Thermodynamics

Chapter 7 | 339The entropy change for each reservoir can be determined from Eq. 7–6since each reservoir undergoes an internally reversible, isothermal process.(a) For the heat transfer process to a sink at 500 K:¢S source Q sourceT source2000 kJ800 K 2.5 kJ>K SEE TUTORIAL CH. 7, SEC. 3 ON THE DVD.and¢S sink Q sinkT sinkS gen ¢S total ¢S source ¢S sink 12.5 4.02 kJ>K 1.5 kJ/K2000 kJ500 K4.0 kJ>KTherefore, 1.5 kJ/K of entropy is generated during this process. Noting thatboth reservoirs have undergone internally reversible processes, the entireentropy generation took place in the partition.(b) Repeating the calculations in part (a) for a sink temperature of 750 K,we obtain¢S source 2.5 kJ>kand¢S sink 2.7 kJ>KS gen ¢S total 12.5 2.72 kJ>K 0.2 kJ/KThe total entropy change for the process in part (b) is smaller, and thereforeit is less irreversible. This is expected since the process in (b) involves asmaller temperature difference and thus a smaller irreversibility.Discussion The irreversibilities associated with both processes could beeliminated by operating a Carnot heat engine between the source and thesink. For this case it can be shown that S total 0.7–3 ■ ENTROPY CHANGE OF PURE SUBSTANCESEntropy is a property, and thus the value of entropy of a system is fixedonce the state of the system is fixed. Specifying two intensive independentproperties fixes the state of a simple compressible system, and thus thevalue of entropy, as well as the values of other properties at that state. Startingwith its defining relation, the entropy change of a substance can beexpressed in terms of other properties (see Sec. 7–7). But in general, theserelations are too complicated and are not practical to use for hand calculations.Therefore, using a suitable reference state, the entropies of substancesare evaluated from measurable property data following rather involved computations,and the results are tabulated in the same manner as the otherproperties such as v, u, and h (Fig. 7–10).The entropy values in the property tables are given relative to an arbitraryreference state. In steam tables the entropy of saturated liquid s f at 0.01°C isassigned the value of zero. For refrigerant-134a, the zero value is assignedto saturated liquid at 40°C. The entropy values become negative at temperaturesbelow the reference value.INTERACTIVETUTORIAL