:
According to the wiki on vortex tubes, they use compressed air, harnessing the expansion of the air in one direction to create a cold stream. If you intend to use atmospheric pressure air and steam, the vortex tube won't do anything useful.

 

[pranj5]

Kindly try to go deep instead of looking at the superficial information. The compressed gas has been released and it's the speed that creates the forced vortex and that leads to two separate flows. In my very first post, I have clearly mentioned that the steam is at 1 barA pressure but is being blown at high speed by a blower and certainly we can expect the same effect as compressed steam. In fact, blowers are better because that doesn't raise the temperature of the input.

 

[RonL]

I'm pretty sure that pressure is what drives velocity

 

[pranj5]

I'm pretty sure that pressure is what drives velocity

My commons sense tells me that velocity can be achieved by a blower too!

 

[russ_watters]

The compressed gas has been released and it's the speed that creates the forced vortex and that leads to two separate flows. In my very first post, I have clearly mentioned that the steam is at 1 barA pressure but is being blown at high speed by a blower...

Vortex tubes require pressurized inlet (typically 100psi), low or no pressure outlet on the cold side (up to 5psi) and no back-pressure on the hot side in order to generate the internal velocity they have:
http://www.spraybest.nl/datasheet/vortex-buizen/files/assets/common/downloads/publication.pdf

What you are describing is not how vortex tubes work and therefore what you want to do with them is not possible. Thread closed.