ME 200 Termodynamics I Lecture 38: Vapor-compression refrigeration systems Yong Li Sangai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Cuan Road Sangai, 200240, P. R. Cina Email : liyo@sjtu.edu.cn Pone: 86-21-34206056; Fax: 86-21-34206056 1.1
Carnot Refrigeration Cycle 1.2
Coefficient of Performance β Coefficient Of Performance ::: Ratio of te refrigeration effect to te net work input required to acieve tat effect. (COP) Represents te maximum teoretical coefficient of performance of any refrigeration cycle operating between regions at T C and T H. 1.3
Departures from te Carnot Cycle Temperature difference T. Compression two-pase liquid vapor mixture. Wet compression. In actual systems, compressor andles vapor only. Dry compression. Te work output of te turbine is normally sacrificed by using a simple trottling valve for te expansion turbine, saving in initial and maintenance costs. 1.4
Departures from te Carnot Cycle T Dry compression Trottling valve 1.5
Cycle Analysis Q cond Trottling Valve 3 4 Condenser Evaporator 2 Compressor 1 W Q evap T T cond subcooling T sc 3 2 T H P p cond 3 2 T evap 4 T-s diagram 1 s T L supereat: T s p evap 4 1 P- diagram 1.6
Ideal Vapor-Compression Refrigeration Cycle Process 1 2s: Isentropic compression of te refrigerant from state 1 to te condenser pressure at state 2s. Process 2s 3: Heat transfer from te refrigerant as it flows at constant pressure troug te condenser. Te refrigerant exits as a liquid at state 3. Process 3 4: Trottling process from state 3 to a two-pase liquid vapor mixture at 4. All of te processes in te above cycle are internally reversible except for te trottling process. Process 4 1: Heat transfer to te refrigerant as it flows at constant pressure troug te evaporator to complete te cycle. 1.7
Analyzing Rankine Cycle---I Assumptions: 1) Steady state. 2) ΔKE=ΔPE=0. 3)stray Q=0 Turbine de dt cv Q cv W cv i m ii Vi 2 2 gz i e m e e Ve 2 2 gz e Condenser Pump Boiler 1.8
Vapor-compression refrigeration cycle de dt cv Q cv W cv i Vi m ii 2 2 gz i e Ve m ee 2 2 gz e Evaporator: Boiler Te eat transfer rate is referred to as te refrigeration capacity. ( kw).» Anoter unit for te refrigeration capacity is te ton of refrigeration, = 211 kj/min. Compressor Condenser Turbine Condenser Pump Trottling process 1.9
Example 1: Ideal Vapor- Compression Refrigeration Cycle Known: R-134a as te working fluid and operates on an ideal vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. mass flow rate of te refrigerant is 0.05 kg/s, Find: Q in, W in, mass flow rate, Q out Assumptions: 1) Steady state. 2) ΔKE=ΔPE=0. Analysis: 1.10
Actual cycle Te eat transfers between te refrigerant and te warm and cold regions are not accomplised reversibly: T evaporator < T C, T condenser > T H.» Te COP as T evaporator and COP as T H. Adiabatic irreversible compression 1.11
Cycle Analysis Second Law Efficiency: COP COP actual Carnot T T cond Irreversibilities 3 2 T H T evap 4 1 T L s 1.12
Compressor Analysis Overall isentropic Efficiency:::Ratio of isentropic compressor power input to actual compressor power input: o,is m r 2s 1 W comp 1.13
Houseold Refrigerator 1.14
Vapor Compression Cycles Best possible performance for a eat pump cycle:» Carnot Cycle! Wat is te difference between a.) Refrigeration/Air Conditioning, and b.) Heat Pumping? 1.15
Refrigerator & Heat pump Te objective of a refrigerator is to remove eat (QL) from te cold medium; Te objective of a eat pump is to supply eat (QH) to a warm medium 1.16
A eat pump can be used to eat a ouse in winter and to cool it in summer 1.17
Example 2:Vapor Compression Cycles Consider a 300 kj/min refrigeration system tat operates on an vapor-compression cycle wit R134a as te working fluid. Te refrigerant enters te compressor as supereated vapor at 140 kpa and is compressed to 800 kpa. If te isentropic efficiency of te compressor is 0.85, and te supereat and subcooling are 5 o C, determine: a) te quality of te refrigerant at te end of te trottling process, b) te coefficient of performance, c) te power input to te compressor (kw). Given: p 1 = p 4 = 140 kpa, T supereat = 5 o C, p 2 = p 3 = 800 kpa, T supcooling = 5 o C, Q L = 300 kj/min Find: x 4, COP R, W in Assumptions:» Steady state, steady flow, Neglect KE and PE» Adiabatic compression» Constant pressure eat transfer processes» No work or eat interactions in te expansion process 1.18
Example 2 p 1 = p 4 = 140 kpa, T supereat = 5 o C, p 2 = p 3 = 800 kpa, T supcooling = 5 o C, Q L = 300 kj/min Solution:» Quality at te end of te trottling process: x 3 4 4 4 4 f 4 3 3 3 T T 800kPa T T 3 3 p sat fg 26.31 o p p, T C subcooling» From R134a Tables: 3 f (T 3 ) = 87.85 kj/kg f (p 4 ) = 27.08 kj/kg fg (p 4 ) = 212.08 kj/kg x 4 0.287 1.19
Example 2 p 1 = p 4 = 140 kpa, T supereat = 5 o C, p 2 = p 3 = 800 kpa, T supcooling = 5 o C, Q L = 300 kj/min Continue Solution:» Coefficient of Performance cooling: 1 4 COPR 1 1 1 2 1 T T p T 1 sat 1 supereat o o o T 18.77 C 5 C 13.77 C 1 1 p, T 242.12 kj kg Compressor analysis: p, s s 2s 2 2s 2s 1 2s s 0.9604 280.41 kj kg K kj kg K 1.20
Example 2 p 1 = p 4 = 140 kpa, T supereat = 5 o C, p 2 = p 3 = 800 kpa, T supcooling = 5 o C, Q L = 300 kj/min Continue Solution:» Coefficient of Performance cooling: C R 2s 1 COP COP 2 1 R 3.425 C 1 4 2s 1» Power input: W in m 2 1 Q in m 1 4 1.21
Example 2 p 1 = p 4 = 140 kpa, T supereat = 5 o C, p 2 = p 3 = 800 kpa, T supcooling = 5 o C, Q L = 300 kj/min Continue Solution:» Power input: m in in 1 4 in 2 1 1 4 Q 300 kj in min 1min Win COP 3.425 60 s in Q Q W W R 1.46kW 1.22
Cycle Analysis Compressor Inlet State p 1 = p evap 1 = (T evap +T s, p evap ) Compressor Outlet State p 2 = p cond 2 1 isentropic specific work ( ) 2s 1 s T T cond T evap 2 2s 1 p 2 isentropic efficiency 2s = (p 2, s 2 = s 1 ) s 1.23
Reading material 1.24
Cycle Analysis Condenser Outlet State p 3 = p cond 3 = (T cond +T sc, p cond ) P P cond P evap 3 4 Evaporator Inlet State p 4 = p evap 4 = 3 1.25
Cycle Analysis Coefficient of Performance COP = refrigerating effect 1 4 2 1 specific work Volumetric Capacity P p cond p evap 3 4 1 2 q = v refrigerating effect 1 4 v 1 compressor inlet specific volume indicator of te compressor displacement requirement 1.26
Cycle Analysis Sources for Irreversibilities Trottling Losses Compressor Losses T T cond 3 3s 2 T T cond 3 2s 2 T H T H T evap 4 1 T L T evap 4 1 T L s s 1.27
Cycle Analysis Sources for Irreversibilities Desupereating Losses Heat Excange Losses T T cond 3 2s T T cond 3 3s 2s T H T H T evap 4 1 T L T evap 4 1 T L s s» also includes losses due to pressure drops in eat excangers and distribution piping 1.28