Question

The company would like you to estimate the boundary work per kg that is needed to supply this process. Assume the process is adiabatic (no heat transfer) and is treated as a closed system. Also, estimate the energy and volume needed for this cylinder if the typical mass compressed in the cylinder is 15.0 kg. Assume the compressor uses R134a.

Problem: Pistons for Ultra Low Refrigeration Applications A cryogenics company is really excited to be expanding as more and more clients are coming to them with the belief that living forever may be possible with modern medical advances that may occur in the future. They want to freeze themselves cryogenically and be awoken when those medical advances occur. However, this cryogenic company needs specialized compressors in part of their proprietary process for removing heat from part of the system. The compressor will be used in vapor-compression refrigeration cycle (only vapor is allowed in the compressor). The process start at the lowest temperature but not at the cryogenic temperatures. The cryogenic company is only willing to reveal the types of operating conditions they expect the compressor to see which only one part of their cryogenic process. The compressor is meant to take in saturated vapor at -40 °C, -20°C, -10 °C, and 0 °C and compress it to the following superheated conditions, P = 0.80 MPa and 60°C, and 1.0 MPa and 60°C, and 2.0 MPa and and 60°C. The company would like you to estimate the boundary work per kg that is need to supply this process. Assume the process is adiabatic (no heat transfer) and is treated like a closed system. Also, estimate the energy and volume needed for this cylinder if the typical mass compressed in the cylinder is 15.0 kg. Assume the compressor uses R134a. -Refrigerant Piston Connecting rod Crank shaft Rotated by an electric motor Figure: Typical compressor and its mode of operation

Answer 1

-40°C CO ieces -20°C 2 co o'smpa, 60°c Impa, 60°C 2mpa 1600 cm ampreggongo -10° 3 0°c (Superheared Condition saturated Vapour massa 15 kg Inlet -40°C at I 0.3608 m3 / If Teg ( Saturated Vapour) 0, his hy = 125-86 Kylky V = VG kg @ -20°c (saturated Vapour) at mlet ha = hy = 238.41 V2 - Ug = 0.147029 - 014709 m3lky -0°c (saturated Vapour) 244151 KI kg Vg E Vg 20.099516 mg/kg At inlet ③ 0°C ( saturated vapour) V4= 0.069959 milky hu= 450.45 bylky. 3) at mlet ha = hy aroutlet , ( supher heated stage) Loispa, 60c] 33 lkg 15-0002 997 3 ho = 296.81 kilky

ar outlet , imla, Impa, 6°C ( Super heated stage) V6 20021 796 milky ho = 28.2. 74 at outlet drapa, fic (super Kulky. heated Stage) m3 kg 1 병 9.4 6 8 10 ht = 287-3KT /icg Forms ist law of thermo dynamics ^ [hothetha) tom m [Chi) thetha that time OPE 30. Since OKE=0 Swimpressor m [ hithe the thy-he-he-ha] weimpressos = 154 | 285-867 838-414 244-57 +250-45 296.81–282.74–287.3 ] KT 1385.7 KJ Wempressure 1•3857 MI minimum volume Requiner Chenge in the volume of Refrigerenz R-1349. bylhe compressor

Volume of compressora Change in volume = in initial rolume- final volume] 515[ Wit Vet Vzut Ves -V5-V6 V7 [av] cump compressov = 15 [019608470 14 7 296 0-3608170.14 72940'099516 to 069259 0 029973 -0021796–9146310 M93 11 9.36234 volume of cempressor 1385,7 kJ Weempressor 1.3857 MJ