FERRER, John Emmanuel E. Assignment 1 ME141-2/C1 CN#12 3.2 A simple saturation ammonia compression system has a high pressure of 1.35 MN/m2 and a low pressure of 0.19 MN/m2. Find per 400,000 kJ/h of refrigerating capacity, the power consumption of the compressor and COP of the cycle. Vapor Compression Cycle LIQUID LINE DISCHARGE LINE CONDENSER β2 = 1710.92 β3 = 346.48 ππ½ ππ HIGH PRESSURE SIDE β1 = 1417.69 EXPANSION DEVICE β4 = 346.48 ππ½ ππ π ππ COMPRESSOR Suction Line EVAPORATOR EXPANSION LINE ππ½ ππ LOW PRESSURE SIDE Pressure-Enthalpy Diagram of Ammonia 3 2 P 4 β3 = β4 = 346.48 ππ½ /ππ 1 β1 = 1417.69 ππ½ β = 1710.92 ππ½ 2 ππ ππ Given Data: P1 = 0.19 MN/m2 = 0.19 MPa = 1.9 bar P2 = 1.35 MN/m2 = 1.35 MPa = 13.5 bar Ref. Capacity = 400,000 kJ/h Getting for enthalpies: Interpolate β1 = β ππ‘ 1.9 πππ 1434.4−β1 1.7441−1.9 ππ± values, = → π = ππππ. ππ π 1434.4−1437.2 1.7441−1.9074 ππ π 1 = π ππ‘ 1.9 πππ Interpolate values, 361.2−β 5.9336−π 1 5.9336−5.9025 13.139−13.5 1.7441−1.9 ππ± = 1.7441−1.9074 → ππ = π. ππππ πππ² ππ± Interpolate values, 361.2−366.1 = 13.139−13.522 → ππ = ππ = πππ. ππ ππ Pressure Enthalpy Chart for R717(Ammonia): According to chart, h2 = 1730 kJ/kg Solution: π = π(β2 − β1 ) ππ = πππ → π = ππ ππ ππ = β1 − β4 ππ½ Finding the value of qo, ππ = β1 − β4 = 1437.07 − 365.82 = 1071.25 ππ The value for Qo in kJ/min, ππ = π 6666.67 ππ½ β 400000 60 πππ ππ½ = 6666.67 πππ ππ The value for m, π = ππ = 1071.25 = 6.22 πππ π a.) Getting the value of Power Consumption (W): π = π(β2 − β1 ) = 6.22 ππ ππ½ ππ½ ππ± ππ± (1730 − 1437.07 ) = ππππ. ππ = ππ. ππ ππ ππΎ min ππ ππ πππ π b.) Getting the value of COP: β1 −β4 πΆππ = β 2 −β1 = ππ½ ππ½ −365.82 ππ ππ ππ½ ππ½ 1730 −1437.07 ππ ππ 1437.07 = π. ππ ______________________________________________________________________ 3.4 In a vapor compression cycle saturated liquid Refrigerant 22 leaving the condenser at 40°C is required to expand to the evaporator temperature of 0°C in a cold storage plant. (a) Determine the percentage saving in network of the cycle per kg of the refrigerant if an isentropic expander could be used to expand the refrigerant in place of the throttling device. (b) Also determine the percentage increase in refrigerating effect per kg of refrigerant as a result of use of the expander. Assume that compression is isentropic from saturated vapor state at 0°C to the condenser pressure. Vapor Compression Cycle LIQUID LINE CONDENSER DISCHARGE LINE β2 = 432 ππβ β3 = 249.08 ππ½ ππ ππ½ ππ HIGH PRESSURE SIDE EXPANSION DEVICE β1 = 405.36 ππ½ ππ COMPRESSOR Suction Line β4 = 249.08 ππ½ ππ EVAPORATOR EXPANSION LINE πβ LOW PRESSURE SIDE Pressure-Enthalpy Diagram of Ammonia 3 40β 2 P 4 β3 = β4 1 0β β1 β2 Given Data: T1 = 0β°C T3 = 40β°C Enthalpies: ππ½ ππ ππ½ π 1 = π π ππ‘ 0β = 1.7518 πππΎ β1 = βπ ππ‘ 0β = 405.36 β3 = β4 = βπ ππ‘ 40β = 249.08 π3 = π2 = 15.335 πππ ππ½ ππ Pressure Enthalpy Chart for R22: According to chart, h2 = 432 kJ/kg 3.6 A standard vapor compression cycle using Freon 22 operates on simple saturation cycle at the following conditions: Refrigerating capacity 15 TR Condensing temperature 40°C Evaporating temperature 5°C Calculate: (a) Refrigerant circulation rate in kg/s. (b) Power required by the compressor in kW. (c) Coefficient of performance. (d) Volume flow rate of the refrigerant at compressor suction. (e) Compressor discharge temperature. (f) Suction vapor volume and power consumption per ton of refrigeration. Refrigeration engineers assume that if this Freon 22 compressor is used with R 134a, its capacity would fall by about 40%. Examine this assumption by doing cycle analysis for R134a with the same compressor. Vapor Compression Cycle LIQUID LINE CONDENSER DISCHARGE LINE β2 = 430 ππβ β3 = 249.08 ππ½ ππ ππ½ ππ HIGH PRESSURE SIDE EXPANSION DEVICE β1 = 407.11 ππ½ ππ COMPRESSOR Suction Line β4 = 249.08 ππ½ ππ EVAPORATOR EXPANSION LINE πβ LOW PRESSURE SIDE Pressure-Enthalpy Diagram of R22 40β 3 2 P 4 1 5β β3 = β4 β1 β2 Given Data: T1 = 5β°C T3 = 40β°C Qo = 15 TR = 15(211 kJ/min) = 3165 kJ/min Getting Enthalpies: Solving h1, v1 and s1 by interpolation, 4−5 406.77 − β1 0.0416 − π£1 1.7460 − π 1 = = = 4 − 6 406.77 − 407.45 0.0416 − 0.0391 1.7460 − 1.7432 ππ½ β1 = βπ @ 5β = 407.11 ππ π3 π£1 = π£π @ 5β = 0.0404 ππ ππ½ π 1 = π π @ 5β = 1.7446 πππΎ ππ½ β3 = β4 = βπ @ 40β = 249.08 ππ π3 = 15.335 πππ Pressure Enthalpy Chart for R22: π΄ππππππππ π‘π πβπππ‘, β2 = 430 ππ½ ππ Solution: a.) Refrigerant circulation rate in kg/s ππ = π(β1 − β4 ) → π = π= ππ½ 3165 min 407.11 ππ β1 − β4 ππ½ ππ½ − 249.08 ππ ππ = 20.03 ππ ππ = π. ππππ πππ π b.) Power required by the compressor in kW π = π(β2 − β1 ) = 0.3338 ππ ππ½ ππ½ (430 − 407.11 ) = π. ππ ππΎ π ππ ππ c.) Coefficient of Performance ππ½ ππ½ β1 − β4 407.11 ππ − 249.08 ππ πΆππ = = = π. πππ ππ½ ππ½ β2 − β1 430 − 407.11 ππ ππ d.) Volume flow rate of the refrigerant at compressor suction ππ π3 ππ π1 = ππ£1 = 0.3338 (0.0404 ) = π. ππππ π ππ π e.) Compressor discharge temperature π»π = ππβ f.) Suction vapor volume and power consumption per ton of refrigeration. π3 1000 πΏ π³ π1 = 0.0404 ( ) = ππ. π ππ 1 π3 ππ 7.64 ππ ππΎ π= = π. ππππ 15 π‘πππ ππππ Refrigeration engineers assume that if this Freon 22 compressor is used with R 134a, its capacity would fall by about 40%. Examine this assumption by doing cycle analysis for R134a with the same compressor. Vapor Compression Cycle LIQUID LINE CONDENSER DISCHARGE LINE β2 = 420 ππβ β3 = 256.41 ππ½ ππ ππ½ ππ HIGH PRESSURE SIDE EXPANSION DEVICE β1 = 401.49 ππ½ ππ COMPRESSOR Suction Line β4 = 256.41 ππ½ ππ EVAPORATOR EXPANSION LINE πβ LOW PRESSURE SIDE Pressure-Enthalpy Diagram of R134a 40β 3 2 P 4 1 5β β3 = β4 β1 β2 Given Data: T1 = 5β°C T3 = 40β°C Qo would fall by 40% = 15TR(0.4)=6TR Qo =15TR – 6TR = 9TR = 9TR(211 kJ/min) =1899 kJ/min Getting Enthalpies: Solving h1, v1 and s1 by interpolation, 4−5 400.92 − β1 0.06039 − π£1 1.7250 − π 1 = = = 4 − 6 400.92 − 402.06 0.06039 − 0.05644 1.7250 − 1.7240 ππ½ β1 = βπ @ 5β = 401.49 ππ π3 π£1 = π£π @ 5β = 0.0584 ππ ππ½ π 1 = π π @ 5β = 1.7245 πππΎ ππ½ β3 = β4 = βπ @ 40β = 256.41 ππ π3 = 1.0166 πππ Pressure Enthalpy Chart for R134a: π΄ππππππππ π‘π πβπππ‘, β2 = 420 ππ½ ππ Solution: a.) Refrigerant circulation rate in kg/s ππ = π(β1 − β4 ) → π = π= ππ½ 1899 min 401.49 ππ β1 − β4 ππ½ ππ½ − 256.41 ππ ππ = 13.09 ππ ππ = π. ππππ πππ π b.) Power required by the compressor in kW π = π(β2 − β1 ) = 0.2182 ππ ππ½ ππ½ (420 − 401.49 ) = π. ππ ππΎ π ππ ππ c.) Coefficient of Performance ππ½ ππ½ β1 − β4 401.49 ππ − 256.41 ππ πΆππ = = = π. πππ ππ½ ππ½ β2 − β1 420 − 401.49 ππ ππ d.) Volume flow rate of the refrigerant at compressor suction ππ π3 ππ π1 = ππ£1 = 0.2182 (0.0584 ) = π. ππππ π ππ π e.) Compressor discharge temperature π»π = ππβ f.) Suction vapor volume and power consumption per ton of refrigeration. π3 (1000 πΏ) π³π π = (0.0584 ) ( ) = ππ. π ππ 1 π3 ππ 4.04 ππ ππΎ π= = π. ππππ 9 π‘πππ ππππ 3.8 A R134a machine operates at –15°C evaporator and 35°C condenser temperatures. Assuming a simple-saturation cycle, calculate the volume of the suction vapor and power consumption per ton of refrigeration and COP of the cycle. Calculate the same if the system has a regenerative heat exchanger with the suction vapor leaving at 20°C from the heat exchanger. Vapor Compression Cycle LIQUID LINE CONDENSER DISCHARGE LINE β2 = 420 ππβ β3 = 249.01 ππ½ ππ ππ½ ππ HIGH PRESSURE SIDE EXPANSION DEVICE β1 = 389.63 ππ½ ππ COMPRESSOR Suction Line β4 = 249.01 ππ½ ππ EVAPORATOR EXPANSION LINE −ππβ LOW PRESSURE SIDE Pressure-Enthalpy Diagram of R22 35β 3 2 P 4 1 −15β β3 = β4 β1 β2 Given Data: T1 = -15β°C T3 = 35β°C Getting Enthalpies: Solving h1, v1 and s1 by interpolation −16 − (−15) 389.02 − β1 0.12551 − π£1 1.7379 − π 1 = = = −16 − (−14) 389.02 − 390.24 0.12551 − 0.11605 1.7379 − 1.7363 ππ½ β1 = βπ @ − 15β = 389.63 ππ π3 π£1 = π£π @ − 15β = 0.12078 ππ ππ½ π 1 = π π @ − 15β = 1.7371 πππΎ Solving h3 = h4 and P3 by interpolation 34 − 35 247.54 − β 0.86263 − π = = 34 − 36 247.54 − 250.48 0.86263 − 0.91185 ππ½ β3 = β4 = βπ @35β = 249.01 ππ π3 = 0.88724 πππ Pressure Enthalpy Chart for R134a: β2 = 426 ππ½ ππ Solution: a.) Volume of the Suction Vapor ππ½ 1 πππ 211 β ππ πππ 60 π = 0.025 ππ π= = β1 − β4 (389.63 − 249.01) π π1 = ππ£1 = 0.025(0.12078) = π. ππππ × ππ −π ππ π b.) π = π(β2 − β1 ) = 0.025(426 − 389.63) = π. πππππ ππΎ c.) COP ππ β1 − β4 (389.63 − 249.01) ππ πΆππ = = = π. ππππ ππ β2 − β1 (426 − 389.63) ππ With Heat Exchangers: Vapor Compression Cycle β3 = 249.01 β2 = 468 ππ½ ππ CONDENSER ππ½ ππ ππβ β1 = 420.34 ππ½ ππ HEAT EXCHANGER β6 = 389.63 β4 = 218.3 ππ½ ππ ππ½ ππ EVAPORATOR −ππβ EV β5 = 218.3 ππ½ ππ COMPRESSOR Pressure Enthalpy Diagram of R134a 35β 4’ P 2’ 3 6 1 −15β 5 20β β4 = β5 β3 β6 β1 β2 Given Data: β1 = β@20β(π π’πππβπππ‘ππ π‘ππππ) = 420.34 π 1 = π @20β(π π’πππβπππ‘ππ π‘ππππ) = 1.8864 πΎπ πππΎ π£1 = π£@20β(π π’πππβπππ‘ππ π‘ππππ) = 0.2304 β3 = βπ @35β = 249.01 ππ½ ππ Heat Balance for Heat Exchanger: β3 + β6 = β1 + β4 249.01 + 389.63 = 420.34 + β4 β4′ = 218.3 ππ½ ππ ππ½ ππ π3 ππ Pressure Enthalpy Chart for R134a: β2′ = 468 ππ½ ππ Solution: a.) Volume of suction vapor ππ½ 1 πππ 211 ∗ ππ ππ ππ 60 π π= = = 0.02053 β6 − β5 389.63 − 218.3 π ππ π = ππ£1 = 0.02053(0.2304) = π. πππππ × ππ π π b.) π = π(β2 − β1 ) = 0.02053(468 − 420.34) = π. ππππ ππΎ c.) COP 389.63 − 218.3 = π. ππππ β2 − β1 468 − 420.34 ______________________________________________________________________ πΆππ = β6 − β5 = 3.10 A simple saturation cycle using Freon 22 is designed for a load of 100 TR. The saturated suction and discharge temperatures are 5°C and 40°C respectively. Calculate: (a) The mass flow rate of refrigerant. (b) The COP and isentropic horsepower. (c) The heat rejected in the condenser. Use the following data: t P hf hg β°C bar kJ/kg kJ/kg 5 5.836 205.9 407.1 40 15.331 249.53 416.4 Specific heat of vapor is 0.65 kJ/(kg.K). sf kJ/(kg.K) 1.02115 1.16659 sg kJ/(kg.K) 1..7447 1.69953 vg m3/kg 0.0404 Vapor Compression Cycle LIQUID LINE DISCHARGE LINE CONDENSER β2 = 433.5 ππβ β3 = 249.53 ππ½ ππ ππ½ ππ HIGH PRESSURE SIDE EXPANSION DEVICE β1 = 407.1 ππ½ ππ COMPRESSOR Suction Line β4 = 249.53 ππ½ ππ EVAPORATOR EXPANSION LINE πβ LOW PRESSURE SIDE Pressure-Enthalpy Diagram of R22 3 40β 2 P 4 β3 = β4 1 5β β1 β2 Given Data: RE = 100 TR = 100(211 kJ/min) = 21100 kJ/min ππ½ ππ ππ½ π 1 = π π @5β = 1.7447 ππ π3 π£1 = π£π @5β = 0.0404 ππ β1 = βπ @5β = 407.1 β3 = β4 = βπ @40β = 249.53 π3 = 15.331 πππ ππ½ ππ Pressure Enthalpy Chart for R22: β2 = 433.5 ππ½ ππ Solution: a.) The mass flow rate of refrigerant ππ½ 1 πππ (21100 min) ( 60 π ) π πΈ ππ π= = = π. ππππ ππ½ β1 − β4 π (407.1 − 249.53) ππ b.) The COP and isentropic horsepower πΆππ = β1 − β4 407.1 − 249.53 = = π. ππππ β2 − β1 433.5 − 407.1 π = π(β2 − β1 ) = 2.2318(433.5 − 407.1) = 58.0268 ππ ( 1 βπ ) = ππ. ππππ ππ 0.746 ππ c.) The heat rejected in the condenser. ππ = π(β2 − β3 ) = 2.318(433.5 − 249.53) = πππ. ππ ππΎ
