Thermodynamics

Chapter 7 | 389temperature change so that the entire boundary of the extended system is atthe surrounding temperature of 25C. The entropy balance for this extendedsystem (system immediate surroundings) yieldsS in S out123Net entropy transferby heat and mass S gen ¢S system123 123EntropygenerationChangein entropyorS gen Q outT b ¢S system Q outT b S gen ¢S sysem600 kJ125 2732 K 11.61 kJ>K2 0.40 kJ /KThe entropy generation in this case is entirely due to irreversible heat transferthrough a finite temperature difference.Note that the entropy change of this extended system is equivalent to theentropy change of water since the piston–cylinder device and the immediatesurroundings do not experience any change of state at any point, and thusany change in any property, including entropy.Discussion For the sake of argument, consider the reverse process (i.e., thetransfer of 600 kJ of heat from the surrounding air at 25C to saturated waterat 100C) and see if the increase of entropy principle can detect theimpossibility of this process. This time, heat transfer will be to the water(heat gain instead of heat loss), and thus the entropy change of water will be1.61 kJ/K. Also, the entropy transfer at the boundary of the extendedsystem will have the same magnitude but opposite direction. This will resultin an entropy generation of 0.4 kJ/K. The negative sign for the entropygeneration indicates that the reverse process is impossible.To complete the discussion, let us consider the case where the surroundingair temperature is a differential amount below 100C (say 99.999 . . . 9C)instead of being 25C. This time, heat transfer from the saturated water tothe surrounding air will take place through a differential temperaturedifference rendering this process reversible. It can be shown that S gen 0 forthis process.Remember that reversible processes are idealized processes, and they canbe approached but never reached in reality.Entropy Generation Associatedwith a Heat Transfer ProcessIn Example 7–21 it is determined that 0.4 kJ/K of entropy is generated duringthe heat transfer process, but it is not clear where exactly the entropygeneration takes place, and how. To pinpoint the location of entropy generation,we need to be more precise about the description of the system, its surroundings,and the system boundary.In that example, we assumed both the system and the surrounding air tobe isothermal at 100°C and 25°C, respectively. This assumption is reasonableif both fluids are well mixed. The inner surface of the wall must also be