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Subcooler (outside air The air temp: 20°C right cools the interior at a rate of 10 kW and has the following state information: R134a refrigerant State State State State State 2 4 3 1 5 400 1600 1600 1600 400 р P2= Throttle (ndiabatic) [kPa] 16 MPa T2-70°C T 8.9 70 57.9 32 8.9 Compressor [°C] PS P1-3 (ndinbatic) P1-400 kPa 1 sat vapor h 403.7 441.4 284.1 244.6 244.6 [kJ/kg] Devap x 1.0 0.0 0.2 លោ = -10°C Notice that the subcooler (process 3->4) is oversized (the outlet temperature is the same as the outdoor temperature). It is designed this way to compensate for the times when the hot water heater cannot accept more energy. At some point, the hot water heater will reach the maximum temperature, and any further heat will create the danger of scalding users in the home. Bypassing the hot water heater results in the following states: State State State State State 1 3 4 5 2 1600 1600 400 1600 400 р [kPa] T 8.9 70 70 57.9 8.9 [°C] 403.7 441.4 441.4 284.11 284.11 h [kJ/kg] 0.0 0.37 X 1.0 a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the evaporator. Is this more or less than the flow rate needed for the original cycle? Qevup, 0.013 (403.7-210.4) = 2 kW. Qevap = (1/1). b) Find the coefficient of performance for both cycles. CUP, 4.55 COP=0.246 c) Is the second cycle effective? exfective veru temp: 32°C) Fan P3 P2 sat. liquid P4=P2 henter 150 liters (max temp: 50°C) var hot-water nir flow 4 AT air

Subcooler (outside air The air temp: 20°C right cools the interior at a rate of 10 kW and has the following state information: R134a refrigerant State State State State State 2 4 3 1 5 400 1600 1600 1600 400 р P2= Throttle (ndiabatic) [kPa] 16 MPa T2-70°C T 8.9 70 57.9 32 8.9 Compressor [°C] PS P1-3 (ndinbatic) P1-400 kPa 1 sat vapor h 403.7 441.4 284.1 244.6 244.6 [kJ/kg] Devap x 1.0 0.0 0.2 លោ = -10°C Notice that the subcooler (process 3->4) is oversized (the outlet temperature is the same as the outdoor temperature). It is designed this way to compensate for the times when the hot water heater cannot accept more energy. At some point, the hot water heater will reach the maximum temperature, and any further heat will create the danger of scalding users in the home. Bypassing the hot water heater results in the following states: State State State State State 1 3 4 5 2 1600 1600 400 1600 400 р [kPa] T 8.9 70 70 57.9 8.9 [°C] 403.7 441.4 441.4 284.11 284.11 h [kJ/kg] 0.0 0.37 X 1.0 a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the evaporator. Is this more or less than the flow rate needed for the original cycle? Qevup, 0.013 (403.7-210.4) = 2 kW. Qevap = (1/1). b) Find the coefficient of performance for both cycles. CUP, 4.55 COP=0.246 c) Is the second cycle effective? exfective veru temp: 32°C) Fan P3 P2 sat. liquid P4=P2 henter 150 liters (max temp: 50°C) var hot-water nir flow 4 AT air

Refrigeration and Air Conditioning Technology (MindTap Course List)
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter46: Room Air Conditioners
Section: Chapter Questions
Problem 15RQ
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not water henter. 150 liters (mnx temp: 50"C) The air conditioning system Home inside right cools the interior at a rate of 10 kW and has the following state information: State Subcooler (outside air temp: 32°C) air temp: 20°C RI34a refrigerant State State State | Ps P2 sat. liquid State 1 3 4 Qhol-water 1600 1600 400 P2= O L6 MPa T2 - 70°C 400 1600 P4 = P2 Throttle (adiabatie) [kPa] T 8.9 70 57.9 32 8.9 Compressor PS = P1 O (ndinbatie) Fan [°C] h Pl= 400 kPa 1) sat. vopor 403.7 441.4 284.1 244.6 244.6 [kJ/kg] nir low Devap 1.0 0.0 0.2 AT nir =-10°C Notice that the subcooler (process 3->4) is oversized (the outlet temperature is the same as the outdoor temperature). It is designed this way to compensate for the times when the hot water heater cannot accept more energy. At some point, the hot water heater will reach the maximum temperature, and any further heat will create the danger of scalding users in the home. Bypassing the hot water heater results in the following states: State State State State State 1 4 400 1600 1600 1600 400 [kPa] T 8.9 70 70 57.9 8.9 [°C] h 403.7 441.4 441.4 284.11 284.11 [kJ/kg] 0.0 0.37 1.0 a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the evaporator. Is this more or less than the flow rate needed for the original cycle? Aevup, - 0.013 - (403.7-210.1)=2 kw, Qevar. = b) Find the coefficient of performance for both cycles. CUP,-4.55 COP-0.746 c) Is the second cycle effective? Not very efective
Transcribed Image Text:not water henter. 150 liters (mnx temp: 50"C) The air conditioning system Home inside right cools the interior at a rate of 10 kW and has the following state information: State Subcooler (outside air temp: 32°C) air temp: 20°C RI34a refrigerant State State State | Ps P2 sat. liquid State 1 3 4 Qhol-water 1600 1600 400 P2= O L6 MPa T2 - 70°C 400 1600 P4 = P2 Throttle (adiabatie) [kPa] T 8.9 70 57.9 32 8.9 Compressor PS = P1 O (ndinbatie) Fan [°C] h Pl= 400 kPa 1) sat. vopor 403.7 441.4 284.1 244.6 244.6 [kJ/kg] nir low Devap 1.0 0.0 0.2 AT nir =-10°C Notice that the subcooler (process 3->4) is oversized (the outlet temperature is the same as the outdoor temperature). It is designed this way to compensate for the times when the hot water heater cannot accept more energy. At some point, the hot water heater will reach the maximum temperature, and any further heat will create the danger of scalding users in the home. Bypassing the hot water heater results in the following states: State State State State State 1 4 400 1600 1600 1600 400 [kPa] T 8.9 70 70 57.9 8.9 [°C] h 403.7 441.4 441.4 284.11 284.11 [kJ/kg] 0.0 0.37 1.0 a) Determine the mass flow rate necessary to maintain a heat transfer rate of 10 kW in the evaporator. Is this more or less than the flow rate needed for the original cycle? Aevup, - 0.013 - (403.7-210.1)=2 kw, Qevar. = b) Find the coefficient of performance for both cycles. CUP,-4.55 COP-0.746 c) Is the second cycle effective? Not very efective
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