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Precept 6: Vapor-compression refrigeration cycle Mechanical Engineering - Energy Systems Authors: •Group xx: xxxxx xxxxx (xxxxxxxx), Fabio Santoro (xxxxxxxx) Lecturer:Prof. Stefano Consonni Teaching Assistants:Riccardo Cremona, Hamidreza Heydari and Nima Razmjoo Academic year:2025-2026Contents Description2 1 Plant scheme,p-handT-sdiagrams3 2 Thermodynamic properties4 3 Compressor volumetric fluid rate and displacement6 4 Thermal power rejected by condenser6 5 Fan and compressor electrical power6 6 Coefficient of performances: cycle and plant7 7 Double-throttling refrigeration configuration8 8 Single vs double throttling comparison14 References14 1 Energy Systems- Report Precept 6 - Group xx Description The topic of this precept is the design of a refrigeration system for fresh food storage. The system consists in a vapour-compressor (reciprocating single action volumetric), one evaporator and one lami- nation valve. The working fluid isR134a, the other general data is collected in Table1. The condenser is a compact tube with fin heat exchanger as represented in Figure1and in Table2we collected the data of the heat exchanger an the ambient air conditions. Extra data will be provided in each request. Table 1:Design parameter of the refrigeration cycleSymbolValueUnitDescription ˙ QRE F18 kW Refrigeration power (evaporation)T E V A-4◦ C Evaporation temperature∆T S H2◦ C Superheating in the evaporator∆T S C3◦ C Condensate subcooling temperatureη is70 % Compressor isoetropic efficiencyη em88 % Electro-mechanical efficiency of the compressor driverη vol90 % Compressor volumetric efficiencyn2970 rpm Compressor rotating speed Figure 1:Condenser: heat exchanger layout Table 2:Design parameter of the tube and fin heat exchanger and ambient air conditionSymbolValueUnitDescription N R3 - Number of tube rowsvel air3 m/s Air velocity∆p air0.208·vel1.87 airmm H2OAir pressure drop (per tube row)2η f an60 % Fans efficiencyγ air1.39 - Air-specific heat ratioM M air28.85 kg/kmol Air molecular mass2 Energy Systems- Report Precept 6 - Group xx 1.Plant scheme,p-handT-sdiagrams The first request is to draw the plant scheme (Figure2) that fits the description provided in the text. Then, draw generic diagrams on the p−h(Figure3) andT−s(Figure4) planes. We assume that the outlet of the compressor lies in the vapour region. This is because, for the working fluid considered, the saturated vapour curve (in theT−sdiagram), although almost vertical, still has a negative slope. Therefore, if the compressor inlet is in superheated vapour conditions, the outlet state will also remain in the superheated vapour region.Figure 2:Plant scheme Figure 3:p−hcycle diagram (R134a) Figure 4:T−scycle diagram (R134a) 3condenser evaporator valve throttling compressor 2 3 4 1 ~ Q EVA Q COND ~ W compr W FAN fan air p h 4 4 sv 3 3 sv 1 p eva p cond Q COND Q EVA log scale 3 sl 2 iso 2 s iso s real T eva T eva T cond T cond T s 4 4 sv 3 3 sl 3 sv 2 iso 2 p cond p eva 1 T eva T cond Δ T SC Δ T SH Q COND Q EVA W compr Energy Systems- Report Precept 6 - Group xx 2.Thermodynamic properties In this section we will compute the thermody- namic properties of the refrigeration cycle for three different cases of condensation tempera- tureTcond: 28◦ C, 35◦ C and 45◦ C. The thermo- dynamic properties for R134a are evaluated with the coolprop package of python [1]. •Point 1 At the compressor inlet the fluid is at super- heated conditions at temperatureT1=T E V A+ ∆TS Hat evaporation pressurep 1=p(T E V A). All other propertiesh1,s 1andv 1are evaluated from temperature and pressure. •Point 2iso The isoentropic compression (s2iso=s 1) of the fluid to the condensation pressurep2iso= pcond=p(T cond). With these we can extract the other propertiesT2iso,h 2isoandv 2iso. •Point 2 The real compression considers the device ineffi- ciency but, we reach the same condensation pres- surep2=p 2iso. We can compute the enthalpy with the formula:h2=h 1+ (h 2iso−h 1)/η is. With pressure and enthalpy we can extract the other propertiesT2,s 2andv 2. •Point 3sv Now the stream enters the condenser at super- heated vapour conditions so, first we reach vapour saturation at the condensation pressure p3sv=p condand the other properties at satu- rated conditionsT3sv=T cond,h 3sv,s 3svandv 3sv considering the vapour quality asx3sv= 1. •Point 3sl In the condenser, before exiting, we reach satu-rated liquid conditions3 sl. We are still at con- densation pressurep3sl=p condand the other properties at saturated conditionsT3sl=T cond, h3sl,s 3slandv 3slconsidering the vapour quality asx3sl= 0. •Point 3 The sub-cooled conditions at the exit of the con- denserp3=p condare at temperatureT 3= Tcond−∆T S C. With these we can extract the other propertiesh3,s 3andv 3. •Point 4 The fluid is throttled to reduce the pressure from condensation to evaporationp4=p E V Aand we fall in the 2 phase regionT4=T E V A. In the throttling valve the stream enthalpy remains constanth4=h 3. We can extract the remaining propertiess4,v 4and vapour qualityx 4. •Point 4sv In the evaporator, the stream reaches saturated vapour conditions before entering the compres- sor.p4sv=p E V AandT 4sv=T E V A. The other properties are extractedh4sv,s 4svandv 4svcon- sidering the vapour quality asx4sv= 1. In Tables3,4and5we collected the numerical results of the thermodynamic properties while in Figures5and6we represented, respectively, the p−handT−srefrigeration cycle diagrams. Points 1 and 4sv are shared for every conden- sation temperature case while, we reach, gener- ally higher enthalpy levels for increasing values ofTcond. In fact, we achieve higher temperature at the compressor outlet and the throttling en- thalpy outlet has higher vapour content. Table 3:Thermodynamic properties withTcond= 28◦ CipThsvxStatus [bar][ ◦ C][kJ/kg][kJ/(kgK)][m 3 /kg]1 2.527 -2 398.01 1.7359 0.08068 - sup-heat vap 2is 7.269 34.082 420.14 1.7359 0.02933 - sup-heat vap 2 7.269 43.446 429.63 1.7663 0.03088 - sup-heat vap 3sv 7.269 28 413.84 1.7152 0.02826 1 sat vap 3sl 7.269 28 238.84 1.1341 0.00084 0 sat liq 3 7.269 25 234.55 1.1198 0.00083 - sub-cool 4 2.527 -4 234.55 1.1286 0.01642 0.198 liq-vap 4sv 2.527 -4 396.25 1.7294 0.07987 1 sat vap 4 Energy Systems- Report Precept 6 - Group xx Table 4:Thermodynamic properties withT cond= 35◦ CipThsvxStatus [bar][ ◦ C][kJ/kg][kJ/(kgK)][m 3 /kg]1 2.527 -2 398.01 1.7359 0.08068 - sup-heat vap 2is 8.87 41.653 424.39 1.7359 0.02405 - sup-heat vap 2 8.87 52.434 435.7 1.7712 0.02559 - sup-heat vap 3sv 8.87 35 417.19 1.7128 0.02303 1 sat vap 3sl 8.87 35 249.01 1.167 0.00086 0 sat liq 3 8.87 32 244.62 1.1527 0.00085 - sub-cool 4 2.527 -4 244.62 1.166 0.02037 0.248 liq-vap 4sv 2.527 -4 396.25 1.7294 0.07987 1 sat vap Table 5:Thermodynamic properties withT cond= 45◦ CipThsvxStatus [bar][ ◦ C][kJ/kg][kJ/(kgK)][m 3 /kg]1 2.527 -2 398.01 1.7359 0.08068 - sup-heat vap 2is 11.599 52.407 430.11 1.7359 0.01831 - sup-heat vap 2 11.599 64.848 443.86 1.7774 0.01976 - sup-heat vap 3sv 11.599 45 421.52 1.7092 0.01734 1 sat vap 3sl 11.599 45 263.94 1.2139 0.00089 0 sat liq 3 11.599 42 259.39 1.1996 0.00088 - sub-cool 4 2.527 -4 259.39 1.2209 0.02617 0.321 liq-vap 4sv 2.527 -4 396.25 1.7294 0.07987 1 sat vap Figure 5:p−hdiagram forT condset to 28, 35 and 45◦ C 5200225250275300325350375400425450 Entalpy h [kJ/kg]234567891020 Pressure p [bar] (logarithm)T cond= 28 CT cond= 35 CT cond= 45 C Energy Systems- Report Precept 6 - Group xx Figure 6:T−sdiagram forT condset to 28, 35 and 45◦ C 3.Compressor volumetric fluid rate and displacement With the thermodynamic properties evaluated in the previous section, we can compute the refrigerant mass flow rate˙mRE Ffrom the heat power rejected in the evaporator˙ QRE F=˙ QE V A. Then, with the mass rate, we can evaluate the volumetric rate at the compressor inlet˙ V1and finally, we can compute the compressor displacement in volumetric termVcil. All numerical results are collected in Table7. ˙ QRE F= ˙m RE F(h 1−h 4)⇒˙m RE F=˙ QRE Fh 1−h 4; ˙ V1= ˙m RE F·v 1;η vol=˙ V1nV cil⇒V cil=˙ V1n·η vol 4.Thermal power rejected by condenser The heat power rejected by the condenser as˙ Qcond= ˙m RE F·(h 2−h 3)with numerical results dependent onTcondare collected in Table7. 5.Fan and compressor electrical power In this section we are require to compute the fan and compressor electrical power consumptions. Here we will present the process, while all the numerical results for each value ofTcondare in Table7. The compressor electrical power is computed as: ˙ Wcompr,el=˙m RE F·(h 2−h 1)η emTable 6:Air temperaturesT condT air,inT air,out[ ◦ C][ ◦ C][ ◦ C]28 18 25 35 21 28 45 35 40 Now, we focus on the fan. In table6we can find the values of the inlet and outlet temperatures of the air that crosses the condenser and absorbs the rejected heat power. With the data provided in Table2 we can evaluate the specific heat capacitycp,airand then, we write the air mass flow rate˙m airformula: cp,air=R uM M air· γ airγ air−1≈1.027 kJkgK  ˙mair=˙ Qcondc p,air·(T air,out−T air,in) 61 :01 :11 :21 :31 :41 :51 :61 :71 :8Entropy s [kJ/(kgK)] 1001020304050607080 Temperature T [ ° C]T cond= 28 CT cond= 35 CT cond= 45 C Energy Systems- Report Precept 6 - Group xx Given the air inlet temperatureT air,infrom Table6and assuming the air inlet pressurep air,inat atmospheric value, with the ideal gas law, we can compute the air densityρair,in. With the density we can compute the volumetric air flow˙ Vair. Then, we can compute the fan power consumption considering a pressure drop through the condenser considering the number of rows (tubes) of the device and finally the total electric consumption. ρair,in=M M air·p air,inR u·T air,in˙ Vair=˙m airρ air,in ˙ Wf an,el=˙ Vair·∆p air·N Rη f an˙ Wtot,el=˙ Wcompr,el+˙ Wf an,el 6.Coefficient of performances: cycle and plant The coefficient of performance COP of the cycle and plant can be evaluated for eachTcond: C OPcycle=˙ QRE F˙ Wcompr= ˙ QRE F˙m RE F(h 2−h 1)C OP plant=˙ QRE F˙ Wtot,el Table 7:Single throttling system results tableT cond˙m RE F˙ V1V cil˙ Qcond˙ Wcompr,el˙ Wcompr˙m air[ ◦ C][kg/min][m 3 /h][cm 3 ][kWe][kWe][kWth][kg/s] 28 6.607 31.984 199.42 21.482 3.957 3.482 2.988 35 7.041 34.083 212.52 22.423 5.026 4.423 3.119 45 7.791 37.715 235.16 23.954 6.766 5.954 4.664 T condρ air,in˙ Vair˙ Wf an,el˙ Wtot,elC OP cycleC OP plant[ ◦ C][kg/m 3 ][m 3 /s][We][kWe] 28 1.208 2.474 196.88 4.154 5.170 4.334 35 1.195 2.609 207.62 5.234 4.070 3.439 45 1.141 4.088 325.29 7.091 3.023 2.538 For increasing condensation temperatureT cond, the power consumption of the compressor and fan increases while the performances decrease. 7 Energy Systems- Report Precept 6 - Group xx 7.Double-throttling refrigeration configuration Now, we change the configuration of the system by adding another throttling valve so, we introduce an intermediate (separa- tion) pressure and we split the compressor in two levels. In Figure7we represented the new scheme. Then, draw generic dia- grams on thep−h(Figure8) andT−s (Figure9) planes. After the first throttling valve, the stream enters a separator, the saturated liquid is throttled in the evapo- rator while the saturated vapour is mixed with the outlet of the low pressure com- pressor before entering in the high pressure compressor. The mass flow rate stream managed bu the evaporator is lower than the one in the condenser and their distri- bution depends on the separation pressure psep. In this section we are required to determine the thermodynamic properties for three values ofpsep: 3, 5 and 7 bar. We will analyze these cases considering the condensation temperatureTcond= 35◦ C.Figure 7:Plant scheme with intermediate pressure levelFigure 8:p−hcycle diagram (R134a) - Intermediate pressure configuration 8condenser evaporator valve HP throttling 4 5 6 Q EVA Q COND ~ W FAN fan air valve LP throttling 7 8 1 2 9 3 ~ W compr,el separator LP HP p h p eva p sep p cond 5 6 7 8 8 sv 1 9 5 sl 5 sv 4 iso 3 4 2 2 iso s 4,iso s 4 s 2 s 2,iso Q COND Q EVA T eva T sep T cond log scale Energy Systems- Report Precept 6 - Group xx Figure 9:T−scycle diagram (R134a) - Intermediate pressure configuration Now we can compute the thermodynamic prop- erties for this system configuration. As done be- fore we will report here just the procedure with- out the results that will be collected in Tables. We keep use the coolprop package of python to evaluate the remaining properties. •Point 1 At the compressor inlet the fluid is at super- heated conditions at temperatureT1=T E V A+ ∆TS Hat evaporation pressurep 1=p(T E V A). All other propertiesh1,s 1andv 1are evaluated from temperature and pressure.•Point 2iso The isoentropic low pressure compression (s2iso=s 1) of the fluid to the separation pres- surep2iso=p sep. With these we can extract the other propertiesT2iso,h 2isoandv 2iso. •Point 2 The real compression considers the device inef- ficiency but, we reach the same separation pres- surep2=p 2iso. We can compute the enthalpy with the formula:h2=h 1+ (h 2iso−h 1)/η is. With pressure and enthalpy we can extract the other propertiesT2,s 2andv 2. •Point 5 The sub-cooled conditions at the exit of the con- denserp5=p(T cond)are at temperatureT 5= Tcond−∆T S C. With these we can extract the other propertiesh5,s 5andv 5.•Point 6 The fluid is throttled to reduce the pressure from condensation to separationp6=p sepbefore en- tering the separator and we fall in the 2 phase regionT6=T(p sep). In the throttling valve the stream enthalpy remains constanth6=h 5. We can extract the remaining propertiess6,v 6and vapour qualityx6. •Point 7 In the separator part of the fluid goes to satu- rated liquid conditions. The temperature and pressure are equal to point 6:T7=T 6and p7=p 6but, we fix the vapour quality tox 7= 0. We can extract the remaining propertiesh7,s 7 andv7. •Point 8 The fluid at saturated liquid conditions goes through the second throttling valve that reduces the pressure to evaporationp8=p E V AatT 8= TE V Ain the two phase region and keeping the enthalpy constanth8=h 7. The other properties s8,v 8and vapour qualityx 8are extracted with python package.•Point 8sv The stream exiting the second throttling valve (8) enters the evaporator and it is heated up to reach saturated vapour conditionsx8sv= 1with p8sv=p E V AatT 8sv=T E V A. We can extract the other propertiess8svandv 8sv. 9T s 5 5 sl 6 7 8 8 sv 9 5 sv 4 iso 4 3 2 1 2 iso T cond T sep T eva p cond p eva p sep Q EVA Q COND W compr,LP W compr,HP Δ T SC Δ T SH Energy Systems- Report Precept 6 - Group xx •Point 9 In the separator the remaining part of the fluid goes to saturated vapour conditions. The tem- perature and pressure are equal to point 6:T9= T6andp 9=p 6but, we fix the vapour quality to x9= 1. We can extract the other propertiesh 9, s9andv 9. •Mass balance In the system we can point out three mass flow streams:˙mE V A,˙m sepand˙m cond. We can eval- uate the first one by performing the energy bal- ance on the evaporator: ˙mE V A=˙ QRE Fh 1−h 8 We can write the mass balance in the separator and considering the vapour qualityx6. First we define which points are related to each mass flow rate: ˙m1= ˙m 2is= ˙m 2= ˙m 8sv= ˙m 8= ˙m E V A ˙m3= ˙m 4is= ˙m 4= ˙m 5sv= ˙m 5sl= ˙m 5= = ˙m6= ˙m cond ˙m9= ˙m sep The with the mass balance we obtain: ˙mcond=˙m E V A1−x 6 ˙msep=x 6·˙m E V A1−x 6 •Point 3 Before going through the high pressure compres- sor, the streams in points 2 and 9 are mixed at the separation pressurep3=p sep. We can evalu- ate the enthalpy by applying the energy balance: h3=x 6·h 9+ (1−x 6)·h 2 The other properties are extracted:T3,s 3and v3. In our numerical cases, the vapour quality x6