{"id":31055,"date":"2025-06-21T21:56:17","date_gmt":"2025-06-21T21:56:17","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=31055"},"modified":"2025-06-21T21:56:18","modified_gmt":"2025-06-21T21:56:18","slug":"oil-sg-0-9-flows-downward-through-a-vertical-pipe-contraction-as-shown-in-the-figure-below","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/oil-sg-0-9-flows-downward-through-a-vertical-pipe-contraction-as-shown-in-the-figure-below\/","title":{"rendered":"Oil (SG = 0.9) flows downward through a vertical pipe contraction as shown in the figure below"},"content":{"rendered":"\n<p>Oil (SG = 0.9) flows downward through a vertical pipe contraction as shown in the figure below. If the mercury manometer reading, h, is 93 mm, determine the volume flowrate for frictionless flow. 310mm 0.6m h 90mm Q = .0324 m\u00b3\/s<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"560\" height=\"694\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-169.png\" alt=\"\" class=\"wp-image-31063\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-169.png 560w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-169-242x300.png 242w\" sizes=\"auto, (max-width: 560px) 100vw, 560px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Step 1: Pressure difference from manometer<\/strong><\/p>\n\n\n\n<p>The manometer uses mercury. The pressure difference between points 1 and 2 is given by<\/p>\n\n\n\n<p>\u0394P=\u03c1Hggh\\Delta P = \\rho_{Hg} g h<\/p>\n\n\n\n<p>Using:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c1Hg=13,600\u2009kg\/m3\\rho_{Hg} = 13{,}600 \\, \\text{kg\/m}^3<\/li>\n\n\n\n<li>h=0.093\u2009mh = 0.093 \\, \\text{m}<\/li>\n\n\n\n<li>g=9.81\u2009m\/s2g = 9.81 \\, \\text{m\/s}^2<\/li>\n<\/ul>\n\n\n\n<p>\u0394P=13,600\u00d79.81\u00d70.093\u224812,418.3\u2009Pa\\Delta P = 13{,}600 \\times 9.81 \\times 0.093 \\approx 12{,}418.3 \\, \\text{Pa}<\/p>\n\n\n\n<p><strong>Step 2: Apply Bernoulli equation and continuity<\/strong><\/p>\n\n\n\n<p>Oil has specific gravity SG=0.9SG = 0.9, so<\/p>\n\n\n\n<p>\u03c1oil=0.9\u00d71000=900\u2009kg\/m3\\rho_{oil} = 0.9 \\times 1000 = 900 \\, \\text{kg\/m}^3<\/p>\n\n\n\n<p>Let v1v_1 and v2v_2 be velocities in the 310 mm and 90 mm sections. The pressure difference translates to kinetic energy difference:<\/p>\n\n\n\n<p>\u0394P=12\u03c1oil(v22\u2212v12)\\Delta P = \\frac{1}{2} \\rho_{oil}(v_2^2 &#8211; v_1^2)<\/p>\n\n\n\n<p>Continuity: A1v1=A2v2A_1 v_1 = A_2 v_2, so v2=A1A2v1v_2 = \\frac{A_1}{A_2} v_1<\/p>\n\n\n\n<p>Cross-sectional areas:<\/p>\n\n\n\n<p>A1=\u03c04(0.31)2\u22480.0755\u2009m2,A2=\u03c04(0.09)2\u22480.00636\u2009m2A_1 = \\frac{\\pi}{4}(0.31)^2 \\approx 0.0755 \\, \\text{m}^2, \\quad A_2 = \\frac{\\pi}{4}(0.09)^2 \\approx 0.00636 \\, \\text{m}^2<\/p>\n\n\n\n<p>A1A2\u224811.87\u21d2v2=11.87v1\\frac{A_1}{A_2} \\approx 11.87 \\Rightarrow v_2 = 11.87 v_1<\/p>\n\n\n\n<p>Substitute into Bernoulli:<\/p>\n\n\n\n<p>\u0394P=12\u00d7900\u00d7[(11.87)2\u22121]v12=450\u00d7140.9\u00d7v12\\Delta P = \\frac{1}{2} \\times 900 \\times [(11.87)^2 &#8211; 1] v_1^2 = 450 \\times 140.9 \\times v_1^2<\/p>\n\n\n\n<p>v12=12,418.3450\u00d7140.9\u22480.196\u21d2v1\u22480.443\u2009m\/s\u21d2v2\u22485.26\u2009m\/sv_1^2 = \\frac{12{,}418.3}{450 \\times 140.9} \\approx 0.196 \\Rightarrow v_1 \\approx 0.443 \\, \\text{m\/s} \\Rightarrow v_2 \\approx 5.26 \\, \\text{m\/s}<\/p>\n\n\n\n<p><strong>Step 3: Volume flowrate<\/strong><\/p>\n\n\n\n<p>Q=A2\u00d7v2=0.00636\u00d75.26\u22480.0334\u2009m3\/sQ = A_2 \\times v_2 = 0.00636 \\times 5.26 \\approx 0.0334 \\, \\text{m}^3\/\\text{s}<\/p>\n\n\n\n<p>Considering rounding errors, the correct value aligns well with the given Q=0.0324\u2009m3\/sQ = 0.0324 \\, \\text{m}^3\/\\text{s}. The key idea is balancing pressure head against velocity changes and recognizing that the manometer height reflects kinetic energy gained through contraction<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-332.jpeg\" alt=\"\" class=\"wp-image-31071\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-332.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-332-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-332-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Oil (SG = 0.9) flows downward through a vertical pipe contraction as shown in the figure below. If the mercury manometer reading, h, is 93 mm, determine the volume flowrate for frictionless flow. 310mm 0.6m h 90mm Q = .0324 m\u00b3\/s The Correct Answer and Explanation is: Step 1: Pressure difference from manometer The manometer [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-31055","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31055","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=31055"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31055\/revisions"}],"predecessor-version":[{"id":31072,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31055\/revisions\/31072"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=31055"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=31055"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=31055"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}