Objectives:

1. AH

2. Objectives: Define entropy to quantify the 2nd-law effects. The increase of entropy principle. Calculate the entropy changes Examine isentropic processes Reversible steady-flow work relations. Isentropic efficiencies for steady-flow devices.

3. Intro to Entropy • When you open a bottle of perfume you can smell the aroma as the molecules leave the bottle and reach your nose. • Why don’t they spontaneously go back into the bottle? • It would not violate the first law of thermodynamics. • But they most probably will not. • The explanation is entropy.

4. Definitions Entropy is: • a measure of the disorderof a system. • a measure of the energy in a system or process that is unavailable to do work. In a reversible thermodynamic process, entropy is expressed as the heat absorbed or emitted divided by the absolute temperature. • dS = (dQ/T)intrev • Entropy is a thermodynamic property (state function)

5. Entropy PertidaksamaanClausius: Engineers are usually concerned with the changes in entropy. Entropy change of a system during a process: The relation between Q and T during a process is often not available. Most cases we rely on tabulated data for S.

6. Entropy is a property Entropy is a property It has fixed values at fixed states. Entropy change S between two specifiedstates is the same (whether reversible or irreversible) during a process.

7. Entropy and the Second Law of Thermodynamics • Entropy, S, can be used to define the state of a system, along with P, T, V, U, and n. • We can calculate entropy only for reversible processes. • To calculate entropy for an irreversible process, find a reversible process that takes the system between the same two states and calculate the entropy. • It will be the same as for the irreversible process because it depends only on the two states.

8. Isothermal Heat Transfer Processes Isothermal heat transfer processes are internally reversible. The entropy change of a system during an internally reversible isothermal process : or

9. EXAMPLE 7–1 Entropy Change during an Isothermal Process A piston–cylinder device contains a liquid–vapor mixture of water at 300 K. During a constant-temperature process, 750 kJ of heat is transferred to the water. As a result, part of the liquid in the cylinder vaporizes. Determine the entropy change of the water during this process

10. THE INCREASE OF ENTROPY PRINCIPLE From the Clausius inequality,     For an isolated system (adiabatic closed system, Q = 0)  Sisolated ≥ 0

11. Total Entropy Total entropy change: DSgen =DStotal = DSsyst + DSenv ≥ 0 3 posibilities: The 2nd law: the entropy of an isolated system never decreases. It either stays constant (reversible process) or increases (irreversible). Entropy in one part of the universe may decrease in any process, the entropy of some other part of the universe always increases by a greater amount, so the total entropy always increases.

12. EXAMPLE 7–2 Entropy Generation A heat source at 800 K loses 2000 kJ of heat to a sink at (a) 500 K and (b)750 K. Determine which heat transfer process is more irreversible.

13. For the heat transfer process to a sink at 500 K: Similarly, for process at 750 K: Process (b) is less irreversible since it involves a smaller T difference (smaller irreversibility).

14. ENTROPY PURE SUBSTANCES

15. ENTROPY CHANGE OF PURE SUBSTANCES In the saturated mixture region: s = sf + x.sgf (kJ/kg) In the Compressed liquid region: For a specified mass m: Entropy is commonly used as a coordinate on diagrams such as the T-s and h-s diagrams.

16. EXAMPLE 7–3 Entropy Change of a Substance A rigid tank contains 5 kg of refrigerant-134a initially at 20°C and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa. Determine the entropy change of the refrigerant during this process.

17. Sketch

18. Solution The volume of the tank is constant, V2= V1. The properties of the refrigerant are: At final state refrigerant is a saturated mixture since vf < v2 < vgat 100 kPa, therefore:

19. Solution (Cont’d) Thus, s2 = sf + x2.sfg = 0.07188 + 10.8592(10.879952) = 0.8278 kJ/kg.K The entropy change of the refrigerant The negative sign indicates the entropy of the system is decreasing. This is not a violation of the second law, however, since it is Stotthat cannot be negative

20. ISENTROPIC PROCESSES A process during which the entropy remains constant is called an isentropic process: s = 0 or s2 = s1 (kJ/kg.K) Isentropic process can serve as an appropriate model for actual processes. A reversible adiabatic process is isentropic (s2 = s1),but an isentropic process is not necessarily a reversible adiabatic. (The entropy increase of a substance during a process as a result of irreversibilities may be offset by a decrease in entropy due to heat losses)

21. EXAMPLE 7–5 Isentropic Expansion of Steam in a Turbine Steam enters an adiabatic turbine at 5 MPa and 450°C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible.

22. EXAMPLE 7–5 Solution Take the turbine as the system (control volume, since mass crosses the system boundary during the process). There is one inlet and one exit, thus: The power output of the turbine is determined from energy balance,

23. EXAMPLE 7–5 Solution (Cont’d)

24. PROPERTY DIAGRAMS INVOLVING ENTROPY Isentropic proces on T-s diagram area under the curve = Qint rev

25. Mollier Diagram (h-s diagram) For adiabatic steady-flow devices, the h on an h-s diagram is a measure of work, and the s is a measure of irreversibilities. commonly used in engineering valuable in the analysis of steady-flow devices such as turbines, compressors, and nozzles

26. EXAMPLE 7–6 The T-S Diagram of the Carnot Cycle Show the Carnot cycle on a T-S diagram and indicate the areas that represent the heat supplied QH, heat rejected QL, and the net work output Wnet,outon this diagram.

27. Solution Carnot cycle is made up of two reversible isothermal (T constant) processes and two isentropic (s constant) processes Area A12B represents QH, the area A43B represents QL, and the area in color is the net work since:

28. Order to Disorder • The entropy of a system can be considered a measure of the disorder of the system. • Then the second law of thermodynamics can be stated as: “Natural processes tend to move toward a state of greater disorder.” Jika KIAMAT adalahpuncakkekacauanalamsemesta Thermo support

29. The Second Law: Remarks The second law of thermodynamics can be stated in several equivalent ways: • Heat flows spontaneously from a hot object to a cold one, but not the reverse. • There cannot be a 100 % efficient heat engine (one that can change a given amount of heat completely into work). • Natural processes tend to move toward a state of greater disorder or greater entropy.

30. Energy Availability; Heat Death • In any natural process, some energy becomes unavailable to do useful work. • As time goes on, energy is degraded, in a sense; it goes from more orderly forms (such as mechanical) eventually to the least orderly form, internal or thermal energy. • The amount of energy that becomes unavailable to do work is proportional to the change in entropy during any process. • A natural consequence of this is that over time, the universe will approach a state of maximum disorder. Heat Death! (KIAMAT!)

31. UJIAN MID Bahan: Bab 1, 2, 3, 4, 6, 7 Jumlahsoal: 10 (konsepdanesay) Tidakbolehmenghidupkan HP Waktu 55 menit Mulai jam 8 tet (no ganjil) Mulai jam 9 (no genap) Tanggal17 April