{"id":1835,"date":"2021-07-27T20:21:56","date_gmt":"2021-07-28T00:21:56","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/6-5-entropy-and-entropy-generation\/"},"modified":"2022-08-11T19:31:18","modified_gmt":"2022-08-11T23:31:18","slug":"6-5-entropy-and-entropy-generation","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/6-5-entropy-and-entropy-generation\/","title":{"raw":"6.5 Entropy and entropy generation","rendered":"6.5 Entropy and entropy generation"},"content":{"raw":"

6.5.1 The inequality of Clausius<\/strong><\/h2>\r\nThe inequality of Clausius states that for any cycle, reversible or irreversible, there exists the following relation:\r\n\r\n \r\n

[latex]\\displaystyle\\oint\\dfrac{\\delta Q}{T}\\le0\\ \\ \\ \\ \\ \\ \\ \\ \\mathrm{(``=\"\\ for\\ reversible\\ cycles;\\ ``<\"\\ for\\ irreversible\\ cycles)}[\/latex]<\/p>\r\n \r\n\r\nwhere\u00a0 [latex] \\delta Q [\/latex]\u00a0 represents the differential amount of heat transfer into or out of a system through an infinitesimal part of the system boundary. [latex] \\delta Q [\/latex]\u00a0\u00a0is positive for heat transfer into the system and is negative for heat transfer out of the system. [latex]T[\/latex] is the absolute temperature at the infinitesimal part of the system boundary, where the heat transfer occurs. \u00a0The cyclic integral symbol [latex]\\oint [\/latex] indicates that the integration must be done for the entire cycle. In other words, all heat transfer into and out of the system, as well as their corresponding boundary temperatures, must be considered in the integral.\r\n\r\n \r\n\r\nThe inequality of Clausius applies to all cycles. We will prove it by using the heat engine cycle, Figure 6.1.3,<\/a> as an example. For a reversible heat engine cycle operating between a heat source at a constant temperature of [latex]T_H[\/latex] and a heat sink at a constant temperature of [latex]T_L[\/latex], the cyclic integral can be written as\r\n\r\n \r\n

[latex]\\displaystyle\\oint\\left(\\dfrac{\\delta Q}{T}\\right)_{rev} = \\left(\\dfrac{Q_H}{T_H}\\right)_{rev} + \\left(\\dfrac{-Q_L}{T_L}\\right)_{rev} [\/latex]<\/p>\r\n \r\n\r\nNote that for a reversible cycle,\r\n\r\n \r\n

[latex]\\left(\\displaystyle\\dfrac{T_H}{T_L}\\right)_{rev} = \\left(\\displaystyle\\dfrac{Q_H}{Q_L}\\right)_{rev}[\/latex]<\/p>\r\n \r\n\r\nTherefore, the following equation exists for a reversible cycle.\r\n\r\n \r\n

[latex]\\displaystyle\\oint\\left(\\dfrac{\\delta Q}{T}\\right)_{rev} =0 [\/latex]<\/p>\r\n \r\n\r\nFor an irreversible cycle operating between the same two heat reservoirs at constant temperatures of [latex]T_H[\/latex] and [latex]T_L[\/latex], we assume that the heat absorbed from the heat source, [latex]Q_H[\/latex], remains the same as that in the reversible cycle,\r\n\r\n \r\n

[latex] \\because Q_{H, rev} = Q_{H, irrev} = Q_{H} [\/latex]\u00a0\u00a0\u00a0 and\u00a0 \u00a0\u00a0 [latex]W_{rev} > W_{irrev} [\/latex]<\/p>\r\n \r\n

[latex]\\therefore Q_{L, rev} < Q_{L, irrev} [\/latex]<\/p>\r\n \r\n

[latex] \\therefore \\left(\\dfrac{Q_H}{Q_L}\\right)_{irrev} < \\left(\\dfrac{Q_H}{Q_L}\\right)_{rev} [\/latex]\u00a0\u00a0\u00a0 and\u00a0\u00a0 [latex] \\left(\\dfrac{Q_H}{Q_L}\\right)_{rev}=\\left(\\dfrac{T_H}{T_L}\\right)_{rev} [\/latex]<\/p>\r\n \r\n

[latex] \\therefore \\left(\\dfrac{Q_H}{Q_L}\\right)_{irrev} < \\left(\\dfrac{T_H}{T_L}\\right)_{rev} [\/latex] \u00a0 \u00a0 and\u00a0\u00a0 [latex] \\left(\\dfrac{T_H}{T_L}\\right)_{rev}=\\left(\\dfrac{T_H}{T_L}\\right)_{irrev} [\/latex]<\/p>\r\n \r\n

[latex] \\therefore \\left(\\dfrac{Q_H}{T_H}\\right)_{irrev} < \\left(\\dfrac{Q_L}{T_L}\\right)_{irrev} [\/latex]<\/p>\r\n \r\n\r\nTherefore,\r\n\r\n \r\n

[latex]\\displaystyle\\oint\\left(\\dfrac{\\delta Q}{T}\\right)_{irrev} = \\left(\\dfrac{Q_H}{T_H}+\\dfrac{-Q_L}{T_L}\\right)_{irrev} = \\left(\\dfrac{Q_H}{T_H}\\right)_{irrev} - \\left(\\dfrac{Q_L}{T_L}\\right)_{irrev}\u00a0 <\u00a0 0[\/latex]<\/p>\r\n \r\n\r\nNow, we have proven the inequality of Clausius for heat engine cycles. A similar procedure may be applied to prove the inequality of Clausius for refrigerator and heat pump cycles.\r\n

6.5.2 Definition of entropy<\/strong><\/h2>\r\nWhy is the inequality of Clausius important? The cyclic integral is either equal to or less than zero depending on the nature of the cycle: reversible or irreversible. The inequality of Clausius provides a basis for introducing the concepts of entropy and entropy generation. Both concepts are important in the second law of <\/a>thermodynamics.\r\n\r\n \r\n\r\n[caption id=\"attachment_2261\" align=\"aligncenter\" width=\"300\"]\"A<\/a> Figure 6.5.1 <\/strong>A reversible cycle consisting of path A and path B<\/em>[\/caption]\r\n\r\nLet us apply the inequality of Clausius to a reversible cycle consisting of two reversible processes 1[latex]\\to[\/latex]2 via path A and 2[latex]\\to[\/latex]1 via path B, see Figure 6.5.1<\/a>.\r\n\r\n \r\n

[latex]\\begin{align*} \\because \\displaystyle\\oint \\left(\\dfrac{\\delta Q}{T}\\right)_{rev} &= \\int_{1}^{2}\\left(\\dfrac{\\delta Q}{T}\\right)_{path A} + \\int_{2}^{1}\\left(\\dfrac{\\delta Q}{T}\\right)_{path B} \\\\ &= \\int_{1}^{2}\\left(\\dfrac{\\delta Q}{T}\\right)_{path A} - \\int_{1}^{2}\\left(\\dfrac{\\delta Q}{T}\\right)_{path B}=0 \\end{align*}[\/latex]<\/p>\r\n \r\n

[latex]\\therefore \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right)_{path A} = \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right)_{path B} [\/latex]<\/p>\r\n \r\n\r\nThe above equation indicates that the integral between the two states 1 and 2 of any reversible processes depends only on the two states, not on the paths; therefore, the integral [latex]\\displaystyle\\int_{1}^{2}\\left(\\dfrac{\\delta Q}{T}\\right)_{rev}[\/latex] is a state function and must be related to a thermodynamic property. We define such thermodynamic property as [pb_glossary id=\"814\"]entropy[\/pb_glossary], <\/strong>[latex]S[\/latex], and the change in entropy between two states can be expressed as\r\n\r\n \r\n

[latex]\\displaystyle\\Delta S = S_2-S_1=\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right)_{rev}[\/latex]<\/p>\r\n \r\n\r\nThe infinitesimal change of entropy in a reversible process can thus be written as\r\n\r\n \r\n

[latex]dS=\\left(\\dfrac{\\delta Q}{T}\\right)_{rev}[\/latex]<\/p>\r\n \r\n\r\nwhere [latex]S[\/latex] is the entropy and [latex]T[\/latex] is the absolute temperature. The common SI units for entropy are kJ\/K or J\/K. It is important to note that entropy, [latex]S[\/latex], is a state function; [latex]\\Delta S[\/latex] in a process depends on the initial and final states, not on the path of the process.\r\n\r\n \r\n\r\nEntropy is an extensive property; its corresponding intensive property is called [pb_glossary id=\"815\"]specific entropy[\/pb_glossary], [latex] s = \\dfrac{S}{m}[\/latex], and its common SI units are kJ\/kgK or J\/kgK.\r\n\r\n \r\n

\r\n

Example 1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider a reversible process and an irreversible process from states 1 to 2, as shown in the T-S<\/em> diagram, Figure 6.5.e1<\/a>. Answer the following questions\r\n\r\n(1) Is the change in entropy, \u0394S, <\/em>the same or different in these two processes?\r\n\r\n(2) Is it possible to show the heat transfer of the reversible process in the T-S diagram?\r\n\r\n(3) Is it possible to show the heat transfer of the irreversible process in the T-S diagram<\/a>?\r\n\r\n[caption id=\"attachment_2941\" align=\"aligncenter\" width=\"300\"]\"T-S<\/a> Figure 6.5.e1<\/strong> T-S<\/em> diagram for a reversible process and an irreversible process with the same initial and final states[\/caption]\r\n\r\nSolution<\/span>:<\/em>\r\n\r\n(1) Entropy is a state function. The two processes have the same initial and final states, therefore, the same \u0394S<\/em>.\r\n\r\n(2) From the definition of entropy, the heat transfer in the reversible process can be found from\r\n

[latex]Q_{rev}=\\displaystyle\\int_{1}^{2} {\\left(\\delta Q\\right)}_{rev} = \\int_{1}^{2} \\left(TdS\\right)_{rev}[\/latex]<\/p>\r\nThis integral can be shown graphically as the shaded area under the T-S<\/em> curve of the reversible process, <\/a>see Figure 6.5.e2<\/a>.\r\n\r\n[caption id=\"attachment_2943\" align=\"aligncenter\" width=\"300\"]\"T-S<\/a> Figure 6.5.e2<\/strong> T-S diagram: the shaded area represents the heat transfer of a reversible process.<\/em>[\/caption]\r\n\r\n(3) The heat transfer of the irreversible process cannot be simply calculated without additional information, and it cannot be shown in the T-S<\/em> diagram.\r\n\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n

\r\n

Example 2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nA reversible<\/strong> process from states 1[latex]\\to[\/latex]2 in a piston-cylinder is shown in Figure 6.5.e3<\/a>. Determine whether the change in specific internal energy [latex]\\Delta u=u_2-u_1[\/latex], specific work w<\/em>, and specific heat transfer q<\/em> are positive, zero, or <\/a>negative.\r\n\r\n[caption id=\"attachment_2555\" align=\"aligncenter\" width=\"458\"]\"P-v<\/a> Figure 6.5.e3<\/strong><\/em> P-v and T-s diagrams of a reversible process<\/em>[\/caption]\r\n\r\nSolution:<\/em><\/span>\r\n\r\nThe [latex]P-v[\/latex] diagram shows an expansion process in the piston-cylinder. Its specific boundary work can be shown as the shaded area in the [latex]P-v[\/latex] diagram, see Figure 6.5.e3<\/a>. It can also be expressed as\r\n

[latex]{}_{1}w_{2} = \\displaystyle\\int_{1}^{2} Pdv > 0 [\/latex]<\/p>\r\nFrom the definition of entropy,\r\n

[latex]dS=\\left(\\displaystyle\\frac{\\delta Q}{T}\\right)_{rev}[\/latex]<\/p>\r\nThe process is reversible; therefore,\r\n

[latex](\\delta Q)_{rev} = TdS [\/latex]<\/p>\r\n

[latex] \\therefore \\; {}_{1}Q_{2} = Q_{rev} =\u00a0 \\displaystyle\\int_{1}^{2} (\\delta Q)_{rev} = \\displaystyle\\int_{1}^{2} TdS\u00a0 [\/latex]<\/p>\r\nThe specific heat transfer is\r\n

[latex]{}_{1}q_{2} = \\dfrac{{}_{1}Q_{2}}{m} = \\displaystyle\\int_{1}^{2} Tds [\/latex]<\/p>\r\nFrom the [latex]T-s[\/latex] diagram in Figure 6.5.e3<\/a>, [latex] s_1 = s_2 [\/latex]; therefore,\r\n

[latex]{}_{1}q_{2} = 0 [\/latex]<\/p>\r\nThe specific heat transfer in a reversible process can be shown graphically as the area under the process line in the [latex]T-s[\/latex] diagram. Note: this statement is only true for reversible processes; it is not valid for irreversible processes<\/em>!\r\n\r\n \r\n\r\nApply the first law of thermodynamics to the piston-cylinder (closed system),\r\n

[latex] \\Delta u = {}_{1}q_{2} - {}_{1}w_{2} = 0 - {}_{1}w_{2} < 0 [\/latex]<\/p>\r\nIn conclusion, the reversible expansion process illustrated in the [latex]P-v[\/latex] and [latex]T-s[\/latex] diagrams in Figure 6.5.e3<\/a> has a positive boundary work and zero heat transfer (adiabatic). The specific internal energy decreases in the process.\r\n\r\n<\/div>\r\n<\/div>\r\n

6.5.3 Entropy generation, Sgen<\/sub><\/strong><\/h2>\r\n \r\n\r\nEntropy generation is another important concept in the second law of thermodynamics. Let us consider a cycle consisting of two processes; process 2[latex]\\to[\/latex]1 is a reversible process and process 1[latex]\\to[\/latex]2 can be any process, either reversible or irreversible, see Figure 6.5.2<\/a>. We will apply the definition of entropy and the inequality of Clausius in the following derivations.\r\n\r\n \r\n

[latex]\\because \\Delta S = S_2-S_1=\\displaystyle\\int_{1}^{2}\\left(\\dfrac{\\delta Q}{T}\\right)_{rev}[\/latex]<\/p>\r\n \r\n

[latex]\\begin{align*} \\therefore\\displaystyle\\oint \\dfrac{\\delta Q}{T} &= \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right) + \\displaystyle\\int_{2}^{1}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right)_{rev} \\\\&= \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right) - \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right)_{rev} \\\\ &= \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right) - (S_2 - S_1) \\le 0 \\end{align*}[\/latex]<\/p>\r\n \r\n

[latex] \\therefore (S_2-S_1) \\ge \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right) [\/latex]<\/p>\r\n \r\n\r\nWe may change the above inequality to an equation by introducing entropy generation, [latex] S_{gen} \\ge 0 [\/latex], to the right side; therefore,\r\n\r\n \r\n

[latex] \\Delta S = (S_2-S_1) = \\displaystyle\\int_{1}^{2}\\left(\\displaystyle\\frac{\\delta Q}{T}\\right) + S_{gen} \\hspace{3em}\u00a0 \\left( S_{gen} \\ge 0\\right) [\/latex]<\/p>\r\n \r\n\r\nThis relation is valid for all processes with the \"[latex]=[\/latex]\" sign for reversible processes, and the \"[latex]>[\/latex]\" sign for irreversible processes. The differential form of the relation can be expressed <\/a>as\r\n\r\n \r\n

[latex]dS=\\displaystyle\\frac{\\delta Q}{T}+\\delta S_{gen} \\hspace{3em}\u00a0 \\left( \\delta S_{gen} \\ge 0\\right)[\/latex]<\/p>\r\n \r\n\r\n[caption id=\"attachment_2722\" align=\"aligncenter\" width=\"400\"]\"A<\/a> Figure 6.5.2<\/strong><\/em>\u00a0A cycle consisting of reversible and\/or irreversible processes for introducing entropy generation <\/em>[\/caption]\r\n\r\nIt is important to note that [latex]S_{gen}[\/latex] and [latex]\\Delta S[\/latex] are different concepts. \r\n<\/strong>\r\n