Thermodynamics

Chapter 8 | 463The maximum power output (reversible power) is determined from the rateform of the exergy balance applied on the extended system (system + immediatesurroundings), whose boundary is at the environment temperature ofT 0 , and by setting the exergy destruction term equal to zero,X # in X # out X # destroyed dX system >dt 0S 0 (reversible)S 0 (steady)⎫ ⎪⎬⎪⎭⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎪⎬⎪⎪⎪⎭Rate of net exergy transfer Rate of exergy Rate of changeby heat, work, and mass destruction in exergyX # in X # outm # 1c 1 m # 2c 2 W # rev,out X # Q 0heat m # 3c 3W # rev,out m # 1c 1 m # 2c 2 m # 3c 3Note that exergy transfer by heat is zero when the temperature at the pointof transfer is the environment temperature T 0 , and the kinetic and potentialenergies are negligible. Therefore,W # rev,out m # 1 1h 1 T 0 s 1 2 m # 2 1h 2 T 0 s 2 2 m # 3 1h 3 T 0 s 3 2That is, we could have produced work at a rate of 4588 Btu/min if we ran aheat engine between the hot and the cold fluid streams instead of allowingthem to mix directly.The exergy destroyed is determined fromX # destroyed W # rev,out W # u→0 T 0 S # genThus, 1300 lbm>min2318.07 Btu>lbm 1530 R210.03609 Btu>lbm # R24 122.7 lbm>min231162.3 Btu>lbm 1530 R211.7406 Btu>lbm # R24 1322.7 lbm>min2397.99 Btu>lbm 1530 R210.18174 Btu>lbm # R24 4588 Btu/minX # destroyed W # rev,out 4588 Btu/minsince there is no actual work produced during the process (Fig. 8–47).Discussion The entropy generation rate for this process was determined inExample 7–20 to be S . gen 8.65 Btu/min · R. Thus the exergy destroyedcould also be determined from the second part of the above equation:X # destroyed T 0 S # gen 1530 R218.65 Btu>min # R2 4585 Btu>minThe slight difference between the two results is due to roundoff error.EXAMPLE 8–17Charging a Compressed Air Storage SystemA 200-m 3 rigid tank initially contains atmospheric air at 100 kPa and 300 Kand is to be used as a storage vessel for compressed air at 1 MPa and 300 K(Fig. 8–48). Compressed air is to be supplied by a compressor that takes inatmospheric air at P 0 100 kPa and T 0 300 K. Determine the minimumwork requirement for this process.Solution Air is to be compressed and stored at high pressure in a largetank. The minimum work required is to be determined.FIGURE 8–47For systems that involve no actualwork, the reversible work andirreversibility are identical.© Reprinted with special permission of KingFeatures Syndicate.AIRV = 200 m 3100 kPa → 1 MPa300 KFIGURE 8–48Schematic for Example 8–17.Compressor100 kPa300 K