{"id":220726,"date":"2025-05-28T17:15:30","date_gmt":"2025-05-28T17:15:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220726"},"modified":"2025-05-28T17:15:32","modified_gmt":"2025-05-28T17:15:32","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-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/a-reducing-elbow-in-a-horizontal-pipe-is-used-to-deflect-water-flow-by-an-angle-45-from-the-flow-direction-while-accelerating-it-3\/","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 <strong>anchoring force<\/strong> required to hold the reducing elbow in place, we apply the <strong>linear momentum equation<\/strong> to a control volume encompassing the elbow.<\/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>Flow deflection angle: \u03b8=45\u2218\\theta = 45^\\circ<\/li>\n\n\n\n<li>Inlet area, A1=150\u2009cm2=0.015\u2009m2A_1 = 150 \\, \\text{cm}^2 = 0.015 \\, \\text{m}^2<\/li>\n\n\n\n<li>Exit area, A2=25\u2009cm2=0.0025\u2009m2A_2 = 25 \\, \\text{cm}^2 = 0.0025 \\, \\text{m}^2<\/li>\n\n\n\n<li>Height difference, \u0394z=z2\u2212z1=0.40\u2009m\\Delta z = z_2 &#8211; z_1 = 0.40 \\, \\text{m}<\/li>\n\n\n\n<li>Mass of elbow + water: m=50\u2009kgm = 50 \\, \\text{kg}<\/li>\n\n\n\n<li>Discharge to atmosphere: P2=Patm=0\u2009gauge\u00a0pressureP_2 = P_{\\text{atm}} = 0 \\, \\text{gauge pressure}<\/li>\n<\/ul>\n\n\n\n<p>Assume:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Steady, incompressible flow<\/li>\n\n\n\n<li>Water density, \u03c1=1000\u2009kg\/m3\\rho = 1000 \\, \\text{kg\/m}^3<\/li>\n\n\n\n<li>Atmospheric pressure at outlet<\/li>\n\n\n\n<li>Neglect viscous effects and friction<\/li>\n<\/ul>\n\n\n\n<p>Let\u2019s denote:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>v1v_1: velocity at the inlet<\/li>\n\n\n\n<li>v2v_2: velocity at the outlet<\/li>\n\n\n\n<li>QQ: volumetric flow rate<\/li>\n\n\n\n<li>m\u02d9\\dot{m}: mass flow rate<\/li>\n\n\n\n<li>F\u20d7\\vec{F}: net anchoring force<\/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>Q=A1v1=A2v2\u21d2v2=A1A2v1=0.0150.0025v1=6v1Q = A_1 v_1 = A_2 v_2 \\Rightarrow v_2 = \\frac{A_1}{A_2} v_1 = \\frac{0.015}{0.0025} v_1 = 6v_1<\/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: Apply Bernoulli\u2019s Equation (from inlet to outlet)<\/strong><\/h3>\n\n\n\n<p>P1\u03c1g+v122g+z1=v222g+z2\\frac{P_1}{\\rho g} + \\frac{v_1^2}{2g} + z_1 = \\frac{v_2^2}{2g} + z_2 P1\u03c1g=(v22\u2212v12)2g+\u0394z=(36v12\u2212v12)2g+0.4=35v122g+0.4\u21d2P1=\u03c1(35v122+0.4g)\\frac{P_1}{\\rho g} = \\frac{(v_2^2 &#8211; v_1^2)}{2g} + \\Delta z = \\frac{(36v_1^2 &#8211; v_1^2)}{2g} + 0.4 = \\frac{35v_1^2}{2g} + 0.4 \\Rightarrow P_1 = \\rho \\left(\\frac{35v_1^2}{2} + 0.4g\\right)<\/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: Momentum Equation (Control Volume)<\/strong><\/h3>\n\n\n\n<p>In x-direction (horizontal): Fx=m\u02d9(v2x\u2212v1x)+P1A1F_x = \\dot{m}(v_{2x} &#8211; v_{1x}) + P_1 A_1<\/p>\n\n\n\n<p>Inlet: horizontal<br>Outlet: deflected 45\u00b0 up v1x=v1,v2x=v2cos\u2061(45\u2218)=6v1\u22c522v_{1x} = v_1, \\quad v_{2x} = v_2 \\cos(45^\\circ) = 6v_1 \\cdot \\frac{\\sqrt{2}}{2} m\u02d9=\u03c1Q=\u03c1A1v1\\dot{m} = \\rho Q = \\rho A_1 v_1 Fx=\u03c1A1v1(6v1\u22c522\u2212v1)+P1A1=\u03c1A1v12(32\u22121)+P1A1F_x = \\rho A_1 v_1 \\left(6v_1 \\cdot \\frac{\\sqrt{2}}{2} &#8211; v_1\\right) + P_1 A_1 = \\rho A_1 v_1^2 \\left(3\\sqrt{2} &#8211; 1\\right) + P_1 A_1<\/p>\n\n\n\n<p>Substitute P1P_1 from Bernoulli: Fx=\u03c1A1v12[(32\u22121)+(35v122+0.4g)\/v12]F_x = \\rho A_1 v_1^2 \\left[(3\\sqrt{2} &#8211; 1) + \\left(\\frac{35v_1^2}{2} + 0.4g\\right)\/v_1^2 \\right]<\/p>\n\n\n\n<p>Now pick a flow rate to get a number.<\/p>\n\n\n\n<p>Assume: v1=2\u2009m\/s\u21d2v2=12\u2009m\/sv_1 = 2 \\, \\text{m\/s} \\Rightarrow v_2 = 12 \\, \\text{m\/s} P1=1000(35(4)2+0.4\u22c59.81)=1000(70+3.924)=73,924\u2009PaP_1 = 1000 \\left(\\frac{35(4)}{2} + 0.4 \\cdot 9.81 \\right) = 1000(70 + 3.924) = 73,924 \\, \\text{Pa}<\/p>\n\n\n\n<p>Now compute: Fx=1000\u22c50.015\u22c54\u22c5(32\u22121)+73,924\u22c50.015=60\u22c5(32\u22121)+1,108.86\u224860\u22c5(3\u22c51.414\u22121)+1,108.86=60\u22c5(4.242\u22121)+1,108.86=60\u22c53.242+1,108.86=194.52+1,108.86\u22481,303.38\u2009N&nbsp;(horizontal)F_x = 1000 \\cdot 0.015 \\cdot 4 \\cdot (3\\sqrt{2} &#8211; 1) + 73,924 \\cdot 0.015 = 60 \\cdot (3\\sqrt{2} &#8211; 1) + 1,108.86 \\approx 60 \\cdot (3 \\cdot 1.414 &#8211; 1) + 1,108.86 = 60 \\cdot (4.242 &#8211; 1) + 1,108.86 = 60 \\cdot 3.242 + 1,108.86 = 194.52 + 1,108.86 \\approx \\boxed{1,303.38 \\, \\text{N (horizontal)}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Vertical Force (y-direction)<\/strong><\/h3>\n\n\n\n<p>Fy=m\u02d9(v2y\u2212v1y)\u2212mg=\u03c1A1v1(v2sin\u2061\u03b8\u22120)\u221250\u22c59.81=1000\u22c50.015\u22c52\u22c512\u22c522\u2212490.5\u2248254.56\u2212490.5\u2248\u2212235.94\u2009N&nbsp;(downward)F_y = \\dot{m} (v_{2y} &#8211; v_{1y}) &#8211; mg = \\rho A_1 v_1 (v_2 \\sin \\theta &#8211; 0) &#8211; 50 \\cdot 9.81 = 1000 \\cdot 0.015 \\cdot 2 \\cdot 12 \\cdot \\frac{\\sqrt{2}}{2} &#8211; 490.5 \\approx 254.56 &#8211; 490.5 \\approx \\boxed{-235.94 \\, \\text{N (downward)}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Result: Anchoring Force<\/strong><\/h3>\n\n\n\n<p>F\u20d7anchor={Fx\u22481303\u2009N&nbsp;(to&nbsp;the&nbsp;left)Fy\u2248236\u2009N&nbsp;(upward)\\boxed{ \\vec{F}_{\\text{anchor}} = \\begin{cases} F_x \\approx 1303 \\, \\text{N (to the left)} \\\\ F_y \\approx 236 \\, \\text{N (upward)} \\end{cases} }<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>In fluid systems, when flow is redirected or accelerated, forces are exerted on the conduits due to momentum change. A reducing elbow with water flowing through it changes both the speed and direction of the water, resulting in dynamic forces. To prevent movement or damage, an external anchoring force must be applied.<\/p>\n\n\n\n<p>The given problem involves water accelerating from a larger inlet area (150 cm\u00b2) to a smaller exit area (25 cm\u00b2) and being deflected by 45\u00b0. The water speed increases by a factor of 6 due to the area reduction, leading to a higher outlet velocity. Using the continuity equation, we relate inlet and outlet velocities. Next, Bernoulli\u2019s principle helps determine the inlet pressure by accounting for elevation gain and velocity change.<\/p>\n\n\n\n<p>To find the anchoring force, the linear momentum equation is applied in both horizontal (x) and vertical (y) directions. In the x-direction, the force is due to the momentum change between inlet and outlet velocities and the pressure force on the inlet. In the y-direction, the change in vertical velocity and the weight of the elbow system are considered.<\/p>\n\n\n\n<p>We assumed a typical inlet velocity (2 m\/s) to compute numerical results. The anchoring force turns out to be approximately <strong>1303 N horizontally<\/strong> and <strong>236 N vertically<\/strong>. These values indicate the elbow tends to move in the direction of water exit and downward due to momentum and gravity, respectively. The anchoring system must counteract these forces to keep the elbow stationary and ensure safe and stable operation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-75.jpeg\" alt=\"\" class=\"wp-image-220727\"\/><\/figure>\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-220726","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220726","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=220726"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220726\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220726"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220726"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220726"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}