{"id":1876,"date":"2021-07-30T19:15:26","date_gmt":"2021-07-30T23:15:26","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/thermo1\/?post_type=chapter&#038;p=1876"},"modified":"2022-08-12T02:29:34","modified_gmt":"2022-08-12T06:29:34","slug":"applications-of-the-second-law-of-thermodynamics-in-open-systems","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/applications-of-the-second-law-of-thermodynamics-in-open-systems\/","title":{"raw":"6.10 Applications of the second law of thermodynamics in open systems","rendered":"6.10 Applications of the second law of thermodynamics in open systems"},"content":{"raw":"When solving problems in a control volume, the first and second laws of thermodynamics are often used together with the continuity equation and the thermodynamic tables or the ideal gas equation. The following strategy may be adapted:\r\n<ol>\r\n \t<li>Set up a proper control volume to enclose the device of interest and identify the flow condition, i.e., steady or transient flow, through the control volume.<\/li>\r\n \t<li>Determine the relations among the mass flow rates at the inlet(s) and outlet(s) by using the continuity equation.<\/li>\r\n \t<li>Find the fluid properties at the inlet(s) and outlet(s), such as, [latex]P, T, v, h, s[\/latex], by using the thermodynamic tables or equations, e.g., for an ideal gas, solid or liquid.<\/li>\r\n \t<li>Determine the rate of heat transfer, [latex]\\dot Q[\/latex], or other unknowns by applying the first law of thermodynamics for open systems;<\/li>\r\n \t<li>Determine the rate of entropy generation, [latex]\\dot S_{gen}[\/latex], or other unknowns by applying the second law of thermodynamics for open systems.<\/li>\r\n<\/ol>\r\n&nbsp;\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSteam is used to provide heating to air in a building through a well-insulated heat exchanger, see <a href=\"#6.10.e1\">Figure 6.10.e1<\/a>.\r\n<ul>\r\n \t<li>Saturated steam at 100<sup>o<\/sup>C enters the heat exchanger at a mass flow rate of 0.5 kg\/s and leaves the heat exchanger as a saturated liquid at 100<sup>o<\/sup><span style=\"font-size: 1em\">C<\/span><\/li>\r\n \t<li>Air enters the heat exchanger at 5<sup>o<\/sup>C, 101 kPa, and leaves at 25<sup>o<\/sup>C, 101 kPa.<\/li>\r\n<\/ul>\r\nAssume air is an ideal gas. <a id=\"6.10.e1\"><\/a>Determine the mass flow rate of air and [latex]\\dot{S}_{gen}[\/latex] in this process.\r\n\r\n[caption id=\"attachment_2897\" align=\"aligncenter\" width=\"400\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex.png\" target=\"_blank\" rel=\"noopener\"><img class=\"wp-image-2897\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-1024x358.png\" alt=\"Well-insulated heat exchanger to allow heat transfer between steam and air\" width=\"400\" height=\"140\" \/><\/a> <em><strong>Figure 6.10.e1<\/strong> Well-insulated heat exchanger<\/em>[\/caption]\r\n\r\n<em><span style=\"text-decoration: underline\">Solution<\/span>:<\/em>\r\n\r\nAnalysis\r\n<ul>\r\n \t<li>The mass flow rate of air can be found by using the first law of thermodynamics<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 80px\">[latex] \\because \\dot{m}_{w}(h_1 - h_2) = \\dot{m}_{a}(h_4 - h_3) = \\dot{m}_{a}C_p(T_4 - T_3) [\/latex]<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\therefore \\dot{m}_{a} = \\dot{m}_{w}\u00a0 \\dfrac{h_1 - h_2}{C_p(T_4 - T_3)} [\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>The rate of entropy generation can be found by using the second law of thermodynamics. Note that the heat exchanger is well-insulated.<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 80px\">[latex] \\because \\sum\\dot{m}_{e}s_{e} - \\sum\\dot{m}_{i}s_{i} = \\sum\\dfrac{\\dot{Q}_{k}}{T_k} + \\dot{S}_{gen} [\/latex]\u00a0\u00a0\u00a0\u00a0 and \u00a0 \u00a0 [latex]\\dot{Q}_{k} = 0 [\/latex]<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*}\\therefore \\dot{S}_{gen} &amp;=\u00a0 \\sum\\dot{m}_{e}s_{e} - \\sum\\dot{m}_{i}s_{i} \\\\&amp;= (\\dot{m}_{w}s_{2} + \\dot{m}_{a}s_{4}) -\u00a0 (\\dot{m}_{w}s_{1} + \\dot{m}_{a}s_{3}) \\\\&amp;= \\dot{m}_{w}(s_{2}-s_{1}) + \\dot{m}_{a}(s_{4} - s_{3}) \\end{align*}[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>The following properties are needed in order to complete the calculations:\r\n<ul>\r\n \t<li>For the steam-water stream: [latex]h_1 \\ , \\ h_2 \\ , \\ s_1 \\ , \\ s_2[\/latex]<\/li>\r\n \t<li>For the air stream: [latex]s_3 \\ , \\ s_4[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nNow, let us solve the problem in detail.\r\n\r\nFrom <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/appendix-a-thermodynamic-properties-of-water#TA1\" target=\"_blank\" rel=\"noopener\">Table A1<\/a>, we can find the specific enthalpies and specific entropies of the saturated steam and saturated liquid water at 100<sup>o<\/sup>C.\r\n<p style=\"padding-left: 40px\"><em>h<sub>1<\/sub> = h<sub>g<\/sub><\/em> = 2675.57 kJ\/kg ,\u00a0\u00a0 <em>h<sub>2<\/sub> = h<sub>f<\/sub><\/em> = 419.17 kJ\/kg<\/p>\r\n<p style=\"padding-left: 40px\"><em>s<sub>1<\/sub> = s<sub>g<\/sub><\/em> = 7.3541 kJ\/kgK,\u00a0\u00a0\u00a0\u00a0 <em>s<sub>2<\/sub> = s<sub>f<\/sub><\/em> = 1.3072 kJ\/kgK<\/p>\r\nFrom <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/__unknown__-2#TG1\" target=\"_blank\" rel=\"noopener\">Table G1<\/a> for air:\u00a0 <em>C<sub>p<\/sub><\/em> = 1.005 kJ\/kgK, <em>R<\/em> = 0.287 kJ\/kgK\r\n<p style=\"padding-left: 40px\">[latex] \\because s_4-s_3=C_pln\\displaystyle\\frac{T_4}{T_3}-Rln\\frac{P_4}{P_3}[\/latex]\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0 [latex]P_4=P_3[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">[latex] \\therefore s_4-s_3= 1.005ln\\displaystyle\\frac{273.15 + 25}{273.15 + 5} = 0.06978 \\ \\rm{kJ\/kgK} [\/latex]<\/p>\r\nNow, we can complete the calculations. The mass flow rate of air is\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\dot{m}_{a} &amp;= \\dot{m}_{w} \\dfrac{ h_1 - h_2}{C_p (T_4 -T_3)} \\\\&amp;= 0.5 \\times \\dfrac{ 2675.57 - 419.17}{1.005 (25-5)} = 56.13 \\ \\rm{kg\/s} \\end{align*}[\/latex]<\/p>\r\nThe rate of entropy generation is\r\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{S}_{gen}&amp;= \\dot{m}_{w}(s_{2}-s_{1}) + \\dot{m}_{a}(s_{4} - s_{3}) \\\\&amp;= 0.5 \\times (1.3072 - 7.3541) + 56.13 \\times 0.06978 \\\\&amp;= 0.8934 \\ \\rm{kW\/K} \\ &gt; 0 \\end{align*}[\/latex]<\/p>\r\nComment:\r\n\r\nThe heat transfer process is irreversible; therefore, the rate of entropy generation is greater than zero.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nR134a at 300 kPa, 20\u00b0C is heated to 50\u00b0C isobarically in a heat exchanger. The mass flow rate of R134a is 1.5 kg\/s. The heat at a rate of [latex] \\dot{Q}_{H}[\/latex] is supplied by a heat pump, which absorbs heat at a rate of [latex] \\dot{Q}_{L}[\/latex] from the ambient at 280 K, see <a href=\"#6.10.e2\">Figure 6.10.e2<\/a>. If [latex]COP_{HP}=5[\/latex], and there is no heat loss in the heat exchanger, find the power input, [latex] \\dot{W}_{HP}[\/latex], and the rate of heat transfer, [latex] \\dot{Q}_{L}[\/latex]. Is this setup <a id=\"6.10.e2\"><\/a>consisting of the heat exchanger and heat pump possible?\r\n\r\n[caption id=\"attachment_2893\" align=\"aligncenter\" width=\"500\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex.png\" target=\"_blank\" rel=\"noopener\"><img class=\"wp-image-2893\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-300x190.png\" alt=\"A device consisting of a heat exchanger and a heat pump\" width=\"500\" height=\"317\" \/><\/a> <em><strong>Figure 6.10.e2<\/strong> A device consisting of a heat exchanger and a heat pump<\/em>[\/caption]\r\n\r\n<span style=\"text-decoration: underline\"><em>Solution:<\/em><\/span>\r\n\r\nThe control volume is set up to enclose both the heat pump and heat exchanger, as shown in <a href=\"#6.10.e2\">Figure 6.10.e2<\/a>.\r\n\r\nFrom <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/appendix-c-thermodynamic-properties-of-r134a#TC1\" target=\"_blank\" rel=\"noopener\">Table C1<\/a>, we can tell that, at the given pressures and temperatures, R134a remains a superheated vapour at both the inlet and the outlet of the heat exchanger. The specific enthalpies and specific entropies of R134a can be found in <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/appendix-c-thermodynamic-properties-of-r134a#TC2\" target=\"_blank\" rel=\"noopener\">Table C2<\/a>.\r\n<ul>\r\n \t<li>At the inlet, <em>P<sub>1<\/sub><\/em> = 300 kPa, <em>T<sub>1<\/sub><\/em>= 20<sup>o<\/sup>C; therefore, <em>h<sub>1<\/sub><\/em> = 416.24 kJ\/kg, <em>s<sub>1<\/sub><\/em> = 1.7876 kJ\/kgK<\/li>\r\n \t<li>At the outlet, <em>P<sub>2<\/sub><\/em> = 300 kPa, <em>T<sub>2<\/sub><\/em> = 50<sup>o<\/sup>C; therefore, <em>h<sub>2<\/sub><\/em> = 443.31 kJ\/kg, <em>s<sub>1<\/sub><\/em> = 1.8755 kJ\/kgK<\/li>\r\n<\/ul>\r\nApply the first law of thermodynamics to the control volume.\r\n<p style=\"padding-left: 40px\">[latex] \\dot{m}h_1 + \\dot{Q}_L + \\dot{W}_{HP} = \\dot{m}h_2 [\/latex]<\/p>\r\nFor the heat pump\r\n<p style=\"padding-left: 40px\">[latex] COP_{HP} = \\dfrac{\\dot{Q}_{H}}{\\dot{W}_{HP}} [\/latex]\u00a0\u00a0 and \u00a0 [latex] \\dot{Q}_{L} + \\dot{W}_{HP} = \\dot{Q}_{H}[\/latex]<\/p>\r\nCombine the above three relations, we can derive\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\dot{W}_{HP} &amp;= \\dfrac{\\dot{m}(h_2 - h_1)}{COP_{HP}} \\\\&amp;= \\dfrac{ 1.5 \\times (443.31 - 416.24)}{5} = 8.121 \\ \\rm{kW}\\end{align*} [\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\dot{Q}_{L} &amp;= (COP_{HP} - 1)\\dot{W}_{HP} \\\\&amp;=(5-1) \\times 8.121 = 32.484 \\ \\rm{kW}\\end{align*}[\/latex]<\/p>\r\nApply the second law of thermodynamics to the control volume.\r\n<p style=\"padding-left: 40px\">[latex]\\because \\dot{m}(s_2 - s_1) = \\sum\\dfrac{\\dot{Q}_k}{T_k}+\\dot{S}_{gen}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\therefore \\dot{S}_{gen} &amp;= \\dot{m}(s_2 - s_1) - \\dfrac{\\dot{Q}_L}{T_{surr}} \\\\&amp;=1.5 \\times (1.8755 - 1.7876) - \\dfrac{32.484}{280} = 0.0158 \\ \\rm{kW\/K} \\ &gt; 0 \\end{align*}[\/latex]<\/p>\r\nComment:\r\n\r\nThe rate of entropy generation in this control volume (consisting of both heat exchanger and heat pump) is positive; therefore, it is very likely that this setup can work in reality, and the processes in the heat exchanger and heat pump are irreversible.\r\n\r\nTo further verify this, we can calculate the rate of entropy generation for each individual device, i.e., heat exchanger and heat pump, separately. However, with the given information, there exists a challenge to evaluate the amount of entropy transfer due to the heat transfer provided by the heat pump to the heat exchanger. Strictly speaking, [latex] \\displaystyle\\int \\dfrac{\\dot{Q}}{T_b}[\/latex] should be evaluated because the temperature of the heat exchanger varies, leading to a possible variation of temperature, [latex]T_b[\/latex], at the boundary between the heat pump and the heat exchanger. The relation between the instantaneous rate of heat transfer from the heat pump to the heat exchanger, [latex]\\dot{Q}[\/latex], and the boundary temperature, [latex]T_b[\/latex], must be obtained to enable a detailed analysis.\r\n\r\nHere, let us perform a simplified calculation to demonstrate the concept. Assume that [latex]T_b[\/latex] remains constant as the average temperature between the inlet and the outlet of the heat exchanger, thus [latex]T_b=35^ \\rm{o} \\rm{C}=308.15 \\ \\rm{K}[\/latex].\r\n\r\n&nbsp;\r\n\r\nApply the second law of thermodynamics to the heat pump. Note that [latex] \\dot{Q}_H=\\dot{Q}_L+\\dot{W}_{HP}[\/latex], and [latex]\\Delta s=0[\/latex] for the heat pump cycle.\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\dot{S}_{gen, HP} &amp;= \\dot{m} \\Delta s - \\dfrac{\\dot{Q}_L}{T_{surr}} + \\dfrac{\\dot{Q}_H}{T_b}\\\\&amp;=0 - \\dfrac{32.484}{280} + \\dfrac{32.484+8.121}{308.15}= 0.015755 \\ \\rm{kW\/K} \\ &gt; 0 \\end{align*}[\/latex]<\/p>\r\nFurthermore, the <em>COP<\/em> of the Carnot cycle operating between the heat sink of 280 K and the heat source of 308.15 K is\r\n<p style=\"padding-left: 40px\">[latex]\u00a0 COP_{HP, rev} = \\dfrac{308.15}{308.15-280}=10.9 &gt; 5=COP_{HP}[\/latex]<\/p>\r\nBecause [latex]\\dot{S}_{gen, HP}&gt;0[\/latex] and [latex]COP_{HP} &lt; COP_{HP, rev}[\/latex], this heat pump cycle is an irreversible, realistic cycle.\r\n<!--more-->\r\n\r\nSimilarly, let us apply the second law of thermodynamics to the heat exchanger.\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\dot{S}_{gen, HE} &amp;= \\dot{m}(s_2-s_1) - \\dfrac{\\dot{Q}_H}{T_b} \\\\&amp;=1.5 \\times (1.8755 - 1.7876) - \\dfrac{32.484+8.121}{308.15} \\\\&amp;= 8 \\times 10^{-5} \\ \\rm{kW\/K} \\approx 0 \\end{align*}[\/latex]<\/p>\r\nBecause [latex]\\dot{S}_{gen, HE} \\approx 0[\/latex], the process in this heat exchanger is almost ideal and reversible. It is noted that heat losses are typically unavoidable in real devices, but they are not considered in this example. Due to heat losses from the heat exchanger, the real process will have [latex]\\dot{S}_{gen, HE}&gt;0[\/latex]. Consequently, the outlet temperature may not be able to reach 50<sup>o<\/sup>C.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<p style=\"text-align: justify\">Consider an adiabatic two-stage compressor having an intercooler, as shown in <a href=\"#6.10.e3\">Figure 6.10.e3<\/a>.<\/p>\r\n\r\n<ul style=\"text-align: justify\">\r\n \t<li>Air at 10\u00b0C, 101 kPa enters the first stage of the compressor and is compressed to 300 kPa, 150\u00b0C.<\/li>\r\n \t<li>After being cooled to 30\u00b0C in a well-insulated intercooler, the air then enters the second stage cylinder and is compressed to 900 kPa.<\/li>\r\n<\/ul>\r\n<p style=\"text-align: justify\">Assume that (1) the pressure drop in the intercooler is negligible, (2) air is an ideal gas, and (3) the two compression processes are polytropic with the same polytropic exponent. Determine<\/p>\r\n\r\n<ol>\r\n \t<li style=\"text-align: justify\">The total specific work of the two-stage compressor.<\/li>\r\n \t<li style=\"text-align: justify\">The specific entropy generation in each of the two stages.<\/li>\r\n \t<li style=\"text-align: justify\">The total specific entropy generation in the system consisting of the two-stage compressor and the intercooler, assuming a constant temperature of 30\u00b0C at the system boundary.<\/li>\r\n \t<li style=\"text-align: justify\">Assume there is no intercooler. The air from the first stage enters directly to the second stage and is compressed to 900 kPa, what is the air temperature at the exit of the second stage?\u00a0 What is the total specific work of the compressor? Assume the polytropic exponent remains <a id=\"6.10.e3\"><\/a>the same in both stages.<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1211\" align=\"aligncenter\" width=\"500\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/06\/Fig.-6-22.jpg\" target=\"_blank\" rel=\"noopener\"><img class=\"wp-image-1211\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/06\/Fig.-6-22.jpg\" alt=\"Two stage compressor with an intercooler\" width=\"500\" height=\"333\" \/><\/a> <em><strong>Figure 6.10.e3<\/strong> Two stage compressor with an intercooler<\/em>[\/caption]\r\n\r\n<em><span style=\"text-decoration: underline\">Solution:<\/span><\/em>\r\n\r\n(1) Assume air is an ideal gas. From <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/__unknown__-2#TG1\" target=\"_blank\" rel=\"noopener\">Table G1<\/a> for air: <em>R<\/em> = 0.287 kJ\/kgK, <em>C<sub>p<\/sub><\/em> = 1.005 kJ\/kgK.\r\n\r\nThe compressor is adiabatic with the following inlet and outlet conditions for each stage:\r\n<ul>\r\n \t<li>Stage 1\r\n<ul>\r\n \t<li>Inlet <em>P<sub>1<\/sub><\/em> = 101 kPa, <em>T<sub>1<\/sub><\/em> = 10\u00b0C<\/li>\r\n \t<li>Outlet <em>P<sub>2<\/sub><\/em> = 300 kPa, <em>T<sub>2<\/sub><\/em> =150\u00b0C<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px\">Therefore, the specific work of stage 1 is<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*} w_1 &amp;= h_2 - h_1 = C_p(T_2 - T_1) \\\\&amp;= 1.005 \\times (150 - 10) = 140.7 \\ \\rm{kJ\/kg} \\end{align*}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">The polytropic exponent can be found by combining the polytropic relation and the ideal gas law.<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\because P_{1}v_{1}^{n} = P_{2}v_{2}^{n} [\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0 [latex] Pv = RT [\/latex]<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\therefore \\dfrac{P_2}{P_1} = \\left(\\dfrac{T_2}{T_1} \\right)^{\\dfrac{n}{n-1}} [\/latex]<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*} \\therefore n &amp;= \\dfrac{ln\\dfrac{P_2}{P_1}}{ln\\left(\\dfrac{P_{2}T_{1}}{P_{1}T_{2}}\\right)} \\\\&amp;= \\dfrac{ln\\dfrac{300}{101}}{ln\\left[ \\dfrac{300 \\times (273.15 + 10)}{101 \\times (273.15 + 150)}\\right]}=1.58486 \\end{align*}[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>Stage 2\r\n<ul>\r\n \t<li>Inlet P<sub>3<\/sub> = 300 kPa, T<sub>3<\/sub> = 30\u00b0C<\/li>\r\n \t<li>Outlet P<sub>4<\/sub> = 900 kPa<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px\">The air temperature at the outlet of stage 2 can be found from the polytropic relation.<\/p>\r\n<p style=\"padding-left: 80px\">[latex]\\because \\dfrac{P_4}{P_3} = \\left(\\dfrac{T_4}{T_3}\\right)^{\\dfrac{n}{n-1}} [\/latex]<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*} \\therefore T_4 &amp;= T_3 \\left(\\dfrac{P_4}{P_3}\\right)^{\\dfrac{n-1}{n}} \\\\&amp;= (273.15 + 30) \\times \\dfrac{900}{300}^{\\dfrac{1.58486 -1}{1.58486}} \\\\&amp;= 454.7 \\ \\rm{K} = 181.56^\u00a0 \\rm{o} \\rm{C}\\end{align*}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">Therefore, the specific work of stage 2 is<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*} w_2 &amp;= h_4 - h_3 = C_p(T_4 - T_3) \\\\&amp;= 1.005(181.56 - 30) = 152.31 \\ \\rm{kJ\/kg} \\end{align*}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">The total specific work of the two stage compressor is<\/p>\r\n<p style=\"padding-left: 80px\">[latex] w_{tot} = w_1 + w_2 = 140.7 + 152.31 = 293 \\ \\rm{kJ\/kg}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n(2) Both stages are adiabatic; therefore, for each stage\r\n<p style=\"padding-left: 80px\">[latex]\\because \\dot{m}(s_e - s_i) = \\displaystyle\\sum\\dfrac{\\dot{Q}_k}{T_k}+\\dot{S}_{gen}[\/latex]\u00a0\u00a0\u00a0\u00a0 and \u00a0\u00a0 [latex] \\dot{Q}_{k} = 0 [\/latex]<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\therefore s_{gen} = \\dfrac{\\dot{S}_{gen}}{\\dot{m}} = s_e - s_i = C_{p}ln\\dfrac{T_e}{T_i} - Rln\\dfrac{P_e}{P_i}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">For stage 1,<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*} s_{gen, 1} &amp;= C_{p}ln\\dfrac{T_2}{T_1} - Rln\\dfrac{P_2}{P_1}\\\\&amp;= 1.005ln\\dfrac{(273.15 + 150)}{(273.15 + 10)} - 0.287ln\\dfrac{300}{101} \\\\&amp;= 0.0913 \\ \\rm{kJ\/kgK} \\end{align*}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"padding-left: 40px\">For stage 2,<\/p>\r\n<p style=\"padding-left: 80px\">[latex] \\begin{align*} s_{gen, 2} &amp;= C_{p}ln\\dfrac{T_4}{T_3} - Rln\\dfrac{P_4}{P_3} \\\\&amp;= 1.005ln\\dfrac{454.7}{(273.15 + 30)} - 0.287ln\\dfrac{900}{300} \\\\&amp;= 0.0921 \\ \\rm{kJ\/kgK} \\end{align*}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n(3) To calculate the total specific entropy generation in the system consisting of the compressor and the heat exchanger, we can set up the control volume <a id=\"6.10.e4\"><\/a>as shown in <a href=\"#6.10.e4\">Figure 6.10.e4<\/a>.\r\n\r\n[caption id=\"attachment_2896\" align=\"aligncenter\" width=\"500\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1.png\" target=\"_blank\" rel=\"noopener\"><img class=\"wp-image-2896\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-300x169.png\" alt=\"A two-stage compressor with an intercooler\" width=\"500\" height=\"281\" \/><\/a> <em><strong>Figure 6.10.e4<\/strong> Schematic of a two-stage compressor with an intercooler<\/em>[\/caption]\r\n\r\nThe specific heat transfer rejected from the cooler to the surroundings is\r\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} q &amp;= \\dfrac{\\dot{Q}}{\\dot{m}} = h_2 - h_3 = C_p(T_2 - T_3) \\\\&amp;=1.005 \\times (150 - 30) = 120.6 \\ \\rm{kJ\/kg} \\end{align*}[\/latex]<\/p>\r\nApply the second law of thermodynamics to the control volume as outlined in <a href=\"#6.10.e4\">Figure 6.10.e4<\/a>, we can find the total specific entropy generation.\r\n<p style=\"padding-left: 40px\">[latex]\\because \\dot{m}(s_e - s_i) = \\displaystyle\\sum\\dfrac{\\dot{Q}_k}{T_k}+\\dot{S}_{gen}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">[latex]\\therefore \\dot{m}(s_4 - s_1) = \\dfrac{-\\dot{Q}}{T_{surr}}+\\dot{S}_{gen}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">[latex] \\therefore s_{gen} = \\dfrac{\\dot{S}_{gen}}{\\dot{m}} = (s_4 - s_1) - \\dfrac{-q}{T_{surr}} [\/latex]<\/p>\r\nSince\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} s_4 - s_1 &amp;= C_{p}ln\\dfrac{T_4}{T_1} - Rln\\dfrac{P_4}{P_1} \\\\&amp;=1.005 ln\\dfrac{454.7}{(273.15 + 10)} - 0.287ln\\dfrac{900}{101} \\\\&amp;= -0.1517 \\ \\rm{kJ\/kgK} \\end{align*}[\/latex]<\/p>\r\nTherefore, the total specific entropy generation is\r\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} s_{gen}&amp;= (s_4 - s_1) - \\dfrac{-q}{T_{surr}} \\\\&amp;= -0.1517 + \\dfrac{120.6}{(273.15 + 30)} =0.246 \\ \\rm{kJ\/kgK} \\ &gt; 0 \\end{align*}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nComment:\r\n\r\nCompare the results from part (2) and part (3), [latex] s_{gen} &gt; s_{gen,1} + s_{gen,2} [\/latex], because the heat transfer process in the heat exchanger is irreversible and generates additional entropy.\r\n\r\n&nbsp;\r\n\r\n(4) If the air is directly compressed from <em>P<sub>1<\/sub><\/em> = 101 kPa to <em>P<sub>4<\/sub><\/em> = 900 kPa in a polytropic process with <em>n<\/em> = 1.58486, then its final temperature at the exit of stage 2 will be\r\n<p style=\"padding-left: 40px\">[latex] \\because \\dfrac{P_4}{P_1} = \\left(\\dfrac{T_{4'}}{T_1}\\right)^{\\dfrac{n}{n-1}}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px\">[latex] \\begin{align*} \\therefore T_{4'} &amp;= T_1 \\left(\\dfrac{P_4}{P_1}\\right)^{\\dfrac{n-1}{n}} \\\\&amp;= (273.15 + 10) \\left(\\dfrac{900}{101}\\right)^{\\dfrac{0.58486}{1.58486}} \\\\&amp;= 634.7 \\ \\rm{K} = 361.5^{\u00b0} \\rm{C} \\end{align*}[\/latex]<\/p>\r\nThe specific work of the compressor will be\r\n<p style=\"padding-left: 40px\">[latex] w_{tot'} = C_p(T_{4'} - T_1) = 1.005 \\times (361.5 - 10) = 353.3 \\ \\rm{kJ\/kg} [\/latex]<\/p>\r\n&nbsp;\r\n\r\nComment:\r\n\r\nIntercoolers are commonly used in multiple-stage compressors. Without intercoolers, the air temperature in the compressor may increase significantly, which may reduce the efficiency of the compressor. In addition, by comparing the specific work in part 1 and part 4, [latex]w_{tot'}&gt;w_{tot}[\/latex], it is apparent that without intercoolers, the compressor will need to consume much more power in order to achieve the same final pressure.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Practice Problems<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n[h5p id=\"49\"]\r\n\r\n<\/div>\r\n<\/div>","rendered":"<p>When solving problems in a control volume, the first and second laws of thermodynamics are often used together with the continuity equation and the thermodynamic tables or the ideal gas equation. The following strategy may be adapted:<\/p>\n<ol>\n<li>Set up a proper control volume to enclose the device of interest and identify the flow condition, i.e., steady or transient flow, through the control volume.<\/li>\n<li>Determine the relations among the mass flow rates at the inlet(s) and outlet(s) by using the continuity equation.<\/li>\n<li>Find the fluid properties at the inlet(s) and outlet(s), such as, [latex]P, T, v, h, s[\/latex], by using the thermodynamic tables or equations, e.g., for an ideal gas, solid or liquid.<\/li>\n<li>Determine the rate of heat transfer, [latex]\\dot Q[\/latex], or other unknowns by applying the first law of thermodynamics for open systems;<\/li>\n<li>Determine the rate of entropy generation, [latex]\\dot S_{gen}[\/latex], or other unknowns by applying the second law of thermodynamics for open systems.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Steam is used to provide heating to air in a building through a well-insulated heat exchanger, see <a href=\"#6.10.e1\">Figure 6.10.e1<\/a>.<\/p>\n<ul>\n<li>Saturated steam at 100<sup>o<\/sup>C enters the heat exchanger at a mass flow rate of 0.5 kg\/s and leaves the heat exchanger as a saturated liquid at 100<sup>o<\/sup><span style=\"font-size: 1em\">C<\/span><\/li>\n<li>Air enters the heat exchanger at 5<sup>o<\/sup>C, 101 kPa, and leaves at 25<sup>o<\/sup>C, 101 kPa.<\/li>\n<\/ul>\n<p>Assume air is an ideal gas. <a id=\"6.10.e1\"><\/a>Determine the mass flow rate of air and [latex]\\dot{S}_{gen}[\/latex] in this process.<\/p>\n<figure id=\"attachment_2897\" aria-describedby=\"caption-attachment-2897\" style=\"width: 400px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex.png\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2897\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-1024x358.png\" alt=\"Well-insulated heat exchanger to allow heat transfer between steam and air\" width=\"400\" height=\"140\" srcset=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-1024x358.png 1024w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-300x105.png 300w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-768x268.png 768w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-65x23.png 65w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-225x79.png 225w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex-350x122.png 350w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.1ex.png 1282w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a><figcaption id=\"caption-attachment-2897\" class=\"wp-caption-text\"><em><strong>Figure 6.10.e1<\/strong> Well-insulated heat exchanger<\/em><\/figcaption><\/figure>\n<p><em><span style=\"text-decoration: underline\">Solution<\/span>:<\/em><\/p>\n<p>Analysis<\/p>\n<ul>\n<li>The mass flow rate of air can be found by using the first law of thermodynamics<\/li>\n<\/ul>\n<p style=\"padding-left: 80px\">[latex]\\because \\dot{m}_{w}(h_1 - h_2) = \\dot{m}_{a}(h_4 - h_3) = \\dot{m}_{a}C_p(T_4 - T_3)[\/latex]<\/p>\n<p style=\"padding-left: 80px\">[latex]\\therefore \\dot{m}_{a} = \\dot{m}_{w}\u00a0 \\dfrac{h_1 - h_2}{C_p(T_4 - T_3)}[\/latex]<\/p>\n<ul>\n<li>The rate of entropy generation can be found by using the second law of thermodynamics. Note that the heat exchanger is well-insulated.<\/li>\n<\/ul>\n<p style=\"padding-left: 80px\">[latex]\\because \\sum\\dot{m}_{e}s_{e} - \\sum\\dot{m}_{i}s_{i} = \\sum\\dfrac{\\dot{Q}_{k}}{T_k} + \\dot{S}_{gen}[\/latex]\u00a0\u00a0\u00a0\u00a0 and \u00a0 \u00a0 [latex]\\dot{Q}_{k} = 0[\/latex]<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*}\\therefore \\dot{S}_{gen} &=\u00a0 \\sum\\dot{m}_{e}s_{e} - \\sum\\dot{m}_{i}s_{i} \\\\&= (\\dot{m}_{w}s_{2} + \\dot{m}_{a}s_{4}) -\u00a0 (\\dot{m}_{w}s_{1} + \\dot{m}_{a}s_{3}) \\\\&= \\dot{m}_{w}(s_{2}-s_{1}) + \\dot{m}_{a}(s_{4} - s_{3}) \\end{align*}[\/latex]<\/p>\n<ul>\n<li>The following properties are needed in order to complete the calculations:\n<ul>\n<li>For the steam-water stream: [latex]h_1 \\ , \\ h_2 \\ , \\ s_1 \\ , \\ s_2[\/latex]<\/li>\n<li>For the air stream: [latex]s_3 \\ , \\ s_4[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Now, let us solve the problem in detail.<\/p>\n<p>From <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/appendix-a-thermodynamic-properties-of-water#TA1\" target=\"_blank\" rel=\"noopener\">Table A1<\/a>, we can find the specific enthalpies and specific entropies of the saturated steam and saturated liquid water at 100<sup>o<\/sup>C.<\/p>\n<p style=\"padding-left: 40px\"><em>h<sub>1<\/sub> = h<sub>g<\/sub><\/em> = 2675.57 kJ\/kg ,\u00a0\u00a0 <em>h<sub>2<\/sub> = h<sub>f<\/sub><\/em> = 419.17 kJ\/kg<\/p>\n<p style=\"padding-left: 40px\"><em>s<sub>1<\/sub> = s<sub>g<\/sub><\/em> = 7.3541 kJ\/kgK,\u00a0\u00a0\u00a0\u00a0 <em>s<sub>2<\/sub> = s<sub>f<\/sub><\/em> = 1.3072 kJ\/kgK<\/p>\n<p>From <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/__unknown__-2#TG1\" target=\"_blank\" rel=\"noopener\">Table G1<\/a> for air:\u00a0 <em>C<sub>p<\/sub><\/em> = 1.005 kJ\/kgK, <em>R<\/em> = 0.287 kJ\/kgK<\/p>\n<p style=\"padding-left: 40px\">[latex]\\because s_4-s_3=C_pln\\displaystyle\\frac{T_4}{T_3}-Rln\\frac{P_4}{P_3}[\/latex]\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0 [latex]P_4=P_3[\/latex]<\/p>\n<p style=\"padding-left: 40px\">[latex]\\therefore s_4-s_3= 1.005ln\\displaystyle\\frac{273.15 + 25}{273.15 + 5} = 0.06978 \\ \\rm{kJ\/kgK}[\/latex]<\/p>\n<p>Now, we can complete the calculations. The mass flow rate of air is<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{m}_{a} &= \\dot{m}_{w} \\dfrac{ h_1 - h_2}{C_p (T_4 -T_3)} \\\\&= 0.5 \\times \\dfrac{ 2675.57 - 419.17}{1.005 (25-5)} = 56.13 \\ \\rm{kg\/s} \\end{align*}[\/latex]<\/p>\n<p>The rate of entropy generation is<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{S}_{gen}&= \\dot{m}_{w}(s_{2}-s_{1}) + \\dot{m}_{a}(s_{4} - s_{3}) \\\\&= 0.5 \\times (1.3072 - 7.3541) + 56.13 \\times 0.06978 \\\\&= 0.8934 \\ \\rm{kW\/K} \\ > 0 \\end{align*}[\/latex]<\/p>\n<p>Comment:<\/p>\n<p>The heat transfer process is irreversible; therefore, the rate of entropy generation is greater than zero.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>R134a at 300 kPa, 20\u00b0C is heated to 50\u00b0C isobarically in a heat exchanger. The mass flow rate of R134a is 1.5 kg\/s. The heat at a rate of [latex]\\dot{Q}_{H}[\/latex] is supplied by a heat pump, which absorbs heat at a rate of [latex]\\dot{Q}_{L}[\/latex] from the ambient at 280 K, see <a href=\"#6.10.e2\">Figure 6.10.e2<\/a>. If [latex]COP_{HP}=5[\/latex], and there is no heat loss in the heat exchanger, find the power input, [latex]\\dot{W}_{HP}[\/latex], and the rate of heat transfer, [latex]\\dot{Q}_{L}[\/latex]. Is this setup <a id=\"6.10.e2\"><\/a>consisting of the heat exchanger and heat pump possible?<\/p>\n<figure id=\"attachment_2893\" aria-describedby=\"caption-attachment-2893\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex.png\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2893\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-300x190.png\" alt=\"A device consisting of a heat exchanger and a heat pump\" width=\"500\" height=\"317\" srcset=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-300x190.png 300w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1024x650.png 1024w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-768x487.png 768w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-65x41.png 65w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-225x143.png 225w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-350x222.png 350w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex.png 1242w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a><figcaption id=\"caption-attachment-2893\" class=\"wp-caption-text\"><em><strong>Figure 6.10.e2<\/strong> A device consisting of a heat exchanger and a heat pump<\/em><\/figcaption><\/figure>\n<p><span style=\"text-decoration: underline\"><em>Solution:<\/em><\/span><\/p>\n<p>The control volume is set up to enclose both the heat pump and heat exchanger, as shown in <a href=\"#6.10.e2\">Figure 6.10.e2<\/a>.<\/p>\n<p>From <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/appendix-c-thermodynamic-properties-of-r134a#TC1\" target=\"_blank\" rel=\"noopener\">Table C1<\/a>, we can tell that, at the given pressures and temperatures, R134a remains a superheated vapour at both the inlet and the outlet of the heat exchanger. The specific enthalpies and specific entropies of R134a can be found in <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/appendix-c-thermodynamic-properties-of-r134a#TC2\" target=\"_blank\" rel=\"noopener\">Table C2<\/a>.<\/p>\n<ul>\n<li>At the inlet, <em>P<sub>1<\/sub><\/em> = 300 kPa, <em>T<sub>1<\/sub><\/em>= 20<sup>o<\/sup>C; therefore, <em>h<sub>1<\/sub><\/em> = 416.24 kJ\/kg, <em>s<sub>1<\/sub><\/em> = 1.7876 kJ\/kgK<\/li>\n<li>At the outlet, <em>P<sub>2<\/sub><\/em> = 300 kPa, <em>T<sub>2<\/sub><\/em> = 50<sup>o<\/sup>C; therefore, <em>h<sub>2<\/sub><\/em> = 443.31 kJ\/kg, <em>s<sub>1<\/sub><\/em> = 1.8755 kJ\/kgK<\/li>\n<\/ul>\n<p>Apply the first law of thermodynamics to the control volume.<\/p>\n<p style=\"padding-left: 40px\">[latex]\\dot{m}h_1 + \\dot{Q}_L + \\dot{W}_{HP} = \\dot{m}h_2[\/latex]<\/p>\n<p>For the heat pump<\/p>\n<p style=\"padding-left: 40px\">[latex]COP_{HP} = \\dfrac{\\dot{Q}_{H}}{\\dot{W}_{HP}}[\/latex]\u00a0\u00a0 and \u00a0 [latex]\\dot{Q}_{L} + \\dot{W}_{HP} = \\dot{Q}_{H}[\/latex]<\/p>\n<p>Combine the above three relations, we can derive<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{W}_{HP} &= \\dfrac{\\dot{m}(h_2 - h_1)}{COP_{HP}} \\\\&= \\dfrac{ 1.5 \\times (443.31 - 416.24)}{5} = 8.121 \\ \\rm{kW}\\end{align*}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{Q}_{L} &= (COP_{HP} - 1)\\dot{W}_{HP} \\\\&=(5-1) \\times 8.121 = 32.484 \\ \\rm{kW}\\end{align*}[\/latex]<\/p>\n<p>Apply the second law of thermodynamics to the control volume.<\/p>\n<p style=\"padding-left: 40px\">[latex]\\because \\dot{m}(s_2 - s_1) = \\sum\\dfrac{\\dot{Q}_k}{T_k}+\\dot{S}_{gen}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\therefore \\dot{S}_{gen} &= \\dot{m}(s_2 - s_1) - \\dfrac{\\dot{Q}_L}{T_{surr}} \\\\&=1.5 \\times (1.8755 - 1.7876) - \\dfrac{32.484}{280} = 0.0158 \\ \\rm{kW\/K} \\ > 0 \\end{align*}[\/latex]<\/p>\n<p>Comment:<\/p>\n<p>The rate of entropy generation in this control volume (consisting of both heat exchanger and heat pump) is positive; therefore, it is very likely that this setup can work in reality, and the processes in the heat exchanger and heat pump are irreversible.<\/p>\n<p>To further verify this, we can calculate the rate of entropy generation for each individual device, i.e., heat exchanger and heat pump, separately. However, with the given information, there exists a challenge to evaluate the amount of entropy transfer due to the heat transfer provided by the heat pump to the heat exchanger. Strictly speaking, [latex]\\displaystyle\\int \\dfrac{\\dot{Q}}{T_b}[\/latex] should be evaluated because the temperature of the heat exchanger varies, leading to a possible variation of temperature, [latex]T_b[\/latex], at the boundary between the heat pump and the heat exchanger. The relation between the instantaneous rate of heat transfer from the heat pump to the heat exchanger, [latex]\\dot{Q}[\/latex], and the boundary temperature, [latex]T_b[\/latex], must be obtained to enable a detailed analysis.<\/p>\n<p>Here, let us perform a simplified calculation to demonstrate the concept. Assume that [latex]T_b[\/latex] remains constant as the average temperature between the inlet and the outlet of the heat exchanger, thus [latex]T_b=35^ \\rm{o} \\rm{C}=308.15 \\ \\rm{K}[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>Apply the second law of thermodynamics to the heat pump. Note that [latex]\\dot{Q}_H=\\dot{Q}_L+\\dot{W}_{HP}[\/latex], and [latex]\\Delta s=0[\/latex] for the heat pump cycle.<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{S}_{gen, HP} &= \\dot{m} \\Delta s - \\dfrac{\\dot{Q}_L}{T_{surr}} + \\dfrac{\\dot{Q}_H}{T_b}\\\\&=0 - \\dfrac{32.484}{280} + \\dfrac{32.484+8.121}{308.15}= 0.015755 \\ \\rm{kW\/K} \\ > 0 \\end{align*}[\/latex]<\/p>\n<p>Furthermore, the <em>COP<\/em> of the Carnot cycle operating between the heat sink of 280 K and the heat source of 308.15 K is<\/p>\n<p style=\"padding-left: 40px\">[latex]\u00a0 COP_{HP, rev} = \\dfrac{308.15}{308.15-280}=10.9 > 5=COP_{HP}[\/latex]<\/p>\n<p>Because [latex]\\dot{S}_{gen, HP}>0[\/latex] and [latex]COP_{HP} < COP_{HP, rev}[\/latex], this heat pump cycle is an irreversible, realistic cycle.\n<!--more --><\/p>\n<p>Similarly, let us apply the second law of thermodynamics to the heat exchanger.<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\dot{S}_{gen, HE} &= \\dot{m}(s_2-s_1) - \\dfrac{\\dot{Q}_H}{T_b} \\\\&=1.5 \\times (1.8755 - 1.7876) - \\dfrac{32.484+8.121}{308.15} \\\\&= 8 \\times 10^{-5} \\ \\rm{kW\/K} \\approx 0 \\end{align*}[\/latex]<\/p>\n<p>Because [latex]\\dot{S}_{gen, HE} \\approx 0[\/latex], the process in this heat exchanger is almost ideal and reversible. It is noted that heat losses are typically unavoidable in real devices, but they are not considered in this example. Due to heat losses from the heat exchanger, the real process will have [latex]\\dot{S}_{gen, HE}>0[\/latex]. Consequently, the outlet temperature may not be able to reach 50<sup>o<\/sup>C.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p style=\"text-align: justify\">Consider an adiabatic two-stage compressor having an intercooler, as shown in <a href=\"#6.10.e3\">Figure 6.10.e3<\/a>.<\/p>\n<ul style=\"text-align: justify\">\n<li>Air at 10\u00b0C, 101 kPa enters the first stage of the compressor and is compressed to 300 kPa, 150\u00b0C.<\/li>\n<li>After being cooled to 30\u00b0C in a well-insulated intercooler, the air then enters the second stage cylinder and is compressed to 900 kPa.<\/li>\n<\/ul>\n<p style=\"text-align: justify\">Assume that (1) the pressure drop in the intercooler is negligible, (2) air is an ideal gas, and (3) the two compression processes are polytropic with the same polytropic exponent. Determine<\/p>\n<ol>\n<li style=\"text-align: justify\">The total specific work of the two-stage compressor.<\/li>\n<li style=\"text-align: justify\">The specific entropy generation in each of the two stages.<\/li>\n<li style=\"text-align: justify\">The total specific entropy generation in the system consisting of the two-stage compressor and the intercooler, assuming a constant temperature of 30\u00b0C at the system boundary.<\/li>\n<li style=\"text-align: justify\">Assume there is no intercooler. The air from the first stage enters directly to the second stage and is compressed to 900 kPa, what is the air temperature at the exit of the second stage?\u00a0 What is the total specific work of the compressor? Assume the polytropic exponent remains <a id=\"6.10.e3\"><\/a>the same in both stages.<\/li>\n<\/ol>\n<figure id=\"attachment_1211\" aria-describedby=\"caption-attachment-1211\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/06\/Fig.-6-22.jpg\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1211\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/06\/Fig.-6-22.jpg\" alt=\"Two stage compressor with an intercooler\" width=\"500\" height=\"333\" \/><\/a><figcaption id=\"caption-attachment-1211\" class=\"wp-caption-text\"><em><strong>Figure 6.10.e3<\/strong> Two stage compressor with an intercooler<\/em><\/figcaption><\/figure>\n<p><em><span style=\"text-decoration: underline\">Solution:<\/span><\/em><\/p>\n<p>(1) Assume air is an ideal gas. From <a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/back-matter\/__unknown__-2#TG1\" target=\"_blank\" rel=\"noopener\">Table G1<\/a> for air: <em>R<\/em> = 0.287 kJ\/kgK, <em>C<sub>p<\/sub><\/em> = 1.005 kJ\/kgK.<\/p>\n<p>The compressor is adiabatic with the following inlet and outlet conditions for each stage:<\/p>\n<ul>\n<li>Stage 1\n<ul>\n<li>Inlet <em>P<sub>1<\/sub><\/em> = 101 kPa, <em>T<sub>1<\/sub><\/em> = 10\u00b0C<\/li>\n<li>Outlet <em>P<sub>2<\/sub><\/em> = 300 kPa, <em>T<sub>2<\/sub><\/em> =150\u00b0C<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p style=\"padding-left: 40px\">Therefore, the specific work of stage 1 is<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*} w_1 &= h_2 - h_1 = C_p(T_2 - T_1) \\\\&= 1.005 \\times (150 - 10) = 140.7 \\ \\rm{kJ\/kg} \\end{align*}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">The polytropic exponent can be found by combining the polytropic relation and the ideal gas law.<\/p>\n<p style=\"padding-left: 80px\">[latex]\\because P_{1}v_{1}^{n} = P_{2}v_{2}^{n}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0 [latex]Pv = RT[\/latex]<\/p>\n<p style=\"padding-left: 80px\">[latex]\\therefore \\dfrac{P_2}{P_1} = \\left(\\dfrac{T_2}{T_1} \\right)^{\\dfrac{n}{n-1}}[\/latex]<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*} \\therefore n &= \\dfrac{ln\\dfrac{P_2}{P_1}}{ln\\left(\\dfrac{P_{2}T_{1}}{P_{1}T_{2}}\\right)} \\\\&= \\dfrac{ln\\dfrac{300}{101}}{ln\\left[ \\dfrac{300 \\times (273.15 + 10)}{101 \\times (273.15 + 150)}\\right]}=1.58486 \\end{align*}[\/latex]<\/p>\n<ul>\n<li>Stage 2\n<ul>\n<li>Inlet P<sub>3<\/sub> = 300 kPa, T<sub>3<\/sub> = 30\u00b0C<\/li>\n<li>Outlet P<sub>4<\/sub> = 900 kPa<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p style=\"padding-left: 40px\">The air temperature at the outlet of stage 2 can be found from the polytropic relation.<\/p>\n<p style=\"padding-left: 80px\">[latex]\\because \\dfrac{P_4}{P_3} = \\left(\\dfrac{T_4}{T_3}\\right)^{\\dfrac{n}{n-1}}[\/latex]<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*} \\therefore T_4 &= T_3 \\left(\\dfrac{P_4}{P_3}\\right)^{\\dfrac{n-1}{n}} \\\\&= (273.15 + 30) \\times \\dfrac{900}{300}^{\\dfrac{1.58486 -1}{1.58486}} \\\\&= 454.7 \\ \\rm{K} = 181.56^\u00a0 \\rm{o} \\rm{C}\\end{align*}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">Therefore, the specific work of stage 2 is<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*} w_2 &= h_4 - h_3 = C_p(T_4 - T_3) \\\\&= 1.005(181.56 - 30) = 152.31 \\ \\rm{kJ\/kg} \\end{align*}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">The total specific work of the two stage compressor is<\/p>\n<p style=\"padding-left: 80px\">[latex]w_{tot} = w_1 + w_2 = 140.7 + 152.31 = 293 \\ \\rm{kJ\/kg}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>(2) Both stages are adiabatic; therefore, for each stage<\/p>\n<p style=\"padding-left: 80px\">[latex]\\because \\dot{m}(s_e - s_i) = \\displaystyle\\sum\\dfrac{\\dot{Q}_k}{T_k}+\\dot{S}_{gen}[\/latex]\u00a0\u00a0\u00a0\u00a0 and \u00a0\u00a0 [latex]\\dot{Q}_{k} = 0[\/latex]<\/p>\n<p style=\"padding-left: 80px\">[latex]\\therefore s_{gen} = \\dfrac{\\dot{S}_{gen}}{\\dot{m}} = s_e - s_i = C_{p}ln\\dfrac{T_e}{T_i} - Rln\\dfrac{P_e}{P_i}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">For stage 1,<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*} s_{gen, 1} &= C_{p}ln\\dfrac{T_2}{T_1} - Rln\\dfrac{P_2}{P_1}\\\\&= 1.005ln\\dfrac{(273.15 + 150)}{(273.15 + 10)} - 0.287ln\\dfrac{300}{101} \\\\&= 0.0913 \\ \\rm{kJ\/kgK} \\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"padding-left: 40px\">For stage 2,<\/p>\n<p style=\"padding-left: 80px\">[latex]\\begin{align*} s_{gen, 2} &= C_{p}ln\\dfrac{T_4}{T_3} - Rln\\dfrac{P_4}{P_3} \\\\&= 1.005ln\\dfrac{454.7}{(273.15 + 30)} - 0.287ln\\dfrac{900}{300} \\\\&= 0.0921 \\ \\rm{kJ\/kgK} \\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>(3) To calculate the total specific entropy generation in the system consisting of the compressor and the heat exchanger, we can set up the control volume <a id=\"6.10.e4\"><\/a>as shown in <a href=\"#6.10.e4\">Figure 6.10.e4<\/a>.<\/p>\n<figure id=\"attachment_2896\" aria-describedby=\"caption-attachment-2896\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1.png\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2896\" src=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-300x169.png\" alt=\"A two-stage compressor with an intercooler\" width=\"500\" height=\"281\" srcset=\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-300x169.png 300w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-1024x576.png 1024w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-768x432.png 768w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-65x37.png 65w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-225x127.png 225w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1-350x197.png 350w, https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/02\/6.10.3ex-1.png 1500w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a><figcaption id=\"caption-attachment-2896\" class=\"wp-caption-text\"><em><strong>Figure 6.10.e4<\/strong> Schematic of a two-stage compressor with an intercooler<\/em><\/figcaption><\/figure>\n<p>The specific heat transfer rejected from the cooler to the surroundings is<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} q &= \\dfrac{\\dot{Q}}{\\dot{m}} = h_2 - h_3 = C_p(T_2 - T_3) \\\\&=1.005 \\times (150 - 30) = 120.6 \\ \\rm{kJ\/kg} \\end{align*}[\/latex]<\/p>\n<p>Apply the second law of thermodynamics to the control volume as outlined in <a href=\"#6.10.e4\">Figure 6.10.e4<\/a>, we can find the total specific entropy generation.<\/p>\n<p style=\"padding-left: 40px\">[latex]\\because \\dot{m}(s_e - s_i) = \\displaystyle\\sum\\dfrac{\\dot{Q}_k}{T_k}+\\dot{S}_{gen}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">[latex]\\therefore \\dot{m}(s_4 - s_1) = \\dfrac{-\\dot{Q}}{T_{surr}}+\\dot{S}_{gen}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">[latex]\\therefore s_{gen} = \\dfrac{\\dot{S}_{gen}}{\\dot{m}} = (s_4 - s_1) - \\dfrac{-q}{T_{surr}}[\/latex]<\/p>\n<p>Since<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} s_4 - s_1 &= C_{p}ln\\dfrac{T_4}{T_1} - Rln\\dfrac{P_4}{P_1} \\\\&=1.005 ln\\dfrac{454.7}{(273.15 + 10)} - 0.287ln\\dfrac{900}{101} \\\\&= -0.1517 \\ \\rm{kJ\/kgK} \\end{align*}[\/latex]<\/p>\n<p>Therefore, the total specific entropy generation is<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} s_{gen}&= (s_4 - s_1) - \\dfrac{-q}{T_{surr}} \\\\&= -0.1517 + \\dfrac{120.6}{(273.15 + 30)} =0.246 \\ \\rm{kJ\/kgK} \\ > 0 \\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Comment:<\/p>\n<p>Compare the results from part (2) and part (3), [latex]s_{gen} > s_{gen,1} + s_{gen,2}[\/latex], because the heat transfer process in the heat exchanger is irreversible and generates additional entropy.<\/p>\n<p>&nbsp;<\/p>\n<p>(4) If the air is directly compressed from <em>P<sub>1<\/sub><\/em> = 101 kPa to <em>P<sub>4<\/sub><\/em> = 900 kPa in a polytropic process with <em>n<\/em> = 1.58486, then its final temperature at the exit of stage 2 will be<\/p>\n<p style=\"padding-left: 40px\">[latex]\\because \\dfrac{P_4}{P_1} = \\left(\\dfrac{T_{4'}}{T_1}\\right)^{\\dfrac{n}{n-1}}[\/latex]<\/p>\n<p style=\"padding-left: 40px\">[latex]\\begin{align*} \\therefore T_{4'} &= T_1 \\left(\\dfrac{P_4}{P_1}\\right)^{\\dfrac{n-1}{n}} \\\\&= (273.15 + 10) \\left(\\dfrac{900}{101}\\right)^{\\dfrac{0.58486}{1.58486}} \\\\&= 634.7 \\ \\rm{K} = 361.5^{\u00b0} \\rm{C} \\end{align*}[\/latex]<\/p>\n<p>The specific work of the compressor will be<\/p>\n<p style=\"padding-left: 40px\">[latex]w_{tot'} = C_p(T_{4'} - T_1) = 1.005 \\times (361.5 - 10) = 353.3 \\ \\rm{kJ\/kg}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Comment:<\/p>\n<p>Intercoolers are commonly used in multiple-stage compressors. Without intercoolers, the air temperature in the compressor may increase significantly, which may reduce the efficiency of the compressor. In addition, by comparing the specific work in part 1 and part 4, [latex]w_{tot'}>w_{tot}[\/latex], it is apparent that without intercoolers, the compressor will need to consume much more power in order to achieve the same final pressure.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Practice Problems<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div id=\"h5p-49\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-49\" class=\"h5p-iframe\" data-content-id=\"49\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"S_6.10_Q\"><\/iframe><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"media-attributions clear\" prefix:cc=\"http:\/\/creativecommons.org\/ns#\" prefix:dc=\"http:\/\/purl.org\/dc\/terms\/\"><h2>Media Attributions<\/h2><ul><li about=\"https:\/\/commons.wikimedia.org\/wiki\/File:Inter_Cooler.jpg\"><a rel=\"cc:attributionURL\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Inter_Cooler.jpg\" property=\"dc:title\">Two stage compressor with intercooler<\/a>  &copy;  Brian S. Elliott    is licensed under a  <a rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA (Attribution ShareAlike)<\/a> license<\/li><\/ul><\/div>","protected":false},"author":175,"menu_order":11,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1876","chapter","type-chapter","status-publish","hentry"],"part":286,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/chapters\/1876","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/wp\/v2\/users\/175"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/chapters\/1876\/revisions"}],"predecessor-version":[{"id":3374,"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/chapters\/1876\/revisions\/3374"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/parts\/286"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/chapters\/1876\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/wp\/v2\/media?parent=1876"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/pressbooks\/v2\/chapter-type?post=1876"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/wp\/v2\/contributor?post=1876"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-json\/wp\/v2\/license?post=1876"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}