Entropy Generation Example
Example: Show that the heat can not transfer from the low-temperature sink to the high-temperature source based on the increase of entropy principle.
DS(source) = 2000/800 = 2.5 (kJ/K)
DS(sink) = -2000/500 = -4 (kJ/K)
Sgen= DS(source)+ DS(sink) = -1.5(kJ/K) < 0
It is impossible based on the entropy increase principle
Sgen?0, therefore, the heat can not transfer from low-temp. to high-temp. without external work input
- If the process is reversed, 2000 kJ of heat is transferred from the source to the sink, Sgen=1.5 (kJ/K) > 0, and the process can occur according to the second law
- If the sink temperature is increased to 700 K, how about the entropy generation? DS(source) = -2000/800 = -2.5(kJ/K)
DS(sink) = 2000/700 = 2.86 (kJ/K)
Sgen= DS(source)+ DS(sink) = 0.36 (kJ/K) < 1.5 (kJ/K)
Entropy generation is less than when the sink temperature is 500 K, less irreversibility. Heat transfer between objects having large temperature difference generates higher degree of irreversibilities