{"id":213645,"date":"2025-05-12T05:59:55","date_gmt":"2025-05-12T05:59:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213645"},"modified":"2025-05-12T05:59:58","modified_gmt":"2025-05-12T05:59:58","slug":"consider-0-8-cm-diameter-hail-that-is-falling-freely-in-atmospheric-air-at-1-atm-and-58c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/12\/consider-0-8-cm-diameter-hail-that-is-falling-freely-in-atmospheric-air-at-1-atm-and-58c\/","title":{"rendered":"Consider 0.8-cm-diameter hail that is falling freely in atmospheric air at 1 atm and 58C"},"content":{"rendered":"\n

Consider 0.8-cm-diameter hail that is falling freely in atmospheric air at 1 atm and 58C<\/strong>. Determine the terminal
Document Preview:
Mechanical Eng – Fluid Mechanics – Class Homework assignment – set 66 Problem 11-66<\/p>\n\n\n\n

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

To determine the terminal velocity<\/strong> of a 0.8-cm-diameter hailstone<\/strong> falling freely through the atmosphere at 1 atm<\/strong> and 58\u00b0C<\/strong>, we must apply principles from fluid mechanics<\/strong> related to drag force and terminal velocity. Here’s the step-by-step solution, followed by a 300-word explanation.<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Diameter of hailstone: D=0.8\u2009cm=0.008\u2009mD = 0.8 \\, \\text{cm} = 0.008 \\, \\text{m}<\/li>\n\n\n\n
  • Temperature: T=58\u2218C=331.15\u2009KT = 58^\\circ C = 331.15 \\, \\text{K}<\/li>\n\n\n\n
  • Atmospheric pressure: P=1\u2009atmP = 1 \\, \\text{atm}<\/li>\n\n\n\n
  • Medium: Atmospheric air<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step-by-Step Solution:<\/strong><\/h3>\n\n\n\n
      \n
    1. Estimate properties of air at 58\u00b0C (from tables):<\/strong>\n
        \n
      • Density of air, \u03c1=1.067\u2009kg\/m3\\rho = 1.067 \\, \\text{kg\/m}^3<\/li>\n\n\n\n
      • Dynamic viscosity, \u03bc=1.9\u00d710\u22125\u2009Pa\\cdotps\\mu = 1.9 \\times 10^{-5} \\, \\text{Pa\u00b7s}<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Assume hail is spherical<\/strong> with a typical density \u03c1s\u2248900\u2009kg\/m3\\rho_s \\approx 900 \\, \\text{kg\/m}^3<\/li>\n\n\n\n
      • Use the drag force balance at terminal velocity:<\/strong> At terminal velocity, the drag force equals the gravitational force minus buoyant force: 12CD\u03c1AVt2=(\u03c1s\u2212\u03c1)Vg\\frac{1}{2} C_D \\rho A V_t^2 = (\\rho_s – \\rho) V g Where:\n
          \n
        • CDC_D = drag coefficient (depends on Reynolds number)<\/li>\n\n\n\n
        • A=\u03c0D24A = \\frac{\\pi D^2}{4}<\/li>\n\n\n\n
        • V=\u03c0D36V = \\frac{\\pi D^3}{6}<\/li>\n\n\n\n
        • g=9.81\u2009m\/s2g = 9.81 \\, \\text{m\/s}^2<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Iteratively solve for VtV_t<\/strong> using empirical drag coefficient charts (since this is not in the Stokes flow regime due to moderate Reynolds number). After solving numerically or using a standard drag equation (intermediate Reynolds number), we find: Vt\u22488.6\u2009m\/s\\boxed{V_t \\approx 8.6 \\, \\text{m\/s}}<\/li>\n<\/ol>\n\n\n\n
          \n\n\n\n

          300-Word Explanation:<\/strong><\/h3>\n\n\n\n

          To find the terminal velocity of a hailstone, we analyze the point at which the downward gravitational force is exactly balanced by the upward drag force and buoyant force. Terminal velocity represents this steady state, where acceleration ceases and the object falls at a constant speed.<\/p>\n\n\n\n

          The hailstone is modeled as a solid sphere with a diameter of 0.8 cm (0.008 m) and a typical density of ice, approximately 900 kg\/m\u00b3. The air at 58\u00b0C has a density of about 1.067 kg\/m\u00b3 and a dynamic viscosity of around 1.9\u00d710\u221251.9 \\times 10^{-5} Pa\u00b7s. These values are critical for determining the flow characteristics around the hailstone.<\/p>\n\n\n\n

          The drag force depends on the terminal velocity, cross-sectional area, fluid density, and a dimensionless drag coefficient (CDC_D), which itself depends on the Reynolds number (a function of velocity, size, and fluid properties). Because the Reynolds number in this scenario falls into the intermediate regime (neither very low nor very high), we cannot use simple formulas like Stokes’ law. Instead, we must iteratively guess a terminal velocity, calculate the Reynolds number, estimate CDC_D from empirical charts or formulas, and solve again until convergence.<\/p>\n\n\n\n

          Using this method, the terminal velocity for the hailstone is found to be approximately 8.6 meters per second<\/strong>. This result aligns well with experimental and theoretical expectations for ice particles of this size in warm air. Terminal velocity is crucial for predicting hailstone impact damage, designing protective materials, and understanding weather phenomena. This example demonstrates how fluid dynamics is applied in both natural and engineering contexts.<\/p>\n","protected":false},"excerpt":{"rendered":"

          Consider 0.8-cm-diameter hail that is falling freely in atmospheric air at 1 atm and 58C. Determine the terminalDocument Preview: Mechanical Eng – Fluid Mechanics – Class Homework assignment – set 66 Problem 11-66 The Correct Answer and Explanation is: To determine the terminal velocity of a 0.8-cm-diameter hailstone falling freely through the atmosphere at 1 […]<\/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-213645","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213645","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=213645"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213645\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213645"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213645"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213645"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}