1. ENZO ZANCHINI Università di Bologna enzo.zanchini@unibo.it +39 051 20 93295 AVAILABILITY FUNCTIONS AND THERMODYNAMIC EFFICIENCY 3. Availability functions for closed systems and flow availability Bologna - September 2013 1

2. AVAILABILITY FUNCTIONS We call availability function for a system A a property of A whose decrease gives the highest work (or, the highest shaft work) which can be obtained in a process A 1 → A 2 of A which fulfils prefixed conditions. For simplicity, we will consider only simple systems or unions of simple systems ADIABATIC AVAILABILITY We call adiabatic availability of a system A a property of A whose value gives, for every initial state A 1, the maximum work obtainable by a weight process for A starting from A 1, such that the region of space occupied by A in the final state coincides with that occupied in the initial state A 1. By Theorem 1, the adiabatic availability is zero for every initial stable equilibrium state. 2

3. 3 E S A1A1 AsAs Adiabatic availability of A 1 V

4. We call available energy of a system AR, where A is a closed system and R is a thermal reservoir, the property Ω = E - T 0 S, where E and S are the energy and the entropy of A, and T 0 is the temperature of the reservoir R. AVAILABLE ENERGY Given the states A 1 and A 2 of a system A and a thermal reservoir R, the highest work obtainable by a weight process for AR from A 1 to A 2 is Ω 1 – Ω 2, and is obtained only by any reversible process. THEOREM 3.1 4

5. PROOF Let us consider the energy and the entropy balance for AR; then, we combine algebrically the two balance equations; X is a system which performs a cycle. Energy balance (3.1) Entropy balance (3.2) (3.3) Summing (3.1) and(3.3): (3.4) The Gibbs equation for R gives From (3.4): (3.5) W A X R A 1 → A 2 T0T0 5

6. PROBLEM A closed system A composed of a mass m = 100 kg of liquid water, is kept at constant volume (the pressure is p = 1.013 bar at the temperature T = 58.5 °C). In the initial state, the temperature is T 1 e =e363 K (  90 °C). The external environment can be considered as a thermal reservoir R with temperature T 0 = 300 K (  27 °C). Find the maximum work that can be obtained by a weight process for AR (adiabatic availability of AR, in the initial state). A X R W T0T0 H 2 O, m = 100 kg V = cost To obtain the maximum work, one must bring A in mutual stable equilibrium with R, namely at the temperature T 0. SOLUTION By Theorem 3.1: 6

7. Since the volume is constant: Then: http://webbook.nist.gov/chemistry/fluid/c v = 3.984 kJ/(kg K) c v evaluated at pressure 1.013 bar and temperature 58.5 °C 7

8. KEENAN’S AVAILABILITY FUNCTION,  Let A and B be two closed simple systems, such that B crosses only stable equilibrium states with the same values of temperature T 0 and pressure p 0, and the volume of AB is fixed. We call Keenan’s availability function  of AB the property  = E – T 0 S + p 0 V, where E, S and V are the energy, the entropy and the volume of A. Comment. System B has a behaviour very similar to that of a thermal reservoir, but its volume can change, with changes that are opposite to those of A. A B T 0, p 0 8

9. THEOREM 3.2 For given states A 1 and A 2, the maximum work obtainable in a weight process for AB from A 1 to A 2, with fixed volume of AB, is  1 –  2, and is obtained only by any reversible process. PROOF A B T 0, p 0 X W A1→A2A1→A2 Energy balance Entropy balance  T 0 Sum (3.6) 9

10. Gibbs equation for B: (3.7) By substituting (3.8) in (3.6) one obtains: (3.9) (3.8) From (3.7): 10

11. PROBLEM SOLUTION p 1, T 0 p 0, T 0 Since air has the same composition as the environment, the system can be considered as closed. The availability function, per mol, is 11 A student decides to build a rocket which works by compressed air. The mass of the rocket, without the air, is m = 0.5 kg. The vessel which contains the compressed air has a volume V = 0.5 dm 3. The compressed air has the same temperature as the environment air, T 0 = 300 K. The compressor compresses the air from the environment pressure p 0 = 1 bar to p 1 = 10 bar. Which is the theoretical maximum high which could be reached by the rocket in the absence of the viscous drag of the environment air?

12. T 1 = T 0 = 300 K p 1 = 10 bar p 0 = 1 bar V 1 = 0.5 dm 3 12

13. 13

14. We call control volume a fixed region of space crossed by a fluid in motion (portion of a duct). The fluid is considered as the union of infinite infinitely small simple systems, called fluid elements, which move in a uniform gravity field g. The specific energy (per unit mass) of a fluid element is The energy of the fluid that, at an instant , is contained within the control volume, which occupies the region of space V, is given by (3.10) (3.11) (internal energy + potential energy + kinetic energy) CONTROL VOLUME 14

15. Let C be a control volume, with inlet section 1 and outlet section 2, crossed by a fluid in motion. We call shaft work performed by the fluid, in the time interval [  1,  2 ], the difference between the total work performed and the work performed against the external fluid (upstream and downstream). (3.12) ENERGY BALANCE FOR A CONTROL VOLUME By neglecting the work against viscous forces: 15 To introduce in C the infinitesimal volume dV 1 of fluid, pressure p 1 performs on C the work p 1 dV 1 (work per unit mass p 1 v 1 ); to expel from C the infinitesimal volume dV 2 of fluid, system C performs the work p 2 dV 2 against the pressure p 2 of the external fluid (work per unit mass p 2 v 2 ). 1 2 V C QQ  W sh dV1dV1 dV2dV2 p1p1 p2p2 dV 1 = v 1 dm dV 2 = v 2 dm

16. 16 The energy balance equation per unit time is 1 2 V C (3.13) “Specific energy” of fluid at inlet: e 1 +p 1 v 1, of fluid at outlet e 2 +p 2 v 2. This is not a rigorous proof, but a simple way to get the result

17. If C is in a steady state, the left hand side of Eq. (3.13) is zero; moreover (3.14) From Eq. (3.13) with vanishing left hand side, by employing Eq. (3.14), one gets (3.16) = heat received by every kg of fluid which crosses C; = = shaft work performed by every kg of fluid which crosses C. (3.13) (3.15) CONTROL VOLUME IN STEADY STATE By dividing both sides of Eq. (3.15) by, one obtains 17

18. (3.16) Mean specific energy e in a cross section: fluid elements have the mean velocity W and are placed at the high z of the centre of the cross section (3.17) (3.18) (3.19) By introducing the specific enthalpy 18

19. 1 2 R0R0 T0T0 C  C IN STEADY STATE ENTROPY BALANCE FOR A CONTROL VOLUME (3.20) (3.21) 19

20. Let C be a control volume in a stationary state, crossed by a fluid in motion. The states 1 and 2 of the fluid in sections 1 and 2 are fixed, and C can exchange heat with only one thermal reservoir R 0, with temperatureT 0 (which represents the external environment). 1 2 R0R0 T0T0 C TEOREM 3.3: FLOW AVAILABILITY The maximum shaft work that can be obtained from every kg of fluid which goes from section 1 to section 2 of C is, where  = h – T 0 s is called flow availability, and is obtained only by any reversible process. 20

21. Energy balance (by neglecting the changes in kinetic and potential energy): (3.22) Entropy balance:   (3.23) By substituting Eq. (3.23) in Eq. (3.22) one obtains: (3.24) PROOF 21

22. GIBBS FREE ENERGY AS A SPECIAL CASE OF FLOW AVAILABILITY 1 2 R T 0 = T C T T If in sections 1 and 2 the fluid has the same temperature T, which coincides with the temperature T 0 of R, one has : Since,, if T = T 0 one has  = g (3.25)Per kg, or mol: For a mass m or n moles:(3.26) 22