{"id":187991,"date":"2025-02-06T09:05:26","date_gmt":"2025-02-06T09:05:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187991"},"modified":"2025-02-06T09:05:28","modified_gmt":"2025-02-06T09:05:28","slug":"using-the-tables-for-water-determine-the-specified-property-data-at-the-indicated-states","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/using-the-tables-for-water-determine-the-specified-property-data-at-the-indicated-states\/","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.<br>(a) At p = 3 bar, v = 0.35m ^ 3 \/ k * g find Tin \u00b0C and u in kJ\/kg.<br>(b) At T = 320 deg * C v = 0.13m ^ 3 \/ k * g find p in MPa and u in kJ\/kg.<br>(c) At , T = 400 deg * C find v in m\u00b3\/kg and h in kJ\/kg. p = 28MPi<br>(d) At T = 10 deg * C v = 100m ^ 3 \/ k * g find pin kPa and h 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 determine the specified thermodynamic properties of water at the given states, we refer to standard steam tables, which provide detailed data on water&#8217;s properties across various temperatures and pressures. Here&#8217;s the analysis for each part:<\/p>\n\n\n\n<p><strong>(a) At ( p = 3 \\, \\text{bar} ) and ( v = 0.35 \\, \\text{m}^3\/\\text{kg} ):<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the Saturation Temperature:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At ( p = 3 \\, \\text{bar} ) (which is equivalent to 300 kPa), the saturation temperature ( T_{\\text{sat}} ) is approximately 133.5\u00b0C.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Compare Specific Volume with Saturated Values:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>From steam tables:\n<ul class=\"wp-block-list\">\n<li>Specific volume of saturated liquid (( v_f )) at 3 bar: approximately ( 0.001017 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n\n\n\n<li>Specific volume of saturated vapor (( v_g )) at 3 bar: approximately ( 0.6058 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Given ( v = 0.35 \\, \\text{m}^3\/\\text{kg} ), which lies between ( v_f ) and ( v_g ), indicating a mixture of liquid and vapor (i.e., a saturated mixture).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate Quality (x):<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Quality ( x ) represents the mass fraction of vapor in the mixture and is calculated as:<br>[<br>x = \\frac{v &#8211; v_f}{v_g &#8211; v_f} = \\frac{0.35 &#8211; 0.001017}{0.6058 &#8211; 0.001017} \\approx 0.577<br>]<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Temperature and Specific Internal Energy (u):<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For a saturated mixture at 3 bar:\n<ul class=\"wp-block-list\">\n<li>Temperature ( T ) is ( T_{\\text{sat}} = 133.5^\\circ\\text{C} ).<\/li>\n\n\n\n<li>Specific internal energy ( u ) is calculated as:<br>[<br>u = u_f + x \\times (u_g &#8211; u_f)<br>]<br>where:<\/li>\n\n\n\n<li>( u_f ) (internal energy of saturated liquid) \u2248 561.7 kJ\/kg.<\/li>\n\n\n\n<li>( u_g ) (internal energy of saturated vapor) \u2248 2550.5 kJ\/kg.<\/li>\n\n\n\n<li>Thus:<br>[<br>u = 561.7 + 0.577 \\times (2550.5 &#8211; 561.7) \\approx 1615.5 \\, \\text{kJ\/kg}<br>]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>(b) At ( T = 320^\\circ\\text{C} ) and ( v = 0.13 \\, \\text{m}^3\/\\text{kg} ):<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the Saturation Pressure:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At ( T = 320^\\circ\\text{C} ), the saturation pressure ( p_{\\text{sat}} ) is approximately 11.29 MPa.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Compare Specific Volume with Saturated Values:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>From steam tables:\n<ul class=\"wp-block-list\">\n<li>( v_f ) at 320\u00b0C: approximately ( 0.001307 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n\n\n\n<li>( v_g ) at 320\u00b0C: approximately ( 0.01803 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Given ( v = 0.13 \\, \\text{m}^3\/\\text{kg} ), which is significantly greater than ( v_g ), indicating a superheated vapor state.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Pressure and Specific Internal Energy:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For superheated steam at 320\u00b0C and ( v = 0.13 \\, \\text{m}^3\/\\text{kg} ):\n<ul class=\"wp-block-list\">\n<li>Using superheated steam tables, interpolate to find the corresponding pressure and internal energy. However, since ( v ) is much larger than ( v_g ), this suggests a low-pressure superheated state.<\/li>\n\n\n\n<li>Estimating from available data, the pressure ( p ) is approximately 0.5 MPa (5 bar).<\/li>\n\n\n\n<li>Specific internal energy ( u ) at this state is approximately 2700 kJ\/kg.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>(c) At ( T = 400^\\circ\\text{C} ) and ( p = 28 \\, \\text{MPa} ):<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Compare with Critical Point:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The critical pressure of water is 22.064 MPa. Since 28 MPa > 22.064 MPa, the water is in a compressed liquid state at 400\u00b0C.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Specific Volume and Enthalpy:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Compressed liquid data at such high pressures are limited in standard steam tables. However, for approximation:\n<ul class=\"wp-block-list\">\n<li>Specific volume ( v ) is slightly less than that of saturated liquid at the same temperature, so ( v \\approx v_f ) at 400\u00b0C, which is approximately ( 0.00146 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n\n\n\n<li>Enthalpy ( h ) can be approximated as ( h_f ) at 400\u00b0C, which is approximately 1738 kJ\/kg.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>(d) At ( T = 10^\\circ\\text{C} ) and ( v = 100 \\, \\text{m}^3\/\\text{kg} ):<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Saturation Pressure:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At ( T = 10^\\circ\\text{C} ), the saturation pressure ( p_{\\text{sat}} ) is approximately 1.23 kPa.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Compare Specific Volume with Saturated Values:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>From steam tables:\n<ul class=\"wp-block-list\">\n<li>( v_f ) at 10\u00b0C: approximately ( 0.001 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n\n\n\n<li>( v_g ) at 10\u00b0C: approximately ( 14.67 \\, \\text{m}^3\/\\text{kg} ).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Given ( v = 100 \\, \\text{m}^3\/\\text{kg} ), which is much greater than ( v_g ), indicating a superheated vapor state.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Pressure and Enthalpy:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For superheated steam at 10\u00b0C and ( v = 100 \\, \\text{m}^3\/\\text{kg} ):\n<ul class=\"wp-block-list\">\n<li>Using the ideal gas equation ( p = \\frac{RT}{v} ):<\/li>\n\n\n\n<li>( R ) (specific gas constant for water vapor) \u2248 0.4615 kJ\/(kg\u00b7K).<\/li>\n\n\n\n<li>( T ) in Kelvin = 10 + 273.15 = 283.15 K.<\/li>\n\n\n\n<li>Thus:<br>[<br>p = \\frac{0.4615 \\times 283.15}{100} \\approx 1.31 \\, \\text{kPa}<br>]<\/li>\n\n\n\n<li>This pressure is close to the saturation pressure, confirming the superheated state.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Using the tables for water, determine the specified property data at the indicated states.(a) At p = 3 bar, v = 0.35m ^ 3 \/ k * g find Tin \u00b0C and u in kJ\/kg.(b) At T = 320 deg * C v = 0.13m ^ 3 \/ k * g find p in MPa [&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 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