{"id":213276,"date":"2025-05-10T07:07:41","date_gmt":"2025-05-10T07:07:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213276"},"modified":"2025-05-10T07:07:44","modified_gmt":"2025-05-10T07:07:44","slug":"air-enters-a-compressor-operating-at-steady-state-with-a-pressure-of-14-7-lbf-in2-a-temperature-of-80-degrees-fahrenheit-and-a-volumetric-flow-rate-of-18-ft3-s","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/10\/air-enters-a-compressor-operating-at-steady-state-with-a-pressure-of-14-7-lbf-in2-a-temperature-of-80-degrees-fahrenheit-and-a-volumetric-flow-rate-of-18-ft3-s\/","title":{"rendered":"Air enters a compressor operating at steady state with a pressure of 14.7 lbf\/in2, a temperature of 80 degrees Fahrenheit, and a volumetric flow rate of 18 ft3\/s."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Air enters a compressor operating at steady state with a pressure of 14.7 lbf\/in2, a temperature of 80 degrees Fahrenheit, and a volumetric flow rate of 18 ft3\/s. The air exits the compressor at a pressure of 90 lbf\/in2. Heat transfer from the compressor to its surroundings occurs at a rate of 9.7 Btu per lb of air flowing. The compressor power input is 90 hp. Neglecting kinetic and potential energy effects and modeling the air as an ideal gas, determine the exit temperature, in degrees Fahrenheit.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">To solve this problem, we apply the <strong>First Law of Thermodynamics for a steady-flow process<\/strong>: Q\u02d9\u2212W\u02d9=m\u02d9(h2\u2212h1)\\dot{Q} &#8211; \\dot{W} = \\dot{m}(h_2 &#8211; h_1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Q\u02d9\\dot{Q}: heat transfer rate per unit mass (Btu\/lb)<\/li>\n\n\n\n<li>W\u02d9\\dot{W}: work input rate per unit mass (Btu\/lb)<\/li>\n\n\n\n<li>h1,h2h_1, h_2: specific enthalpy at inlet and outlet (Btu\/lb)<\/li>\n\n\n\n<li>m\u02d9\\dot{m}: mass flow rate (lb\/s)<\/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 given values to consistent units<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Given:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Inlet pressure: P1=14.7\u2009lbf\/in2P_1 = 14.7 \\, \\text{lbf\/in}^2<\/li>\n\n\n\n<li>Inlet temperature: T1=80\u2218F=540\u2009\u00b0RT_1 = 80^\\circ F = 540 \\, \\text{\u00b0R}<\/li>\n\n\n\n<li>Volumetric flow rate: V\u02d9=18\u2009ft3\/s\\dot{V} = 18 \\, \\text{ft}^3\/\\text{s}<\/li>\n\n\n\n<li>Exit pressure: P2=90\u2009lbf\/in2P_2 = 90 \\, \\text{lbf\/in}^2<\/li>\n\n\n\n<li>Heat transfer: Q\u02d9=\u22129.7\u2009Btu\/lb\\dot{Q} = -9.7 \\, \\text{Btu\/lb}<\/li>\n\n\n\n<li>Compressor power: W\u02d9=90\u2009hp=90\u00d72545=229050\u2009ft\\cdotplbf\/s\\dot{W} = 90 \\, \\text{hp} = 90 \\times 2545 = 229050 \\, \\text{ft\u00b7lbf\/s}<\/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 2: Find the mass flow rate<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Using the ideal gas law: m\u02d9=P1V\u02d9RT1\\dot{m} = \\frac{P_1 \\dot{V}}{R T_1}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where R=53.35\u2009ft\\cdotplbf\/(lbm\\cdotp\u00b0R)R = 53.35 \\, \\text{ft\u00b7lbf}\/(\\text{lbm\u00b7\u00b0R})<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Convert P1=14.7\u00d7144=2116.8\u2009lbf\/ft2P_1 = 14.7 \\times 144 = 2116.8 \\, \\text{lbf\/ft}^2 m\u02d9=2116.8\u00d71853.35\u00d7540=38102.428809\u22481.323\u2009lbm\/s\\dot{m} = \\frac{2116.8 \\times 18}{53.35 \\times 540} = \\frac{38102.4}{28809} \\approx 1.323 \\, \\text{lbm\/s}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Apply the First Law<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Rewriting the energy balance: Q\u02d9total\u2212W\u02d9totalm\u02d9=h2\u2212h1=cp(T2\u2212T1)\\frac{\\dot{Q}_{total} &#8211; \\dot{W}_{total}}{\\dot{m}} = h_2 &#8211; h_1 = c_p(T_2 &#8211; T_1)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Q\u02d9total=m\u02d9\u22c5(\u22129.7)\\dot{Q}_{total} = \\dot{m} \\cdot (-9.7)<\/li>\n\n\n\n<li>W\u02d9total=229050\u2009ft\\cdotplbf\/s=229050778.17\u2248294.4\u2009Btu\/s\\dot{W}_{total} = 229050 \\, \\text{ft\u00b7lbf\/s} = \\frac{229050}{778.17} \\approx 294.4 \\, \\text{Btu\/s}<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So: \u22129.7\u22c51.323\u2212294.41.323=cp(T2\u2212T1)\\frac{-9.7 \\cdot 1.323 &#8211; 294.4}{1.323} = c_p (T_2 &#8211; T_1) \u221212.83\u2212294.41.323=cp(T2\u2212T1)\\frac{-12.83 &#8211; 294.4}{1.323} = c_p (T_2 &#8211; T_1) \u2212232.1=cp(T2\u2212T1)-232.1 = c_p (T_2 &#8211; T_1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Use cp=0.24\u2009Btu\/lbm\\cdotp\u00b0Rc_p = 0.24 \\, \\text{Btu\/lbm\u00b7\u00b0R} T2\u2212T1=\u2212232.10.24\u2248\u2212967.1\u2009\u00b0RT_2 &#8211; T_1 = \\frac{-232.1}{0.24} \\approx -967.1 \\, \\text{\u00b0R} T2=T1+\u0394T=540\u2212967.1=\u2212427.1\u2009\u00b0RT_2 = T_1 + \\Delta T = 540 &#8211; 967.1 = -427.1 \\, \\text{\u00b0R}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This is <strong>not physically possible<\/strong>, so let\u2019s <strong>recheck the signs<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since <strong>work is input<\/strong> (positive into the system) and <strong>heat is lost<\/strong>, we must set: Q\u02d9=\u22129.7\u22c5m\u02d9(out)W\u02d9=+294.4(in)\\dot{Q} = -9.7 \\cdot \\dot{m} \\quad (\\text{out}) \\\\ \\dot{W} = +294.4 \\quad (\\text{in})<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So: h2\u2212h1=\u221212.83+294.41.323=281.571.323=212.87\u2009Btu\/lbh_2 &#8211; h_1 = \\frac{-12.83 + 294.4}{1.323} = \\frac{281.57}{1.323} = 212.87 \\, \\text{Btu\/lb} T2\u2212T1=212.870.24\u2248886.98\u2009\u00b0RT_2 &#8211; T_1 = \\frac{212.87}{0.24} \\approx 886.98 \\, \\text{\u00b0R} T2=540+886.98=1426.98\u2009\u00b0R=1426.98\u2212459.67=967.3\u2218FT_2 = 540 + 886.98 = 1426.98 \\, \\text{\u00b0R} = 1426.98 &#8211; 459.67 = \\boxed{967.3^\\circ F}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">967.3\u2218F\\boxed{967.3^\\circ \\text{F}?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Air enters a compressor operating at steady state with a pressure of 14.7 lbf\/in2, a temperature of 80 degrees Fahrenheit, and a volumetric flow rate of 18 ft3\/s. The air exits the compressor at a pressure of 90 lbf\/in2. Heat transfer from the compressor to its surroundings occurs at a rate of 9.7 Btu per [&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-213276","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213276","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=213276"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213276\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213276"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213276"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213276"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}