Thermodynamics

Chapter 8 | 455(c) The wasted work potential is equivalent to the exergy destroyed, which canbe determined from X destroyed T 0 S gen or by performing an exergy balance onthe system. The second approach is more convenient in this case since theinitial and final exergies of the system are already evaluated.X in X out X destroyed ¢X system⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭⎫⎪⎬⎪⎭Net exergy transfer Exergy Changeby heat, work, and mass destruction in exergy0 X destroyed X 2 X 1X destroyed X 1 X 2 315 95.6 219.4 kJDiscussion Note that 219.4 kJ of work could have been produced as theiron was cooled from 350 to 31.7°C and water was heated from 30 to31.7°C, but was not.EXAMPLE 8–14Exergy Destructionduring Heat Transfer to a GasA frictionless piston–cylinder device, shown in Fig. 8–41, initially contains0.01 m 3 of argon gas at 400 K and 350 kPa. Heat is now transferred to theargon from a furnace at 1200 K, and the argon expands isothermally untilits volume is doubled. No heat transfer takes place between the argon andthe surrounding atmospheric air, which is at T 0 300 K and P 0 100 kPa.Determine (a) the useful work output, (b) the exergy destroyed, and (c) thereversible work for this process.ArgonFurnaceT R = 1200 K400 KSolution Argon gas in a piston–cylinder device expands isothermally as aQ R 350 kParesult of heat transfer from a furnace. The useful work output, the exergydestroyed, and the reversible work are to be determined.Assumptions 1 Argon at specified conditions can be treated as an ideal gasFIGURE 8–41since it is well above its critical temperature of 151 K. 2 The kinetic andpotential energies are negligible.Schematic for Example 8–14.Analysis We take the argon gas contained within the piston–cylinder deviceas the system. This is a closed system since no mass crosses the systemboundary during the process. We note that heat is transferred to the systemfrom a source at 1200 K, but there is no heat exchange with the environmentat 300 K. Also, the temperature of the system remains constant during theexpansion process, and its volume doubles, that is, T 2 T 1 and V 2 2V 1 .(a) The only work interaction involved during this isothermal process is thequasi-equilibrium boundary work, which is determined from2W W b P dV P 1 V 1 ln V 2 1350 kPa2 10.01 m 3 0.02 m32 lnV 1 0.01 m 31 2.43 kPa # m 3 2.43 kJThis is the total boundary work done by the argon gas. Part of this work isdone against the atmospheric pressure P 0 to push the air out of the way, andit cannot be used for any useful purpose. It is determined from Eq. 8–3:W surr P 0 1V 2 V 1 2 1100 kPa2310.02 0.012 m 3 1 kJ4a1 kPa # b 1 kJm3T 0 = 300 KP 0 = 100 kPa