1. AH

2.  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. 3  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. 4 Entropy is:  a measure of the disorder of 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) int rev  Entropy is a thermodynamic property (state function)

5.  Pertidaksamaan Clausius:  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  It has fixed values at fixed states.  Entropy change  S between two specified states is the same (whether reversible or irreversible) during a process.

7. 7  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 are internally reversible.  The entropy change of a system during an internally reversible isothermal process : or

9.  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.  From the Clausius inequality,       For an isolated system (adiabatic closed system, Q = 0)  S isolated ≥ 0

11.  Total entropy change:  S gen =  S total =  S syst +  S env ≥ 0  3 posibilities:  The 2 nd 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.  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. (a) For the heat transfer process to a sink at 500 K: (b) Similarly, for process at 750 K: Process (b) is less irreversible since it involves a smaller T difference (smaller irreversibility).

15.  In the saturated mixture region: s = s f + x.s gf (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.  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.

18.  The volume of the tank is constant, V 2 = V 1.  The properties of the refrigerant are:  At final state refrigerant is a saturated mixture since v f < v 2 < v g at 100 kPa, therefore:

19.  Thus, s 2 = s f + x 2.s fg  = 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 S tot that cannot be negative

20.  A process during which the entropy remains constant is called an isentropic process:  s = 0 or s 2 = s 1 (kJ/kg.K)  Isentropic process can serve as an appropriate model for actual processes.  A reversible adiabatic process is isentropic (s 2 = s 1 ), 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. 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.  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,

24. area under the curve = Q int rev Isentropic proces on T-s diagram

25.  commonly used in engineering  valuable in the analysis of steady-flow devices such as turbines, compressors, and nozzles 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.

26.  Show the Carnot cycle on a T-S diagram and indicate the areas that represent the heat supplied Q H, heat rejected Q L, and the net work output W net,out on this diagram.

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

28.  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 adalah puncak kekacauan alam semesta  Thermo support

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

30. 30  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.  Bahan: Bab 1, 2, 3, 4, 6, 7  Jumlah soal: 10 (konsep dan esay)  Tidak boleh menghidupkan HP  Waktu 55 menit  Mulai jam 8 tet (no ganjil)  Mulai jam 9 (no genap)  Tanggal 17 April