{"id":1315,"date":"2020-06-23T16:09:10","date_gmt":"2020-06-23T20:09:10","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chbe220\/?post_type=chapter&#038;p=1315"},"modified":"2020-08-20T13:56:07","modified_gmt":"2020-08-20T17:56:07","slug":"introduction-to-energy-balances","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chbe220\/chapter\/introduction-to-energy-balances\/","title":{"raw":"Introduction to Energy Balances","rendered":"Introduction to Energy Balances"},"content":{"raw":"<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nBy the end of this section, you should be able to:\r\n<p id=\"Identify:\"><strong>Identify <\/strong>relevant terms for energy balances for open and closed systems<\/p>\r\n<p id=\"Use:\"><strong>Use <\/strong>thermodynamic data tables to identify enthalpy, internal energy, and other thermodynamic properties using system temperatures and pressures<\/p>\r\n<p id=\"Solve:\"><strong>Solve <\/strong>energy balance problems using thermodynamic data<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<h2 id=\"Forms-of-Energy\">Forms of Energy<\/h2>\r\nSystems are typically divided into three main categories depending on how the system interacts with its surroundings:\r\n<ul>\r\n \t<li>Isolated - No energy or mass transfer between system and surroundings, energy may change form within the system<\/li>\r\n \t<li>Closed - Energy, but no mass transfer between system and surroundings<\/li>\r\n \t<li>Open - Energy and mass transfer between systems and surroundings, typically use a [latex]\\dot{}[\/latex] with quantities that change over time in these open systems to denote the flow rate of energy or mass.<\/li>\r\n<\/ul>\r\n<img class=\"wp-image-1016 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems-300x169.png\" alt=\"\" width=\"467\" height=\"263\" \/>\r\n<div style=\"text-align: center\">Image by <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Diagram_Systems.png\">Alkh.Alwa<\/a> \/ <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\">CC BY-SA<\/a><\/div>\r\n<div class=\"prompt input_prompt\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h3 id=\"Kinetic-Energy---$E_{k}$\">Kinetic Energy - [latex]E_{k}[\/latex]<\/h3>\r\n<strong>Kinetic Energy<\/strong> is energy associated with motion, which can be described as translational or rotational energy.\r\n<p style=\"text-align: center\">[latex] E_{k} = \\frac{1}{2} mu^{2} [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] \\dot{E}_{k} = \\frac{1}{2} \\dot{m} u^{2} [\/latex]<\/p>\r\nwhere:\r\n<blockquote>[latex]m[\/latex] is mass\r\n\r\n[latex]u[\/latex] is velocity relative to a reference. Generally we refer to earth's surface as stationary.<\/blockquote>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h3 id=\"Potential-Energy---$E_{p}$\">Potential Energy - [latex]E_{p}[\/latex]<\/h3>\r\n<strong>Potential energy<\/strong> can be described as energy present due to position in a field, such as gravitational position or magnetic position\r\n\r\nUsually for chemical processes, we consider the potential energy change due to the gravitational position of the process equipments.\r\n<p style=\"text-align: center\">[latex] E_{p} = m g z [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] \\dot{E}_{p} = \\dot{m} g z [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta E_{p} = E_{p2} - E_{p1} = m g (z_{2} - z_{1})[\/latex]<\/p>\r\nwhere:\r\n<blockquote>[latex]m[\/latex] is mass\r\n\r\n[latex]g[\/latex] is the gravitational acceleration (approximately 9.8 [latex]m\/s^{2} [\/latex])\r\n\r\n[latex]z[\/latex] is the height about the point of reference<\/blockquote>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h3 id=\"Internal-Energy---$U$\">Internal Energy - [latex]U[\/latex]<\/h3>\r\nInternal energy can be described as all other energy present in a system, including motion, and molecular interaction.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h3 id=\"Heat---$Q$\">Heat - [latex]Q[\/latex]<\/h3>\r\n<ul>\r\n \t<li>Heat is the energy flow due to temperature difference<\/li>\r\n \t<li>Heat flows from higher temperatures to lower temperatures<\/li>\r\n \t<li>Heat is generally defined as positive when it is transferred from the surroundings to the system<\/li>\r\n<\/ul>\r\n<h3 id=\"Work---$W$\">Work - [latex]W[\/latex]<\/h3>\r\n<ul>\r\n \t<li>Work is energy resulting from driving forces (not temperature) such as force, torque, or voltage<\/li>\r\n \t<li>We will define work as positive when work is done by the surroundings on the system, this is a typical convention in chemistry. With this convention, we would write \"Q + W\" in our energy balances. However, historically work has also been defined in physics as positive when work is done by the system on the surroundings. In this other case, the energy balance would be written with \"Q - W\". Both can be used and this is accounted for in the sign we use in front of the work term in energy balances.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\">\r\n<h2 id=\"Energy-Transfer-in-Closed-Systems\" style=\"margin-top: 2.14286em;margin-bottom: 1.42857em\">Energy Transfer in Closed Systems<\/h2>\r\n<p style=\"text-align: start\">Closed systems are defined as systems with no mass transfer across the system's boundaries. All the energy forms described above are applicable to closed systems.<\/p>\r\n\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Energy Balance Sign Conventions<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div id=\"notebook-container\" class=\"container\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<div class=\"alert alert-block alert-warning\">\r\n\r\nConsider a system that consists of a stirred tank reactor where an exothermic reaction is taking place, where an external motor is mixing the contents in the reactors. What are the signs of [latex]Q[\/latex] and [latex]W[\/latex] for this system?\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"prompt input_prompt\">\r\n<div class=\"textbox\">\r\n<h3 class=\"prompt input_prompt\">Solution<\/h3>\r\n<div>For an exothermic reaction, heat is produced by the system. Therefore, [latex]Q[\/latex]is <strong>negative<\/strong>.\r\nFor an external motor that is mixing the contents in the reactor, work is being done by the surroundings on the system. Therefore, [latex]W[\/latex] is <strong>positive<\/strong>.<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h3 id=\"Energy-Balances-on-Closed-Systems\">Energy Balances on Closed Systems<\/h3>\r\nEnergy in closed systems follows the Law of Conservation of Energy\r\n<p style=\"text-align: center\">[latex] Accumulation = Input - Output [\/latex]<\/p>\r\n&nbsp;\r\n\r\nIn terms of general energy:\r\n<p style=\"text-align: center\">[latex]E_{system,final} - E_{system,initial} = E_{system,transferred}[\/latex]<\/p>\r\n\r\n<blockquote>The initial energy in the system can be defined as:\r\n<p style=\"text-align: center\">[latex] U_{i} + E_{k,i} + E_{p,i} [\/latex]<\/p>\r\nThe final energy in the system can be defined as:\r\n<p style=\"text-align: center\">[latex] U_{f} + E_{k,f} + E_{p,f} [\/latex]<\/p>\r\nThe energy transfer of the system can be defined as:\r\n<p style=\"text-align: center\">[latex] Q + W [\/latex]<\/p>\r\nThis yields the following closed system energy balance, defined as\r\n<strong>the First Law of Thermodynamics<\/strong>:\r\n<p style=\"text-align: center\">[latex] (U_{f} - U_{i}) + (E_{k,f} - E_{k,i}) + (E_{p,f} - E_{p,i}) = Q + W [\/latex]<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 51.1065%;height: 38px\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 38px\">\r\n<td style=\"width: 100%;text-align: center;height: 38px\"><span style=\"color: #ff0000;font-size: 16px\">[latex] \\Delta U + \\Delta E_{k} + \\Delta E_{p} = Q + W [\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<blockquote>Assumptions made by the First Law of Thermodynamics\r\n<ul>\r\n \t<li>If no acceleration exists in the system, the change in the kinetic energy term will be 0, and can be ommitted from the balance<\/li>\r\n \t<li>If no change in height (or other fields) exist in the system, the change in the potential energy term will be 0, and can be ommitted from the balance<\/li>\r\n \t<li>Internal energy depends on chemical composition, state (solid, liquid, or gas) and temperature; Pressure's effect is negligible.<\/li>\r\n \t<li>If the system has the same temperature as its surroundings or is adiabatic, the heat term will be 0, and can be ommitted from the balance<\/li>\r\n \t<li>If there are no moving parts, electrical currents, or radiation in the system, the work term will be 0 and can be omitted from the balance<\/li>\r\n<\/ul>\r\n<\/blockquote>\r\n<\/blockquote>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h2 id=\"Work-in-Open-Systems\">Work in Open Systems<\/h2>\r\n<h3 id=\"Open-Systems:\">Open Systems:<\/h3>\r\nOpen systems are defined as systems where both mass and energy cross the system's boundaries.\r\n\r\nTwo types of work are typically observed in these systems:\r\n<ul>\r\n \t<li><strong>Shaft Work - [latex]W_{s}[\/latex]or [latex]\\dot{W}_{s}[\/latex]<\/strong>\r\nShaft work is work done on process fluid by a moving part, such as a pump, rotor, or a stirrer.<\/li>\r\n \t<li><strong>Flow Work - [latex]W_{fl}[\/latex] or [latex]\\dot{W}_{fl}[\/latex]<\/strong>\r\nFlow work is work done on process fluid (inlet minus outlet). For the work flow in, the surroundings do work on the system, therefore it is positive. For the work flow out, the system does work on the surroundings, therefore it is negative.<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center\">[latex] \\dot{W}_{fl} = \\dot{W}_{fl-in} - \\dot{W}_{fl-out} = P_{in}\\dot{V}_{in} - P_{out}\\dot{V}_{out} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h2 id=\"Steady-State-Open-System-Energy-Balance\">Steady-State Open-System Energy Balance<\/h2>\r\n<h3 id=\"Energy-Conservation-for-a-Steady-State-System\">Energy Conservation for a Steady-State System<\/h3>\r\nFor stream 'j' in a system:\r\n<p style=\"text-align: center\">[latex] \\Sigma_{in} \\dot{E}_{j} + \\dot{Q} + \\dot{W} = \\Sigma_{out} \\dot{E}_{j} [\/latex]<\/p>\r\nRearranging the energy terms, we get:\r\n<p style=\"text-align: center\">[latex] \\dot{Q} + \\dot{W} = \\Sigma_{out} \\dot{E}_{j} - \\Sigma_{in} \\dot{E}_{j} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Energy Flow in a System<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider the following system:\r\n\r\n<img class=\" wp-image-1005 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System-300x118.jpg\" alt=\"\" width=\"361\" height=\"142\" \/>\r\n\r\nThere are 2 streams with energy entering the system (streams 1 and 2), and 2 streams with energy exiting the system (streams 3 and 4).\r\n\r\nFor this system:\r\n<p style=\"text-align: center\">[latex] \\dot{E}_{1} + \\dot{E}_{2} + \\dot{Q} + \\dot{W} =\u00a0 \\dot{E}_{3} + \\dot{E}_{4} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nRecall the three forms of energy:\r\n<p style=\"text-align: center\">[latex] \\dot{E}_{j} = \\dot{U}_{j} + \\dot{E}_{k,j} + \\dot{E}_{p,j} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<blockquote>Each energy flow term can be further separated into: [latex] \\dot{U}_{j} = \\dot{m} * \\hat{U}_{j} [\/latex]\r\n\r\n<strong>Specific Property<\/strong> \"[latex]\\hat{ }[\/latex]\" : This denotes an intensive property obtained by dividing an extensive property by a total amount of flow rate (can be total number of moles or total mass)\r\n<p style=\"text-align: center\">[latex] \\hat{V} = \\frac{V}{n} [\/latex]<\/p>\r\n<p style=\"text-align: center\">or<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{U}=\\frac{\\dot{U}}{\\dot{m}}[\/latex]<\/p>\r\n<\/blockquote>\r\nCombining all these terms:\r\n<p style=\"text-align: center\">[latex]\\dot{Q} + \\dot{W} = \\Sigma_{out} \\dot{m}_{j} * (\\hat{U}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j}) - \\Sigma_{in} \\dot{m}_{j} * (\\hat{U}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j}) [\/latex]<\/p>\r\nRecall the work terms expansion:\r\n<p style=\"text-align: center\">[latex] \\dot{W} = \\dot{W}_{fl} + \\dot{W}_s [\/latex]<\/p>\r\n\r\n<blockquote>where flow work is dependant on system pressure and volume\r\n<p style=\"text-align: center\">[latex] \\dot{W}_{fl} = \\Sigma_{in} \\dot{m}_{j} P_{j} \\hat{V}_{j} - \\Sigma_{out} \\dot{m}_{j} P_{j} \\hat{V}_{j} [\/latex]<\/p>\r\n<\/blockquote>\r\nNow we have:\r\n<p style=\"text-align: center\">[latex] \\dot{Q} + \\dot{W}_{s} = \\Sigma_{out} \\dot{m}_{j} * (\\hat{U}_{j} + P_{j} \\hat{V}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j}) - \\Sigma_{in} \\dot{m}_{j}*(\\hat{U}_{j} + P_{j}\\hat{V}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j}) [\/latex]<\/p>\r\nBecause [latex] \\hat{U} + P\\hat{V} [\/latex] usually appear together in the energy balances, we define them to be \"enthalpy\" ([latex]\\hat{H} [\/latex]):\r\n<p style=\"text-align: center\">[latex]\\hat{H} = \\hat{U} + P\\hat{V}[\/latex]<\/p>\r\n\r\n<blockquote>where [latex]\\hat{U}[\/latex] is the internal energy and [latex]P\\hat{V}[\/latex] is the flow work<\/blockquote>\r\nThe following terms are defined:\r\n<ul>\r\n \t<li>[latex]\\Delta\\dot{H} = \\Sigma_{out}\\dot{m}_{j}*\\hat{H}_{j} - \\Sigma_{in}\\dot{m}_{j}*\\hat{H}_{j} [\/latex]<\/li>\r\n \t<li>[latex]\\Delta\\dot{E}_{k} = \\Sigma_{out}\\dot{m}_{j}*\\hat{E}_{k,j} - \\Sigma_{in}\\dot{m}_{j}*\\hat{E}_{k,j} [\/latex]<\/li>\r\n \t<li>[latex]\\Delta\\dot{E}_{p} = \\Sigma_{out}\\dot{m}_{j}*\\hat{E}_{p,j} - \\Sigma_{in}\\dot{m}_{j}*\\hat{E}_{p,j} [\/latex]<\/li>\r\n<\/ul>\r\nFinally, an open system steady-state energy balance is defined:\r\n<p style=\"text-align: center\">[latex]\\dot{Q} + \\dot{W}_{s} = \\Delta\\dot{H} + \\Delta\\dot{E}_{k} + \\Delta\\dot{E}_{p}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"prompt input_prompt\">\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Heat for an Ideal Gas<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<div class=\"alert alert-block alert-warning\">\r\n\r\nPrior to entering a furnace, air is heated from [latex]25^{\\circ}C[\/latex] to [latex]150^{\\circ}C[\/latex] and the change in specific enthalpy for the whole heating process is 3640 J\/mol. The flow rate of air at the outlet of the heater is [latex]1.5 m^3\/min[\/latex] and the air pressure at this point is 150 kPa absolute.\r\n\r\nCalculate the heat needed for the process in kW. Assume the ideal gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n<strong>Step 1<\/strong>: Calculate the molar flowrate using the ideal gas law.\r\n\r\n\\begin{align*}\r\n\\dot{n} &amp;\u00a0 = \\frac{1.5m^{3}}{min}*\\frac{273K}{150+273K}*\\frac{150kPa}{101.3kPa}*\\frac{1mol}{22.4L}*\\frac{10^{3}L}{1m^{3}}\\\\\r\n\\dot{n} &amp; =64.0\\frac{mol}{min}\r\n\\end{align*}\r\n\r\n<strong>Step 2<\/strong>: Calculate the heat using the specific enthalpy. Since the potential and kinetic energy changes are zero, the following calculations are made\r\n\r\n\\begin{align*}\r\n\\dot{Q} &amp; =\\Delta\\dot{H}= \\dot{n}\\Delta\\hat{H}\\\\\r\n\\dot{Q} &amp; =\\frac{64.0 mol}{min}*\\frac{1min}{60s}*\\frac{3640J}{mol}*\\frac{kW}{10^{3} J\/s} \\\\\r\n\\dot{Q} &amp; =3.88kW\r\n\\end{align*}\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"prompt input_prompt\">\r\n<h2 id=\"Reference-States\">Reference States<\/h2>\r\n<strong>Reference State:<\/strong> a substance at some pressure, temperature, and state of aggregation (solid, liquid, gas; pure or mixture).\r\n\r\nIt is much easier to estimate the energy of a system as a change from a reference state rather than the absolute energy.\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Cooling in a Heat Exchanger<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<div class=\"alert alert-block alert-warning\">\r\n<p style=\"text-align: left\">Water is used to cool a liquid in a heat exchanger. Water enters the heat exchanger at [latex]10^{\\circ}C[\/latex] and exits at [latex]100^{\\circ}C[\/latex]. Using the table below, find the change in enthalpy of water in its liquid state.<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 68.4422%;height: 60px\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\"><strong>Entry #<\/strong><\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\"><strong>[latex]T (^{\\circ}C)[\/latex]<\/strong><\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\"><strong>[latex]\\hat{H}_{L} (\\frac{kJ}{kg})[\/latex]<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">1<\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">5<\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">21.02<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">2<\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">10<\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">42.02<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">3<\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">100<\/td>\r\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">419.17<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n<div>\r\n\r\n<strong>Step 1<\/strong>: Determine which reference state you are going to use. In this case, we are using [latex]10^{\\circ}C[\/latex] as the reference state.\r\n<p style=\"text-align: center\">[latex]\\Delta\\hat{H} = \\hat{H}_{100^{\\circ}C} - \\hat{H}_{10^{\\circ}C}[\/latex]<\/p>\r\n<strong>Step 2<\/strong>: Find the change in enthalpy by taking the difference of the system's specific enthalpies at different temperatures.\r\n\r\n\\begin{align*}\r\n\\Delta\\hat{H} &amp; = \\hat{H}_{3} - \\hat{H}_{2}\\\\\r\n\\Delta\\hat{H} &amp; = (419.17 - 42.02) kJ\/kg\\\\\r\n\\Delta\\hat{H} &amp; = 377.15 kJ\/kg\r\n\\end{align*}\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h2 id=\"Steam-Tables\">Steam Tables<\/h2>\r\nSince water is a commonly used resource in processes for heating and cooling, detailed information on its state properties at different temperatures and pressures is available.\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\">\r\n<div class=\"textbox shaded\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h3 id=\"How-to-Access-Steam-Tables-on-NIST\">How to Access Steam Tables on NIST<\/h3>\r\n<ul>\r\n \t<li>Steam tables can be found on <a href=\"https:\/\/webbook.nist.gov\/chemistry\/fluid\/\">NIST<\/a><\/li>\r\n<\/ul>\r\nA. Select 'Water' from the 'Please select the species of interest:' drop-down menu\r\n\r\nB. Choose the steam table units you'd like to work with, in step 2.\r\n\r\nC. In Step 3, choose what kind of data you're looking to obtain. For an isothermal system, select 'Isothermal properties'. For a constant pressure system, select 'Isobaric properties'.\r\n\r\nD. Select the desired standard state convention. This course will most likely only use the 'Default for fluid' convention.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Steam Tables<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<span style=\"color: #333333;font-size: 1em\">Superheated steam at 40 bar absolute and [latex]500^{\\circ}C[\/latex] flowing at a rate of [latex]200 kg\/min[\/latex] is sent to an adiabatic turbine which expands to 5 bar. The turbine outputs [latex]1250kW[\/latex]. The expanded steam is then sent to a heat exchanger where isobaric heating occurs, resulting in the stream being reheated to its initial temperature. Assume no changes in kinetic energy. Write the energy balance for the turbine and determine the outlet stream temperature.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n<div><strong>Step 1:<\/strong> Determine the enthalpy values for water vapor at\u00a0 [latex]500^{\\circ}C[\/latex] and [latex]40 bar[\/latex] and [latex]5 bar[\/latex].<\/div>\r\n<div>From the steam tables:<\/div>\r\n<div><\/div>\r\n<ul>\r\n \t<li>For water vapor at [latex]500^{\\circ}C[\/latex] and [latex]40 bar[\/latex], the specific enthalpy\u00a0 is [latex]3445 \\;kJ\/kg[\/latex]<\/li>\r\n \t<li>\u00a0For water vapor at [latex]500^{\\circ}C[\/latex] and [latex]5 bar[\/latex], the specific enthalpy\u00a0 is [latex]3484 \\;kJ\/kg[\/latex]<\/li>\r\n<\/ul>\r\n<div><\/div>\r\n<div><strong>Step 2:<\/strong> Write the energy balance. Since there are no changes in potential and kinetic energy and no heat transfer, the change in enthalpy will be equal to a negative shaft work.<\/div>\r\n<div>\\begin{align*}\r\n\\Delta\\dot{H} &amp;=-\\dot{W}_{s}\\\\\r\n\\Delta\\dot{H} &amp;=\\dot{m}*(\\hat{H}_{2}-\\hat{H}_{1})\r\n\\end{align*}<\/div>\r\n<div>\\begin{align*}\r\n\\hat{H}_{2} &amp;=\\hat{H}_{1}-\\frac{\\dot{W}_{s}}{\\dot{m}}\\\\\r\n&amp;=\\frac{3445kJ}{kg}-\\frac{1250kJ}{s}*\\frac{min}{200kg}*\\frac{60s}{1min}\\\\\r\n&amp;=3070\\frac{kJ}{kg}\r\n\\end{align*}<\/div>\r\n<div><\/div>\r\n<div><strong>Step 3:<\/strong> Determine the temperature of the steam corresponding to [latex]5bar[\/latex] and [latex]3070 kJ\/kg[\/latex].<\/div>\r\n<div>From the steam tables:<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: center\">[latex]T=302^{\\circ}C[\/latex]<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>By the end of this section, you should be able to:<\/p>\n<p id=\"Identify:\"><strong>Identify <\/strong>relevant terms for energy balances for open and closed systems<\/p>\n<p id=\"Use:\"><strong>Use <\/strong>thermodynamic data tables to identify enthalpy, internal energy, and other thermodynamic properties using system temperatures and pressures<\/p>\n<p id=\"Solve:\"><strong>Solve <\/strong>energy balance problems using thermodynamic data<\/p>\n<\/div>\n<\/div>\n<h2 id=\"Forms-of-Energy\">Forms of Energy<\/h2>\n<p>Systems are typically divided into three main categories depending on how the system interacts with its surroundings:<\/p>\n<ul>\n<li>Isolated &#8211; No energy or mass transfer between system and surroundings, energy may change form within the system<\/li>\n<li>Closed &#8211; Energy, but no mass transfer between system and surroundings<\/li>\n<li>Open &#8211; Energy and mass transfer between systems and surroundings, typically use a [latex]\\dot{}[\/latex] with quantities that change over time in these open systems to denote the flow rate of energy or mass.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1016 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems-300x169.png\" alt=\"\" width=\"467\" height=\"263\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems-300x169.png 300w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems-65x37.png 65w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems-225x127.png 225w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems-350x197.png 350w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Diagram_Systems.png 387w\" sizes=\"auto, (max-width: 467px) 100vw, 467px\" \/><\/p>\n<div style=\"text-align: center\">Image by <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Diagram_Systems.png\">Alkh.Alwa<\/a> \/ <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\">CC BY-SA<\/a><\/div>\n<div class=\"prompt input_prompt\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h3 id=\"Kinetic-Energy---$E_{k}$\">Kinetic Energy &#8211; [latex]E_{k}[\/latex]<\/h3>\n<p><strong>Kinetic Energy<\/strong> is energy associated with motion, which can be described as translational or rotational energy.<\/p>\n<p style=\"text-align: center\">[latex]E_{k} = \\frac{1}{2} mu^{2}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\dot{E}_{k} = \\frac{1}{2} \\dot{m} u^{2}[\/latex]<\/p>\n<p>where:<\/p>\n<blockquote><p>[latex]m[\/latex] is mass<\/p>\n<p>[latex]u[\/latex] is velocity relative to a reference. Generally we refer to earth&#8217;s surface as stationary.<\/p><\/blockquote>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h3 id=\"Potential-Energy---$E_{p}$\">Potential Energy &#8211; [latex]E_{p}[\/latex]<\/h3>\n<p><strong>Potential energy<\/strong> can be described as energy present due to position in a field, such as gravitational position or magnetic position<\/p>\n<p>Usually for chemical processes, we consider the potential energy change due to the gravitational position of the process equipments.<\/p>\n<p style=\"text-align: center\">[latex]E_{p} = m g z[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\dot{E}_{p} = \\dot{m} g z[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta E_{p} = E_{p2} - E_{p1} = m g (z_{2} - z_{1})[\/latex]<\/p>\n<p>where:<\/p>\n<blockquote><p>[latex]m[\/latex] is mass<\/p>\n<p>[latex]g[\/latex] is the gravitational acceleration (approximately 9.8 [latex]m\/s^{2}[\/latex])<\/p>\n<p>[latex]z[\/latex] is the height about the point of reference<\/p><\/blockquote>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h3 id=\"Internal-Energy---$U$\">Internal Energy &#8211; [latex]U[\/latex]<\/h3>\n<p>Internal energy can be described as all other energy present in a system, including motion, and molecular interaction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h3 id=\"Heat---$Q$\">Heat &#8211; [latex]Q[\/latex]<\/h3>\n<ul>\n<li>Heat is the energy flow due to temperature difference<\/li>\n<li>Heat flows from higher temperatures to lower temperatures<\/li>\n<li>Heat is generally defined as positive when it is transferred from the surroundings to the system<\/li>\n<\/ul>\n<h3 id=\"Work---$W$\">Work &#8211; [latex]W[\/latex]<\/h3>\n<ul>\n<li>Work is energy resulting from driving forces (not temperature) such as force, torque, or voltage<\/li>\n<li>We will define work as positive when work is done by the surroundings on the system, this is a typical convention in chemistry. With this convention, we would write &#8220;Q + W&#8221; in our energy balances. However, historically work has also been defined in physics as positive when work is done by the system on the surroundings. In this other case, the energy balance would be written with &#8220;Q &#8211; W&#8221;. Both can be used and this is accounted for in the sign we use in front of the work term in energy balances.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\">\n<h2 id=\"Energy-Transfer-in-Closed-Systems\" style=\"margin-top: 2.14286em;margin-bottom: 1.42857em\">Energy Transfer in Closed Systems<\/h2>\n<p style=\"text-align: start\">Closed systems are defined as systems with no mass transfer across the system&#8217;s boundaries. All the energy forms described above are applicable to closed systems.<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Energy Balance Sign Conventions<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div id=\"notebook-container\" class=\"container\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<div class=\"alert alert-block alert-warning\">\n<p>Consider a system that consists of a stirred tank reactor where an exothermic reaction is taking place, where an external motor is mixing the contents in the reactors. What are the signs of [latex]Q[\/latex] and [latex]W[\/latex] for this system?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"prompt input_prompt\">\n<div class=\"textbox\">\n<h3 class=\"prompt input_prompt\">Solution<\/h3>\n<div>For an exothermic reaction, heat is produced by the system. Therefore, [latex]Q[\/latex]is <strong>negative<\/strong>.<br \/>\nFor an external motor that is mixing the contents in the reactor, work is being done by the surroundings on the system. Therefore, [latex]W[\/latex] is <strong>positive<\/strong>.<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h3 id=\"Energy-Balances-on-Closed-Systems\">Energy Balances on Closed Systems<\/h3>\n<p>Energy in closed systems follows the Law of Conservation of Energy<\/p>\n<p style=\"text-align: center\">[latex]Accumulation = Input - Output[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>In terms of general energy:<\/p>\n<p style=\"text-align: center\">[latex]E_{system,final} - E_{system,initial} = E_{system,transferred}[\/latex]<\/p>\n<blockquote><p>The initial energy in the system can be defined as:<\/p>\n<p style=\"text-align: center\">[latex]U_{i} + E_{k,i} + E_{p,i}[\/latex]<\/p>\n<p>The final energy in the system can be defined as:<\/p>\n<p style=\"text-align: center\">[latex]U_{f} + E_{k,f} + E_{p,f}[\/latex]<\/p>\n<p>The energy transfer of the system can be defined as:<\/p>\n<p style=\"text-align: center\">[latex]Q + W[\/latex]<\/p>\n<p>This yields the following closed system energy balance, defined as<br \/>\n<strong>the First Law of Thermodynamics<\/strong>:<\/p>\n<p style=\"text-align: center\">[latex](U_{f} - U_{i}) + (E_{k,f} - E_{k,i}) + (E_{p,f} - E_{p,i}) = Q + W[\/latex]<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 51.1065%;height: 38px\">\n<tbody>\n<tr style=\"height: 38px\">\n<td style=\"width: 100%;text-align: center;height: 38px\"><span style=\"color: #ff0000;font-size: 16px\">[latex]\\Delta U + \\Delta E_{k} + \\Delta E_{p} = Q + W[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<blockquote><p>Assumptions made by the First Law of Thermodynamics<\/p>\n<ul>\n<li>If no acceleration exists in the system, the change in the kinetic energy term will be 0, and can be ommitted from the balance<\/li>\n<li>If no change in height (or other fields) exist in the system, the change in the potential energy term will be 0, and can be ommitted from the balance<\/li>\n<li>Internal energy depends on chemical composition, state (solid, liquid, or gas) and temperature; Pressure&#8217;s effect is negligible.<\/li>\n<li>If the system has the same temperature as its surroundings or is adiabatic, the heat term will be 0, and can be ommitted from the balance<\/li>\n<li>If there are no moving parts, electrical currents, or radiation in the system, the work term will be 0 and can be omitted from the balance<\/li>\n<\/ul>\n<\/blockquote>\n<\/blockquote>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h2 id=\"Work-in-Open-Systems\">Work in Open Systems<\/h2>\n<h3 id=\"Open-Systems:\">Open Systems:<\/h3>\n<p>Open systems are defined as systems where both mass and energy cross the system&#8217;s boundaries.<\/p>\n<p>Two types of work are typically observed in these systems:<\/p>\n<ul>\n<li><strong>Shaft Work &#8211; [latex]W_{s}[\/latex]or [latex]\\dot{W}_{s}[\/latex]<\/strong><br \/>\nShaft work is work done on process fluid by a moving part, such as a pump, rotor, or a stirrer.<\/li>\n<li><strong>Flow Work &#8211; [latex]W_{fl}[\/latex] or [latex]\\dot{W}_{fl}[\/latex]<\/strong><br \/>\nFlow work is work done on process fluid (inlet minus outlet). For the work flow in, the surroundings do work on the system, therefore it is positive. For the work flow out, the system does work on the surroundings, therefore it is negative.<\/li>\n<\/ul>\n<p style=\"text-align: center\">[latex]\\dot{W}_{fl} = \\dot{W}_{fl-in} - \\dot{W}_{fl-out} = P_{in}\\dot{V}_{in} - P_{out}\\dot{V}_{out}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h2 id=\"Steady-State-Open-System-Energy-Balance\">Steady-State Open-System Energy Balance<\/h2>\n<h3 id=\"Energy-Conservation-for-a-Steady-State-System\">Energy Conservation for a Steady-State System<\/h3>\n<p>For stream &#8216;j&#8217; in a system:<\/p>\n<p style=\"text-align: center\">[latex]\\Sigma_{in} \\dot{E}_{j} + \\dot{Q} + \\dot{W} = \\Sigma_{out} \\dot{E}_{j}[\/latex]<\/p>\n<p>Rearranging the energy terms, we get:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} + \\dot{W} = \\Sigma_{out} \\dot{E}_{j} - \\Sigma_{in} \\dot{E}_{j}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Energy Flow in a System<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider the following system:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1005 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System-300x118.jpg\" alt=\"\" width=\"361\" height=\"142\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System-300x118.jpg 300w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System-65x26.jpg 65w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System-225x89.jpg 225w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System-350x138.jpg 350w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Energy-Flow-in-System.jpg 474w\" sizes=\"auto, (max-width: 361px) 100vw, 361px\" \/><\/p>\n<p>There are 2 streams with energy entering the system (streams 1 and 2), and 2 streams with energy exiting the system (streams 3 and 4).<\/p>\n<p>For this system:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{E}_{1} + \\dot{E}_{2} + \\dot{Q} + \\dot{W} =\u00a0 \\dot{E}_{3} + \\dot{E}_{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>Recall the three forms of energy:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{E}_{j} = \\dot{U}_{j} + \\dot{E}_{k,j} + \\dot{E}_{p,j}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<blockquote><p>Each energy flow term can be further separated into: [latex]\\dot{U}_{j} = \\dot{m} * \\hat{U}_{j}[\/latex]<\/p>\n<p><strong>Specific Property<\/strong> &#8220;[latex]\\hat{ }[\/latex]&#8221; : This denotes an intensive property obtained by dividing an extensive property by a total amount of flow rate (can be total number of moles or total mass)<\/p>\n<p style=\"text-align: center\">[latex]\\hat{V} = \\frac{V}{n}[\/latex]<\/p>\n<p style=\"text-align: center\">or<\/p>\n<p style=\"text-align: center\">[latex]\\hat{U}=\\frac{\\dot{U}}{\\dot{m}}[\/latex]<\/p>\n<\/blockquote>\n<p>Combining all these terms:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} + \\dot{W} = \\Sigma_{out} \\dot{m}_{j} * (\\hat{U}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j}) - \\Sigma_{in} \\dot{m}_{j} * (\\hat{U}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j})[\/latex]<\/p>\n<p>Recall the work terms expansion:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{W} = \\dot{W}_{fl} + \\dot{W}_s[\/latex]<\/p>\n<blockquote><p>where flow work is dependant on system pressure and volume<\/p>\n<p style=\"text-align: center\">[latex]\\dot{W}_{fl} = \\Sigma_{in} \\dot{m}_{j} P_{j} \\hat{V}_{j} - \\Sigma_{out} \\dot{m}_{j} P_{j} \\hat{V}_{j}[\/latex]<\/p>\n<\/blockquote>\n<p>Now we have:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} + \\dot{W}_{s} = \\Sigma_{out} \\dot{m}_{j} * (\\hat{U}_{j} + P_{j} \\hat{V}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j}) - \\Sigma_{in} \\dot{m}_{j}*(\\hat{U}_{j} + P_{j}\\hat{V}_{j} + \\hat{E}_{k,j} + \\hat{E}_{p,j})[\/latex]<\/p>\n<p>Because [latex]\\hat{U} + P\\hat{V}[\/latex] usually appear together in the energy balances, we define them to be &#8220;enthalpy&#8221; ([latex]\\hat{H}[\/latex]):<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H} = \\hat{U} + P\\hat{V}[\/latex]<\/p>\n<blockquote><p>where [latex]\\hat{U}[\/latex] is the internal energy and [latex]P\\hat{V}[\/latex] is the flow work<\/p><\/blockquote>\n<p>The following terms are defined:<\/p>\n<ul>\n<li>[latex]\\Delta\\dot{H} = \\Sigma_{out}\\dot{m}_{j}*\\hat{H}_{j} - \\Sigma_{in}\\dot{m}_{j}*\\hat{H}_{j}[\/latex]<\/li>\n<li>[latex]\\Delta\\dot{E}_{k} = \\Sigma_{out}\\dot{m}_{j}*\\hat{E}_{k,j} - \\Sigma_{in}\\dot{m}_{j}*\\hat{E}_{k,j}[\/latex]<\/li>\n<li>[latex]\\Delta\\dot{E}_{p} = \\Sigma_{out}\\dot{m}_{j}*\\hat{E}_{p,j} - \\Sigma_{in}\\dot{m}_{j}*\\hat{E}_{p,j}[\/latex]<\/li>\n<\/ul>\n<p>Finally, an open system steady-state energy balance is defined:<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} + \\dot{W}_{s} = \\Delta\\dot{H} + \\Delta\\dot{E}_{k} + \\Delta\\dot{E}_{p}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"prompt input_prompt\">\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Heat for an Ideal Gas<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<div class=\"alert alert-block alert-warning\">\n<p>Prior to entering a furnace, air is heated from [latex]25^{\\circ}C[\/latex] to [latex]150^{\\circ}C[\/latex] and the change in specific enthalpy for the whole heating process is 3640 J\/mol. The flow rate of air at the outlet of the heater is [latex]1.5 m^3\/min[\/latex] and the air pressure at this point is 150 kPa absolute.<\/p>\n<p>Calculate the heat needed for the process in kW. Assume the ideal gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<p><strong>Step 1<\/strong>: Calculate the molar flowrate using the ideal gas law.<\/p>\n<p>\\begin{align*}<br \/>\n\\dot{n} &amp;\u00a0 = \\frac{1.5m^{3}}{min}*\\frac{273K}{150+273K}*\\frac{150kPa}{101.3kPa}*\\frac{1mol}{22.4L}*\\frac{10^{3}L}{1m^{3}}\\\\<br \/>\n\\dot{n} &amp; =64.0\\frac{mol}{min}<br \/>\n\\end{align*}<\/p>\n<p><strong>Step 2<\/strong>: Calculate the heat using the specific enthalpy. Since the potential and kinetic energy changes are zero, the following calculations are made<\/p>\n<p>\\begin{align*}<br \/>\n\\dot{Q} &amp; =\\Delta\\dot{H}= \\dot{n}\\Delta\\hat{H}\\\\<br \/>\n\\dot{Q} &amp; =\\frac{64.0 mol}{min}*\\frac{1min}{60s}*\\frac{3640J}{mol}*\\frac{kW}{10^{3} J\/s} \\\\<br \/>\n\\dot{Q} &amp; =3.88kW<br \/>\n\\end{align*}<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"prompt input_prompt\">\n<h2 id=\"Reference-States\">Reference States<\/h2>\n<p><strong>Reference State:<\/strong> a substance at some pressure, temperature, and state of aggregation (solid, liquid, gas; pure or mixture).<\/p>\n<p>It is much easier to estimate the energy of a system as a change from a reference state rather than the absolute energy.<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Cooling in a Heat Exchanger<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<div class=\"alert alert-block alert-warning\">\n<p style=\"text-align: left\">Water is used to cool a liquid in a heat exchanger. Water enters the heat exchanger at [latex]10^{\\circ}C[\/latex] and exits at [latex]100^{\\circ}C[\/latex]. Using the table below, find the change in enthalpy of water in its liquid state.<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 68.4422%;height: 60px\">\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"width: 33.3333%;height: 15px;text-align: center\"><strong>Entry #<\/strong><\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\"><strong>[latex]T (^{\\circ}C)[\/latex]<\/strong><\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\"><strong>[latex]\\hat{H}_{L} (\\frac{kJ}{kg})[\/latex]<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">1<\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">5<\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">21.02<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">2<\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">10<\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">42.02<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">3<\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">100<\/td>\n<td style=\"width: 33.3333%;height: 15px;text-align: center\">419.17<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<div>\n<p><strong>Step 1<\/strong>: Determine which reference state you are going to use. In this case, we are using [latex]10^{\\circ}C[\/latex] as the reference state.<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\hat{H} = \\hat{H}_{100^{\\circ}C} - \\hat{H}_{10^{\\circ}C}[\/latex]<\/p>\n<p><strong>Step 2<\/strong>: Find the change in enthalpy by taking the difference of the system&#8217;s specific enthalpies at different temperatures.<\/p>\n<p>\\begin{align*}<br \/>\n\\Delta\\hat{H} &amp; = \\hat{H}_{3} &#8211; \\hat{H}_{2}\\\\<br \/>\n\\Delta\\hat{H} &amp; = (419.17 &#8211; 42.02) kJ\/kg\\\\<br \/>\n\\Delta\\hat{H} &amp; = 377.15 kJ\/kg<br \/>\n\\end{align*}<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div><\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h2 id=\"Steam-Tables\">Steam Tables<\/h2>\n<p>Since water is a commonly used resource in processes for heating and cooling, detailed information on its state properties at different temperatures and pressures is available.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\">\n<div class=\"textbox shaded\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h3 id=\"How-to-Access-Steam-Tables-on-NIST\">How to Access Steam Tables on NIST<\/h3>\n<ul>\n<li>Steam tables can be found on <a href=\"https:\/\/webbook.nist.gov\/chemistry\/fluid\/\">NIST<\/a><\/li>\n<\/ul>\n<p>A. Select &#8216;Water&#8217; from the &#8216;Please select the species of interest:&#8217; drop-down menu<\/p>\n<p>B. Choose the steam table units you&#8217;d like to work with, in step 2.<\/p>\n<p>C. In Step 3, choose what kind of data you&#8217;re looking to obtain. For an isothermal system, select &#8216;Isothermal properties&#8217;. For a constant pressure system, select &#8216;Isobaric properties&#8217;.<\/p>\n<p>D. Select the desired standard state convention. This course will most likely only use the &#8216;Default for fluid&#8217; convention.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Steam Tables<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><span style=\"color: #333333;font-size: 1em\">Superheated steam at 40 bar absolute and [latex]500^{\\circ}C[\/latex] flowing at a rate of [latex]200 kg\/min[\/latex] is sent to an adiabatic turbine which expands to 5 bar. The turbine outputs [latex]1250kW[\/latex]. The expanded steam is then sent to a heat exchanger where isobaric heating occurs, resulting in the stream being reheated to its initial temperature. Assume no changes in kinetic energy. Write the energy balance for the turbine and determine the outlet stream temperature.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<div><strong>Step 1:<\/strong> Determine the enthalpy values for water vapor at\u00a0 [latex]500^{\\circ}C[\/latex] and [latex]40 bar[\/latex] and [latex]5 bar[\/latex].<\/div>\n<div>From the steam tables:<\/div>\n<div><\/div>\n<ul>\n<li>For water vapor at [latex]500^{\\circ}C[\/latex] and [latex]40 bar[\/latex], the specific enthalpy\u00a0 is [latex]3445 \\;kJ\/kg[\/latex]<\/li>\n<li>\u00a0For water vapor at [latex]500^{\\circ}C[\/latex] and [latex]5 bar[\/latex], the specific enthalpy\u00a0 is [latex]3484 \\;kJ\/kg[\/latex]<\/li>\n<\/ul>\n<div><\/div>\n<div><strong>Step 2:<\/strong> Write the energy balance. Since there are no changes in potential and kinetic energy and no heat transfer, the change in enthalpy will be equal to a negative shaft work.<\/div>\n<div>\\begin{align*}<br \/>\n\\Delta\\dot{H} &amp;=-\\dot{W}_{s}\\\\<br \/>\n\\Delta\\dot{H} &amp;=\\dot{m}*(\\hat{H}_{2}-\\hat{H}_{1})<br \/>\n\\end{align*}<\/div>\n<div>\\begin{align*}<br \/>\n\\hat{H}_{2} &amp;=\\hat{H}_{1}-\\frac{\\dot{W}_{s}}{\\dot{m}}\\\\<br \/>\n&amp;=\\frac{3445kJ}{kg}-\\frac{1250kJ}{s}*\\frac{min}{200kg}*\\frac{60s}{1min}\\\\<br \/>\n&amp;=3070\\frac{kJ}{kg}<br \/>\n\\end{align*}<\/div>\n<div><\/div>\n<div><strong>Step 3:<\/strong> Determine the temperature of the steam corresponding to [latex]5bar[\/latex] and [latex]3070 kJ\/kg[\/latex].<\/div>\n<div>From the steam tables:<\/div>\n<div><\/div>\n<div style=\"text-align: center\">[latex]T=302^{\\circ}C[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":949,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1315","chapter","type-chapter","status-publish","hentry"],"part":1313,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1315","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/users\/949"}],"replies":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/comments?post=1315"}],"version-history":[{"count":11,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1315\/revisions"}],"predecessor-version":[{"id":2806,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1315\/revisions\/2806"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/parts\/1313"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1315\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/media?parent=1315"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapter-type?post=1315"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/contributor?post=1315"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/license?post=1315"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}