Question

120; = ? 2 (P2 = 1000 kPa; T2 = 65 °C) Problem 5. A compression refrigeration cycle (see Figure) has R-134a as the refrigerant. The mass flow rate is 3 kg/min. The refrigerant goes through isothermal evaporation in the evaporator and leaves the evaporator at -20 °C as saturated vapor. It enters the condenser with a pressure of 1 MPa and a temperature of 65 °C. Assume no losses in the pipelines connecting different components. Also assume steady state and the changes in kinetic energy and potential energy are negligible. Please answer the following questions: Condense (isobaric) | 3 (x3=0) I Expansion valve W, = ? Compressort 4 (X4 = 0.2) 1 (T1 = -20 °C; X1 = 1) Evaporator Kisothermal (a) If the pipe entering the compressor has a diameter of 0.05 m, what is the flow velocity there (at location 1)? (b) If the compressor works adiabatically, what is the rate of work input? Please include the sign convention in your answer. (c) If refrigerant undergoes isobaric condensation to saturated liquid in the condenser, what is the rate of heat transfer in the condenser? Please include the sign convention in your answer. (d) The expansion valve expands the refrigerant to a quality of x4 = 0.2. What is rate of heat transfer to or from the valve? (e) What is COP of this refrigeration cycle? What is theoretical maximum possible COP of this cycle? Is this cycle possible?

Answer 1

Qaz DATEIDID condense isobaric) 72 (p=10ookPa, 3 +(*3=0) T = 656) Expansion A valve +4(X4=0:2) Εναλογα - 1 (T, =20, X2 = 1) Given - Refrigerant is R134a I mass flow rate, m = 3kg/min TE-20c 1 P2 = 1 MPa = 10bar T2 = 65°c - 2 2 s & procey 2:2 - I sentropic compression. Process 2-3 i-Constant Pressure heat rejection process 3-4 r Iso-enthalpic expansion.

Process 4-1:- constant temperature heat addition, I From refrigerant table R134a . at T = zic e h, 386.55 kitke T2 = 65C 62 = 427.82 kolice. at P = P₂ = 10bar, h = 256.41 kille = h4 at T = T: 20¢ ha-173-64 kitleg. (a) Flow Velocity at location 1 ? Given diameter of pipe , d = 0.05m. lic Area of pipe , A = Td ² - 1.96810 m² we know that, discharge through pipe, Q== Ax 1 s 1 Also discharge, Q = ngxt *FP-6.7845 (60 sec 6.7845light at T,--doc = (3kg on substituting in een 6 . 70.34 m2 =(0.96xb} no =) ( = 173 m/sec INC

b) rate of work input to the compressor, Rate of work input to compressor = compressor work x mass flow rate ..5 (427.82 – 386.55) K X3kg) 60 sec - = 2:06 KTL 962 Joel = 10 52.06 kwi 1. Rate of work input to compressor = -2.06ko negative sign is included because workis done on the system... c) Rate of heat transfer in the conden for - = (Heat rejected & mass flow rate of refrigerant = (h2 - by x nig = (427.82-256.41) kettligt 3 kg = 8.67 ksfisec = 8.67 ko 4. Rate of heat transfer in the condensor ___ = 48.67KW.. negative sign is included because heat is released in the process. g 60 see

(d) Heat transfer in expansion Value. As the expansion of liquid refrigerant I is happening iso- enthalpically, there 1 iu no heat transfer to or from the I Talwe. (e) cop of refrigeration cycle cop = Heat extracted compressor work -56-64) . Ziarovas (ha-hi). (386.55-256-41) (427.82-386.55) <cop = 3.15 )

In order to solve the problem, I have taken the values of specific enthalpy at different locations and density of refrigerant from R134a property table.