{"id":193109,"date":"2025-02-19T05:11:01","date_gmt":"2025-02-19T05:11:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=193109"},"modified":"2025-02-19T05:11:03","modified_gmt":"2025-02-19T05:11:03","slug":"using-the-tables-for-water-determine-the-specified-property-data-at-the-indicated-states-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/19\/using-the-tables-for-water-determine-the-specified-property-data-at-the-indicated-states-2\/","title":{"rendered":"Using the tables for water, determine the specified property data at the indicated states"},"content":{"rendered":"\n<p>Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on sketches of the p\u2013v and T\u2013v diagrams.<\/p>\n\n\n\n<p>(a) At p = 3 bar, T = 240\u00b0C, find v in m3 \/kg and u in kJ\/kg.<\/p>\n\n\n\n<p>(b) At p = 3 bar, v = 0.5 m3 \/kg, find T in \u00b0C and u in kJ\/kg.<\/p>\n\n\n\n<p>(c) At T = 400\u00b0C, p = 10 bar, find v in m3 \/kg and h in kJ\/kg.<\/p>\n\n\n\n<p>(d) At T = 320\u00b0C, v = 0.03 m3 \/kg, find p in MPa and u in kJ\/kg.<\/p>\n\n\n\n<p>(e) At p = 28 MPa, T = 520\u00b0C, find v in m3 \/kg and h in kJ\/kg.<\/p>\n\n\n\n<p>(f) At T = 100\u00b0C, x = 60%, find p in bar and v in m3 \/kg.<\/p>\n\n\n\n<p>(g) At T = 10\u00b0C, v = 100 m3 \/kg, find p in kPa and h in kJ\/kg.<\/p>\n\n\n\n<p>(h) At p = 4 MPa, T = 160\u00b0C, find v in m3 \/kg and u in kJ\/kg.<\/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 these thermodynamics problems, we will use steam tables to determine the properties of water at the specified states. We can use the p-v (pressure-specific volume) and T-v (temperature-specific volume) diagrams to help visualize the states. Here are the steps to determine the specified properties for each part of the question:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(a) At p = 3 bar, T = 240\u00b0C, find v in m\u00b3\/kg and u in kJ\/kg.<\/strong><\/h3>\n\n\n\n<p>From the steam tables:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At 3 bar and T = 240\u00b0C, water is in the superheated region (since the temperature is above the saturated temperature for 3 bar, which is approximately 133.5\u00b0C).<\/li>\n\n\n\n<li>Using the superheated steam tables, we find:<\/li>\n\n\n\n<li><strong>v (specific volume)<\/strong> \u2248 0.2577 m\u00b3\/kg<\/li>\n\n\n\n<li><strong>u (specific internal energy)<\/strong> \u2248 2837.5 kJ\/kg<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(b) At p = 3 bar, v = 0.5 m\u00b3\/kg, find T in \u00b0C and u in kJ\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At p = 3 bar and v = 0.5 m\u00b3\/kg, check if the state is in the saturated or superheated region:<\/li>\n\n\n\n<li>The saturated volume at 3 bar is between 0.00113 m\u00b3\/kg (saturated liquid) and 12.67 m\u00b3\/kg (saturated vapor).<\/li>\n\n\n\n<li>The value 0.5 m\u00b3\/kg is much higher than the saturated liquid volume but lower than the saturated vapor volume, so this is a superheated state.<\/li>\n\n\n\n<li>From the superheated steam tables for 3 bar and v = 0.5 m\u00b3\/kg, the corresponding:<\/li>\n\n\n\n<li><strong>T \u2248 500\u00b0C<\/strong><\/li>\n\n\n\n<li><strong>u \u2248 3470 kJ\/kg<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(c) At T = 400\u00b0C, p = 10 bar, find v in m\u00b3\/kg and h in kJ\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At T = 400\u00b0C and p = 10 bar, this state is in the superheated region (T is higher than the saturated temperature for 10 bar, which is approximately 180.4\u00b0C).<\/li>\n\n\n\n<li>Using the superheated steam tables:<\/li>\n\n\n\n<li><strong>v \u2248 0.2572 m\u00b3\/kg<\/strong><\/li>\n\n\n\n<li><strong>h (specific enthalpy) \u2248 3755.5 kJ\/kg<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(d) At T = 320\u00b0C, v = 0.03 m\u00b3\/kg, find p in MPa and u in kJ\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At T = 320\u00b0C and v = 0.03 m\u00b3\/kg, this state is in the superheated region.<\/li>\n\n\n\n<li>From the superheated steam tables, we find:<\/li>\n\n\n\n<li><strong>p \u2248 4.5 MPa<\/strong><\/li>\n\n\n\n<li><strong>u \u2248 3015 kJ\/kg<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(e) At p = 28 MPa, T = 520\u00b0C, find v in m\u00b3\/kg and h in kJ\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At p = 28 MPa and T = 520\u00b0C, this state is in the superheated region.<\/li>\n\n\n\n<li>From the superheated steam tables:<\/li>\n\n\n\n<li><strong>v \u2248 0.0562 m\u00b3\/kg<\/strong><\/li>\n\n\n\n<li><strong>h \u2248 3535 kJ\/kg<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(f) At T = 100\u00b0C, x = 60%, find p in bar and v in m\u00b3\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At T = 100\u00b0C and a quality (x) of 60%, the state is in the saturated mixture region.<\/li>\n\n\n\n<li>Using the saturated steam tables for 100\u00b0C, we find:<\/li>\n\n\n\n<li><strong>p \u2248 1.013 bar<\/strong><\/li>\n\n\n\n<li><strong>v \u2248 0.001043 + 0.6 * 1.672 = 1.004 m\u00b3\/kg<\/strong> (the specific volume of the mixture is calculated as (v = v_f + x(v_fg))).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(g) At T = 10\u00b0C, v = 100 m\u00b3\/kg, find p in kPa and h in kJ\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At T = 10\u00b0C and v = 100 m\u00b3\/kg, this state is in the subcooled liquid region, where the specific volume is larger than that of saturated liquid.<\/li>\n\n\n\n<li>From the steam tables:<\/li>\n\n\n\n<li><strong>p \u2248 50 kPa<\/strong><\/li>\n\n\n\n<li><strong>h \u2248 42 kJ\/kg<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(h) At p = 4 MPa, T = 160\u00b0C, find v in m\u00b3\/kg and u in kJ\/kg.<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At p = 4 MPa and T = 160\u00b0C, the state is in the superheated region.<\/li>\n\n\n\n<li>From the superheated steam tables:<\/li>\n\n\n\n<li><strong>v \u2248 0.07425 m\u00b3\/kg<\/strong><\/li>\n\n\n\n<li><strong>u \u2248 3450 kJ\/kg<\/strong><\/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 of Process:<\/strong><\/h3>\n\n\n\n<p>For each of the given states, we used the steam tables, which list the thermodynamic properties of water at various temperatures and pressures. These tables allow us to find properties such as specific volume (v), internal energy (u), enthalpy (h), and others, based on given conditions (pressure, temperature, or quality).<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Saturated liquid and vapor<\/strong> states are used when the temperature or pressure corresponds to the saturation point (i.e., the state where liquid and vapor phases coexist).<\/li>\n\n\n\n<li><strong>Superheated steam<\/strong> states occur when the temperature or pressure exceeds the saturation temperature or pressure, and the water is entirely in the vapor phase.<\/li>\n\n\n\n<li><strong>Quality (x)<\/strong> refers to the fraction of the vapor phase in a saturated mixture (0 is all liquid, 1 is all vapor).<\/li>\n<\/ol>\n\n\n\n<p>These values can be confirmed by plotting them on p-v and T-v diagrams, where the state would lie in the appropriate regions of the diagram based on whether it&#8217;s in the saturated, superheated, or subcooled regions.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on sketches of the p\u2013v and T\u2013v diagrams. (a) At p = 3 bar, T = 240\u00b0C, find v in m3 \/kg and u in kJ\/kg. (b) At p = 3 bar, v = 0.5 [&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-193109","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193109","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=193109"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193109\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193109"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193109"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}