EXERGY Basic Definitions Exergy: is property used to determine the useful work potential of a given amount of energy at some specified state. It does not represent the amount of work that a work-producing device will actually deliver upon installation. Rather, it represents the upper limit on the amount of work a device can deliver without violating any thermodynamic laws. A system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment (dead state) Dead State: a system is said to be in dead state when it is in thermodynamic equilibrium with its environment. Also it has no potential or kinetic energy. And it is chemically inert (no reaction with the environment) Exergy of potential energy: (work potential) of a system is equal to the potential energy itself regardless of the temperature and pressure of the environment. Exergy of kinetic energy: (work potential) of a system is equal to the kinetic energy itself regardless of the temperature and pressure of the environment. (not necessarily true) REVERSIBLE WORK AND IRREVERSIBILITY: At difference to exergy, actual processes do not occur from an initial point to a final point equal to the dead state. On the other hand isentropic efficiencies are limited to adibatic processes. Surrounding Work: is the work done by or against the surroundings during a process. It has significance only for a process where boundary work occurs (closed system). Wsurr = P0(V2 – V1) Then the useful work would be: Wu = W – Wsurr = W – P0(V2 – V1) How is Wsurr for a rigid tank? Reversible Work: maximum amount of useful work that can be produced (or the minimum work that needs to be supplied) as a system undergoes a process between the specified initial and final states. Whats is the difference between exergy and reversible work? Any difference between reversible and useful work is due to irreversibilities. I = Wrev,out – Wu,out I = Wu,in – Wrev,in SECOND LAW EFFICIENCY rev, A TL 1 TH 300 K 1 50% 600 K A T H 300 K rev, B 1 L 1 70% T 1000 K B In cases like this the first-law efficiency alone is not a realistic measure of performance of engineering devices. The second-law efficiency is defined as the ratio of the thermal efficiency to the maximum possible thermal efficiency under the same conditions th II rev Then, in the example: II , A 0.30 0.60 0.50 II , B 0.30 0.43 0.70 In other words: Wu II WRe v Work producing devices WRe v II Wu Work consuming devices II COP COPRe v Refrigerators and Heat Pumps In general II Exergy Re cov ered Exergy Destroyed 1 Exergy Supplied Exergy Supplied Closed System Total useful work delivered in a reversible process to the dead state: Wuseful U U 0 P0 V V0 T0 S S 0 The total exergy for a closed process would be given by: V2 X U U 0 P0 V V0 T0 S S0 m mgz 2 (kJ) The total exergy per unit mass: V2 u u 0 P0 v v0 T0 s s0 gz 2 (kJ/kg) EXERGY CHANGE FOR A CLOSED SYSTEM V22 V12 X m( 2 1 ) U 2 U 1 P0 V2 V1 T0 S 2 S1 m mg ( z 2 z1 ) 2 X m(2 1 ) E2 E1 P0 V2 V1 T0 S2 S1 Per unit mass: V22 V12 (2 1 ) u2 u1 P0 v2 v1 T0 s2 s1 g ( z2 z1 ) 2 (2 1 ) e2 e1 P0 v2 v1 T0 s2 s1 Exergy of a flow stream x flow Pv P0v ( P P0 )v x flowing xno _ flowing x flow V2 u u0 P0 v v0 T0 s s0 gz ( P P0 )v 2 V2 h h0 T0 s s0 gz 2 Exergy change of a flow stream V22 V12 2 1 ) h2 h1 T0 s2 s1 g ( z2 z1 ) 2 Decrease or Exergy Principle (Exergy Destruction) W Q Isolated system System (closed or open) T0 S gen X 0 Exergy decreases X destroyed T0 S gen Exergy destruction Exergy Balance Total Total Total Change in the exergy exergy exergy total exergy entering leaving destroyed of the system Exergy transfer by heat, work and mass By Heat: By Work: By mass: T X heat 1 0 Q T X work W Wsur W X mass m If boundary work No boundary work X heat X work X destroyed X T0 1 T k Closed system (no mass flowing) Qk W P0 (V2 V1 ) T0 S gen X T0 dV dX 1 Q W P T S gen T k 0 dt 0 dt k Sol. V2 X U U 0 P0 V V0 T0 S S0 m mgz 2 X mu u0 P0 mv v0 T0 ms s0 From table A6 & A4, for water: u=2594.7 kJ/kg @ 180°C 800 kPa v=0.2472 m3/kg s=6.7155 kJ/kg.K u0 = 104.83 kJ/kh @ 25°C 100 kPa v0 = 0.001003 m3/kg s0 = 0.3672 kJ/kg.K From table A13 (R-134a superheated vapor): u=386.99 kJ/kg @ 180°C 800 kPa v=0.044554 m3/kg s=1.3327 kJ/kg.K u0 = 252.615 kJ/kh @ 25°C 100 kPa v0 = 0.23803 m3/kg s0 = 1.10605 kJ/kg.K kJ m3 X w 1kg761.92 104.83 100kPa.1kg0.001127 0.001003 kg kg kJ 298K .1kg2.1392 0.3672 kg.K X w 622.7kJ kJ m3 X R 1kg386.99 252.615 100kPa.1kg0.044554 0.23803 kg kg kJ 298K .1kg1.3327 1.10605 kg.K X R 47.5kJ Sol. X in X out X destroyed X X destroyed 0 For a reversible process, therefore Wrev=X2-X1 2 1 u2 u1 P0 v2 v1 T0 s2 s1 u2 u1 CvT2 T1 Cv = 0.164 Btu/lbm.R and but s2 s1 Cv ln T2 R ln v2 T1 R = 0.0621 Btu/lbm.R v1 Btu 985 535R 2 1 0.164 lbm.R Btu 985 Btu 1.5 535R 0.164 ln 0 . 0621 ln lbm.R 535 lbm.R 12 ft 3 Btu 14.7 psia 1.5 12 3 lbm 5.4039 psi. ft 2 1 60.77 Btu lbm Solution: P (kPa) 1’ 180 100 2 First the process is at constant volume until the pressure is enough to move the piston (1-1’), then the process is at constant pressure (1’-2) 1 v (m3/kg) States 1 and 2 are in the region of superheated vapor, the summarized data from table is: P1=140kPa v1=0.1652 m3/kg P2=180kPa u1=246.01 kJ/kg T1=20°C s1=1.0532 kJ/kgK v2=0.17563 m3/kg u2=331.96 kJ/kg T2=120°C s2=1.3118 kJ/kgK Also at (1’) v1’ = v1 and P1’=P2 a) The work done is the boundary work, from (1) to (1’) is zero since it is a constant volume process; from (1’) to (2) is a constant pressure process. Then the boundary work is given by: Wb = P2.m.(v2-v1) = (180)(1.4)(0.17563 – 0.1652) = 2.63 kJ b) Doing a energy balance we obtain: Q-W = m(u1’ – u1) + m(u2 – u1’) = m(u2 – u1) Then Q = m(u2 – u1) + W = (1.4)(331.96 – 246.01) + 2.63 kJ Q = 122.96 kJ Since there is no kinetic nor potential energy involved the exergy change can be expressed by: X mu2 u1 mP0 v2 v1 mT0 s2 s1 X 1.4331.96 246.01 (1.4)(100)0.17563 0.1652 (1.4)(298)1.3118 1.0532) X 13.90kJ The useful work at the exit is given by the boundary work minus the work against the environment: Wu = Wb – m.P0(v2-v1) = 2.63kJ – (1.4)(100)(0.17563–0.1652) Wu = 1.17 kJ From the total exergy change the only amount of useful work is 1.17kJ everything else is the exergy destroyed, therefore: X destroyed X Wu 13.90kJ 1.17kJ 12.73kJ d) The second law efficiency is given by: II Wu 1.17 0.084 X 13.9 Exergy Balance Open System X heat X work X mass X destroyed X Notice that now we are including the exergy entering and leaving with mass, then: T0 Qk W P0 V2 V1 m m X destroyed X 2 X 1 in out k 1 T (kJ) In rate form: T0 dV dX 1 Q W P m m X destroyed T k 0 dt dt in out k (kW) Fortunately we usually have to deal with steady flow devices, then our equation becomes: T0 m m X destroyed 0 1 T Q k W in out k for a single stream this last equation becomes: T0 1 T Q k W m 2 1 X destroyed 0 k (kW) Per unit mass: T0 1 T qk w 2 1 xdestroyed 0 k (kJ/kg) Previous equations can be used to determine the reversible work by making the exergy destruction term equal to zero since (i.e. no irreversibilities implies no exergy destruction) Wrev X destroyed 0 Then: T0 m m 0 1 T Q k W rev in out k or: T0 m m 1 Q k 0 out in Tk W rev Single stream (one inlet - one outlet): W rev Example: T0 m 2 1 1 Q k 0 Tk Solution: We need to determine the exergy destroyed during this process. In this case the easiest way is by: xdestroyed T0 s gen This equation leads us to find the entropy generated which is given by doing an entropy balance: q s gen s T s gen q ( s2 s1 ) T From table at 200 psia: State (1) sat. liquid: h1=355.46 Btu/lbm s1=0.54379Btu/lbm.R State (2) sat. vapor: h2=1198.8 Btu/lbm s2=1.5460 Btu/lbm.R At environment conditions (P0=14.7 psia, To=80°F comp. liq.): h0=48.07 Btu/lbm s0=0.09328 Btu/lbm.R We need to determine q in the previous equation, we obtain this by doing an energy balance: q – w = (h2 – h1) since there is no work we have: q = 1198.8 – 355.46 = 843.34 Btu/lbm Now we can determine the entropy generated: s gen (1.5460 0.54379) 843.34 Btu / lbm.R Btu 0.124 960 R lbm.R 960R is the absolute gas temperature (500°F) And the exergy destroyed will be: Btu Btu xdestroyed T0 s gen (540 R) 0.124 66.96 lbm.R lbm The exergy 9or work potential) of the steam is given by: h2 h1 T0 s2 s1 Therefore the temperature of the gases does not affect the exergy of the steam. However it does affect sgen and therefore xdestroyed too. a) We find the actual work by doing an ener4gy balance: Q W a mh2 h1 KE PE Kinetic and potential energy changes are assumed to be zero. Consider specific heats for the enthalpy change. Solving for work we have: W a Q m CpT2 T1 Cp=1.134 kJ/kg.K (from table) Q = -30kW (lost) Substitue in the previous equation to obtain: Wa 30kW 3.4 kg kJ 903 1023K 432.67kW 1.134 s kgK We have an expression for the reversible work from the exergy balance for a single stream: W rev T0 m 2 1 1 Q k 0 Tk The ideal situation for a turbine occurs when there are no heat losses, therefore: W rev m 2 1 mh2 h1 T0 [ s2 s1 ] The entropy change is obtained from: s2 s1 Cp ln T 2 R ln P2 T1 P1 1.134 kJ 903 kJ 500 kJ ln 0.287 ln 0.11 kg.K 1023 kg.K 1200 kg.K Then the rev. work is: W rev 3.4 kg kJ kJ 1.134 [903 1023]K 298K [0.11 ] 574kJ s kg.K kg.K The exergy destroyed will be given by: X destroyed W rev W a 574.12 432.67 141.45kJ Finally, the second law efficiency would be: II Wa W rev 432.67 0.754 574.12