Chapter 7 Entropy PDF

Chapter 7 ENTROPY I n Chap. 6, we introduced the second law of thermody- namics and applied it to cycles and cyclic...

Chapter 7 ENTROPY I n Chap. 6, we introduced the second law of thermody- namics and applied it to cycles and cyclic devices. In this chapter, we apply the second law to processes. The first law of thermodynamics deals with the property energy and the conservation of it. The second law leads to the definition Objectives The objectives of Chapter 7 are to: Apply the second law of thermodynamics to processes. of a new property called entropy. Entropy is a somewhat Define a new property called entropy to quantify the abstract property, and it is difficult to give a physical descrip- second-law effects. tion of it without considering the microscopic state of the sys- Establish the increase of entropy principle. tem. Entropy is best understood and appreciated by studying Calculate the entropy changes that take place during its uses in commonly encountered engineering processes, processes for pure substances, incompressible substances, and this is what we intend to do. and ideal gases. This chapter starts with a discussion of the Clausius Examine a special class of idealized processes, called inequality, which forms the basis for the definition of entropy, isentropic processes, and develop the property relations for and continues with the increase of entropy principle. Unlike these processes. energy, entropy is a nonconserved property, and there is no such thing as conservation of entropy. Next, the entropy Derive the reversible steady-flow work relations. changes that take place during processes for pure sub- Develop the isentropic efficiencies for various steady-flow stances, incompressible substances, and ideal gases are dis- devices. cussed, and a special class of idealized processes, called Introduce and apply the entropy balance to various isentropic processes, is examined. Then, the reversible systems. steady-flow work and the isentropic efficiencies of various engineering devices such as turbines and compressors are considered. Finally, entropy balance is introduced and applied to various systems. | 331 332 | Thermodynam ics 7–1  ENTROPY The second law of thermodynamics often leads to expressions that involve SEE TUTORIAL CH. 7, SEC. 1 ON THE DVD. inequalities. An irreversible (i.e., actual) heat engine, for example, is less efficient than a reversible one operating between the same two thermal energy reservoirs. Likewise, an irreversible refrigerator or a heat pump has a lower coefficient of performance (COP) than a reversible one operating between the same temperature limits. Another important inequality that has major consequences in thermodynamics is the Clausius inequality. It was first stated by the German physicist R. J. E. Clausius (1822–1888), one of the founders of thermodynamics, and is expressed as dQ ф T ≤0 That is, the cyclic integral of dQ/T is always less than or equal to zero. This inequality is valid for all cycles, reversible or irreversible. The symbol ф (inte- gral symbol with a circle in the middle) is used to indicate that the Thermal reservoir integration TR is to be performed over the entire cycle. Any heat transfer to or from a system  QR can be considered to consist of differential amounts of heat transfer. Then the cyclic integral of dQ/T can be viewed as the sum of all these differential amounts of heat transfer divided by the temperature at the boundary. Reversible To demonstrate the validity of the Clausius inequality, consider a system cyclic  Wrev connected to a thermal energy reservoir at a constant thermodynamic (i.e., device absolute) temperature of TR through a reversible cyclic device (Fig. 7–1). The cyclic device receives heat dQR from the reservoir and supplies heat dQ Q to the system whose temperature at that part of the boundary is T (a vari- T able) while producing work dWrev. The system produces work dWsys as a result of this heat transfer. Applying the energy balance to the combined System  Wsys system identified by dashed lines yields dWC = dQR — dEC Combined system where dWC is the total work of the combined system (dWrev + dWsys ) and (system and cyclic device) dEC is the change in the total energy of the combined system. Considering that the cyclic device is a reversible one, we have FIGURE 7–1 dQR dQ = The system considered in the TR T development of the Clausius inequality. where the sign of dQ is determined with respect to the system (positive if to the system and negative if from the system) and the sign of dQR is deter- mined with respect to the reversible cyclic device. Eliminating dQR from the two relations above yields dQ dWC = TR — dEC T We now let the system undergo a cycle while the cyclic device undergoes an integral number of cycles. Then the preceding relation becomes WC = TR ф dQT since the cyclic integral of energy (the net change in the energy, which is a property, during a cycle) is zero. Here WC is the cyclic integral of dWC, and it represents the net work for the combined cycle.  Chapter 7 | 333 It appears that the combined system is exchanging heat with a single ther- mal energy reservoir while involving (producing or consuming) work WC during a cycle. On the basis of the Kelvin–Planck statement of the second law, which states that no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir, we reason that WC cannot be a work output, and thus it cannot be a positive quantity. Considering that TR is the thermodynamic temperature and thus a positive quantity, we must have dQ ф T ≤0 (7–1) which is the Clausius inequality. This inequality is valid for all thermody- namic cycles, reversible or irreversible, including the refrigeration cycles. If no irreversibilities occur within the system as well as the reversible cyclic device, then the cycle undergone by the combined system is inter- nally reversible. As such, it can be reversed. In the reversed cycle case, all the quantities have the same magnitude but the opposite sign. Therefore, the work WC, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case. Then it follows that WC,int rev = 0 since it cannot be a positive or negative quantity, and therefore dQ ф aT b int rev =0 (7–2) for internally reversible cycles. Thus, we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones. To develop a relation for the definition of entropy, let us examine Eq. 7–2 more closely. Here we have a quantity whose cyclic integral is zero. Let us think for a moment what kind of quantities can have this characteristic. We know that the cyclic integral of work is not zero. (It is a good thing that it is not. Otherwise, heat engines that work on a cycle such as steam power plants would produce zero net work.) Neither is the cyclic integral of heat. Now consider the volume occupied by a gas in a piston–cylinder device undergoing a cycle, as shown in Fig. 7–2. When the piston returns to its ini- tial position at the end of a cycle, the volume of the gas also returns to its initial value. Thus the net change in volume during a cycle is zero. This is  dV = V cycle =0 also expressed as FIGURE 7–2 (7–3) The net change in volume (a property) ф dV = 0 during a cycle is always zero. That is, the cyclic integral of volume (or any other property) is zero. Con- versely, a quantity whose cyclic integral is zero depends on the state only and not the process path, and thus it is a property. Therefore, the quantity (dQ/T )int rev must represent a property in the differential form. Clausius realized in 1865 that he had discovered a new thermodynamic property, and he chose to name this property entropy. It is designated S and is defined as dQ dS = a b 1kJ>K (7–4) T int rev 2 334 | Thermodynam ics Entropy is an extensive property of a system and sometimes is referred to as total entropy. Entropy per unit mass, designated s, is an intensive property and has the unit kJ/kg · K. The term entropy is generally used to refer to both total entropy and entropy per unit mass since the context usually clari- fies which one is meant. The entropy change of a system during a process can be determined by integrating Eq. 7–4 between the initial and the final states: 2 dQ ¢S = S2 — S1 = ∫ a T b 1kJ>K 2 (7–5) 1 int rev Notice that we have actually defined the change in entropy instead of entropy itself, just as we defined the change in energy instead of the energy itself when we developed the first-law relation. Absolute values of entropy are determined on the basis of the third law of thermodynamics, which is discussed later in this chapter. Engineers are usually concerned with the changes in entropy. Therefore, the entropy of a substance can be assigned a zero value at some arbitrarily selected reference state, and the entropy val- ues at other states can be determined from Eq. 7–5 by choosing state 1 to be T the reference state (S = 0) and state 2 to be the state at which entropy is to S = S2 – S1 = 0.4 kJ/K be determined. Irreversible To perform the integration in Eq. 7–5, one needs to know the relation process between Q and T during a process. This relation is often not available, and 2 the integral in Eq. 7–5 can be performed for a few cases only. For the majority of cases we have to rely on tabulated data for entropy. Note that entropy is a property, and like all other properties, it has fixed 1 values at fixed states. Therefore, the entropy change ΔS between two speci- fied states is the same no matter what path, reversible or irreversible, is fol- Reversible process lowed during a process (Fig. 7–3). Also note that the integral of dQ/T gives us the value of entropy change 0.3 0.7 S, kJ/K only if the integration is carried out along an internally reversible path between the two states. The integral of dQ/T along an irreversible path is FIGURE 7–3 not a property, and in general, different values will be obtained when the The entropy change between two integration is carried out along different irreversible paths. Therefore, even specified states is the same whether for irreversible processes, the entropy change should be determined by carry- the process is reversible or ing out this integration along some convenient imaginary internally irreversible. reversible path between the specified states. A Special Case: Internally Reversible Isothermal Heat Transfer Processes Recall that isothermal heat transfer processes are internally reversible. Therefore, the entropy change of a system during an internally reversible isothermal heat transfer process can be determined by performing the inte- gration in Eq. 7–5: 2 dQ 2 dQ 1 2 ¢S = ∫ 1 a T b int rev = ∫ 1 a T0 b int rev = T0 ∫ 1 1dQ 2 int rev which reduces to Q ¢S = 1kJ>K (7–6) T0 2  Chapter 7 | 335 where T0 is the constant temperature of the system and Q is the heat transfer for the internally reversible process. Equation 7–6 is particularly useful for determining the entropy changes of thermal energy reservoirs that can absorb or supply heat indefinitely at a constant temperature. Notice that the entropy change of a system during an internally reversible isothermal process can be positive or negative, depending on the direction of heat transfer. Heat transfer to a system increases the entropy of a system, whereas heat transfer from a system decreases it. In fact, losing heat is the only way the entropy of a system can be decreased. 7–1 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. Solution Heat is transferred to a liquid–vapor mixture of water in a piston– cylinder device at constant pressure. The entropy change of water is to be determined. Assumptions No irreversibilities occur within the system boundaries during the process. T = 300 K = const. Analysis We take the entire water (liquid + vapor) in the cylinder as the system (Fig. 7–4). This is a closed system since no mass crosses the system Q Ssys = = 2.5 kJ boundary during the process. We note that the temperature of the system T K remains constant at 300 K during this process since the temperature of a pure substance remains constant at the saturation value during a phase- change process at constant pressure. Q = 750 kJ The system undergoes an internally reversible, isothermal process, and thus its entropy change can be determined directly from Eq. 7–6 to be FIGURE 7–4 Q 750 kJ Schematic for Example 7–1. ¢S sys,isothermal = = = 2.5 kJ/K Tsys 300 K Discussion Note that the entropy change of the system is positive, as expected, since heat transfer is to the system. 7–2 THE INCREASE OF ENTROPY PRINCIPLE  Consider a cycle that is made up of two processes: process 1-2, which is arbitrary (reversible or irreversible), and process 2-1, which is internally SEE TUTORIAL CH. 7, SEC. 2 ON THE DVD. reversible, as shown in Figure 7–5. From the Clausius inequality, dQ ф T ≤0 or 2 dQ 1 dQ ∫ 1 T + ∫ 2 a b T int rev ≤0 336 | Thermodynam Process 1-2ics 2 The second integral in the previous relation is recognized as the entropy (reversible or irreversible) change S1 — S2. Therefore, 2 dQ ∫ 1 T + S1 — S2 ≤ 0 1 which can be rearranged as Process 2-1 (internally 2 dQ reversible) S2 — S1 ≥ ∫ T (7–7) FIGURE 7–5 1 A cycle composed of a reversible and It can also be expressed in differential form as an irreversible process. dQ dS ≥ (7–8) T where the equality holds for an internally reversible process and the inequality for an irreversible process. We may conclude from these equa- tions that the entropy change of a closed system during an irreversible process is greater than the integral of dQ/T evaluated for that process. In the limiting case of a reversible process, these two quantities become equal. We again emphasize that T in these relations is the thermodynamic temperature at the boundary where the differential heat dQ is transferred between the system and the surroundings. The quantity ΔS = S2 — S1 represents the entropy change of the system. For a reversible process, it becomes equal to ʃ2 dQ/T, which represents the 1 entropy transfer with heat. The inequality sign in the preceding relations is a constant reminder that the entropy change of a closed system during an irreversible process is always greater than the entropy transfer. That is, some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibilities. The entropy generated during a process is called entropy generation and is denoted by Sgen. Noting that the difference between the entropy change of a closed system and the entropy transfer is equal to entropy generation, Eq. 7–7 can be rewritten as an equality as 2 dQ ¢Ssys = S2 — S1 = ∫ 1 T + Sgen (7–9) Note that the entropy generation Sgen is always a positive quantity or zero. Its value depends on the process, and thus it is not a property of the system. Also, in the absence of any entropy transfer, the entropy change of a system is equal to the entropy generation. Equation 7–7 has far-reaching implications in thermodynamics. For an isolated system (or simply an adiabatic closed system), the heat transfer is zero, and Eq. 7–7 reduces to ¢Sisolated ≥ 0 (7–10) This equation can be expressed as the entropy of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant. In other words, it never decreases. This is known as the increase of entropy principle. Note that in the absence of any heat transfer, entropy change is due to irreversibilities only, and their effect is always to increase entropy.  Chapter 7 | 337 Entropy is an extensive property, and thus the total entropy of a system is (Isolated) equal to the sum of the entropies of the parts of the system. An isolated sys- tem may consist of any number of subsystems (Fig. 7–6). A system and its Subsystem 1 N surroundings, for example, constitute an isolated system since both can be Stotal = Si > 0 enclosed by a sufficiently large arbitrary boundary across which there is no Subsystem i =1 2 heat, work, or mass transfer (Fig. 7–7). Therefore, a system and its sur- roundings can be viewed as the two subsystems of an isolated system, and Subsystem Subsystem the entropy change of this isolated system during a process is the sum of the 3 N entropy changes of the system and its surroundings, which is equal to the entropy generation since an isolated system involves no entropy transfer. FIGURE 7–6 That is, The entropy change of an isolated Sgen = ¢Stotal = ¢Ssys + ¢Ssurr ≥ 0 (7–11) system is the sum of the entropy where the equality holds for reversible processes and the inequality for irre- changes of its components, and is never less than zero. versible ones. Note that ΔS surr refers to the change in the entropy of the sur- roundings as a result of the occurrence of the process under consideration. Isolated system m=0 Since no actual process is truly reversible, we can conclude that some boundary Q=0 entropy is generated during a process, and therefore the entropy of the uni- W=0 verse, which can be considered to be an isolated system, is continuously increasing. The more irreversible a process, the larger the entropy generated during that process. No entropy is generated during reversible processes (Sgen = 0). Entropy increase of the universe is a major concern not only to engineers but also to philosophers, theologians, economists, and environmentalists since entropy is viewed as a measure of the disorder (or “mixed-up-ness”) System in the universe. m Q, W The increase of entropy principle does not imply that the entropy of a sys- tem cannot decrease. The entropy change of a system can be negative dur- ing a process (Fig. 7–8), but entropy generation cannot. The increase of Surroundings entropy principle can be summarized as follows: 7 0 Irreversible process Sgen = 0 Reversible process FIGURE 7–7 6 0 Impossible process A system and its surroundings form an isolated system. This relation serves as a criterion in determining whether a process is reversible, irreversible, or impossible. Things in nature have a tendency to change until they attain a state of equi- librium. The increase of entropy principle dictates that the entropy of an iso- lated system increases until the entropy of the system reaches a maximum value. At that point, the system is said to have reached an equilibrium state since the increase of entropy principle prohibits the system from undergoing any change of state that results in a decrease in entropy. Some Remarks about Entropy In light of the preceding discussions, we draw the following conclusions: 1. Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, Sgen ≥ 0. A process that violates this princi- ple is impossible. This principle often forces chemical reactions to come to a halt before reaching completion. 338 | Thermodynam ics 2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. Entropy is conserved during the ide- Surroundings alized reversible processes only and increases during all actual processes. 3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities present during that process. The greater the extent Ssys = –2 kJ/K of irreversibilities, the greater the entropy generation. Therefore, SYSTEM entropy generation can be used as a quantitative measure of irreversibil- Q ities associated with a process. It is also used to establish criteria for the performance of engineering devices. This point is illustrated further in Example 7–2.  Ssurr = 3 kJ/K Sgen =  Stotal =  Ssys +  Ssurr = 1 kJ/K FIGURE 7–8 7–2 The entropy change of a system can be negative, but the entropy generation cannot. 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. Solution Heat is transferred from a heat source to two heat sinks at differ- ent temperatures. The heat transfer process that is more irreversible is to be determined. Analysis A sketch of the reservoirs is shown in Fig. 7–9. Both cases involve heat transfer through a finite temperature difference, and therefore both are Source Source irreversible. The magnitude of the irreversibility associated with each process 800 K 800 K can be determined by calculating the total entropy change for each case. The total entropy change for a heat transfer process involving two reservoirs 2000 kJ (a source and a sink) is the sum of the entropy changes of each reservoir since the two reservoirs form an adiabatic system. Or do they? The problem statement gives the impression that the two reservoirs are in direct contact during the heat transfer process. But this Sink A Sink B cannot be the case since the temperature at a point can have only one value, 500 K 750 K and thus it cannot be 800 K on one side of the point of contact and 500 K on the other side. In other words, the temperature function cannot have a (a) (b) jump discontinuity. Therefore, it is reasonable to assume that the two reser- voirs are separated by a partition through which the temperature drops from FIGURE 7–9 800 K on one side to 500 K (or 750 K) on the other. Therefore, the entropy Schematic for Example 7–2. change of the partition should also be considered when evaluating the total entropy change for this process. However, considering that entropy is a prop- erty and the values of properties depend on the state of a system, we can argue that the entropy change of the partition is zero since the partition appears to have undergone a steady process and thus experienced no change in its properties at any point. We base this argument on the fact that the temperature on both sides of the partition and thus throughout remains con- stant during this process. Therefore, we are justified to assume that ΔSpartition = 0 since the entropy (as well as the energy) content of the partition remains constant during this process.  Chapter 7 | 339 The entropy change for each reservoir can be determined from Eq. 7–6 since each reservoir undergoes an internally reversible, isothermal process. (a) For the heat transfer process to a sink at 500 K: Q source —2000 kJ ¢S source = = = —2.5 kJ>K Tsource 800 K Q sink 2000 kJ ¢S sink = = = + 4.0 kJ>K Tsink 500 K and Therefore, 1.5 kJ/K of entropy is generated during this process. Noting that both reservoirs have undergone internally reversible processes, the entire entropy generation took place in the partition. (b) Repeating the calculations in part (a) for a sink temperature of 750 K, we obtain ¢Ssource = —2.5 kJ>k and Sgen = ¢Stotal = 1—2.5 + 2.72 kJ>K = 0.2 kJ/K The total entropy change for the process in part (b) is smaller, and therefore it is less irreversible. This is expected since the process in (b) involves a smaller temperature difference and thus a smaller irreversibility. Discussion The irreversibilities associated with both processes could be eliminated by operating a Carnot heat engine between the source and the sink. For this case it can be shown that ΔStotal = 0. 7–3 ENTROPY CHANGE OF PURE SUBSTANCES  Entropy is a property, and thus the value of entropy of a system is fixed once the state of the system is fixed. Specifying two intensive independent SEE TUTORIAL CH. 7, SEC. 3 ON THE DVD. properties fixes the state of a simple compressible system, and thus the value of entropy, as well as the values of other properties at that state. Start- ing with its defining relation, the entropy change of a substance can be expressed in terms of other properties (see Sec. 7–7). But in general, these relations are too complicated and are not practical to use for hand calcula- tions. Therefore, using a suitable reference state, the entropies of substances are evaluated from measurable property data following rather involved com- putations, and the results are tabulated in the same manner as the other properties such as v, u, and h (Fig. 7–10). The entropy values in the property tables are given relative to an arbitrary reference state. In steam tables the entropy of saturated liquid sf at 0.01°C is assigned the value of zero. For refrigerant-134a, the zero value is assigned to saturated liquid at —40°C. The entropy values become negative at tem- peratures below the reference value. 340 | Thermodynam T ics The value of entropy at a specified state is determined just like any other P1 property. In the compressed liquid and superheated vapor regions, it can be s s T1 1 ƒ@T1 T3 obtained directly from the tables at the specified state. In the saturated mix- s P3 3 ture region, it is determined from 1kJ>kg. K 2 Compressed Superheated liquid 2 vapor s = sf + xsfg 1 Saturated 3 where x is the quality and sf and sfg values are listed in the saturation tables. liquid–vapor mixture In the absence of compressed liquid data, the entropy of the compressed liq- T2 s = s ƒ + x2sƒg uid can be approximated by the entropy of the saturated liquid at the given x2 2 temperature: s@ T,P ≅ sf @ T 1kJ>kg. K 2 The entropy change of a specified mass m (a closed system) during a s process is simply ΔS = mΔs = m 1s2 — s1 2 1kJ>K (7–12) FIGURE 7–10 2 The entropy of a pure substance is which is the difference between the entropy values at the final and initial determined from the tables (like other states. properties). When studying the second-law aspects of processes, entropy is commonly used as a coordinate on diagrams such as the T-s and h-s diagrams. The general characteristics of the T-s diagram of pure substances are shown in Fig. 7–11 using data for water. Notice from this diagram that the constant- volume lines are steeper than the constant-pressure lines and the constant- pressure lines are parallel to the constant-temperature lines in the saturated liquid–vapor mixture region. Also, the constant-pressure lines almost coin- cide with the saturated liquid line in the compressed liquid region. T, C 500 Critical 400 state 300 Saturated liquid line 200 Saturated vapor line 100 FIGURE 7–11 Schematic of the T-s diagram for water. 0 1 2 3 4 5 6 7 8 s, kJ/kg K  Chapter 7 | 341 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. Solution The refrigerant in a rigid tank is cooled while being stirred. The entropy change of the refrigerant is to be determined. Assumptions The volume of the tank is constant and thus v2 = v1. Analysis We take the refrigerant in the tank as the system (Fig. 7–12). This is a closed system since no mass crosses the system boundary during the process. We note that the change in entropy of a substance during a process is simply the difference between the entropy values at the final and initial states. The initial state of the refrigerant is completely specified. Recognizing that the specific volume remains constant during this process, the properties of the refrigerant at both states are State 1: P1 = 140 kPa s1 = 1.0624 kJ>kg. K f T1 = 20°C v1 = 0.16544 m3>kg P2 = 100 kPa vf = 0.0007259 m3>kg State 2: f 1v2 = v1 2 vg = 0.19254 m3>kg The refrigerant is a saturated liquid–vapor mixture at the final state since vf < v2 < vg at 100 kPa pressure. Therefore, we need to determine the quality first: v2 — vf 0.16544 — 0.0007259 x2 = = = 0.859 vfg 0.19254 — 0.0007259 Thus, s2 = sf + x2sfg = 0.07188 + 10.8592 10.879952 = 0.8278 kJ>kg. K Then the entropy change of the refrigerant during this process is ΔS = m 1s2 — s1 2 = 15 kg 2 10.8278 — 1.06242 kJ>kg. K = —1.173 kJ/K Discussion The negative sign indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation Sgen that cannot be negative. T m = 5 kg 1 Refrigerant-134a T1 = 20C 2 P1 = 140 kPa FIGURE 7–12 S = ? Heat Schematic and T-s diagram for s2 s1 s Example 7–3. 342 | Thermodynam ics EXAMPLE 7–4 Entropy Change during a Constant-Pressure Process A piston–cylinder device initially contains 3 lbm of liquid water at 20 psia and 70°F. The water is now heated at constant pressure by the addition of 3450 Btu of heat. Determine the entropy change of the water during this process. Solution Liquid water in a piston–cylinder device is heated at constant pressure. The entropy change of water is to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero, ΔKE = ΔPE = 0. 2 The process is quasi-equilibrium. 3 The pressure remains constant during the process and thus P2 = P1. Analysis We take the water in the cylinder as the system (Fig. 7–13). This is a closed system since no mass crosses the system boundary during the process. We note that a piston–cylinder device typically involves a moving boundary and thus boundary work Wb. Also, heat is transferred to the system. Water exists as a compressed liquid at the initial state since its pressure is greater than the saturation pressure of 0.3632 psia at 70°F. By approximat- ing the compressed liquid as a saturated liquid at the given temperature, the properties at the initial state are State 1: P1 = 20 psia s1 ≅ sf @ 70°F = 0.07459 Btu>lbm. R f T1 = 70°F h1 ≅ hf @ 70°F = 38.08 Btu>lbm At the final state, the pressure is still 20 psia, but we need one more prop- erty to fix the state. This property is determined from the energy balance, Ein — Eout = ΔEsystem           Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc., energies Q in — Wb = ΔU Q in = ΔH = m 1h2 — h1 2 3450 Btu = 13 lbm 2 1h2 — 38.08 Btu>lbm 2 h2 = 1188.1 Btu>lbm since ΔU + Wb = ΔH for a constant-pressure quasi-equilibrium process. Then, P2 = 20 psia s2 = 1.7761 Btu>lbm. R State 2: f h2 = 1188.1 Btu> lbm 1Table A-6E, interpolation 2 T 2 H2O Qin P1 = 20 psia 1 FIGURE 7–13 T1 = 70F Schematic and T-s diagram for Example 7–4. s1 s2 s  Chapter 7 | 343 Steam Therefore, the entropy change of water during this process is s1 = 5.105 Btu/R No irreversibilities (internally reversible) 7–4 ISENTROPIC PROCESSES  No heat transfer We mentioned earlier that the entropy of a fixed mass can be changed by (adiabatic) s2 = s1 (1) heat transfer and (2) irreversibilities. Then it follows that the entropy of a fixed mass does not change during a process that is internally reversible FIGURE 7–14 and adiabatic (Fig. 7–14). A process during which the entropy remains During an internally reversible, constant is called an isentropic process. It is characterized by adiabatic (isentropic) process, the Isentropic process: Δs = 0 or s2 = s 1 1kJ>kg. K2 (7–13) entropy remains constant. That is, a substance will have the same entropy value at the end of the process as it does at the beginning if the process is carried out in an isen- tropic manner. Many engineering systems or devices such as pumps, turbines, nozzles, and diffusers are essentially adiabatic in their operation, and they perform SEE TUTORIAL CH. 7, SEC. 4 ON THE DVD. best when the irreversibilities, such as the friction associated with the process, are minimized. Therefore, an isentropic process can serve as an appropriate model for actual processes. Also, isentropic processes enable us T to define efficiencies for processes to compare the actual performance of these devices to the performance under idealized conditions. It should be recognized that a reversible adiabatic process is necessarily isentropic (s2 = s1), but an isentropic process is not necessarily a reversible 1 Isentropic adiabatic process. (The entropy increase of a substance during a process as expansion a result of irreversibilities may be offset by a decrease in entropy as a result of heat losses, for example.) However, the term isentropic process is cus- 1.4 MPa 2 tomarily used in thermodynamics to imply an internally reversible, adia- batic process. s2 = s1 s 7–5 P1 = 5 MPa T1 = 450°C Steam enters an adiabatic turbine at 5 MPa and 450°C and leaves at a pres- sure of 1.4 MPa. Determine the work output of the turbine per unit mass of wout = ? steam if the process is reversible. STEAM TURBINE Solution Steam is expanded in an adiabatic turbine to a specified pressure in a reversible manner. The work output of the turbine is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus ΔmCV = 0, ΔECV = 0, and ΔSCV = 0. 2 The process is reversible. 3 Kinetic and potential energies are negligible. 4 The P2 = 1.4 MPa turbine is adiabatic and thus there is no heat transfer. s2 = s1 Analysis We take the turbine as the system (Fig. 7–15). This is a control volume since mass crosses the system boundary dur ing the proce ss. We note FIGURE 7–15... that there is only one inlet and one exit, and thus m1 = m2 = m. Schematic and T-s diagram for Example 7–5. 344 | Thermodynam ics The power output of the turbine is determined from the rate form of the energy balance, 0 (steady) Ein — Eout = dEsystem/dt =0        Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc., energies Ein = Eout. m h1 = Wout 2 1since Q = 0, ke ≅ pe ≅ 02 Wout = m 1h1 — h2 2 The inlet state is completely specified since two properties are given. But only one property (pressure) is given at the final state, and we need one more property to fix it. The second property comes from the observation that the process is reversible and adiabatic, and thus isentropic. Therefore, s2 = s1, and P1 = 5 MPa h1 = 3317.2 kJ State 1: T1 = 450°C s1 = 6.8210 kJ P2 = 1.4 MPa State 2: h2 = 2967.4 kJ s2 = s1 Then the work output of the turbine per unit mass of the steam becomes wout = h1 — h2 = 3317.2 — 2967.4 = 349.8 kJ/kg 7–5 PROPERTY DIAGRAMS INVOLVING ENTROPY  Property diagrams serve as great visual aids in the thermodynamic analysis SEE TUTORIAL CH. 7, SEC. 5 ON THE DVD. of processes. We have used P-v and T-v diagrams extensively in previous chapters in conjunction with the first law of thermodynamics. In the second- T law analysis, it is very helpful to plot the processes on diagrams for which one of the coordinates is entropy. The two diagrams commonly used in the Internally second-law analysis are the temperature-entropy and the enthalpy-entropy reversible diagrams. process Consider the defining equation of entropy (Eq. 7–4). It can be dA = T dS = Q rearranged as dQint rev = T dS 1kJ 2 (7–14) As shown in Fig. 7–16, dQrev int corresponds to a differential area on a T-S T dS = Q 2 Area = 1 diagram. The total heat transfer during an internally reversible process is determined by integration to be 2 S FIGURE 7–16 Q int rev = ∫ 1 T dS 1kJ 2 (7–15) On a T-S diagram, the area under the which corresponds to the area under the process curve on a T-S diagram. process curve represents the heat Therefore, we conclude that the area under the process curve on a T-S dia- transfer for internally reversible gram represents heat transfer during an internally reversible process. This processes. is somewhat analogous to reversible boundary work being represented by  Chapter 7 | 345 the area under the process curve on a P-V diagram. Note that the area under the process curve represents heat transfer for processes that are internally T (or totally) reversible. The area has no meaning for irreversible processes. Equations 7–14 and 7–15 can also be expressed on a unit-mass basis as 1 dqint rev = T ds 1kJ>kg (7–16) Isentropic 2 process and 2 2 qint rev = ∫ 1 T ds 2 1kJ>kg (7–17) To perform the integrations in Eqs. 7–15 and 7–17, one needs to know the relationship between T and s during a process. One special case for which these integrations can be performed easily is the internally reversible s2 = s1 s isothermal process. It yields FIGURE 7–17 Q int rev = T0 ¢S 1kJ 2 (7–18) The isentropic process appears as a or vertical line segment on a T-s diagram. qint rev = T0 ¢s 1kJ>kg (7–19) 2 where T0 is the constant temperature and ΔS is the entropy change of the system during the process. h An isentropic process on a T-s diagram is easily recognized as a vertical- line segment. This is expected since an isentropic process involves no 1 heat transfer, and therefore the area under the process path must be zero (Fig. 7–17). The T-s diagrams serve as valuable tools for visualizing the h second-law aspects of processes and cycles, and thus they are frequently used in thermodynamics. The T-s diagram of water is given in the appendix 2 in Fig. A–9. s Another diagram commonly used in engineering is the enthalpy-entropy diagram, which is quite valuable in the analysis of steady-flow devices such as turbines, compressors, and nozzles. The coordinates of an h-s diagram s represent two properties of major interest: enthalpy, which is a primary property in the first-law analysis of the steady-flow devices, and entropy, FIGURE 7–18 which is the property that accounts for irreversibilities during adiabatic For adiabatic steady-flow devices, the processes. In analyzing the steady flow of steam through an adiabatic tur- vertical distance Δh on an h-s diagram bine, for example, the vertical distance between the inlet and the exit states is a measure of work, and the Δh is a measure of the work output of the turbine, and the horizontal dis- horizontal distance Δs is a measure of tance Δs is a measure of the irreversibilities associated with the process irreversibilities. (Fig. 7–18). The h-s diagram is also called a Mollier diagram after the German scien- tist R. Mollier (1863–1935). An h-s diagram is given in the appendix for steam in Fig. A–10. 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 repre- sent the heat supplied QH, heat rejected QL, and the net work output Wnet,out on this diagram. 346 | Thermodynam ics T Solution The Carnot cycle is to be shown on a T-S diagram, and the areas that represent QH, QL, and Wnet,out are to be indicated. 1 2 TH Analysis Recall that the Carnot cycle is made up of two reversible isother- mal (T = constant) processes and two isentropic (s = constant) processes. Wnet On a T-S diagram, the area under the process curve represents the heat TL 3 4 transfer for that process. Thus the area A12B represents QH, the area A43B represents QL, and the difference between these two (the area in color) rep- resents the net work since A B Wnet,out = QH — QL S1 = S4 S2 = S3 S Therefore, the area enclosed by the path of a cycle (area 1234) on a T-S dia- FIGURE 7–19 gram represents the net work. Recall that the area enclosed by the path of a The T-S diagram of a Carnot cycle cycle also represents the net work on a P-V diagram. (Example 7–6). 7–6  WHAT IS ENTROPY? It is clear from the previous discussion that entropy is a useful property and SEE TUTORIAL CH. 7, SEC. 6 ON THE DVD. serves as a valuable tool in the second-law analysis of engineering devices. But this does not mean that we know and understand entropy well. Because we do not. In fact, we cannot even give an adequate answer to the question, What is entropy? Not being able to describe entropy fully, however, does not take anything away from its usefulness. We could not define energy either, but it did not interfere with our understanding of energy transforma- tions and the conservation of energy principle. Granted, entropy is not a household word like energy. But with continued use, our understanding of entropy will deepen, and our appreciation of it will grow. The next discus- Entropy, kJ/kg K sion should shed some light on the physical meaning of entropy by consid- ering the microscopic nature of matter. Entropy can be viewed as a measure of molecular disorder, or molecular randomness. As a system becomes more disordered, the positions of the mol- ecules become less predictable and the entropy increases. Thus, it is not sur- GAS prising that the entropy of a substance is lowest in the solid phase and highest in the gas phase (Fig. 7–20). 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 LIQUID about at random, collide with each other, and change direction, making it extremely difficult to predict accurately the microscopic state of a system at SOLID any instant. Associated with this molecular chaos is a high value of entropy. When viewed microscopically (from a statistical thermodynamics point of view), an isolated system that appears to be at a state of equilibrium may FIGURE 7–20 exhibit a high level of activity because of the continual motion of the mole- The level of molecular disorder cules. To each state of macroscopic equilibrium there corresponds a large (entropy) of a substance increases as it number of possible microscopic states or molecular configurations. The melts or evaporates. entropy of a system is related to the total number of possible microscopic  Chapter 7 | 347 states of that system, called thermodynamic probability p, by the Boltz- mann relation, expressed as S = k ln p (7–20) where k = 1.3806 × 10 —23 J/K is the Boltzmann constant. Therefore, from a microscopic point of view, the entropy of a system increases whenever the molecular randomness or uncertainty (i.e., molecular probability) of a sys- tem increases. Thus, entropy is a measure of molecular disorder, and the molecular disorder of an isolated system increases anytime it undergoes a Pure crystal process. T=0K As mentioned earlier, the molecules of a substance in solid phase continu- Entropy = 0 ally oscillate, creating an uncertainty about their position. These oscilla- tions, however, fade as the temperature is decreased, and the molecules supposedly become motionless at absolute zero. This represents a state of FIGURE 7–21 ultimate molecular order (and minimum energy). Therefore, the entropy of a A pure crystalline substance at pure crystalline substance at absolute zero temperature is zero since there is absolute zero temperature is in no uncertainty about the state of the molecules at that instant (Fig. 7–21). perfect order, and its entropy is zero This statement is known as the third law of thermodynamics. The third (the third law of thermodynamics). law of thermodynamics provides an absolute reference point for the deter- mination of entropy. The entropy determined relative to this point is called absolute entropy, and it is extremely useful in the thermodynamic analysis LOAD of chemical reactions. Notice that the entropy of a substance that is not pure crystalline (such as a solid solution) is not zero at absolute zero tempera- ture. This is because more than one molecular configuration exists for such substances, which introduces some uncertainty about the microscopic state FIGURE 7–22 of the substance. Molecules in the gas phase possess a considerable amount of kinetic Disorganized energy does not create much useful effect, no matter how energy. However, we know that no matter how large their kinetic energies large it is. are, the gas molecules do not rotate a paddle wheel inserted into the con- tainer and produce work. This is because the gas molecules, and the energy they possess, are disorganized. Probably the number of molecules trying to Wsh rotate the wheel in one direction at any instant is equal to the number of molecules that are trying to rotate it in the opposite direction, causing the wheel to remain motionless. Therefore, we cannot extract any useful work directly from disorganized energy (Fig. 7–22). Now consider a rotating shaft shown in Fig. 7–23. This time the energy of the molecules is completely organized since the molecules of the shaft are rotating in the same direction together. This organized energy can readily be used to perform useful tasks such as raising a weight or generating electric- ity. Being an organized form of energy, work is free of disorder or random- ness and thus free of entropy. There is no entropy transfer associated with energy transfer as work. Therefore, in the absence of any friction, the WEIGHT process of raising a weight by a rotating shaft (or a flywheel) does not pro- duce any entropy. Any process that does not produce a net entropy is FIGURE 7–23 reversible, and thus the process just described can be reversed by lowering In the absence of friction, raising a the weight. Therefore, energy is not degraded during this process, and no weight by a rotating shaft does not potential to do work is lost. create any disorder (entropy), and thus Instead of raising a weight, let us operate the paddle wheel in a container energy is not degraded during this filled with a gas, as shown in Fig. 7–24. The paddle-wheel work in this case process. 348 | Thermodynam ics is converted to the internal energy of the gas, as evidenced by a rise in gas GAS temperature, creating a higher level of molecular disorder in the container. Wsh This process is quite different from raising a weight since the organized paddle-wheel energy is now converted to a highly disorganized form of T energy, which cannot be converted back to the paddle wheel as the rota- tional kinetic energy. Only a portion of this energy can be converted to work FIGURE 7–24 by partially reorganizing it through the use of a heat engine. Therefore, energy is degraded during this process, the ability to do work is reduced, The paddle-wheel work done on a gas molecular disorder is produced, and associated with all this is an increase in increases the level of disorder entropy. (entropy) of the gas, and thus energy is The quantity of energy is always preserved during an actual process (the degraded during this process. first law), but the quality is bound to decrease (the second law). This decrease in quality is always accompanied by an increase in entropy. As an example, consider the transfer of 10 kJ of energy as heat from a hot medium HOT BODY to a cold one. At the end of the process, we still have the 10 kJ of energy, Heat COLD BODY but at a lower temperature and thus at a lower quality. 80C 20C Heat is, in essence, a form of disorganized energy, and some disorganiza- (Entropy (Entropy tion (entropy) flows with heat (Fig. 7–25). As a result, the entropy and the decreases) increases) level of molecular disorder or randomness of the hot body decreases with the entropy and the level of molecular disorder of the cold body increases. FIGURE 7–25 The second law requires that the increase in entropy of the cold body be During a heat transfer process, the net greater than the decrease in entropy of the hot body, and thus the net entropy increases. (The increase in the entropy of the combined system (the cold body and the hot body) increases. entropy of the cold body more than That is, the combined system is at a state of greater disorder at the final offsets the decrease in the entropy of state. Thus we can conclude that processes can occur only in the direction the hot body.) of increased overall entropy or molecular disorder. That is, the entire uni- verse is getting more and more chaotic every day. Entropy and Entropy Generation in Daily Life The concept of entropy can also be applied to other areas. Entropy can be viewed as a measure of disorder or disorganization in a system. Likewise, entropy generation can be viewed as a measure of disorder or disorganiza- tion generated during a process. The concept of entropy is not used in daily life nearly as extensively as the concept of energy, even though entropy is readily applicable to various aspects of daily life. The extension of the entropy concept to nontechnical fields is not a novel idea. It has been the topic of several articles, and even some books. Next we present several ordi- nary events and show their relevance to the concept of entropy and entropy generation. Efficient people lead low-entropy (highly organized) lives. They have a place for everything (minimum uncertainty), and it takes minimum energy for them to locate something. Inefficient people, on the other hand, are dis- organized and lead high-entropy lives. It takes them minutes (if not hours) FIGURE 7–26 to find something they need, and they are likely to create a bigger disorder The use of entropy (disorganization, as they are searching since they will probably conduct the search in a disor- uncertainty) is not limited to ganized manner (Fig. 7–26). People leading high-entropy lifestyles are thermodynamics. always on the run, and never seem to catch up. © Reprinted with permission of King Features You probably noticed (with frustration) that some people seem to learn Syndicate. fast and remember well what they learn. We can call this type of learning  Chapter 7 | 349 organized or low-entropy learning. These people make a conscientious effort to file the new information properly by relating it to their existing knowledge base and creating a solid information network in their minds. On the other hand, people who throw the information into their minds as they study, with no effort to secure it, may think they are learning. They are bound to discover otherwise when they need to locate the information, for example, during a test. It is not easy to retrieve information from a database that is, in a sense, in the gas phase. Students who have blackouts during tests should reexamine their study habits. A library with a good shelving and indexing system can be viewed as a low- entropy library because of the high level of organization. Likewise, a library with a poor shelving and indexing system can be viewed as a high-entropy library because of the high level of disorganization. A library with no indexing system is like no library, since a book is of no value if it cannot be found. Consider two identical buildings, each containing one million books. In the first building, the books are piled on top of each other, whereas in the second building they are highly organized, shelved, and indexed for easy reference. There is no doubt about which building a student will prefer to go to for checking out a certain book. Yet, some may argue from the first-law point of view that these two buildings are equivalent since the mass and knowledge content of the two buildings are identical, despite the high level of disorganization (entropy) in the first building. This example illustrates that any realistic comparisons should involve the second-law point of view. Two textbooks that seem to be identical because both cover basically the same topics and present the same information may actually be very different depending on how they cover the topics. After all, two seemingly identical cars are not so identical if one goes only half as many miles as the other one on the same amount of fuel. Likewise, two seemingly identical books are not so identical if it takes twice as long to learn a topic from one of them as it does from the other. Thus, comparisons made on the basis of the first law only may be highly misleading. Having a disorganized (high-entropy) army is like having no army at all. It is no coincidence that the command centers of any armed forces are among the primary targets during a war. One army that consists of 10 divi- sions is 10 times more powerful than 10 armies each consisting of a single division. Likewise, one country that consists of 10 states is more powerful than 10 countries, each consisting of a single state. The United States would not be such a powerful country if there were 50 independent countries in its place instead of a single country with 50 states. The European Union has the potential to be a new economic and political superpower. The old cliché “divide and conquer” can be rephrased as “increase the entropy and conquer.” We know that mechanical friction is always accompanied by entropy generation, and thus reduced performance. We can generalize this to daily life: friction in the workplace with fellow workers is bound to generate FIGURE 7–27 entropy, and thus adversely affect performance (Fig. 7–27). It results in reduced productivity. As in mechanical systems, friction in We also know that unrestrained expansion (or explosion) and uncontrolled the workplace is bound to generate electron exchange (chemical reactions) generate entropy and are highly irre- entropy and reduce performance. versible. Likewise, unrestrained opening of the mouth to scatter angry words © Vol. 26/PhotoDisc 350 | Thermodynam ics is highly irreversible since this generates entropy, and it can cause consider- able damage. A person who gets up in anger is bound to sit down at a loss. Hopefully, someday we will be able to come up with some procedures to quantify entropy generated during nontechnical activities, and maybe even pinpoint its primary sources and magnitude. 7–7 THE T ds RELATIONS  Recall that the quantity (dQ/T )int rev corresponds to a differential change in SEE TUTORIAL CH. 7, SEC. 7 ON THE DVD. the property entropy. The entropy change for a process, then, can be evalu- ated by integrating dQ/T along some imaginary internally reversible path between the actual end states. For isothermal internally reversible processes, this integration is straightforward. But when the temperature varies during the process, we have to have a relation between dQ and T to perform this integration. Finding such relations is what we intend to do in this section. The differential form of the conservation of energy equation for a closed stationary system (a fixed mass) containing a simple compressible substance can be expressed for an internally reversible process as dQ int rev — dWint rev,out = dU (7–21) But dQ int rev = T dS dWint rev,out = P dV Thus, T dS = dU + P dV 1kJ 2 (7–22) or T ds = du + P dv 1kJ>kg2 (7–23) This equation is known as the first T ds, or Gibbs, equation. Notice that the only type of work interaction a simple compressible system may involve as it undergoes an internally reversible process is the boundary work. The second T ds equation is obtained by eliminating du from Eq. 7–23 by using the definition of enthalpy (h = u + Pv): h = u + Pv ¡ dh = du + P dv + v dP fT ds = dh — v dP (7–24) 1Eq. 7–232 ¡ T ds = du + P dv Equations 7–23 and 7–24 are extremely valuable since they relate entropy Closed CV changes of a system to the changes in other properties. Unlike Eq. 7–4, they system are property relations and therefore are independent of the type of the processes. T ds = du + P dv These T ds relations are developed with an internally reversible process in T ds = dh – v dP mind since the entropy change between two states must be evaluated along a reversible path. However, the results obtained are valid for both reversible and irreversible processes since entropy is a property and the change in a FIGURE 7–28 property between two states is independent of the type of process the sys- The T ds relations are valid for both tem undergoes. Equations 7–23 and 7–24 are relations between the proper- reversible and irreversible processes ties of a unit mass of a simple compressible system as it undergoes a change and for both closed and open systems. of state, and they are applicable whether the change occurs in a closed or an open system (Fig. 7–28).  Chapter 7 | 351 Explicit relations for differential changes in entropy are obtained by solv- ing for ds in Eqs. 7–23 and 7–24: du P dv ds = + (7–25) T T and dh v dP ds = — (7–26) T T The entropy change during a process can be determined by integrating either of these equations between the initial and the final states. To perform these integrations, however, we must know the relationship between du or dh and the temperature (such as du = cv dT and dh = cp dT for ideal gases) as well as the equation of state for the substance (such as the ideal-gas equation of state Pv = RT ). For substances for which such relations exist, the integration of Eq. 7–25 or 7–26 is straightforward. For other substances, we have to rely on tabulated data. The T ds relations for nonsimple systems, that is, systems that involve more than one mode of quasi-equilibrium work, can be obtained in a similar manner by including all the relevant quasi-equilibrium work modes. 7–8 ENTROPY CHANGE OF LIQUIDS AND SOLIDS  Recall that liquids and solids can be approximated as incompressible sub- stances since their specific volumes remain nearly constant during a process. SEE TUTORIAL CH. 7, SEC. 8 ON THE DVD. Thus, dv ≅ 0 for liquids and solids, and Eq. 7–25 for this case reduces to du c dT ds = = (7–27) T T since cp = cv = c and du = c dT for incompressible substances. Then the entropy change during a process 2 is determined by integration to be dT T2. Liquids, solids: s2 — s1 = ∫ 1 c 1T 2 T ≅ cavg ln T1 1kJ>kg K2 (7–28) where cavg is the average specific heat of the substance over the given tem- perature interval. Note that the entropy change of a truly incompressible substance depends on temperature only and is independent of pressure. Equation 7–28 can be used to determine the entropy changes of solids and liquids with reasonable accuracy. However, for liquids that expand consider- ably with temperature, it may be necessary to consider the effects of volume change in calculations. This is especially the case when the temperature change is large. A relation for isentropic processes of liquids and solids is obtained by set- ting the entropy change relation above equal to zero. It gives Isentropic: s —s =c ln T2 = 0 → T =T (7–29) 2 1 avg 2 1 T1 That is, the temperature of a truly incompressible substance remains con- stant during an isentropic process. Therefore, the isentropic process of an incompressible substance is also isothermal. This behavior is closely approximated by liquids and solids. 352 | Thermodynam ics 7–7 Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K (or —82°C), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given in Table 7–1. Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa (a) using tabulated properties and (b) approximating liquid methane as an incompressible sub- stance. What is the error involved in the latter case? Solution Liquid methane undergoes a process between two specified states. The entropy change of methane is to be determined by using actual data and by assuming methane to be incompressible. Analysis (a) We consider a unit mass of liquid methane (Fig. 7–29). The P2 = 5 MPa properties of the methane at the initial and final states are T2 = 120 K