Thermodynamics

Chapter 7 | 357P 2 = 600 kPaT 2 = 330 KAIRCOMPRESSORT2P 2 = 600 kPaP 1 = 100 kPaFIGURE 7–34Schematic and T-s diagram forExample 7–9.1P 1 = 100 kPaT 1 = 290 KsEXAMPLE 7–9Entropy Change of an Ideal GasAir is compressed from an initial state of 100 kPa and 17°C to a final stateof 600 kPa and 57°C. Determine the entropy change of air during this compressionprocess by using (a) property values from the air table and (b) averagespecific heats.Solution Air is compressed between two specified states. The entropychange of air is to be determined by using tabulated property values andalso by using average specific heats.Assumptions Air is an ideal gas since it is at a high temperature and lowpressure relative to its critical-point values. Therefore, entropy change relationsdeveloped under the ideal-gas assumption are applicable.Analysis A sketch of the system and the T-s diagram for the process aregiven in Fig. 7–34. We note that both the initial and the final states of airare completely specified.(a) The properties of air are given in the air table (Table A–17). Reading s°values at given temperatures and substituting, we findP 2s 2 s 1 s° 2 s° 1 R lnP 1 311.79783 1.668022 kJ>kg # K4 10.287 kJ>kg # K2 ln600 kPa 0.3844 kJ/kg # K(b) The entropy change of air during this process can also be determinedapproximately from Eq. 7–34 by using a c p value at the average temperatureof 37°C (Table A–2b) and treating it as a constant:T 2s 2 s 1 c p,avg ln R lnT 1330 K 11.006 kJ>kg # K2 ln290 K 10.287 kJ>kg 600 kPa# K2 ln100 kPa 0.3842 kJ/kg # KP 2P 1100 kPaDiscussion The two results above are almost identical since the change intemperature during this process is relatively small (Fig. 7–35). When thetemperature change is large, however, they may differ significantly. For thosecases, Eq. 7–39 should be used instead of Eq. 7–34 since it accounts forthe variation of specific heats with temperature.AIRT 1 = 290 KT 2 = 330 KPs 2 – s 1 = s 2 ° – s 1 ° – R ln–– 2P 1= – 0.3844 kJ/kg . K––P 2s ––T22 – s 1 = C p,avgln – R lnT 1 P 1= – 0.3842 kJ/kg . KFIGURE 7–35For small temperature differences, theexact and approximate relations forentropy changes of ideal gases givealmost identical results.