Evaluating entropy changes

1. Evaluating entropy changes • Since entropy is a property, the entropy change between two specified states does not depend on the process (reversible or irreversible) that occurs between the two states. • However, to evaluate entropy change by the above definition a reversible process has to be imagined connecting the initial and final states and integration has to be carried along its path.

2. Using entropy change and thermodynamic temperature to graphically interpret the heat supplied in reversible processes P T 1 2 S V

3. Showing a Carnot cycle on a T-s diagram Qnet,in Isotherms: horizontal lines Isentropes:vertical lines 2 1 T Application: Carnot engine 3 4 s

4. Isobars and isochores on the T-s diagram for an ideal gas (Sec. 7.9) v P Isochores: Isochores: Isobars: T s

5. T-s diagram of a cycle: try yourself • The Brayton cycle used in a gas turbine engine consists of following steps: • isentropic compression (1-2). • isobaric heat addition (2-3). • isentropic expansion (3-4). • isobaric heat rejection (4-1). Plot this process on a T-s and P-v diagram.

6. Use of the Tds relations in calculating entropy changes Integrating between initial (1) and the final state (2): Even though the Tds equations are derived using an internally reversible process, the change in entropy during an irreversible process occuring between the same two equilibrium states can be also calculated using the integrals on the left. First TdS relation Second TdS relation On an intensive (per unit mass) basis:

7. Entropy change of liquids/solids (Sec. 7.8) if temperature variation of specific heat (c)can be neglected. from First Tds relation • Liquid/solid can be approximated as incompressible substances (v=constant): i.e. specific volume ( or density) remains constant even when other properties change (dv=0) Internal energy of a solid/liquid is a function of temperature alonecp=cv=c (one specific heat) For solids/liquids, the isentropic process is also isothermal.

8. Second law for a closed system undergoing a process Clausius inequality: Tb=T for an internally reversible process b: boundary > for irreversible process 1-2 = internally reversible process 1-2

9. Second law for an isolated system Also, across the boundaries of an isolated system no energy (heat/work) is transferred (by definition) The universe (any system + its surroundings) can be considered an isolated system. “The total entropy of the universe is increasing.”

10. Second law for an adiabatic process undergone by a closed system Second law for a system undergoing adiabatic (irreversible/reversible) process (dQ=0)

11. The entropy balance for closed systems entropy generation within the system entropy change entropy transfer accompanying heat transfer Sgen>0 if irreversibilities present inside the system Sgen=0 for no irreversibilities inside the system The value of Sgen is a measure of the extent of irreversibilities within the system. More irreversibilities Sgen↑

12. Entropy generation (Sgen>0) within the system is due to irreversible processes within the system boundary • Irreversible processes occurring within the system boundary result in entropy generation. • Examples: • Friction (solid-solid, solid-fluid) may be present between parts of the system. • Hot and cold zones may be present within the system which may be interacting irreversibly through heat transfer. • Non-quasi-equilibrium compression/expansion occuring between parts of the system. • Other: mixing between substances having different chemical composition, chemical reaction.

13. Example of entropy generation due to an irreversible process in an isolated system Pf,Tf,(m1+m2) Initial state (1) Final state (2) Isolated system has Q=0 To show a process is irreversible )to show Sgen>0  Remove separator • Objective: • To show that the process is irreversible.

14. Example (contd.): The final state of the process P1,V1, T, m1 P2,V2, T, m2 Final temperature: Pf,Tf,(m1+m2) High pressure Low pressure • From the first law Uf=Ui. • At the final state: • .

15. Example of entropy generation due to an irreversible process in an isolated system P1,,,V1, T, m1 P2,V2, T, m2 Entropy balance Entropy change of the system is the entropy change of its parts Pf,Tf,(m1+m2) Low pressure High pressure

16. Example of entropy generation due to an irreversible process in an isolated system P1,,,V1, T, m1, P2,V2, T, m2 from integration of Tds=dh-vdP Similarly Pf,Tf,(m1+m2) Second law (entropy balance)

17. Example of entropy generation due to an irreversible process in an isolated system P1,,,V1, T, m1, P2,V2, T, m2 Pf,T,(m1+m2)

18. Calculating work done in a reversiblesteady flow process • Reversible cycles develop the maximum work. • Reversible cycles consist of reversible processes through each device. • At every stage of the internally reversible process: • qrev-wrev=dh +d(ke)+d(pe) • Combining: wrev=-vdP-d(ke)-d(pe) • Integrating: First law for control volumes

19. Graphical representation of reversible steady flow work on a p-v diagram if changes in KE and PE can be neglected (e.g. in turbines, compressor and pumps, but not in nozzles)

20. Since specific volume of a liquid is nearly independent of pressure: (since pump is an adiabatic device) Alternatively similar to compressor Reversible steady flow work for a pump (PUMP HANDLES LIQUIDS)

21. In steam power plants, how big is “turbine work out” compared to “pump work in”? v Specific volume of vapors are orders of magnitude larger than the specific volume of liquid (e.g. vf' 10-3 m3/kg vg' 2m3/kg at 100 kPa)) Gas turbine pumps consume a significant part of work developed in turbines to run compressor