{"id":217366,"date":"2025-05-21T19:12:42","date_gmt":"2025-05-21T19:12:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217366"},"modified":"2025-05-21T19:12:46","modified_gmt":"2025-05-21T19:12:46","slug":"13-47-methane-ch4-at-25c-enters-the-combustor-of-a-simple-open-gas-turbine-power-plant-and-burns-completely-with-400-of-theoretical-air-entering-the-compressor-at-25c-1-atm-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/21\/13-47-methane-ch4-at-25c-enters-the-combustor-of-a-simple-open-gas-turbine-power-plant-and-burns-completely-with-400-of-theoretical-air-entering-the-compressor-at-25c-1-atm-2\/","title":{"rendered":"13.47 Methane (CH4), at 25\u00b0C, enters the combustor of a simple open gas turbine power plant and burns completely with 400% of theoretical air entering the compressor at 25\u00b0C, 1 atm"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">13.47 Methane (CH4), at 25\u00b0C, enters the combustor of a simple open gas turbine power plant and burns completely with 400% of theoretical air entering the compressor at 25\u00b0C, 1 atm. Products of combustion exit the turbine at 577\u00b0C, 1 atm. The rate of heat transfer from the gas turbine is estimated as 10% of the net power developed. Determine the net power output, in MW, if the fuel mass flow rate is 1200 kg\/h. Kinetic and potential energy effects are negligible.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The Correct Answer and Explanation is:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Let&#8217;s analyze the problem step-by-step and then find the net power output.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given Data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Fuel: Methane (CH\u2084)<\/li>\n\n\n\n<li>Fuel mass flow rate, $\\dot{m}_{fuel} = 1200\\, \\text{kg\/h}$<\/li>\n\n\n\n<li>Temperature of fuel and air entering: 25\u00b0C<\/li>\n\n\n\n<li>Pressure at inlet: 1 atm<\/li>\n\n\n\n<li>Air-fuel ratio: 400% theoretical air (4 times the stoichiometric air)<\/li>\n\n\n\n<li>Combustion products exit turbine at 577\u00b0C, 1 atm<\/li>\n\n\n\n<li>Heat loss from turbine: 10% of net power output<\/li>\n\n\n\n<li>Assume kinetic and potential energy effects negligible<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert mass flow rate to kg\/s<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\dot{m}_{fuel} = \\frac{1200\\, \\text{kg}}{3600\\, \\text{s}} = 0.3333\\, \\text{kg\/s}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Stoichiometric combustion of methane<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The balanced reaction of methane combustion with air (assuming air as 21% O\u2082 and 79% N\u2082 by volume) is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>CH_4 + 2(O_2 + 3.76 N_2) \\to CO_2 + 2H_2O + 7.52 N_2<br>$$<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Stoichiometric oxygen needed: 2 mol O\u2082 per mol CH\u2084<\/li>\n\n\n\n<li>Stoichiometric air = 2 \u00d7 (1 + 3.76) = 9.52 mol air per mol CH\u2084<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Air supplied is 400% theoretical air<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This means actual air is 4 times stoichiometric air:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Actual air} = 4 \\times 9.52 = 38.08\\, \\text{mol air per mol CH}_4<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Determine heat input rate<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Heat input is based on fuel mass flow and the <strong>Lower Heating Value (LHV)<\/strong> of methane:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>LHV of methane $\\approx 50,000\\, \\text{kJ\/kg}$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So, heat input rate:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>Q_{in} = \\dot{m}_{fuel} \\times \\text{LHV} = 0.3333\\, \\text{kg\/s} \\times 50,000\\, \\text{kJ\/kg} = 16,666.5\\, \\text{kW} = 16.67\\, \\text{MW}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Calculate enthalpy change of the gas<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Assuming ideal gas, the enthalpy change of combustion gases is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta h = c_p (T_{exit} &#8211; T_{inlet})<br>$$<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$T_{exit} = 577^\\circ C = 850 K$<\/li>\n\n\n\n<li>$T_{inlet} = 25^\\circ C = 298 K$<\/li>\n\n\n\n<li>$c_p$ of combustion products \u2248 1.15 kJ\/kg\u00b7K (typical value for combustion gases with excess air)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Total mass flow rate of the gas = mass of fuel + mass of air<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 6: Calculate mass flow of air<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Molecular weights:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$CH_4 = 16\\, \\text{kg\/kmol}$<\/li>\n\n\n\n<li>Air \u2248 29 kg\/kmol<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Moles of fuel per second:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\frac{0.3333\\, \\text{kg\/s}}{16\\, \\text{kg\/kmol}} = 0.02083\\, \\text{kmol\/s}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Moles of air per second:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>38.08 \\times 0.02083 = 0.793\\, \\text{kmol\/s}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Mass flow of air:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>0.793 \\times 29 = 22.997\\, \\text{kg\/s}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 7: Total mass flow rate of gas<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\dot{m}<em>{gas} = \\dot{m}<\/em>{fuel} + \\dot{m}_{air} = 0.3333 + 22.997 = 23.33\\, \\text{kg\/s}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 8: Calculate power output (isenthalpic assumption)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Power developed by turbine is energy extracted as gas cools from inlet temp to outlet temp (assuming all fuel energy is converted to raising temp, then work extracted):<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Power output} = \\dot{m}<em>{gas} \\times c_p \\times (T<\/em>{exit} &#8211; T_{inlet})<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>= 23.33\\, \\text{kg\/s} \\times 1.15\\, \\text{kJ\/kg\u00b7K} \\times (850 &#8211; 298) K<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>= 23.33 \\times 1.15 \\times 552 = 14809\\, \\text{kW} = 14.81\\, \\text{MW}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 9: Account for heat loss<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Heat transfer from turbine is 10% of net power output:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>Q_{loss} = 0.1 \\times P_{net}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, net power output $P_{net}$ satisfies:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P_{net} + Q_{loss} = \\text{Power from turbine} = 14.81\\, \\text{MW}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P_{net} + 0.1 P_{net} = 14.81 \\Rightarrow 1.1 P_{net} = 14.81<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P_{net} = \\frac{14.81}{1.1} = 13.46\\, \\text{MW}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Net power output = 13.46 MW<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In this problem, methane burns completely with four times the theoretical air, ensuring excess oxygen for complete combustion. The stoichiometric combustion equation defines the theoretical air-fuel ratio; actual air supplied is four times that, leading to a large air mass flow, which affects the temperature and mass of combustion gases.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The fuel mass flow rate is converted to molar flow to calculate the total mass flow of air and fuel. This total gas mass flow is crucial because the turbine power depends on the mass flow rate times the enthalpy drop of the gas. The specific heat capacity of combustion products (approximate average value) and temperature difference between inlet and turbine exit allow calculation of the turbine&#8217;s power output.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Heat loss from the turbine reduces the net power output by 10%. This heat loss is modeled as a percentage of net power, so algebraic manipulation yields the net output power.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This process shows how thermodynamic properties, stoichiometry, and energy balances combine in a practical gas turbine scenario to estimate the net mechanical power generated from a given fuel flow rate under defined combustion conditions.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>13.47 Methane (CH4), at 25\u00b0C, enters the combustor of a simple open gas turbine power plant and burns completely with 400% of theoretical air entering the compressor at 25\u00b0C, 1 atm. Products of combustion exit the turbine at 577\u00b0C, 1 atm. The rate of heat transfer from the gas turbine is estimated as 10% of [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-217366","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217366","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=217366"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217366\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217366"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217366"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217366"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}