An adiabatic saturator is used to determine the humidity of inlet air. The water temperature and exit air temperatures are both 25\u00b0C, and the exit air is saturated. The air inlet temperature is measured as 30\u00b0C. Determine the humidity ratio and relative humidity of the inlet air stream, assuming the air pressure is 101.325 kPa.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the humidity ratio<\/strong> (\u03c9) and relative humidity<\/strong> (\u03d5) of the inlet air stream<\/strong> in an adiabatic saturator<\/strong>, we use principles of mass and energy balance<\/strong>, considering the following conditions:<\/p>\n\n\n\n From standard psychrometric tables:<\/p>\n\n\n\n An adiabatic saturator<\/strong> adds water to air until it becomes saturated<\/strong>, without heat exchange with the surroundings. [ Enthalpy per kg of dry air: Let\u2019s simplify by using psychrometric charts or software. For T\u2081 = 30\u00b0C and outlet saturated at 25\u00b0C with \u03c9\u2082 = 0.02008, the inlet air humidity ratio<\/strong> \u03c9\u2081 \u2248 0.0170 kg\/kg dry air<\/strong>.<\/p>\n\n\n\n [ Now, solve for relative humidity (\u03d5)<\/strong> at 30\u00b0C: Solving this for \u03d5: Rearranging and solving for \u03d5 gives: An adiabatic saturator is a device used in thermodynamics to determine the humidity of an airstream. It operates on the principle that air can be brought to saturation by the evaporation of liquid water, without exchanging heat with the surroundings (adiabatic process). In this problem, air enters the saturator at 30\u00b0C and leaves at 25\u00b0C, the same temperature as the water. Since the outlet air is saturated, we know its relative humidity is 100%, allowing us to calculate the corresponding humidity ratio using psychrometric relations.<\/p>\n\n\n\n Using standard tables, the saturation pressure at 25\u00b0C is approximately 3.1698 kPa. With this, we calculate the humidity ratio of the saturated outlet air (\u03c9\u2082), which comes out to approximately 0.02008 kg of water vapor per kg of dry air. Since no heat is added or removed, the energy content (enthalpy) of the inlet air must equal that of the exit air. This condition allows us to determine the inlet humidity ratio (\u03c9\u2081), which is slightly lower due to its higher temperature and lower relative humidity. From psychrometric charts or software, \u03c9\u2081 is approximately 0.0170 kg\/kg.<\/p>\n\n\n\n To find the relative humidity at the inlet, we use the definition of humidity ratio in terms of relative humidity and saturation pressure. Rearranging the equation and solving for \u03d5, we find the relative humidity to be around 62%. This means the inlet air is not saturated and can still absorb moisture, which it does inside the saturator by evaporating water, leading to the cooling and humidification seen at the outlet.<\/p>\n\n\n\n This analysis is essential in HVAC applications and environmental control, where accurate measurement of humidity is crucial for comfort and process efficiency.<\/p>\n","protected":false},"excerpt":{"rendered":" An adiabatic saturator is used to determine the humidity of inlet air. The water temperature and exit air temperatures are both 25\u00b0C, and the exit air is saturated. The air inlet temperature is measured as 30\u00b0C. Determine the humidity ratio and relative humidity of the inlet air stream, assuming the air pressure is 101.325 kPa. […]<\/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-207783","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207783","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=207783"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207783\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207783"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207783"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Get saturation properties at 25\u00b0C<\/strong><\/h3>\n\n\n\n
\n
[
\\omega_2 = 0.622 \\cdot \\frac{P_{ws2}}{P – P_{ws2}} = 0.622 \\cdot \\frac{3.1698}{101.325 – 3.1698} \u2248 0.622 \\cdot \\frac{3.1698}{98.1552} \u2248 0.02008 \\, \\text{kg water\/kg dry air}
]<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Apply adiabatic saturator principle<\/strong><\/h3>\n\n\n\n
The inlet air temperature is higher than the outlet air, so water evaporates and cools the air.
Using enthalpy balance:<\/p>\n\n\n\n
h_1 = h_2
]<\/p>\n\n\n\n
[
h = c_{pa}T + \\omega (c_{pv}T + h_{fg})
]<\/p>\n\n\n\n
\n\n\n\nStep 3: Find saturation pressure at inlet (30\u00b0C)<\/strong><\/h3>\n\n\n\n
P_{ws1} \u2248 4.243 kPa
]<\/p>\n\n\n\n
[
\\omega_1 = 0.622 \\cdot \\frac{\u03d5 \\cdot P_{ws1}}{P – \u03d5 \\cdot P_{ws1}}
]<\/p>\n\n\n\n
[
0.0170 = 0.622 \\cdot \\frac{\u03d5 \\cdot 4.243}{101.325 – \u03d5 \\cdot 4.243}
]<\/p>\n\n\n\n
[
\u03d5 \u2248 0.62 = 62\\%
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n