Question

Question 4 5 pts R134A at 30°C, 0.6 MPa is adiabatically throttled to 0°C, 0.2 MPa. The specific entropy generation is most closely equal to: O 0.2 kJ/kg-K O 0.0 kJ/kg-K 2.0 kJ/kg-K O 0.02 kJ/kg-K

Answer 1

Finding the required values of entropy from the properties table of R-134a we can solve this question as below-
as below Sol: The data provided in the question are Initial condition : 30°C Pressure p = 0.6 MPa ; Temperaturet final condition :- pressure, p = 0.2 mpa i TemperatureTze o'c We have to find out specific entropy Jeneration in the throtting process. Assumptions :- Steady state condition. is done Throttling adiabat atically KE and P.E changes are neglected Applying Entropy Principie Entropy Entropy change Entropy transfer - generation of the system ie Sgen =(52-5,) - S đQ T [ Sors, 5920 (Adiabatic process Therefore, specific entropy generation, igen 82-8, Now, found properties of the refrigerant from the R-134a can be out table At 0.6 Mpa f 30°C si 0.9499 kJ/gk. 0.9698 KJ 119k At s foc 0.2 mpa 2

specific entropy generation =82-8, 0.9698 -0.9499 0.0199 KJ Vaki 그 0.02 KJ ligk Ans - Therefore, option d is correct