Can Hamson-Linde cycle liquidize N2?

[jonhswon]

TL;DR Summary

To me it seems linda cycle can not liquidize nitrogen.
Please explain me, can it liquidize N2? Here is how I come to this result:

Hi !

To me it seems linda cycle can not liquidize nitrogen.

Please explain me, can it liquidize N2? Here is how I come to this result:

Imagine compressor take nitrogen 1bar 300k and gives 200bar hot, but cooled to 200bar 315k.

Next n2 is cooled in counterflow heat exchanges (hex) by cool n2 that wasnt liquidized in next step.
I am assuming that heat exchangers are not ideal, but say ~95% efficient, slightly worse than ideal.

Then n2 is passed from throttling valve becomes 1bar and colder, hopefully cold enough that some liquid is produced.

So now to some steady state calculation:

I start by assuming temperature at which n2 leaves hex is 170K, this is highest that can yield any liquid droplets at all.
https://upload.wikimedia.org/wikipedia/commons/9/9d/Throttling_in_Ts_diagram_01.jpg
Look at this T-s diagram of n2.

Expanding n2 from 200bar 170k to 1bar 78k, say maybe 0.01%liquid (this isnt anything at all) and 99.99% gas.

Now back to check if hex works as expected.

For hex, heat lost = heat gained.
Heat gained by cold n2 = Cp*dT = 1.05*(300-78) = 233
Heat lost by hot n2 = Cp*dT = 1.6*dT
(Cp values obatained from nist website, and vary depending on pressure and temperature)

233=1.6*dT
dT=145

gas exiting hex is 315-145=170K, correct! If this was otherwise then I guessed wrong, but lucky.

Final words:
So you see, even using 200bar compressor we are barely getting any liquid at all, close to none.
And increasing pressure doesn't make things better, maybe only tiny bit better. Check t-s diagram.
Do I have a mistake.
Or is linda-hampsons so impractical ?

edit: this can be moved to thermodynamics maybe. [Done.]

 

[Chestermiller]

Please provide a flow diagram.

 

[jonhswon]

Hi. Is this what you required?

 

[Chestermiller]

Yes. I'll redo the calculation my own way and see what I get.

 

[Chestermiller]

According to the TS diagram, you're going to have to cool it to, at most, 160K in order to get any kind of liquid forming. At this inlet temperature, about 15% of the n2 would condense, and 85 % would remain vapor. From the TS diagram, the change in enthalpy of heating this vapor from 78 K to 300 K would be about 245 kJ/kg. (Where did you get your heat capacities from?) But you would only have a mass flow rate to heat that is 85% of the mass flow rate being cooled.

Try the calculation again and see if you can get closure on the energy balances with the 160K temperature entering the flash.

 

[jonhswon]

Chestermiller,

You are correct. Any formation of liquid will reduce amount cold gas leaving the dewar, thus making incoming gas hotter, which will again yield less liquid.

The thing is, I have already done calculations for 15% liquid formation, and there is not enough cold gas to cool incoming 315k gas to 160K. So I was forced to assume a higher 170K leaving hex, which is stable but yields almost no liquid. And these calculations show that hampson-linda cycle impractical/ very uneconomical for producing liquid n2 or air, maybe only for demostrations purposes. Even considering oxygens higher 90K boiling point present in air, better yield yes, but still impractical. For simplicity I assumed air to be of only nitrogen.

Of course modern producers use better cycles, much better than linda. But for some reason I wasnt expecting it to be so bad.

Heat capacities I got from nist website and sharing here now below.

 

[Chestermiller]

Chestermiller,

You are correct. Any formation of liquid will reduce amount cold gas leaving the dewar, thus making incoming gas hotter, which will again yield less liquid.

The thing is, I have already done calculations for 15% liquid formation, and there is not enough cold gas to cool incoming 315k gas to 160K. So I was forced to assume a higher 170K leaving hex, which is stable but yields almost no liquid. And these calculations show that hampson-linda cycle impractical/ very uneconomical for producing liquid n2 or air, maybe only for demostrations purposes. Even considering oxygens higher 90K boiling point present in air, better yield yes, but still impractical. For simplicity I assumed air to be of only nitrogen.

Of course modern producers use better cycles, much better than linda. But for some reason I wasnt expecting it to be so bad.

Heat capacities I got from nist website and sharing here now below.

Well, it sounds like you're right on top of this. So, are we done here?

 

[jonhswon]

Yes. Thanks.