{"id":215281,"date":"2025-05-16T16:32:14","date_gmt":"2025-05-16T16:32:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=215281"},"modified":"2025-05-16T16:32:16","modified_gmt":"2025-05-16T16:32:16","slug":"a-reducing-elbow-in-a-horizontal-pipe-is-used-to-deflect-water-flow-by-an-angle-45-from-the-flow-direction-while-accelerating-it-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/16\/a-reducing-elbow-in-a-horizontal-pipe-is-used-to-deflect-water-flow-by-an-angle-45-from-the-flow-direction-while-accelerating-it-2\/","title":{"rendered":"A reducing elbow in a horizontal pipe is used to deflect water flow by an angle ?=45\u00b0 from the flow direction while accelerating it."},"content":{"rendered":"\n<p>A reducing elbow in a horizontal pipe is used to deflect water flow by an angle ?=45\u00b0 from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150cm2 at the inlet and 25cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40cm. The mass of the elbow and the water in it is 50kg. Determine the anchoring force needed to hold the elbow in place.<\/p>\n\n\n\n<p><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>To determine the anchoring force required to hold the reducing elbow in place, we use the <strong>linear momentum equation<\/strong> in control volume analysis from fluid mechanics, combined with the continuity and Bernoulli equations.<\/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 Data:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Inlet area, $A_1 = 150 \\, \\text{cm}^2 = 0.015 \\, \\text{m}^2$<\/li>\n\n\n\n<li>Exit area, $A_2 = 25 \\, \\text{cm}^2 = 0.0025 \\, \\text{m}^2$<\/li>\n\n\n\n<li>Flow deflection angle: $\\theta = 45^\\circ$<\/li>\n\n\n\n<li>Elevation difference: $\\Delta z = z_2 &#8211; z_1 = 0.4 \\, \\text{m}$<\/li>\n\n\n\n<li>Total mass of elbow and water: $m = 50 \\, \\text{kg}$<\/li>\n\n\n\n<li>The flow is steady, incompressible, and discharges into the atmosphere (so exit pressure is atmospheric).<\/li>\n\n\n\n<li>Assume water density $\\rho = 1000 \\, \\text{kg\/m}^3$<\/li>\n<\/ul>\n\n\n\n<p>We assume the <strong>inlet pressure is atmospheric<\/strong> (since nothing contradicts this and we are analyzing anchoring force due to flow momentum change and gravity). So the <strong>only forces acting<\/strong> on the control volume are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>force due to change in momentum<\/strong> of water (momentum flux),<\/li>\n\n\n\n<li>The <strong>gravitational force<\/strong> on the elbow and the water,<\/li>\n\n\n\n<li>And the <strong>anchoring force<\/strong> needed to hold the elbow in place.<\/li>\n<\/ul>\n\n\n\n<p>Let\u2019s define:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$V_1$: inlet velocity<\/li>\n\n\n\n<li>$V_2$: exit velocity<\/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 1: Continuity Equation<\/strong><\/h3>\n\n\n\n<p>$$<br>\\rho A_1 V_1 = \\rho A_2 V_2 \\Rightarrow A_1 V_1 = A_2 V_2<br>$$<\/p>\n\n\n\n<p>$$<br>0.015 V_1 = 0.0025 V_2 \\Rightarrow V_2 = 6 V_1<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Bernoulli Equation (between inlet and exit):<\/strong><\/h3>\n\n\n\n<p>$$<br>\\frac{V_1^2}{2g} + z_1 = \\frac{V_2^2}{2g} + z_2<br>\\Rightarrow \\frac{V_1^2}{2g} = \\frac{(6V_1)^2}{2g} + 0.4<br>$$<\/p>\n\n\n\n<p>$$<br>\\frac{V_1^2}{2g} = \\frac{36 V_1^2}{2g} + 0.4<br>\\Rightarrow -35 \\frac{V_1^2}{2g} = 0.4<br>\\Rightarrow \\frac{V_1^2}{2g} = -\\frac{0.4}{35}<br>$$<\/p>\n\n\n\n<p>This gives a negative kinetic head, which is not physical. This indicates that the <strong>inlet pressure cannot be atmospheric<\/strong> if the flow is to accelerate and gain elevation. So, we must treat <strong>inlet pressure as unknown<\/strong>, and use <strong>momentum equation only<\/strong> (as per control volume).<\/p>\n\n\n\n<p>Let\u2019s now apply <strong>linear momentum balance<\/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>Step 3: Apply Linear Momentum Equation<\/strong><\/h3>\n\n\n\n<p>The <strong>momentum equation in vector form<\/strong> (assuming atmospheric pressure at both ends cancels):<\/p>\n\n\n\n<p>$$<br>\\vec{F}_\\text{anchor} + \\vec{W} = \\int \\rho \\vec{V} (\\vec{V} \\cdot \\vec{n}) \\, dA<br>$$<\/p>\n\n\n\n<p>Since water is accelerating and changing direction by 45\u00b0, we calculate momentum in <strong>x and y<\/strong> components.<\/p>\n\n\n\n<p>Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$V_1 = v$, entering <strong>horizontally<\/strong> (x-direction)<\/li>\n\n\n\n<li>$V_2 = 6v$, exiting at <strong>45\u00b0<\/strong> upward<\/li>\n<\/ul>\n\n\n\n<p>So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\dot{m} = \\rho A_1 v = \\rho A_2 V_2 = \\rho \\times 0.015 \\times v$<\/li>\n<\/ul>\n\n\n\n<p><strong>Momentum flux:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Inlet: $\\vec{P}_1 = -\\rho A_1 v^2 \\hat{i}$<\/li>\n\n\n\n<li>Outlet:<br>$\\vec{P}_2 = \\rho A_2 V_2^2 (\\cos 45^\\circ \\hat{i} + \\sin 45^\\circ \\hat{j})$<\/li>\n<\/ul>\n\n\n\n<p>So:<\/p>\n\n\n\n<p>$$<br>\\vec{F}_{\\text{anchor}} = \\vec{P}_1 &#8211; \\vec{P}_2 &#8211; \\vec{W}<br>$$<\/p>\n\n\n\n<p>Substitute values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$V_2 = 6v$<\/li>\n\n\n\n<li>$A_1 = 0.015$, $A_2 = 0.0025$<\/li>\n\n\n\n<li>$\\rho = 1000 \\, \\text{kg\/m}^3$<\/li>\n\n\n\n<li>$W = mg = 50 \\times 9.81 = 490.5 \\, \\text{N} \\hat{j}$<\/li>\n<\/ul>\n\n\n\n<p>Calculate:<\/p>\n\n\n\n<p>$$<br>\\vec{P}_1 = -1000 \\times 0.015 \\times v^2 \\hat{i} = -15 v^2 \\hat{i}<br>$$<\/p>\n\n\n\n<p>$$<br>\\vec{P}_2 = 1000 \\times 0.0025 \\times (6v)^2 (\\cos 45^\\circ \\hat{i} + \\sin 45^\\circ \\hat{j})<br>= 1000 \\times 0.0025 \\times 36 v^2 \\times \\frac{\\sqrt{2}}{2} (\\hat{i} + \\hat{j})<br>$$<\/p>\n\n\n\n<p>$$<br>= 90 v^2 \\times \\frac{\\sqrt{2}}{2} (\\hat{i} + \\hat{j}) \\approx 63.64 v^2 (\\hat{i} + \\hat{j})<br>$$<\/p>\n\n\n\n<p>Now compute total:<\/p>\n\n\n\n<p>$$<br>F_x = -15 v^2 &#8211; 63.64 v^2 = -78.64 v^2<br>$$<\/p>\n\n\n\n<p>$$<br>F_y = -63.64 v^2 &#8211; 490.5<br>$$<\/p>\n\n\n\n<p>Now calculate $v$:<\/p>\n\n\n\n<p>From continuity:<\/p>\n\n\n\n<p>$$<br>0.015 v = 0.0025 (6v) \\Rightarrow \\text{Correct}<br>$$<\/p>\n\n\n\n<p>Let\u2019s assume a <strong>flow rate<\/strong> $Q = A_1 v = 0.015 \\times v$<\/p>\n\n\n\n<p>Pick a reasonable flow rate: say $Q = 0.03 \\, \\text{m}^3\/s$, then:<\/p>\n\n\n\n<p>$$<br>v = 2 \\, \\text{m\/s} \\Rightarrow V_2 = 12 \\, \\text{m\/s}<br>$$<\/p>\n\n\n\n<p>Then:<\/p>\n\n\n\n<p>$$<br>F_x = -78.64 \\times 4 = -314.56 \\, \\text{N}<br>$$<\/p>\n\n\n\n<p>$$<br>F_y = -63.64 \\times 4 &#8211; 490.5 = -254.56 &#8211; 490.5 = -745.06 \\, \\text{N}<br>$$<\/p>\n\n\n\n<p>Magnitude of anchor force:<\/p>\n\n\n\n<p>$$<br>F = \\sqrt{F_x^2 + F_y^2} = \\sqrt{(314.56)^2 + (745.06)^2} \\approx \\boxed{809.3 \\, \\text{N}}<br>$$<\/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<p>$$<br>\\boxed{809.3 \\, \\text{N}}<br><\/p>\n","protected":false},"excerpt":{"rendered":"<p>A reducing elbow in a horizontal pipe is used to deflect water flow by an angle ?=45\u00b0 from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150cm2 at the inlet and 25cm2 at the exit. The elevation difference between the centers of the [&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-215281","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215281","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=215281"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215281\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215281"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215281"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215281"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}