To solve this problem, we’ll perform an energy balance around the gas turbine system using the First Law of Thermodynamics. Here’s a step-by-step explanation followed by the correct numerical solution.<\/p>\n\n\n\n
\n\n\n\n
Given Data:<\/strong><\/h3>\n\n\n\n\n- Fuel: Methane (CH\u2084)<\/li>\n\n\n\n
- Fuel flow rate: 1200 kg\/h = 0.3333 kg\/s<\/li>\n\n\n\n
- Air supply: 400% of theoretical air<\/li>\n\n\n\n
- Inlet conditions: 25\u00b0C, 1 atm<\/li>\n\n\n\n
- Exit conditions: 577\u00b0C, 1 atm<\/li>\n\n\n\n
- Heat loss = 10% of net power output<\/li>\n\n\n\n
- Kinetic and potential energy changes are negligible<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 1: Combustion Reaction<\/strong><\/h3>\n\n\n\nComplete combustion of methane with stoichiometric air: 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
For 400% theoretical air, we multiply all air species by 4: CH4+8(O2+3.76N2)\u2192CO2+2H2O+8O2+30.08N2\\text{CH}_4 + 8(\\text{O}_2 + 3.76\\text{N}_2) \\rightarrow \\text{CO}_2 + 2\\text{H}_2\\text{O} + 8\\text{O}_2 + 30.08\\text{N}_2<\/p>\n\n\n\n
Total products (per mole CH\u2084):<\/p>\n\n\n\n
\n- 1 CO\u2082<\/li>\n\n\n\n
- 2 H\u2082O<\/li>\n\n\n\n
- 8 O\u2082<\/li>\n\n\n\n
- 30.08 N\u2082<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Enthalpy Change<\/strong><\/h3>\n\n\n\nUse average specific heat capacities CpC_p (ideal gas assumption) and \u0394T = 577\u00b0C \u2212 25\u00b0C = 552 K:<\/p>\n\n\n\n
Approximate CpC_p values at average temperature (~300\u2013800 K):<\/p>\n\n\n\n
\n- CO\u2082: 1.10 kJ\/kg\u00b7K, MW = 44<\/li>\n\n\n\n
- H\u2082O(g): 2.00 kJ\/kg\u00b7K, MW = 18<\/li>\n\n\n\n
- O\u2082: 1.00 kJ\/kg\u00b7K, MW = 32<\/li>\n\n\n\n
- N\u2082: 1.04 kJ\/kg\u00b7K, MW = 28<\/li>\n<\/ul>\n\n\n\n
Calculate enthalpy change per kmol CH\u2084: \u0394H=\u2211niCp,i\u0394T\\Delta H = \\sum n_i C_{p,i} \\Delta T \u0394H=(1)(1.10)(552)+(2)(2.00)(552)+(8)(1.00)(552)+(30.08)(1.04)(552)\\Delta H = (1)(1.10)(552) + (2)(2.00)(552) + (8)(1.00)(552) + (30.08)(1.04)(552) \u0394H=607.2+2208+4416+17286.3=24517.5 kJ\/kmol CH\u2084\\Delta H = 607.2 + 2208 + 4416 + 17286.3 = 24517.5 \\text{ kJ\/kmol CH\u2084}<\/p>\n\n\n\n
\n\n\n\nStep 3: Convert to per second basis<\/strong><\/h3>\n\n\n\nMolecular weight of CH\u2084 = 16 g\/mol = 16 kg\/kmol
So moles\/s of CH\u2084 = 0.333316=0.02083\\frac{0.3333}{16} = 0.02083 kmol\/s<\/p>\n\n\n\n
Energy per second (total output before losses): W\u02d9gross=0.02083\u00d724517.5=510.5 kW\\dot{W}_{gross} = 0.02083 \\times 24517.5 = 510.5 \\text{ kW}<\/p>\n\n\n\n
\n\n\n\nStep 4: Account for Heat Loss<\/strong><\/h3>\n\n\n\nOnly 90% of this is net power output: W\u02d9net=510.51.1=464.1 kW=0.464 MW\\dot{W}_{net} = \\frac{510.5}{1.1} = 464.1 \\text{ kW} = \\boxed{0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\nNet power output=0.464 MW\\boxed{\\text{Net power output} = 0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nExplanation ):<\/strong><\/h3>\n\n\n\nThis problem involves analyzing a gas turbine power plant where methane is combusted with excess air. The energy released from combustion is partially converted to useful mechanical work, with some losses. The First Law of Thermodynamics (energy conservation) is applied here in a control volume around the combustor and turbine.<\/p>\n\n\n\n
We first established the combustion equation, accounting for 400% theoretical air, meaning five times the stoichiometric air amount. The combustion produces CO\u2082, H\u2082O, unused O\u2082, and N\u2082. The total enthalpy change of the products, from inlet to exit temperature, determines the energy made available for work. We estimated this enthalpy change using average specific heat capacities over the temperature range from 25\u00b0C to 577\u00b0C.<\/p>\n\n\n\n
Using the fuel flow rate, we calculated the number of kilomoles of methane combusted per second. This allowed us to find the total energy released per second (or power) using the enthalpy of the exhaust gases.<\/p>\n\n\n\n
However, not all of this power is converted to net mechanical power, as 10% is lost via heat transfer. Therefore, we corrected for this loss to find the net useful power output.<\/p>\n\n\n\n
Ultimately, this thermodynamic analysis yielded a net power output of 0.464 MW for a methane fuel flow rate of 1200 kg\/h. This kind of analysis is essential in designing efficient power generation systems and estimating performance from basic operational parameters.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner6-128.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"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. 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 […]<\/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-221316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316","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=221316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 1: Combustion Reaction<\/strong><\/h3>\n\n\n\nComplete combustion of methane with stoichiometric air: 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
For 400% theoretical air, we multiply all air species by 4: CH4+8(O2+3.76N2)\u2192CO2+2H2O+8O2+30.08N2\\text{CH}_4 + 8(\\text{O}_2 + 3.76\\text{N}_2) \\rightarrow \\text{CO}_2 + 2\\text{H}_2\\text{O} + 8\\text{O}_2 + 30.08\\text{N}_2<\/p>\n\n\n\n
Total products (per mole CH\u2084):<\/p>\n\n\n\n
\n- 1 CO\u2082<\/li>\n\n\n\n
- 2 H\u2082O<\/li>\n\n\n\n
- 8 O\u2082<\/li>\n\n\n\n
- 30.08 N\u2082<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Enthalpy Change<\/strong><\/h3>\n\n\n\nUse average specific heat capacities CpC_p (ideal gas assumption) and \u0394T = 577\u00b0C \u2212 25\u00b0C = 552 K:<\/p>\n\n\n\n
Approximate CpC_p values at average temperature (~300\u2013800 K):<\/p>\n\n\n\n
\n- CO\u2082: 1.10 kJ\/kg\u00b7K, MW = 44<\/li>\n\n\n\n
- H\u2082O(g): 2.00 kJ\/kg\u00b7K, MW = 18<\/li>\n\n\n\n
- O\u2082: 1.00 kJ\/kg\u00b7K, MW = 32<\/li>\n\n\n\n
- N\u2082: 1.04 kJ\/kg\u00b7K, MW = 28<\/li>\n<\/ul>\n\n\n\n
Calculate enthalpy change per kmol CH\u2084: \u0394H=\u2211niCp,i\u0394T\\Delta H = \\sum n_i C_{p,i} \\Delta T \u0394H=(1)(1.10)(552)+(2)(2.00)(552)+(8)(1.00)(552)+(30.08)(1.04)(552)\\Delta H = (1)(1.10)(552) + (2)(2.00)(552) + (8)(1.00)(552) + (30.08)(1.04)(552) \u0394H=607.2+2208+4416+17286.3=24517.5 kJ\/kmol CH\u2084\\Delta H = 607.2 + 2208 + 4416 + 17286.3 = 24517.5 \\text{ kJ\/kmol CH\u2084}<\/p>\n\n\n\n
\n\n\n\nStep 3: Convert to per second basis<\/strong><\/h3>\n\n\n\nMolecular weight of CH\u2084 = 16 g\/mol = 16 kg\/kmol
So moles\/s of CH\u2084 = 0.333316=0.02083\\frac{0.3333}{16} = 0.02083 kmol\/s<\/p>\n\n\n\n
Energy per second (total output before losses): W\u02d9gross=0.02083\u00d724517.5=510.5 kW\\dot{W}_{gross} = 0.02083 \\times 24517.5 = 510.5 \\text{ kW}<\/p>\n\n\n\n
\n\n\n\nStep 4: Account for Heat Loss<\/strong><\/h3>\n\n\n\nOnly 90% of this is net power output: W\u02d9net=510.51.1=464.1 kW=0.464 MW\\dot{W}_{net} = \\frac{510.5}{1.1} = 464.1 \\text{ kW} = \\boxed{0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\nNet power output=0.464 MW\\boxed{\\text{Net power output} = 0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nExplanation ):<\/strong><\/h3>\n\n\n\nThis problem involves analyzing a gas turbine power plant where methane is combusted with excess air. The energy released from combustion is partially converted to useful mechanical work, with some losses. The First Law of Thermodynamics (energy conservation) is applied here in a control volume around the combustor and turbine.<\/p>\n\n\n\n
We first established the combustion equation, accounting for 400% theoretical air, meaning five times the stoichiometric air amount. The combustion produces CO\u2082, H\u2082O, unused O\u2082, and N\u2082. The total enthalpy change of the products, from inlet to exit temperature, determines the energy made available for work. We estimated this enthalpy change using average specific heat capacities over the temperature range from 25\u00b0C to 577\u00b0C.<\/p>\n\n\n\n
Using the fuel flow rate, we calculated the number of kilomoles of methane combusted per second. This allowed us to find the total energy released per second (or power) using the enthalpy of the exhaust gases.<\/p>\n\n\n\n
However, not all of this power is converted to net mechanical power, as 10% is lost via heat transfer. Therefore, we corrected for this loss to find the net useful power output.<\/p>\n\n\n\n
Ultimately, this thermodynamic analysis yielded a net power output of 0.464 MW for a methane fuel flow rate of 1200 kg\/h. This kind of analysis is essential in designing efficient power generation systems and estimating performance from basic operational parameters.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"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. 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 […]<\/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-221316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316","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=221316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 2: Enthalpy Change<\/strong><\/h3>\n\n\n\nUse average specific heat capacities CpC_p (ideal gas assumption) and \u0394T = 577\u00b0C \u2212 25\u00b0C = 552 K:<\/p>\n\n\n\n
Approximate CpC_p values at average temperature (~300\u2013800 K):<\/p>\n\n\n\n
\n- CO\u2082: 1.10 kJ\/kg\u00b7K, MW = 44<\/li>\n\n\n\n
- H\u2082O(g): 2.00 kJ\/kg\u00b7K, MW = 18<\/li>\n\n\n\n
- O\u2082: 1.00 kJ\/kg\u00b7K, MW = 32<\/li>\n\n\n\n
- N\u2082: 1.04 kJ\/kg\u00b7K, MW = 28<\/li>\n<\/ul>\n\n\n\n
Calculate enthalpy change per kmol CH\u2084: \u0394H=\u2211niCp,i\u0394T\\Delta H = \\sum n_i C_{p,i} \\Delta T \u0394H=(1)(1.10)(552)+(2)(2.00)(552)+(8)(1.00)(552)+(30.08)(1.04)(552)\\Delta H = (1)(1.10)(552) + (2)(2.00)(552) + (8)(1.00)(552) + (30.08)(1.04)(552) \u0394H=607.2+2208+4416+17286.3=24517.5 kJ\/kmol CH\u2084\\Delta H = 607.2 + 2208 + 4416 + 17286.3 = 24517.5 \\text{ kJ\/kmol CH\u2084}<\/p>\n\n\n\n
\n\n\n\nStep 3: Convert to per second basis<\/strong><\/h3>\n\n\n\nMolecular weight of CH\u2084 = 16 g\/mol = 16 kg\/kmol
So moles\/s of CH\u2084 = 0.333316=0.02083\\frac{0.3333}{16} = 0.02083 kmol\/s<\/p>\n\n\n\n
Energy per second (total output before losses): W\u02d9gross=0.02083\u00d724517.5=510.5 kW\\dot{W}_{gross} = 0.02083 \\times 24517.5 = 510.5 \\text{ kW}<\/p>\n\n\n\n
\n\n\n\nStep 4: Account for Heat Loss<\/strong><\/h3>\n\n\n\nOnly 90% of this is net power output: W\u02d9net=510.51.1=464.1 kW=0.464 MW\\dot{W}_{net} = \\frac{510.5}{1.1} = 464.1 \\text{ kW} = \\boxed{0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\nNet power output=0.464 MW\\boxed{\\text{Net power output} = 0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nExplanation ):<\/strong><\/h3>\n\n\n\nThis problem involves analyzing a gas turbine power plant where methane is combusted with excess air. The energy released from combustion is partially converted to useful mechanical work, with some losses. The First Law of Thermodynamics (energy conservation) is applied here in a control volume around the combustor and turbine.<\/p>\n\n\n\n
We first established the combustion equation, accounting for 400% theoretical air, meaning five times the stoichiometric air amount. The combustion produces CO\u2082, H\u2082O, unused O\u2082, and N\u2082. The total enthalpy change of the products, from inlet to exit temperature, determines the energy made available for work. We estimated this enthalpy change using average specific heat capacities over the temperature range from 25\u00b0C to 577\u00b0C.<\/p>\n\n\n\n
Using the fuel flow rate, we calculated the number of kilomoles of methane combusted per second. This allowed us to find the total energy released per second (or power) using the enthalpy of the exhaust gases.<\/p>\n\n\n\n
However, not all of this power is converted to net mechanical power, as 10% is lost via heat transfer. Therefore, we corrected for this loss to find the net useful power output.<\/p>\n\n\n\n
Ultimately, this thermodynamic analysis yielded a net power output of 0.464 MW for a methane fuel flow rate of 1200 kg\/h. This kind of analysis is essential in designing efficient power generation systems and estimating performance from basic operational parameters.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"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. 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 […]<\/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-221316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316","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=221316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Calculate enthalpy change per kmol CH\u2084: \u0394H=\u2211niCp,i\u0394T\\Delta H = \\sum n_i C_{p,i} \\Delta T \u0394H=(1)(1.10)(552)+(2)(2.00)(552)+(8)(1.00)(552)+(30.08)(1.04)(552)\\Delta H = (1)(1.10)(552) + (2)(2.00)(552) + (8)(1.00)(552) + (30.08)(1.04)(552) \u0394H=607.2+2208+4416+17286.3=24517.5 kJ\/kmol CH\u2084\\Delta H = 607.2 + 2208 + 4416 + 17286.3 = 24517.5 \\text{ kJ\/kmol CH\u2084}<\/p>\n\n\n\n
\n\n\n\n
Step 3: Convert to per second basis<\/strong><\/h3>\n\n\n\nMolecular weight of CH\u2084 = 16 g\/mol = 16 kg\/kmol
So moles\/s of CH\u2084 = 0.333316=0.02083\\frac{0.3333}{16} = 0.02083 kmol\/s<\/p>\n\n\n\n
Energy per second (total output before losses): W\u02d9gross=0.02083\u00d724517.5=510.5 kW\\dot{W}_{gross} = 0.02083 \\times 24517.5 = 510.5 \\text{ kW}<\/p>\n\n\n\n
\n\n\n\nStep 4: Account for Heat Loss<\/strong><\/h3>\n\n\n\nOnly 90% of this is net power output: W\u02d9net=510.51.1=464.1 kW=0.464 MW\\dot{W}_{net} = \\frac{510.5}{1.1} = 464.1 \\text{ kW} = \\boxed{0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\nNet power output=0.464 MW\\boxed{\\text{Net power output} = 0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nExplanation ):<\/strong><\/h3>\n\n\n\nThis problem involves analyzing a gas turbine power plant where methane is combusted with excess air. The energy released from combustion is partially converted to useful mechanical work, with some losses. The First Law of Thermodynamics (energy conservation) is applied here in a control volume around the combustor and turbine.<\/p>\n\n\n\n
We first established the combustion equation, accounting for 400% theoretical air, meaning five times the stoichiometric air amount. The combustion produces CO\u2082, H\u2082O, unused O\u2082, and N\u2082. The total enthalpy change of the products, from inlet to exit temperature, determines the energy made available for work. We estimated this enthalpy change using average specific heat capacities over the temperature range from 25\u00b0C to 577\u00b0C.<\/p>\n\n\n\n
Using the fuel flow rate, we calculated the number of kilomoles of methane combusted per second. This allowed us to find the total energy released per second (or power) using the enthalpy of the exhaust gases.<\/p>\n\n\n\n
However, not all of this power is converted to net mechanical power, as 10% is lost via heat transfer. Therefore, we corrected for this loss to find the net useful power output.<\/p>\n\n\n\n
Ultimately, this thermodynamic analysis yielded a net power output of 0.464 MW for a methane fuel flow rate of 1200 kg\/h. This kind of analysis is essential in designing efficient power generation systems and estimating performance from basic operational parameters.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"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. 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 […]<\/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-221316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316","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=221316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
So moles\/s of CH\u2084 = 0.333316=0.02083\\frac{0.3333}{16} = 0.02083 kmol\/s<\/p>\n\n\n\n
Only 90% of this is net power output: W\u02d9net=510.51.1=464.1 kW=0.464 MW\\dot{W}_{net} = \\frac{510.5}{1.1} = 464.1 \\text{ kW} = \\boxed{0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\n
Final Answer:<\/strong><\/h3>\n\n\n\nNet power output=0.464 MW\\boxed{\\text{Net power output} = 0.464 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nExplanation ):<\/strong><\/h3>\n\n\n\nThis problem involves analyzing a gas turbine power plant where methane is combusted with excess air. The energy released from combustion is partially converted to useful mechanical work, with some losses. The First Law of Thermodynamics (energy conservation) is applied here in a control volume around the combustor and turbine.<\/p>\n\n\n\n
We first established the combustion equation, accounting for 400% theoretical air, meaning five times the stoichiometric air amount. The combustion produces CO\u2082, H\u2082O, unused O\u2082, and N\u2082. The total enthalpy change of the products, from inlet to exit temperature, determines the energy made available for work. We estimated this enthalpy change using average specific heat capacities over the temperature range from 25\u00b0C to 577\u00b0C.<\/p>\n\n\n\n
Using the fuel flow rate, we calculated the number of kilomoles of methane combusted per second. This allowed us to find the total energy released per second (or power) using the enthalpy of the exhaust gases.<\/p>\n\n\n\n
However, not all of this power is converted to net mechanical power, as 10% is lost via heat transfer. Therefore, we corrected for this loss to find the net useful power output.<\/p>\n\n\n\n
Ultimately, this thermodynamic analysis yielded a net power output of 0.464 MW for a methane fuel flow rate of 1200 kg\/h. This kind of analysis is essential in designing efficient power generation systems and estimating performance from basic operational parameters.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"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. 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 […]<\/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-221316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316","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=221316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
This problem involves analyzing a gas turbine power plant where methane is combusted with excess air. The energy released from combustion is partially converted to useful mechanical work, with some losses. The First Law of Thermodynamics (energy conservation) is applied here in a control volume around the combustor and turbine.<\/p>\n\n\n\n
We first established the combustion equation, accounting for 400% theoretical air, meaning five times the stoichiometric air amount. The combustion produces CO\u2082, H\u2082O, unused O\u2082, and N\u2082. The total enthalpy change of the products, from inlet to exit temperature, determines the energy made available for work. We estimated this enthalpy change using average specific heat capacities over the temperature range from 25\u00b0C to 577\u00b0C.<\/p>\n\n\n\n
Using the fuel flow rate, we calculated the number of kilomoles of methane combusted per second. This allowed us to find the total energy released per second (or power) using the enthalpy of the exhaust gases.<\/p>\n\n\n\n
However, not all of this power is converted to net mechanical power, as 10% is lost via heat transfer. Therefore, we corrected for this loss to find the net useful power output.<\/p>\n\n\n\n
Ultimately, this thermodynamic analysis yielded a net power output of 0.464 MW for a methane fuel flow rate of 1200 kg\/h. This kind of analysis is essential in designing efficient power generation systems and estimating performance from basic operational parameters.<\/p>\n\n\n\n 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 […]<\/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-221316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316","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=221316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}<\/figure>\n","protected":false},"excerpt":{"rendered":"