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THERMODYNAMICS - CASE STUDY SOLUTION

   

Refrigerant R-134a vaporizes in the evaporator and absorbs heat from the cooled space. Entropy changes of the Refrigerant R-134a, the cooled space and the whole process need to be determined.

     
Saturated R-134a
Temperature Table
Saturated R-134a Pressure Table<
< Superheated Steam Table
 

Refrigerant R-134a enters the evaporator as a saturated liquid-vapor mixture and leaves the evaporator as saturated vapor. Hence the temperature in the evaporator remains constant at the saturation temperature of 200 kPa, which is,

      Tsat. @ 200 kPa = -10.09oC

     

Heat Transferred into System A
 

(1) Determine the entropy change of refrigerant R-134a

Take refrigerant R-134a as a system (system A). It enters the evaporator as a saturated liquid-vapor mixture, which is state 1.

      P1 = 200 kPa
      T1 = Tsat. @ 200 kPa = -10.09oC

Refrigerant R-134a leaves the evaporator as saturated vapor, which is state 2.

      P2 = 200 kPa
      T2 = Tsat. @ 200 kPa = -10.09oC

During this process, refrigerant R-134 absorbs 150 kJ heat.

      Q = 150 kJ

Hence, the entropy change for refrigerant R-134a from process 1-2 is

      

Temperature T is a constant in this case, and the above equation can be integrated as

      

Substituting Q and T into the above equation yields,

      ΔSA = 150(1000)/(-10.09 + 273) = 570.56 J/K

     

Heat Transferred out of System B
 

(2) Determine the entropy change of the cooled space

Take the cooled space as a system (system B). During the heat transfer process, the space is maintained at -4oC and it dissipates 150 kJ heat to refrigerant R-134a. The entropy change of the cooled space can be determined as

      ΔSB =Q/T = -150(1000)/(-4 + 273) = -557.6 J/K

     


System C is an Isolated System
 

(3) Determine the total entropy change of the process

Take refrigerant R-134a and the cooled space together as a system (system C). Then refrigerant R-134a and the cooled space are subsystems of system C. The entropy change of system C is the sum of the entropy changes of its two subsystems.

      ΔSC = ΔSA+ ΔSB
             = 570.56 + (-557.6) = 12.44 J/K

System C is an isolated system since no heat and mass transfer out of its boundaries. The entropy change of system C is greater than 0, which satisfies the increase of entropy principle.