Liquefaction of N2 by Joule–Thomson Throttling​
Housam Binous and Ahmed Bellagi

This problem is reproduced from Example 5-10 of Ref. 1.​Consider the liquefaction of N2 by Joule–Thomson throttling. If the inlet to the expansion valve is at T3 = -122°C and P3 = 100 bar and the outlet is at P5=P6=1 bar. We determine the percentage of N2 that is liquefied.​​
​​​References:​[1] Milo D. Koretsky, Engineering and Chemical Thermodynamics, NY: Wiley, 2013.
Thermophysical Properties of Nitrogen from: https://webbook.nist.gov/cgi/fluid.cgi?ID=C7727379&Action=Page
saturation properties - temperature increment = 5 °C​
Temperature (°C), Pressure (bar), Enthalpy (kJ/kg), Entropy (J/g*K), Enthalpy (kJ/kg), Entropy (J/g*K)
In[]:=
datasat={{-210.00,0.12520,-462.15,-5.5162,-246.64,-2.1035},{-205.00,0.29159,-452.12,-5.3636,-241.97,-2.2800},{-200.00,0.59850,-442.01,-5.2210,-237.62,-2.4268},{-195.00,1.1118,-431.80,-5.0869,-233.66,-2.5515},{-190.00,1.9069,-421.45,-4.9597,-230.20,-2.6597},{-185.00,3.0663,-410.90,-4.8383,-227.33,-2.7559},{-180.00,4.6771,-400.08,-4.7213,-225.18,-2.8437},{-175.00,6.8303,-388.89,-4.6074,-223.88,-2.9263},{-170.00,9.6204,-377.18,-4.4952,-223.63,-3.0066},{-165.00,13.147,-364.75,-4.3826,-224.74,-3.0880},{-160.00,17.517,-351.22,-4.2668,-227.73,-3.1754},{-155.00,22.855,-335.81,-4.1417,-233.74,-3.2778},{-150.00,29.331,-316.02,-3.9883,-246.41,-3.4231}};
In[]:=
tblhl=Table[{datasat[[i]][[2]],datasat[[i]][[3]]},{i,1,Length[datasat]}];​​tblsl=Table[{datasat[[i]][[2]],datasat[[i]][[4]]},{i,1,Length[datasat]}];
In[]:=
fhl=Interpolation[tblhl];fsl=Interpolation[tblsl];
In[]:=
tblhv=Table[{datasat[[i]][[2]],datasat[[i]][[5]]},{i,1,Length[datasat]}];​​tblsv=Table[{datasat[[i]][[2]],datasat[[i]][[6]]},{i,1,Length[datasat]}];
In[]:=
fhv=Interpolation[tblhv];fsv=Interpolation[tblsv];
In[]:=
tbltp=Table[{datasat[[i]][[2]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
fpt=Interpolation[tbltp];
In[]:=
H6l=fhl[1]
Out[]=
-434.658
In[]:=
H5v=fhv[1]
Out[]=
-234.861
isobaric properties - pressure = 100 bar​
{Temperature (°C), Pressure (bar), Enthalpy (kJ/kg), Entropy (kJ/kg*K), State)}
In[]:=
In[]:=
tblhCOMP=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,1,Length[dataisobaric]}];​​tblsCOMP=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,1,Length[dataisobaric]}];
In[]:=
fhCOMP=Interpolation[tblhCOMP];fsCOMP=Interpolation[tblsCOMP];
In[]:=
H3=fhCOMP[-122]
Out[]=
-259.614
We get the fraction of the nitrogen, which is liquefied = 12.39 % and a temperature T4 = T5 = -196.41 °C.​​​The above values closely matches their counterparts (13.83% and -195.9 °C) obtained using HYSYS and the Peng-Robinson EoS:​​
​​​However, in Ref. 1, the reported result is different (roughly 28% of nitrogen is liquified).
In[]:=
z=(H3−H6l)/(H5v-H6l)
Out[]=
0.876111
In[]:=
1-z
Out[]=
0.123889
In[]:=
fpt[1]
Out[]=
-196.41