# Some points about a vortex tube

• T C
In summary: The whole calculation above is based on one assumption that the process is both isentropic and isenthalpic. I am sharing my thoughts here just to understand whether I am right or wrong. The temperatures are specified at a pressure of 1 atm after exit from the hot air restriction, or from the cold air vent. Pressure changes will vary through the process, so temperatures will also vary through the process. The extreme temperatures you calculate will not necessarily occur anywhere internally. The biggest temperature step will probably be across the hot air restriction. It is interesting that the temperature inside a cow, 100°F, or a bucket of water with freezing salt, 0°F, are still used in engineering.

#### T C

in this link here you can see the hot exhaust temperature and cold exhaust temperature at different ratio (hot to cold) with input air/gas at different pressure level. I want to discuss the top here i.e. the data where the input air/gas is at 20 PSIG pressure. You can see that with a 80:20 ratio (cold to hot) the temperature of the hot exhaust is 107°F above the input temperature.

But there is one point. The input air/gas is released from 20 PSIG pressure to atmospheric i.e. 1 barA pressure and that means the temperature of the air flow inside the vortex tube is at around -39°C considering the process to be adiabatic in nature and the temperature of the input air/gas to be at 300°K or 27°C.

Now, the hot exhaust temperature here is to be above 107°F considering the input temperature to be at 27°C but in reality the rise in temperature would be from -39°C, not from 27°C. Therefore, the real rise in temperature should be (59.44 (107°F)+ 27+ 39)°C i.e. 125.44°C.

The whole calculation above is based on one assumption that the process is both isentropic and isenthalpic.
I am sharing my thoughts here just to understand whether I am right or wrong.

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The temperatures are specified at a pressure of 1 atm after exit from the hot air restriction, or from the cold air vent. Pressure changes will vary through the process, so temperatures will also vary through the process. The extreme temperatures you calculate will not necessarily occur anywhere internally. The biggest temperature step will probably be across the hot air restriction.

It is interesting that the temperature inside a cow, 100°F, or a bucket of water with freezing salt, 0°F, are still used in engineering.

russ_watters
Baluncore said:
The extreme temperatures you calculate will not necessarily occur anywhere internally.
When compressed air/gas is released, then the velocity of that comes at the cost of internal enthalpy i.e. the pressure and temperature will fall and that decrease in enthalpy will be converted into speed. Do you want to mean that this phenomenon doesn't occur inside a vortex tube?

T C said:
Do you want to mean that this phenomenon doesn't occur inside a vortex tube?
No, I don't see how that could work.

Energy exchange happens over the path taken by the flow. The pressure and temperature are not changed instantaneously, so the hypothetical steps and extreme peaks you predict are actually counter-flow ramps.

russ_watters
When air/gas at 20 PSIG pressure and at 300°K or 27°C.temperature when releases at 1 barA pressure, then the temperature will fall to -39°C if the process is adiabatic. And in this case, as it occurs quickly then there is a high chance that the process is almost adiabatic. This change in temperature and pressure means that the enthalpy has been converted into energy i.e. velocity in this case. This high speed air/gas when enters the vortex tube, the linear motion is converted into angular velocity and that gives rise to all the effects inside vortex tube.
Considering the rotation to be like a solid body, the low speed molecules concentrate at the centre and high speed molecules concentrate at the periphery. That's why the more we move away from the centre, enthalpy increases and that means increase in pressure and temperature. That's why the flow coming from the centre is cold and from the periphery is hot.
Kindly explain why this doesn't work.

T C said:
in this link here you can see the hot exhaust temperature and cold exhaust temperature at different ratio (hot to cold) with input air/gas at different pressure level. I want to discuss the top here i.e. the data where the input air/gas is at 20 PSIG pressure. You can see that with a 80:20 ratio (cold to hot) the temperature of the hot exhaust is 107°F above the input temperature.

But there is one point. The input air/gas is released from 20 PSIG pressure to atmospheric i.e. 1 barA pressure and that means the temperature of the air flow inside the vortex tube is at around -39°C
How did you calculate that number (-39C)? As @Baluncore said, the temperature inside the tube isn't consistent and I'm not sure it is even useful here...
Now, the hot exhaust temperature here is to be above 107°F considering the input temperature to be at 27°C but in reality the rise in temperature would be from -39°C, not from 27°C. Therefore, the real rise in temperature should be (59.44 (107°F)+ 27+ 39)°C i.e. 125.44°C.
The choice of reference temperature for the temperature rise is arbitrary and not very meaningful - it is chosen for usefulness. And I'd say that comparing the outlet temperatures to the inlet temperature is very useful. So why would they want to use a different reference temperature?

russ_watters said:
How did you calculate that number (-39C)? As @Baluncore said, the temperature inside the tube isn't consistent and I'm not sure it is even useful here...
Very easy! Simply by using the P,T related formula for adiabatic process.
russ_watters said:
The choice of reference temperature for the temperature rise is arbitrary and not very meaningful - it is chosen for usefulness. And I'd say that comparing the outlet temperatures to the inlet temperature is very useful. So why would they want to use a different reference temperature?
Why arbitrary? The temperature inside is -39°C while the temperature of the hot exhaust is 59.44°C (107°F) above the input temperature that is 27°C. It's nothing but basic addition. And regarding you second point, there is also a pressure in addition to the temperature of the input air/gas. When the air/gas is released, the temperature will fall.

T C said:
Why arbitrary?
Arbitrary: based on random choice or personal whim, rather than any reason or system.

The reference temperature can be literally whatever temperature you want, for whatever reason you want. So why bother with the reference temperature you chose? What usefulness does it have over the one they use?

russ_watters said:
Arbitrary: based on random choice or personal whim, rather than any reason or system.
Is there any problem regarding that part? What kind of system do you suggest to start for?
russ_watters said:
The reference temperature can be literally whatever temperature you want, for whatever reason you want. So why bother with the reference temperature you chose? What usefulness does it have over the one they use?
In the page, the reference temperature given isn't much away from the temperature that I have chosen. The chosen temperature is 70°F. How far it is from the temperature that I have chosen?

T C said:
In the page, the reference temperature given isn't much away from the temperature that I have chosen. The chosen temperature is 70°F. How far it is from the temperature that I have chosen?
That's not an answer to my question.

This thread needs to get a point, rapidly, or it will be closed.

russ_watters said:
That's not an answer to my question.
I don't think that you are looking for an answer. You are just looking for a "reason" to close this thread as per your habit. That's what I am fearing from the very moment when you have entered this discussion.

T C said:
I don't think that you are looking for an answer. You are just looking for a "reason" to close this thread as per your habit. That's what I am fearing from the very moment when you have entered this discussion.
And you've given me one by not answering. But please feel free to PM me with an answer and explanation of the point of this thread and I will consider re-opening it.