Termodinâmica II - Solução de Exercícios do Capítulo 10 sobre Refrigeração por Compressão de Vapor

1 CANKAYA UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE MECHANICAL ENGINEERING DEPARTMENT ME 212 THERMODYNAMICS II CHAPTER 10 EXAMPLES SOLUTION 1 An ideal vaporcompression refrigerant cycle operates at steady state with Refrigerant 134a as the working fluid Saturated vapor enters the compressor at 100C and saturated liquid leaves the condenser at 280C The mass flow rate of refrigerant is 5 kgmin Determine a The compressor power in kW b The refrigerating capacity in tons c The coefficient of performance Solution Known Refrigerant 134a is the working fluid in an ideal vaporcompression refrigerant cycle Operating data are known Schematic Given Data 2 Analysis First fix each of the principle states Compressor Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 h mh W m m m m m m h m h W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2 c 2 1 e i 2 2 1 1 c e e 2 e e e i i 2 i i i vc vc vc c W 2 1 3 1 2 2 1 2 2 1 1 cv i e e e i i j j j cv s s m m m m s m s 0 m s m s T Q dt dS State 1 T1 10oC saturated vapor h1 24135 kJkg s1 09253 kJkgK State 2 P2 Psat28 o C72675 bars s2 s1 Double interpolation gives h2 2679 kJkg Expansion valve Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 Neglect work w0 4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 vc 0 vc vc h h m m m m m m h m h 0 gz 2 V h m gz 2 V h m W Q dt dE State 3 T3 28oC saturated liquid h3 8861 kJkg State 4 Throttling process h4 h3 8861 kJkg 4 3 4 Evaporator Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect work w 0 h mh Q m m m m m m h m h Q 0 gz 2 V h m gz 2 V h m W Q dt dE 4 1 in 4 1 e i 1 1 4 4 in e e 2 e e e i i 2 i i i 0 vc vc vc a The compressor power is 2 212kW kJ s 1 kW 1 kg 24135 kJ 267 9 60s min 1 min 5 kg h mh W 1 2 c b The refrigeration capacity is 3 62tons kJ min 211 ton 1 kg 8861 kJ min 24135 5 kg h mh Q 4 1 in c The coefficient of performance is 5 75 24135kJ kg 267 9 8861kJ kg 24135 h h h h W Q 1 2 4 1 c in Q in 1 4 5 2 A vaporcompression refrigeration system circulates Refrigerant 134a at rate of 6 kgmin The refrigerant enters the compressor at 100C 14 bar and exits at 7 bar The isentropic compressor efficiency is 67 There are no appreciable pressure drops as the refrigerant flows through the condenser and evaporator The refrigerant leaves the condenser at 7 bar 240C Ignoring heat transfer between the compressor and its surroundings determine a The coefficient of performance b The refrigerating capacity in tons c The irreversibility rates of the compressor and expansion valve each in kW d The changes in specific flow availability of the refrigerant passing through the evaporator and condenser respectively each in kJkg Let To 21oC po 1 bar Solution Known A vaporcompression refrigeration system circulates Refrigerant 134a with a known mass flow rate Data are given at various locations the compressor efficiency is specified Schematic Given Data Analysis First fix each of the principal states 6 Compressor Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 h mh W m m m m m m h m h W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2s c 2 1 e i 2s 2 1 1 c e e 2 e e e i i 2 i i i vc vc vc 1 s 2 2 1 2s 2 1 1 cv i e e e i i j j j cv s s m m m m s m s 0 m s m s T Q dt dS State 1 T1 10oC P1 14 bars h1 24340 kJkg s1 09606 kJkgK State 2 For isentropic compression P2 7 bars s2s s1 h2s 27806 kJkg Using the compressor efficiency 29513kJ kg 67 0 24340 27806 24340 h h h h h h h h c 1 s 2 1 2 1 2 1 s 2 C s2 10135 kJkgK c W 2 1 7 State 3 P3 7 bars T 24oC h3 hf24 o C 8290 kJkg s3 03113 kJkgK Expansion valve Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 Neglect work w0 4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 vc 0 vc vc h h m m m m m m h m h 0 gz 2 V h m gz 2 V h m W Q dt dE State 4 Throttling process h4 h3 8290 kJkg s4 033011 kJkgK Evaporator Assumptions Q in 1 4 4 3 8 steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect work w 0 h mh Q m m m m m m h m h Q 0 gz 2 V h m gz 2 V h m W Q dt dE 4 1 in 4 1 e i 1 1 4 4 in e e 2 e e e i i 2 i i i 0 vc vc vc a The coefficient of performance is 3 10 h h h h 1 2 4 1 b The refrigerating capacity is 4 564tons kJ min 211 ton 1 kg 8290 kJ min 24340 6 kg h mh Q 4 1 in a For the compressor comp 2 1 0 J j s m s T Q 0 Thus 1 55kW kJ s 1 kW 1 kg 0 9606 kJ 1 0135 60s min 1 min 294K 6 kg s T m s T I 1 2 0 comp o comp Similarly for the valve 0 5530kW kJ s 1 kW 1 kg 0 3113 kJ 0 33011 60s min 1 min 294K 6 kg s T m s I 3 4 o valve d The change in specific flow availability for refrigerant passing through the evaporator is 9 2486kJ kg kgK 0 33011 kJ 294K 0 9606 kg 8290 kJ 24340 s T s h h e e 4 1 o 4 1 f 4 f1 For the condenser kg 5 78 kJ kgK 1 0135 kJ 294 K 0 3113 kgK 29513 kJ 8290 s T s h h e e 2 3 o 2 3 f 2 f 3 Comment Although there is heat transfer to the refrigerant passing through the evaporator the specific flow availability decreases This can be explained by noting that the state of the working fluid moves closer to the dead state as it is heated at a temperature below To 10 3 An ideal vaporcompression heat pump cycle with Refrigerant 134a as the working fluid provides 15 kW to maintain a building at 200C when the outside temperature is 50C Saturated vapor at 24 bar leaves the evaporator and saturated liquid at 8 bar leaves the condenser Calculate a The power input to the compressor in kW b The coefficient of performance c The coefficient of performance of a reversible heat pump cycle operating between thermal reservoirs at 20 and 50C Solution Known An ideal vaporcompression heat pump cycle uses Refrigerant 134a as the working fluid and provides a known energy output to heat a building Data are known at various locations Schematic and Given Data Analysis First fix each of the principal states Compressor 11 Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 h mh W m m m m m m h m h W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2 c 2 1 e i 2 2 1 1 c e e 2 e e e i i 2 i i i vc vc vc 2 1 2 1 2 2 1 1 cv i e e e i i j j j cv s s m m m m s m s 0 m s m s T Q dt dS Condenser Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Q out 2 3 Condenser c W 2 1 12 Neglect work w 0 h mh Q m m m m m m h m h Q 0 gz 2 V h m gz 2 V h m W Q dt dE 3 2 in 3 2 e i 3 3 2 2 out e e 2 e e e i i 2 i i i 0 vc vc vc State 1 0 9222 kJ kgK 24409 kJ kg s h bars saturated vapor 42 p 1 1 1 State 2 26897 kJ kg h s 8 bars s p 2 1 2 2 State 3 9342 kJ kg h 8 bars saturated liquid p 3 3 Expansion valve Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 Neglect work w0 4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 vc 0 vc vc h h m m m m m m h m h 0 gz 2 V h m gz 2 V h m W Q dt dE 4 3 13 State 4 9342 kJ kg h h Throttling process 4 3 a To determine the compressor first find the mass flow rate from 126 kW 2 kJ s 1 kW 1 kg 24409 kJ 26897 s 0 08544 kg h mh W Thus 0 08544 kg s kW 1 kJ s 1 kg 9342 kJ 26897 kW 15 h h Q m or h mh Q 1 2 c 3 2 out 3 2 out b The coefficient of performance is 7 055 2 126 15 W Q c out c For a reversible heat pump operating between reservoirs at 1953 278 293 293 T T T 278 K 293 K and T T C H H max C H 14 4 A vaporcompression heat pump with a heating capacity of 500 kJmin is driven by a power cycle with a thermal efficiency of 25 For the heat pump Refrigerant 134a is compressed from saturated vapor at 100C to the condenser pressure of 10 bar The isentropic compressor efficiency is 80 Liquid enters the expansion valve at 96 bar 340C For the power cycle 80 of the heat rejected is transferred to the heated space a Determine the power input to the heat pump compressor in kW b Evaluate the ratio of the total rate that heat is delivered to the heated space to the rate of heat input to the power cycle Discuss Solution Known Refrigerant 134a is the working fluid in a vaporcompression heat pump driven by a power cycle Operating data are specified for the heat pump and the power cycle Analysis First fix each state of the heat pump cycle Compressor 15 Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 h mh W h mh W m m m m m m h m h W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2 a c 1 2s s c 2 1 e i 2s 2 1 1 s c e e 2 e e e i i 2 i i i vc vc vc 1 2 1 s 2 a c s c h h h h W W Evaporator Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect work w 0 Q in 4 1 Evaporator 2 1 W net 16 h mh Q m m m m m m h m h Q 0 gz 2 V h m gz 2 V h m W Q dt dE 4 1 in 4 1 e i 1 1 4 4 in e e 2 e e e i i 2 i i i 0 vc vc vc Expansion valve Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect heat transfer q0 Neglect work w0 4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 vc 0 vc vc h h m m m m m m h m h 0 gz 2 V h m gz 2 V h m W Q dt dE Condenser 4 3 17 Assumptions steady state steady flow processSSSF open system ΔKE ΔPE 0 Neglect work w 0 h mh Q m m m m m m h m h Q 0 gz 2 V h m gz 2 V h m W Q dt dE 3 2 in 3 2 e i 2 2 3 3 in e e 2 e e e i i 2 i i i 0 vc vc vc State 1 0 9253 kJ7kgK 24134 kJ kg s h 10 C saturated vapor T 1 1 0 1 State 2 For isentropic compression 27463 kJ kg h s 10 bars s p 2s 1 2s 2 Using the compressor efficiency 28295 kJ kg 80 24124kJ kg 27463kJ kg kg 24134 kJ h h h h h h h h c 1 s 2 1 2 1 2 1 s 2 c State 3 9731 kJ kg h 34 C compressed liquid h 34 C bars T 69 p 0 f 3 0 3 3 State 4 Throttling process 9731 kJ kg h h 3 4 a The mass flow rate of refrigerant is 0 04489 kg s 60 s min 1 9731 kJ kg 28295 kJ min 500 h h Q m 3 2 out1 The compressor power becomes Q in 3 2 Condenser 18 868 kW 1 kJ s 1 kW 1 kg 24134 kJ 28295 s 0 04489 kg h mh W 1 2 c b For the power cycle c 0 25 With 1 868 kW W W heat pump power cycle 7 472 W Q cycle in2 The total rejected is 4 483 kW 80 Q Q Thus 5 604 kW 1 868 kW 7 472 kW Q rej 2 out rej Finally 1 715 472kW 7 4 483kW 1kJ kW 1 60s min 1 min kJ 500 Q Q Q 2 in out2 out 1 COMMENT The enginedriven heat pump delivers more energy to the heated space than could be obtained by burning fuel directly 19 5 Air enters the compressor of an ideal Brayton refrigeration cycle at 140 kPa 270 K with a volumetric flow rate of 1 m3s and is compressed to 420 kPa The temperature at the turbine inlet is 320 K Determine a The net power input in kW b The refrigerating capacity in kW c The coefficient of performance d The coefficient of performance of a reversible refrigeration cycle operating between reservoirs at TC 270 K and TH 320 K Solution Known Air is the refrigerant in an ideal Brayton refrigeration cycle Data are known at various locations and the volumetric flow rate at the compressor inlet is given Schematic and Given Data Analysis First fix each of the principal states State 1 0 9590 27011 kJ kg p h 270 K T r1 1 1 20 State 2 37010 kJ kg h 2 877 kPa 140 420kPa 0 9590 p p p p 2 1 2 r1 r2 State 3 1 7375 32029 kJ kg p h 320 K T r3 3 3 State 4 23361 kJ kg h 0 5792 kPa 420 1 7375 140kPa p p p p 4 3 4 r3 r4 a The mass flow rate is 1 807 kg s Nm 10 kJ 1 kPa 1 N m 10 97 kgK 270K 28 314 kJ 8 m s140kPa 1 RT A p v A m 3 2 3 3 1 1 1 1 1 Thus the net power is 05 kW 24 kJ s 1 kW 1 kg 23361 kJ 32029 27011 370 1 s 807 kg 1 h h h m h W W W W 4 3 1 2 cycle p t cycle b The refrigerating capacity is 6596 kW kg 23361 kJ 27011 s 1 807 kg h mh Q 4 1 in c The coefficient of performance is 2 743 05 24 96 65 W Q cycle in d For a reversible cycle operating between thermal reservoirs at 270 K and TH 320 K is 45 270 320 270 T T T C H C max