Thermodynamics

386 | Thermodynamics(c) The entropy generated during this process can be determined by applyingan entropy balance on an extended system that includes the iron block andits immediate surroundings so that the boundary temperature of the extendedsystem is at 285 K at all times:S in S out123Net entropy transferby heat and mass S gen ¢S system123Entropygeneration123Changein entropyorS gen Q outT b ¢S system Q outT bDiscussion The entropy generated can also be determined by taking the ironblock and the entire lake as the system, which is an isolated system, andapplying an entropy balance. An isolated system involves no heat or entropytransfer, and thus the entropy generation in this case becomes equal to thetotal entropy change,S gen ¢S total ¢S system ¢S lake 12.65 16.97 4.32 kJ>Kwhich is the same result obtained above. S gen ¢S system4838 kJ285 K 112.65 kJ>K2 4.32 kJ /KEXAMPLE 7–20Entropy Generation in a Mixing Chamber180 Btu/minT 1 = 50°F300 lbm/min MixingchamberP = 20 psiaT 2 = 240°FFIGURE 7–68Schematic for Example 7–20.T 3 = 130°FWater at 20 psia and 50F enters a mixing chamber at a rate of 300 lbm/minwhere it is mixed steadily with steam entering at 20 psia and 240F. Themixture leaves the chamber at 20 psia and 130F, and heat is lost to thesurrounding air at 70F at a rate of 180 Btu/min. Neglecting the changes inkinetic and potential energies, determine the rate of entropy generationduring this process.Solution Water and steam are mixed in a chamber that is losing heat at aspecified rate. The rate of entropy generation during this process is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0, E CV 0, and S CV 0. 2 There areno work interactions involved. 3 The kinetic and potential energies arenegligible, ke pe 0.Analysis We take the mixing chamber as the system (Fig. 7–68). This is acontrol volume since mass crosses the system boundary during the process.We note that there are two inlets and one exit.