Example 11-5 reproduced from Ref. 1

Consider a two-stage compression refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. The working fluid is refrigerant-134a. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.32 MPa. Part of the refrigerant evaporates during this flashing process, and this vapor is mixed with the refrigerant leaving the low-pressure compressor. The mixture is then compressed to the condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves the evaporator as a saturated vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser, and (c) the coefficient of performance.
Thermophysical Properties from: https://webbook.nist.gov/
cycle 1 = UPPER cycle with R134
saturation properties - pressure increment = 1 bar​Temperature (C) Pressure (MPa) Enthalpy (kJ/kg) Entropy (J/g*K) Enthalpy (kJ/kg) Entropy (J/g*K)
datasat=
;
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,InterpolationOrder1];fsl=Interpolation[tblsl,InterpolationOrder1];
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,InterpolationOrder1];fsv=Interpolation[tblsv,InterpolationOrder1];
In[]:=
tbltp=Table[{datasat[[i]][[1]],datasat[[i]][[2]]},{i,1,Length[datasat]}];
In[]:=
ftp=Interpolation[tbltp,InterpolationOrder1];
In[]:=
FindRoot[ftp[x]==0.8,{x,1}]
Out[]=
{x31.3275}
In[]:=
FindRoot[ftp[x]==0.32,{x,1}]
Out[]=
{x2.41497}
In[]:=
FindRoot[ftp[x]==0.14,{x,1}]
Out[]=
{x-18.9838}
isobaric properties - pressure = 0.8 MPa
Temperature (C) Pressure (MPa) Enthalpy (kJ/kg) Entropy (J/g*K) Phase
dataisobaric=
;
In[]:=
pos=Position[dataisobaric,31.328]
Out[]=
{{284,1},{285,1}}
In[]:=
tblhsv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[3]]},{i,285,Length[dataisobaric]}];​​tblssv=Table[{dataisobaric[[i]][[1]],dataisobaric[[i]][[4]]},{i,285,Length[dataisobaric]}];​​tblshsv=Table[{dataisobaric[[i]][[4]],dataisobaric[[i]][[3]]},{i,285,Length[dataisobaric]}];
In[]:=
fshsv=Interpolation[tblshsv];
In[]:=
tblhl1=Table[{datasat[[i]][[3]],datasat[[i]][[2]]},{i,1,Length[datasat]}];​​tblsl1=Table[{datasat[[i]][[4]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
tblhv1=Table[{datasat[[i]][[5]],datasat[[i]][[2]]},{i,1,Length[datasat]}];​​tblsv1=Table[{datasat[[i]][[6]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
fhsv=Interpolation[tblhsv,InterpolationOrder1];
In[]:=
fssv=Interpolation[tblssv,InterpolationOrder1];
isobaric properties - pressure = 0.32
Temperature (C) Pressure (MPa) Enthalpy (kJ/kg) Entropy (J/g*K) Phase
dataisobaric2=
;
In[]:=
pos=Position[dataisobaric2,2.4768]
Out[]=
{{224,1},{225,1}}
In[]:=
tblhsv=Table[{dataisobaric2[[i]][[1]],dataisobaric2[[i]][[3]]},{i,225,Length[dataisobaric]}];​​tblssv=Table[{dataisobaric2[[i]][[1]],dataisobaric2[[i]][[4]]},{i,225,Length[dataisobaric]}];​​tblshsv=Table[{dataisobaric2[[i]][[4]],dataisobaric2[[i]][[3]]},{i,225,Length[dataisobaric]}];
In[]:=
fshsv2=Interpolation[tblshsv];
In[]:=
tblhl1=Table[{datasat[[i]][[3]],datasat[[i]][[2]]},{i,1,Length[datasat]}];​​tblsl1=Table[{datasat[[i]][[4]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
tblhv1=Table[{datasat[[i]][[5]],datasat[[i]][[2]]},{i,1,Length[datasat]}];​​tblsv1=Table[{datasat[[i]][[6]],datasat[[i]][[1]]},{i,1,Length[datasat]}];
In[]:=
fhsv2=Interpolation[tblhsv];
In[]:=
fssv2=Interpolation[tblssv];
First, let us determine the unknown variables for points {1, 2, 3, 4,5,6,7,8,9} of the Table below:
In[]:=
H5=fhl[0.8]
Out[]=
95.5006
In[]:=
S5=fsl[0.8]
Out[]=
0.354137
In[]:=
H3=fhv[0.32]
Out[]=
251.854
In[]:=
S3=fsv[0.32]
Out[]=
0.930194
In[]:=
H6=H5
Out[]=
95.5006
Point 6 is in the two-phase region
In[]:=
H6
Out[]=
95.5006
In[]:=
P6=0.32
Out[]=
0.32
In[]:=
H3=fhv[P6]
Out[]=
251.854
In[]:=
S3=fsv[P6]
Out[]=
0.930194
In[]:=
H7=fhl[P6]
Out[]=
55.1149
In[]:=
S7=fsl[P6]
Out[]=
0.216153
In[]:=
x6=(H6-H7)/(H3-H7)
Out[]=
0.205275
In[]:=
S6=x6(S3-S7)+S7
Out[]=
0.362728
In[]:=
P1=0.14;
In[]:=
H1=fhv[P1]
Out[]=
239.026

References

[1] Çengel, Yunus A., Boles, Michael A., Kanoğlu, Mehmet, Thermodynamics: An Engineering Approach, Ninth Edition, NY: McGraw-Hill, 2019.

CITE THIS NOTEBOOK

A two-stage refrigeration cycle with a flash chamber​
by Housam Binous and Ahmed Bellagi
Wolfram Community, STAFF PICKS, February 26, 2025
​https://community.wolfram.com/groups/-/m/t/3400496