IS A PARTICULAR PROCESS / REACTION FEASIBLE? TO WHAT EXTENT DOES THE PROCESS / REACTION PROCEED?

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1 Limitations of First Law of Thermodynamics The first law of thermodynamics is a law of conservation of energy. It does not specify the direction of the process. All spontaneous processes processed in one direction only. The first law of thermodynamics does not deny the feasibility of a process reversing itself. The first law of thermodynamics does not provide answers to the following questions. IS A PARTICULAR PROCESS / REACTION FEASIBLE? TO WHAT EXTENT DOES THE PROCESS / REACTION PROCEED? IS COMPLETE CONVERSION OF INTERNAL ENERGY INTO WORK POSSIBLE? There exists a law which determines the direction in which a spontaneous process proceeds. The law, known as the second law of thermodynamics, is a principle of wide generality and provides answer to the above questions. It is essential to understand the meaning of the following terms in order to discuss the second law of thermodynamics: Thermal reservoir is a large body from which a finite quantity of energy can be extracted or to which a finite quantity of energy can be added as heat without changing its temperature. A source is a thermal reservoir at high temperature from which a heat engine receives the energy as heat. A sink is a low temperature thermal reservoir to which a heat engine rejects energy as heat.

2 Heat Engine A heat engine is a device which converts the energy it receives at heat, into work. It is a cyclically operating device. It receives energy as heat form a high temperature body, converts part of it into work and rejects the rest to a low temperature body. A thermal power plant is an example of a heat engine. Figure 16.1 Figure 16.1 describes a basic arrangement of a thermal power plant In the boiler, the working fluid receives a certain amount of heat from the hot combustion products. The superheated steam enters a turbine where it undergoes expansion performing the shaft work. The low pressure steam enters a condenser where it exchange energy as heat at constant pressure with the cooling water and emerges as the condensate. The condensate rejects a certain amount of heat to the cooling water.

3 The low pressure condensate from the condenser enters the pump. Work is done on the pump to elevate the condensate to the boiler pressure and return it to the boiler. In the above example, Work done by the system Energy absorbed as heat by the system Energy rejected as heat by the system According to first law of thermodynamics, the heat and work interaction are related by the equation. (1) Finally, the thermal efficiency of a heat engine can be expressed as or, (2) Heat Pump Heat Pump is cyclically operating device which absorbs energy form a low temperature reservoir and reject energy as heat to a high temperature reservoir when work is performed on the device. Its objective is to reject energy as heat to a

4 high temperature body (space heating in winter). The atmosphere acts as the low temperature reservoir. Refrigerator A refrigerator is a cyclically operating device which absorbs energy as heat from a low temperature body and rejects energy as heat to a high temperature body when work is performed on the device. The objective of this device is to refrigerate a body at low temperature. Usually it uses atmosphere as the high temperature reservoir. Figure 16.2 Refer to figure Let and represents the amount of energy absorbed as heat from the low temperature reservoir and the energy rejected as heat to the high temperature reservoir respectively, Let W be the work done on the device to accomplish the task. (3) Therefore,

5 (4) and (5) Heat engine and the refrigerator (/heat pump) can be represented as shown in Figure The efficiency of a heat engine is given by (6) since (heat) transferred to the system cannot be completely converted to work in a cycle. Therefore is less than unity. A heat engine can never be 100 efficient. Therefore i.e., there has always to be a heat rejection. Thus a heat engine has to exchange heat with two reservoirs, the source and the sink. This experience leads to the proposition of the second law of thermodynamics which has been stated in several different ways. KELVIN PLANCK STATEMENT It is impossible to construct a cyclically operating device such that it produces no other effect than the absorption of energy as heat from a single thermal reservoir and performs an equivalent amount of work. The only option then is that the engine converts part of the energy it receives as heat into work and rejects the rest to another thermal reservoir the temperature of which is less than the temperature of

6 the source. Two thermal reservoirs, one of high temperature (source), from which the working fluid receives energy as heat, and the other of low temperature (sink), to which the working fluid rejects energy as heat, are needed for a heat engine. Once the heat engine rejects a part of the energy it receives, its efficiency becomes less than one. Thus the Kelvin Planck statement further implies that no heat engine can have a thermal efficiency of one (hundred percent). This does not violate the first law of thermodynamics either. Kelvin - Planck Statement (continued) Second law restricts the thermal efficiency of a heat engine to less than one. It stipulates that some portion of the energy absorbed as heat from a source must always be rejected to a low temperature sink. Wilhelm Ostwald introduced the concept of perpetual motion machine of the second kind (PMMSK or PMM2), that is, of a device which would perform work solely by absorbing energy as heat from a body. Such a device does not violate the first law of thermodynamics. Figure 17.1 A PMMSK is a hypothetical device (Figure 17.1) which working cyclically,

7 receives energy as heat from a single thermal reservoir, and delivers as equivalent amount of work. The Kelvin-Planck statement of the second law tells us that it is impossible to constructs a perpetual motion machine of the second kind. Clausius Statement of the Second Law Heat always flows from a body at higher temperature to a body at a lower temperature. The reverse process never occurs spontaneously. Clausius' statement of the second law gives: It is impossible to construct a device which, operating in a cycle, will produce no effect other than the transfer of heat from a low-temperature body to a high temperature body. This statement tells us that it is impossible for any device, unaided by an external agency, to transfer energy as heat from a cooler body to a hotter body. Consider the case of a refrigerator or a heat pump (Figure 17.2) Figure 17.2 When and

8 It is impossible to construct a refrigerator or a heat pump whose COP is infinity. Consider a domestic refrigerator, this device extracts energy as heat from the substance to be coded and transfers it to the surroundings. The refrigerator is supplied with electric power. Energy transfer as heat from a high temperature body to a low temperature body is a spontaneous process. The Clausius statement of the second law of thermodynamics tells that this spontaneous process cannot proceed in the reverse direction. Apparently, the Kelvin Planck statement and the Clausius statement of the second law of thermodynamics are altogether different. They are not! Instead, they are equivalent. A violation of Kelvin Planck statement leads to a violation of the Clausis statement too and vice-versa. Clausius Statement of the Second Law (contd...) Let us suppose that the Kelven Planck statement is incorrect. Figure 17.3 Refer to Figure 17.3 that is, it is possible to construct a device I which, working cyclically, absorbs energy a heat from a source at temperature and performs an equivalent amount of work. Next consider a device II which absorbs amount of energy from a low temperature body at and delivers energy as heat to a high temperature reservoir at. To accomplish this, work is done on the device. The device II does not violate the Clausius statement. For device II,

9 we can write. Now combine I and II. The work delivered by device I is used by device, II. Then (1) (2) This combined device (which is no more aided by any external agency) working cyclically, is not producing any effect other than the transfer of energy as heat from the low temperature reservoir to the high temperature reservoir. This is in violation of the Clausius statement. Figure 17.4 To prove that violation of the Clausius' statement leads to violation of Kelvin Planck statement, let us assume that the Clausius' statement is incorrect. That is, it is possible to constructs a device I (refer to Figure 17.4) such that it transfers energy as heat from a body at lower temperature to a body at higher temperature unaided by any external agency. Consider another device II which receives energy as heat from a body at higher temperature, delivers work and rejects energy as heat to the body at a low temperature. Device II does not violent Kelvin Planck statement. Application of the first law of thermodynamics to device

10 II gives, (3) Now consider the combination of devices I and II as a single device. This combined device, working cyclically, absorbs as heat from the thermal reservoir at temperature amount of energy and delivers work, leaving the thermal reservoir at temperature unaffected. That is, the resulting device is a PMMSK, which is in violation of the Kelvin Planck statement. Thus the Kelvin Planck statement and the Clausius' statement are equivalent. Reversibility, Irreversibility and Carnot cycle The second law of thermodynamics distinguishes between reversible and irreversible processes. If a process can proceed in either direction without violating the second law of thermodynamics, it is reversible process. A reversible process is carried out infinitely slowly with an infinitesimal gradient, so that every state passed through by the system is an equilibrium state. So, a reversible process is a quasi-static process which can proceed in either direction. Given a process, if the attempt to reverse its direction leads to a violation of the second law of thermodynamics, then the given process is irreversible. Any natural process carried out with a finite gradient is an irreversible process. A reversible process which consists of a succession of equilibrium states, is an idealized hypothetical process, approached only as a limit. It is said to be an asymptote to reality, All spontaneous processes are irreversible.

11 Irreversible Processes The example of irreversible processes are: Motion with friction, free expansion, Expansion/ compression with finite pressure difference, Energy transfer as heat with finite, Mixing of matter at different states, Mixing of non-identical gases. Reversible Processes The processes which can be idealized as reversible are: Motion without friction, Expansion/compression with infinitesimal pressure difference, Energy transfer as heat with infinitesimal temperature difference. Carnot Cycle A French engineer Sadi Carnot was the first to introduce the idea of reversible cycle. From the second law, it has been observed that the efficiency of a heat engine is less than unity. If the efficiency of heat engine is less than unity, what is the maximum efficiency of a heat engine? This can be answered by considering the Carnot cycle. The concept of carnot cycle is executed via Carnot engine. Carnot Engine Let us consider the operation of a hypothetical engine which employs the Carnot cycle. The Carnot engine consists of a cylinder-piston assembly in which a certain amount of gas(working fluid) is enclosed. Refer to Figure 18.1 representing the Carnot cycle.

12 Figure 18.1 Reversible Isothermal Heat Addition In the first process, the cylinder head is brought into contact with a source at temperature. The gas inside the cylinder is also at temperature. The gas expands reversibly and isothermally. During this process, the system absorbs energy as heat from 1 to 2 on the where, for an ideal gas, from the source. The system changes its state diagram. (1) Reversible Adiabatic Expansion In the second process, the cylinder head is insulated and the gas is allowed to expand till its temperature is equal to the sink temperature. The system thus reaches state 3. This is a reversible adiabatic process. (2)

13 Reversible Isothermal Heat Rejection In the next process, the system is brought into contact with the sink which is at a temperature. The heat leaves the system and the internal energy further decreases (3) where, only for an ideal gas, Through a reversible isothermal process the system reaches state 4. Reversible Adiabatic Compression In the next process, the gas is compressed reversibly and adiabatically till it reaches the initial state 1, thus, completing the cycle. (4) Summing up all the processes, one can write or,

14 The thermal efficiency, (5) Efficiency of Carnot Engine Using Ideal Gas 1-2: A reversible isothermal expansion with heat addition (6) 2-3: A reversible adiabatic expansion (7) 3-4: A reversible isothermal compression with heat rejection (8) 4-1: A reversible adiabatic compression (9)

15 (10) Energy absorbed as heat (11) Thermal efficiency, (12) Here, for the ideal gases we can write Also, or,

16 (13) So, (14) Two consequences of the second law of thermodynamics are will known as Carnot's principles. Principle I: No heat engine operating between the two given thermal reservoirs, each of which is maintained at a constant temperature, can be more efficient than a reversible engine operating between the same two thermal reservoirs. Refer to Figure 19.1 Figure 19.1 Let two heat engines and operate between the given source at temperature and the given sink at temperature as shown.

17 Let be any heat engine and any reversible heat engine. We are to prove that the efficiency of is more than that of. Let us assume that it is not true. Let the rates of working of the engines be such that (1) Since, (2) Now let the direction of be reversed. Figure 19.2 Refer to Figure Since is a reversible heat engine, the magnitudes of heat and work quantities will remain the same, but their directions will be reversed as shown. Since some part of (equal to ) may be fed to drive the reversed heat engine. Since,, the heat discharged by the reversed may be supplied to.

18 The source may, therefore, be eliminated. The net result is that and together constitute a heat engine which, operating in a cycle, produces net work, while exchanging heat with a single reservoir at. This violates the Kelvin-Planck statement of the second law. Hence the assumption is wrong. Therefore, (3) Principle II: All reversible heat engines operating between the two given thermal reservoirs have the same efficiency. The efficiency of reversible heat engine does not depend on the working fluid, it depends only on the temperature of the reservoirs between which it operates. To prove the proposition, let us assume that the efficiency of the reversible engine engine. is greater than the efficiency of the reversible

19 Figure 19.3 Refer to Figure The engine absorbs energy as heat from the constant temperature thermal reservoir at, does work and rejects energy as heat to the reservoir at. The engine absorbs energy as heat from the reservoir at, does work and rejects energy as heat to the reservoir at. Then and (4) and (5) By assumption, Then, (6) Therefore, Since is a reversible engine, it can be made to execute the cycle in the reversed order. That is, when work is performed on the device, it absorbs energy as heat, from the reservoir at and rejects energy as heat to the reservoir at. Since,, can be run as a heat pump utilizing part of the work done by. The combination of the two devices is also shown in the figure. The net work done by the device is given by

20 (7) The resulting device absorbs energy as heat from the reservoir at. Does not require any interaction with the second reservoir. Delivers an equivalent amount of work. This is in violation of the Kelvin-Planck statement of the second law of thermodynamics. Hence the assumption that incorrect. Therefore, Nov let us assume that the reversible engine, is (8) is more efficient then the reversible engine. Then the reversible engine can be run as a heat pump, utilizing the part of the work done by. By following the similar argument as the earlier case, we can arrive at the result that, (9) Hence, it can be concluded that (10) Stated in works: All reversible engines operating between the two given thermal reservoirs have the same efficiency. Thermodynamic Temperature Scale

21 A temperature scale, which does not depend on the thermodynamic property of the substance can be established by making use of the fact that the efficiency of a reversible heat engine does not depend on the nature of the nature of the working fluid but depends only upon the temperature of the reservoirs between which it operates. The establishment of thermodynamic temperature scale is also a consequence of the second law of thermodynamics. Refer to Figure Figure 20.1 Let, and be the isotherms. Let the two reversible adiabatics intersect the three isotherms. Let the reversible engine 1 absorb energy as heat, from the reservoir at and reject energy as heat, to the reservoir at. Let the reversible engine 2 absorb energy as heat, from the reservoir and reject heat, to the reservoir at. Let and represent the work done by engines 1 and 2, respectively. Let a third Carnot engine absorb energy as heat from the reservoir at and reject energy as heat to the reservoir at. The work done by the engine is given by (1)

22 Also, (2) or, (3) or, (4) Now, (5) or, (6) or, (7) Similarly,

23 (8) or, (9) or, (10) Then, (11) or, (12) In the above equation, does not appear on the left hand side. Therefore, on the right hand side, the dependence on should cancel out. Hence, the nature of function F is such that, (13) Then,

24 (14) Therefore, or, (15) So, (16) There is no further restriction on the function. The choice of the function now defines the temperature scale of interest. If we choose (17) (18) or, (19) Recall, (20)

25 or, (21) The efficiency of no heat engine can be greater than one. The lowest possible temperature in this scale is zero. Figure 20.2 Refer to Figure Hence it is called the absolute temperature scale. The absolute temperature scale is also known as Kelvin temperature scale. In defining the Kelvin temperature scale also, the triple point of water is taken as the standard reference point. For a Carnot engine operating between reservoirs at temperature and being the triple point of water arbitrarily assigned the value of, or, (22)

26 If this equation is compared with the perfect gas temperature scale [remember, we have obtained where, it is seen that in the Kelvin Scale Q plays the role of the thermometric property. Consider a Carnot engine using a perfect gas as a working fluid. Let the high temperature reservoir be at and low temperature reservoir be at (23) or, But the absolute temperature scale is defined by the relation (24) Therefore, Since the numerical value assigned to the triple point of water can be chosen to be same for both the perfect gas scale and the absolute temperature scale,. Hereafter, the common symbol may be used to represent the temperature.

27 Substitution of a reversible process by the reversible isothermal and reversible adiabatic processes: Figure 21.1 Let a system be taken from an equilibrium state to another equilibrium state by following the reversible path. Refer to figure 21.1 Let a reversible adiabatic and another reversible be drawn in such a manner that and can be joined by a reversible isotherm. Also, these reversible adiabatic and reversible isothermal lines are such that the area under is equal to the area under. Applying the first law for, Applying the first law for, (1) (2) We know (area under the curves are same) From Eqn. (21.1) and (21.2) (3)

28 Since and are zero (4) Thus, any reversible path may be substituted by a reversible adiabatic, a reversible isotherm and a reversible adiabatic between the same end states such that the heat transferred during the isothermal process is the same as that transferred during the original process. Clausius Inequality Consider a system undergoing a reversible cycle. The given cycle may be sub-divided by drawing a family of reversible adiabatic lines. Every two adjacent adiabatic lines may be joined by two reversible isotherms (refers to Figure 21.2) Figure 21.2 Now, and (5) Also, is a Carnot cycle which receives heat during

29 the process and rejects heat during the process. Let the heat addition be at temperature and the heat rejection be at temperature. Then it is possible to write, and (6) or, (7) Since is negative, it reduces to (8) Similarly for the cycle (9) If similar equations are written for all the elementary cycles, then (10) or, (11)

30 Let us go back to the cycle (12) Now where, and this is not equal to. For the irreversible cycle, (13) or, and (14) because is negative. Similarly, for the irreversible cycle (15) Summing up all elementary cycles (16) The above two conclusions about reversible and irreversible cycles can be generalized as

31 (17) The equality holds good for a reversible cycle and the inequality holds good for an irreversible cycles. The complete expression is known as Clausius Inequality. Concept of Entropy Clausius inequality can be used to analyze the cyclic process in a quantitative manner. The second law became a law of wider applicability when Clausius introduced the property called entropy. By evaluating the entropy change, one can explain as to why spontaneous processes occur only in one direction. Figure 22.1 Consider a system in initial state 1. Let the system be taken from state 1 to state 2 along a reversible path 1-A-2, and then be restored to its initial state by following another reversible path 2BI (Figure 22.1). Then the two paths put together form a reversible cycle 1A2BI. Apply the Clausius inequality to this reversible cycle and

32 obtain (1) or, (2) or, (3) Since path 2B1 is reversible, the limits of the integral can be reversed. That is, has the same value whether the path followed is 1A2 or 1B2. It is possible to connect the states 1 and 2 by several reversible paths and see that has the same value irrespective of the path as long as the paths are reversible. Therefore is an exact differential of some function which we identify as entropy. Hence it can be said that there exists a function S, called entropy, the change in entropy is expressed as This follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore Calculation of Entropy Change (4) (for reversible process only) (5) The following facts should be kept in mind while calculating the change in entropy for a process 1. for a reversible process

33 2. Entropy is a state function. The entropy change of a system is determined by its initial and final states only, irrespective of how the system has changed its state. 3. In analyzing irreversible proceses, it is not necessary to make a direct analysis of the actual process. One can substitute the actual process by a reversible process connecting the final state to the initial state, and the entropy change for the imaginary reversible process can be evaluated. Entropy change for some elementary processes a. Absorption of energy by a constant temperature reservoir A certain amount of heat is added to a constant temperature reservoir. The actual process can be replaced by a reversible path in which an equivalent amount of energy is added to the reservoir. Then, the entropy change of the reservoir is given by b. Heating or Cooling of matter The heating can be carried out either at constant pressure or at constant volume. From the first law of thermodynamics for constant volume heating/cooling process for constant pressure heating/cooling process (6), for a constant pressure process (7) or,, for a constant pressure process (8) Similarly,

34 , for a constant volume process (9) or, ln, for a constant volume process (10) (c) phase change at constant temperature and pressure Melting : [ solid to liquid] Evaporation: [ liquid to vapor] (d) Change of state for an ideal gas If an ideal gas undergoes a change of state from to (11) or, (12) or, (13) or, or, ln (14)

35 Again, or, (15) or, (16) or, (22.17) or, (18) ln ln (19) For a constant temperature process is either ln or, ln. (e) Mixing of non-identical gases A rigid insulated container is divided into two compartments by a partition.

36 Figure 22.2 Refer to Figure One compartment contains mol of an ideal gas A at pressure P and temperature T while the second compartment contains mol of an ideal gas B at the same pressure and temperature. The partition is removed and the gases are allowed to mix. To evaluate the entropy change on mixing, let us device the following reversible process. Let the gases A and B expand reversible and isothermally to their final partial pressures and respectively. Then the gases are introduced through semi-permeable membranes into a container in which the partial pressure of gas A is and the partial pressure of gas B is. The reversible mixing process is, therefore, accomplished in the following two steps. Reversible isothermal expansion of gases to their respective partial pressures. Forcing the gases reversibility through semi permeable membranes. For gas A, ln (20)

37 Similarly, for gas B, ln (21) Where, and Here and are the mole fractions of gases A and B respectively, in the final mixture. Hence, ln ln and (22) Therefore, the entropy change associated with the mixing of non-identical gases is or, In - R In (23) In In (24) In (25) Where is the molar entropy. Since, is always positive.

38 Figure 23.1 Refer to Figure Let a system change from state 1 to state 2 by a reversible process A and return to state 1 by another reversible process B. Then 1A2B1 is a reversible cycle. Therefore, the Clausius inequality gives: If the system is restored to the initial state 1 from state 2 by an irreversible process C, then 1A2C1 is an irreversible cycle. Then the Clausius inequality gives: Subtracting (23.2) from (23,1) (1) (2) (3) Since the process 2B1 is reversible (4)

39 or, (5) This can be generalized as Where the equality sign holds good for a reversible process and the inequality sign holds good for an irreversible process. Now let us apply the above results to evaluate the entropy of the universe when a system interacts with the surroundings. Refer to Figure (6) Figure 23.2 Let Temperature of the surroundings Temperature of the system and, (7) If the energy exchange takes place, surroundings to the system. will be the energy transfer from the

40 (8) (9) (10) [since ] (11) So, (12) If the system is isolated, that is, when there is no interaction between the system and the surroundings, then (13) Now let us redefine a system which includes our earlier system and its surroundings. Then for an isolated system (14) Through generalization we can write for an isolated system (15) Examine the statement critically. For a reversible process it is equal to zero.

41 For an irreversible process it is greater than zero. Similarly it is possible to write. The equality sign holds good if the process undergone is reversible; the inequality sign holds good if the process undergone is irreversible. Clausius summarized these as: Die Energie der Welt ist konstant. Die Entropie der Welt strebt einem Maximum. This means: The energy of the universe in constant (first law). The entropy of the universe tends towards a maximum (second law) The lecture contains Temperature Entropy Diagram Second law analysis of a control volume Steady-state steady-flow processes TdS Equations Entropy change of an incompressible substance Temperature Entropy Diagram Entropy change of a system is given by. During the reversible process, the energy transfer as heat to the system from the surroundings is given by (1)

42 Figure 24.1 Refer to figure Here T and S are chosen as independent variables. The gives Therefore, is the area under the curve. The first law of thermodynamics. Also for a reversible process, we can write, and (2) For a cyclic process, the above equation reduces to For a cyclic process, the above equation reduces to For a cyclic process represents the net heat interaction which is equal to the net work done by the system. Hence the area enclosed by a cycle on a T S diagram represents the net work done by a system. For a reversible adiabatic process, we know that or, (3) (4) (5)

43 (6) or, (7) Hence a reversible adiabatic process is also called an isentropic process. On a T S diagram, the Carnot cycle can be represented as shown in Fig The area under the curve 1-2 represents the energy absorbed as heat by the system during the isothermal process. The area under the curve 3-4 is the energy rejected as heat by the system. The shaded area represents the net work done by the system. We have already seen that the efficiency of a Carnot cycle operating between two thermal reservoirs at temperatures T 1 and T 2 by is given (8) This was derived assuming the working fluid to be an ideal gas. The advantage of T S diagram can be realized by a presentation of the Carnot cycle on the T S diagram. Let the system change its entropy from to and, and, during the isothermal expansion process 1-2. Then, (9) (10)

44 or, (11) This demonstrates the utility of T S diagram. Second law analysis of a control volume It was shown that the change in entropy of a system is given by (12) Where the equality sign holds good for the reversible processes and the inequality sign holds for all irreversible processes. This can be expressed as (13) Where represents entropy generation in the system and it cannot take a negative value. for a reversible process. for an irreversible process. Consider a control volume through which material flows (Figure 24.2)

45 Figure 24.2 Let us identify a system such that its mass remains constant during a given process. At time t, the system constitutes both the mass inside the control volume and the mass about the enter the control volume during a small interval dt (that is, the mass occupying the region A). At time (t+dt), the system is the mass inside the control volume as well as the mass in region B. During the time interval dt, the system undergoes a change in configuration and receives energy as heat from the surroundings. At time t, the entropy of the system (14) At time, the entropy of the system (15) Then, (16) or, (17)

46 or, (18) This can be rearranged as (19) In words: The rate of accumulation of entropy = Rate of inflow of entropy Rate of outflow of entropy + Rate of generation of entropy Steady-state steady-flow processes The above equation can be simplified if one assumes steady flow and steady state conditions. Under these conditions, there is no change in the entropy of the control volume with respect to time, that is. The steady flow assumption implies that and that is a constant. Then we can write, (20) or, (21)

47 For an adiabatic process therefore. However, if the process is reversible and adiabatic then Summary In this lecture we have shown that The Carnot cycle on a T S diagram and identify the heat transfer at both the high and low temperatures, and the work output from the cycle. 1-2, reversible isothermal heat transfer area 1-2-B-A. 2-3, reversible adiabatic expansion isentropic process, S = constant. 3-4, reversible isothermal heat transfer area 3-4-a-B. 4-1, reversible adiabatic compression isentropic process,. Net work, the area enclosed by , the shaded area. We have also done a Second Law Analysis of the Flow processes (control volume based method) TdS Equations For a closed system contaning a pure compressible substance undergoing a reversible process. (22) or (per unit mass) (23)

48 This is the famous Gibbsian equation. Eliminate du by using the definition of enthalpy (24) Thus, (25) (26) Also, (27) Important: These equations relate the entropy change of a system to the changes in otherproperties :. Therefore, they are independent of the process. These relations can be used for reversible as well as irreversible processes. Example Problem-1 Consider steam is undergoing a phase transition from liquid to vapor at a constant temperature of 20 0 C. Determine the entropy change using the Gibbsian equations and compare the value to that read directly from the thermodynamic table.

49 Corresponding to change from liquid to vapor From table: It compares favourably with the tabulated value Entropy change of an incompressible substance For most liquids and all solids the density is not changed as pressure changes, that is,. Gibbsian equation states that,, for an incompressible substance is a function of temperature only. Therefore, Integrate to determine the entropy change during a process where is the averaged specific heat of the substance over the given temperature range. Specific heat for some common liquids and solids can be found in thermodynamic tables.

50 Example Problem 2 An 1-kg metal bar initially at 1000 K is removed from an oven and quenched by immersing in a closed tank containing 20 kg of water initially at 300 K. Assume both substance are incompressible and c (water) = 4 (kj/kg K), c (metal) = 0.4 (kj/kg K). Neglect heat transfer between the tank and its surroundings. (a) Determine the final temperature of the metal bar, (b) entropy generation during the process. Solution a. Energy balance from the first law: no heat transfer and no work done, both bar and water reach final temperature. b. No heat transfer with surroundings. The entropy balance of the system (generation) = In In

51 In In ( kj / K ) The total entropy of the system increases, thus satisfies the second law. Example Problem 3 Steam enters an adiabatic turbine at 5 MPa and C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass flowing through the turbine if we can assume the process is reversible and neglect all changes of KE and PE. Refer to Figure 24.3 Figure 24.3 State 1: table: F From State 2: at 1.4M Pa (from table) State 2 is superheated. From table,

52 Available Energy In order to determine that part of the energy which can be converted into work by an engine, we require consider a thermal reservoir at constant temperature T from which a quantity of energyq is being absorbed as heat. A Carnot engine may be operated between the reservoir at T and the surroundings at temperature T 0 to convert part of the energy into work. The amount or work that can be obtained is given by (1) Where is the change in the entropy of the reservoir that supplies the energy. The Carnot cycle is shown on a diagram (Figure 25.1) Figure 25.1 The area 1234 represents the portion of energy delivered as work by

53 the reversible engine and is called available energy. The area 43BA represents the portion of the energy which is rejected to the atmosphere. The portion of energy is not available for conversion into work is called unavailable energy. Therefore, the available energy represents the portion of the energy supplied as heat which can be converted into work by means of a reversible engine. It is also known as EXERGY. The result is true also for the case of a finite body, which acts as a source, the temperature of which varies during the interaction. Let us consider a finite body from which energy in the form of heat can be absorbed by a heat engine. As the energy is with drawn from the finite body, the temperature of the body keeps on decreasing. This process can be carried out using a large number of differential Carnot engine. For a differential Carnot cycle, let represent the energy absorbed at temperature T. The differential amount of work done by the engine is given by (2) Figure 25.2

54 If the Carnot engine works till the temperature of the body attains a value (Figure 25.2) the total work done by the reversible engine is given by (3) Therefore, if a certain amount of energy, Q is absorbed as heat from a finite body, then represents the available portion of the energy which can be converted into work. The unavailable energy is also known as ANERGY. Loss of Available Energy due to Heat Transfer through a Finite Temperature Difference Suppose a certain amount of energy Q is transferred as heat from a body at constant temperature temperature. to a body at constant Because of the finite temperature difference between the two interacting bodies, the energy transfer is an irreversible process. This irreversibility leads to a loss in the available energy. This loss in the available energy would not have occurred, had a reversible engine been employed to transfer the energy from the body at to the body at. Suppose the same quantity of energy Q is absorbed by a reversible engine. Before the energy transfer takes place the energy is available at temperature available at a lower temperature. After the energy transfer the energy is. Let the ambient be at a

55 temperature. Then the initial available energy (4) the final available energy (5) Therefore the loss of available energy (6) (7) Where that whenever is the change in the entropy of the universe. It implies, there is a loss in the available energy. The second law of thermodynamics states that,, where the equality is true for reversible processes and the inequality is true for all Spontaneous or irreversible processes. Therefore, all spontaneous or irreversible processes are associated with a loss of available energy. Note that though the same amount of energy as heat is available before and after the irreversible energy transfer, the available energy is less at compared to the available energy at. That is, there is a degradation in the quality of the energy. The internal energy at lower temperature is considered as degraded energy. We have seen that all the energy absorbed as heat cannot be converted into work. That is, the internal energy of a reservoir or a body is a low quantity energy. On the other hand, work is considered as the highest quality of energy, because it is interchangeable to the other forms

56 with hundred percent energy conversion efficiency. It is possible to convert work completely into internal energy, but not viceversa. The entropy generation associated with a heat transfer process can be well explained by Figure 25.3 (a) and (b). Figure 25.3 In Figure 25.3 (a) the presence of a well is ignored. The quantity signifies entropy transfer. It is clear from both the cases, that whenever energy is transferred through a finite temperature difference, entropy generation is brought about. Entropy generation is an obvious outcome of an Irreversible process. Avavialibility and IRREVERSIBILITY Introduction An engineer is interested in knowing the maximum useful work that can be obtained from a system when it is in a given state. Obviously, the work obtained will be the maximum only when the final state of the system is in equilibrium with the surroundings. In other words, till it reaches a state

57 when there is no possibility of obtaining further work from the system due to interaction with the surroundings. The availability of a given system is defined as the maximum useful work that can be obtained in a process in which the system comes to equilibrium with the surroundings or attains a dead state. Clearly, the availability of a system depends on the condition of the system as well as those of the surroundings Availability Function for a non-flow Process Let the ambient pressure be and the temperature be. Since the system comes toequilibrium with the ambient conditions, its final pressure is. Let and represent the initial and final volumes of system respectively. Then the change in volume of the system is (1) This change in volume of the system is resisted by the ambient pressure. Then the work done in pushing back the ambient pressure is. Hence the useful work that can be obtained from the system is the maximum work less the work done in pushing back the ambient atmosphere. Then Availability = (3) To determine the availability of a system, let us consider a system which interacts with the ambient at We can write (2)

58 or (4) or (5) Now, (6) We can also write (7) Where is the Availability Function, given by If a system undergoes a change of state from state-1(where Availability to the state-2 (where Availability, the change in Availability is given by, (8) (9) Availability Function of Flow Processes

59 Maximum work that can be obtained in a steady flow process while the control volume exchanges energy as heat with the ambience at T 0 can be calculate in the following way. For a steady flow process, we can write or, (10) The second law of thermodynamics applied to the control volume for reversible process gives or From (26.10) and (26.11) (11) (12) or Since no part of the shaft work is spent in pushing back the atmosphere, there is no difference between the Maximum Useful Work and the Maximum Work. The dead state of flow process implies kinetic energy is zero relative to surroundings, in addition to Thermal and mechanical equilibrium.

60 The stream Availability is given by (13) Where the subscript 0 refers to dead state. If the CV undergoes a change from initial state-1 to final state-2, then the change in availability is (14) Many industrial processes involve energy interaction between two bodies. We can consider the combination of both the interacting bodies as our system and evaluate the change in the availability of the system. Example In a desalination plant, drinking water is produced by condensing steam in a heat exchanger. Steam at 0.1 MPa and C enters the heat exchanger. Shown in fig at the rate of kg/h and emerges as a liquid at 40 0 C. Cooling water enters the heat exchanger at 25 0 C and leaves at 35 0 C. Estimate the loss in availability in the heat exchange process. The ambient temperature is 300K Figure 26.1 First law:

61 = kj/ kg = kj/ kg K = kj/ kg = kj/ kg K = kj/ kg = kj/ kg K = kj/ kg = kj/ kg K 10 4 (2775.8) + (104.77) = 10 4 (167.45) + (146.56) or Change in availability of system Change in availability of cooling water Net change in availability But

62 Hence net change in the availability kj Therefore the loss in availability kj Irreversibility The irreversibility of a process is defined as the difference between the useful work and the actual work (15) (16) Consider the system ant it surroundings as a compositive system, the work done by the composite system under adiabatic condition or (17) Application of the two laws of thermodynamics to the surroundings gives (18) (19)

63 Substituting (26.19) in (26.16) or (20) Net change in the volume of the composite system is zero (21) For an irreversible process. and for a reversible process, Irreversibility (Alternative Method) Figure 26.2 Refer to Figure 26.2

64 (22) Thermodynamic Relations Gibbs and Helmholtz functions, Maxwell Relations Introduction In thermodynamics, an important task is to develop relationships which express the nonmeasurable properties like U,H,A,G and S in terms of measurable properties P,V & T The number of partial derivatives of the type among the above eight properties are If there exists a relation S=S (T,P) among the three variables S,T,P then (1) Dividing the above relation with dt while V is held constant, we obtain

65 (2) There exists one relationship among four partial derivatives (3) Therefore a large number of relationships is possible among 336 partial derivatives. Important Mathematical Relations Suppose there exists a relationship among the there variables and z. Then the following relation can be obtained. The chain rule of partial differentiation (4) Reciprocal relation, (5) Cyclic relation, (6) Jacobian Methods

66 (7) In general can be written as Let This is a special case (8) Similarly (9) 1. Properties: (10) 2. Transposition (11) 3. Inversion (12) 4. Chain rule

67 (13) 5. Cyclic rule (4) (5) (16) or (17) Let us consider (18) (19) In Jacobian form (20) again

68 Let us consider (21) Earlier we have obtained the Tds equations. These can be rewritten as (22) (23) In addition, we know the availability function and Gibbs function (24) (25) These equations can be generated through Mnemonic diagram (figure-32.1)

69 Figure 32.1 Differential (property) = differential of (independent variables) Sign convention: going away from independent variable positive Coming in toward independent variable negative Equations (32.23) (32.25) can now be expressed in the form Let b be the dummy variable (26) (27) (28) Maxwell Relations Maxwell relations are links between the independent variables. We require focus on the independent variables and the directions of the arrows only. From figure 32.1 we can write (30)

70 Since both the arrow are in the same direction (upward), there is no change in sign. For the next Maxwell relation, we require. To rotate the Mnemonic diagram by 90 0 (Figure 32.2) Figure 32.2 (31) For next relation, we require to rotate it by 90 0 again (Figure 32.3) Figure 32.3 (32)

71 Similarly for the other relation we require to rotate another 90 0 (Figure 32.4) And the relation is Figure 32.4 (33) Equations (32.30) - (32.33) can be re-written as (34) (35) (36) (37) All four relations can be summarized as

72 or (38) Other measurable quantities, such as Coefficient of volume expansion, (39) Compressibility (40) Specific heat at constant pressure (41) Specific heat at constant volume (42) Therfore (43) Examples: Thermodynamics Relations involving Entropy The entropy of a substance can be expressed in terms of independent variable (T,P) or (T,v) and these relations are quite useful in estimating the entropy change when the system changes its state from to (a) Entropy as a function of T and P

73 or Maxwell relation and or (44) (b) Entropy a as function of T and v using

74 or (45) Clapeyron Equations Let us study v T diagram of a system. The system consists of a liquid phase at state A and a vapour phase at state B (figure 33.1) FIGURE 33.1 The line AB corresponds to the pressure P and temperature T. Gibbs free energy can be conveniently used in analyzing the problem. Using Gibbs equation we can write (1) Along AB both dp and dt vanish; therefore dg=0 Hence, Gibbs function at a given pressure and temp has the same value for the saturated liquid and the saturated vapour. The same result holds good for solid-liquid transition. Next, consider a neighbouring line corresponding to pressure and temperature The value of the specific Gibbs function on the line may be denoted by Then assuming etc to be small, the change along is given by (2) While the same along is given by

75 (33.3) Hence form (33.2) and (33.3) we get, (.5) Diving by and proceeding to the limit (5) As transition from A to B is a constant pressure and constant temperature process, we get (6) Where L or is the latent heat of evaporation at temperature T Hence (7) This is known as Clapeyron Equation. Also valid for solid - liquid transition. is always positive so will depend on Liquid vapour: is always positive. At higher pressure, boiling point is higher. But for solid liquid: sometimes is negative; Notably for water: Melting is accompanied by contraction. A rise is pressure will lower down the melting point. (figure 33.2) This explains the phenomenon of relegation.

76 FIGURE 33.2 The Clapeyron equation can be put in a simple form if we make certain approximations. Let us consider the liquid vapour phase transition at low pressures. Then the vapour phase may be approximated as an ideal gas. Moreover the volume if the liquid phase is negligible compared to the volume of the vapour phase Hence (8) Using these condition, Clapeyron equation becomes (9) or (10) Which is known as Clausius Clapeyron equation further, if we assume that over a small temperature range, the above equation may be integrated to get is constant (11) or

77 Constant (12) Therefore a plot of InP versus 1/T yields a straight line the slope of which is equal to The vapour pressure of most the substances agree fairly well with this equation. Change of Latent Heat with Temperature The latent heat (13) In going from AB to (14) Again (15) or (16) or (17) or or

78 (18) or (19) This is known as Kirchoff equation. This can be rewritten as: (20) TWO SPECIAL CASES: a. In solid to liquid transition, the specific volume and specific heat are constant in each phase and the value of in each phase is negligible; consequently, (21) Where h sf is the latent heat of Fusion b. If the transition is from liquid to ideal gas, then Also, so that (.22) This equation is also applicable in the case of sublimation with subscript f (for liquid) replaced by subscript s (for solid). Phase Equilibrium If two phases are in a state of equilibrium, the temperature and pressure of both the phases must be identical. The chemical potential of component must be identical in both the

79 phases. That is for all (23) Where and refer to two different phases. The criterion of equilibrium at constant temperature and pressure is given by