Question

R-134a is contained in a frictionless piston-cylinder device. The initial temperaure of the mixture is 39.37 oC. Over an hour 400 kJ of thermal energy is transferred to the roon which is maintained at a constant temperature of 22 oC. The condensation process is internally reversible. Determine the total entropy generation during this thermal energy process

Answer 1

Solution :

Given :

Initial temperature of the refrigerant is :

$T_i=39.37 ^ \circ C$

   = ( 39.37 + 273 ) K

  = 312.3 K

Room which is maintained at constant temperature is :

$T_o=22 ^ \circ C$

    = (22+273) K

    = 295 K

The thermal energy transferred to the room is :

Q = 400 kJ

   = $400 \times 10^3 \ J$

Therefore, the total entropy generation during the thermal energy process is :

$\Delta S =\left[\frac{-Q}{T_i}+ \frac{+Q}{T_i}\right]$

Here, -Q = heat is leaving the system maintained at a temperature of $T_i$ K.

 +Q = heat is entering the system maintained at a temperature of $T_o$ K.    

Therefore, substituting the values :

$\Delta S =\left[\frac{-400\times 10^3}{312.3}+ \frac{400\times 10^3}{295}\right]$

      = [-1280.8197 + 1355.9322]

      = 75.1125 J/K

       = 0.0751125 kJ/K

      = 0.075 kJ/K