{"id":216891,"date":"2025-05-20T15:19:16","date_gmt":"2025-05-20T15:19:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=216891"},"modified":"2025-05-20T15:19:18","modified_gmt":"2025-05-20T15:19:18","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","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/20\/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\/","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>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><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we&#8217;ll follow a step-by-step thermodynamic analysis involving:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Combustion stoichiometry with excess air<\/li>\n\n\n\n<li>Energy balance for the combustion and expansion process<\/li>\n\n\n\n<li>Use of enthalpy values (from ideal gas properties)<\/li>\n\n\n\n<li>Power and heat transfer calculations<\/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\"><strong>Step 1: Fuel Flow Rate<\/strong><\/h3>\n\n\n\n<p>m\u02d9CH4=1200\u2009kg\/h=12003600=0.3333\u2009kg\/s\\dot{m}_{\\text{CH}_4} = 1200 \\, \\text{kg\/h} = \\frac{1200}{3600} = 0.3333 \\, \\text{kg\/s}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Theoretical Air Requirement<\/strong><\/h3>\n\n\n\n<p>The stoichiometric combustion of methane: CH4+2(O2+3.76N2)\u2192CO2+2H2O+7.52N2\\text{CH}_4 + 2(\\text{O}_2 + 3.76\\text{N}_2) \\rightarrow \\text{CO}_2 + 2\\text{H}_2\\text{O} + 7.52\\text{N}_2<\/p>\n\n\n\n<p>The theoretical air-fuel ratio (AFRstoich_{stoich}) by mass:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>O\u2082 required = 2 moles (64 g)<\/li>\n\n\n\n<li>Air = 2 \u00d7 (1 + 3.76) = 8.52 moles air<\/li>\n\n\n\n<li>Air molar mass \u2248 28.97 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>AFRstoich=8.52\u00d728.9716\u224815.42\\text{AFR}_{\\text{stoich}} = \\frac{8.52 \\times 28.97}{16} \\approx 15.42<\/p>\n\n\n\n<p>With 400% of theoretical air (i.e., 5 \u00d7 stoichiometric): AFRactual=5\u00d715.42=77.1\\text{AFR}_{\\text{actual}} = 5 \\times 15.42 = 77.1<\/p>\n\n\n\n<p>Air mass flow rate: m\u02d9air=77.1\u00d70.3333=25.7\u2009kg\/s\\dot{m}_{\\text{air}} = 77.1 \\times 0.3333 = 25.7 \\, \\text{kg\/s}<\/p>\n\n\n\n<p>Total mass flow rate of combustion gases \u2248 0.3333 + 25.7 \u2248 <strong>26.03 kg\/s<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Energy Balance Using Enthalpy<\/strong><\/h3>\n\n\n\n<p>Assuming ideal gas behavior, negligible KE\/PE, and complete combustion.<\/p>\n\n\n\n<p>Use <strong>Lower Heating Value (LHV)<\/strong> of methane = <strong>50,000 kJ\/kg<\/strong><\/p>\n\n\n\n<p>Fuel energy input: Qin=m\u02d9CH4\u00d7LHV=0.3333\u00d750000=16,665\u2009kWQ_{\\text{in}} = \\dot{m}_{\\text{CH}_4} \\times \\text{LHV} = 0.3333 \\times 50000 = 16,665 \\, \\text{kW}<\/p>\n\n\n\n<p>Let net power output = WnetW_{\\text{net}}<\/p>\n\n\n\n<p>Given that 10% of the power is lost as heat from the turbine: Qloss=0.1\u00d7Wnet\u21d2Useful&nbsp;energy=Qin=Wnet+0.1Wnet=1.1WnetQ_{\\text{loss}} = 0.1 \\times W_{\\text{net}} \\Rightarrow \\text{Useful energy} = Q_{\\text{in}} = W_{\\text{net}} + 0.1W_{\\text{net}} = 1.1W_{\\text{net}}<\/p>\n\n\n\n<p>Solve: 1.1Wnet=16665\u21d2Wnet=166651.1=15,150\u2009kW=15.15\u2009MW1.1 W_{\\text{net}} = 16665 \\Rightarrow W_{\\text{net}} = \\frac{16665}{1.1} = 15,150 \\, \\text{kW} = \\boxed{15.15 \\, \\text{MW}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>15.15\u2009MW\\boxed{15.15 \\, \\text{MW}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300+ words):<\/strong><\/h3>\n\n\n\n<p>This problem involves analyzing a methane-fueled gas turbine using principles of thermodynamics and combustion. The system is operating with excess air (400% of theoretical), which affects the combustion products and thermal capacity of the flow.<\/p>\n\n\n\n<p>We begin by converting the given fuel flow rate to kg\/s for consistency in SI units. The stoichiometric reaction of methane with oxygen requires 2 moles of O\u2082, which translates into about 15.42 kg of air per kg of methane when accounting for nitrogen in the air. With 400% theoretical air, or 5 times stoichiometric, the actual air-fuel ratio becomes 77.1. Multiplying this by the fuel flow rate gives the air mass flow rate, which is critical for determining the total exhaust flow.<\/p>\n\n\n\n<p>The next key concept is the energy input. Methane\u2019s lower heating value (LHV), which excludes latent heat of vaporization of water, is used because water vapor remains in gaseous form during expansion. The fuel supplies energy at a rate of 16,665 kW.<\/p>\n\n\n\n<p>According to the problem, 10% of the power is lost from the turbine to the surroundings, meaning that the net useful power is 90% of the total turbine output. Rearranging the energy balance formula: Qin=Wnet+0.1Wnet=1.1WnetQ_{\\text{in}} = W_{\\text{net}} + 0.1W_{\\text{net}} = 1.1W_{\\text{net}}<\/p>\n\n\n\n<p>Solving for WnetW_{\\text{net}}, we find that the gas turbine delivers <strong>15.15 MW<\/strong> of usable mechanical power. No changes in kinetic or potential energy were assumed, simplifying the energy balance. This analysis demonstrates the importance of air-to-fuel ratio, fuel energy content, and system efficiency in determining gas turbine performance.<\/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-216891","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216891","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=216891"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216891\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216891"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216891"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216891"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}