{"id":79,"date":"2017-03-09T20:48:21","date_gmt":"2017-03-10T01:48:21","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/tpps\/?post_type=chapter&#038;p=79"},"modified":"2018-02-05T15:39:04","modified_gmt":"2018-02-05T20:39:04","slug":"combustion-analysis","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/tpps\/chapter\/combustion-analysis\/","title":{"raw":"Combustion Analysis","rendered":"Combustion Analysis"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nOperate the Plant at 80% capacity burning coal to\r\n<ul>\r\n \t<li>Perform combustion analyses for two types of coal,<\/li>\r\n \t<li>Compare results.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>Theory<\/h1>\r\nIn the <a href=\"https:\/\/pressbooks.bccampus.ca\/tpps\/chapter\/boiler-efficiency\/\" target=\"_blank\" rel=\"noopener\">Boiler Efficiency<\/a> lab, we stated that Combustion Efficiency is defined as the ratio of the burner\u2019s capability to burn fuel completely to the unburned fuel and excess air in the exhaust. In this lab, we will perform a combustion analysis.\r\n\r\nFossil fuels may be classified into solid, liquid and gaseous fuels. The vast majority of fuels are based on carbon (C), hydrogen (H<sub>2<\/sub>) or some combination of carbon and hydrogen called hydrocarbons.\r\n\r\nDuring combustion, oxygen (O<sub>2<\/sub>) combines rapidly with C, H<sub>2<\/sub>, sulphur (S<sub>2<\/sub>) and their compounds in solid, liquid and gaseous fuels and results in the liberation of energy. Except for special applications such as oxyacetylene welding, in which a high-temperature flame is required, the O<sub>2 <\/sub>necessary for combustion is obtained from air. Air contains O<sub>2<\/sub> and nitrogen (N<sub>2<\/sub>), plus negligible amounts of other gasses and for engineering purposes, may be considered to have the following percentage composition by mass:\r\n<p style=\"text-align: center\">O<sub>2<\/sub>: 23%\r\nN<sub>2<\/sub>: 77%<\/p>\r\nThe proportions in which the elements enter into the combustion reaction by mass are dependent upon the relative molecular weights as shown below:\r\n<table style=\"width: 370.267px\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 112px\"><strong>Element<\/strong><\/td>\r\n<td style=\"width: 112px\"><strong>Symbol<\/strong><\/td>\r\n<td style=\"width: 112px\"><strong>Molecular Weight<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 112px\">Carbon<\/td>\r\n<td style=\"width: 112px\">C<\/td>\r\n<td style=\"width: 112.267px\">12<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 112px\">Sulphur<\/td>\r\n<td style=\"width: 112px\">S<sub>2<\/sub><\/td>\r\n<td style=\"width: 112.267px\">32<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 112px\">Hydrogen<\/td>\r\n<td style=\"width: 112px\">H<sub>2<\/sub><\/td>\r\n<td style=\"width: 112.267px\">2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 112px\">Oxygen<\/td>\r\n<td style=\"width: 112px\">O<sub>2<\/sub><\/td>\r\n<td style=\"width: 112.267px\">32<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 112px\">Nitrogen<\/td>\r\n<td style=\"width: 112px\">N<sub>2<\/sub><\/td>\r\n<td style=\"width: 112.267px\">28<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h1>Stoichiometric Combustion Theory<\/h1>\r\nComplete combustion of simple hydrocarbon fuels forms carbon dioxide (C0<sub>2<\/sub>) from the carbon and water (H<sub>2<\/sub>0) from the hydrogen, so for a hydrocarbon fuel with the general composition C<sub>n<\/sub>H<sub>m<\/sub>, the combustion equation on a molar basis is as flows:\r\n<p style=\"text-align: center\"><span>[latex]C_{n}H_{m}+yO_{2}\\rightarrow aCO_{2}+bH_{2}O[\/latex]<\/span><\/p>\r\nWhere the balance should be satisfied following the moles for any mathematcial\u00a0 equation:\r\n\r\nCarbon balance:\r\n<p style=\"text-align: center\"><span>[latex]a=n[\/latex] kmol CO<sub>2<\/sub>\/ kmol fuel<\/span><\/p>\r\n<p style=\"text-align: left\">Hydrogen balance:<\/p>\r\n<p style=\"text-align: center\"><span>[latex]2b=m[\/latex]\u00a0<\/span><\/p>\r\n<p style=\"text-align: center\"><span>[latex]b = \\frac{m}{2}[\/latex]\u00a0<\/span>kmol H<sub>2<\/sub>O\/ kmol fuel<\/p>\r\nOxygen balance:\r\n<p style=\"text-align: center\"><span>[latex]2y = 2a+b[\/latex]<\/span><\/p>\r\n<p style=\"text-align: center\"><span>[latex]y = a+\\frac{b}{2}[\/latex]\u00a0kmol O<sub>2<\/sub>\/ kmol fuel\u00a0<\/span><\/p>\r\nConsidering that combustion occurs in air rather than in pure oxygen, the nitrogen in the air may react in the combustion process to produce nitrogen oxides. Beside, some fuels contain elements other than carbon, and these elements may react with oxygen during combustion. Also, combustion is not always complete, and the exhaust gases contain unburned and partially burned products in addition to C0<sub>2<\/sub> and H<sub>2<\/sub>O.\r\n\r\nAir is composed of oxygen, nitrogen, and small amounts of carbon dioxide, argon, and other trace components. For the purposes of the further calculation it is perfectly reasonable to consider air as a mixture of 21% (mole basis) 02 and 79 % (mole basis) N<sub>2<\/sub>.\u00a0 Nitrogen will be considered as an \u201cinert\" gas in the combustion calculations. The stoichiometric relation for complete combustion of a hydrocarbon fuel, C<sub>n<\/sub>H<sub>m<\/sub>, becomes\r\n<p style=\"text-align: center\"><span>[latex]C_{n}H_{m}+y(O_{2}+\\frac{79}{21}N_{2}\\rightarrow aCO_{2}+bH_{2}O+cN_{2}[\/latex]<\/span><\/p>\r\nThe balance equations are:\r\n\r\nCarbon balance:\r\n<p style=\"text-align: center\"><span>[latex]a=n[\/latex] kmol CO<sub>2<\/sub>\/ kmol fuel<\/span><\/p>\r\n<p style=\"text-align: left\">Hydrogen balance:<\/p>\r\n<p style=\"text-align: center\"><span>[latex]2b=m[\/latex]\u00a0<\/span><\/p>\r\n<p style=\"text-align: center\"><span>[latex]b = \\frac{m}{2}[\/latex]\u00a0<\/span>kmol H<sub>2<\/sub>O\/ kmol fuel<\/p>\r\nOxygen balance:\r\n<p style=\"text-align: center\"><span>[latex]2y = 2a+b[\/latex]<\/span><\/p>\r\n<p style=\"text-align: center\"><span>[latex]y = a+\\frac{b}{2}[\/latex]\u00a0kmol O<sub>2<\/sub>\/ kmol fuel\u00a0<\/span><\/p>\r\nNitrogen balance:\r\n<p style=\"text-align: center\"><span>[latex]y\\frac{79}{21}=c[\/latex]\u00a0kmol N<sub>2<\/sub>\/kmol fuel<\/span><\/p>\r\nFlue gas compositions are presented in terms of mole fractions as kmol of product per kmol of fuel.\r\n\r\nExample 1.\u00a0 Combustion of Octane in Air\r\n\r\nDetermine the stoichiometric air\/ fuel mass ratio and product gas composition for combustion of octane ( C<sub>8<\/sub>H<sub>18<\/sub>)\u00a0 in air.\r\n<p style=\"text-align: center\"><span>[latex]C_{8}H_{18}+y(O_{2}+\\frac{79}{21}N_{2}\\rightarrow aCO_{2}+bH_{2}O+cN_{2}[\/latex]<\/span><\/p>\r\nCarbon balance:\r\n<p style=\"text-align: center\"><span>[latex]a=n=8[\/latex] kmol CO<sub>2<\/sub>\/ kmol fuel<\/span><\/p>\r\n<p style=\"text-align: left\">Hydrogen balance:<\/p>\r\n<p style=\"text-align: center\"><span>[latex]2b=18[\/latex]\u00a0<\/span><\/p>\r\n<p style=\"text-align: center\"><span>[latex]b = \\frac{18}{2}=9[\/latex]\u00a0<\/span>kmol H<sub>2<\/sub>O\/ kmol fuel<\/p>\r\nOxygen balance:\r\n<p style=\"text-align: center\"><span>[latex]2y = 2(8)+b[\/latex]<\/span><\/p>\r\n<p style=\"text-align: center\"><span>[latex]y = 8+\\frac{9}{2}=12.5[\/latex]\u00a0kmol O<sub>2<\/sub>\/ kmol fuel\u00a0<\/span><\/p>\r\nNitrogen balance:\r\n<p style=\"text-align: center\"><span>[latex]y\\frac{79}{21}=c=12.5\\frac{79}{21}=47[\/latex]\u00a0kmol N<sub>2<\/sub>\/kmol fuel<\/span><\/p>\r\nThe combustion equation becomes:\r\n<p style=\"text-align: center\"><span>[latex]C_{8}H_{18}+12.5(O_{2}+\\frac{79}{21}N_{2}\\rightarrow 8CO_{2}+9H_{2}O+47N_{2}[\/latex]<\/span><\/p>\r\n<p style=\"text-align: left\">Air\/ fuel mass based ratio considering that 1 kmol fuel is 114 kg of fuel ( 8*12 + 18 *(1) = 114):<\/p>\r\n<p style=\"text-align: center\"><span>[latex]12.5\\frac{kmol\\ O_{2}}{kmol\\ fuel}\\frac{1\\ kmol\\ fuel}{114\\ kg\\ fuel}32\\frac{kg\\ O_{2}}{kmol\\ O_{2}}\\frac{100\\ kg\\ air}{23\\ kg\\ O_{2}}=15.25\\frac{kg\\ air}{kg\\ fuel}[\/latex]<\/span><\/p>\r\n<p style=\"text-align: left\">Flue gas composition on molar basis is:<\/p>\r\nTotal number of kmol of flue gasses = kmol CO<sub>2<\/sub> \u00a0+ kmol H<sub>2<\/sub>O + kmol N<sub>2<\/sub> = 8 +9+ 47 = 64 kmol flue gasses\/ kmol fuel\r\n\r\nCO<sub>2<\/sub> \u00a0= 8\/64 = 12.5 %\r\n\r\nH<sub>2<\/sub>O = 9\/64 = 14 %\r\n<p style=\"text-align: left\">N<sub>2<\/sub> = 47\/64 = 73.5%<\/p>\r\nOther components and impurities in the fuel make the calculation process more complicated. For example if sulfur exists in fuel it is usually combust into sulfur dioxide (SO<sub>2<\/sub>). Ash, the noncombustible inorganic (mineral) impurities in the fuel, undergoes a number of transformations at combustion temperatures, will be neglected in the further calculation (ash will be assumed to be inert).\r\n\r\nFor most solid and liquid fuels, the chemical composition is on a mass basis, as determined in the ultimate analysis.\r\n<h2>Mass Based Chemistry of Combustion<\/h2>\r\nThe combustion reactions are written following the stoichiometric rules as defined above. The quantity of matter entering into a reaction is equal to the quantity of matter in the products of the reaction.\r\n\r\nThe reaction for the complete combustion of C may be written as follows:\r\n<p style=\"text-align: center\">C+O<sub>2<\/sub>=CO<sub>2<\/sub><\/p>\r\nor, if molecular weights are used,\r\n<p style=\"text-align: center\">12+32=44<\/p>\r\n<p style=\"text-align: center\">1 kg C + 2\u2154 kg O<sub>2<\/sub> = 3\u2154 kg CO<sub>2<\/sub><\/p>\r\nThe complete combustion of H<sub>2<\/sub> occurs as follows:\r\n<p style=\"text-align: center\">2H<sub>2<\/sub> +O<sub>2<\/sub> =2H<sub>2<\/sub>O<\/p>\r\n<p style=\"text-align: center\">4+32=36<\/p>\r\n<p style=\"text-align: center\">1 kg H<sub>2<\/sub> + 8 kg O<sub>2<\/sub> = 9 kg H<sub>2<\/sub>O<\/p>\r\n<p style=\"text-align: left\">Sulphur burns as follows:<\/p>\r\n<p style=\"text-align: center\">S +O<sub>2<\/sub> =SO<sub>2<\/sub><\/p>\r\n<p style=\"text-align: center\">32+32=64<\/p>\r\n<p style=\"text-align: center\">1 kg S + 1 kg O<sub>2<\/sub> = 2 kg SO<sub>2<\/sub><\/p>\r\nIn addition, 1 kg of O<sub>2<\/sub> (<strong>stoichiometric mass of O<sub>2<\/sub><\/strong>) is contained in 1\/0.232=4.3 kg air which is the <strong>stoichiometric mass of air<\/strong>. This air will contain 4.3-1=3.3 kg N<sub>2<\/sub>. Therefore we can write:\r\n<p style=\"text-align: center\">1 kg S + 4.3 kg air = 2 kg SO<sub>2<\/sub>+3.3 kg N<sub>2<\/sub><\/p>\r\n\r\n<h1 style=\"text-align: left\">Procedure<\/h1>\r\n<p style=\"text-align: left\">If the analysis of fuel is given by mass, follow the steps below:<\/p>\r\n\r\n<ol>\r\n \t<li style=\"text-align: left\"><strong>Total O<sub>2<\/sub> required:<\/strong> Determine the mass of O<sub>2<\/sub> required for each constituent and find the total mass of O<sub>2 <\/sub>(Subtract any O<sub>2<\/sub> which may be in the fuel)<\/li>\r\n \t<li style=\"text-align: left\"><strong>Stoichiometric air: <\/strong>Stoichiometric mass of air = O<sub>2<\/sub> required\/0.232<\/li>\r\n \t<li style=\"text-align: left\"><strong>Total<\/strong><strong> mass of combustion products:<\/strong> Determine the mass of each combustion product. For example, given C content of 84.9%, CO<sub>2<\/sub>=84.9\/100*3\u2154=3.11% and find the total mass of combustion products.<\/li>\r\n \t<li style=\"text-align: left\"><strong>Analysis of combustion products by mass: <\/strong>Suppose the total mass of combustion products in step 3 has been found as 12.09 kg\/kg fuel, then CO<sub>2<\/sub>=3.11\/12.09*100=25.74. This means that 25.74% of the flue gas is CO<sub>2<\/sub>. Repeat this calculation for each constituent.<\/li>\r\n<\/ol>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Lab Instructions<\/h3>\r\nRun the initial condition I14 80% Coal and setup trends for the following variables:\r\n<p style=\"text-align: center\">X00820 X00821 X00822 X00823 X00824 E23356<\/p>\r\n<p style=\"text-align: center\">G02196 G02197 X32419 X02419 G00831 H00830<\/p>\r\n\r\n<ol>\r\n \t<li><strong>Burning default coal:<\/strong> After 10 minutes of running the simulator, freeze simulator and print the two trends. This is the reference point for the next step.<\/li>\r\n \t<li><strong>Burning poor quality<\/strong> <strong>coal:<\/strong> Switch to run mode and access Variable List page 0111 on MD180. Set the new values as shown below. After 10 minutes, freeze simulator and print the two trends.\r\n<ul>\r\n \t<li>X00820: 70.40<\/li>\r\n \t<li>X00821: 5.10<\/li>\r\n \t<li>X00822: 1.10<\/li>\r\n \t<li>X00823: 12.50<\/li>\r\n \t<li>X00824: 1.60<\/li>\r\n \t<li>X00825: 9.30<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><strong>Computation:<\/strong> Compute the following values for both types of fuels:\r\n<ul>\r\n \t<li>Total O<sub>2<\/sub> required<\/li>\r\n \t<li>Stoichiometric air<\/li>\r\n \t<li>Total mass of combustion products<\/li>\r\n \t<li>Analysis of combustion products by mass [%]: CO<sub>2<\/sub>, H<sub>2<\/sub>O, SO<sub>2<\/sub>, N<sub>2<\/sub><\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><strong>Comparison:<\/strong> Compare your findings based on the following data:\r\n<ul>\r\n \t<li>Furnace outlet SO<sub>x<\/sub> flow<\/li>\r\n \t<li>Furnace outlet NO<sub>x<\/sub> flow<\/li>\r\n \t<li>CO content in flue gas<\/li>\r\n \t<li>Oxygen content in flue gas<\/li>\r\n \t<li>Theoretical combustion air<\/li>\r\n \t<li>Coal Heat Value<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2 style=\"text-align: left\">Hints &amp; Tips<\/h2>\r\nIn this lab, you are carrying out two combustion analyses. For data collection, set up your trends, a sample trend plot is shown below (Make sure your trend printouts are labeled properly otherwise, data analysis will be very confusing):\r\n\r\n[caption id=\"attachment_55\" align=\"aligncenter\" width=\"658\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal.jpg\" alt=\"Sample data for coal\" width=\"658\" height=\"520\" class=\"size-full wp-image-55\" \/><\/a> Sample data for coal[\/caption]\r\n\r\nTo change the fuel composition use the Variable List Page#: 0111 on MD180:\r\n\r\n[caption id=\"attachment_49\" align=\"aligncenter\" width=\"711\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel.jpg\" alt=\"Chemical composition of coal\" width=\"711\" height=\"351\" class=\"size-full wp-image-49\" \/><\/a> Chemical composition of coal[\/caption]\r\n\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Deliverables<\/h3>\r\nYour lab report is to include the following:\r\n<ul>\r\n \t<li><strong>Trend plots:<\/strong> Supply all plots taken for each of the 2 fuels,<\/li>\r\n \t<li><strong>Computation:<\/strong> <span>As per lab instructions above, perform combustion analyses u<\/span>sing MATLAB or MS Excel.<\/li>\r\n \t<li><span><strong>Conclusion:<\/strong> Write a summary (max. 500 words<span itemscope=\"\" itemtype=\"http:\/\/schema.org\/WebPage\">, in a text box if using Excel<\/span>) comparing your results and suggestions for further study.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n\r\nFurther Reading:\r\n<ul>\r\n \t<li>Basic Engineering Thermodynamics in SI Units by R. Joel: Combustion.<\/li>\r\n \t<li>Thermal Engineering by H.L. Solberg, O.C. Cromer and A.R. Spalding: Fossil fuels and their combustion.<\/li>\r\n \t<li>\r\n<p id=\"firstHeading\" class=\"firstHeading\" lang=\"en\"><a href=\"https:\/\/en.wikipedia.org\/wiki\/Stoichiometry\" target=\"_blank\" rel=\"noopener\">Stoichiometry. <\/a><\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>Operate the Plant at 80% capacity burning coal to<\/p>\n<ul>\n<li>Perform combustion analyses for two types of coal,<\/li>\n<li>Compare results.<\/li>\n<\/ul>\n<\/div>\n<h1>Theory<\/h1>\n<p>In the <a href=\"https:\/\/pressbooks.bccampus.ca\/tpps\/chapter\/boiler-efficiency\/\" target=\"_blank\" rel=\"noopener\">Boiler Efficiency<\/a> lab, we stated that Combustion Efficiency is defined as the ratio of the burner\u2019s capability to burn fuel completely to the unburned fuel and excess air in the exhaust. In this lab, we will perform a combustion analysis.<\/p>\n<p>Fossil fuels may be classified into solid, liquid and gaseous fuels. The vast majority of fuels are based on carbon (C), hydrogen (H<sub>2<\/sub>) or some combination of carbon and hydrogen called hydrocarbons.<\/p>\n<p>During combustion, oxygen (O<sub>2<\/sub>) combines rapidly with C, H<sub>2<\/sub>, sulphur (S<sub>2<\/sub>) and their compounds in solid, liquid and gaseous fuels and results in the liberation of energy. Except for special applications such as oxyacetylene welding, in which a high-temperature flame is required, the O<sub>2 <\/sub>necessary for combustion is obtained from air. Air contains O<sub>2<\/sub> and nitrogen (N<sub>2<\/sub>), plus negligible amounts of other gasses and for engineering purposes, may be considered to have the following percentage composition by mass:<\/p>\n<p style=\"text-align: center\">O<sub>2<\/sub>: 23%<br \/>\nN<sub>2<\/sub>: 77%<\/p>\n<p>The proportions in which the elements enter into the combustion reaction by mass are dependent upon the relative molecular weights as shown below:<\/p>\n<table style=\"width: 370.267px\">\n<tbody>\n<tr>\n<td style=\"width: 112px\"><strong>Element<\/strong><\/td>\n<td style=\"width: 112px\"><strong>Symbol<\/strong><\/td>\n<td style=\"width: 112px\"><strong>Molecular Weight<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 112px\">Carbon<\/td>\n<td style=\"width: 112px\">C<\/td>\n<td style=\"width: 112.267px\">12<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 112px\">Sulphur<\/td>\n<td style=\"width: 112px\">S<sub>2<\/sub><\/td>\n<td style=\"width: 112.267px\">32<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 112px\">Hydrogen<\/td>\n<td style=\"width: 112px\">H<sub>2<\/sub><\/td>\n<td style=\"width: 112.267px\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 112px\">Oxygen<\/td>\n<td style=\"width: 112px\">O<sub>2<\/sub><\/td>\n<td style=\"width: 112.267px\">32<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 112px\">Nitrogen<\/td>\n<td style=\"width: 112px\">N<sub>2<\/sub><\/td>\n<td style=\"width: 112.267px\">28<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h1>Stoichiometric Combustion Theory<\/h1>\n<p>Complete combustion of simple hydrocarbon fuels forms carbon dioxide (C0<sub>2<\/sub>) from the carbon and water (H<sub>2<\/sub>0) from the hydrogen, so for a hydrocarbon fuel with the general composition C<sub>n<\/sub>H<sub>m<\/sub>, the combustion equation on a molar basis is as flows:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-b80726d7f0de3f62e96ced05ccde22e8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#67;&#95;&#123;&#110;&#125;&#72;&#95;&#123;&#109;&#125;&#43;&#121;&#79;&#95;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#32;&#97;&#67;&#79;&#95;&#123;&#50;&#125;&#43;&#98;&#72;&#95;&#123;&#50;&#125;&#79;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"240\" style=\"vertical-align: -4px;\" \/><\/span><\/p>\n<p>Where the balance should be satisfied following the moles for any mathematcial\u00a0 equation:<\/p>\n<p>Carbon balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-218bd0726d317ef049496a67493b6515_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#110;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"44\" style=\"vertical-align: 0px;\" \/> kmol CO<sub>2<\/sub>\/ kmol fuel<\/span><\/p>\n<p style=\"text-align: left\">Hydrogen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-52bd181f139c085ef0d24d25fe849137_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#98;&#61;&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"55\" style=\"vertical-align: 0px;\" \/>\u00a0<\/span><\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-7327346b10360f550a91e7d834c77b0b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#98;&#32;&#61;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#109;&#125;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"46\" style=\"vertical-align: -6px;\" \/>\u00a0<\/span>kmol H<sub>2<\/sub>O\/ kmol fuel<\/p>\n<p>Oxygen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-5893cc1de8304bd1a9825170a77a1d36_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#121;&#32;&#61;&#32;&#50;&#97;&#43;&#98;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"90\" style=\"vertical-align: -4px;\" \/><\/span><\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-12dce3c71b1c7cbdd4dcdee17ba777df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#121;&#32;&#61;&#32;&#97;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#98;&#125;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"73\" style=\"vertical-align: -6px;\" \/>\u00a0kmol O<sub>2<\/sub>\/ kmol fuel\u00a0<\/span><\/p>\n<p>Considering that combustion occurs in air rather than in pure oxygen, the nitrogen in the air may react in the combustion process to produce nitrogen oxides. Beside, some fuels contain elements other than carbon, and these elements may react with oxygen during combustion. Also, combustion is not always complete, and the exhaust gases contain unburned and partially burned products in addition to C0<sub>2<\/sub> and H<sub>2<\/sub>O.<\/p>\n<p>Air is composed of oxygen, nitrogen, and small amounts of carbon dioxide, argon, and other trace components. For the purposes of the further calculation it is perfectly reasonable to consider air as a mixture of 21% (mole basis) 02 and 79 % (mole basis) N<sub>2<\/sub>.\u00a0 Nitrogen will be considered as an \u201cinert&#8221; gas in the combustion calculations. The stoichiometric relation for complete combustion of a hydrocarbon fuel, C<sub>n<\/sub>H<sub>m<\/sub>, becomes<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-96714411f252ebf2def5e587679d208b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#67;&#95;&#123;&#110;&#125;&#72;&#95;&#123;&#109;&#125;&#43;&#121;&#40;&#79;&#95;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#55;&#57;&#125;&#123;&#50;&#49;&#125;&#78;&#95;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#32;&#97;&#67;&#79;&#95;&#123;&#50;&#125;&#43;&#98;&#72;&#95;&#123;&#50;&#125;&#79;&#43;&#99;&#78;&#95;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"358\" style=\"vertical-align: -7px;\" \/><\/span><\/p>\n<p>The balance equations are:<\/p>\n<p>Carbon balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-218bd0726d317ef049496a67493b6515_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#110;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"44\" style=\"vertical-align: 0px;\" \/> kmol CO<sub>2<\/sub>\/ kmol fuel<\/span><\/p>\n<p style=\"text-align: left\">Hydrogen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-52bd181f139c085ef0d24d25fe849137_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#98;&#61;&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"55\" style=\"vertical-align: 0px;\" \/>\u00a0<\/span><\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-7327346b10360f550a91e7d834c77b0b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#98;&#32;&#61;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#109;&#125;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"46\" style=\"vertical-align: -6px;\" \/>\u00a0<\/span>kmol H<sub>2<\/sub>O\/ kmol fuel<\/p>\n<p>Oxygen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-5893cc1de8304bd1a9825170a77a1d36_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#121;&#32;&#61;&#32;&#50;&#97;&#43;&#98;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"90\" style=\"vertical-align: -4px;\" \/><\/span><\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-12dce3c71b1c7cbdd4dcdee17ba777df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#121;&#32;&#61;&#32;&#97;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#98;&#125;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"73\" style=\"vertical-align: -6px;\" \/>\u00a0kmol O<sub>2<\/sub>\/ kmol fuel\u00a0<\/span><\/p>\n<p>Nitrogen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-4e7b51c1d67d3e3a983de0610e0406a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#121;&#92;&#102;&#114;&#97;&#99;&#123;&#55;&#57;&#125;&#123;&#50;&#49;&#125;&#61;&#99;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"59\" style=\"vertical-align: -7px;\" \/>\u00a0kmol N<sub>2<\/sub>\/kmol fuel<\/span><\/p>\n<p>Flue gas compositions are presented in terms of mole fractions as kmol of product per kmol of fuel.<\/p>\n<p>Example 1.\u00a0 Combustion of Octane in Air<\/p>\n<p>Determine the stoichiometric air\/ fuel mass ratio and product gas composition for combustion of octane ( C<sub>8<\/sub>H<sub>18<\/sub>)\u00a0 in air.<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-679170c81d5ad2600a5e138cd5770f0f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#67;&#95;&#123;&#56;&#125;&#72;&#95;&#123;&#49;&#56;&#125;&#43;&#121;&#40;&#79;&#95;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#55;&#57;&#125;&#123;&#50;&#49;&#125;&#78;&#95;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#32;&#97;&#67;&#79;&#95;&#123;&#50;&#125;&#43;&#98;&#72;&#95;&#123;&#50;&#125;&#79;&#43;&#99;&#78;&#95;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"358\" style=\"vertical-align: -7px;\" \/><\/span><\/p>\n<p>Carbon balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-cf1f5109f086d0e5fb4b57c03d49363a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#110;&#61;&#56;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"76\" style=\"vertical-align: 0px;\" \/> kmol CO<sub>2<\/sub>\/ kmol fuel<\/span><\/p>\n<p style=\"text-align: left\">Hydrogen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-622d5cc260cb708ccdb4f8449c676b7b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#98;&#61;&#49;&#56;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"58\" style=\"vertical-align: -1px;\" \/>\u00a0<\/span><\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-9d2169b65ff1a6c863549067c619cac8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#98;&#32;&#61;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#56;&#125;&#123;&#50;&#125;&#61;&#57;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"82\" style=\"vertical-align: -6px;\" \/>\u00a0<\/span>kmol H<sub>2<\/sub>O\/ kmol fuel<\/p>\n<p>Oxygen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-7dad087bfcac524419f02b3fb4fb48df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#121;&#32;&#61;&#32;&#50;&#40;&#56;&#41;&#43;&#98;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"103\" style=\"vertical-align: -4px;\" \/><\/span><\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-37074a66a8771aa75696923800617578_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#121;&#32;&#61;&#32;&#56;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#57;&#125;&#123;&#50;&#125;&#61;&#49;&#50;&#46;&#53;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"129\" style=\"vertical-align: -6px;\" \/>\u00a0kmol O<sub>2<\/sub>\/ kmol fuel\u00a0<\/span><\/p>\n<p>Nitrogen balance:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-a801db2fda2fea88f56da4d727760dc2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#121;&#92;&#102;&#114;&#97;&#99;&#123;&#55;&#57;&#125;&#123;&#50;&#49;&#125;&#61;&#99;&#61;&#49;&#50;&#46;&#53;&#92;&#102;&#114;&#97;&#99;&#123;&#55;&#57;&#125;&#123;&#50;&#49;&#125;&#61;&#52;&#55;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"173\" style=\"vertical-align: -7px;\" \/>\u00a0kmol N<sub>2<\/sub>\/kmol fuel<\/span><\/p>\n<p>The combustion equation becomes:<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-348efb91b6bcb49d9cfe0d1bef33ed1a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#67;&#95;&#123;&#56;&#125;&#72;&#95;&#123;&#49;&#56;&#125;&#43;&#49;&#50;&#46;&#53;&#40;&#79;&#95;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#55;&#57;&#125;&#123;&#50;&#49;&#125;&#78;&#95;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#32;&#56;&#67;&#79;&#95;&#123;&#50;&#125;&#43;&#57;&#72;&#95;&#123;&#50;&#125;&#79;&#43;&#52;&#55;&#78;&#95;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"391\" style=\"vertical-align: -7px;\" \/><\/span><\/p>\n<p style=\"text-align: left\">Air\/ fuel mass based ratio considering that 1 kmol fuel is 114 kg of fuel ( 8*12 + 18 *(1) = 114):<\/p>\n<p style=\"text-align: center\"><span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/ql-cache\/quicklatex.com-fe911d97390656f1cd391f4911326c62_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#50;&#46;&#53;&#92;&#102;&#114;&#97;&#99;&#123;&#107;&#109;&#111;&#108;&#92;&#32;&#79;&#95;&#123;&#50;&#125;&#125;&#123;&#107;&#109;&#111;&#108;&#92;&#32;&#102;&#117;&#101;&#108;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#92;&#32;&#107;&#109;&#111;&#108;&#92;&#32;&#102;&#117;&#101;&#108;&#125;&#123;&#49;&#49;&#52;&#92;&#32;&#107;&#103;&#92;&#32;&#102;&#117;&#101;&#108;&#125;&#51;&#50;&#92;&#102;&#114;&#97;&#99;&#123;&#107;&#103;&#92;&#32;&#79;&#95;&#123;&#50;&#125;&#125;&#123;&#107;&#109;&#111;&#108;&#92;&#32;&#79;&#95;&#123;&#50;&#125;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#48;&#48;&#92;&#32;&#107;&#103;&#92;&#32;&#97;&#105;&#114;&#125;&#123;&#50;&#51;&#92;&#32;&#107;&#103;&#92;&#32;&#79;&#95;&#123;&#50;&#125;&#125;&#61;&#49;&#53;&#46;&#50;&#53;&#92;&#102;&#114;&#97;&#99;&#123;&#107;&#103;&#92;&#32;&#97;&#105;&#114;&#125;&#123;&#107;&#103;&#92;&#32;&#102;&#117;&#101;&#108;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"436\" style=\"vertical-align: -9px;\" \/><\/span><\/p>\n<p style=\"text-align: left\">Flue gas composition on molar basis is:<\/p>\n<p>Total number of kmol of flue gasses = kmol CO<sub>2<\/sub> \u00a0+ kmol H<sub>2<\/sub>O + kmol N<sub>2<\/sub> = 8 +9+ 47 = 64 kmol flue gasses\/ kmol fuel<\/p>\n<p>CO<sub>2<\/sub> \u00a0= 8\/64 = 12.5 %<\/p>\n<p>H<sub>2<\/sub>O = 9\/64 = 14 %<\/p>\n<p style=\"text-align: left\">N<sub>2<\/sub> = 47\/64 = 73.5%<\/p>\n<p>Other components and impurities in the fuel make the calculation process more complicated. For example if sulfur exists in fuel it is usually combust into sulfur dioxide (SO<sub>2<\/sub>). Ash, the noncombustible inorganic (mineral) impurities in the fuel, undergoes a number of transformations at combustion temperatures, will be neglected in the further calculation (ash will be assumed to be inert).<\/p>\n<p>For most solid and liquid fuels, the chemical composition is on a mass basis, as determined in the ultimate analysis.<\/p>\n<h2>Mass Based Chemistry of Combustion<\/h2>\n<p>The combustion reactions are written following the stoichiometric rules as defined above. The quantity of matter entering into a reaction is equal to the quantity of matter in the products of the reaction.<\/p>\n<p>The reaction for the complete combustion of C may be written as follows:<\/p>\n<p style=\"text-align: center\">C+O<sub>2<\/sub>=CO<sub>2<\/sub><\/p>\n<p>or, if molecular weights are used,<\/p>\n<p style=\"text-align: center\">12+32=44<\/p>\n<p style=\"text-align: center\">1 kg C + 2\u2154 kg O<sub>2<\/sub> = 3\u2154 kg CO<sub>2<\/sub><\/p>\n<p>The complete combustion of H<sub>2<\/sub> occurs as follows:<\/p>\n<p style=\"text-align: center\">2H<sub>2<\/sub> +O<sub>2<\/sub> =2H<sub>2<\/sub>O<\/p>\n<p style=\"text-align: center\">4+32=36<\/p>\n<p style=\"text-align: center\">1 kg H<sub>2<\/sub> + 8 kg O<sub>2<\/sub> = 9 kg H<sub>2<\/sub>O<\/p>\n<p style=\"text-align: left\">Sulphur burns as follows:<\/p>\n<p style=\"text-align: center\">S +O<sub>2<\/sub> =SO<sub>2<\/sub><\/p>\n<p style=\"text-align: center\">32+32=64<\/p>\n<p style=\"text-align: center\">1 kg S + 1 kg O<sub>2<\/sub> = 2 kg SO<sub>2<\/sub><\/p>\n<p>In addition, 1 kg of O<sub>2<\/sub> (<strong>stoichiometric mass of O<sub>2<\/sub><\/strong>) is contained in 1\/0.232=4.3 kg air which is the <strong>stoichiometric mass of air<\/strong>. This air will contain 4.3-1=3.3 kg N<sub>2<\/sub>. Therefore we can write:<\/p>\n<p style=\"text-align: center\">1 kg S + 4.3 kg air = 2 kg SO<sub>2<\/sub>+3.3 kg N<sub>2<\/sub><\/p>\n<h1 style=\"text-align: left\">Procedure<\/h1>\n<p style=\"text-align: left\">If the analysis of fuel is given by mass, follow the steps below:<\/p>\n<ol>\n<li style=\"text-align: left\"><strong>Total O<sub>2<\/sub> required:<\/strong> Determine the mass of O<sub>2<\/sub> required for each constituent and find the total mass of O<sub>2 <\/sub>(Subtract any O<sub>2<\/sub> which may be in the fuel)<\/li>\n<li style=\"text-align: left\"><strong>Stoichiometric air: <\/strong>Stoichiometric mass of air = O<sub>2<\/sub> required\/0.232<\/li>\n<li style=\"text-align: left\"><strong>Total<\/strong><strong> mass of combustion products:<\/strong> Determine the mass of each combustion product. For example, given C content of 84.9%, CO<sub>2<\/sub>=84.9\/100*3\u2154=3.11% and find the total mass of combustion products.<\/li>\n<li style=\"text-align: left\"><strong>Analysis of combustion products by mass: <\/strong>Suppose the total mass of combustion products in step 3 has been found as 12.09 kg\/kg fuel, then CO<sub>2<\/sub>=3.11\/12.09*100=25.74. This means that 25.74% of the flue gas is CO<sub>2<\/sub>. Repeat this calculation for each constituent.<\/li>\n<\/ol>\n<div class=\"textbox learning-objectives\">\n<h3>Lab Instructions<\/h3>\n<p>Run the initial condition I14 80% Coal and setup trends for the following variables:<\/p>\n<p style=\"text-align: center\">X00820 X00821 X00822 X00823 X00824 E23356<\/p>\n<p style=\"text-align: center\">G02196 G02197 X32419 X02419 G00831 H00830<\/p>\n<ol>\n<li><strong>Burning default coal:<\/strong> After 10 minutes of running the simulator, freeze simulator and print the two trends. This is the reference point for the next step.<\/li>\n<li><strong>Burning poor quality<\/strong> <strong>coal:<\/strong> Switch to run mode and access Variable List page 0111 on MD180. Set the new values as shown below. After 10 minutes, freeze simulator and print the two trends.\n<ul>\n<li>X00820: 70.40<\/li>\n<li>X00821: 5.10<\/li>\n<li>X00822: 1.10<\/li>\n<li>X00823: 12.50<\/li>\n<li>X00824: 1.60<\/li>\n<li>X00825: 9.30<\/li>\n<\/ul>\n<\/li>\n<li><strong>Computation:<\/strong> Compute the following values for both types of fuels:\n<ul>\n<li>Total O<sub>2<\/sub> required<\/li>\n<li>Stoichiometric air<\/li>\n<li>Total mass of combustion products<\/li>\n<li>Analysis of combustion products by mass [%]: CO<sub>2<\/sub>, H<sub>2<\/sub>O, SO<sub>2<\/sub>, N<sub>2<\/sub><\/li>\n<\/ul>\n<\/li>\n<li><strong>Comparison:<\/strong> Compare your findings based on the following data:\n<ul>\n<li>Furnace outlet SO<sub>x<\/sub> flow<\/li>\n<li>Furnace outlet NO<sub>x<\/sub> flow<\/li>\n<li>CO content in flue gas<\/li>\n<li>Oxygen content in flue gas<\/li>\n<li>Theoretical combustion air<\/li>\n<li>Coal Heat Value<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<h2 style=\"text-align: left\">Hints &amp; Tips<\/h2>\n<p>In this lab, you are carrying out two combustion analyses. For data collection, set up your trends, a sample trend plot is shown below (Make sure your trend printouts are labeled properly otherwise, data analysis will be very confusing):<\/p>\n<figure id=\"attachment_55\" aria-describedby=\"caption-attachment-55\" style=\"width: 658px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal.jpg\" alt=\"Sample data for coal\" width=\"658\" height=\"520\" class=\"size-full wp-image-55\" srcset=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal.jpg 658w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal-300x237.jpg 300w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal-65x51.jpg 65w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal-225x178.jpg 225w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/TrendGroupCoal-350x277.jpg 350w\" sizes=\"auto, (max-width: 658px) 100vw, 658px\" \/><\/a><figcaption id=\"caption-attachment-55\" class=\"wp-caption-text\">Sample data for coal<\/figcaption><\/figure>\n<p>To change the fuel composition use the Variable List Page#: 0111 on MD180:<\/p>\n<figure id=\"attachment_49\" aria-describedby=\"caption-attachment-49\" style=\"width: 711px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel.jpg\" alt=\"Chemical composition of coal\" width=\"711\" height=\"351\" class=\"size-full wp-image-49\" srcset=\"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel.jpg 711w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel-300x148.jpg 300w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel-65x32.jpg 65w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel-225x111.jpg 225w, https:\/\/pressbooks.bccampus.ca\/tpps\/wp-content\/uploads\/sites\/134\/2017\/03\/PoorQualityFuel-350x173.jpg 350w\" sizes=\"auto, (max-width: 711px) 100vw, 711px\" \/><\/a><figcaption id=\"caption-attachment-49\" class=\"wp-caption-text\">Chemical composition of coal<\/figcaption><\/figure>\n<div class=\"textbox learning-objectives\">\n<h3>Deliverables<\/h3>\n<p>Your lab report is to include the following:<\/p>\n<ul>\n<li><strong>Trend plots:<\/strong> Supply all plots taken for each of the 2 fuels,<\/li>\n<li><strong>Computation:<\/strong> <span>As per lab instructions above, perform combustion analyses u<\/span>sing MATLAB or MS Excel.<\/li>\n<li><span><strong>Conclusion:<\/strong> Write a summary (max. 500 words<span itemscope=\"itemscope\" itemtype=\"http:\/\/schema.org\/WebPage\">, in a text box if using Excel<\/span>) comparing your results and suggestions for further study.<\/span><\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<p>Further Reading:<\/p>\n<ul>\n<li>Basic Engineering Thermodynamics in SI Units by R. Joel: Combustion.<\/li>\n<li>Thermal Engineering by H.L. Solberg, O.C. Cromer and A.R. Spalding: Fossil fuels and their combustion.<\/li>\n<li>\n<p id=\"firstHeading\" class=\"firstHeading\" lang=\"en\"><a href=\"https:\/\/en.wikipedia.org\/wiki\/Stoichiometry\" target=\"_blank\" rel=\"noopener\">Stoichiometry. <\/a><\/p>\n<\/li>\n<\/ul>\n<\/div>\n","protected":false},"author":84,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-79","chapter","type-chapter","status-publish","hentry"],"part":30,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/chapters\/79","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/wp\/v2\/users\/84"}],"version-history":[{"count":20,"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/chapters\/79\/revisions"}],"predecessor-version":[{"id":293,"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/chapters\/79\/revisions\/293"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/parts\/30"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/chapters\/79\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/wp\/v2\/media?parent=79"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/pressbooks\/v2\/chapter-type?post=79"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/wp\/v2\/contributor?post=79"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/tpps\/wp-json\/wp\/v2\/license?post=79"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}