{"id":218297,"date":"2025-05-23T14:41:20","date_gmt":"2025-05-23T14:41:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218297"},"modified":"2025-05-23T14:41:22","modified_gmt":"2025-05-23T14:41:22","slug":"what-differences-are-there-between-the-flow-in-a-circular-tube-of-radius-r-and-the-flow-in-the-same-tube-with-a-thin-wire-placed-along-the-axis","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/what-differences-are-there-between-the-flow-in-a-circular-tube-of-radius-r-and-the-flow-in-the-same-tube-with-a-thin-wire-placed-along-the-axis\/","title":{"rendered":"What differences are there between the flow in a circular tube of radius R and the flow in the same tube with a thin wire placed along the axis"},"content":{"rendered":"\n

Show that the Hagen-Poiseuille formula is dimensionally consistent? What differences are there between the flow in a circular tube of radius R and the flow in the same tube with a thin wire placed along the axis? Under what conditions would you expect the analysis in to be inapplicable?<\/p>\n\n\n\n

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

\n\n\n\n

1. Dimensional Consistency of the Hagen-Poiseuille Formula<\/strong><\/h3>\n\n\n\n

The Hagen-Poiseuille equation describes the volumetric flow rate QQ of a viscous fluid through a circular pipe: Q=\u03c0\u0394PR48\u03bcLQ = \\frac{\\pi \\Delta P R^4}{8 \\mu L}<\/p>\n\n\n\n

where<\/p>\n\n\n\n

    \n
  • QQ = volumetric flow rate [L3T\u22121][L^3 T^{-1}],<\/li>\n\n\n\n
  • \u0394P\\Delta P = pressure difference across the length [ML\u22121T\u22122][M L^{-1} T^{-2}],<\/li>\n\n\n\n
  • RR = pipe radius [L][L],<\/li>\n\n\n\n
  • \u03bc\\mu = dynamic viscosity [ML\u22121T\u22121][M L^{-1} T^{-1}],<\/li>\n\n\n\n
  • LL = length of the pipe [L][L].<\/li>\n<\/ul>\n\n\n\n

    Dimensional analysis:<\/strong><\/p>\n\n\n\n

    Right side dimensions: \u03c0\u0394PR48\u03bcL\u223c[ML\u22121T\u22122]\u00d7[L4][ML\u22121T\u22121]\u00d7[L]=ML3T\u22122ML0T\u22121=L3T\u22121\\frac{\\pi \\Delta P R^4}{8 \\mu L} \\sim \\frac{[M L^{-1} T^{-2}] \\times [L^4]}{[M L^{-1} T^{-1}] \\times [L]} = \\frac{M L^{3} T^{-2}}{M L^{0} T^{-1}} = L^{3} T^{-1}<\/p>\n\n\n\n

    which matches the dimension of QQ, volumetric flow rate. So the equation is dimensionally consistent.<\/p>\n\n\n\n

    \n\n\n\n

    2. Differences in Flow: Circular Tube vs. Tube with a Thin Wire Along the Axis<\/strong><\/h3>\n\n\n\n
      \n
    • Flow Profile Without Wire:<\/strong>
      The flow in a circular tube is laminar, with a parabolic velocity profile: maximum at the centerline and zero at the walls due to the no-slip condition.<\/li>\n\n\n\n
    • Flow Profile With Wire Along Axis:<\/strong>
      Introducing a thin wire along the pipe axis changes the geometry \u2014 the fluid now flows through an annulus (ring-shaped cross-section) between the pipe wall and the wire.<\/li>\n<\/ul>\n\n\n\n

      Effects:<\/strong><\/p>\n\n\n\n

        \n
      • The flow area reduces, so for the same pressure drop, the flow rate QQ decreases.<\/li>\n\n\n\n
      • The velocity profile changes; it\u2019s no longer a simple parabola but an annular profile, which depends on the wire radius.<\/li>\n\n\n\n
      • The shear stress distribution changes due to the new boundary at the wire surface.<\/li>\n\n\n\n
      • Increased frictional resistance due to the additional surface area.<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        3. Conditions Where Hagen-Poiseuille Analysis is Inapplicable<\/strong><\/h3>\n\n\n\n
          \n
        • Turbulent Flow:<\/strong>
          The formula assumes laminar, steady, incompressible flow. At high Reynolds numbers (Re > 2000), turbulence invalidates the model.<\/li>\n\n\n\n
        • Non-Newtonian Fluids:<\/strong>
          Hagen-Poiseuille applies for Newtonian fluids with constant viscosity. For shear-thinning or shear-thickening fluids, the formula fails.<\/li>\n\n\n\n
        • Non-Circular or Changing Geometry:<\/strong>
          If the pipe cross-section is non-circular or the wire significantly changes geometry, the simple formula doesn’t hold.<\/li>\n\n\n\n
        • Compressible Fluids or Significant Temperature Variations:<\/strong>
          If fluid density changes significantly or viscosity varies with temperature, the assumptions break down.<\/li>\n\n\n\n
        • Entrance\/Exit Effects:<\/strong>
          The formula assumes fully developed flow; near pipe entrances or exits, the velocity profile isn’t fully developed.<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Summary<\/h3>\n\n\n\n

          The Hagen-Poiseuille formula is dimensionally consistent because the units on both sides match volumetric flow rate dimensions. When a wire is introduced, the flow changes from a full circular profile to an annular flow with different velocity and shear stress distributions, reducing flow rate for the same pressure gradient. The formula only applies under laminar, steady, Newtonian flow conditions with a simple circular cross-section and fully developed flow; outside these conditions (turbulence, complex geometry, non-Newtonian fluids), it becomes inaccurate.<\/p>\n\n\n\n\n\n

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

          Show that the Hagen-Poiseuille formula is dimensionally consistent? What differences are there between the flow in a circular tube of radius R and the flow in the same tube with a thin wire placed along the axis? Under what conditions would you expect the analysis in to be inapplicable? The Correct Answer and Explanation is: […]<\/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-218297","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218297","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=218297"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218297\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}