Chapter 7

1 7-11 Chapter 7 Example Refrigerant 134a is the working fluid in an ideal vapor-compression refrigeration cycle that communicates thermally with a cold region at 0 C and a warm region at 26 C. Saturated vapor enters the compressor at 0 C and saturated liquid leaves the condenser at 26 C. The mass flow rate of the refrigerant is kg/s. Determine (a) the compressor power, in kW, (b) the refrigeration capacity, in tons, (c) the coefficient of performance, and (d) the coefficient of performance of a Carnot refrigeration cycle operating between warm and cold regions at 26 and 0 C, respectively. Solution ---------------------------------------- ---------------------------------------- ---------- CondenserEvaporatorExpansionvalveCompres sor4123 QLQHPLPHTsLiquidVaporIsenthalp1234 Wabc26 Co0 Co Specific Internal Specific Specific Type Temp Pressure Volume Energy Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg kJ/kg/K 1 R-134a 0 1 Saturated Vapor 3 R-134a 26 236 0 Saturated Liquid 2 R-134a Superheated Vapor (a) The compressor work is give by cW = m (h2 - h1) = ( kg/s)( - ) kJ/kg = kW (b) The refrigeration capacity, in tons, is LQ = m (h1 - h4) = ( kg/s)( - 236) kJ/kg = kW LQ = ( kJ/s)

2 (60 s/min)1 ton211 kJ/min = ton 6 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 539 7-12 (c) Tthe coefficient of performance is b = LcQW = = (d) The coefficient of performance of a Carnot refrigeration cycle operating between warm and cold regions at 26 and 0 C, respectively b = LcQW = LHLQQQ- = LHLTTT- = 273299273- = ---------------------------------------- ---------------------------------------- ---------- CondenserEvaporatorExpansionvalveCompres sor4123 QLQHPLPHTsLiquidVaporIsenthalp1234 WTHTL2s Figure A vapor-compression refrigeration with irreversibilities.

3 Figure shows several features existed in actual vapor compression systems. The heat transfers between the refrigerant and the warm and cold regions are not accomplished reversibly: the refrigerant temperature in the evaporator is less than the low reservoir temperature, TL, and the refrigerant temperature in the condenser is greater than the high reservoir temperature, TH. The compressor will not have 100% efficiency so that the fluid leaving the compressor will be at state (2), which is at higher entropy than the isentropic compression state (2s). The coefficient of performance decreases as the average temperature of the refrigerant in the evaporator decreases and as the average temperature of the refrigerant in the condenser increases.

4 If the thermal efficiency of the compressor is known, the enthapy h2 at state (2) can be determined from the following expression: hc = ()()//cscWmWm = 2121shhhh-- Due to frictions, there will be pressure drops as the refrigerant flows through the evaporator, condenser, and piping connecting the various components. The pressure drops are ignored in subsequent calcualtions for simplicity. 7-13 Example Refrigerant 134a is the working fluid in an ideal vapor-compression refrigeration cycle that communicates thermally with a cold region at - 10 C. Saturated vapor enters the compressor at - 10 C and saturated liquid leaves the condenser at 9 bar.

5 The mass flow rate of the refrigerant is kg/s. Determine (a) the compressor power, in kW, (b) the refrigeration capacity, in tons, (c) the coefficient of performance. Solution ---------------------------------------- ---------------------------------------- ---------- CondenserEvaporatorExpansionvalveCompres sor4123 QLQHPLPHTsLiquidVaporIsenthalp1234 Wabc9 bar-10 Co Specific Internal Specific Specific Type Temp Pressure Volume Energy Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg kJ/kg/K 1 R-134a -10 1 Saturated Vapor 2 R-134a Superheated Vapor 3 R-134a 249 0 Saturated Liquid (a) The compressor work is give by cW = m (h2 - h1) = ( kg/s)(423 - ) kJ/kg = kW (b) The refrigeration capacity, in tons, is LQ = m (h1 - h4) = ( kg/s)( - ) kJ/kg = kW LQ = ( kJ/s)

6 (60 s/min)1 ton211 kJ/min = ton (c) The coefficient of performance is b = LcQW = = 7 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 541 7-14 Example Refrigerant 134a is the working fluid in an ideal vapor-compression refrigeration cycle that communicates thermally with a cold region at - 10 C. Saturated vapor enters the compressor at - 10 C and liquid leaves the condenser at 9 bar and 30oC. The compressor has an efficiency of 80%. The mass flow rate of the refrigerant is kg/s. Determine (a) the compressor power, in kW, (b) the refrigeration capacity, in tons, (c) the coefficient of performance.

7 Solution ---------------------------------------- ---------------------------------------- ---------- CondenserEvaporatorExpansionvalveCompres sor4123 QLQHPLP= 9 barH TsLiquidVaporIsenthalp1234W2s30 Co-10 Co Specific Specific Specific Type Temp Pressure Volume Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg/K 1 R-134a -10 1 Saturated Vapor 2s R-134a Superheated Vapor 3 R-134a 30 Compressed Liquid 4 R-134a -10 Liquid Vapor Mixture (a) The compressor work is give by cW = m (h2 - h1) = m 21shhh- = ( kg/s) kJ/kg = kW (b) The refrigeration capacity, in tons, is LQ = m (h1 - h4) = ( kg/s)( - ) kJ/kg = kW LQ = ( kJ/s)(60 s/min)1 ton211 kJ/min = ton (c) Tthe coefficient of performance is b = LcQW = = 8 Moran, M.

8 J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 543 7-15 Example ---------------------------------------- ---------------------------------------- -- Design a refrigeration system, which supplies two levels of ammonia refrigerant at 10oC and at - 10oC to two exchangers requiring duties of 850 and 2,500 kW, respectively. Draw a process flow diagram (PFD) showing major equipment, flow rates in kg/s of refrigerant, duty of each heat exchanger, duty, and horsepower of each compressor. Also, show temperatures and pressures of all streams. There are no inter-coolers between compressors.

9 The compressed gas from the last stage compressor can be cooled to saturated liquid at 30oC. Use 75% adiabatic efficiency for all compressors. Solution ---------------------------------------- ---------------------------------------- ---------- E1F1C1E3T , P53T , Pi3TP11TP11TP11E2F2C2TP22TP22TP22T , P31T , P41V1V2 Figure A two level refrigeration system. A design of a two level refrigeration system is shown in Figure Saturated liquid ammonia at Ti is partially vaporized through a let-down valve V1 to T1 and P1. Part of the liquid separated by the flash drum F1 is used to provide the first level refrigeration through the heat exchanger E1.

10 The remaining liquid passes through a second valve where its temperature and pressure is reduced to T2 and P2. The liquid stream from the second flash drum F2 is used for the second level refrigeration through the heat exchanger E2 where it becomes saturated vapor. This stream is combined with the saturated vapor from the second flash drum and compressed back to pressure P1. The problem can be solved either with a Pressure-Enthalpy diagram of ammonia as shown in Figure or with a program solving an equation of state to obtain thermodynamic properties of a pure substance given any two other properties (for example, temperature and pressure or saturation temperature).