A 05m3 rigid tank contains refrigerant134a initially at 200

A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 35 degree C until the pressure rises to 400 kPa. Determine the entropy change of the refrigerant, the entropy change of the heat source, and the total entropy change for this process.

Solution

a) at 200 kPa and x = 0.4

    v1 = 0.0404 m^3 / kg

    h1 = 121 kJ/kg

    s1 = 468 J/kg-K

    v2 = v1 = 0.0404 m^3 /kg (because mass of R-134a is a constant)

    we get s2 = 781 J/kg-K , h2 = 215 kJ/kg-K

    Entropy change for refrigerant = mass * (s2 - s1)

                                                = 0.5 / 0.0404 * (781 - 468)   = 3.873 kJ/K

b) Entropy change for the heat source   =   - Q / T_source

                                                           = - mass * (h2 - h1) / T_source

                                                           = - 0.5 / 0.0404 * (215 - 121) / (273 + 35)    = -3.77716 kJ/K

c) entropy change for the process = 3.873 - 3.77716 = 0.095836 kJ/K