Content. Entropy and principle of increasing entropy. Change of entropy in an ideal gas.

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1 Entropy

2 Content Entropy and principle of increasing entropy. Change of entropy in an ideal gas.

3 Entropy Entropy can be viewed as a measure of molecular disorder, or molecular randomness. As a system becomes more disordered, the positions of the molecules become less predictable and the entropy increases. Thus, it is not surprising that the entropy of a substance is lowest in the solid phase and highest in the gas phase. In the solid phase, the molecules of a substance continually oscillate about their equilibrium positions, but they cannot move relative to each other, and their position at any instant can be predicted with good certainty. In the gas phase, however, the molecules move about at random, collide with each other, and change direction, making it extremely difficult to predict accurately the microscopic state of a system at any instant. Associated with this molecular chaos is a high value of entropy.

4 Clausius Inequality Consider an extremely reversible heat engine 1 as shown in the figure, which operates between the temperature limits T c and T f. Using equations of Carnot cycles we have,

5 Clausius Inequality Similarly, now consider a thermal machine 2 that operates between the same limits of temperature T c and T f, as shown in Figure. According to the principle of Carnot, W 1 > W 2 then, Q f1 < Q f2 And, Therefore, to the machine 2,

6 Clausius Inequality From the above equations it follows that The above equation is known as Clausius inequality. Since any reversible cycle can be replaced by a series of cycles of Carnot, Clausius inequality is valid for any thermal machine (or refrigerator) reversible or irreversible, where equality is preserved in reversible cycles and inequality in irreversible cycles. Note that with increasing the irreversibility in a given machine, the cyclic integral of dq/t becomes more negative.

7 Definition of Entropy Once established Clausius inequality, consider a reversible cycle consisting of the processes A and B as shown in schematic form in Fig. According to the Clausius inequality, Similarly, for the reversible cycle consisting of processes A and C,

8 Definition of Entropy Comparing the above equations, In the above equation shows that (dq/t) rev becomes the same value along any reversible path between state 2 and state 1. Accordingly, this amount depends solely on the initial and final states of the process, and is therefore a thermodynamic property. Indeed, the entropy S is defined by the equation,

9 Definition of Entropy That is, the entropy change between any two thermodynamic states can be determined using the expression, The above equation, valid for any closed system or constant mass, clearly stated that integration must be performed along a reversible path, if you want to evaluate the entropy difference between two states, and also requires knowledge of ratio dq/t throughout the reversible process. However, since the entropy is a thermodynamic property, the difference S between two states is the same regardless if the process is reversible or irreversible.

10 Definition of Entropy Furthermore, if the process in question is irreversible, the entropy difference between the states 1 and 2 can not be evaluated directly by using the above equation, therefore, the equation that applies generally to any process is where equality is preserved in reversible processes and the inequality in irreversible processes.

11 Definition of Entropy It follows from the integration of (dq / T) along an irreversible trajectory in a given process, do not result in the entropy difference between the initial state and the final state. Similarly, it can be added to all reversible adiabatic isentropic process. That is, all isentropic process is adiabatic, but not all adiabatic process is isentropic; only reversible adiabatic processes are isentropic. Furthermore, since entropy is a property, the difference S between two thermodynamic states can be evaluated using the entropy equation for reversible processes, as long as one or more reversible paths connecting the initial and final states of the process (reversible or irreversible) are selected.

12 The Increase of Entropy Principle Consider an isolated system, namely, a closed system where no energy transfer with the surroundings. According to the First Law of Thermodynamics, this system can only acquire those states in which the total internal energy of the system remains constant. On the other hand, the Second Law of Thermodynamics establishes, that the isolated state can only acquire those states that the entropy of this increase or remain constant. Namely, S 0 where equality is preserved in reversible processes and the inequality in irreversible processes.

13 The Increase of Entropy Principle The above equation is known as the Increase of Entropy Principle and is a quantitative way to establish the second law of thermodynamics. In fact, this axiom states that "The entropy of an isolated system increases in all irreversible processes, and the limit remains constant for reversible processes" Unlike the energy, entropy is not conserved except reversible processes. In other words, the first law states that energy is neither created nor destroyed, while the second law states that entropy is not destroyed, only created.

14 The Increase of Entropy Principle Since an isolated system can always be formed with any system (open or closed) and its surroundings, the above equation also indicates that S S 0 Since the surroundings are everything that surrounds a system, often is called universe the set formed by the system and its surroundings. Accordingly, S 0

15 Entropy as a Function of Other Properties For ideal gases:, ln ln, ln ln (kj/kgk) Constant Specific Heat (kj/kgk) Constant Specific Heat ln has absolute zero as the reference temperature (kj/kgk) Variable Specific Heat Tables

Examples (textbook) EXAMPLE 7 1 Entropy Change during an Isothermal Process A piston cylinder device contains a liquid vapor mixture of water at 300 K.

16 Examples (textbook) 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 pressure 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. Data: = 300 K, = 750 kj, =? 750 kj 300 K. /

17 Examples (textbook) EXAMPLE 7 2 Entropy Generation during Heat Transfer Processes 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. a) = 800 K, = 500 K, = 2000 kj, = 2000 kj, =? 2000 kj 2.5 kj/k 800 K 2000 kj 4.0 kj/k 500 K 2.5 kj/k 4.0 kj/k. /

18 Examples (textbook) b) = 800 K, = 750 K, = 2000 kj, = 2000 kj, =? 2000 kj 2.5 kj/k 800 K 2000 kj 2.7 kj/k 750 K 2.5 kj/k 2.7 kj/k. / The process in part (b) is smaller, and therefore it is less irreversible

19 Examples (textbook) EXAMPLE 7 3 Entropy Change of a Substance in a Tank 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. Data: = 5 kg, R134a, = 20 C, = 140 kpa, = 100 kpa, kj/kgk Table A m 3 /kg Table A13

0.16544 m 3 /kg Examples (textbook) @ 100 kpa 0.0007258 m 3 /kg 0.19255 m 3 /kg Table A12 @ 100 kpa 0.07182 kj/kg 0.

20 m 3 /kg Examples 100 kpa m 3 /kg m 3 /kg Table 100 kpa kj/kg kj/kg Table A kj/kg kj/kg kj/kg 5kg kj/kgk. /

21 Examples (textbook) 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. Data: = 5 MPa, = 450 C, = 1.4 MPa, = constant, =? kj/kg Table kj/kgk Table A6

Examples (textbook) 6.810 kj/kgk @ 1.4 MPa 2.2835 kj/kgk 6.

22 Examples (textbook) MPa kj/kgk kj/kgk Table kj/kg Table A6 Interpolation kj/kg kj/kg. /

23 Examples (textbook) EXAMPLE 7 8 Economics of Replacing a Valve by a Turbine A cryogenic manufacturing facility handles liquid methane at 115 K and 5 MPa at a rate of m 3 /s. A process requires dropping the pressure of liquid methane to 1 MPa, which is done by throttling the liquid methane by passing it through a flow resistance such as a valve. A recently hired engineer proposes to replace the throttling valve by a turbine in order to produce power while dropping the pressure to 1 MPa. Using data from Table 7 1, determine the maximum amount of power that can be produced by such a turbine. Also, determine how much this turbine will save the facility from electricity usage costs per year if the turbine operates continuously (8760 h/yr) and the facility pays $0.075/kWh for electricity.

24 Examples (textbook)

25 Examples (textbook) Data: Liquid, = 115 K, = 5 MPa, = m 3 /s, = 1 MPa, = constant, 8760 h/yr, $0.075/kWh, =?, $/yr kj/kg (Table kj/kgk (Table kg/m 3 (Table kj/kg (Table 7 1 Interpolation) kg/m kg/s m 3 /s kg/s kj/kg,

26 Examples (textbook) 1123 kw 8760 h kwh/yr Saving $/kwh kwh/yr $/kwh $, /

27 Examples (textbook) EXAMPLE 7 9 Entropy Change of an Ideal Gas Air is compressed from an initial state of 100 kpa and 17 C to a final state of 600 kpa and 57 C. Determine the entropy change of air during this compression process by using (a) property values from the air table and (b) average specific heats. a) = 100 kpa, = 17 C, = 600 kpa, = 57 C, =? Using Table A17 ln

28 Examples (textbook) 17 C K 57 C kj/kgk Table A kj/kgk Table A kj/kgk Table A1 ln 600 kpa kj/kgk kj/kgk ln 100 kpa. /

29 Examples (textbook) b) = 100 kpa, = 17 C, = 600 kpa, = 57 C, =? Using Table A2b, ln ln C 37 C K kj/kgk Table A2b interpolation kj/kgk ln. / K K kj/kgk ln 600 kpa 100 kpa

30 Homework 4b Problems from the textbook (Thermodynamics, Yunus, 8th ed.): Choose 5 conceptual problems and answer them (those who you consider to provide better understanding to the subject seen in this section) Chapter 7, problems: 1 18, 28, Choose 5 problems and answer them (those who you consider to provide better understanding to the subject seen in this section) Chapter 7, problems: 19 27, 29 58, 70 98

31 Entropy

Lecture 29-30: Closed system entropy balance

ME 200 Thermodynamics I Spring 2016 Lecture 29-30: Closed system entropy balance Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 200240, P.