Question

A Refrigeration System Using R-134A In a refrigeration system, the refrigerant R-134A begins as saturated vapor at -15°(State 1). It then goes through a reversible adiabatic compressor to reach State 2. After flowing through the condenser (a heat exchanger), the refrigerant exits as saturated liquid at 70°C (State 3). It is then throttled by going through an expansion valve, to reach State 4. It finishes the cycle by going through another heat exchanger (the evaporator), to return to State 1. Condenser Throttle Evaporator 4 Write the energy equation for the compression process from State 1 to State 2. What is the pressure in State 1? What are the pressure and entropy in State 2? What is the temperature of the refrigerant in State 2? What is the specific work wc required to drive the compressor? Write the energy equation for the heat transfer qH from the condenser, State 2 to State3 What is the heat transfer out of the refrigerant, per kilogram, in the condenser? Write the energy equation for the throttling process from State 3 to State 4 What are the enthalpy, pressure, and quality of the refrigerant in State 4? Write the energy equation for the heat transfer qu into the evaporator, State 4 to State 1. What is the heat transfer into the evaporator? What is the coefficient of performance of the refrigeration system? What COP would a Carnot heat pump achieve between the same minimum and maximum temperatures?

Answer 1

2 s Sphopy 3 enthalpy /kg /. 639y 8ol9389.63 0.92559737/ A1-168 Зоу.2842g.CSI. 33321,6956 State 21.168 bor = 1.6394 bar at stak 2

T2 2 ろ43 Te= 1.03073 ろ43 nge-st 389.3 = 1.3658×353, STEアー389.28 ,-93.2377 ky爪 Energy equadion for ⑥ Heat trongfer out Hefrigeront m the Condenser =1.3658X 35 3.5567-y304. 28

9,,- hfe-,s.ct hfge-s.cxx, 59.25% vapour ondvetandiquid uiatian = 387-63-3 y.es 85.35 ю/s CoP