Presentation on theme: "Evaluating entropy changes"— Presentation transcript:
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 processesPT12SV2
3 Showing a Carnot cycle on a T-s diagram Isotherms: horizontal linesIsentropes:vertical lines21TQnet,inApplication:Carnot engine34s
4 Isobars and isochores on the T-s diagram for an ideal gas (Sec. 7.9) PIsochores:Isochores:vIsobars:Ts
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 On an intensive (per unit mass) basis:First TdS relationSecond TdS relationIntegrating 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 bealso calculated using the integrals on the left.
7 Entropy change of liquids/solids (Sec. 7.8) 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 alonecp=cv=c (one specific heat)from First Tdsrelationif temperature variation of specific heat (c)can be neglected.For solids/liquids, the isentropic process is also isothermal.
8 Second law for a closed system undergoing a process Clausius inequality:b: boundaryTb=T foran internallyreversibleprocess> for irreversible process 1-2= internally reversible process 1-288
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.”99
10 Second law for an adiabatic process undergone by a closed system Second law for a system undergoing adiabatic (irreversible/reversible) process(dQ=0)1010
11 The entropy balance for closed systems entropy generationwithin thesystementropy changeentropy transfer accompanyingheat transferSgen>0 if irreversibilities present inside the systemSgen=0 for no irreversibilities inside the systemThe 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 boundaryIrreversible 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.1212
13 Pf,Tf, (m1+m2) ) to show Sgen>0 Example of entropy generation due to an irreversible process in an isolated systemInitial state (1)Pf,Tf, (m1+m2)Final state (2)Remove separatorObjective:To show that the process is irreversible.Isolated system has Q=0To show a process is irreversible) to show Sgen>0
14 Example (contd.): The final state of the process Pf,Tf, (m1+m2)P1,V1, T, m1P2,V2, T, m2High pressureLow pressureFinal temperature: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, m1P2,V2, T, m2Pf,Tf, (m1+m2)Low pressureHigh pressureEntropy balanceEntropy change of the system is the entropy change of its parts
16 Example of entropy generation due to an irreversible process in an isolated system P1,,,V1, T, m1,P2,V2, T, m2Pf,Tf, (m1+m2)Second law (entropy balance)from integration of Tds=dh-vdPSimilarly
17 Example of entropy generation due to an irreversible process in an isolated system P1,,,V1, T, m1,P2,V2, T, m2Pf,T, (m1+m2)17
18 Calculating work done in a reversible steady 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 PEcan be neglected (e.g. inturbines, compressor andpumps, but not in nozzles)
20 Reversible steady flow work for a pump (PUMP HANDLES LIQUIDS) Since specific volume of a liquid is nearly independent of pressure:(since pump is an adiabatic device)similar to compressorAlternatively
21 In steam power plants, how big is “turbine work out” compared to “pump work in”? vvf=1.7 m^3/kg and vf=10^-3 m^3/kg at 100 kPaSpecific volume of vapors are orders of magnitude larger than thespecific volume of liquid (e.g. vf' 10-3 m3/kg vg' 2 m3/kg at 100 kPa))Gas turbine pumps consume asignificant part of work developedin turbines to run compressor