“Compress the vapor, cool it; throttle thevapor, heat it” Refrigeration Cycles Review of Refrigeration Cycles Lecture 1 10 January 2018 ME 437 – HVAC VCC: Reversed Carnot T-s cycle Engr. Mustafa Mogri 2 10 January 2018 ME 437 – HVAC VCC: Reversed Rankine cycle Engr. Mustafa Mogri 3 10 January 2018 ME 437 – HVAC VCC: Stages of cycle Engr. Mustafa Mogri 4 10 January 2018 ME 437 – HVAC VCC: Stages of cycle (contd.) Engr. Mustafa Mogri 5 10 January 2018 ME 437 – HVAC Vapor Compression Problem A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -35°C by rejecting waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 50°C and leaves at the same pressure subcooled by 5°C. If the compressor consumes 3.3 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the refrigeration load, (c) the COP, and (d) the minimum power input to the compressor for the same refrigeration load. Engr. Mustafa Mogri 6 10 January 2018 ME 437 – HVAC Refrigeration Problem Solution: Problem Statement: Find the mass flow rate; refrigeration load; COP; minimum work input Data: TL = -35°C; Condenser: State 1: T1 = 18°C or 291K; State 2: T2 = 26°C or 299K; mc = 0.25 kg/s; Refrigerant: State 1: T1 = 50°C or 323K; P1 = 1.2 MPa; State 2: P2 = P1 = 1.2 MPa; T2 = Tsat - 5K = 41.3°C or 314.3K; Win = 3.3 kW; Engr. Mustafa Mogri 7 10 January 2018 ME 437 – HVAC Refrigeration Problem Energy balance of the condenser 𝐸ሶ 𝑖𝑛 = 𝐸ሶ 𝑜𝑢𝑡 𝑚ሶ 𝑐 ℎ1 + 𝑄ሶ 𝑐,𝑖𝑛 = 𝑚ሶ 𝑐 ℎ2 𝑄ሶ 𝑐,𝑖𝑛 = 𝑚ሶ 𝑐 ℎ2 − ℎ1 = 8.367 𝑘𝑊 Energy balance of the refrigerant 𝐸ሶ 𝑖𝑛 = 𝐸ሶ 𝑜𝑢𝑡 𝑚ሶ 𝑅 ℎ1 = 𝑚ሶ 𝑅 ℎ2 + 𝑄ሶ 𝑅,𝑜𝑢𝑡 𝑄ሶ 𝑅,𝑜𝑢𝑡 = 𝑚ሶ 𝑅 ℎ1 − ℎ2 Engr. Mustafa Mogri 8 10 January 2018 ME 437 – HVAC Refrigeration Problem Overall energy balance 𝑄ሶ 𝑅,𝑜𝑢𝑡 = 𝑄ሶ 𝑐,𝑖𝑛 𝑚ሶ 𝑅 ℎ1 − ℎ2 = 𝑚ሶ 𝑐 ℎ2 − ℎ1 ⟹ 𝑚ሶ 𝑅 = 0.0498 𝑘𝑔/𝑠 𝑄ሶ 𝑐,𝑖𝑛 = 𝑄ሶ 𝐻 𝑄ሶ 𝐻 = 𝑄ሶ 𝐿 + 𝑊ሶ 𝑖𝑛 ⟹ 𝑄ሶ 𝐿 = 5.07 𝑘𝑊 Engr. Mustafa Mogri 9 10 January 2018 ME 437 – HVAC Refrigeration Problem 𝑄ሶ 𝐿 𝐶𝑂𝑃 = = 1.54 𝑊ሶ 𝑖𝑛 1 𝐶𝑂𝑃𝑅,𝑟𝑒𝑣 = = 4.49 𝑇𝐻 ൗ𝑇 − 1 𝐿 𝑄ሶ 𝐿 𝐶𝑂𝑃 = ⟹ 𝑊ሶ 𝑚𝑖𝑛 = 1.13 𝑘𝑊 𝑊ሶ 𝑚𝑖𝑛 Engr. Mustafa Mogri 10 10 January 2018 ME 437 – HVAC Cascade Refrigeration Systems o Some industrial applications require moderately low temperatures, and the temperature range they involve may be too large for a single vapor compression refrigeration cycle to be practical. o A large temperature range also means a large pressure range in the cycle and a poor performance for a reciprocating compressor. o One way of dealing with such situations is to perform the refrigeration process in stages, that is, to have two or more refrigeration cycles that operate in series. o Such refrigeration cycles are called cascade refrigeration cycles. Engr. Mustafa Mogri 11 10 January 2018 ME 437 – HVAC Cascade Refrigeration Systems Engr. Mustafa Mogri 12 10 January 2018 ME 437 – HVAC Problem 11-57 [1] Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counter-flow heat exchanger where both streams enter at about 0.55 MPa. If the mass flow rate of the refrigerant through the upper cycle is 0.24 kg/s, determine (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space and the power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator. Engr. Mustafa Mogri 13 10 January 2018 ME 437 – HVAC Problem 11-57 [1] Solution: Problem Statement: Find the mass flow rate; rate of heat removal; COP Assumptions: Steady flow conditions exist Schematic: Engr. Mustafa Mogri 14 10 January 2018 ME 437 – HVAC Problem 11-57 [1] Data: The enthalpies of the refrigerant at each state point is determined from Tables A-11 thru A-13 State 1: Saturated vapor; P1 = 0.14 MPa; state 1 is completely defined. h1 =
[email protected] = 239.19 kJ/kg; s1 = sg; State 2: P2 = 0.55 MPa; s2s = s2 = s1; state 2 is completely defined. h2 = 267.37 kJ/kg; State 3: P3 = 0.55 MPa; saturated liquid; state 3 is completely defined. h3 =
[email protected] = 77.54 kJ/kg; State 4: P4 = 0.14 MPa; Liquid-vapor mixture; h3 = h4 = 77.54 kJ/kg; Engr. Mustafa Mogri 15 10 January 2018 ME 437 – HVAC Problem 11-57 [1] State 5: P5 = 1.4 MPa; Saturated vapor; state 5 is completely defined. h5 =
[email protected] = 260.98 kJ/kg; State 6: P6 = 0.8 MPa; s6s = s6 = s5; h6 = 268.71 kJ/kg; State 7: P7 = 0.8 MPa; Saturated liquid; state 7 is completely defined. h7 =
[email protected] = 95.48 kJ/kg; State 8: P8 = 0.55 MPa; Liquid-vapor mixture; h8 = h7 = 95.48 kJ/kg; Engr. Mustafa Mogri 16 10 January 2018 ME 437 – HVAC Problem 11-57 [1] State 5: P5 = 1.4 MPa; Saturated vapor; state 5 is completely defined. h5 =
[email protected] = 260.98 kJ/kg; State 6: P6 = 0.8 MPa; s6s = s6 = s5; h6 = 268.71 kJ/kg; State 7: P7 = 0.8 MPa; Saturated liquid; state 7 is completely defined. h7 =
[email protected] = 95.48 kJ/kg; State 8: P8 = 0.55 MPa; Liquid-vapor mixture; h8 = h7 = 95.48 kJ/kg; Engr. Mustafa Mogri 17 10 January 2018 ME 437 – HVAC Problem 11-57 [1] 𝑚ሶ 𝑒 ℎ𝑒 = 𝑚ሶ 𝑖 ℎ𝑖 𝑚ሶ 𝐴 ℎ5 − ℎ8 = 𝑚ሶ 𝐵 ℎ2 − ℎ3 ⟹ 𝑚ሶ 𝐵 = 0.2092 𝑘𝑔/𝑠 𝑄ሶ 𝐿 = 𝑚ሶ 𝐵 ℎ1 − ℎ4 = 33.9 𝑘𝑊 𝑄ሶ 𝐿 = 𝑊ሶ 𝑐𝑜𝑚𝑝𝐼,𝑖𝑛 + 𝑊ሶ 𝑐𝑜𝑚𝑝𝐼𝐼,𝑖𝑛 = 𝑚ሶ 𝐴 ℎ6 − ℎ5 + 𝑚ሶ 𝐵 ℎ2 − ℎ1 = 7.75 𝑘𝑊 𝑄ሶ 𝐿 𝐶𝑂𝑃𝑅 = = 4.36 𝑊ሶ 𝑛𝑒𝑡,𝑖𝑛 Engr. Mustafa Mogri 18