APPLIED THERMODYNAMICS

["6.10. PROBABILISTIC APPROACH TO ENTROPY 301 1 pn 1 pn 1 pn <n> = 1 <n> = 2 <n> = 3 S = 0kB S = 1.36747kB S = 1.74851kB n n n 0123456 0123456 0123456 1 pn <n> = 3.5 S = 1.79176kB n 0123456 1 pn 1 pn 1 pn <n> = 4 <n> = 5 <n> = 6 S = 1.74851kB S = 1.36747kB S = 0kB n n n 0123456 0123456 0123456 Figure 6.36: Probability distribution functions for a six-sided die weighted so as to achieve a variety of mean values <n> = 1, 2, 3, 7\/2, 4, 5, 6 while maximizing entropy. 2 S\/kB 1 0 1 2 3 4 5 6<n> Figure 6.37: Scaled entropy as a function of mean value of the six-sided die. CC BY-NC-ND. 15 May 2021, J. M. Powers.","302 CHAPTER 6. ENTROPY As an aside, let us reconsider Eq. (6.529), that itself holds for any value of <n>. Because 6 pn = n=1 1, we can form 66 pn = 1 = exp (\u22121 \u2212 \u03bb\u22171) exp (\u2212n\u03bb\u22172) . (6.543) n=1 n=1 Now, let us divide Eq. (6.529) by the left and right sides of Eq. (6.543) so as to get pn = exp (\u2212n\u03bb2\u2217) . (6.544) (6.545) 6 exp (\u2212n\u03bb2\u2217 ) n=1 We next de\ufb01ne the so-called partition function, Z(\u03bb\u22172), as 6 Z(\u03bb2\u2217) = exp (\u2212n\u03bb\u22172) . n=1 Now, rewriting Eq. (6.527), we have ln pn = \u22121 \u2212 \u03bb1\u2217 \u2212 n\u03bb2\u2217. (6.546) Now, take the logarithm of Eq. (6.544) and employ the de\ufb01nition of the partition function, Eq. (6.545), to get ln pn = \u2212n\u03bb\u22172 \u2212 ln Z. (6.547) Combining Eqs. (6.546) and (6.547), we get 1 + \u03bb\u22171 = ln Z. (6.548) This relates the \ufb01rst Lagrange multiplier to the second, that is embodied within Z. Now, let us uncover a non-obvious, but useful relation through the following sequence of operations. Let us prove that d d\u03bb2\u2217 <n> = \u2212 ln Z. (6.549) We can show this by employing our de\ufb01nitions combined with a series of operations and see if we are led to an identity. First use the chain rule <n> = \u2212 dZ d ln Z, (6.550) d\u03bb2\u2217 dZ (6.551) (6.552) = \u2212 dZ 1 , (6.553) d\u03bb\u22172 Z (6.554) = \u2212 d 6 exp (\u2212n\u03bb\u22172) 1 , (6.555) d\u03bb2\u2217 n=1 Z = 6 1 , Z n exp (\u2212n\u03bb\u22172) n=1 = 6 n exp (\u2212n\u03bb2\u2217) . n=1 6 exp (\u2212n\u03bb\u22172) n=1 Now, we also know by de\ufb01nition of the average and the probability pn that 6 <n> = npn. n=1 CC BY-NC-ND. 15 May 2021, J. M. Powers.","6.10. PROBABILISTIC APPROACH TO ENTROPY 303 Using Eq. (6.544) to remove pn from Eq. (6.555), we recover exactly Eq. (6.554). Thus our assertion of Eq. (6.549) is true. Lastly, we see that given <n>, it is possible to solve Eq. (6.554) to \ufb01nd \u03bb\u22172. With \u03bb\u22172, one can then use Eq. (6.544) to \ufb01nd each of the pn. In fact, the problem reduces to solving a polynomial equation. If we take x = exp(\u2212\u03bb\u22172), Eq. (6.554) reduces to <n> = x + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 . (6.556) x + x2 + x3 + x4 + x5 + x6 A factor of x cancels, and after rearrangement, we \ufb01nd a \ufb01fth order polynomial in x: (6 \u2212 <n>)x5 + (5 \u2212 <n>)x4 + (4 \u2212 <n>)x3 + (3 \u2212 <n>)x2 + (2 \u2212 <n>)x + (1 \u2212 <n>) = 0. (6.557) There are \ufb01ve roots to this equation, and at least one real root. It is likely possible to prove that in cases of interest there is only one real root. Let us check for the special case in which <n> = 3.5 = 7\/2. In that case, we get 5x5 + 3x4 + x3 \u2212 x2 \u2212 3x \u2212 5 = 0, (6.558) (x \u2212 1)(5x4 + 8x3 + 9x2 + 8x + 5) = 0. (6.559) There is obviously one real root, x = 1. The other four roots are complex. For x = exp(\u2212\u03bb2\u2217) = 1, we must have \u03bb\u22172 = 0, as we have seen to be the case for the uniform distribution pn = 1\/6 associated with <n> = 7\/2. Unaware of Boltzmann\u2019s theory, electrical engineer and mathematician Claude Shannon published in 1948 what amounts to precisely the same theory in the context of data com- munication. He arrived at a similar formula as Gibbs for the so-called information entropy, Sinf ormation : N (6.560) Sinformation = \u2212 pi ln pi. n=1 This is e\ufb00ectively a dimensionless version of the Gibbs entropy function, Eq. (6.488), di\ufb00ering only by a factor of kB. When applied in such context, Boltzmann\u2019s theory is known as information theory. Information theory was constructed to quantify data lost in telephone line signals. The theory and its author has had a seminal e\ufb00ect on modern computer and communication technologies.1112 Shannon as well as scenes from a public park dedicated to him in his hometown of Gaylord, Michigan, is depicted in Fig. 6.38. 11C. E. Shannon, 1948, \u201cA mathematical theory of communication,\u201d Bell System Technical Journal, 27(3): 379-423. 12C. E. Shannon, 1948, \u201cA mathematical theory of communication,\u201d Bell System Technical Journal, 27(4): 623-656. CC BY-NC-ND. 15 May 2021, J. M. Powers.","304 CHAPTER 6. ENTROPY Figure 6.38: Claude Elwood Shannon (1916-2001), American electrical engi- neer and mathematician whose \u201cinformation entropy\u201d is mathematically iden- tical to Boltzmann\u2019s statistical de\ufb01nition of entropy; leftmost image from https:\/\/en.wikipedia.org\/wiki\/Claude Elwood Shannon. 6.11 Summary statement of thermodynamics We have now covered the major ideas of the natural philosophy that is thermodynamics. We can summarize for an isolated universe by stating \u2022 The energy of the universe is constant, and its entropy is increasing. This is a loose translation of the statement of Clausius,13 \u2022 Die Energie der Welt ist konstant. Die Entropie der Welt strebt einem Maximum zu. It is a pessimistic set of principles! Thinking cosmologically, these laws would suggest that a large concentrated mass-energy complex, initially in a highly structured state, would ultimately deteriorate into spatially homogeneous space dust at a state of \ufb01nal equilibrium, the so-called heat death scenario posed initially in the nineteenth century by Kelvin and others. Present non-equilibrium thermodynamic theory would admit local structures to self- organize into ordered units such as a solar system or living beings. Such structures could potentially draw the energy necessary for self-organization from residual energy from the initial state. Though some disagree, it is claimed here that the science of thermodynamics is incapable of de\ufb01nitively answering theological questions that often arise regarding the origin of the universe, its ultimate fate, and the evolution of structures within it. It simply gives a framework for what is admissible with a given set of assumptions. In that it can illuminate some aspects of theology by identifying those parts of it that are in and out of agreement with 13R. Clausius, 1865, \u201cUeber verschiedene fu\u00a8r die Anwendung bequeme Formen der Hauptgleichungen der mechanischen Wa\u00a8rmetheorie,\u201d Annalen der Physik und Chemie, 125(7): 353-390. CC BY-NC-ND. 15 May 2021, J. M. Powers.","6.11. SUMMARY STATEMENT OF THERMODYNAMICS 305 empirical observations and their consequences, it may be of some use to such disciplines that are outside its realm. More generally, many scientists follow the train of thought popularized by the Austrian philosopher of science Karl Popper (1902-1994) who restricted scienti\ufb01c theories to those that are empirically testable, or more speci\ufb01cally, \u201cfalsi\ufb01able.\u201d Statements need not be falsi\ufb01able to be true, that thus admits the possibility of theological fact; they simply are not science. So, if our universe is formally isolated, we can look forward to \u201cheat death\u201d and the ultimate equilibrium, \ufb01rst suggested by Thomson.14 If it is not isolated, there is more uncertainty, and perhaps less reason for pessimism. The so-called laws of thermodynamics are simply an e\ufb03cient re\ufb02ection of present-day empirical data. Science is in that sense radically pragmatic; if unimpeachable data is found that contradict our present axioms of thermodynamics, science resorts to new and improved axioms. 14W. Thomson, 1852, \u201cOn a universal tendency in nature to the dissipation of mechanical energy,\u201d Trans- actions of the Royal Society of Edinburgh, 20(3): 139-142. CC BY-NC-ND. 15 May 2021, J. M. Powers.","306 CHAPTER 6. ENTROPY CC BY-NC-ND. 15 May 2021, J. M. Powers.","Chapter 7 Second law analysis for a control volume Read BS, Chapter 7 In this chapter, we will apply notions from control volume analysis to problems that involve the second law of thermodynamics. Recall that the fundamental description of our axioms is written for systems. We simply modify these axioms when applying them to control volumes. We shall omit most of the details of the reduction of the second law to control volume formulation. It is not unlike that done for mass and energy conservation. 7.1 Irreversible entropy production First recall an important form of the second law for a system, Eq. (6.31): S2 \u2212 S1 \u2265 2 \u03b4Q . (7.1) 1 T Let us introduce a convenient variable, the \u2022 Irreversible entropy production: a quantity that characterizes that portion of entropy production that is irreversible. We note that entropy can be produced by reversible heat transfer as well, that we segregate and do not consider here. We adopt the common notation of 1\u03c32 for irreversible entropy production, with units of kJ\/K. Our 1\u03c32 is equivalent to 1S2 gen of Eq. (7.11) of BS (p. 283), but is more aligned with the notation of non-equilibrium thermodynamics. We give it the subscripts to emphasize that it is path-dependent. Mathematically, we recast the second law for a system by the following two equations: S2 \u2212 S1 = 2 \u03b4Q + 1\u03c32, (7.2) 1 T (7.3) 1\u03c32 \u2265 0. 307","308 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Clearly, this is just a notational convenience that moves the inequality from one equation to another. On a di\ufb00erential basis, we can say for a system dS = \u03b4Q + \u03b4\u03c3. (7.4) T And for time-dependent processes, we say for a system dS = 1 \u03b4Q + \u03b4\u03c3 . (7.5) dt T dt dt Now, let us expand to an unsteady control volume, that should be similar to that for a system, with corrections for inlets and exits. We get dScv = Q\u02d9 j + m\u02d9 isi \u2212 m\u02d9 ese + \u03c3\u02d9 cv. (7.6) dt Tj j i e Our irreversible entropy production rate, \u03c3\u02d9 cv, with units of kW\/K, is equivalent to S\u02d9gen of Eq. (7.10) of BS (p. 283). Let us study this equation in some common limits. First, if the problem is in steady state, then 0= Q\u02d9 j + m\u02d9 isi \u2212 m\u02d9 ese + \u03c3\u02d9 cv, (7.7) Tj j i e steady state limit. If it is in steady state and there is one entrance and one exit, then mass conservation gives m\u02d9 i = m\u02d9 e = m\u02d9 , and 0= Q\u02d9 j + m\u02d9 (si \u2212 se) + \u03c3\u02d9 cv, (7.8) Tj j steady state, one entrance, one exit. We can rearrange to say se \u2212 si = 1 Q\u02d9 j + \u03c3\u02d9 cv . (7.9) m\u02d9 Tj m\u02d9 (7.10) j If there is no heat transfer to the control volume, then se \u2212 si = \u03c3\u02d9 cv , m\u02d9 no heat transfer to control volume, steady state, one entrance\/exit. CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.1. IRREVERSIBLE ENTROPY PRODUCTION 309 Example 7.1 Let us revisit an earlier example problem from p. 191 involving throttling and \ufb01nd the irreversible entropy production \ufb02ux. Consider calorically perfect ideal air \ufb02owing in a duct at P1 = 100000 Pa, T1 = 300 K, v1 = 10 m\/s. Take cP = 1000 J\/kg\/K. The air is throttled down by a valve to P2 = 90000 Pa. We \ufb01rst repeat some of the earlier analysls. From the ideal gas law we get \u03c11 = P1 = 100000 Pa = 1.16144 kg . (7.11) RT1 m3 287 J (300 K) kg K We write the mass, energy, thermal and caloric state equations as \u03c12v2 = \u03c11v1, (7.12) h2 + v22 = h1 + v12 , (7.13) 2 2 (7.14) (7.15) P2 = \u03c12RT2, h2 \u2212 h1 = cP (T2 \u2212 T1). This simpli\ufb01es somewhat to \u03c12v2 = \u03c11v1, (7.16) (7.17) cP T2 + v22 = cP T1 + v12 , (7.18) 2 2 P2 = \u03c12RT2. Substituting numbers, we get the system of non-linear algebraic equations \u03c12v2 = 1.16144 kg 10 m = 11.6144 kg s , (7.19) m3 s m2 (7.20) J v22 J 10 m 2 J kg K 2 kg K s kg 1000 T2 + = 1000 (300 K) + = 300050 , 2 90000 Pa = \u03c12 287 J T2. (7.21) kg K This forms three equations in the three unknowns \u03c12, T2, v2. Detailed manipulation can reduce this to a quadratic equation, with two roots. The \ufb01rst root yields \u03c12 = 1.04533 kg , (7.22) m3 (7.23) (7.24) T2 = 299.988 K, v2 = 11.1107 m . s This is the physical root. From Eq. (6.87), we have s2 \u2212 s1 = cP ln T2 \u2212 R ln P2 , (7.25) T1 P1 (7.26) = 1000 J ln 299.988 K \u2212 287 J ln 90000 Pa , kg K 300 K kg K 100000 Pa \u22120.0390793 J\/kg\/K \u221230.2385 J\/kg\/K = 30.1994 J . (7.27) kg K CC BY-NC-ND. 15 May 2021, J. M. Powers.","310 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME We can specialize Eq. (7.10) to say \u03c3\u02d9 cv = s2 \u2212 s1, (7.28) m\u02d9 (7.29) (7.30) \u03c3\u02d9 cv = \u03c11v1(s2 \u2212 s1), (7.31) A = 1.16144 kg 10 m 30.1994 J , m3 s kg K = 350.748 W . m2 K As required by the second law, this is a positive quantity. Note that because all the entropy change is associated with irreversibility, and none with heat transfer, the area under the curve on a T \u2212 s diagram, while non-zero, does not represent any heat transfer. One could say \u2022 If there is reversible heat transfer, there will be area under the curve in a T \u2212 s diagram. \u2022 If there is area under a curve in a T \u2212 s diagram, it could be due to either irreversibility or heat transfer. Example 7.2 A steam turbine has an inlet condition of P1 = 30 bar, T1 = 400 \u25e6C, v1 = 160 m\/s. Its exhaust condition is T2 = 100 \u25e6C, v2 = 100 m\/s, x2 = 1. The work for the turbine is wcv = 540 kJ\/kg. Find \u03c3\u02d9 cv\/m\u02d9 . The surroundings are at 500 K. See Fig. 7.1. control volume steam turbine T2 = 100 oC P1 = 30 bar v = 100 m\/s T1 = 400 oC 2 v = 160 m\/s x2 = 1 1 Tsurr = 500 K qcv wcv = 540 kJ\/kg Figure 7.1: Steam turbine schematic. Mass conservation tells us dmcv = m\u02d9 1 \u2212 m\u02d9 2, (7.32) dt (7.33) =0 m\u02d9 1 = m\u02d9 2 = m\u02d9 . CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.1. IRREVERSIBLE ENTROPY PRODUCTION 311 Energy conservation tells us dEcv = Q\u02d9 cv \u2212 W\u02d9 cv + m\u02d9 h1 + v12 \u2212 m\u02d9 h2 + v22 , (7.34) dt 2 2 (7.35) =0 (7.36) Q\u02d9 cv = W\u02d9 cv + (h2 \u2212 h1) + 1 v22 \u2212 v12 , m\u02d9 m\u02d9 2 qcv = wcv + (h2 \u2212 h1) + 1 v22 \u2212 v12 . 2 We \ufb01nd from the tables that P2 = 101.3 kPa, h2 = 2676.05 kJ\/kg, and h1 = 3230.82 kJ\/kg. So qcv = 540 kJ + 2676.05 kJ \u2212 3230.82 kJ kg kg kg 1 m 2 m 2 kJ 2 s s + 100 \u2212 160 kg m2 , (7.37) s2 (7.38) 1000 = \u221222.6 kJ . kg This represents a loss of heat to the surroundings. The second law, Eq. (7.9), tells us s2 \u2212 s1 = qcv,j + \u03c3\u02d9 cv , (7.39) Tj m\u02d9 (7.40) \u03c3\u02d9 cv = s2 \u2212 s1 \u2212 qcv . m\u02d9 T From the tables, we \ufb01nd s1 = 6.9212 kJ\/kg\/K, s2 = 7.3549 kJ\/kg\/K. So \u03c3\u02d9 cv = 7.3549 kJ \u2212 6.9212 kJ \u2212 \u221222.6 kJ , (7.41) m\u02d9 kg K kg K kg (7.42) 500 K = 0.4789 kJ > 0. kg K The process is sketched in Fig. 7.2. Example 7.3 Steam is \ufb02owing in a di\ufb00user. At the entrance it has P1 = 0.2 MPa, T1 = 200 \u25e6C, v1 = 700 m\/s. At the exhaust it has v2 = 70 m\/s. Assume an adiabatic reversible process. Find the \ufb01nal pressure and temperature. See Fig. 7.3. Because the process is reversible and adiabatic, the second law simply reduces to (7.43) s2 = s1. We go to the tables and \ufb01nd s1 = 7.5066 kJ\/kg\/K. So s2 = 7.5066 kJ\/kg\/K. CC BY-NC-ND. 15 May 2021, J. M. Powers.","312 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME T P 1 1 T=4 00 Co P=3000 kPa P=101.3 kPa 2 2 T=100 oC sv Figure 7.2: T \u2212 s and P \u2212 v diagrams for steam turbine problem. control volume P1 = 0.2 MPa v = 70 m\/s T1 = 200 oC 2 v = 700 m\/s 1 Figure 7.3: Steam di\ufb00user schematic. Now, the \ufb01rst law tells us dEcv = Q\u02d9 cv \u2212 W\u02d9 cv +m\u02d9 h1 + 1 v12 + gz1 \u2212 m\u02d9 h2 + 1 v22 + gz2 . (7.44) dt 2 2 =0 =0 =0 Now, we cannot neglect kinetic energy changes in a di\ufb00user. We can neglect potential energy changes. We can also neglect unsteady e\ufb00ects as well as control volume work. We were told it is adiabatic, so heat transfer can be neglected. Thus, we get 0 = m\u02d9 h1 + 1 v12 \u2212 m\u02d9 h2 + 1 v22 , (7.45) 2 2 (7.46) h2 = h1 + 1 v12 \u2212 v22 . 2 The tables give us h1 = 2870.5 kJ\/kg. We thus can get kJ 1 m 2 m 2 kJ kJ kg 2 s s kg h2 = 2870.5 + 700 \u2212 70 kg = 3113.05 . (7.47) 1000 m2 s2 Now, we know two properties, h2 and s2. To \ufb01nd the \ufb01nal state, we have to double interpolate the CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.2. BERNOULLI PRINCIPLE 313 superheated steam tables. Doing so, we \ufb01nd T2 = 324.1 \u25e6C, P2 = 542 kPa. (7.48) See Fig. 7.4 for a diagram of the process. Note the temperature rises in this process. The kinetic TP P=542 kPa 2 2 T=324.1 oC P=200 kPa 1 1 T=200 oC s v Figure 7.4: Steam di\ufb00user schematic. energy is being converted to thermal energy. 7.2 Bernoulli principle Let us consider our thermodynamics in appropriate limit to develop the well known \u2022 Bernoulli principle: a useful equation in thermal science, often used beyond its realm of validity, relating pressure, \ufb02uid velocity, density, and \ufb02uid height, valid only in the limit in which mechanical energy is conserved. The principle was \ufb01rst elucidated, though not without considerable turmoil within his proli\ufb01c family, by Daniel Bernoulli,1 depicted in Fig. 7.5. For the Bernoulli principle to be formally valid requires some restrictive assumptions. We shall make them here in the context of thermodynamics. The same assumptions allow one to equivalently obtain the principle from an analysis of the linear momentum equation developed in \ufb02uid mechanics courses. In such a \ufb02uids development, we would need to make several additional, but roughly equivalent, assumptions. This equivalence is obtained because we shall develop the equation in the limit that mechanical energy is not dissipated. For our analysis here, we will make the following assumptions: \u2022 the \ufb02ow is steady, 1D. Bernoulli, 1738, Hydrodynamica, sive de Viribus et Motibus Fluidorum Commentarii, J. H. Deckeri, Strasbourg. CC BY-NC-ND. 15 May 2021, J. M. Powers.","314 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Figure 7.5: Daniel Bernoulli (1700-1782), Dutch-born mathematician and physicist; image from http:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Bernoulli Daniel.html. \u2022 all processes are fully reversible, \u2022 there is one inlet and exit, and \u2022 there is contact with one thermal reservoir in which thermal energy is transferred reversibly. Though we will not study it, there is another important version of the Bernoulli principle for unsteady \ufb02ows. Our second law, Eq. (7.6), reduces in the limits we study to dScv = Q\u02d9 cv + m\u02d9 s1 \u2212 m\u02d9 s2 + \u03c3\u02d9 cv , (7.49) dt T =0 (7.50) =0 (7.51) (7.52) 0 = Q\u02d9 cv + m\u02d9 (s1 \u2212 s2), T Q\u02d9 cv m\u02d9 (s2 \u2212 s1) = T , m\u02d9 T (s2 \u2212 s1) = Q\u02d9 cv. Now, let us non-rigorously generalize this somewhat and allow for di\ufb00erential heat transfer at a variety of temperatures so as to get 2 (7.53) m\u02d9 T ds = Q\u02d9 cv. 1 In a more formal analysis from continuum mechanics, this step is much cleaner, but would require signi\ufb01cant development. What we really wanted was a simpli\ufb01cation for Q\u02d9 cv that we CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.2. BERNOULLI PRINCIPLE 315 could use in the relevant energy equation, Eq. (4.182), considered next: dEcv = Q\u02d9 cv \u2212W\u02d9 cv + m\u02d9 h1 + 1 v12 + gz1 \u2212 m\u02d9 h2 + 1 v22 + gz2 , (7.54) dt 2 2 (7.55) =0 =m\u02d9 2 T ds 1 W\u02d9 cv m\u02d9 = 2 ds + (h1 \u2212 h2) + 1 (v12 \u2212 v22) + g(z1 \u2212 z2). 2 wcv T 1 Now, one form of the Gibbs equation, Eq. (6.65), has T ds = dh \u2212 v dP , so 2 22 (7.56) (7.57) T ds = dh \u2212 v dP, (7.58) 1 11 2 = h2 \u2212 h1 \u2212 v dP, 1 22 T ds + (h1 \u2212 h2) = \u2212 v dP. 11 Now, substitute Eq. (7.58) into Eq. (7.55) to get wcv = 2 ds + (h1 \u2212 h2) + 1 (v12 \u2212 v22) + g(z1 \u2212 z2), (7.59) 2 (7.60) T 1 =\u2212 2 v dP 1 = \u2212 2 v dP + 1 (v12 \u2212 v22) + g(z1 \u2212 z2). 1 2 Now, if wcv = 0, we get a rarely used, but valuable generalization of the Bernoulli principle: 0= 2 v dP + 1 (v22 \u2212 v12) + g(z2 \u2212 z1). (7.61) 1 2 7.2.1 Incompressible limit In the important limit for liquids in which v is approximately constant, we recall that \u03c1 = 1\/v and write Eq. (7.61) as 0 = v 2 dP + 1 (v22 \u2212 v12) + g(z2 \u2212 z1), (7.62) 1 2 (7.63) (7.64) = v(P2 \u2212 P1) + 1 (v22 \u2212 v12) + g(z2 \u2212 z1), 2 = P2 \u2212 P1 + 1 (v22 \u2212 v12) + g(z2 \u2212 z1). \u03c1 2 CC BY-NC-ND. 15 May 2021, J. M. Powers.","316 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME We can rewrite this as P + 1 v2 + gz = constant, (7.65) \u03c1 2 Bernoulli principle for an incompressible liquid. In a di\ufb00erent limit, that in which changes in kinetic and potential energy can be neglected, Eq. (7.60) reduces to 2 (7.66) wcv = \u2212 v dP. 1 This integral is not the area under the curve in P \u2212 v space. It is the area under the curve 2 in v \u2212 P space instead. Contrast this with the system result where we get 1w2 = 1 P dv. In the important operation of pumping liquids, v is nearly constant, and we can say wpump = \u2212v(P2 \u2212 P1), (7.67) W\u02d9 pump = m\u02d9 v(P1 \u2212 P2). (7.68) 7.2.2 Calorically perfect ideal gas limit Let us consider the Bernoulli principle for a CPIG undergoing a reversible adiabatic process. For such a process, we have from Eq. (6.170) that P vk = C. Thus, v = (C\/P )1\/k. Let us consider Eq. (7.61) for this case: 2 1 1 C k 1 (v22 v12) P 2 0= dP + \u2212 + g(z2 \u2212 z1), (7.69) (7.70) 1 P2 1 (7.71) = k P Ck 2 k\u22121 P + (v22 \u2212 v12) + g(z2 \u2212 z1), P1 = k k 1 (P2v2 \u2212 P1v1) + 1 (v22 \u2212 v12) + g(z2 \u2212 z1). \u2212 2 Thus, for a CPIG obeying the Bernoulli principle, we can say, taking v = 1\/\u03c1, kP + 1 v2 + gz = constant, (7.72) k\u22121 \u03c1 2 Bernoulli principle for a CPIG. While this is \ufb01ne and very similar to the form given for incompressible \ufb02ow, it really should be simpli\ufb01ed more because P is directly related to \u03c1 for this compressible \ufb02ow. Using, Eq. (6.170), P vk = C = Povok and \u03c1o = 1\/vo, we can rewrite Eq. (7.72) as k P k\u22121 Po 1 v2 k k \u2212 1 Po + + gz = constant. (7.73) \u03c1o 2 CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.2. BERNOULLI PRINCIPLE 317 Similarly for isentropic pumps or turbines using CPIGs with negligible changes in kinetic and potential energies, Eq. (7.60) reduces to 2 wcv = \u2212 v dP, (7.74) (7.75) 1 (7.76) (7.77) = \u2212 k k 1 (P2v2 \u2212 P1v1), \u2212 (7.78) = \u2212 k k 1 R(T2 \u2212 T1), \u2212 = \u2212 kRT1 T2 \u2212 1 , k\u22121 T1 = \u2212cP T1 P2 k\u22121 . k P1 \u22121 An isothermal pump or compressor using a CPIG has 2 wcv = \u2212 v dP, (7.79) (7.80) 1 (7.81) (7.82) = \u2212 2 RT dP, 1P = \u2212RT1 ln P2 , P1 = \u2212P1v1 ln P2 . P1 7.2.3 Torricelli\u2019s formula Let us consider a special case of the Bernoulli principle, known as Torricelli\u2019s formula, de- veloped by Evangelista Torricelli,2 the inventor of the barometer, and for whom the pressure unit torr is named (1 torr = 133.322 Pa = 1\/760 atm.) Torricelli is sketched in Fig. 7.6. Consider the scenario of Fig. 7.7. Here, a \ufb02uid is in an open container. The container has a small hole near its bottom. The \ufb02uid at the top of the container, z = z1, is at P1 = Patm. The leaking \ufb02uid exhausts at the same pressure P2 = Patm. The \ufb02uid leaks at velocity v2 at a hole located at z = z2. The \ufb02uid at the top of the container barely moves; so, it has negligible velocity, v1 \u223c 0. The \ufb02uid exists in a constant gravitational \ufb01eld with gravitational acceleration g, as sketched. Assume the \ufb02uid is incompressible and all of the restrictions of the Bernoulli principle are present. Let us apply Eq. (7.65): P1 + 1 v12 +gz1 = P2 + 1 v22 + gz2. (7.83) \u03c1 2 \u03c1 2 \u223c0 2E. Torricelli, 1643, De Motu Gravium Naturaliter Accelerato, Firenze. CC BY-NC-ND. 15 May 2021, J. M. Powers.","318 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Figure 7.6: Evangelista Torricelli (1608-1647), Italian physicist and mathematician; image from https:\/\/en.wikipedia.org\/wiki\/Evangelista Torricelli. Setting P1 = P2 = Patm and ignoring v1 gives gz1 = 1 v22 + gz2, (7.84) 2 v2 = 2g(z1 \u2212 z2), (7.85) Torricelli\u2019s formula. Notice rearranging Torricelli\u2019s formula gives 1 v22 = g(z1 \u2212 z2) . (7.86) 2 kinetic energy potential energy It represents a balance of kinetic and potential energy of the \ufb02uid, and thus is concerned only with mechanical energy. Example 7.4 Let us design a liquid water fountain by cutting a hole in a high pressure water pipe. See Fig. 7.8. We desire the \ufb01nal height of the water jet to be 30 m. The jet rises against a gravitational \ufb01eld with g = 9.81 m\/s2. The atmospheric pressure is 100 kPa. Water has density \u03c1 = 997 kg\/m3. Find the necessary pipe gauge pressure P1 and jet exit velocity v2. Let us apply the Bernoulli principle between states 1 and 3, the pipe interior and the peak of the height of the fountain. We will estimate the velocity of the water in the pipe to be small, v1 \u223c 0 m\/s. We will also estimate the velocity at the apex of the motion to be negligible, v3 = 0 m\/s. CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.2. BERNOULLI PRINCIPLE 319 z1 P1 = Patm 1 v1~ 0 g v 2 2 z2 P2 = Patm Figure 7.7: Fluid container with hole. Let us apply Eq. (7.65): P1 + 1 v12 +gz1 = P3 + 1 v32 +gz3, (7.87) \u03c1 2 \u03c1 2 (7.88) (7.89) \u223c0 =0 (7.90) P1 = Patm + \u03c1g(z3 \u2212 z1), (7.91) P1 \u2212 Patm = \u03c1g(z3 \u2212 z1). (7.92) (7.93) =Pgauge (7.94) Substituting numbers, we \ufb01nd Pgauge = 997 kg 9.81 m (30 m) = 2.934 \u00d7 105 Pa = 293.4 kPa. m3 s2 We can use the same principle to estimate the exit velocity, v2. Here, we take z1 \u223c z2. P1 + 1 v12 +gz1 = P2 + 1 v22 + g z2 , \u03c1 2 \u03c1 2 \u223c0 \u223cz1 P1 = Patm + 1 v22 , \u03c1 \u03c1 2 v2 = 2(P1 \u2212 Patm) , \u03c1 v2 = 2Pgauge . \u03c1 Substituting numbers, we \ufb01nd v2 = 2(2.934 \u00d7 105 Pa) = 24.26 m . (7.95) s 997 kg m3 CC BY-NC-ND. 15 May 2021, J. M. Powers.","320 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME v ~ 0 m\/s, P3 = 100 kPa 3 30 m g = 9.81 m\/s2 v = ?, P2 = 100 kPa 2 = 997 kg\/m3, P1 = ?, v ~ 0 m\/s 1 Figure 7.8: Sketch of simple water fountain. Figure 7.9: The University of Notre Dame\u2019s War Memorial Fountain, 4 June 2010. One could use a similar analysis to estimate the necessary pressure to generate the jet of the University of Notre Dame\u2019s War Memorial Fountain, depicted in Fig. 7.9. Example 7.5 Perform a similar calculation for the problem sketched in Fig. 7.8, but account for mass conservation. Take the cross-sectional area of the pipe to be A1 = A4 = 1 m2, and that of the hole to be A2 = 0.01 m2. We measure v1 = 1 m\/s. See Fig. 7.10. Assume we have the same Pgauge = 293.4 kPa as calculated earlier. Find the new height of the fountain, z3, and the new exit velocity v2. CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.2. BERNOULLI PRINCIPLE 321 v = 1 m\/s A2 = 0.01 m2 1 m. 2 A1 = 1 m2 A4 = 1 m2 . m. 1 m4 Figure 7.10: Schematic of mass balance for water fountain problem. The mass balance gives us dmcv = m\u02d9 1 \u2212 m\u02d9 2 \u2212 m\u02d9 4, (7.96) We recall that m\u02d9 = \u03c1vA, so dt (7.97) =0 (7.98) 0 = m\u02d9 1 \u2212 m\u02d9 2 \u2212 m\u02d9 4, m\u02d9 4 = m\u02d9 1 \u2212 m\u02d9 2. \u03c14v4A4 = \u03c11v1A1 \u2212 \u03c12v2A2. (7.99) We assume incompressible \ufb02ow, so \u03c11 = \u03c12 = \u03c14 = \u03c1, and we have A1 = A4, so \u03c1v4A1 = \u03c1v1A1 \u2212 \u03c1v2A2, (7.100) (7.101) v4 = v1 \u2212 v2 A2 . A1 This is nice, but not that useful. It simply predicts a lessening of velocity downstream of the hole. The Bernoulli principle, Eq. (7.65), applied between 1 and 2 gives P1 + 1 v12 + gz1 = P2 + 1 v22 + g z2 , (7.102) \u03c1 2 \u03c1 2 (7.103) (7.104) \u223cz1 (7.105) P1 + 1 v12 = Patm + 1 v22, \u03c1 2 \u03c1 2 v2 = 2(P1 \u2212 Patm) + v12, \u03c1 = 2Pgauge + v12. \u03c1 With numbers, we get v2 = 2(2.934 \u00d7 105 Pa) + 1 m 2 24.28 m . (7.106) s s 997 kg = m3 The exit velocity is barely changed from our earlier analysis. CC BY-NC-ND. 15 May 2021, J. M. Powers.","322 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Now, determine the new height. Let us again apply Eq. (7.65): P1 + 1 v12 + gz1 = P3 + 1 v32 +gz3, (7.107) \u03c1 2 \u03c1 2 (7.108) (7.109) =0 (7.110) P1 + 1 v12 + gz1 = Patm + gz3, (7.111) \u03c1 2 \u03c1 P1 \u2212 Patm + 1 v12 = g(z3 \u2212 z1), \u03c1 2 z3 \u2212 z1 = Pgauge + 1 v12 . \u03c1g 2g Substituting numbers, we get z3 \u2212 z1 = 2.934 \u00d7 105 Pa + 1 1 m 2 30.05 m. 2 s kg m 9.81 m = m3 s s2 997 9.81 The extra boost in height comes from accounting for the initial kinetic energy of the water. 7.3 Component e\ufb03ciency Recall for cycles, as shown in the analysis surrounding Eq. (4.315), we de\ufb01ned a thermal e\ufb03ciency as what you want\/what you pay for, which for a power cycle is \u03b7 = Wnet\/QH . We can further de\ufb01ne e\ufb03ciencies for components. For a component, we will take an e\ufb03ciency for some devices to be what we get best we could \u03b7component = the get . (7.112) For other devices, we will take an e\ufb03ciency to be \u03b7component = the least required . (7.113) what is actually required The optimal is generally an isentropic device. So, for example, for a turbine, we say \u03b7turbine = work done actual work turbine = w. (7.114) by an isentropic ws Here, the subscript \u201cs\u201d denotes isentropic. For a nozzle, we would like to maximize the kinetic energy of the working \ufb02uid, so we say \u03b7nozzle = v2 . (7.115) vs2 CC BY-NC-ND. 15 May 2021, J. M. Powers.","7.3. COMPONENT EFFICIENCY 323 However for pumps and compressors, the isentropic pump requires the least work input. So we take instead ws w \u03b7pump,compressor = . (7.116) Example 7.6 N2 is adiabatically compressed from T1 = 300 K, P1 = 100 kPa to P2 = 1000 kPa. The compressor e\ufb03ciency is \u03b7c = 0.9. Find the \ufb01nal state and the compression work per unit mass. Assume N2 is a CIIG. The \ufb01rst law for adiabatic compression gives w = h2 \u2212 h1. (7.117) Table A.8 from BS tells us that at T1 = 300 K, h1 = 311.67 kJ\/kg, soT1 = 6.8463 kJ\/kg\/K. But we are not sure what state 2 is, and the \ufb01rst law does not help yet, as we do not know either w or h2. Let us calculate state 2 assuming an isentropic process, and then use our knowledge of compressor e\ufb03ciency to correct for real e\ufb00ects. We \ufb01rst note for N2 that R= R = 8.31451 kJ K = 0.2968 kJ . (7.118) M kmole kg K 28.013 kg kmole From Eq. (6.98), we can conclude that for an isentropic process in which s2 = s1 that s2 \u2212 s1 = 0 = sTo 2 \u2212 soT1 \u2212 R ln P2 , (7.119) P1 (7.120) (7.121) soT2 = soT1 + R ln P2 , (7.122) P1 (7.123) = 6.8463 kJ + 0.2968 kJ ln 1000 kPa , kg K kg K 100 kPa = 7.52971 kJ . kg K Knowing soT2 , we next interpolate Table A.8 from BS to get T2s = 576.133 K, h2s = 601.712 kJ . kg (Note the CPIG assumption would have yielded T2s = (300 K)(10)0.286 = 579.209 K). So the work for the isentropic compressor is ws = h2s \u2212 h1 = 601.712 kJ \u2212 311.67 kJ = 290.042 kJ . (7.124) kg kg kg Now, consider the compressor e\ufb03ciency: ws , (7.125) \u03b7c = w (7.126) w= ws \u03b7c , (7.127) = 290.042 kJ , 0.9 kg = 322.269 kJ . (7.128) kg CC BY-NC-ND. 15 May 2021, J. M. Powers.","324 CHAPTER 7. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Thus, we have the actual work per unit mass. So the actual enthalpy at state 2 can be derived from the \ufb01rst law: h2 = h1 + w = 311.67 kJ + 322.269 kJ = 633.939 kJ . (7.129) kg kg kg Now, knowing h2, we can again interpolate Table A.8 of BS to \ufb01nd the \ufb01nal temperature T2 to be T2 = 606.263 K. (7.130) We had to add more energy to achieve the non-isentropic compression relative to the isentropic com- pression. CC BY-NC-ND. 15 May 2021, J. M. Powers.","Chapter 8 Cycles Read BS, Chapters 9, 10 In this chapter, we will delve more deeply into some thermodynamic cycles. 8.1 Rankine Large electric power plants typically utilize a vapor power cycle. Regardless of the heat source, be it nuclear or combustion of coal, oil, natural gas, wood chips, etc., the remaining details of these plants are similar. Typically a pure working \ufb02uid, usually water, is circulated through a cycle, and that \ufb02uid trades heat and work with its surroundings. We sketch a typical power plant cycle for electricity generation in Fig. 8.1. The ideal Rankine cycle was \ufb01rst described in 1859 by William John Macquorn Rankine, long after the steam engine was in wide usage. Expanding on our earlier discussion of Ch. 4.5, the cycle has the following steps: \u2022 1 \u2192 2: isentropic compression in a pump, \u2022 2 \u2192 3: isobaric heating in a boiler, \u2022 3 \u2192 4: isentropic expansion in a turbine, and \u2022 4 \u2192 1: isobaric cooling in a condenser. Two variants of the T \u2212 s diagram are given in Fig. 8.2. The \ufb01rst is more e\ufb03cient as it has the appearance of a Carnot cycle. However, it is impractical, as it induces liquid water in the turbine, that can damage its blades. So the second is more common. The thermal e\ufb03ciency, consistent with what was \ufb01rst introduced in Eq. (4.316), is \u03b7 = W\u02d9 net = W\u02d9 turbine + W\u02d9 pump . (8.1) Q\u02d9 H Q\u02d9 boiler 325","326 CHAPTER 8. CYCLES combustion exhaust 3 turbine generator .. + - fuel 4 air cold water boiler condenser hot water pump cooling 21 tower work in Figure 8.1: Rankine cycle schematic. This reduces to \u03b7 = m\u02d9 ((h3 \u2212 h4) + (h1 \u2212 h2)) , (8.2) m\u02d9 (h3 \u2212 h2) (8.3) (8.4) = 1 \u2212 h4 \u2212 h1 , h3 \u2212 h2 qout,condenser = 1 \u2212 qin,boiler . This is consistent with the earlier Eq. (5.8). Note that because the Rankine cycle is not a Carnot cycle, we have qout,condenser\/qin,boiler = T1\/T3. Power plants are sometimes characterized by their \u2022 back work ratio: bwr, the ratio of pump work to turbine work. Here, bwr = |pump work| = h2 \u2212 h1 . (8.5) |turbine work| h3 \u2212 h4 We model the pump work as an isentropic process. Recall our analysis for isentropic pumps that generated Eq. (7.67). The Gibbs equation, Eq. (6.66), gives dh = T ds + v dP . If ds = 0, we have dh = v dP, (8.6) CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.1. RANKINE 327 T T 3 3 4 2 2 14 1 s s Figure 8.2: T \u2212 s for two Rankine cycles. Thus, for the pump h2 \u2212 h1 = v(P2 \u2212 P1), (8.7) because v is nearly constant, so the integration is simple. It might be tempting to make the Rankine cycle into a Carnot cycle as sketched in Fig. 8.3. However, it is practically di\ufb03cult to build a pump to handle two-phase mixtures. T 23 14 s Figure 8.3: Rankine-Carnot cycle. The gas phase can seriously damage the pump. Some features that could be desirable for a Rankine cycle include CC BY-NC-ND. 15 May 2021, J. M. Powers.","328 CHAPTER 8. CYCLES \u2022 high power output: One can enhance this by raising the \ufb02uid to a high temperature during the combustion process or by pumping the \ufb02uid to a high pressure. Both strategies soon run into material limits; turbine blades melt and pipes burst. Another strategy is to lower the condenser pressure. That means that one must maintain a vacuum, and this can be di\ufb03cult. \u2022 high thermal e\ufb03ciency: The key design strategy here lies in 1) increasing component e\ufb03ciencies, and 2) rendering the overall cycle as much like a Carnot cycle as is feasible. Modern power plants have had revolutionary increases in overall thermal e\ufb03ciency because of enhancements that make the process more Carnot-like. There are some important loss mechanisms in the Rankine cycle that inhibit e\ufb03ciency. They include \u2022 Turbine losses: These are the major losses. To avoid these losses requires detailed consideration of \ufb02uid mechanics, material science, and heat transfer and is beyond the scope of classical thermodynamics. Thermodynamics develops broad measures of turbine e\ufb03ciency such as \u03b7turbine = (h3 \u2212 h4)\/(h3 \u2212 h4s). \u2022 Pump losses: Again, \ufb02uid mechanics, machine design, and material science are required to analyze how to actually avoid these losses. Thermodynamics characterizes them by pump e\ufb03ciency, \u03b7pump = (h2s \u2212 h1)\/(h2 \u2212 h1). \u2022 Heat transfer losses from components. \u2022 Pressure drop in pipes. \u2022 Incomplete fuel combustion. \u2022 Pollution removal devices. \u2022 Loss of heat to surroundings in the condenser. One simple design strategy to make the system more Carnot-like is to use \u2022 Reheat: a design strategy in which steam is extracted from the turbine before it is fully expanded, then sent to the boiler again, and re-expanded through the remainder of the turbine. This has the e\ufb00ect of making the system more like a Carnot cycle. A schematic and T \u2212 s diagram for the Rankine cycle with reheat is given in Fig. 8.4. Example 8.1 Consider water in a Rankine power cycle with reheat. The \ufb01rst turbine has water enter at P3 = 8000 kPa, T3 = 500 \u25e6C. The water expands to 600 kPa, undergoes reheat, and then expands again to 10 kPa. The mass \ufb02ow rate is m\u02d9 = 2.63 \u00d7 105 kg\/hr. We have \u03b7t = 0.88 for each turbine, and \u03b7p = 0.80 for the pump. Find the net power generated, \u03b7, and the heat transfer to the condenser. CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.1. RANKINE 329 s boiler T 35 4 2 3 6 pump turbine 4 5 62 1 condenser 1 Figure 8.4: Rankine cycle with reheat schematic and T \u2212 s diagram. Let us consider the big picture \ufb01rst. The net speci\ufb01c power will be the positive e\ufb00ect of the two turbines and the negative e\ufb00ect of the pump: wnet = (h3 \u2212 h4) + (h5 \u2212 h6) + (h1 \u2212 h2) . (8.8) turbine 1 turbine 2 pump Now, the heat input for the reheat is in two stages: qin = (h3 \u2212 h2) + (h5 \u2212 h4). (8.9) Lastly, the heat rejection in the condenser is qout = h6 \u2212 h1. (8.10) Let us start at the entrance of the \ufb01rst turbine, at 3. We are given P3 and T3, so we consult the tables and \ufb01nd kJ kJ kg kg K h3 = 3398.27 , s3 = 6.7239 . (8.11) We are given P4 = 600 kPa. Now, let us get the ideal behavior of the turbine: s4s = s3 = 6.7239 kJ\/kg\/K. At this condition, we \ufb01nd state 4 is a two-phase mixture. At 600 kPa, we \ufb01nd sf = 1.9311 kJ\/kg\/K, sg = 6.7600 kJ\/kg\/K. So s4s \u2212 sf 6.7239 kJ \u2212 1.9311 kJ sg \u2212 sf kg K kg K x4s = = = 0.992524. (8.12) kJ kJ 6.7600 kg K \u2212 1.9311 kg K We can thus get h4s by consulting the tables to \ufb01nd h4s = hf + x4shfg = 670.54 kJ + (0.992524) 2086.26 kJ = 2741.2 kJ . (8.13) kg kg kg Now, \u03b7t = (h3 \u2212 h4)\/(h3 \u2212 h4s), so h4 = h3 \u2212 \u03b7t(h3 \u2212 h4s), (8.14) (8.15) = 3398.27 kJ \u2212 (0.88) 3398.27 kJ \u2212 2741.2 kJ , kg kg kg (8.16) = 2820.05 kJ . kg CC BY-NC-ND. 15 May 2021, J. M. Powers.","330 CHAPTER 8. CYCLES Now, state 5 is after the reheat, that was isobaric at P4 = P5 = 600 kPa, and the reheating returns the temperature to T5 = 500 \u25e6C. From the superheat tables, we \ufb01nd h5 = 3482.75 kJ\/kg, s5 = 8.0020 kJ\/kg\/K. After expansion in the second turbine, we have s6s = s5 = 8.0020 kJ\/kg. And we were given P6 = 10 kPa. We consult the saturation tables to \ufb01nd at this pressure sf = 0.6492 kJ\/kg\/K, sg = 8.1501 kJ\/kg\/K. Thus, s6s \u2212 sf 8.0020 kJ \u2212 0.6492 kJ sg \u2212 sf kg K kg K x6s = = = 0.980256. (8.17) kJ kJ 8.1501 kg K \u2212 0.6492 kg K (8.18) The tables then give the necessary information to compute h6s: (8.19) (8.20) h6s = hf + x6shfg = 191.81 kJ + (0.980256) 2392.82 kJ = 2537.39 kJ . (8.21) kg kg kg (8.22) Now, the actual h6 is found via (8.23) h6 = h5 \u2212 \u03b7t(h5 \u2212 h6s), (8.24) (8.25) = 3482.75 kJ \u2212 (0.88) 3482.75 kJ \u2212 2537.39 kJ , (8.26) kg kg kg (8.27) = 2650.83 kJ . kg Now, the tables give us h1 = hf = 191.81 kJ , v1 = vf = 0.001010 m3 . kg kg For the pump, we have P1 = P6 = 10 kPa and P2 = P3 = 8000 kPa. So \u03b7p = ws , wp wp = ws , \u03b7p = v1(P2 \u2212 P1) , \u03b7p 0.001010 m3 ((8000 kPa) \u2212 (10 kPa)) kg 0.8 , = = 10.0874 kJ . kg So h2 = h1 + wp, (8.28) (8.29) = 191.81 kJ + 10.0874 kJ , kg kg (8.30) = 201.897 kJ . kg CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.1. RANKINE 331 Now, substitute all these values into Eq. (8.8) and get (8.31) (8.32) wnet = 3398.27 kJ \u2212 2820.05 kJ (8.33) kg kg (8.34) turbine 1 (8.35) (8.36) + 3482.75 kJ \u2212 2650.83 kJ (8.37) kg kg (8.38) turbine 2 + 191.81 kJ \u2212 201.897 kJ , kg kg pump = 1400.05 kJ . kg On a mass basis, we have W\u02d9 = m\u02d9 wnet = 2.63 \u00d7 105 kg hr 1400.05 kJ = 1.02282 \u00d7 105 kW. hr 3600 s kg From Eq. (8.9), the heat added is qin = 3398.27 kJ \u2212 201.897 kJ kg kg \ufb01rst boiling + 3482.75 kJ \u2212 2820.05 kJ , kg kg second boiling = 3859.07 kJ . kg So the cycle\u2019s thermal e\ufb03ciency is \u03b7 = wnet = 1400.05 kJ = 0.362795. qin 3859.07 kg kJ kg The heat per unit mass rejected in the condenser is from Eq. (8.10): qout = 2650.83 kJ \u2212 191.81 kJ = 2459.02 kJ . kg kg kg So the power rejected as heat is Q\u02d9 L = m\u02d9 qout = 2.63 \u00d7 105 kg hr 2459.02 kJ = 1.79645 \u00d7 105 kW. hr 3600 s kg Example 8.2 Repeat the previous analysis without reheat. Find the thermal e\ufb03ciency. CC BY-NC-ND. 15 May 2021, J. M. Powers.","332 CHAPTER 8. CYCLES In this case state 4 would be taken down to 10 kPa. We would have s4s \u2212 sf 6.7239 kJ \u2212 0.6492 kJ sg \u2212 sf kg K kg K x4s = = = 0.809863. (8.39) kJ kJ 8.1501 kg K \u2212 0.6492 kg K We can thus get h4s by consulting the tables to \ufb01nd h4s = hf + x4shfg = 191.81 kJ + (0.809863) 2392.82 kJ = 2129.67 kJ . (8.40) kg kg kg Now h4 = h3 \u2212 \u03b7t(h3 \u2212 h4s), (8.41) (8.42) = 3398.27 kJ \u2212 (0.88) 3398.27 kJ \u2212 2129.67 kJ , kg kg kg (8.43) = 2281.90 kJ . kg We then get wnet = h3 \u2212 h4 \u2212 wp = 3398.27 kJ \u2212 2281.90 kJ \u2212 10.0874 kJ = 1106.28 kJ . (8.44) kg kg kg kg We also get qin = h3 \u2212 h2 = 3398.27 kJ \u2212 201.897 kJ = 3196.37 kJ . (8.45) kg kg kg So kJ kg \u03b7 = wnet 1106.28 = 0.346106. (8.46) qin = kJ kg 3196.37 The thermal e\ufb03ciency without reheat (0.346106) is less than that with reheat (0.362795). The reheat altered the topology of the T \u2212 s diagram to make it more Carnot-like, and thus generated a more e\ufb03cient use of resources. Example 8.3 A Rankine power cycle with water as the working \ufb02uid has \u03b7t = \u03b7p = 0.88. The turbine inlet pressure and temperature are at P3 = 1200 psia, T3 = 1000 \u25e6F. The condenser pressure is at P4 = 1 psia. The steam generator provides Q\u02d9 H = 2\u00d7109 Btu\/hr. In the condenser the cooling water enters at 60 \u25e6F, and we wish to keep the exit cooling water temperature at 80 \u25e6F. Find the net power, the thermal e\ufb03ciency, and the mass \ufb02ow rate of cooling water, m\u02d9 cw. We interpolate the steam tables to \ufb01nd h3 = 1499.6 Btu , s3 = 1.6297 Btu . (8.47) lbm lbm \u25e6R CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.1. RANKINE 333 For an isentropic turbine, we have s4s = s3 = 1.6297 Btu\/lbm\/ \u25e6R. At P4 = 1 psia, we \ufb01nd state 4s is a two-phase mixture: x4s = s4 \u2212 sf = 1.6297 Btu \u2212 0.1327 Btu = 0.811. (8.48) sg \u2212 sf lbm \u25e6R lbm \u25e6R 1.9779 Btu \u2212 0.1327 Btu lbm \u25e6R lbm \u25e6R Thus, h4s = hf + x4shfg = 69.74 Btu + (0.811) 1036 Btu = 909.9 Btu . (8.49) lbm lbm lbm Now, for the actual turbine, we get h4 = h3 \u2212 \u03b7t(h3 \u2212 h4s), (8.50) (8.51) = 1499.7 Btu \u2212 (0.88) 1499.7 Btu \u2212 909.9 Btu , lbm lbm lbm (8.52) = 979.9 Btu . lbm Now, after the condenser, we take x1 = 0, so h1 = hf , s1 = sf , and v1 = vf , all at P1 = 1 psia. These yield h1 = 69.74 Btu , s1 = 0.1327 Btu , v1 = 0.01614 ft3 . (8.53) lbm lbm \u25e6R lbm Now, wp = ws , (8.54) \u03b7p (8.55) = v(P4 \u2212 P3) , (8.56) \u03b7p (8.57) 0.01614 ft3 1200 lbf \u2212 1 lbf 144 in2 Btu lbm in2 in2 ft2 778 ft lbf = , 0.88 = 4.07 Btu . lbm Now, h2 = h1 + wp = 69.74 Btu + 4.07 Btu = 73.81 Btu . (8.58) lbm lbm lbm In the boiler, we have Q\u02d9 H = m\u02d9 (h3 \u2212 h2), (8.59) (8.60) m\u02d9 = Q\u02d9 H , (8.61) h3 \u2212 h2 (8.62) = 2 \u00d7 109 Btu hr s, hr 3600 1499.7 Btu \u2212 73.81 Btu lbm lbm = 390 lbm . s We also note Q\u02d9 H = 2 \u00d7 109 Btu hr s = 5.556 \u00d7 105 Btu . (8.63) hr 3600 s CC BY-NC-ND. 15 May 2021, J. M. Powers.","334 CHAPTER 8. CYCLES Now, the net power is the sum of the turbine and pump work: W\u02d9 net = m\u02d9 ((h3 \u2212 h4) + (h1 \u2212 h2)) , (8.64) = 390 lbm 1499.6 Btu \u2212 979.9 Btu + 69.74 Btu \u2212 73.81 Btu , (8.65) s lbm lbm lbm lbm = 2.01 \u00d7 105 Btu . (8.66) s The thermal e\ufb03ciency is thus \u03b7 = W\u02d9 net = 2.01 \u00d7 105 Btu = 0.3618. (8.67) Q\u02d9 H 5.556 \u00d7 105 s Btu s The cooling water and the water in the Rankine cycle exchange heat in the condenser. This is sketched in Fig. 8.5. The \ufb01rst law for the heat exchanger is .. mh4 Rankine cycle water mh1 m. cw m. cw 60 oF Thot = 80 oF Tcold = cooling water Figure 8.5: Rankine cycle condenser\/heat exchanger. dEcv = Q\u02d9 cv \u2212 W\u02d9 cv +m\u02d9 (h4 \u2212 h1) + m\u02d9 cwcP (Tcold \u2212 Thot), (8.68) dt (8.69) =0 =0 =0 (8.70) (8.71) 0 = m\u02d9 (h4 \u2212 h1) + m\u02d9 cwcP (Tcold \u2212 Thot), (8.72) m\u02d9 cw = m\u02d9 (h4 \u2212 h1) , cP (Thot \u2212 Tcold) = 390 lbm 979 Btu \u2212 69.74 Btu , s lbm lbm 1.00 Btu ((80 \u25e6F) \u2212 (60 \u25e6F)) lbm \u25e6R = 17730 lbm . s 8.2 Brayton Gas turbine power plants, both stationary and those for jet engines operate on the Brayton cycle. The cycle is named after George Brayton, an American mechanical engineer. Brayton CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON 335 Figure 8.6: George Brayton (1830-1892), American mechanical engineer from Exeter, New Hampshire; image from https:\/\/en.wikipedia.org\/wiki\/George Brayton. fuel P T 2 combustion 2 isobar 3 3 chamber 3 w isentrope isobar isentrope isentrope compressor turbine isentrope 2 isobar isobar 4 1 air products 4 1 1 environmental exhaust return 4 v s Figure 8.7: Schematic of Brayton cycle along with P \u2212 v and T \u2212 s diagrams. is depicted in Fig. 8.6. It has many similarities to the Rankine cycle. A schematic and T \u2212 s and P \u2212 v diagrams for the Brayton cycle for a power plant are illustrated in Fig. 8.7. The Brayton cycle is outlined as follows: \u2022 1 \u2192 2: isentropic compression (W added), \u2022 2 \u2192 3: isobaric heat addition (Q added), \u2022 3 \u2192 4: isentropic expansion (W extracted), and \u2022 4 \u2192 1: isobaric heat rejection (passive exhaust). Note, the work extracted is greater than the work added, i.e. |h3 \u2212 h4| > |h2 \u2212 h1|. (8.73) CC BY-NC-ND. 15 May 2021, J. M. Powers.","336 CHAPTER 8. CYCLES Often we will be dealing with a CPIG, in which case \u2206h = cP \u2206T . If so, then we can say |T3 \u2212 T4| > |T2 \u2212 T1|. (8.74) The reason for this is that \u2022 isobars diverge in T \u2212 s space as s increases. This is easy to understand when we recall the Gibbs equation, Eq. (6.65): T ds = dh \u2212 v dP. On an isobar, we have dP = 0, so T ds = dh, on isobar, (8.75) (8.76) = cP dT, if IG, (8.77) \u2202T = T . \u2202s P cP Because at a given s, a high T isobar sits above a low T isobar, and the slope of the isobar is proportional to T , it is easily seen how they must diverge. This is illustrated in Fig. 8.8. T highhigphreerssTu,rehiigsohbearrslope lolowwperreTss,ulroewiesrobslaorpe s Figure 8.8: Sketch of diverging isobars for CPIG in T \u2212 s plane. There are other classes of Brayton cycle plants. Schematics are shown next. \u2022 Turbojet. In the turbojet, the kinetic energy of the \ufb02uid becomes important at two points in the cycle. In the compression, the freestream \ufb02uid, entering the compressor at the \ufb02ight speed, has its pressure increased by the so-called \u201cram e\ufb00ect\u201d where the \ufb02uid decelerates. Second, the point of the turbojet is to produce thrust, that requires a signi\ufb01cant exit velocity. The turbine work is used solely to power the compressor. See Fig. 8.9. \u2022 Turbojet with afterburners. We are limited in an ordinary turbojet by how much heat can be added to the \ufb02ow in combustion because such \ufb02ow typically must pass through the turbine blades, that su\ufb00er material degradation if the \ufb02uid is too hot. However, CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON T 337 qin qin wt CT v2\/2 wc v2\/2 s Figure 8.9: Sketch of turbojet schematic and associated T \u2212 s plane. we can add heat after the turbine in so-called afterburners. This releases chemical energy, turns it into \ufb02uid potential energy in the form of high P\/\u03c1, and then converts to kinetic energy in the nozzle. This can enhance the thrust, though it can be shown it is not particularly e\ufb03cient. A sketch is given in Fig. 8.10. qin T qin v2\/2 qin wt qin CT wc v2\/2 s Figure 8.10: Sketch of turbojet with afterburners schematic and associated T \u2212 s plane. \u2022 Ramjet. A ramjet is much simpler. The compressor and turbine are removed. We rely on the ram compression e\ufb00ect alone for compression and convert as much of the thermal energy as possible into mechanical energy used to generate thrust force. A sketch is given in Fig. 8.11. Let us consider an \u2022 Air standard analysis: a common set of assumptions used for idealized cyclic devices. CC BY-NC-ND. 15 May 2021, J. M. Powers.","338 CHAPTER 8. CYCLES qin T qin v2\/2 v2\/2 s Figure 8.11: Sketch of ramjet schematic and associated T \u2212 s plane. The air standard make many compromises in order to admit some simple analysis tools to be used to make simple estimates for the performance of a variety of devices. Actual design calculations would have to remedy the many shortcomings. But it is useful for a framework of understanding. We take the air standard to entail \u2022 The working \ufb02uid is air. This ignores any e\ufb00ect of the properties of the fuel or any other \ufb02uid that is mixed with the air. \u2022 The working \ufb02uid is an ideal gas. We will often assume it is a CPIG, but sometimes not. \u2022 We will ignore all details of the combustion process and treat it as a simple heat addition. Often in cycle analysis, the formal sign convention is ignored. We take the following \u2022 Turbine work: wt = h3 \u2212 h4. Here, the sign convention is maintained. \u2022 Compressor work: wc = h2 \u2212 h1. Here, the sign convention is ignored. \u2022 Heat addition: qin = h3 \u2212 h2. Here, the sign convention is maintained. \u2022 Heat rejection: qout = h4 \u2212 h1. Here, the sign convention is ignored. The cycle e\ufb03ciency is (h3 \u2212 h4) \u2212 (h2 \u2212 h1) \u03b7 = wnet = turbine \u2212 compressor . (8.78) qin h3 h2 combustor CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON 339 Rearranging Eq. (8.78), we can also say h4 \u2212 h1 \u03b7 = 1 \u2212 heat rejected . (8.79) h3 \u2212 h2 heat added The back work ratio, bwr, is the ratio of compressor to turbine work: bwr = wc = h2 \u2212 h1 . (8.80) wt h3 \u2212 h4 Note the back work ratio will be seen to be much larger for gas phase power cycles than it was for vapor cycles. For Brayton cycles, we may see bwr \u223c 0.4. For Rankine cycles, we usually see bwr \u223c 0.01. Now, if we have a CPIG, we get \u2206h = cP dT to reduce to \u2206h = cP \u2206T . So Eq. (8.78) reduces to \u03b7 = cP (T3 \u2212 T4) \u2212 cP (T2 \u2212 T1) , (8.81) cP (T3 \u2212 T2) (8.82) (8.83) = T3 \u2212 T4 \u2212 T2 + T1 , (8.84) T3 \u2212 T2 = 1 \u2212 T4 \u2212 T1 , T3 \u2212 T2 = 1 \u2212 T1 T4 \u22121 . T2 T1 T3 \u22121 T2 Now, 1 \u2192 2 is isentropic. Recall from Eq. (6.169) for a CPIG that is isentropic that T2\/T1 = (P2\/P1)(k\u22121)\/k. We also have 3 \u2192 4 to be isentropic, so T3\/T4 = (P3\/P4)(k\u22121)\/k. But P2 = P3 and P1 = P4. So T2 = T3 , (8.85) T1 T4 (8.86) T4 = T3 . T1 T2 So \u03b7 = 1 \u2212 T1 . (8.87) T2 With the temperature ratio \u03b8 across the compressor de\ufb01ned as \u03b8 = T2 , (8.88) T1 CC BY-NC-ND. 15 May 2021, J. M. Powers.","340 CHAPTER 8. CYCLES we have \u03b7 = 1 \u2212 1 . (8.89) \u03b8 In terms of the pressure ratio, the e\ufb03ciency is \u03b7 = 1\u2212 1 k\u22121 . (8.90) P2 k P1 A plot of \u03b7 versus the temperature ratio \u03b8 is plotted in Fig. 8.12. As the temperature 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 =T2\/T1 Figure 8.12: Thermal e\ufb03ciency versus temperature ratio across the compressor for an air standard Brayton cycle. ratio rises, the thermal e\ufb03ciency increases for the Brayton cycle. For practical application, materials have temperature limits. Note the e\ufb03ciency looks like that for a Carnot cycle, but it is not. The highest temperature in the Brayton cycle is T3, so the equivalent Carnot e\ufb03ciency would be 1 \u2212 T1\/T3. Example 8.4 Consider a CPIG air standard Brayton cycle with \ufb01xed inlet conditions P1 and T1. We also \ufb01x the maximum temperature as the metallurgical limit of the turbine blades, Tmax. Find the temperature ratio \u03b8 that maximizes the net work. Then \ufb01nd the temperature ratio that maximizes the thermal e\ufb03ciency. We have P2 k\u22121 P4 k\u22121 T2 = T1\u03b8, P1 P3 T2 = T1 k T3 = Tmax, T4 = T3 k (8.91) , . CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON 341 We also have P4 = P1 and P2 = P3. So P1 k\u22121 P2 1\u2212k Tmax P2 P1 \u03b8 k k T4 = Tmax = Tmax = . (8.92) Now, the net work is (8.93) (8.94) wnet = (h3 \u2212 h4) \u2212 (h2 \u2212 h1), (8.95) (8.96) = cP (T3 \u2212 T4 \u2212 T2 + T1), (8.97) = cP (Tmax \u2212 Tmax\u03b8\u22121 \u2212 T1\u03b8 + T1), (8.98) (8.99) = cP T1 Tmax \u2212 Tmax \u03b8\u22121 \u2212\u03b8 +1 . T1 T1 Note wnet = 0 when \u03b8 = 1 and when \u03b8 = Tmax\/T1. To \ufb01nd the maximum wnet we take dwnet\/d\u03b8 and set to zero: dwnet = cP T1 Tmax \u03b8\u22122 \u2212 1 , d\u03b8 T1 0 = cP T1 Tmax \u03b8\u22122 \u2212 1 , T1 \u03b8 = \u00b1 Tmax . T1 We take the positive root, because a negative temperature ratio does not make sense: \u03b8= Tmax . (8.100) T1 The second derivative tells us whether our critical point is a maximum or a minimum. d2wnet = \u22122cP Tmax\u03b8\u22123. (8.101) d\u03b82 (8.102) When \u03b8 > 0, d2wnet\/d\u03b82 < 0, so we have found a maximum of wnet. The maximum value is (8.103) (8.104) wnet |max = cP Tmax \u2212 Tmax Tmax \u22121\/2 Tmax 1\/2 , (8.105) T1 T1 \u2212 T1 + T1 = cP T1 Tmax \u2212 Tmax Tmax \u22121\/2 Tmax 1\/2 , T1 T1 T1 T1 \u2212 +1 = cP T1 Tmax \u22122 Tmax 1\/2 , T1 T1 +1 = cP T1 Tmax 1\/2 2 T1 \u22121 . Now, at what value of \u03b8 is the thermal e\ufb03ciency maximum? Consider \u03b7 = 1 \u2212 \u03b8\u22121, (8.106) (8.107) d\u03b7 = 1 . d\u03b8 \u03b82 CC BY-NC-ND. 15 May 2021, J. M. Powers.","342 CHAPTER 8. CYCLES At a maximum, we must have d\u03b7\/d\u03b8 = 0. So we must have \u03b8 \u2192 \u221e in order to have \u03b7 reach a maximum. But we are limited to \u03b8 \u2264 Tmax\/T1. So the achievable \u03b8 that maximizes \u03b7 is \u03b8 = Tmax . (8.108) T1 So the e\ufb03ciency at our highest allowable \u03b8 is \u03b7 = 1\u2212 1 = 1 \u2212 T1 . (8.109) Tmax Tmax T1 But at the value of peak e\ufb03ciency, the net work is approaching zero! So while this is highly e\ufb03cient, it is not highly useful! Lastly, what is the e\ufb03ciency at the point where we maximize work? \u03b7 =1\u2212 T1 . (8.110) Tmax A plot of scaled net work, wnet\/cP \/T1 ver\u221asus temperature ratio is given for Tmax\/T1 = 10 in Fig. 8.13. Here, the \u03b8 that maximizes wnet is \u03b8 = 10 = 3.162. At that value of \u03b8, we \ufb01nd \u03b7 = 1 \u2212 1\/10 = wnet\/(cPT1) 4.675 5 4 3 2 1 2 4 6 8 10 3.162 = T2\/T1 Figure 8.13: Scaled net work versus temperature ratio for Brayton cycle with Tmax\/T1 = 10. 0.683772 and wnet\/cP \/T1 = \u221a \u2212 1)2 = 4.675. ( 10 Example 8.5 Consider the Brayton power cycle for a spacecraft sketched in Fig. 8.14. The working \ufb02uid is argon, that is well modeled as a CPIG over a wide range of T and P . We take the pressure in the heating process to be isobaric, P2 = P3 = 140 kPa, and the pressure in the cooling process to be isobaric, P4 = P1 = 35 kPa. We are given that T1 = 280 K, T3 = 1100 K. The compressor and turbine both have component e\ufb03ciencies of \u03b7t = \u03b7c = 0.8. We are to \ufb01nd the net work, the thermal e\ufb03ciency, and a plot of the process on a T \u2212 s diagram. CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON 343 P2 = 140 kPa 2 heat source T3 = 1100 K qH P3 = 140 kPa heater 3 compressor wcompressor turbine c = 0.8 t = 0.8 wnet 1 cooler 4 P1 = 35 kPa qL P4 = 35 kPa T1 = 280 K low temperature reservoir Figure 8.14: Schematic of Brayton power cycle for spacecraft. For argon, we have R = 0.20813 kJ , cP = 0.5203 kJ , k = 5 \u223c 1.667. (8.111) kg K kg K 3 Note that cP = kR\/(k \u2212 1). Let us start at state 1. We \ufb01rst assume an isentropic compressor. We will quickly relax this to account for the compressor e\ufb03ciency. But for an isentropic compressor, we have for the CPIG P2 k\u22121 T2s P1 T1 k = . (8.112) (8.113) Here, T2s is the temperature that would be realized if the process were isentropic. We \ufb01nd k\u22121 5\/3\u22121 T2s = T1 P2 k = (280 K) 140 kPa 5\/3 P1 35 kPa = 487.5 K. Now, \u03b7c = ws\/wcompressor, so ws h2s \u2212 h1 cP (T2s \u2212 T1) 0.5203 kJ (487.5 K \u2212 280 K) kJ \u03b7c \u03b7c \u03b7c kg K 0.8 kg wcompressor = = = = = 135.0 . (8.114) (8.115) Now wcompressor = h2 \u2212 h1 = cP (T2 \u2212 T1), so T2 = T1 + wcompressor = (280 K) + 135.0 kJ = 539.5 K. cP kg 0.5203 kJ kg K CC BY-NC-ND. 15 May 2021, J. M. Powers.","344 CHAPTER 8. CYCLES Notice that T2 > T2s. The ine\ufb03ciency (like friction) is manifested in more work being required to achieve the \ufb01nal pressure than that which would have been required had the process been ideal. In the heater, we have qH = h3 \u2212 h2 = cP (T3 \u2212 T2) = 0.5203 kJ ((1100 K) \u2212 (539.5 K)) = 291.6 kJ . (8.116) kg K kg Now, consider an ideal turbine: T4s P4 k\u22121 T3 = P3 k (8.117) (8.118) , (8.119) P4 k\u22121 (8.120) P3 T4s = T3 k , 5\/3\u22121 = (1100 K) 35 kPa 5\/3 140 kPa , = 631.7 K. But for the real turbine, \u03b7t = wturbine , (8.121) ws (8.122) wturbine = \u03b7tws, (8.123) (8.124) = \u03b7t(h3 \u2212 h4s), (8.125) = \u03b7tcP (T3 \u2212 T4s), = (0.8) 0.5203 kJ ((1100 K) \u2212 (631.7 K)) = 194.9 kJ . kg K kg Thus, because wturbine = h3 \u2212 h4 = cP (T3 \u2212 T4), we get T4 = T3 \u2212 wturbine = (1100 K) \u2212 194.9 kJ = 725.4 K. (8.126) cP kg 0.5203 kJ kg K Note that T4 is higher than would be for an isentropic process. This indicates that we did not get all the possible work out of the turbine. Note also that some of the turbine work was used to drive the compressor, and the rest wnet is available for other uses. We \ufb01nd wnet = wturbine \u2212 wcompressor = 194.9 kJ \u2212 135.0 kJ = 59.9 kJ . (8.127) kg kg kg Now, for the cooler, qL = h4 \u2212 h1 = cP (T4 \u2212 T1) = 0.5203 kJ ((725.4 K) \u2212 (280 K)) = 231.7 kJ . (8.128) kg K kg CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON 345 We are now in a position to calculate the thermal e\ufb03ciency for the cycle. \u03b7 = wnet , (8.129) qH (8.130) (8.131) = wturbine \u2212 wcompressor , (8.132) qH (8.133) (8.134) = cP ((T3 \u2212 T4) \u2212 (T2 \u2212 T1)) , cP (T3 \u2212 T2) = (T3 \u2212 T4) \u2212 (T2 \u2212 T1) , T3 \u2212 T2 = ((1100 K) \u2212 (725.4 K)) \u2212 ((539.5 K) \u2212 (280 K)) , (1100 K) \u2212 (539.6 K) = 0.205. If we had been able to employ a Carnot cycle operating between the same temperature bounds, we would have found the Carnot e\ufb03ciency to be \u03b7C arnot = 1\u2212 T1 = 1\u2212 280 K = 0.745 > 0.205. (8.135) T3 1100 K A plot of the T \u2212 s diagram for this Brayton cycle is shown in Fig. 8.15. Note that from 1 to 2 (as T 3 2 P = 140 kPa 2s P = 35 kPa 4s 4 1 s Figure 8.15: T \u2212s diagram of Brayton power cycle for spacecraft with turbine and compressor ine\ufb03ciencies. well as 3 to 4) there is area under the curve in the T \u2212 s diagram. But the process is adiabatic! Recall that isentropic processes are both adiabatic and reversible. The 1-2 process is an example of a process that is adiabatic but irreversible. So the entropy change is not due to heat addition e\ufb00ects but instead is due to other e\ufb00ects. CC BY-NC-ND. 15 May 2021, J. M. Powers.","346 CHAPTER 8. CYCLES Example 8.6 We are given a turbojet \ufb02ying with a \ufb02ight speed of 300 m\/s. The compression ratio of the compressor is 7. The ambient air is at Ta = 300 K, Pa = 100 kPa. The turbine inlet temperature is 1500 K. The mass \ufb02ow rate is m\u02d9 = 10 kg\/s. All of the turbine work is used to drive the compressor. Find the exit velocity and the thrust force generated. Assume an air standard with a CPIG; k = 1.4, cP = 1.0045 kJ\/kg\/K. A plot of the T \u2212 s diagram for this Brayton cycle is shown in Fig. 8.16. We \ufb01rst calculate the ram T3 qin wt 4 2 v2\/2 wc 5 1 v2\/2 a s Figure 8.16: T \u2212 s diagram of Brayton cycle in a turbojet engine. compression e\ufb00ect: h1 + 1 v12 = ha + 1 va2 . (8.136) 2 2 \u223c0 We typically neglect the kinetic energy of the \ufb02ow once it has been brought to near rest within the engine. So we get h1 \u2212 ha = 1 va2 , (8.137) 2 (8.138) (8.139) cP (T1 \u2212 Ta) = 1 va2 , (8.140) 2 (8.141) T1 = Ta + va2 , 2cP 300 m 2 kJ s = (300 K) + , kJ 1000 m2 2 1.0045 kg K s2 = 344.8 K. CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.2. BRAYTON 347 Now, consider the isentropic compression in the compressor. For this, we have (8.142) (8.143) \uf8eb \uf8f6 k\u22121 (8.144) k (8.145) T2 = \uf8ec P2 \uf8f7 , (8.146) T1 \uf8ec \uf8f7 (8.147) \uf8ed P1 \uf8f8 =7 T2 = (344.8 K)(7) 1.4\u22121 , 1.4 = 601.27 K. Let us calculate P2\/Pa, that we will need later. From the isentropic relations, P2 T2 k Pa Ta = k\u22121 , = 601.27 K 1.4 300 K 1.4\u22121 , = 11.3977. We were given the turbine inlet temperature, T3 = 1500 K. Now, the compressor work must equal the turbine work. This amounts to, ignoring the sign convention, wc = wt, (8.148) h2 \u2212 h1 = h3 \u2212 h4, (8.149) cP (T2 \u2212 T1) = cP (T3 \u2212 T4), (8.150) T2 \u2212 T1 = T3 \u2212 T4, (8.151) (8.152) T4 = T3 \u2212 T2 + T1, (8.153) = (1500 K) \u2212 (601.27 K) + (344.8 K), (8.154) = 1243.5 K. Now, we use the isentropic relations to get T5. Process 3 to 5 is isentropic with P5\/P3 = Pa\/P2 = 1\/11.3977 , so we have P5 T5 k P3 T3 = k\u22121 (8.155) , (8.156) (8.157) P5 k\u22121 (8.158) P3 T5 = T3 k , 1 1.4\u22121 = (1500 K) 11.3977 1.4 , = 748.4 K. CC BY-NC-ND. 15 May 2021, J. M. Powers.","348 CHAPTER 8. CYCLES Now, we need to calculate the exhaust velocity. Take an energy balance through the nozzle to get h4 + 1 v42 = h5 + 1 v52, (8.159) 2 2 (8.160) \u223c0 (8.161) (8.162) h4 = h5 + 1 v52, 2 v5 = 2(h4 \u2212 h5), = 2cP (T4 \u2212 T5), = 2 1.0045 kJ ((1243.5 K) \u2212 (748.4 1000 m2 , (8.163) kg K K)) kJ s2 (8.164) kg = 997.3 m . s Now, Newton\u2019s second law for a control volume can be derived by applying the methods of Sec. 4.1. Leaving out details, it can be shown for one dimensional \ufb02ow with one inlet and exit to be d (\u03c1v) = Fcv + m\u02d9 vi \u2212 m\u02d9 ve. (8.165) dt It says the time rate of change of momentum in the control volume is the net force acting on the control volume plus the momentum brought in minus the momentum that leaves. Often terms of the form m\u02d9 v are known as momentum \ufb02ux.. We will take the problem to be steady and take the force to be the thrust force. So Fcv = m\u02d9 (ve \u2212 vi), (8.166) (8.167) = m\u02d9 (v5 \u2212 va), (8.168) = 10 kg 997.3 m \u2212 300 m , (8.169) s s s = 6973 N. 8.3 Refrigeration A simple way to think of a refrigerator is a cyclic heat engine operating in reverse. Rather than extracting work from heat transfer from a high temperature reservoir and rejecting heat to a low temperature reservoir, the refrigerator takes a work input to move heat from a low temperature reservoir to a high temperature reservoir. A common refrigerator is based on a vapor-compression cycle. This is a Rankine cycle in reverse. While one could employ a turbine to extract some work, it is often impractical. Instead the high pressure gas is simply irreversibly throttled down to low pressure. One can outline the vapor-compression refrigeration cycle as follows: \u2022 1 \u2192 2: isentropic compression CC BY-NC-ND. 15 May 2021, J. M. Powers.","8.3. REFRIGERATION 349 \u2022 2 \u2192 3: isobaric heat transfer to high temperature reservoir in condenser, \u2022 3 \u2192 4: adiabatic expansion in throttling valve, and \u2022 4 \u2192 1: isobaric (and often isothermal) heat transfer from a low temperature reservoir to an evaporator. A schematic and associated T \u2212 s diagram for the vapor-compression refrigeration cycle is shown in Fig. 8.17. One goal in design of refrigerators is low work input. There are two . TH QH 3 condenser 2 expansion Compressor . valve compressor W 4 evaporator 1 T . valve3 condenser 2compressor QL TL evaporator temperature of surroundings refrigerator 4 temperature inside . refrigerator Qin 1 s Figure 8.17: Schematic and T \u2212 s diagrams for the vapor-compression refrigeration cycle. main strategies in this: \u2022 Design the best refrigerator to minimize Q\u02d9 in. This really means reducing the conductive heat \ufb02ux through the refrigerator walls. One can use a highly insulating material. One can also use thick walls. Thick walls will reduce available space for storage however. This is an example of a design trade-o\ufb00. \u2022 For a given Q\u02d9 in, design the optimal thermodynamic cycle to minimize the work nec- essary to achieve the goal. In practice, this means making the topology of the cycle as much as possible resemble that of a Carnot refrigerator. Our vapor compression refrigeration cycle is actually close to a Carnot cycle. CC BY-NC-ND. 15 May 2021, J. M. Powers.","350 CHAPTER 8. CYCLES The e\ufb03ciency does not make sense for a refrigerator as 0 \u2264 \u03b7 \u2264 1. Instead, much as our earlier analysis for Carnot refrigerators, a coe\ufb03cient of performance, \u03b2, is de\ufb01ned as \u03b2 = what one wants , (8.170) what one pays for (8.171) = qL . wc Note that a heat pump is e\ufb00ectively the same as a refrigerator, except one desires qH rather than qL. So for a heat pump, the coe\ufb03cient of performance, \u03b2\u2032, is de\ufb01ned as \u03b2\u2032 = qH . (8.172) wc Example 8.7 R-134a, a common refrigerant, enters a compressor at x1 = 1, T1 = \u221215 \u25e6C. At the compressor inlet, the volume \ufb02ow rate is 1 m3\/min. The R-134a leaves the condenser at T3 = 35 \u25e6C, P3 = 1000 kPa. Analyze the system. We have the state at 1, knowing x1 and T1. The tables then give h1 = 389.20 kJ , s1 = 1.7354 kJ , v1 = 0.12007 m3 . (8.173) kg kg K kg The process 2 to 3 is along an isobar. We know P3 = 1000 kPa, so P2 = 1000 kPa. We assume an isentropic compression to state 2, where P2 = 1000 kPa. We have s2 = s1 = 1.7354 kJ\/kg\/K. We interpolate the superheat tables to get h2 = 426.771 kJ . (8.174) kg State 3 is a subcooled liquid, and we have no tables for it. Let us approximate h3 as hf at T3 = 35 \u25e6C, that is kJ kg h3 \u223c 249.10 . (8.175) In the expansion valve, we have kJ kg h4 = h3 = 249.10 . (8.176) Now, m\u02d9 = \u03c1Av = Av = 1 m3 min = 0.138808 kg . (8.177) v1 min 60 s s 0.12007 m3 kg Note the term Av has units m3\/s and is a volume \ufb02ow rate. Now, the compressor power is W\u02d9 = m\u02d9 (h2 \u2212 h1) = 0.138808 kg 426.771 kJ \u2212 389.20 kJ = 5.2152 kW. (8.178) s kg kg The refrigerator capacity is Q\u02d9 in = m\u02d9 (h1 \u2212 h4) = 0.138808 kg 389.20 kJ \u2212 249.10 kJ = 19.447 kW. (8.179) s kg kg CC BY-NC-ND. 15 May 2021, J. M. Powers."]