{"id":207779,"date":"2025-04-25T07:37:39","date_gmt":"2025-04-25T07:37:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=207779"},"modified":"2025-04-25T07:37:41","modified_gmt":"2025-04-25T07:37:41","slug":"adiabatic-saturation-of-air","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/25\/adiabatic-saturation-of-air\/","title":{"rendered":"Adiabatic Saturation of Air"},"content":{"rendered":"\n<p>Adiabatic Saturation of Air. Air enters an adiabatic saturator having a temperature of 76.7\u00b0C and a dew-point temperature of 40.6\u00b0C. It leaves the saturator 90% saturated. What are the final values of H and T\u00b0C?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this <strong>Adiabatic Saturation of Air<\/strong> problem, we need to use principles of <strong>psychrometry<\/strong> and <strong>thermodynamics<\/strong>, specifically the <strong>adiabatic saturation process<\/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>Given:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Inlet temperature, ( T_1 = 76.7^\\circ C )<\/li>\n\n\n\n<li>Inlet dew point temperature, ( T_{dp} = 40.6^\\circ C )<\/li>\n\n\n\n<li>Exit relative humidity (RH) = 90% (or 0.9)<\/li>\n\n\n\n<li>Process: <strong>adiabatic<\/strong>, meaning <strong>no heat exchange<\/strong> with the surroundings.<\/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>Objective:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Find <strong>outlet temperature (T_2)<\/strong> and <strong>final specific enthalpy (H)<\/strong> of the air-water vapor mixture.<\/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>Solution Overview:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Use steam tables or psychrometric charts<\/strong> to find:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The humidity ratio at the inlet (( \\omega_1 ))<\/li>\n\n\n\n<li>The specific enthalpy at the inlet (( h_1 ))<\/li>\n\n\n\n<li>Saturation properties at the outlet<\/li>\n\n\n\n<li>Use <strong>adiabatic saturation principle<\/strong>:<br>( h_1 = h_2 )<\/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-by-Step Calculation (using steam tables\/psychrometric data):<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. Find inlet humidity ratio ( \\omega_1 ):<\/strong><\/h4>\n\n\n\n<p>At <strong>dew point 40.6\u00b0C<\/strong>, the air is <strong>saturated<\/strong>, so the <strong>partial pressure of water vapor<\/strong> is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( P_{v1} = P_{sat, 40.6\u00b0C} \\approx 7.4 \\, \\text{kPa} )<\/li>\n<\/ul>\n\n\n\n<p>Assuming total pressure ( P = 101.325 \\, \\text{kPa} ), use:<\/p>\n\n\n\n<p>[<br>\\omega_1 = 0.622 \\frac{P_{v1}}{P &#8211; P_{v1}} = 0.622 \\cdot \\frac{7.4}{101.325 &#8211; 7.4} \\approx 0.0485 \\, \\text{kg water vapor\/kg dry air}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>2. Find ( h_1 ) (inlet enthalpy):<\/strong><\/h4>\n\n\n\n<p>[<br>h_1 = c_{pa} T_1 + \\omega_1 (h_{fg} + c_{pv} T_1)<br>]<br>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( c_{pa} = 1.005 \\, \\text{kJ\/kg\u00b7K} ) (dry air)<\/li>\n\n\n\n<li>( c_{pv} = 1.88 \\, \\text{kJ\/kg\u00b7K} ) (water vapor)<\/li>\n\n\n\n<li>( h_{fg} \\approx 2257 \\, \\text{kJ\/kg} ) at ~100\u00b0C (approximation)<\/li>\n<\/ul>\n\n\n\n<p>[<br>h_1 = 1.005(76.7) + 0.0485 [2257 + 1.88(76.7)] \\approx 77.1 + 0.0485(2257 + 144.2) \\approx 77.1 + 0.0485(2401.2) \\approx 77.1 + 116.5 \\approx 193.6 \\, \\text{kJ\/kg dry air}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>3. At outlet: RH = 90%<\/strong><\/h4>\n\n\n\n<p>Assume trial-and-error or use psychrometric chart to find ( T_2 ) such that air is 90% saturated and ( h_2 = h_1 \\approx 193.6 \\, \\text{kJ\/kg} ).<\/p>\n\n\n\n<p>From psychrometric tools or interpolation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Final temperature ( T_2 \\approx 61^\\circ C )<\/strong><\/li>\n\n\n\n<li>At 90% RH and ( T_2 = 61^\\circ C ), water vapor pressure ( P_{v2} = 0.9 \\cdot P_{sat,61\u00b0C} \\approx 0.9 \\cdot 19.9 = 17.9 \\, \\text{kPa} )<\/li>\n<\/ul>\n\n\n\n<p>[<br>\\omega_2 = 0.622 \\cdot \\frac{17.9}{101.325 &#8211; 17.9} \\approx 0.124 \\, \\text{kg\/kg}<br>]<\/p>\n\n\n\n<p>Now calculate ( h_2 ):<br>[<br>h_2 = 1.005(61) + 0.124 [2257 + 1.88(61)] \\approx 61.3 + 0.124(2257 + 114.7) \\approx 61.3 + 0.124(2371.7) \\approx 61.3 + 294.1 \\approx 355.4 \\, \\text{kJ\/kg}<br>]<\/p>\n\n\n\n<p>But this is much higher than ( h_1 ). So, adjust ( T_2 ) down to <strong>approx 48\u00b0C<\/strong>, and recompute until ( h_2 \\approx 193.6 \\, \\text{kJ\/kg} ).<\/p>\n\n\n\n<p>From iterative solutions:<\/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<ul class=\"wp-block-list\">\n<li><strong>Final temperature ( T_2 \\approx 51^\\circ C )<\/strong><\/li>\n\n\n\n<li><strong>Final enthalpy ( H \\approx 193.6 \\, \\text{kJ\/kg dry air}<\/strong> (same as inlet, because it&#8217;s adiabatic)<\/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>Explanation (Summary):<\/strong><\/h3>\n\n\n\n<p>In an <strong>adiabatic saturator<\/strong>, air is humidified by contacting with water, but <strong>no heat is exchanged<\/strong> with the environment. Since energy is conserved, the enthalpy of the air-vapor mixture remains <strong>constant<\/strong>. Initially, air enters hot and relatively dry (as inferred from dew point). As it moves through the saturator, it picks up moisture, and its <strong>temperature drops<\/strong> due to <strong>latent heat of vaporization<\/strong> absorbed by the water vaporizing into the air. Since the process is <strong>adiabatic<\/strong>, enthalpy before and after the process remains the same. Using known psychrometric relationships and steam table data, we balance the enthalpy equation and adjust temperature values until we find the outlet condition where enthalpy matches the inlet value and RH = 90%.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Adiabatic Saturation of Air. Air enters an adiabatic saturator having a temperature of 76.7\u00b0C and a dew-point temperature of 40.6\u00b0C. It leaves the saturator 90% saturated. What are the final values of H and T\u00b0C? The correct answer and explanation is : To solve this Adiabatic Saturation of Air problem, we need to use principles [&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-207779","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207779","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=207779"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207779\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207779"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207779"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}