Problem Summary (13.47)<\/strong><\/h3>\n\n\n\n\n- Fuel<\/strong>: Methane (CH\u2084), mass flow rate = 1200 kg\/h = 0.3333 kg\/s<\/li>\n\n\n\n
- Inlet conditions<\/strong>: Fuel and air enter at 25\u00b0C (298 K), 1 atm<\/li>\n\n\n\n
- Excess air<\/strong>: 400% of theoretical air<\/li>\n\n\n\n
- Combustion products exit<\/strong>: 577\u00b0C (850 K), 1 atm<\/li>\n\n\n\n
- Heat loss<\/strong>: 10% of net power developed<\/li>\n\n\n\n
- Neglect<\/strong>: Kinetic and potential energy effects<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nAssumptions and Constants<\/strong><\/h3>\n\n\n\n\n- HHV of CH\u2084<\/strong> \u2248 55,500 kJ\/kg<\/li>\n\n\n\n
- LHV of CH\u2084<\/strong> \u2248 50,000 kJ\/kg (we use LHV since water stays vapor)<\/li>\n\n\n\n
- Cp of products<\/strong> \u2248 1.10 kJ\/kg\u00b7K (approximate for combustion gases)<\/li>\n\n\n\n
- T\u2081 (inlet temp)<\/strong> = 298 K<\/li>\n\n\n\n
- T\u2082 (exit temp)<\/strong> = 850 K<\/li>\n<\/ul>\n\n\n\n
We will use energy balance<\/strong> on the turbine and combustion process:<\/p>\n\n\n\n
\n\n\n\nStep-by-Step Solution<\/strong><\/h3>\n\n\n\n1. Determine the heat released by combustion per second<\/strong><\/h4>\n\n\n\nm\u02d9fuel=1200 kg\/h3600=0.3333 kg\/s\\dot{m}_{\\text{fuel}} = \\frac{1200 \\text{ kg\/h}}{3600} = 0.3333 \\text{ kg\/s} Q\u02d9in=m\u02d9fuel\u00d7LHVCH4=0.3333\u00d750,000=16,666.7 kW\\dot{Q}_{\\text{in}} = \\dot{m}_{\\text{fuel}} \\times \\text{LHV}_{CH_4} = 0.3333 \\times 50,000 = 16,666.7 \\text{ kW}<\/p>\n\n\n\n
\n\n\n\n2. Apply energy balance considering heat loss<\/strong><\/h4>\n\n\n\nLet WnetW_{\\text{net}} be the net power output.<\/p>\n\n\n\n
Heat loss = 10% of net power, so: Q\u02d9loss=0.10\u00d7Wnet\\dot{Q}_{\\text{loss}} = 0.10 \\times W_{\\text{net}}<\/p>\n\n\n\n
Energy balance: Heat in=Net work out+Heat loss\\text{Heat in} = \\text{Net work out} + \\text{Heat loss} 16,666.7=Wnet+0.10Wnet=1.10Wnet16,666.7 = W_{\\text{net}} + 0.10 W_{\\text{net}} = 1.10 W_{\\text{net}} Wnet=16,666.71.10=15,151.5 kW=15.15 MWW_{\\text{net}} = \\frac{16,666.7}{1.10} = 15,151.5 \\text{ kW} = \\boxed{15.15 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nFinal Answer<\/strong><\/h3>\n\n\n\nNet Power Output = 15.15 MW<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\nThis problem involves calculating the net power output of a simple open gas turbine system fueled by methane. The combustion of methane is complete, and it takes place with an excess of air (400% of the theoretical requirement). The energy released by the combustion of methane is used to generate power, but 10% of this power is lost as heat to the surroundings.<\/p>\n\n\n\n
We begin by calculating the mass flow rate of the methane fuel in seconds, since power is in kW (kJ\/s). Then, we estimate the energy input rate using the lower heating value (LHV) of methane, which is suitable here because the water produced by combustion remains in vapor form. The LHV of methane is approximately 50,000 kJ\/kg. Multiplying this by the mass flow rate gives the heat released per second (kW), which represents the thermal energy made available by fuel combustion.<\/p>\n\n\n\n
Next, we apply an energy balance to the turbine system. Since the problem specifies that 10% of the power is lost as heat transfer, we set up the equation such that the input energy equals the useful net work output plus the heat loss. Solving this equation yields the net power output of the turbine.<\/p>\n\n\n\n
Kinetic and potential energy effects are negligible, simplifying the energy balance further. This problem demonstrates a practical application of the first law of thermodynamics (energy conservation) to a real-world engineering system. It highlights how combustion efficiency and losses influence the actual usable power output of a turbine.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner6-121.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-221240","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221240","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=221240"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221240\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221240"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221240"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Assumptions and Constants<\/strong><\/h3>\n\n\n\n\n- HHV of CH\u2084<\/strong> \u2248 55,500 kJ\/kg<\/li>\n\n\n\n
- LHV of CH\u2084<\/strong> \u2248 50,000 kJ\/kg (we use LHV since water stays vapor)<\/li>\n\n\n\n
- Cp of products<\/strong> \u2248 1.10 kJ\/kg\u00b7K (approximate for combustion gases)<\/li>\n\n\n\n
- T\u2081 (inlet temp)<\/strong> = 298 K<\/li>\n\n\n\n
- T\u2082 (exit temp)<\/strong> = 850 K<\/li>\n<\/ul>\n\n\n\n
We will use energy balance<\/strong> on the turbine and combustion process:<\/p>\n\n\n\n
\n\n\n\nStep-by-Step Solution<\/strong><\/h3>\n\n\n\n1. Determine the heat released by combustion per second<\/strong><\/h4>\n\n\n\nm\u02d9fuel=1200 kg\/h3600=0.3333 kg\/s\\dot{m}_{\\text{fuel}} = \\frac{1200 \\text{ kg\/h}}{3600} = 0.3333 \\text{ kg\/s} Q\u02d9in=m\u02d9fuel\u00d7LHVCH4=0.3333\u00d750,000=16,666.7 kW\\dot{Q}_{\\text{in}} = \\dot{m}_{\\text{fuel}} \\times \\text{LHV}_{CH_4} = 0.3333 \\times 50,000 = 16,666.7 \\text{ kW}<\/p>\n\n\n\n
\n\n\n\n2. Apply energy balance considering heat loss<\/strong><\/h4>\n\n\n\nLet WnetW_{\\text{net}} be the net power output.<\/p>\n\n\n\n
Heat loss = 10% of net power, so: Q\u02d9loss=0.10\u00d7Wnet\\dot{Q}_{\\text{loss}} = 0.10 \\times W_{\\text{net}}<\/p>\n\n\n\n
Energy balance: Heat in=Net work out+Heat loss\\text{Heat in} = \\text{Net work out} + \\text{Heat loss} 16,666.7=Wnet+0.10Wnet=1.10Wnet16,666.7 = W_{\\text{net}} + 0.10 W_{\\text{net}} = 1.10 W_{\\text{net}} Wnet=16,666.71.10=15,151.5 kW=15.15 MWW_{\\text{net}} = \\frac{16,666.7}{1.10} = 15,151.5 \\text{ kW} = \\boxed{15.15 \\text{ MW}}<\/p>\n\n\n\n
\n\n\n\nFinal Answer<\/strong><\/h3>\n\n\n\nNet Power Output = 15.15 MW<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\nThis problem involves calculating the net power output of a simple open gas turbine system fueled by methane. The combustion of methane is complete, and it takes place with an excess of air (400% of the theoretical requirement). The energy released by the combustion of methane is used to generate power, but 10% of this power is lost as heat to the surroundings.<\/p>\n\n\n\n
We begin by calculating the mass flow rate of the methane fuel in seconds, since power is in kW (kJ\/s). Then, we estimate the energy input rate using the lower heating value (LHV) of methane, which is suitable here because the water produced by combustion remains in vapor form. The LHV of methane is approximately 50,000 kJ\/kg. Multiplying this by the mass flow rate gives the heat released per second (kW), which represents the thermal energy made available by fuel combustion.<\/p>\n\n\n\n
Next, we apply an energy balance to the turbine system. Since the problem specifies that 10% of the power is lost as heat transfer, we set up the equation such that the input energy equals the useful net work output plus the heat loss. Solving this equation yields the net power output of the turbine.<\/p>\n\n\n\n
Kinetic and potential energy effects are negligible, simplifying the energy balance further. This problem demonstrates a practical application of the first law of thermodynamics (energy conservation) to a real-world engineering system. It highlights how combustion efficiency and losses influence the actual usable power output of a turbine.<\/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-221240","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221240","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=221240"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221240\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221240"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221240"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
We will use energy balance<\/strong> on the turbine and combustion process:<\/p>\n\n\n\n m\u02d9fuel=1200 kg\/h3600=0.3333 kg\/s\\dot{m}_{\\text{fuel}} = \\frac{1200 \\text{ kg\/h}}{3600} = 0.3333 \\text{ kg\/s} Q\u02d9in=m\u02d9fuel\u00d7LHVCH4=0.3333\u00d750,000=16,666.7 kW\\dot{Q}_{\\text{in}} = \\dot{m}_{\\text{fuel}} \\times \\text{LHV}_{CH_4} = 0.3333 \\times 50,000 = 16,666.7 \\text{ kW}<\/p>\n\n\n\n Let WnetW_{\\text{net}} be the net power output.<\/p>\n\n\n\n Heat loss = 10% of net power, so: Q\u02d9loss=0.10\u00d7Wnet\\dot{Q}_{\\text{loss}} = 0.10 \\times W_{\\text{net}}<\/p>\n\n\n\n Energy balance: Heat in=Net work out+Heat loss\\text{Heat in} = \\text{Net work out} + \\text{Heat loss} 16,666.7=Wnet+0.10Wnet=1.10Wnet16,666.7 = W_{\\text{net}} + 0.10 W_{\\text{net}} = 1.10 W_{\\text{net}} Wnet=16,666.71.10=15,151.5 kW=15.15 MWW_{\\text{net}} = \\frac{16,666.7}{1.10} = 15,151.5 \\text{ kW} = \\boxed{15.15 \\text{ MW}}<\/p>\n\n\n\n Net Power Output = 15.15 MW<\/strong><\/p>\n\n\n\n This problem involves calculating the net power output of a simple open gas turbine system fueled by methane. The combustion of methane is complete, and it takes place with an excess of air (400% of the theoretical requirement). The energy released by the combustion of methane is used to generate power, but 10% of this power is lost as heat to the surroundings.<\/p>\n\n\n\n We begin by calculating the mass flow rate of the methane fuel in seconds, since power is in kW (kJ\/s). Then, we estimate the energy input rate using the lower heating value (LHV) of methane, which is suitable here because the water produced by combustion remains in vapor form. The LHV of methane is approximately 50,000 kJ\/kg. Multiplying this by the mass flow rate gives the heat released per second (kW), which represents the thermal energy made available by fuel combustion.<\/p>\n\n\n\n Next, we apply an energy balance to the turbine system. Since the problem specifies that 10% of the power is lost as heat transfer, we set up the equation such that the input energy equals the useful net work output plus the heat loss. Solving this equation yields the net power output of the turbine.<\/p>\n\n\n\n Kinetic and potential energy effects are negligible, simplifying the energy balance further. This problem demonstrates a practical application of the first law of thermodynamics (energy conservation) to a real-world engineering system. It highlights how combustion efficiency and losses influence the actual usable power output of a turbine.<\/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-221240","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221240","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=221240"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221240\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221240"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221240"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep-by-Step Solution<\/strong><\/h3>\n\n\n\n
1. Determine the heat released by combustion per second<\/strong><\/h4>\n\n\n\n
\n\n\n\n2. Apply energy balance considering heat loss<\/strong><\/h4>\n\n\n\n
\n\n\n\nFinal Answer<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"