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Nitrogen Joule Thomson Expansion Temperature Calculation

nitrogen joule thomson expansion isenthalpic temperature expand

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#1 Daniel89

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Posted 15 April 2016 - 03:59 PM

Hi, I am to calculate the temperature of a pure nitrogen gas after it has expanded;

 

The initial values is as follows: 460 Bar(g) and 90°C , it will go through a choke valve and expand to atmospheric pressure, this I believe will be an isenthalpic, non-reversible adiabatic process, correct? 

 

Can anyone help me to calculate the joule thomson coefficent for this example, in Kelvin drop per bar ? I can't get it right for some reason,I must be misunderstanding at some point,  would be nice to see an example calculated by hand or explained step by step.

 

I do hope for a fast reply, thx a lot in advance :)

 

Best Regards,

 

Daniel


Edited by Daniel89, 16 April 2016 - 12:15 AM.


#2 MrShorty

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Posted 15 April 2016 - 04:30 PM

I'm sure we can help you, but we should probably have some idea of where you are starting and where you are getting stuck at. Along those lines, I would ask what equation or expression are you using for the JT coefficient? Are you using the same expression as Wikipedia uses (https://en.wikipedia....29_coefficient ) or some other expression? What values are using for each of the quantities in your chosen expression for the JT coefficient?



#3 Daniel89

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Posted 15 April 2016 - 05:11 PM

Thx for a fast reply, yes I am using the formula from wikipedia:

 

The rate of change of temperature b9ece18c950afbfa6b0fdbfa4ff731d3.png with respect to pressure 44c29edb103a2872f519ad0c9a0fdaaa.png in a Joule–Thomson process (that is, at constant enthalpy c1d9f50f86825a1a2302ec2449c17196.png) is the Joule–Thomson (Kelvin) coefficient 7398e4caac7770b4ece8d0fce37ce120.png. This coefficient can be expressed in terms of the gas's volume 5206560a306a2e085a437fd258eb57ce.png, its heat capacity at constant pressure fa5fa493f3f31a5fdb9ed5886631cc06.png, and its coefficient of thermal expansion bccfc7022dfb945174d9bcebad2297bb.png as

169bce1c258be66611f0ae1b8f45a0f0.png

For Cp= I am using 1.04 (kJ/(kg K))

 

Coefficient of thermal expansion is: -197.75392 [ 10-3 (1 / K) ]

 

But i do not understand where the V comes from, I can't understand where the volume plays out its role when it comes to defining the temperature drop per pressure unit? can you clarify and push me in the right direction? And as mentioned earlier I would really appreciate a step by step explanation. thank you very much :)



#4 Art Montemayor

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Posted 15 April 2016 - 07:05 PM

Why don't you just resort to using the NIST free database for nitrogen, where you can find the resulting temperature almost immediately?

 

Or are you assigned the task of calculating the J/T expansion temperature by thermodynamic equations?



#5 Daniel89

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Posted 15 April 2016 - 11:16 PM

Hi, yes I am assigned the task of calculating it by thermodynamic equations, but it would off coarse be helpful to know the correct answer in advance, as I could flip the equation , put in the value and maybe understand where I did put in tye wrong values? Can you send a link to the page you are referring to?

#6 breizh

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Posted 16 April 2016 - 12:19 AM

hi Daniel,

 

http://webbook.nist....hemistry/fluid/

 

This is the link Art is wiliing to offer to you.

 

Take a look you may find pointers 

 

http://chemistry.tcd...1_annotated.pdf

 

Good luck

 

Breizh


Edited by breizh, 16 April 2016 - 12:32 AM.


#7 Daniel89

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Posted 16 April 2016 - 06:18 AM

Thank you very much Breizh, I must say I am struggling with this, would be very helpful if someone could calculate the above mentioned example, according to the posted formula, so I can see how its done....



#8 breizh

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Posted 16 April 2016 - 07:19 PM

Daniel ,

I've attached good material to support your work .

 

Can you confirm the data 460 barg and 90 C (supercritical zone if I refer to NIST) ?

 

Anyway I'm adding document and excel sheet I prepared a couple of years ago using PR EOS .

 

Hope this helps ,

 

Good luck

 

Breizh



#9 Daniel89

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Posted 17 April 2016 - 12:52 AM

Breizh thank you very much, this was exactly what I was looking for, now I understand where I earlier misunderstood, and I have an spreadsheet where I can control my own calculations step by step. This was very helpful, thank you!






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