{"id":211799,"date":"2025-05-05T07:56:09","date_gmt":"2025-05-05T07:56:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=211799"},"modified":"2025-05-05T07:56:11","modified_gmt":"2025-05-05T07:56:11","slug":"if-the-viscosity-of-water-at-70-c-is-0-00402-poise-and-its-specific-gravity-is-0-978-determine-its-absolute-viscosity-in-pa-s-and-its-kinematic-viscosity-in-m%c2%b2-s-and-in-stokes","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/05\/if-the-viscosity-of-water-at-70-c-is-0-00402-poise-and-its-specific-gravity-is-0-978-determine-its-absolute-viscosity-in-pa-s-and-its-kinematic-viscosity-in-m%c2%b2-s-and-in-stokes\/","title":{"rendered":"If the viscosity of water at 70 \u00b0C is 0.00402 poise and its specific gravity is 0.978 determine its absolute viscosity in Pa-s and its kinematic viscosity in m\u00b2\/s and in stokes"},"content":{"rendered":"\n

If the viscosity of water at 70 \u00b0C is 0.00402 poise and its specific gravity is 0.978 determine its absolute viscosity in Pa-s and its kinematic viscosity in m\u00b2\/s and in stokes.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To solve this problem, we need to convert the given viscosity values from one unit system to another and understand their relationships.<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Viscosity (dynamic)<\/strong> in Poise:
    $\\mu = 0.00402 \\, \\text{P}$<\/li>\n\n\n\n
  • Specific gravity (SG):<\/strong>
    $SG = 0.978$<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    1. Convert Dynamic (Absolute) Viscosity to SI Units (Pa\u00b7s):<\/strong><\/h3>\n\n\n\n

    1 Poise = 0.1 Pa\u00b7s<\/p>\n\n\n\n

    $$
    \\mu = 0.00402 \\, \\text{P} \\times 0.1 = 0.000402 \\, \\text{Pa\u00b7s}
    $$<\/p>\n\n\n\n

    \n\n\n\n

    2. Find the Density of Water at 70 \u00b0C:<\/strong><\/h3>\n\n\n\n

    Specific gravity is the ratio of the fluid’s density to water at 4 \u00b0C (which is 1000 kg\/m\u00b3):<\/p>\n\n\n\n

    $$
    \\rho = SG \\times 1000 = 0.978 \\times 1000 = 978 \\, \\text{kg\/m}^3
    $$<\/p>\n\n\n\n

    \n\n\n\n

    3. Calculate Kinematic Viscosity (\u03bd):<\/strong><\/h3>\n\n\n\n

    Kinematic viscosity is the ratio of dynamic viscosity to density:<\/p>\n\n\n\n

    $$
    \\nu = \\frac{\\mu}{\\rho} = \\frac{0.000402}{978} \\approx 4.11 \\times 10^{-7} \\, \\text{m}^2\/\\text{s}
    $$<\/p>\n\n\n\n

    \n\n\n\n

    4. Convert Kinematic Viscosity to Stokes:<\/strong><\/h3>\n\n\n\n

    1 Stokes = $1 \\times 10^{-4} \\, \\text{m}^2\/\\text{s}$<\/p>\n\n\n\n

    $$
    \\nu = \\frac{4.11 \\times 10^{-7}}{1 \\times 10^{-4}} = 0.00411 \\, \\text{St}
    $$<\/p>\n\n\n\n

    \n\n\n\n

    Final Answers:<\/strong><\/h3>\n\n\n\n
      \n
    • Absolute viscosity:<\/strong> $0.000402 \\, \\text{Pa\u00b7s}$<\/li>\n\n\n\n
    • Kinematic viscosity:<\/strong><\/li>\n\n\n\n
    • $4.11 \\times 10^{-7} \\, \\text{m}^2\/\\text{s}$<\/li>\n\n\n\n
    • $0.00411 \\, \\text{St}$<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Explanation (300 Words):<\/strong><\/h3>\n\n\n\n

      Viscosity is a key property in fluid mechanics that describes a fluid\u2019s resistance to flow. It comes in two forms: dynamic (absolute)<\/strong> viscosity, which measures the internal resistance of a fluid to shear forces, and kinematic<\/strong> viscosity, which considers the effect of fluid density.<\/p>\n\n\n\n

      The unit Poise (P) is part of the CGS system, commonly used in older texts and industry, but SI units (Pa\u00b7s for dynamic and m\u00b2\/s for kinematic) are more standard today. To convert from Poise to Pascal-seconds, we use the relation: $1 \\, \\text{P} = 0.1 \\, \\text{Pa\u00b7s}$. Thus, the given dynamic viscosity of 0.00402 P equals $0.000402 \\, \\text{Pa\u00b7s}$.<\/p>\n\n\n\n

      Kinematic viscosity ($\\nu$) is calculated by dividing the dynamic viscosity ($\\mu$) by the fluid’s density ($\\rho$). The density of water at 70 \u00b0C is not exactly 1000 kg\/m\u00b3 due to thermal expansion; it decreases with temperature. Given a specific gravity of 0.978, the density becomes 978 kg\/m\u00b3.<\/p>\n\n\n\n

      By applying the formula $\\nu = \\mu \/ \\rho$, we calculate the kinematic viscosity in SI units: $4.11 \\times 10^{-7} \\, \\text{m}^2\/\\text{s}$. For completeness, we convert this into Stokes (a CGS unit), resulting in $0.00411 \\, \\text{St}$, using the conversion $1 \\, \\text{St} = 10^{-4} \\, \\text{m}^2\/\\text{s}$.<\/p>\n\n\n\n

      These conversions are crucial in engineering to ensure compatibility of data, accurate modeling of fluid systems, and effective communication across international and interdisciplinary boundaries.<\/p>\n","protected":false},"excerpt":{"rendered":"

      If the viscosity of water at 70 \u00b0C is 0.00402 poise and its specific gravity is 0.978 determine its absolute viscosity in Pa-s and its kinematic viscosity in m\u00b2\/s and in stokes. The correct answer and explanation is : To solve this problem, we need to convert the given viscosity values from one unit system […]<\/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-211799","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211799","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=211799"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211799\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211799"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211799"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211799"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}