Refrigeration Cycles HW Solutions

ME470 Refrigeration Cycles HW Solutions Chapter 11, Solution 18. A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), P1 = 0.14 MPa  h1 = 246.36 kJ/kg  s = 0.97236 kJ/kg ⋅ K T1 = −10°C  1 T P2 = 0.7 MPa   h2 = 288.53 kJ/kg T2 = 50°C  P2 s = 0.7 MPa   h2 s = 281.16 kJ/kg s 2 s = s1  P3 = 0.65 MPa   h3 = h f T3 = 24°C  @ 24°C 0.65 MPa 24°C 2s · QH 2 0.7 MPa 50°C · Win 3 = 84.98 kJ/kg h4 ≅ h3 = 84.98 kJ/kg (throttling ) 0.15 MPa 4 · QL 1 0.14 MPa -10°C s Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from Q L = m (h1 − h4 ) = (0.12 kg/s )(246.36 − 84.98) kJ/kg = 19.4 kW and W in = m (h2 − h1 ) = (0.12 kg/s )(288.53 − 246.36 ) kJ/kg = 5.06 kW (b) The adiabatic efficiency of the compressor is determined from ηC = h2 s − h1 281.16 − 246.36 = = 82.5% 288.53 − 246.36 h2 − h1 (c) The COP of the refrigerator is determined from its definition, Q 19.4 kW COPR = L = = 3.83 W in 5.06 kW Chapter 11, Solution 31. A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. The power input to the heat pump is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 320 kPa  h1 = h g @320 kPa = 251.88 kJ/kg s = s sat. vapor g @ 320 kPa = 0.93006 kJ/kg ⋅ K  1 T P2 = 1.4 MPa   h2 = 282.54 kJ/kg s 2 = s1  P3 = 1.4 MPa   h3 = h f sat. liquid  h4 ≅ h3 = 127.22 kJ/kg @ 1.4 MPa House 3 1.4 MPa = 127.22 kJ/kg (throttling ) · QH 2 · Win 0.32 MPa The heating load of this heat pump is determined from 1 · QL 4 Q H = [m c(T2 − T1 )]water s = (0.12 kg/s )(4.18 kJ/kg ⋅ °C )(45 − 15)°C = 15.05 kW and m R = Then, Q H Q H 15.05 kJ/s = = = 0.09688 kg/s qH h2 − h3 (282.54 − 127.22 ) kJ/kg W in = m R (h2 − h1 ) = (0.09688 kg/s )(282.54 − 251.88) kJ/kg = 2.97 kW Chapter 11, Solution 32. A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the heat source. The power input to the heat pump, the rate of heat absorption from the water, and the increase in electric power input if an electric resistance heater is used instead of a heat pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), P1 = 280 kPa   h1 = 250.83 kJ/kg T1 = 0°C  P2 = 1.0 MPa  h = 293.38 kJ/kg T2 = 60°C  2 P3 = 1.0 MPa   h3 ≅ h f T3 = 30°C  @ 30°C T 2 House · QH · Win 30°C 1 MPa 3 = 93.58 kJ/kg h4 ≅ h3 = 93.58 kJ/kg (throttling ) The mass flow rate of the refrigerant is Q Q H 60,000/3,600 kJ/s m R = H = = = 0.08341 kg/s qH h2 − h3 (293.38 − 93.58) kJ/kg Then the power input to the compressor becomes W in = m (h2 − h1 ) = (0.08341 kg/s )(293.38 − 250.83) kJ/kg = 3.55 kW (b) The rate of hat absorption from the water is Q L = m (h1 − h4 ) = (0.08341 kg/s )(250.83 − 93.58) kJ/kg = 13.12 kW 0.28 MPa 4 60°C · QL Water, 8°C 1 0°C s (c) The electrical power required without the heat pump is W = Q = 60,000 / 3600 kJ/s = 16.67 kW e H Thus, W increase = W e − W in = 16.67 − 3.55 = 13.12 kW Chapter 11, Solution 42. A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vaporcompression cycle with refrigerant-134a as the working fluid. The mass flow rate of refrigerant through the lower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, and the COP of this cascade refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compression refrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressor as a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturated liquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be h1 = 239.16 kJ/kg, h2 = 260.58 kJ/kg h3 = 63.94 kJ/kg, h4 = 63.94 kJ/kg h5 = 255.55 kJ/kg, h6 = 269.91 kJ/kg h7 = 95.47 kJ/kg, h8 = 95.47 kJ/kg The mass flow rate of the refrigerant through the lower cycle is determined from an energy balance on the heat exchanger: E in − E out = ∆E system 0 (steady) = 0 E in = E out T 0.8 MPa 6 7 3 8 A 5 0.14 MPa B 4 · QL ∑ m h = ∑ m h e e 0.4 MPa 2 i i 1 s m A (h5 − h8 ) = m B (h2 − h3 ) m B = h5 − h8 255.55 − 95.47 (0.24 kg/s ) = 0.1954 kg/s m A = 260.58 − 63.94 h2 − h3 (b) The rate of heat removed by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors: Q L = m B (h1 − h4 ) = (0.1954 kg/s )(239.16 − 63.94) kJ/kg = 34.24 kW W in = W compI,in + W compII,in = m A (h6 − h5 ) + m B (h2 − h1 ) = (0.24 kg/s )(269.91 − 255.55) kJ/kg + (0.1954 kg/s )(260.58 − 239.16) kJ/kg = 7.63 kW (c) The COP of this refrigeration system is determined from its definition, Q L 34.24 kW COPR = = = 4.49  7.63 kW Wnet,in COPR = Q L 34.24 kW = = 4.49  7.63 kW Wnet,in Chapter 11, Solution 55. An ideal-gas refrigeration cycle with air as the working fluid is considered. The maximum and minimum temperatures in the cycle, the COP, and the rate of refrigeration are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 250 K  → T1 = 300 K  → h1 = 250.05 kJ / kg Pr 1 = 0.7329 . kJ / kg h3 = 30019 Pr 3 = 1.386 Thus, Pr2 = P2 Pr = (3)(0.7329 ) = 2.1987  → T2 = Tmax = 342.2 K P1 1 h2 = 342.60 kJ/kg Pr4 = P4 1 → T4 = Tmin = 219.0 K Pr =  (1.386) = 0.462  P3 3  3  h4 = 218.97 kJ/kg T 27°C -23°C · QH 2 3 4 1 · QRefrig s (b) The COP of this ideal gas refrigeration cycle is determined from COPR = qL wnet, in = qL wcomp, in − w turb, out where q L = h1 − h4 = 250.05 − 218.97 = 31.08 kJ / kg wcomp, in = h2 − h1 = 342.60 − 250.05 = 92.55 kJ / kg . − 218.97 = 81.22 kJ / kg w turb, out = h3 − h4 = 30019 Thus, COPR = 31.08 = 2.74 92.55 − 81.22 (c) The rate of refrigeration is determined to be Q refrig = m (qL ) = (0.08 kg/s)(31.08 kJ/kg) = 2.49 kJ/s Chapter 11, Solution 61. An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperature that can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations, P  T2 = T1  2   P1  (k −1) / k P  T5 = T4  5   P4  = (266 K )(4 )0.4 / 1.4 = 395.3 K = 122.3°C (k −1) / k 1 = (258 K )  0.4 / 1.4 4 T = 173.6 K = −99.4°C = Tmin (b) From an energy balance on the regenerator, -7°C ∑ m h = ∑ m h e e i i  → m (h3 − h4 ) = m (h1 − h6 ) -15°C or, 2 3 27°C E in − E out = ∆E system 0 (steady) = 0 E in = E out · QH Qrege 1 4 5 6 · QRefrig s → T3 − T4 = T1 − T6 m c p (T3 − T4 ) = m c p (T1 − T6 )  or, T6 = T1 − T3 + T4 = (− 7°C ) − 27°C + (− 15°C ) = −49°C Then the COP of this ideal gas refrigeration cycle is determined from COPR = qL qL = wnet,in wcomp,in − w turb,out = h6 − h5 (h2 − h1 ) − (h4 − h5 ) = T6 − T5 (T2 − T1 ) − (T4 − T5 ) = − 49°C − (− 99.4°C ) = 1.12 [122.3 − (− 7 )]°C − [− 15 − (− 99.4)]°C (c) The mass flow rate is determined from Q refrig Q refrig Q refrig 12 kJ/s = = = = 0.237 kg/s m = qL h6 − h5 c p (T6 − T5 ) (1.005 kJ/kg ⋅ °C )[− 49 − (− 99.4 )]°C Chapter 11, Solution 71. The conditions at which an absorption refrigeration system operates are specified. The maximum COP this absorption refrigeration system can have is to be determined. Analysis The maximum COP that this refrigeration system can have is  T COPR,max = 1 − 0  Ts  TL   T − T L  0   298 K  273   = 1 −   = 2.64   393 K  298 − 273  