Thermodynamics II. Week 9

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1 hermodynamics II Week 9

2 Example Oxygen gas in a piston cylinder at 300K, 00 kpa with volume o. m 3 is compressed in a reversible adiabatic process to a final temperature of 700K. Find the final pressure and volume using able A.8.

3 Summary Second Law of hermodynamics» entropy change for ideal gases» isentropic relationships for ideal gases constant specific heats variable specific heats ln ) ( v v R d C s v ln ) ( P P R d C s p

4 Entropy Balance Closed System For closed systems, entropy change occurs only due to two mechanisms:» entropy transfer due to heat interactions at the boundary» entropy generation due to irreversibilities within the system Entropy balance for closed system: Change in entropy between given states j j S system j Entropy generation due to irreversibilities within the system Entropy transfer due to heat exchange at all j boundaries

5 Entropy Balance Closed System (contd.) Entropy balance for closed system on rate basis: System and surroundings constitute an isolated system; entropy balance for the combined system for a possible process due to increase of entropy principle: ds dt system j. j j. S S S system surroundings total 0 C6. p.39

6 Ws Chapter 6: Examples, Problems 5.6, 6.4, 6.5, 6.7, 6., 6.3, 6.3, 6.47, 6.55, 6.6, 6.6, 6.76, 6.83, 6.08, 6.6

7 W-uiz # Chapter 6 : Problems 5.6, 6.4, 6.5, 6.7, 6., 6.3, 6.3, 6.47, 6.55, 6.6, 6.6, 6.76, 6.83, 6.08, 6.6 When? uesday, May st What time? :50 am to :0 pm Where? 930 What? Chapter 6 problems 及 Prob. 5.6 任選題 Calculator is allowed(with extra batteries if required; no programming) Closed book, closed notes, no property tables, no homework solutions, no communication devices

8 Exam # When? hursday, May 3 rd, 08 What time? 7:0 to 8:30 pm Where? Room 930 What? Chapters 5 and 6

9 Exam # Review Second law of thermodynamics Entropy

10 Second Law of hermodynamics Useful for determining direction of process, assigning quality to energy and predicting maximum possible theoretical performance eat Engine Cycle Refrigerator/eat Pump Cycle Kelvin Planck Statement: eat engine must interact with two thermal reservoirs i.e. h thermal < 00% Clausius Statement: eat cannot move from low temperature to high temperature spontaneously

11 Second Law of hermodynamics (contd.) Carnot Principles» Maximum performance between the given temperature limits for reversible cycles» Maximum performance independent of working substance Carnot Performance C C cycle thermal W h C C cycle C R W COP C C cycle P W COP C cycle rev C.

12 Entropy Entropy q ds int. rev. Clausius Inequality 0 cycle b Increase in Entropy Principle Entropy Balance S S S 0 total system surroundings S S S gen b Remember that S for a system can be positive, zero or negative; but S for the universe must always be positive (irreversible) or zero (reversible)

13 Entropy Balance Eq. S S S gen b Δ Entropy = + in out + gen wo ways to increase entropy:» ransfer heat to the system» ave an irreversible process Only one way to decrease entropy:» ransfer heat from the system

14 Increase of Entropy C6.0 p.36 Principle of the increase of entropy ds net ds c.m. ds surr S gen 0 he equality holds for reversible processes and the inequality for irreversible processes. he only processes that can take place are those in which the net change in entropy of the control mass plus its surroundings increases. his principle dictates the single direction in which any process can proceed and can be considered a quantitative general statement of the second law

15 Increase of Entropy Principle of the increase of entropy ds dt net ds dt c.m. ds dt surr S gen 0 S net S c.m. S surr S S c.m. surr m s c. m. surr s c. m.

16 Entropy (contd.) Entropy Change» hermal Reservoir S source ( ) or S sink ( ) C C» Pure Substance: Use property tables (CL, SL, SLVM, SV and SV) ** Make sure you know how to use the property tables to find missing properties, see problem 6.4, 6.5, 6.7 **

17 Entropy (contd.) Entropy Change» ds Relationships First Gibb s Equation: ds du Pdv Second Gibb s Equation: ds dh vdp applicable to any process related to closed and open systems

18 -s Diagram P da da ds S dv V int, rev ds int,rev ds W PdV W PdV Area under -s diagram = eat transfer during the internally reversible process Area under P-v diagram = Work done during quasi-equilibrium process

19 Carnot Power Cycle on -s Diagram.. int S S rev L L L.. int S S rev a b Area Area W L L cycle h thermal For internally reversible process»» q ds int,rev w Pdv int,rev

20 Incompressible Substances C6.5 p.4 For solids and liquids, dv = 0 and C v = C p = C First Gibb s Equation For solids entropy change dependent only on temperature; for liquids that expand considerably small pressure effect when temperature change is large isentropic processes of solids and liquids need to be isothermal du ds du ds Pdv Cd s ln C av

21 Ideal Gas: Entropy Change (contd.) For ideal gas: d v d P s C ( ) Rln s C ( ) Rln v p v P Use known dependence of C v and C p on temperature and integrate (reasonably accurate; very tedious) Use constant C v and C p neglecting temperature dependence (usually works on small temperature ranges; otherwise average C v and C p ) Use tabulated data (most accurate; ables A.7-9)

22 Ideal Gas: Entropy Change (contd.) Define: s s 0 0 C p ( ) 0 0 s C ( ) p d d Entropy change for an ideal gas considering the actual dependence of C p on temperature is given by: s 補充 68 P 0 0 s s Rln P

23 Ideal Gas: Isentropic Process For constant specific heats, we can easily show that: k k P P ) ( k v v P P ) ( k v v k = C p /C v = Specific eat Ratio C6.6 p.7 補充 69

24 Ideal Gas: Isentropic Process (contd.) 補充教材 For variable specific heats, we can easily show that: Relative Pressure (p r ) P 0 0 s s Rln 0 P Relative Specific Volume (v r ) p r exp r p r and v r listed in ideal gas table (A7. with known scaling constants, A, A ) 0 s R / A v A p r 補充 7

25 Ws Chapter 5 : Examples, Problems 5., 5.3, 5.5, 5.6, 5., 5.3, 5.6, 5.7, 5.38, 5.5, 5.65, 5.67, 5.90, 5.93 Chapter 6 : Examples, Problems 5.6, 6.4, 6.5, 6.7, 6., 6.3, 6.3, 6.47, 6.55, 6.6, 6.6, 6.76, 6.83, 6.08, 6.6

26 Summary and Key Concepts In every chapter, Summary and Key Concepts and Formulas is a good way to review all important topics!

27 Pop quiz Please write down the first and second Gibbs equations.

28 Pop quiz Please derive the first and second Gibbs equations for a simple, compressible substance inside a stationary closed system.

29 Coverage hus Far Second Law of hermodynamics» entropy balance for closed system j. j j S system j ds dt system j.» entropy changes only due to heat interaction and irreversibilities; no entropy transfer due to work j

30 Entropy Balance Open System C7. p.58 For open systems, entropy change occurs due to three mechanisms:» entropy transfer due to heat interactions at the boundary» entropy transfer due to mass flow in and out of the system» entropy generation due to irreversibilities within the system Entropy balance for open system: m S m S S j j j inlets i i exits e e CV

31 Entropy Balance Open System (contd.) Entropy balance for open system on rate basis: Rate of entropy gain due to incoming mass flow Rate of entropy loss due to outgoing mass flow 補充 j m S m S j inlets i i exits e e j ds dt CV Rate of entropy change of control volume Rate of entropy generation due to irreversibilities within the system Rate of entropy transfer due to heat exchange at all j boundaries

32 Steady-State Process Steady-state: de CV dm CV 0 0 dt dt Continuity equation: First law:. m m in. out. inlets. m i h i V i gz i. W outlets. m e h e V e gz e

33 ransient Process Continuity equation:. m in. m out dm dt CV First law: W m h V gz m h V gz de.... i o CV cv cv i i i o o o inlets outlets dt V V m u gz m u gz cv

34 Applications of First Law of hermodynamics eat Exchanger Nozzle Diffuser hrottle urbine Compressor and Pump Power Plant and Refrigerator.etc.

35 Example (Ex. 7.) Steam enters a steam turbine at a pressure of MPa, a temperature of 300 C, and a velocity of 50 m/s. he steam leaves the turbine at a pressure of 50 kpa and a velocity of 00 m/s. Determine the work per kilogram of steam flowing through the turbine, assuming the process to be reversible and adiabatic. Steady-State Process!

36 Example (Ex. 7.5) A de-superheater works by injecting liquid water into a flow of superheated steam. With kg/s at 300 kpa, 00 C, steam flowing in, what mass flow rate of liquid water at 0 C should be added to generate saturated vapor at 300 kpa? We also want to know the rate of entropy generation in the process. Steady-State Process!

37 Example 3 (Ex. 7.6) Assume an air tank has 40 L of 00 kpa air at ambient temperature 7 C. he adiabatic and reversible compressor is started so that it charges the tank up to a pressure of 000 kpa and then it shuts off. We want to know how hot the air in the tank gets and the total amount of work required to fill the tank. ransient Process!

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,