{"id":240787,"date":"2025-07-03T11:50:51","date_gmt":"2025-07-03T11:50:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=240787"},"modified":"2025-07-03T11:50:53","modified_gmt":"2025-07-03T11:50:53","slug":"a-horizontal-pipe-carries-a-smoothly-flowing-liquid-of-density-1110-kg-m3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/a-horizontal-pipe-carries-a-smoothly-flowing-liquid-of-density-1110-kg-m3\/","title":{"rendered":"A horizontal pipe carries a smoothly flowing liquid of density 1110 kg\/m^3."},"content":{"rendered":"\n

A horizontal pipe carries a smoothly flowing liquid of density 1110 kg\/m^3. At Locations 1 and 2 along the pipe, the diameters are d1 = 5.45 cm and d2 = 2.83 cm, respectively. The flow speed at Location 1 is 2.83 m\/s. What is the pressure difference \u00ce\u201dP between Location 2 and Location 1 (including its sign)? Ignore viscosity.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the pressure difference \u0394P\\Delta P\u0394P between Locations 2 and 1 in the pipe, we can use the Bernoulli equation<\/strong> and the continuity equation<\/strong>.<\/p>\n\n\n\n

Step 1: Apply the Continuity Equation<\/h3>\n\n\n\n

The continuity equation<\/strong> states that for an incompressible fluid (like water or most liquids), the flow rate must remain constant across different sections of the pipe. This is expressed as:A1v1=A2v2A_1 v_1 = A_2 v_2A1\u200bv1\u200b=A2\u200bv2\u200b<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • A1A_1A1\u200b and A2A_2A2\u200b are the cross-sectional areas at Locations 1 and 2, respectively,<\/li>\n\n\n\n
  • v1v_1v1\u200b and v2v_2v2\u200b are the flow speeds at Locations 1 and 2, respectively.<\/li>\n<\/ul>\n\n\n\n

    The cross-sectional area of a pipe is given by the formula A=\u03c0d2\/4A = \\pi d^2 \/ 4A=\u03c0d2\/4, where ddd is the diameter of the pipe.<\/p>\n\n\n\n

    At Location 1:A1=\u03c0d124A_1 = \\frac{\\pi d_1^2}{4}A1\u200b=4\u03c0d12\u200b\u200b<\/p>\n\n\n\n

    At Location 2:A2=\u03c0d224A_2 = \\frac{\\pi d_2^2}{4}A2\u200b=4\u03c0d22\u200b\u200b<\/p>\n\n\n\n

    Now, we can use the continuity equation to find the flow speed v2v_2v2\u200b at Location 2.\u03c0d124v1=\u03c0d224v2\\frac{\\pi d_1^2}{4} v_1 = \\frac{\\pi d_2^2}{4} v_24\u03c0d12\u200b\u200bv1\u200b=4\u03c0d22\u200b\u200bv2\u200b<\/p>\n\n\n\n

    Simplifying:d12v1=d22v2d_1^2 v_1 = d_2^2 v_2d12\u200bv1\u200b=d22\u200bv2\u200b<\/p>\n\n\n\n

    Solving for v2v_2v2\u200b:v2=d12v1d22v_2 = \\frac{d_1^2 v_1}{d_2^2}v2\u200b=d22\u200bd12\u200bv1\u200b\u200b<\/p>\n\n\n\n

    Substituting the values:<\/p>\n\n\n\n

      \n
    • d1=5.45\u2009cm=0.0545\u2009md_1 = 5.45 \\, \\text{cm} = 0.0545 \\, \\text{m}d1\u200b=5.45cm=0.0545m,<\/li>\n\n\n\n
    • d2=2.83\u2009cm=0.0283\u2009md_2 = 2.83 \\, \\text{cm} = 0.0283 \\, \\text{m}d2\u200b=2.83cm=0.0283m,<\/li>\n\n\n\n
    • v1=2.83\u2009m\/sv_1 = 2.83 \\, \\text{m\/s}v1\u200b=2.83m\/s.<\/li>\n<\/ul>\n\n\n\n

      v2=(0.0545)2\u00d72.83(0.0283)2v_2 = \\frac{(0.0545)^2 \\times 2.83}{(0.0283)^2}v2\u200b=(0.0283)2(0.0545)2\u00d72.83\u200bv2\u224811.38\u2009m\/sv_2 \\approx 11.38 \\, \\text{m\/s}v2\u200b\u224811.38m\/s<\/p>\n\n\n\n

      Step 2: Apply the Bernoulli Equation<\/h3>\n\n\n\n

      The Bernoulli equation<\/strong> for an incompressible fluid is:P1+12\u03c1v12=P2+12\u03c1v22P_1 + \\frac{1}{2} \\rho v_1^2 = P_2 + \\frac{1}{2} \\rho v_2^2P1\u200b+21\u200b\u03c1v12\u200b=P2\u200b+21\u200b\u03c1v22\u200b<\/p>\n\n\n\n

      Where:<\/p>\n\n\n\n

        \n
      • P1P_1P1\u200b and P2P_2P2\u200b are the pressures at Locations 1 and 2,<\/li>\n\n\n\n
      • \u03c1\\rho\u03c1 is the fluid density,<\/li>\n\n\n\n
      • v1v_1v1\u200b and v2v_2v2\u200b are the velocities at Locations 1 and 2.<\/li>\n<\/ul>\n\n\n\n

        Rearranging to solve for the pressure difference \u0394P=P2\u2212P1\\Delta P = P_2 – P_1\u0394P=P2\u200b\u2212P1\u200b:P2\u2212P1=12\u03c1(v12\u2212v22)P_2 – P_1 = \\frac{1}{2} \\rho (v_1^2 – v_2^2)P2\u200b\u2212P1\u200b=21\u200b\u03c1(v12\u200b\u2212v22\u200b)<\/p>\n\n\n\n

        Substitute the values:<\/p>\n\n\n\n

          \n
        • \u03c1=1110\u2009kg\/m3\\rho = 1110 \\, \\text{kg\/m}^3\u03c1=1110kg\/m3,<\/li>\n\n\n\n
        • v1=2.83\u2009m\/sv_1 = 2.83 \\, \\text{m\/s}v1\u200b=2.83m\/s,<\/li>\n\n\n\n
        • v2=11.38\u2009m\/sv_2 = 11.38 \\, \\text{m\/s}v2\u200b=11.38m\/s.<\/li>\n<\/ul>\n\n\n\n

          \u0394P=12\u00d71110\u00d7(2.832\u221211.382)\\Delta P = \\frac{1}{2} \\times 1110 \\times \\left( 2.83^2 – 11.38^2 \\right)\u0394P=21\u200b\u00d71110\u00d7(2.832\u221211.382)\u0394P=12\u00d71110\u00d7(8.0089\u2212129.6336)\\Delta P = \\frac{1}{2} \\times 1110 \\times (8.0089 – 129.6336)\u0394P=21\u200b\u00d71110\u00d7(8.0089\u2212129.6336)\u0394P\u224812\u00d71110\u00d7(\u2212121.6247)\\Delta P \\approx \\frac{1}{2} \\times 1110 \\times (-121.6247)\u0394P\u224821\u200b\u00d71110\u00d7(\u2212121.6247)\u0394P\u2248\u221267460.1\u2009Pa\\Delta P \\approx -67460.1 \\, \\text{Pa}\u0394P\u2248\u221267460.1Pa<\/p>\n\n\n\n

          Final Answer:<\/h3>\n\n\n\n

          The pressure difference \u0394P\\Delta P\u0394P between Locations 2 and 1 is approximately -67.46 kPa<\/strong>, meaning the pressure at Location 2 is lower<\/strong> than the pressure at Location 1.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          A horizontal pipe carries a smoothly flowing liquid of density 1110 kg\/m^3. At Locations 1 and 2 along the pipe, the diameters are d1 = 5.45 cm and d2 = 2.83 cm, respectively. The flow speed at Location 1 is 2.83 m\/s. What is the pressure difference \u00ce\u201dP between Location 2 and Location 1 (including […]<\/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-240787","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240787","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=240787"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240787\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=240787"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=240787"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=240787"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}