{"id":29157,"date":"2021-03-22T13:00:19","date_gmt":"2021-03-22T13:00:19","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=29157"},"modified":"2024-09-22T11:43:47","modified_gmt":"2024-09-22T11:43:47","slug":"pressure-of-a-fluid-at-rest","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/pressure-of-a-fluid-at-rest\/","title":{"rendered":"Pressure of a Fluid at Rest"},"content":{"rendered":"<p><center><\/p>\n<h1>The Pressure of Fluids<\/h1>\n<p style=\"text-align:center;\"><em><strong>Summary:<\/strong><br \/>\nThis class will focus on the concept of fluid pressure at rest and how it varies with depth. We will learn that the pressure at a point within a fluid directly depends on its density, gravity, and depth.<br \/>\n<\/em><br \/>\n<strong>Learning Objectives:<\/strong><br \/>\nBy the end of the class, the student will be able to<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Understand<\/strong> the relationship between pressure in a fluid and variables like density, gravity, and depth.<\/li>\n<li><strong>Apply<\/strong> the formula P = \u03c1gh to calculate pressure in fluids at rest.<\/li>\n<li><strong>Explain<\/strong> the difference between gauge, atmospheric, and absolute pressure.<\/li>\n<\/ol>\n<p><strong>INDEX OF CONTENTS<\/strong><br \/>\n<a href=\"#1\">Pressure of fluids at rest<\/a><br \/>\n<a href=\"#1\">Relative Pressure<\/a><br \/>\n<a href=\"#1\">Practical Examples<\/a>\n<\/p>\n<p><\/center><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/Qt4mdRDEc44\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"1\"><\/a><\/p>\n<h2>Pressure of Fluids at Rest<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Qt4mdRDEc44&amp;t=105s\" target=\"_blank\" rel=\"noopener\"><strong>What do we know about the pressure of a fluid at rest?<\/strong><\/a> We know that if we place it inside a container, since the relationship <span class=\"katex-eq\" data-katex-display=\"false\">P=F\/A,<\/span> holds and due to its weight, there will be pressure at every point within it, depending on the depth.<\/p>\n<p style=\"text-align: justify;\"><strong>The pressure of a fluid at rest<\/strong> at a given point is directly proportional to its depth. This is known from the expression:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P = \\rho g h<\/span>\n<p style=\"text-align: justify;\">Where <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> is the fluid&#8217;s density, <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> is the gravitational acceleration, and <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span> is the depth.<\/p>\n<p><center><\/center><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-M64DeuX1K_8\/YGDR2Dr6MLI\/AAAAAAAAEvU\/dsWvpC_qUaUqXYxIY_kdFz_RegsAQ8jNQCLcBGAsYHQ\/s0\/presionProfundidad.PNG\" alt=\"Pressure Depth\" class=\"alignnone size-full lazyload\" width=\"360\" height=\"380\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-M64DeuX1K_8\/YGDR2Dr6MLI\/AAAAAAAAEvU\/dsWvpC_qUaUqXYxIY_kdFz_RegsAQ8jNQCLcBGAsYHQ\/s0\/presionProfundidad.PNG\" alt=\"Pressure Depth\" class=\"alignnone size-full lazyload\" width=\"360\" height=\"380\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">We can demonstrate this by submerging an imaginary cylinder with a horizontal base area <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> at some depth <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span> in the fluid.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-4xXHikobfLg\/YGDR2CIDEWI\/AAAAAAAAEvQ\/vMap5iq_jHoL_vHicMEvWiOJbP_CX3pNACLcBGAsYHQ\/s0\/FuerzaProfundidad.PNG\" alt=\"Pressure Depth\" class=\"alignnone size-full lazyload\" width=\"364\" height=\"448\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-4xXHikobfLg\/YGDR2CIDEWI\/AAAAAAAAEvQ\/vMap5iq_jHoL_vHicMEvWiOJbP_CX3pNACLcBGAsYHQ\/s0\/FuerzaProfundidad.PNG\" alt=\"Pressure Depth\" class=\"alignnone size-full lazyload\" width=\"364\" height=\"448\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">We will see that the disc will be crushed by the weight of the fluid above it and by the normal force below, which is equal and opposite to the weight (since we assume the system is at rest).<\/p>\n<h4>Derivation of the formula for the pressure of a fluid at rest: <span class=\"katex-eq\" data-katex-display=\"false\">{P=\\rho g h}<\/span><\/h4>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Qt4mdRDEc44&amp;t=195s\" target=\"_blank\" rel=\"noopener\">The fluid forms another cylinder<\/a> over the submerged body with the same base area <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span>, but height <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span>, thus having a volume.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">V=A h<\/span>\n<p style=\"text-align: justify;\">From which we infer that<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle A=\\frac{V}{h}<\/span>\n<p style=\"text-align: justify;\">If the fluid has a density <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span>, then the mass of the fluid pressing the disc is <span class=\"katex-eq\" data-katex-display=\"false\">m=\\rho V<\/span>, and it exerts a weight force<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_p=m g = \\rho V g<\/span>\n<p style=\"text-align: justify;\">Similarly, the normal force is exerted on the lower face of the cylinder with the same magnitude but on the opposite face.<\/p>\n<p style=\"text-align: justify;\">Since the weight and the normal are vertical, they do not exert pressure on the lateral sides of the cylinder.<\/p>\n<p style=\"text-align: justify;\">If we consider the cylinder to be flat and light enough, its weight will not contribute anything that needs to be countered by the normal force, and any effect on the sides will be negligible compared to everything else. Thus, the total force on the body will be:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{total}=F_p + F_n = 2\\rho V g<\/span>\n<p style=\"text-align: justify;\">The \u00ab+\u00bb in this force is because the forces are oriented toward the surface.<\/p>\n<p style=\"text-align: justify;\">Thus, the pressure on the submerged body will be:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P = \\frac{F_{total}}{A_{inferior}+A_{superior}}=\\frac{2 \\rho V h}{2A} = \\frac{\\rho V g}{\\frac{V}{h}} = \\rho g h<\/span>\n<p style=\"text-align: justify;\">This shows that the pressure exerted by a fluid (at rest) is the same at all points at the same depth. This applies to both liquids and gases (as long as they are at rest), so if we consider the column of air above us, we can also talk about \u00abatmospheric pressure\u00bb.<\/p>\n<p style=\"text-align: justify;\">The atmospheric pressure at sea level is<\/p>\n<p style=\"text-align: justify;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{atm} = 1[atm] = 101.325,0 [Pa] = 760 [Torr]=0.981[barr].<\/span>\n<p>We must remember that <span class=\"katex-eq\" data-katex-display=\"false\">1[Pa] = 1[N\/m^2].<\/span>\n<p style=\"text-align: justify;\">By combining the atmospheric pressure with the pressure exerted by a fluid due to its own weight, we have the <strong>hydrostatic pressure<\/strong><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P = P_{atm} + \\rho g h<\/span>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Relative Pressure<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Qt4mdRDEc44&amp;t=590s\" target=\"_blank\" rel=\"noopener\"><strong>When we measure pressure, we are usually immersed in a medium.<\/strong><\/a> Sometimes the pressure of the medium is relevant, and other times it is not. For example, when you measure the pressure of your car tires, you don&#8217;t worry about adding the atmospheric pressure because what really matters for their proper functioning is the pressure difference between the inside of the tire and the outside environment:<\/p>\n<p style=\"text-align: justify;\"><em>If it&#8217;s too high, it overinflates; if it&#8217;s too low, it deflates.<\/em><\/p>\n<p style=\"text-align: justify;\">This is why we have different ways to talk about pressure.<\/p>\n<h3>Atmospheric Pressure<\/h3>\n<p style=\"text-align: justify;\">We already discussed this before, and it&#8217;s the pressure of the medium we are immersed in. For example, in the Himalayas, the atmospheric pressure can be as low as one-third of what it is at sea level. It is usually represented by <span class=\"katex-eq\" data-katex-display=\"false\">P_{atm}<\/span> or <span class=\"katex-eq\" data-katex-display=\"false\">P_{0}.<\/span>\n<h3>Absolute Pressure<\/h3>\n<p style=\"text-align: justify;\">When we consider the pressure obtained from the sum of all forces acting on a body, we talk about <strong>Absolute Pressure.<\/strong> The hydrostatic pressure we reviewed earlier is a form of absolute pressure because it considers the sum of the pressures due to the weight of the liquid + the pressure exerted by the atmosphere. Another way to express absolute pressure is as \u00abpressure relative to a vacuum\u00bb. It is represented by <span class=\"katex-eq\" data-katex-display=\"false\">P_{abs}.<\/span>\n<h3>Gauge Pressure and Vacuum Pressure<\/h3>\n<p style=\"text-align: justify;\">When we measure the pressure of a vehicle tire, both the tire and the measuring instrument are being pressed by the surrounding atmosphere. For this reason, what the instrument measures is the pressure difference between the inside and outside. This pressure is called \u00abgauge pressure,\u00bb represented by <span class=\"katex-eq\" data-katex-display=\"false\">P_{man}<\/span>, and it satisfies the relationship:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{man} = P_{abs} - P_{atm}<\/span>\n<p style=\"text-align: justify;\">The pressure we measure considering only the weight of a fluid is an example of gauge pressure. If the absolute pressure is higher than atmospheric pressure, we measure gauge pressure; otherwise, we measure vacuum pressure <span class=\"katex-eq\" data-katex-display=\"false\">P_{vac}<\/span>, defined similarly:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{vac} = P_{atm} - P_{abs}<\/span>\n<p style=\"text-align: justify;\">This occurs, for example, when you pull the air out of a syringe, seal the air inlet\/outlet, and then pull the plunger; atmospheric pressure will try to push the plunger back, and the pressure inside the syringe will be, consequently, vacuum pressure.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Practical Examples<\/h2>\n<p style=\"text-align: justify;\">A man has just built a pool that is 39[feet] long, 26[feet] wide, and 5.2[feet] deep and has the following questions:<\/p>\n<ol style=\"text-align: justify;\">\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=Qt4mdRDEc44&amp;t=814s\" target=\"_blank\" rel=\"noopener\"><strong>He bought a pump for the pool,<\/strong><\/a> but only when he got home did he think to check the manufacturer&#8217;s specifications. These say that the gauge pressure should not exceed 0.193[atm]. Can he install the pump at the bottom of his pool?<\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=Qt4mdRDEc44&amp;t=1101s\" target=\"_blank\" rel=\"noopener\"><strong>Not only did he forget the pump specifications,<\/strong><\/a> but this man has a hearing issue. His eardrums cannot withstand a force greater than 10[N]. If his eardrums have a diameter of 1[cm], almost perfectly circular, can he safely dive to the bottom of his pool?<\/li>\n<\/ol>\n<p style=\"text-align: justify; color: #0000d0;\"><strong>SOLUTION:<\/strong><\/p>\n<ol style=\"text-align: justify;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"text-align: justify;\">\n<li>In this case, the gauge pressure at the bottom of the pool can be determined as the pressure exerted only by the pool&#8217;s water. Therefore, from the formula for the pressure of a fluid at rest, we get:\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{man} = \\rho g h<\/span>\n<p>Taking the water&#8217;s density <span class=\"katex-eq\" data-katex-display=\"false\">\\rho=997[kg\/m^3],<\/span> the depth converted to meters <span class=\"katex-eq\" data-katex-display=\"false\">h=5.2[feet] = 5.2\\cdot 0.3048[m]<\/span> and the gravitational acceleration <span class=\"katex-eq\" data-katex-display=\"false\">g=9.81[m\/s^2],<\/span> we get that the gauge pressure at the bottom of the pool will be<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{man} =997[kg\/m^3]\\cdot 9.81[m\/s^2] \\cdot5.2\\cdot 0.3048[m] \\approx 15.501,81[Pa]<\/span>\n<p>But <span class=\"katex-eq\" data-katex-display=\"false\">1[atm] = 101.325[Pa]<\/span>, so<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle P_{man} \\approx \\frac{15.501,81}{101.325}[atm]\\approx 0.1523[atm]<\/span>\n<p>So, yes. Since the gauge pressure at the bottom of the pool is below the <span class=\"katex-eq\" data-katex-display=\"false\">0.193[atm]<\/span> specified by the manufacturer as the limit for the pump&#8217;s proper function, it will work correctly, and the man hasn&#8217;t wasted his money.<\/li>\n<li>From the previous part, we already calculated the gauge pressure at the bottom of the pool, but now we need the total pressure. No problem with that; we just need to remember that:<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{total} = P_{man} + P_{atm}<\/span>\n<p><span>and we already have both. This gives us that<\/span><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_{total} \\approx 15.501,81[Pa] + 101.325[Pa] = 116.286,81[Pa]<\/span>\n<p><span>Now we need to know the area of this man&#8217;s eardrum. Since the eardrum is approximately circular, we have:<\/span><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle A = \\pi r^2 = \\pi \\left(\\frac{d}{2}\\right)^2 = \\frac{\\pi d^2}{4}<\/span>\n<p><span>Here I&#8217;ve expressed the area of the circle as a function of the diameter <span class=\"katex-eq\" data-katex-display=\"false\">d<\/span>, which is <span class=\"katex-eq\" data-katex-display=\"false\">1[cm]<\/span>. Therefore:<\/span><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle A \\approx \\frac{3.14 \\cdot 1[cm^2]}{4} = \\frac{3.14 \\left[\\frac{m}{100}\\right]^2}{4} = \\frac{3.14}{4\\cdot 10.000}[m^2]=0.785\\cdot 10^{-4}[m^2]<\/span>\n<p><span>Finally, since <span class=\"katex-eq\" data-katex-display=\"false\">P=F\/A<\/span>, the total force applied by the pressure at the bottom of the pool on this man&#8217;s eardrum will be<\/span><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F=PA\\approx 116.826,81[Pa] \\cdot 0.785\\cdot 10^{-4}[m^2] \\approx 9.17[N]<\/span>\n<p><span>Since this doesn&#8217;t exceed <span class=\"katex-eq\" data-katex-display=\"false\">10[N]<\/span>, just barely, he can dive safely to the bottom of the pool. What a lucky man.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Pressure of Fluids Summary: This class will focus on the concept of fluid pressure at rest and how it varies with depth. We will learn that the pressure at a point within a fluid directly depends on its density, gravity, and depth. Learning Objectives: By the end of the class, the student will be [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":29154,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":13,"footnotes":""},"categories":[889,635],"tags":[],"class_list":["post-29157","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fluid-mechanics","category-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Pressure of a Fluid at Rest - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"The pressure of a fluid at rest increases with depth and depends on its density and gravity. 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