{"id":169,"date":"2023-02-13T16:52:44","date_gmt":"2023-02-13T21:52:44","guid":{"rendered":"https:\/\/opentextbc.ca\/foundationsofphysics\/?post_type=chapter&#038;p=169"},"modified":"2023-12-21T17:21:30","modified_gmt":"2023-12-21T22:21:30","slug":"power-work-energy","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/power-work-energy\/","title":{"raw":"Power, Work &amp; Energy","rendered":"Power, Work &amp; Energy"},"content":{"raw":"<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Resources<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ul>\r\n \t<li>Extra Help: <a href=\"http:\/\/www.a-levelphysicstutor.com\/index-mech.php\">A-Level Physics Tutor<\/a><\/li>\r\n \t<li>Article to Read: <a href=\"https:\/\/www.greencarcongress.com\/2019\/02\/20190219-bye.html\">Bye Aerospace\u2019s Sun Flyer 2 completes first flight with Siemens electric propulsion motor<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nEquations Introduced and Used for this Topic:\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]a = \\dfrac{\\Delta v}{t}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\r\n \t<li>[latex]v_f = v_i + a t[\/latex]<\/li>\r\n \t<li>[latex]d = v_i t + \\frac{1}{2} a t^2[\/latex]<\/li>\r\n \t<li>[latex]F = m a[\/latex]<\/li>\r\n \t<li>Work [latex](W) = F \\cdot d[\/latex] or [latex]F d \\cos \u00f8[\/latex]<\/li>\r\n \t<li>[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>Power [latex](P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\r\n \t<li>[latex]P = \\frac{1}{2}\\varrho A v^3[\/latex]<\/li>\r\n \t<li>[latex]\\vec{a} = \\dfrac{\\vec{\\Delta v}}{t}[\/latex]<\/li>\r\n \t<li>[latex]v = \\dfrac{(v_i + v_f)}{2}[\/latex]<\/li>\r\n \t<li>[latex]2 a d = v_f^2 - v_i^2[\/latex]<\/li>\r\n \t<li>[latex]d = \\dfrac{(v+i v_f) t}{2}[\/latex]<\/li>\r\n \t<li>[latex]w = m g[\/latex]<\/li>\r\n \t<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\r\n \t<li>[latex]V[\/latex] or [latex]V_g = g h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n \t<li>[latex]\\text{Efficiency } = \\text{ Output } \\times 100%[\/latex]<\/li>\r\n<\/ul>\r\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 21.1738%;\" scope=\"col\">Equations<\/th>\r\n<th style=\"width: 17.3709%;\" scope=\"col\">a<\/th>\r\n<th style=\"width: 16.2441%;\" scope=\"col\">v<sub>t<\/sub><\/th>\r\n<th style=\"width: 16.6667%;\" scope=\"col\">\u00a0v<sub>i<\/sub><\/th>\r\n<th style=\"width: 15.8217%;\" scope=\"col\">d<\/th>\r\n<th style=\"width: 12.723%;\" scope=\"col\">t<\/th>\r\n<\/tr>\r\n<tr>\r\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]v_t = v_i + at[\/latex]<\/th>\r\n<td style=\"width: 17.3709%;\"><\/td>\r\n<td style=\"width: 16.2441%;\"><\/td>\r\n<td style=\"width: 16.6667%;\"><\/td>\r\n<td style=\"width: 15.8217%;\">Not mentioned<\/td>\r\n<td style=\"width: 12.723%;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]2ad = v_{r}^{2} - v_{i}^{2}[\/latex]<\/th>\r\n<td style=\"width: 17.3709%;\"><\/td>\r\n<td style=\"width: 16.2441%;\"><\/td>\r\n<td style=\"width: 16.6667%;\"><\/td>\r\n<td style=\"width: 15.8217%;\"><\/td>\r\n<td style=\"width: 12.723%;\">Not mentioned<\/td>\r\n<\/tr>\r\n<tr>\r\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]d = v_{i}t + \\dfrac{1}{2} at^2[\/latex]<\/th>\r\n<td style=\"width: 17.3709%;\"><\/td>\r\n<td style=\"width: 16.2441%;\">Not mentioned<\/td>\r\n<td style=\"width: 16.6667%;\"><\/td>\r\n<td style=\"width: 15.8217%;\"><\/td>\r\n<td style=\"width: 12.723%;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]d = \\dfrac{(v_i + v_t)t}{2}[\/latex]<\/th>\r\n<td style=\"width: 17.3709%;\">Not mentioned<\/td>\r\n<td style=\"width: 16.2441%;\"><\/td>\r\n<td style=\"width: 16.6667%;\"><\/td>\r\n<td style=\"width: 15.8217%;\"><\/td>\r\n<td style=\"width: 12.723%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhere...\r\n<ul>\r\n \t<li>[latex]P[\/latex] is the <strong>power<\/strong>, measured in watts (W)<\/li>\r\n \t<li>[latex]W[\/latex] is <strong>work<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]\\varrho[\/latex] is the <strong>density of the moving medium<\/strong>, measured in kilograms per metre cubed (kg\/m<sup>3<\/sup>)<\/li>\r\n \t<li>[latex]A[\/latex] is the <strong>cross sectional area of the turbine<\/strong>, measured in metres squared (m<sup>2<\/sup>)<\/li>\r\n \t<li>[latex]v[\/latex] is the <strong>speed of the medium moving through the turbine<\/strong>, measured in metres per second (m\/s)<\/li>\r\n \t<li>[latex]v[\/latex] is the <strong>speed of the object<\/strong> commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h)<\/li>\r\n \t<li>[latex]\\Delta \\text{ Energy}[\/latex] is the <strong>change in energy<\/strong> and can be any form of energy ([latex]E_p[\/latex] or other forms), measured in joules (J)<\/li>\r\n \t<li>[latex]E_p[\/latex] is <strong>gravitational potential energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_k[\/latex] is <strong>kinetic energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]V[\/latex] or [latex]V_g[\/latex] is the <strong>gravitational potential<\/strong>, measured in newtons per kilogram (N\/kg)<\/li>\r\n \t<li>[latex]\\Delta h[\/latex] is the <strong>change in vertical height<\/strong>, measured in metres (m)<\/li>\r\n \t<li>[latex]F[\/latex] is <strong>force<\/strong>, measured in newtons (N)<\/li>\r\n \t<li>[latex]m[\/latex] is <strong>mass<\/strong>, commonly measured in kilograms (Kg), grams (g) or tonnes (t)<\/li>\r\n \t<li>[latex]g[\/latex] or [latex]a_g[\/latex] is the <strong>acceleration due to gravity<\/strong> and varies from 9.78 m\/s<sup>2<\/sup> (equator) to 9.83 m\/s<sup>2<\/sup> (North or South poles)... The average value of gravity is taken to be either 9.80 m\/s<sup>2<\/sup>\u00a0 or 9.81 m\/s<sup>2<\/sup><\/li>\r\n \t<li>[latex]\\vec{v}[\/latex] is <strong>velocity<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/li>\r\n \t<li>[latex]v_i[\/latex] is <strong>initial speed<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h)<\/li>\r\n \t<li>[latex]\\vec{v_i}[\/latex] is <strong>initial velocity<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/li>\r\n \t<li>[latex]v_f[\/latex] is <strong>final speed<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h)<\/li>\r\n \t<li>[latex]\\vec{v_f}[\/latex] is <strong>final velocity<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/li>\r\n \t<li>[latex]\\Delta d[\/latex] is <strong>distance travelled<\/strong>, commonly measured in metres (m), kilometres (km)<\/li>\r\n \t<li>[latex]\\vec{\\Delta d}[\/latex] is <strong>change in displacement<\/strong>, commonly measured in metres (m), kilometres (km) and includes a direction<\/li>\r\n \t<li>[latex]a[\/latex] is <strong>acceleration<\/strong> (deceleration is negative), measured in metres per second squared (m\/s<sup>2<\/sup>)<\/li>\r\n \t<li>[latex]\\vec{a}[\/latex] is <strong>vector acceleration<\/strong>, measured in metres per second squared (m\/s<sup>2<\/sup>) and includes a direction<\/li>\r\n \t<li>[latex]t[\/latex] is <strong>time<\/strong>, commonly measured in seconds (s) or hours (h)<\/li>\r\n<\/ul>\r\n<h1>12.1 Power<\/h1>\r\n<div class=\"textbox\">\r\n\r\nExtra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-1\/Power\">Power<\/a>\r\n\r\n<\/div>\r\nEquations Introduced or Used for this Section:\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li>Work [latex](W) = F \\cdot d[\/latex] or [latex]F d \\cos \u00f8[\/latex]<\/li>\r\n \t<li>P [latex](P) = \\vec{F} \\cdot \\vec{v}[\/latex]<\/li>\r\n \t<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\r\n \t<li>Power [latex](P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\r\n<\/ul>\r\nPower is the rate at which work is done on a system or the rate of energy transfer between systems.\u00a0 Rate refers to the amount of change in some measure of time.\r\n<p style=\"text-align: center;\">1 watt = 1 joule\/second<\/p>\r\nThe concept of power originated from the studies of James Watt (1736-1819) as a way to compare the work done between steam engines and horses.\u00a0 This comparison was done in the absence of modern SI units and no energy concept (James Joule was 8 months old when Watt passed away).\u00a0 Instead, Watt defined his measure of power as the product of Force and Velocity, as a way of relating his steam engines to horses.\u00a0 As such, the first measure of power used units of the pound to measure force and of inches per second, feet per minute and miles per hour.\r\n<p style=\"text-align: center;\">Power [latex](P) = \\text{ Force } \\times \\text{ Velocity}[\/latex]<\/p>\r\nWatt\u2019s work, therefore, was in defining power in terms of horsepower equivalents, specifically:\r\n<p style=\"text-align: center;\">one horsepower (hp) equaled the power to raise 33<span style=\"margin-left: 0.25em;\">000<\/span> pounds (lbs) 1 foot (ft) per minute (550 pounds 1 foot per second).<\/p>\r\nLegend has it that one of Watt\u2019s first customers to whom he was trying to sell his significantly improved steam engine[footnote]Watt\u2019s improvements on the steam engine efficiency are credited as the start of the expansion of the industrial revolution. The unit of watts to measure power was named in honor of James Watt for his steam engine improvements in 1889.[\/footnote] asked Watt to compare it to a horse, which happened to be his strongest and most powerful horse.\u00a0 Not only was Watt\u2019s steam engine more powerful than this horse, the units of horsepower were then defined in terms of the power of this horse.\r\n\r\n<img class=\"wp-image-814 alignnone\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse.png\" alt=\"\" width=\"333\" height=\"249\" \/><img class=\"wp-image-815 alignnone\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2.png\" alt=\"\" width=\"352\" height=\"255\" \/>\r\n\r\nSince Watt\u2019s original definition of the value of horsepower, two measures are commonly used:\r\n<p style=\"text-align: center;\">Mechanical Horsepower\u00a0 = 745.69987158227022 W (\u2248 746 W)<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">Metric Horsepower = 735.49875 W (\u2248 735 W)<\/p>\r\nIn this textbook, mechanical horsepower will be used, since it is the most common in public usage. Electrical horsepower[footnote]Wikipedia has an extensive list of the various form and measures of horsepower: <a href=\"https:\/\/en.wikipedia.org\/wiki\/Horsepower\">https:\/\/en.wikipedia.org\/wiki\/Horsepower<\/a>[\/footnote]\u00a0is defined as 746 W.\u00a0 Questions concerning power will use the following two equations:\r\n<ul>\r\n \t<li style=\"text-align: left;\">Mechanical Power (Conservative Forces): [latex]P = \\dfrac{\\vec{F} \\cdot \\vec{d}}{t}[\/latex]<\/li>\r\n \t<li>Average Power: [latex]P = \\dfrac{W}{t}[\/latex]<\/li>\r\n<\/ul>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.1.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA 300 hp engine (horsepower) is equivalent to how many watts?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]300 \\text{ hp } = 300 \\text{ hp } \\times 746 \\text{ W\/hp}[\/latex]<\/li>\r\n \t<li>[latex]300 \\text{ hp } = 224\\;000 \\text{ W}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.1.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf 3.0 MJ of work is done in 2 min what is the power involved?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]P = \\text{ Find}[\/latex]<\/li>\r\n \t<li>[latex]W = 3.0 \\times 10^6 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]t = 2 \\text{ min or } 120 \\text{ s}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]P = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{3.0 \\times 10^6 \\text{ J}}{120 \\text{ s}} [\/latex]<\/li>\r\n \t<li>[latex]P = 25\\;000 \\text{ W or } 25 \\text{ kW}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.1.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nHow much work can a 746 W motor do in 8.0 h?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]P = \\text{ Find}[\/latex]<\/li>\r\n \t<li>[latex]W = 3.0 \\times 10^6 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]t = 2 \\text{ min or } 120 \\text{ s}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]746 \\text{ W } = \\dfrac{W}{(8.0)(3600 \\text{ s})}[\/latex]<\/li>\r\n \t<li>[latex]W = (746 \\text{W})(8.0)(3600 \\text{ s})[\/latex]<\/li>\r\n \t<li>[latex]W = 2.148 \\times 10^7 \\text{ J } (\\approx 2.1 \\times 10^7 \\text{ J})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.1.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nWhat minimum amount of time would it take a 200 W electric motor to do 12<span style=\"margin-left: 0.25em;\">000<\/span> J of work?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]P = 200 \\text{ W}[\/latex]<\/li>\r\n \t<li>[latex]W = 12\\;000 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]t = \\text{ Find}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]P = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\r\n \t<li>[latex]200 \\text{ W } = \\dfrac{12\\;00 \\text{ J}}{t}[\/latex]<\/li>\r\n \t<li>[latex]t = 12\\;000 \\text{ J } \\div 200 \\text{ W}[\/latex]<\/li>\r\n \t<li>[latex]t = 60 \\text{ s}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.1.5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>\u00a0<\/strong>If only 84% of the force produced by a 1200 W electric motor is conservative, how much time is required to do 50<span style=\"margin-left: 0.25em;\">000<\/span> J of work?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]P = 1200 \\text{ W}[\/latex]<\/li>\r\n \t<li>[latex]W = 50\\;000 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]t = \\text{ Find}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Efficiency } = 84\\% \\text{ or } 0.84[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]P = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\r\n \t<li>[latex](0.84)(1200 \\text{ W}) = \\dfrac{(50\\;000 \\text{ J})}{t}[\/latex]<\/li>\r\n \t<li>[latex]t = 50\\;000 \\text{ J } \\div 1008 \\text{ W}[\/latex]<\/li>\r\n \t<li>[latex]t = 49.6 \\text{ s } (\\approx 50 \\text{ s})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 12.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>If [latex]3 \\times10^6[\/latex] J of work is done in 5 min what is the power involved?<\/li>\r\n \t<li>How much work can a 25 W motor do in 60 h?<\/li>\r\n \t<li>What is the minimum amount of time that it would take a 500 W electric motor to do 15 kJ of work?<\/li>\r\n \t<li>120 hp (horsepower) is equivalent to how many watts?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>12.2 Power &amp; Change in Energy<\/h1>\r\nEquations Introduced or Used for this Section:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>[latex]V \\text{ or } V_g = g h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n<\/ul>\r\nWhen the force used in power settings is conservative, then the work done by this power results in a change of mechanical energy and as a result changes either the kinetic, potential or both types of mechanical energies in a system.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.2.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA conveyor belt carries a flat of 500 kg of asphalt shingles at a constant speed from the ground to the roof of a building 8.0 m above its initial position in 12.0 s.\u00a0 Find the work done on the object and the power required to accomplish it. (Assume the initial and final speeds of these shingles are zero.)\r\n\r\n<strong>Solution<\/strong>\r\n\r\nFirst, find the Work:\r\n<ul>\r\n \t<li>[latex]W = \\Delta \\text{ Energy } = \\Delta E_p \\text{ or } m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]W = (500 \\text{ kg})(9.8 \\text{ m\/s}^2)(8.0 \\text{ m})[\/latex]<\/li>\r\n \t<li>[latex]W = 39\\;200 \\text{ J}[\/latex]<\/li>\r\n<\/ul>\r\nNext, find the Power:\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = \\dfrac{W}{t}[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{39\\;200 \\text{ J}}{12.0 \\text{ s}}[\/latex]<\/li>\r\n \t<li>[latex]P = 3270 \\text{ W } (\\approx 3300 \\text{ W})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.2.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nWhat is the power of an air cannon that can fire a seed pellet (mass 45.0 g) from rest to a speed of 140 km\/h in 0.125 s?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex] where [latex]\\Delta \\text{ Energy } = E_{kf} - E_{ki}[\/latex] where [latex]E_{ki} = 0\\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{\\frac{1}{2} m v_f^2}{t}[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{\\frac{1}{2} (0.045 \\text{ kg})(38.9 \\text{ m\/s}^2)}{0.125 \\text{ s}}[\/latex]<\/li>\r\n \t<li>[latex]P = 272 \\text{ W}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.2.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nWhat is the expected power output that can be achieved from an 85.0 m high hydroelectric dam[footnote]<a href=\"http:\/\/nyfalls.com\/niagara-falls\/faq5\/\">Niagara Falls FAQ<\/a>[\/footnote] where the rate of water moving through the generators is 900 cubic metres per minute?\u00a0 If the efficiency of electric power production is 32% efficient, what is the power produced?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex] where [latex]\\Delta \\text{ Energy } = \\Delta E_p[\/latex] or [latex]m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{m g \\Delta h}{t}[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{(900 \\text{ m}^3)(1000 \\text{ kg\/m}^3)(9.8 \\text{ m\/s}^2)(85 \\text{ m})}{60 \\text{ s}}[\/latex]<\/li>\r\n \t<li>[latex]P = 1.25 \\times 10^7 \\text{ W}[\/latex] or [latex]12.5 \\text{ MW}[\/latex]<\/li>\r\n<\/ul>\r\nAt 32% efficiency the power generated is:\r\n<ul>\r\n \t<li>[latex]P = (0.32)(12.5 \\text{ MW})[\/latex] or [latex]4.0 \\text{ MW}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nNot covered in this textbook is the <strong>mechanical advantage of pulleys<\/strong>. Using pulleys one can increase the load being raised by using the same power source ... In this case, you can see the amount of rope (displacement) that is being used by smaller forces to do the same amount of work.\u00a0 This means that less powerful engines can be used to do the same amount of work, albeit at a slower rate.\u00a0 In the example shown below, each pulley adds to the displacement that the force acts through.\u00a0 In this case one only needs to consider the amount of energy lost to friction and such.\r\n\r\n<img class=\"alignnone wp-image-968 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight.jpg\" alt=\"\" width=\"551\" height=\"258\" \/>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 12.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>A world-class weight lifter can raise a 200 kg dumbbell from the floor to a height of 2.0 m in about 3.0 s.\u00a0 What is this average power output during the lift?<\/li>\r\n \t<li>What power is required to raise a mass of 47 kg a final height of 12 m in 15 s? (The mass comes to a complete stop when the final height is reached.)<\/li>\r\n \t<li>A mass of 100 kg (at rest) is raised to a final resting position 8.0 m above its initial position in 7.0 s. Find the work done on the object and the power required to accomplish it.<\/li>\r\n \t<li>What is the power of an air cannon that can fire a seed pellet (mass 9.0 g) from rest to a speed of 185 m\/s in 0.0065 s?<\/li>\r\n \t<li>What is the minimum power of an engine that can accelerate a 1600 kg pickup from 0 km\/h to 110 km\/h in 12 s?<\/li>\r\n \t<li>What is the power generated by 145 cubic metres of water falling through a potential difference of 392 m<sup>2<\/sup>\/s<sup>2<\/sup> each minute? (The mass of 1 cubic metre of water is 1 tonne.)\r\n<ol type=\"i\">\r\n \t<li>What is the expected power output that can be achieved from a 120 m high hydroelectric dam, where the rate of water moving through the generators is 43 m<sup>3<\/sup>\/s? (The mass of 1 cubic metre of water is 1000 kg or 1 tonne.)<\/li>\r\n \t<li>If the efficiency of this system is 45%, what is the power produced?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>12.3 Power &amp; Accelerated Motion<\/h1>\r\nEquations Introduced or Used for this Section:\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]a = \\dfrac{\\Delta v}{t}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\r\n \t<li>[latex]v_f = v_i + a t[\/latex]<\/li>\r\n \t<li>[latex]d = v_i t + \\frac{1}{2} a t^2[\/latex]<\/li>\r\n \t<li>[latex]F = m a[\/latex]<\/li>\r\n \t<li>Work [latex](W) = F \\cdot d[\/latex] or [latex]F d \\cos \u00f8[\/latex]<\/li>\r\n \t<li>[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>Power [latex](P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{a} = \\dfrac{\\vec{\\Delta v}}{t}[\/latex]<\/li>\r\n \t<li>[latex]v = \\dfrac{(v_i + v_f)}{2}[\/latex]<\/li>\r\n \t<li>[latex]2 a d = v_f^2 - v_i^2[\/latex]<\/li>\r\n \t<li>[latex]d = \\dfrac{(v+i v_f) t}{2}[\/latex]<\/li>\r\n \t<li>[latex]w = m g[\/latex]<\/li>\r\n \t<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\r\n \t<li>[latex]V[\/latex] or [latex]V_g = g h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n<\/ul>\r\nPower by definition when referring to the rate of work being done by a conservative force implies a change in mechanical energy, be it kinetic or potential (or both). When we restrict ourselves to only examining conservative forces, the system can be accelerating, changing vertical height at a constant velocity, or a more complicated combination of both kinetic and potential energies.\r\n\r\nIn the following examples and in using conservative forces, power is expanded to explore both accelerating and constant velocity situations.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.3.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA motor is rated to deliver 8.4 kW.\u00a0 What is the constant speed that this motor can raise a mass of 2100 kg?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex] where [latex]\\Delta \\text{ Energy } = \\Delta E_p[\/latex] or [latex]m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{m g \\Delta h}{t}[\/latex] or [latex]P = m g v_{\\text{up}}[\/latex] or [latex]P = m g v_y[\/latex]<\/li>\r\n \t<li>[latex]8400 \\text{ W } = (2100 \\text{ kg})(9.8 \\text{ m\/s}^2)(v_y)[\/latex]<\/li>\r\n \t<li>[latex]v_y = 8400 \\text{ W } \\div [(2100 \\text{ kg})(9.8 \\text{ m\/s}^2)][\/latex]<\/li>\r\n \t<li>[latex]v_y = 0.41 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.3.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA jet car used in drag racing was timed to have taken 3.4 seconds to travel 400.0 m from a stationary start.\u00a0 If this car had a mass of 0.95 tonnes:\r\n<ol type=\"i\">\r\n \t<li>Calculate the unbalanced force accelerating this car<\/li>\r\n \t<li>What power was developed?<\/li>\r\n<\/ol>\r\n<strong>Solution<\/strong>\r\n\r\nQuestion (i): First we need to calculate the acceleration\r\n<ul>\r\n \t<li>First find the acceleration:\r\n<ul>\r\n \t<li>[latex]d = v_i t + \\frac{1}{2} a t^2[\/latex]<\/li>\r\n \t<li>[latex]400 \\text{ m } = (0 \\text{ m\/s})(3.4 \\text{ s}) + \\frac{1}{2} (a)(3.4 \\text{ s})^2[\/latex]<\/li>\r\n \t<li>[latex]a = 69.2 \\text{ m\/s}^2[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Second find the force:\r\n<ul>\r\n \t<li>[latex]F = m a[\/latex]<\/li>\r\n \t<li>[latex]F = (950 \\text{ kg})(69.2 \\text{ m\/s}^2)[\/latex]<\/li>\r\n \t<li>[latex]F = 65\\;700 \\text{ N } (\\approx 66\\;000 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nQuestion (ii):\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta F d \\cos \u00f8}{t}[\/latex]<\/li>\r\n \t<li>[latex]P = \\dfrac{(65\\;700 \\text{ N})(400 \\text{ m})\\cos 0^{\\circ}}{3.4 \\text{ s}}[\/latex]<\/li>\r\n \t<li>[latex]P = 7.73 \\times 10^6 \\text{ W or } 7.73 \\text{ MW } (\\approx 7.7 \\text{ MW})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.3.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA machine can produce 120 W of power.\u00a0 If the machine is used for 12 s to raise a 50 kg mass straight upwards from a platform, where it is then released, at what speed would you expect this mass to impact the platform?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nThis problem is simpler than it looks. Since the work using conservative forces is a change in potential energy and in the second part where it is dropped, the potential energy changes to kinetic energy, then one can just use the <strong>work-energy theorem<\/strong> to solve this. There is no need to find out the change in potential energy, which we would then use to find the change in potential energy.\r\n\r\nTherefore:\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = \\dfrac{W}{t}[\/latex] or [latex]W = P t[\/latex]<\/li>\r\n<\/ul>\r\nWhere:\r\n<ul>\r\n \t<li>[latex]W = \\Delta E_k[\/latex] or [latex]W = (E_{kr} - E_{ki})[\/latex]<\/li>\r\n<\/ul>\r\nWe are left with:\r\n<ul>\r\n \t<li>[latex]P t = \\frac{1}{2} m v_f^2 - \\frac{1}{2} m v_i^2[\/latex]<\/li>\r\n \t<li>[latex](120 \\text{ W})(12 \\text{ s}) = \\frac{1}{2}(50 \\text{ kg}) v_f^2 - \\frac{1}{2}(50 \\text{ kg})(0 \\text{ m\/s})^2[\/latex]<\/li>\r\n \t<li>[latex]v_f^2 = 1440 \\text{ Ws } \\div 25 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]v_f= 7.6 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.3.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA 2.0 kW powered conveyer belt machine is angled at 28\u00b0 above the horizontal.\u00a0 At what constant speed can this belt be moving a 120 kg crate?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nFor this problem, we need to use the definition of\u00a0 <strong>Power = Force \u00d7 Velocity cos <\/strong><strong>\u00f8<\/strong><strong>, <\/strong>where the angle \u00f8 in this problem is between the force and velocity, and is<strong> 90<\/strong><strong>\u00b0 <\/strong><strong>- 28<\/strong><strong>\u00b0 <\/strong><strong>or 62<\/strong><strong>\u00b0.<\/strong>\r\n\r\n<img class=\"wp-image-971 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example.png\" alt=\"\" width=\"289\" height=\"284\" \/>\r\n\r\nTherefore:\r\n<ul>\r\n \t<li>[latex]\\text{Power } (P) = F \\cdot v \\cos \u00f8[\/latex] where [latex]F = \\text{ weight of } 120 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]2000 \\text{ W } = (120 \\text{ kg})(9.8 \\text{ m\/s}^2)(v)\\cos 62^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]v = 2000 \\text{ W } \\div 522 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]v = 3.6 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 12.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>A motor is rated to deliver 10 kW.\u00a0 What is the <strong>constant speed<\/strong> that this motor can raise a mass of [latex]2.75 \\times 10^4[\/latex] kg?<\/li>\r\n \t<li>What is the maximum possible constant speed that a 120 hp engine could lift a 4.00 ton (metric) mass straight up?<\/li>\r\n \t<li>What power is needed to run an elevator (maximum mass of 3500 kg, elevator and passengers) that rises at an average speed of 0.4 m\/s?<\/li>\r\n \t<li>A cement pump can lift concrete at the rate of 3500 kg up 20 m every minute.\r\n<ol type=\"i\">\r\n \t<li>What would be the minimum power of this pump?<\/li>\r\n \t<li>If this pump is 25% efficient, what minimum power must it be?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A monster truck used in drag racing was timed to have taken 15.0 seconds to travel 400.0 m from a stationary start. If this truck had a mass of 2.40 tonnes:\r\n<ol type=\"i\">\r\n \t<li>Calculate the unbalanced force accelerating this truck.<\/li>\r\n \t<li>What power was developed?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>An engine can produce 25<span style=\"margin-left: 0.25em;\">000<\/span> W of power. If we assume that the conversion of engine power to mechanical energy of this 1000 kg car is 100% efficient, at what speed could this car be traveling after 15 s of full power, starting from rest?<\/li>\r\n \t<li>A machine can produce 40 W of power. If the machine is used for 15 s to raise a 20 kg mass straight upwards where it is released, at what speed would you expect this mass to impact its starting position?<\/li>\r\n \t<li>If a loaded semi (total mass = 20<span style=\"margin-left: 0.25em;\">000<\/span> kg) is able to maintain a constant speed of 20 km\/h up an 8\u00b0 inclined road, what is the power output of the truck\u2019s engine?<\/li>\r\n \t<li>What is the angle of a hill that a loaded pickup truck (m = 3600 kg) can climb at a constant speed of 45 km\/h, if the 300 hp engine is 32% efficient?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 12.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li><strong>It is claimed that large trees <\/strong>can evaporate as much as 1000 kg of water per day.\u00a0 Assuming the average rise of water to be 10 m from the ground, what amount of energy must be supplied for this?\u00a0 If we assume the source of this energy is from the Sun, what average amount of power must the tree receive over 12 hours of sunlight?<\/li>\r\n \t<li><strong>An average human at rest<\/strong> radiates 75 W of power.\u00a0 What amount of energy is radiated in eight hours?<\/li>\r\n \t<li><strong>Power Output of the Sun...<\/strong>\u00a0 The amount of solar energy striking the Earth is around 1350 watts per metre2.\u00a0 Find the power output of our Sun. <strong>Under Business as Usual projections<\/strong> for global energy consumption, Energy usage is expected to rise from 589 EJ in 2016 to 1070 EJ in 2100 (E stands for [latex]exo \\times 10^21[\/latex])<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">Video to Watch: <a href=\"https:\/\/www.youtube.com\/watch?v=GSVv40M2aks\">NASA - SDO Year 5<\/a><\/div>\r\n<h1>12.4 The Wind Resource<\/h1>\r\n<div class=\"textbox\">\r\n<ul>\r\n \t<li>Article to Read: <a href=\"https:\/\/www.ucsusa.org\/resources\/how-wind-energy-works?print=t\">How Wind Energy Works<\/a><\/li>\r\n \t<li>Research News:<a href=\"https:\/\/www.sciencedaily.com\/releases\/2019\/08\/190814081202.htm\"> Europe has the untapped onshore capacity to meet global energy demand<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\nSeveral factors can affect wind speed and the ability of a turbine to generate more power.\u00a0 For example, wind speed increases as the height from the ground increases.\u00a0 If wind speed at 10 meters off the ground is 6 m\/s it will be about 7.5 m\/s at a height of 50 meters.\u00a0 A 2003 Stanford University study examined wind speeds at higher elevations and found that as much as one quarter of the United States, including areas historically thought to have poor wind potential, is potentially suitable for providing affordable electric power from wind.\u00a0 The rotors of the newest wind turbines can now reach heights up to 70 meters.\u00a0 In addition to height, the power in the wind varies with temperature and altitude, both of which affect the air density.\u00a0 Winter winds in Minnesota State will carry more power than summer winds of the same speed high in the passes of Southern States.\r\n\r\nThe more the wind blows, the more power will be produced, with the main drawback being inconsistent wind.\u00a0 The term used to describe this is \"capacity factor\", which is simply the amount of power a turbine actually produces over a period of time divided by the amount of power it could have produced if it had run at its full-rated capacity over that time period.\r\n\r\nA more precise measurement of output is the \"specific yield\". This measures the annual energy output per square meter of area swept by the turbine blades as they rotate.\u00a0 Overall, wind turbines capture between 20 and 40 percent of the energy in the wind.\u00a0 So, at a site with average wind speeds of 7 m\/s, a typical turbine will produce about 1100 kilowatt-hours (kWh) per square meter of area per year.\u00a0 If the turbine has blades that are 40 meters long, for a total swept area of 5029 square meters, the power output will be about 5.5 million kWh for the year.\u00a0 An increase in blade length, which in turn increases the swept area, can have a significant effect on the amount of power output from a wind turbine.\r\n\r\nAnother factor in the cost of wind power is the distance of the turbines from transmission lines.\u00a0 Although some large windy areas, particularly in rural parts of the High Plains and Rocky Mountains, are too far from power lines to be cost-effective today, the potential for these regions is still enormous.\r\n\r\nA final consideration for a wind resource is the seasonal and daily variation in wind speed.\u00a0 If the wind blows during periods of peak power demand, power from a wind farm will be valued more highly than if it blows in off-peak periods.\u00a0 In California, for example, high temperatures in the central valley and low coastal temperatures near San Francisco cause powerful winds to blow across the Altamont in the summer, a period of high power demand.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.4.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<img class=\"wp-image-972 aligncenter\" style=\"font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.1-example.jpg\" alt=\"\" width=\"231\" height=\"357\" \/>What is the kinetic energy or the air mass that passes through an offshore windmill at the Thorntonbank Wind Farm[footnote]<a href=\"https:\/\/en.wikipedia.org\/wiki\/Thorntonbank_Wind_Farm\">Thorntonbank Wind Farm<\/a>[\/footnote], having a rotor diameter of 126 m, if the wind speed averages 30 km\/hr?\u00a0 If we assume that current technology yields 45% power generation from this moving air mass, what power is produced per turbine?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nKinetic Energy:\r\n<ul>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2}(1.225 \\text{ kg\/m}^3)(8.33 \\text{ m\/s})^2[\/latex]<\/li>\r\n \t<li>[latex]E_k = 42.5 \\text{ J\/m}^3[\/latex]<\/li>\r\n<\/ul>\r\nPower Produced (100% Efficient):\r\n<ul>\r\n \t<li>[latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/li>\r\n \t<li>[latex]P = \\frac{1}{2} (1.225 \\text{ kg\/m}^3)(\\pi)(126 \\text{ m})^2(8.33 \\text{ m\/s})^3[\/latex]<\/li>\r\n \t<li>[latex]P = 1.7 \\times 10^7 \\text{ W or } 17.7 \\text{ MW } (\\approx 18 \\text{ MW})[\/latex]<\/li>\r\n<\/ul>\r\nNow, for 45% efficiency:\r\n<ul>\r\n \t<li>[latex]P = (0.45) 1.77 \\times 10^7 \\text{W}[\/latex]<\/li>\r\n \t<li>[latex]P = 7.95 \\times 10^6[\/latex] or [latex]7.95 \\text{ MW } (\\approx 8.0 \\text{ MW})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<h2>Wind Turbine Power:<\/h2>\r\nEquations Introduced or Used for this Section:\r\n<p style=\"text-align: center;\">[latex]P = \\frac{1}{2} \\varrho A C_P v^2 N_g N_b[\/latex]<\/p>\r\nWhere...\r\n<ul>\r\n \t<li>[latex]P[\/latex] = power in watts<\/li>\r\n \t<li>[latex]\\varrho[\/latex] = air density (about 1.225 kg\/m<sup>3<\/sup> at sea level at15\u00b0C)<\/li>\r\n \t<li>[latex]A[\/latex] = rotor swept area exposed to the wind (m2)<\/li>\r\n \t<li>[latex]C_p[\/latex] = coefficient of performance (0.59 {Betz limit} is the maximum theoretically possible 0.35 for a good design)<\/li>\r\n \t<li>[latex]v[\/latex] = wind speed in m\/s<\/li>\r\n \t<li>[latex]N_g[\/latex]\u00a0= generator efficiency (50% for car alternator, 80% or possibly more for a permanent magnet generator or grid-connected induction generator)<\/li>\r\n \t<li>[latex]N_b[\/latex]\u00a0= gearbox\/bearings efficiency (could be as high as 95% if good)<\/li>\r\n<\/ul>\r\n<table style=\"border-collapse: collapse; width: 100%; height: 265px;\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 31px;\">\r\n<th style=\"width: 50%; height: 31px;\" scope=\"col\">Temperature (\u00b0C)<\/th>\r\n<th style=\"width: 50%; height: 31px;\" scope=\"col\">Density of Air (kg\/m<sup>3<\/sup>)<\/th>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">\u221225<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.423<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">\u221220<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.395<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">\u221215<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.368<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">\u221210<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.342<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">\u22125<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.317<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">0<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.292<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">5<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.269<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">10<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.247<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">15<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.225<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">20<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.204<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">25<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.184<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">30<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.164<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px;\">\r\n<td style=\"width: 50%; height: 18px;\">35<\/td>\r\n<td style=\"width: 50%; height: 18px;\">1.146<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 12.4.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Using the graph and the chart on the previous page and assuming 100% efficiency\r\n<ol type=\"i\">\r\n \t<li>What is the area of a 1.0 m circle?<\/li>\r\n \t<li>If the density of air at 20\u00b0 C is 1.204 kg\/m<sup>3<\/sup> at sea level, what must be the minimum speed of the air for the 1.0 m diameter rotors to extract 0.4 kW?<\/li>\r\n \t<li>Using an air temperature of 20\u00b0C, calculate the minimum number of 11 m diameter rotors needed to generate 20 MW?<\/li>\r\n \t<li>What must be the speed of the air at \u221215\u00b0C to create 0.40 kW?<\/li>\r\n \t<li>Assuming the 10 m rotor produces 39.3 kW at an air temperature of 20\u00b0C.\u00a0 What is the percent increase in power output if the temperature drops to -20\u00b0C?<img class=\"alignnone wp-image-974 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1.jpeg\" alt=\"\" width=\"423\" height=\"273\" \/><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The above graph shows the potential amount of energy from wind at 20\u00b0 C.\r\n<ol type=\"i\">\r\n \t<li>What would this curve look like if the air is hotter?<\/li>\r\n \t<li>\u00a0What would this curve look like if the air is colder?<\/li>\r\n \t<li>Notice the power at 8.0 m\/s is around 300 W and at 10 m\/s around 600 W.\u00a0 Show how these numbers fit the equation [latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/li>\r\n \t<li>How big would the rotor area of the windmill graphed above be if we assume it is 100% efficient?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n(Realistically, [latex]C_p N_g N_b[\/latex] = 0.27 or is around 27%)\r\n\r\n<\/div>\r\n<\/div>\r\n<h1>12.5 Power from Ocean and River Currents<\/h1>\r\n<div class=\"textbox\">Research News: <a href=\"https:\/\/www.sciencedirect.com\/science\/article\/pii\/S2590174519300145\">Hunt, J. D., et al. (2019) Using the jet stream for sustainable airship and balloon transportation of cargo and hydrogen<\/a><\/div>\r\nOcean current speeds are generally lower than wind speeds.\u00a0 This is important because the kinetic energy contained in flowing bodies is proportional to the cube of their velocity.\u00a0 However another more important factor in the power available for extraction from a flowing body is the density of the material.\u00a0 Water is about 835 times denser than wind, so for the same area of flow being intercepted the energy contained in a 19 km\/h water flow is equivalent to that contained in an air mass moving at about 180 km\/h.\u00a0 Thus, ocean currents represent a potentially significant currently untapped reservoir of energy.\r\n\r\nThe total worldwide power in ocean currents has been estimated to be about 5000 GW with power densities of up to 15 kW\/m2.\u00a0 The relatively constant extractable energy density near the surface of the Florida Straits Current is about 1 kW\/m2 of flow area.\u00a0 It has been estimated that capturing just 1\/1000th of the available energy from the Gulf Stream, which has 21 000 times more energy than Niagara Falls in a flow of water that is 50 times the total flow of all the world\u2019s freshwater river, would supply Florida with 35% of its electrical needs.\r\n\r\nCountries that are interested in and pursuing the application of ocean current energy technologies include the European Union, Japan and China.\r\n\r\nEquations Introduced or Used for this Section:\r\n<p style=\"text-align: center;\">[latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/p>\r\nWhere...\r\n<ul>\r\n \t<li>[latex]P[\/latex] is power in watts<\/li>\r\n \t<li>[latex]\\varrho[\/latex] is density of water (fresh water is roughly 1000 kg\/m<sup>3<\/sup>).<\/li>\r\n \t<li>[latex]A[\/latex] is rotor swept area exposed to the water (m<sup>2<\/sup>)<\/li>\r\n \t<li>[latex]v[\/latex] is water speed (m\/s)<\/li>\r\n<\/ul>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.5.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nThe Amazon River can reach an average current speed of 7 km\/h at the height of the wet season.\u00a0 What is the expected power that can be produced by an underwater turbine of diameter 2.0 m, assuming an efficiency of 42% for the power generating turbine? (Density of fresh water is around 1000 kg\/m<sup>3<\/sup>)\r\n\r\n<img class=\"wp-image-976 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1.png\" alt=\"\" width=\"311\" height=\"311\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/li>\r\n \t<li>[latex]P = \\frac{1}{2}(1000 \\text{kg\/m}^3) \\pi (2.0 \\text{ m})^2 (1.94 \\text{ m\/s})^3[\/latex]<\/li>\r\n \t<li>[latex]P = 12\\;200 \\text{ W or } 12.2 \\text{ KW } (\\approx 12 \\text{ KW})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 12.5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>The Fraser River has an average current speed ranging from 0.25 m\/s to 1.25 m\/s. What is the expected range of power that can be produced by an underwater turbine of diameter 1.5 m?\u00a0 (Density of fresh water is around 1000 kg\/m<sup>3<\/sup>)<\/li>\r\n \t<li>One of the strongest ocean currents is the Gulf Stream, which moves at speeds greater than 1.5 m\/s. What is the maximum power that can be extracted by a 3.0 m diameter underwater turbine in this current?\u00a0 How many of these turbines would be needed to replace a 50 MW coal-powered electric generator? (Density of ocean water is around 1003 kg\/m<sup>3<\/sup>)<\/li>\r\n \t<li>What maximum power could be extracted from a 1.0 m diameter turbine in water moving at 20 km\/h? (This is the speed of water falling 1.57 m.) How does this compare to what a similar sized turbine could extract from a Category 2 Hurricane where the wind (20\u00b0 C) is moving at 160 km\/h?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">Article to Read: <a href=\"https:\/\/oilprice.com\/Latest-Energy-News\/World-News\/Scientists-Develop-Wave-Energy-Device-That-Generates-Low-Cost-Electricity.html\">Scientists Develop Wave Energy Device That Generates Low-Cost Electricity<\/a><\/div>\r\n<h1>Exercise Answers<\/h1>\r\n<h2>12.1 Power<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>10<span style=\"margin-left: 0.25em;\">000<\/span> W<\/li>\r\n \t<li>5.4 MJ<\/li>\r\n \t<li>30 s<\/li>\r\n \t<li>89<span style=\"margin-left: 0.25em;\">500<\/span> W or 89.5 kW<\/li>\r\n<\/ol>\r\n<h2>12.2 Power &amp; Change in Energy<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>1300 W<\/li>\r\n \t<li>368 W<\/li>\r\n \t<li>7840 J, 1120 W<\/li>\r\n \t<li>23<span style=\"margin-left: 0.25em;\">700<\/span> W<\/li>\r\n \t<li>62<span style=\"margin-left: 0.25em;\">000<\/span> W<\/li>\r\n \t<li>0.947 MW<\/li>\r\n \t<li>\r\n<ol type=\"i\">\r\n \t<li>50.5 MW<\/li>\r\n \t<li>22.8 MW<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h2>12.3 Power &amp; Accelerated Motion<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>0.0371 m\/s<\/li>\r\n \t<li>2.28 m\/s<\/li>\r\n \t<li>13<span style=\"margin-left: 0.25em;\">700<\/span> W or 13.7 kW\r\n<ol type=\"i\">\r\n \t<li>11<span style=\"margin-left: 0.25em;\">400<\/span> W<\/li>\r\n \t<li>45<span style=\"margin-left: 0.25em;\">700<\/span> W<\/li>\r\n<\/ol>\r\n<ol type=\"i\">\r\n \t<li>8500 N<\/li>\r\n \t<li>[latex]2.28 \\times 10^5[\/latex] W = 228 kW<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>27.4 m\/s<\/li>\r\n \t<li>7.7 m\/s<\/li>\r\n \t<li>152<span style=\"margin-left: 0.25em;\">000<\/span> W = 152 kW<\/li>\r\n \t<li>9.3\u00b0<\/li>\r\n<\/ol>\r\n<h2>12.4 Large Tree Evaporation, Human\u2019s Radiated Heat, Power Output of the Sun<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>\r\n<ol type=\"i\">\r\n \t<li>98<span style=\"margin-left: 0.25em;\">000<\/span> J<\/li>\r\n \t<li>2.27 W<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>[latex]2.2 \\times 10^6[\/latex] J<\/li>\r\n \t<li>[latex]3.8 \\times 10^{26}[\/latex] W<\/li>\r\n<\/ol>\r\n<h2>12.4.1 The Wind Resource<\/h2>\r\n<ol>\r\n \t<li>\r\n<ol class=\"twocolumn\" type=\"i\">\r\n \t<li>0.785 m<sup>2<\/sup><\/li>\r\n \t<li>9.46 m\/s<\/li>\r\n \t<li>421 Rotors<\/li>\r\n \t<li>9.18 m\/s<\/li>\r\n \t<li>45.4 kW<\/li>\r\n<\/ol>\r\n<ol type=\"i\">\r\n \t<li>The curve should flatten out due to decreased air density<\/li>\r\n \t<li>The curve should become steeper climb due to increased air density<\/li>\r\n \t<li>The ratio of 600 to 300 watts is roughly the same as the ratio of (10 m\/s)<sup>3<\/sup> to (8 m\/s)<sup>3<\/sup>. Alternatively, on can look at the ratio of [latex]\\dfrac{P}{v^3}[\/latex] which equals 0.6 for both speeds.<\/li>\r\n \t<li>0.997 m<sup>2<\/sup> or 1 m<sup>2<\/sup><\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h2>12.5 Power from Ocean and River Currents<\/h2>\r\n<ol>\r\n \t<li>13.8 W at 0.25 m\/s to 1.73 kW at 1.25 m\/s<\/li>\r\n \t<li>120 kW maximum power, approximately 4200 turbines to replace 50 MW<\/li>\r\n \t<li>67 kW for water compared to 41 kW from a hurricane<\/li>\r\n<\/ol>\r\n<h3>Media Attributions<\/h3>\r\n<ul>\r\n \t<li>\"The Horse Mill\" by John F. Herring (1795 - 1865) is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Horse_Gin_in_use.jpg\">A Horse Gin In Use<\/a>\" by Roger Griffith is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/en.wikipedia.org\/wiki\/File:Four_pulleys.svg\">Pulley diagram with 4 pulleys<\/a>\" by <a title=\"User:Prolineserver\" href=\"https:\/\/commons.wikimedia.org\/wiki\/User:Prolineserver\">Prolineserver<\/a> and <a title=\"User:Tomia\" href=\"https:\/\/commons.wikimedia.org\/wiki\/User:Tomia\">Tomia<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/3.0\/deed.en\">CC BY-SA 3.0 Unported<\/a> licence.<\/li>\r\n \t<li>\"<a href=\"https:\/\/en.wikipedia.org\/wiki\/File:Amazonrivermap.svg\">Amazonrivermap<\/a>\" by <a class=\"external text\" href=\"https:\/\/en.wikipedia.org\/wiki\/User:Kmusser?rdfrom=commons:User:Kmusser\">Kmusser<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/3.0\/deed.en\">CC BY-SA 3.0 Unported<\/a> licence.<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Resources<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ul>\n<li>Extra Help: <a href=\"http:\/\/www.a-levelphysicstutor.com\/index-mech.php\">A-Level Physics Tutor<\/a><\/li>\n<li>Article to Read: <a href=\"https:\/\/www.greencarcongress.com\/2019\/02\/20190219-bye.html\">Bye Aerospace\u2019s Sun Flyer 2 completes first flight with Siemens electric propulsion motor<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>Equations Introduced and Used for this Topic:<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li>[latex]a = \\dfrac{\\Delta v}{t}[\/latex]<\/li>\n<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\n<li>[latex]v_f = v_i + a t[\/latex]<\/li>\n<li>[latex]d = v_i t + \\frac{1}{2} a t^2[\/latex]<\/li>\n<li>[latex]F = m a[\/latex]<\/li>\n<li>Work [latex](W) = F \\cdot d[\/latex] or [latex]F d \\cos \u00f8[\/latex]<\/li>\n<li>[latex]E_p = m g h[\/latex]<\/li>\n<li>Power [latex](P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\n<li>[latex]P = \\frac{1}{2}\\varrho A v^3[\/latex]<\/li>\n<li>[latex]\\vec{a} = \\dfrac{\\vec{\\Delta v}}{t}[\/latex]<\/li>\n<li>[latex]v = \\dfrac{(v_i + v_f)}{2}[\/latex]<\/li>\n<li>[latex]2 a d = v_f^2 - v_i^2[\/latex]<\/li>\n<li>[latex]d = \\dfrac{(v+i v_f) t}{2}[\/latex]<\/li>\n<li>[latex]w = m g[\/latex]<\/li>\n<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\n<li>[latex]V[\/latex] or [latex]V_g = g h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<li>[latex]\\text{Efficiency } = \\text{ Output } \\times 100%[\/latex]<\/li>\n<\/ul>\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 21.1738%;\" scope=\"col\">Equations<\/th>\n<th style=\"width: 17.3709%;\" scope=\"col\">a<\/th>\n<th style=\"width: 16.2441%;\" scope=\"col\">v<sub>t<\/sub><\/th>\n<th style=\"width: 16.6667%;\" scope=\"col\">\u00a0v<sub>i<\/sub><\/th>\n<th style=\"width: 15.8217%;\" scope=\"col\">d<\/th>\n<th style=\"width: 12.723%;\" scope=\"col\">t<\/th>\n<\/tr>\n<tr>\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]v_t = v_i + at[\/latex]<\/th>\n<td style=\"width: 17.3709%;\"><\/td>\n<td style=\"width: 16.2441%;\"><\/td>\n<td style=\"width: 16.6667%;\"><\/td>\n<td style=\"width: 15.8217%;\">Not mentioned<\/td>\n<td style=\"width: 12.723%;\"><\/td>\n<\/tr>\n<tr>\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]2ad = v_{r}^{2} - v_{i}^{2}[\/latex]<\/th>\n<td style=\"width: 17.3709%;\"><\/td>\n<td style=\"width: 16.2441%;\"><\/td>\n<td style=\"width: 16.6667%;\"><\/td>\n<td style=\"width: 15.8217%;\"><\/td>\n<td style=\"width: 12.723%;\">Not mentioned<\/td>\n<\/tr>\n<tr>\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]d = v_{i}t + \\dfrac{1}{2} at^2[\/latex]<\/th>\n<td style=\"width: 17.3709%;\"><\/td>\n<td style=\"width: 16.2441%;\">Not mentioned<\/td>\n<td style=\"width: 16.6667%;\"><\/td>\n<td style=\"width: 15.8217%;\"><\/td>\n<td style=\"width: 12.723%;\"><\/td>\n<\/tr>\n<tr>\n<th style=\"width: 21.1738%;\" scope=\"row\">[latex]d = \\dfrac{(v_i + v_t)t}{2}[\/latex]<\/th>\n<td style=\"width: 17.3709%;\">Not mentioned<\/td>\n<td style=\"width: 16.2441%;\"><\/td>\n<td style=\"width: 16.6667%;\"><\/td>\n<td style=\"width: 15.8217%;\"><\/td>\n<td style=\"width: 12.723%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Where&#8230;<\/p>\n<ul>\n<li>[latex]P[\/latex] is the <strong>power<\/strong>, measured in watts (W)<\/li>\n<li>[latex]W[\/latex] is <strong>work<\/strong>, measured in joules (J)<\/li>\n<li>[latex]\\varrho[\/latex] is the <strong>density of the moving medium<\/strong>, measured in kilograms per metre cubed (kg\/m<sup>3<\/sup>)<\/li>\n<li>[latex]A[\/latex] is the <strong>cross sectional area of the turbine<\/strong>, measured in metres squared (m<sup>2<\/sup>)<\/li>\n<li>[latex]v[\/latex] is the <strong>speed of the medium moving through the turbine<\/strong>, measured in metres per second (m\/s)<\/li>\n<li>[latex]v[\/latex] is the <strong>speed of the object<\/strong> commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h)<\/li>\n<li>[latex]\\Delta \\text{ Energy}[\/latex] is the <strong>change in energy<\/strong> and can be any form of energy ([latex]E_p[\/latex] or other forms), measured in joules (J)<\/li>\n<li>[latex]E_p[\/latex] is <strong>gravitational potential energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_k[\/latex] is <strong>kinetic energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]V[\/latex] or [latex]V_g[\/latex] is the <strong>gravitational potential<\/strong>, measured in newtons per kilogram (N\/kg)<\/li>\n<li>[latex]\\Delta h[\/latex] is the <strong>change in vertical height<\/strong>, measured in metres (m)<\/li>\n<li>[latex]F[\/latex] is <strong>force<\/strong>, measured in newtons (N)<\/li>\n<li>[latex]m[\/latex] is <strong>mass<\/strong>, commonly measured in kilograms (Kg), grams (g) or tonnes (t)<\/li>\n<li>[latex]g[\/latex] or [latex]a_g[\/latex] is the <strong>acceleration due to gravity<\/strong> and varies from 9.78 m\/s<sup>2<\/sup> (equator) to 9.83 m\/s<sup>2<\/sup> (North or South poles)&#8230; The average value of gravity is taken to be either 9.80 m\/s<sup>2<\/sup>\u00a0 or 9.81 m\/s<sup>2<\/sup><\/li>\n<li>[latex]\\vec{v}[\/latex] is <strong>velocity<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/li>\n<li>[latex]v_i[\/latex] is <strong>initial speed<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h)<\/li>\n<li>[latex]\\vec{v_i}[\/latex] is <strong>initial velocity<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/li>\n<li>[latex]v_f[\/latex] is <strong>final speed<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h)<\/li>\n<li>[latex]\\vec{v_f}[\/latex] is <strong>final velocity<\/strong>, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/li>\n<li>[latex]\\Delta d[\/latex] is <strong>distance travelled<\/strong>, commonly measured in metres (m), kilometres (km)<\/li>\n<li>[latex]\\vec{\\Delta d}[\/latex] is <strong>change in displacement<\/strong>, commonly measured in metres (m), kilometres (km) and includes a direction<\/li>\n<li>[latex]a[\/latex] is <strong>acceleration<\/strong> (deceleration is negative), measured in metres per second squared (m\/s<sup>2<\/sup>)<\/li>\n<li>[latex]\\vec{a}[\/latex] is <strong>vector acceleration<\/strong>, measured in metres per second squared (m\/s<sup>2<\/sup>) and includes a direction<\/li>\n<li>[latex]t[\/latex] is <strong>time<\/strong>, commonly measured in seconds (s) or hours (h)<\/li>\n<\/ul>\n<h1>12.1 Power<\/h1>\n<div class=\"textbox\">\n<p>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-1\/Power\">Power<\/a><\/p>\n<\/div>\n<p>Equations Introduced or Used for this Section:<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li>Work [latex](W) = F \\cdot d[\/latex] or [latex]F d \\cos \u00f8[\/latex]<\/li>\n<li>P [latex](P) = \\vec{F} \\cdot \\vec{v}[\/latex]<\/li>\n<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\n<li>Power [latex](P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\n<\/ul>\n<p>Power is the rate at which work is done on a system or the rate of energy transfer between systems.\u00a0 Rate refers to the amount of change in some measure of time.<\/p>\n<p style=\"text-align: center;\">1 watt = 1 joule\/second<\/p>\n<p>The concept of power originated from the studies of James Watt (1736-1819) as a way to compare the work done between steam engines and horses.\u00a0 This comparison was done in the absence of modern SI units and no energy concept (James Joule was 8 months old when Watt passed away).\u00a0 Instead, Watt defined his measure of power as the product of Force and Velocity, as a way of relating his steam engines to horses.\u00a0 As such, the first measure of power used units of the pound to measure force and of inches per second, feet per minute and miles per hour.<\/p>\n<p style=\"text-align: center;\">Power [latex](P) = \\text{ Force } \\times \\text{ Velocity}[\/latex]<\/p>\n<p>Watt\u2019s work, therefore, was in defining power in terms of horsepower equivalents, specifically:<\/p>\n<p style=\"text-align: center;\">one horsepower (hp) equaled the power to raise 33<span style=\"margin-left: 0.25em;\">000<\/span> pounds (lbs) 1 foot (ft) per minute (550 pounds 1 foot per second).<\/p>\n<p>Legend has it that one of Watt\u2019s first customers to whom he was trying to sell his significantly improved steam engine<a class=\"footnote\" title=\"Watt\u2019s improvements on the steam engine efficiency are credited as the start of the expansion of the industrial revolution. The unit of watts to measure power was named in honor of James Watt for his steam engine improvements in 1889.\" id=\"return-footnote-169-1\" href=\"#footnote-169-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> asked Watt to compare it to a horse, which happened to be his strongest and most powerful horse.\u00a0 Not only was Watt\u2019s steam engine more powerful than this horse, the units of horsepower were then defined in terms of the power of this horse.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-814 alignnone\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse.png\" alt=\"\" width=\"333\" height=\"249\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse.png 454w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-300x225.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-65x49.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-225x169.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-350x262.png 350w\" sizes=\"auto, (max-width: 333px) 100vw, 333px\" \/><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-815 alignnone\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2.png\" alt=\"\" width=\"352\" height=\"255\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2.png 472w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2-300x217.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2-65x47.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2-225x163.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.1-horse-2-350x254.png 350w\" sizes=\"auto, (max-width: 352px) 100vw, 352px\" \/><\/p>\n<p>Since Watt\u2019s original definition of the value of horsepower, two measures are commonly used:<\/p>\n<p style=\"text-align: center;\">Mechanical Horsepower\u00a0 = 745.69987158227022 W (\u2248 746 W)<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">Metric Horsepower = 735.49875 W (\u2248 735 W)<\/p>\n<p>In this textbook, mechanical horsepower will be used, since it is the most common in public usage. Electrical horsepower<a class=\"footnote\" title=\"Wikipedia has an extensive list of the various form and measures of horsepower: https:\/\/en.wikipedia.org\/wiki\/Horsepower\" id=\"return-footnote-169-2\" href=\"#footnote-169-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>\u00a0is defined as 746 W.\u00a0 Questions concerning power will use the following two equations:<\/p>\n<ul>\n<li style=\"text-align: left;\">Mechanical Power (Conservative Forces): [latex]P = \\dfrac{\\vec{F} \\cdot \\vec{d}}{t}[\/latex]<\/li>\n<li>Average Power: [latex]P = \\dfrac{W}{t}[\/latex]<\/li>\n<\/ul>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.1.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A 300 hp engine (horsepower) is equivalent to how many watts?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]300 \\text{ hp } = 300 \\text{ hp } \\times 746 \\text{ W\/hp}[\/latex]<\/li>\n<li>[latex]300 \\text{ hp } = 224\\;000 \\text{ W}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.1.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If 3.0 MJ of work is done in 2 min what is the power involved?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]P = \\text{ Find}[\/latex]<\/li>\n<li>[latex]W = 3.0 \\times 10^6 \\text{ J}[\/latex]<\/li>\n<li>[latex]t = 2 \\text{ min or } 120 \\text{ s}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]P = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\n<li>[latex]P = \\dfrac{3.0 \\times 10^6 \\text{ J}}{120 \\text{ s}}[\/latex]<\/li>\n<li>[latex]P = 25\\;000 \\text{ W or } 25 \\text{ kW}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.1.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>How much work can a 746 W motor do in 8.0 h?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]P = \\text{ Find}[\/latex]<\/li>\n<li>[latex]W = 3.0 \\times 10^6 \\text{ J}[\/latex]<\/li>\n<li>[latex]t = 2 \\text{ min or } 120 \\text{ s}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]746 \\text{ W } = \\dfrac{W}{(8.0)(3600 \\text{ s})}[\/latex]<\/li>\n<li>[latex]W = (746 \\text{W})(8.0)(3600 \\text{ s})[\/latex]<\/li>\n<li>[latex]W = 2.148 \\times 10^7 \\text{ J } (\\approx 2.1 \\times 10^7 \\text{ J})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.1.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>What minimum amount of time would it take a 200 W electric motor to do 12<span style=\"margin-left: 0.25em;\">000<\/span> J of work?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]P = 200 \\text{ W}[\/latex]<\/li>\n<li>[latex]W = 12\\;000 \\text{ J}[\/latex]<\/li>\n<li>[latex]t = \\text{ Find}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]P = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\n<li>[latex]200 \\text{ W } = \\dfrac{12\\;00 \\text{ J}}{t}[\/latex]<\/li>\n<li>[latex]t = 12\\;000 \\text{ J } \\div 200 \\text{ W}[\/latex]<\/li>\n<li>[latex]t = 60 \\text{ s}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.1.5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>\u00a0<\/strong>If only 84% of the force produced by a 1200 W electric motor is conservative, how much time is required to do 50<span style=\"margin-left: 0.25em;\">000<\/span> J of work?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]P = 1200 \\text{ W}[\/latex]<\/li>\n<li>[latex]W = 50\\;000 \\text{ J}[\/latex]<\/li>\n<li>[latex]t = \\text{ Find}[\/latex]<\/li>\n<li>[latex]\\text{Efficiency } = 84\\% \\text{ or } 0.84[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]P = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\n<li>[latex](0.84)(1200 \\text{ W}) = \\dfrac{(50\\;000 \\text{ J})}{t}[\/latex]<\/li>\n<li>[latex]t = 50\\;000 \\text{ J } \\div 1008 \\text{ W}[\/latex]<\/li>\n<li>[latex]t = 49.6 \\text{ s } (\\approx 50 \\text{ s})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 12.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>If [latex]3 \\times10^6[\/latex] J of work is done in 5 min what is the power involved?<\/li>\n<li>How much work can a 25 W motor do in 60 h?<\/li>\n<li>What is the minimum amount of time that it would take a 500 W electric motor to do 15 kJ of work?<\/li>\n<li>120 hp (horsepower) is equivalent to how many watts?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>12.2 Power &amp; Change in Energy<\/h1>\n<p>Equations Introduced or Used for this Section:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]E_p = m g h[\/latex]<\/li>\n<li>[latex]V \\text{ or } V_g = g h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<\/ul>\n<p>When the force used in power settings is conservative, then the work done by this power results in a change of mechanical energy and as a result changes either the kinetic, potential or both types of mechanical energies in a system.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.2.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A conveyor belt carries a flat of 500 kg of asphalt shingles at a constant speed from the ground to the roof of a building 8.0 m above its initial position in 12.0 s.\u00a0 Find the work done on the object and the power required to accomplish it. (Assume the initial and final speeds of these shingles are zero.)<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, find the Work:<\/p>\n<ul>\n<li>[latex]W = \\Delta \\text{ Energy } = \\Delta E_p \\text{ or } m g \\Delta h[\/latex]<\/li>\n<li>[latex]W = (500 \\text{ kg})(9.8 \\text{ m\/s}^2)(8.0 \\text{ m})[\/latex]<\/li>\n<li>[latex]W = 39\\;200 \\text{ J}[\/latex]<\/li>\n<\/ul>\n<p>Next, find the Power:<\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = \\dfrac{W}{t}[\/latex]<\/li>\n<li>[latex]P = \\dfrac{39\\;200 \\text{ J}}{12.0 \\text{ s}}[\/latex]<\/li>\n<li>[latex]P = 3270 \\text{ W } (\\approx 3300 \\text{ W})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.2.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>What is the power of an air cannon that can fire a seed pellet (mass 45.0 g) from rest to a speed of 140 km\/h in 0.125 s?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex] where [latex]\\Delta \\text{ Energy } = E_{kf} - E_{ki}[\/latex] where [latex]E_{ki} = 0\\text{ J}[\/latex]<\/li>\n<li>[latex]P = \\dfrac{\\frac{1}{2} m v_f^2}{t}[\/latex]<\/li>\n<li>[latex]P = \\dfrac{\\frac{1}{2} (0.045 \\text{ kg})(38.9 \\text{ m\/s}^2)}{0.125 \\text{ s}}[\/latex]<\/li>\n<li>[latex]P = 272 \\text{ W}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.2.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>What is the expected power output that can be achieved from an 85.0 m high hydroelectric dam<a class=\"footnote\" title=\"Niagara Falls FAQ\" id=\"return-footnote-169-3\" href=\"#footnote-169-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a> where the rate of water moving through the generators is 900 cubic metres per minute?\u00a0 If the efficiency of electric power production is 32% efficient, what is the power produced?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex] where [latex]\\Delta \\text{ Energy } = \\Delta E_p[\/latex] or [latex]m g \\Delta h[\/latex]<\/li>\n<li>[latex]P = \\dfrac{m g \\Delta h}{t}[\/latex]<\/li>\n<li>[latex]P = \\dfrac{(900 \\text{ m}^3)(1000 \\text{ kg\/m}^3)(9.8 \\text{ m\/s}^2)(85 \\text{ m})}{60 \\text{ s}}[\/latex]<\/li>\n<li>[latex]P = 1.25 \\times 10^7 \\text{ W}[\/latex] or [latex]12.5 \\text{ MW}[\/latex]<\/li>\n<\/ul>\n<p>At 32% efficiency the power generated is:<\/p>\n<ul>\n<li>[latex]P = (0.32)(12.5 \\text{ MW})[\/latex] or [latex]4.0 \\text{ MW}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>Not covered in this textbook is the <strong>mechanical advantage of pulleys<\/strong>. Using pulleys one can increase the load being raised by using the same power source &#8230; In this case, you can see the amount of rope (displacement) that is being used by smaller forces to do the same amount of work.\u00a0 This means that less powerful engines can be used to do the same amount of work, albeit at a slower rate.\u00a0 In the example shown below, each pulley adds to the displacement that the force acts through.\u00a0 In this case one only needs to consider the amount of energy lost to friction and such.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-968\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight.jpg\" alt=\"\" width=\"551\" height=\"258\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight.jpg 1854w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-300x141.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-1024x481.jpg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-768x360.jpg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-1536x721.jpg 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-65x31.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-225x106.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.2-weight-350x164.jpg 350w\" sizes=\"auto, (max-width: 551px) 100vw, 551px\" \/><\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 12.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>A world-class weight lifter can raise a 200 kg dumbbell from the floor to a height of 2.0 m in about 3.0 s.\u00a0 What is this average power output during the lift?<\/li>\n<li>What power is required to raise a mass of 47 kg a final height of 12 m in 15 s? (The mass comes to a complete stop when the final height is reached.)<\/li>\n<li>A mass of 100 kg (at rest) is raised to a final resting position 8.0 m above its initial position in 7.0 s. Find the work done on the object and the power required to accomplish it.<\/li>\n<li>What is the power of an air cannon that can fire a seed pellet (mass 9.0 g) from rest to a speed of 185 m\/s in 0.0065 s?<\/li>\n<li>What is the minimum power of an engine that can accelerate a 1600 kg pickup from 0 km\/h to 110 km\/h in 12 s?<\/li>\n<li>What is the power generated by 145 cubic metres of water falling through a potential difference of 392 m<sup>2<\/sup>\/s<sup>2<\/sup> each minute? (The mass of 1 cubic metre of water is 1 tonne.)\n<ol type=\"i\">\n<li>What is the expected power output that can be achieved from a 120 m high hydroelectric dam, where the rate of water moving through the generators is 43 m<sup>3<\/sup>\/s? (The mass of 1 cubic metre of water is 1000 kg or 1 tonne.)<\/li>\n<li>If the efficiency of this system is 45%, what is the power produced?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>12.3 Power &amp; Accelerated Motion<\/h1>\n<p>Equations Introduced or Used for this Section:<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li>[latex]a = \\dfrac{\\Delta v}{t}[\/latex]<\/li>\n<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\n<li>[latex]v_f = v_i + a t[\/latex]<\/li>\n<li>[latex]d = v_i t + \\frac{1}{2} a t^2[\/latex]<\/li>\n<li>[latex]F = m a[\/latex]<\/li>\n<li>Work [latex](W) = F \\cdot d[\/latex] or [latex]F d \\cos \u00f8[\/latex]<\/li>\n<li>[latex]E_p = m g h[\/latex]<\/li>\n<li>Power [latex](P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex]<\/li>\n<li>[latex]\\vec{a} = \\dfrac{\\vec{\\Delta v}}{t}[\/latex]<\/li>\n<li>[latex]v = \\dfrac{(v_i + v_f)}{2}[\/latex]<\/li>\n<li>[latex]2 a d = v_f^2 - v_i^2[\/latex]<\/li>\n<li>[latex]d = \\dfrac{(v+i v_f) t}{2}[\/latex]<\/li>\n<li>[latex]w = m g[\/latex]<\/li>\n<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\n<li>[latex]V[\/latex] or [latex]V_g = g h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<\/ul>\n<p>Power by definition when referring to the rate of work being done by a conservative force implies a change in mechanical energy, be it kinetic or potential (or both). When we restrict ourselves to only examining conservative forces, the system can be accelerating, changing vertical height at a constant velocity, or a more complicated combination of both kinetic and potential energies.<\/p>\n<p>In the following examples and in using conservative forces, power is expanded to explore both accelerating and constant velocity situations.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.3.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A motor is rated to deliver 8.4 kW.\u00a0 What is the constant speed that this motor can raise a mass of 2100 kg?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta \\text{ Energy}}{t}[\/latex] where [latex]\\Delta \\text{ Energy } = \\Delta E_p[\/latex] or [latex]m g \\Delta h[\/latex]<\/li>\n<li>[latex]P = \\dfrac{m g \\Delta h}{t}[\/latex] or [latex]P = m g v_{\\text{up}}[\/latex] or [latex]P = m g v_y[\/latex]<\/li>\n<li>[latex]8400 \\text{ W } = (2100 \\text{ kg})(9.8 \\text{ m\/s}^2)(v_y)[\/latex]<\/li>\n<li>[latex]v_y = 8400 \\text{ W } \\div [(2100 \\text{ kg})(9.8 \\text{ m\/s}^2)][\/latex]<\/li>\n<li>[latex]v_y = 0.41 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.3.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A jet car used in drag racing was timed to have taken 3.4 seconds to travel 400.0 m from a stationary start.\u00a0 If this car had a mass of 0.95 tonnes:<\/p>\n<ol type=\"i\">\n<li>Calculate the unbalanced force accelerating this car<\/li>\n<li>What power was developed?<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<p>Question (i): First we need to calculate the acceleration<\/p>\n<ul>\n<li>First find the acceleration:\n<ul>\n<li>[latex]d = v_i t + \\frac{1}{2} a t^2[\/latex]<\/li>\n<li>[latex]400 \\text{ m } = (0 \\text{ m\/s})(3.4 \\text{ s}) + \\frac{1}{2} (a)(3.4 \\text{ s})^2[\/latex]<\/li>\n<li>[latex]a = 69.2 \\text{ m\/s}^2[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>Second find the force:\n<ul>\n<li>[latex]F = m a[\/latex]<\/li>\n<li>[latex]F = (950 \\text{ kg})(69.2 \\text{ m\/s}^2)[\/latex]<\/li>\n<li>[latex]F = 65\\;700 \\text{ N } (\\approx 66\\;000 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Question (ii):<\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = \\dfrac{W}{t} = \\dfrac{\\Delta F d \\cos \u00f8}{t}[\/latex]<\/li>\n<li>[latex]P = \\dfrac{(65\\;700 \\text{ N})(400 \\text{ m})\\cos 0^{\\circ}}{3.4 \\text{ s}}[\/latex]<\/li>\n<li>[latex]P = 7.73 \\times 10^6 \\text{ W or } 7.73 \\text{ MW } (\\approx 7.7 \\text{ MW})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.3.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A machine can produce 120 W of power.\u00a0 If the machine is used for 12 s to raise a 50 kg mass straight upwards from a platform, where it is then released, at what speed would you expect this mass to impact the platform?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>This problem is simpler than it looks. Since the work using conservative forces is a change in potential energy and in the second part where it is dropped, the potential energy changes to kinetic energy, then one can just use the <strong>work-energy theorem<\/strong> to solve this. There is no need to find out the change in potential energy, which we would then use to find the change in potential energy.<\/p>\n<p>Therefore:<\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = \\dfrac{W}{t}[\/latex] or [latex]W = P t[\/latex]<\/li>\n<\/ul>\n<p>Where:<\/p>\n<ul>\n<li>[latex]W = \\Delta E_k[\/latex] or [latex]W = (E_{kr} - E_{ki})[\/latex]<\/li>\n<\/ul>\n<p>We are left with:<\/p>\n<ul>\n<li>[latex]P t = \\frac{1}{2} m v_f^2 - \\frac{1}{2} m v_i^2[\/latex]<\/li>\n<li>[latex](120 \\text{ W})(12 \\text{ s}) = \\frac{1}{2}(50 \\text{ kg}) v_f^2 - \\frac{1}{2}(50 \\text{ kg})(0 \\text{ m\/s})^2[\/latex]<\/li>\n<li>[latex]v_f^2 = 1440 \\text{ Ws } \\div 25 \\text{ kg}[\/latex]<\/li>\n<li>[latex]v_f= 7.6 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.3.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A 2.0 kW powered conveyer belt machine is angled at 28\u00b0 above the horizontal.\u00a0 At what constant speed can this belt be moving a 120 kg crate?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>For this problem, we need to use the definition of\u00a0 <strong>Power = Force \u00d7 Velocity cos <\/strong><strong>\u00f8<\/strong><strong>, <\/strong>where the angle \u00f8 in this problem is between the force and velocity, and is<strong> 90<\/strong><strong>\u00b0 <\/strong><strong>&#8211; 28<\/strong><strong>\u00b0 <\/strong><strong>or 62<\/strong><strong>\u00b0.<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-971 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example.png\" alt=\"\" width=\"289\" height=\"284\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example.png 559w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example-300x295.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example-65x64.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example-225x221.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.3.4-example-350x344.png 350w\" sizes=\"auto, (max-width: 289px) 100vw, 289px\" \/><\/p>\n<p>Therefore:<\/p>\n<ul>\n<li>[latex]\\text{Power } (P) = F \\cdot v \\cos \u00f8[\/latex] where [latex]F = \\text{ weight of } 120 \\text{ kg}[\/latex]<\/li>\n<li>[latex]2000 \\text{ W } = (120 \\text{ kg})(9.8 \\text{ m\/s}^2)(v)\\cos 62^{\\circ}[\/latex]<\/li>\n<li>[latex]v = 2000 \\text{ W } \\div 522 \\text{ J}[\/latex]<\/li>\n<li>[latex]v = 3.6 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 12.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>A motor is rated to deliver 10 kW.\u00a0 What is the <strong>constant speed<\/strong> that this motor can raise a mass of [latex]2.75 \\times 10^4[\/latex] kg?<\/li>\n<li>What is the maximum possible constant speed that a 120 hp engine could lift a 4.00 ton (metric) mass straight up?<\/li>\n<li>What power is needed to run an elevator (maximum mass of 3500 kg, elevator and passengers) that rises at an average speed of 0.4 m\/s?<\/li>\n<li>A cement pump can lift concrete at the rate of 3500 kg up 20 m every minute.\n<ol type=\"i\">\n<li>What would be the minimum power of this pump?<\/li>\n<li>If this pump is 25% efficient, what minimum power must it be?<\/li>\n<\/ol>\n<\/li>\n<li>A monster truck used in drag racing was timed to have taken 15.0 seconds to travel 400.0 m from a stationary start. If this truck had a mass of 2.40 tonnes:\n<ol type=\"i\">\n<li>Calculate the unbalanced force accelerating this truck.<\/li>\n<li>What power was developed?<\/li>\n<\/ol>\n<\/li>\n<li>An engine can produce 25<span style=\"margin-left: 0.25em;\">000<\/span> W of power. If we assume that the conversion of engine power to mechanical energy of this 1000 kg car is 100% efficient, at what speed could this car be traveling after 15 s of full power, starting from rest?<\/li>\n<li>A machine can produce 40 W of power. If the machine is used for 15 s to raise a 20 kg mass straight upwards where it is released, at what speed would you expect this mass to impact its starting position?<\/li>\n<li>If a loaded semi (total mass = 20<span style=\"margin-left: 0.25em;\">000<\/span> kg) is able to maintain a constant speed of 20 km\/h up an 8\u00b0 inclined road, what is the power output of the truck\u2019s engine?<\/li>\n<li>What is the angle of a hill that a loaded pickup truck (m = 3600 kg) can climb at a constant speed of 45 km\/h, if the 300 hp engine is 32% efficient?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 12.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li><strong>It is claimed that large trees <\/strong>can evaporate as much as 1000 kg of water per day.\u00a0 Assuming the average rise of water to be 10 m from the ground, what amount of energy must be supplied for this?\u00a0 If we assume the source of this energy is from the Sun, what average amount of power must the tree receive over 12 hours of sunlight?<\/li>\n<li><strong>An average human at rest<\/strong> radiates 75 W of power.\u00a0 What amount of energy is radiated in eight hours?<\/li>\n<li><strong>Power Output of the Sun&#8230;<\/strong>\u00a0 The amount of solar energy striking the Earth is around 1350 watts per metre2.\u00a0 Find the power output of our Sun. <strong>Under Business as Usual projections<\/strong> for global energy consumption, Energy usage is expected to rise from 589 EJ in 2016 to 1070 EJ in 2100 (E stands for [latex]exo \\times 10^21[\/latex])<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox\">Video to Watch: <a href=\"https:\/\/www.youtube.com\/watch?v=GSVv40M2aks\">NASA &#8211; SDO Year 5<\/a><\/div>\n<h1>12.4 The Wind Resource<\/h1>\n<div class=\"textbox\">\n<ul>\n<li>Article to Read: <a href=\"https:\/\/www.ucsusa.org\/resources\/how-wind-energy-works?print=t\">How Wind Energy Works<\/a><\/li>\n<li>Research News:<a href=\"https:\/\/www.sciencedaily.com\/releases\/2019\/08\/190814081202.htm\"> Europe has the untapped onshore capacity to meet global energy demand<\/a><\/li>\n<\/ul>\n<\/div>\n<p>Several factors can affect wind speed and the ability of a turbine to generate more power.\u00a0 For example, wind speed increases as the height from the ground increases.\u00a0 If wind speed at 10 meters off the ground is 6 m\/s it will be about 7.5 m\/s at a height of 50 meters.\u00a0 A 2003 Stanford University study examined wind speeds at higher elevations and found that as much as one quarter of the United States, including areas historically thought to have poor wind potential, is potentially suitable for providing affordable electric power from wind.\u00a0 The rotors of the newest wind turbines can now reach heights up to 70 meters.\u00a0 In addition to height, the power in the wind varies with temperature and altitude, both of which affect the air density.\u00a0 Winter winds in Minnesota State will carry more power than summer winds of the same speed high in the passes of Southern States.<\/p>\n<p>The more the wind blows, the more power will be produced, with the main drawback being inconsistent wind.\u00a0 The term used to describe this is &#8220;capacity factor&#8221;, which is simply the amount of power a turbine actually produces over a period of time divided by the amount of power it could have produced if it had run at its full-rated capacity over that time period.<\/p>\n<p>A more precise measurement of output is the &#8220;specific yield&#8221;. This measures the annual energy output per square meter of area swept by the turbine blades as they rotate.\u00a0 Overall, wind turbines capture between 20 and 40 percent of the energy in the wind.\u00a0 So, at a site with average wind speeds of 7 m\/s, a typical turbine will produce about 1100 kilowatt-hours (kWh) per square meter of area per year.\u00a0 If the turbine has blades that are 40 meters long, for a total swept area of 5029 square meters, the power output will be about 5.5 million kWh for the year.\u00a0 An increase in blade length, which in turn increases the swept area, can have a significant effect on the amount of power output from a wind turbine.<\/p>\n<p>Another factor in the cost of wind power is the distance of the turbines from transmission lines.\u00a0 Although some large windy areas, particularly in rural parts of the High Plains and Rocky Mountains, are too far from power lines to be cost-effective today, the potential for these regions is still enormous.<\/p>\n<p>A final consideration for a wind resource is the seasonal and daily variation in wind speed.\u00a0 If the wind blows during periods of peak power demand, power from a wind farm will be valued more highly than if it blows in off-peak periods.\u00a0 In California, for example, high temperatures in the central valley and low coastal temperatures near San Francisco cause powerful winds to blow across the Altamont in the summer, a period of high power demand.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.4.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-972 aligncenter\" style=\"font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.1-example.jpg\" alt=\"\" width=\"231\" height=\"357\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.1-example.jpg 350w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.1-example-194x300.jpg 194w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.1-example-65x101.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.1-example-225x348.jpg 225w\" sizes=\"auto, (max-width: 231px) 100vw, 231px\" \/>What is the kinetic energy or the air mass that passes through an offshore windmill at the Thorntonbank Wind Farm<a class=\"footnote\" title=\"Thorntonbank Wind Farm\" id=\"return-footnote-169-4\" href=\"#footnote-169-4\" aria-label=\"Footnote 4\"><sup class=\"footnote\">[4]<\/sup><\/a>, having a rotor diameter of 126 m, if the wind speed averages 30 km\/hr?\u00a0 If we assume that current technology yields 45% power generation from this moving air mass, what power is produced per turbine?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Kinetic Energy:<\/p>\n<ul>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2}(1.225 \\text{ kg\/m}^3)(8.33 \\text{ m\/s})^2[\/latex]<\/li>\n<li>[latex]E_k = 42.5 \\text{ J\/m}^3[\/latex]<\/li>\n<\/ul>\n<p>Power Produced (100% Efficient):<\/p>\n<ul>\n<li>[latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/li>\n<li>[latex]P = \\frac{1}{2} (1.225 \\text{ kg\/m}^3)(\\pi)(126 \\text{ m})^2(8.33 \\text{ m\/s})^3[\/latex]<\/li>\n<li>[latex]P = 1.7 \\times 10^7 \\text{ W or } 17.7 \\text{ MW } (\\approx 18 \\text{ MW})[\/latex]<\/li>\n<\/ul>\n<p>Now, for 45% efficiency:<\/p>\n<ul>\n<li>[latex]P = (0.45) 1.77 \\times 10^7 \\text{W}[\/latex]<\/li>\n<li>[latex]P = 7.95 \\times 10^6[\/latex] or [latex]7.95 \\text{ MW } (\\approx 8.0 \\text{ MW})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h2>Wind Turbine Power:<\/h2>\n<p>Equations Introduced or Used for this Section:<\/p>\n<p style=\"text-align: center;\">[latex]P = \\frac{1}{2} \\varrho A C_P v^2 N_g N_b[\/latex]<\/p>\n<p>Where&#8230;<\/p>\n<ul>\n<li>[latex]P[\/latex] = power in watts<\/li>\n<li>[latex]\\varrho[\/latex] = air density (about 1.225 kg\/m<sup>3<\/sup> at sea level at15\u00b0C)<\/li>\n<li>[latex]A[\/latex] = rotor swept area exposed to the wind (m2)<\/li>\n<li>[latex]C_p[\/latex] = coefficient of performance (0.59 {Betz limit} is the maximum theoretically possible 0.35 for a good design)<\/li>\n<li>[latex]v[\/latex] = wind speed in m\/s<\/li>\n<li>[latex]N_g[\/latex]\u00a0= generator efficiency (50% for car alternator, 80% or possibly more for a permanent magnet generator or grid-connected induction generator)<\/li>\n<li>[latex]N_b[\/latex]\u00a0= gearbox\/bearings efficiency (could be as high as 95% if good)<\/li>\n<\/ul>\n<table style=\"border-collapse: collapse; width: 100%; height: 265px;\">\n<tbody>\n<tr style=\"height: 31px;\">\n<th style=\"width: 50%; height: 31px;\" scope=\"col\">Temperature (\u00b0C)<\/th>\n<th style=\"width: 50%; height: 31px;\" scope=\"col\">Density of Air (kg\/m<sup>3<\/sup>)<\/th>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">\u221225<\/td>\n<td style=\"width: 50%; height: 18px;\">1.423<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">\u221220<\/td>\n<td style=\"width: 50%; height: 18px;\">1.395<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">\u221215<\/td>\n<td style=\"width: 50%; height: 18px;\">1.368<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">\u221210<\/td>\n<td style=\"width: 50%; height: 18px;\">1.342<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">\u22125<\/td>\n<td style=\"width: 50%; height: 18px;\">1.317<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">0<\/td>\n<td style=\"width: 50%; height: 18px;\">1.292<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">5<\/td>\n<td style=\"width: 50%; height: 18px;\">1.269<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">10<\/td>\n<td style=\"width: 50%; height: 18px;\">1.247<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">15<\/td>\n<td style=\"width: 50%; height: 18px;\">1.225<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">20<\/td>\n<td style=\"width: 50%; height: 18px;\">1.204<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">25<\/td>\n<td style=\"width: 50%; height: 18px;\">1.184<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">30<\/td>\n<td style=\"width: 50%; height: 18px;\">1.164<\/td>\n<\/tr>\n<tr style=\"height: 18px;\">\n<td style=\"width: 50%; height: 18px;\">35<\/td>\n<td style=\"width: 50%; height: 18px;\">1.146<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 12.4.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Using the graph and the chart on the previous page and assuming 100% efficiency\n<ol type=\"i\">\n<li>What is the area of a 1.0 m circle?<\/li>\n<li>If the density of air at 20\u00b0 C is 1.204 kg\/m<sup>3<\/sup> at sea level, what must be the minimum speed of the air for the 1.0 m diameter rotors to extract 0.4 kW?<\/li>\n<li>Using an air temperature of 20\u00b0C, calculate the minimum number of 11 m diameter rotors needed to generate 20 MW?<\/li>\n<li>What must be the speed of the air at \u221215\u00b0C to create 0.40 kW?<\/li>\n<li>Assuming the 10 m rotor produces 39.3 kW at an air temperature of 20\u00b0C.\u00a0 What is the percent increase in power output if the temperature drops to -20\u00b0C?<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-974\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1.jpeg\" alt=\"\" width=\"423\" height=\"273\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1.jpeg 1950w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-300x194.jpeg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-1024x661.jpeg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-768x495.jpeg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-1536x991.jpeg 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-65x42.jpeg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-225x145.jpeg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/12.4.4-omafra-electricity-generation-small-wind-turbines-home-farm-use-figure-7-en-1950x1258-2018-02-13-v1-350x226.jpeg 350w\" sizes=\"auto, (max-width: 423px) 100vw, 423px\" \/><\/li>\n<\/ol>\n<\/li>\n<li>The above graph shows the potential amount of energy from wind at 20\u00b0 C.\n<ol type=\"i\">\n<li>What would this curve look like if the air is hotter?<\/li>\n<li>\u00a0What would this curve look like if the air is colder?<\/li>\n<li>Notice the power at 8.0 m\/s is around 300 W and at 10 m\/s around 600 W.\u00a0 Show how these numbers fit the equation [latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/li>\n<li>How big would the rotor area of the windmill graphed above be if we assume it is 100% efficient?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>(Realistically, [latex]C_p N_g N_b[\/latex] = 0.27 or is around 27%)<\/p>\n<\/div>\n<\/div>\n<h1>12.5 Power from Ocean and River Currents<\/h1>\n<div class=\"textbox\">Research News: <a href=\"https:\/\/www.sciencedirect.com\/science\/article\/pii\/S2590174519300145\">Hunt, J. D., et al. (2019) Using the jet stream for sustainable airship and balloon transportation of cargo and hydrogen<\/a><\/div>\n<p>Ocean current speeds are generally lower than wind speeds.\u00a0 This is important because the kinetic energy contained in flowing bodies is proportional to the cube of their velocity.\u00a0 However another more important factor in the power available for extraction from a flowing body is the density of the material.\u00a0 Water is about 835 times denser than wind, so for the same area of flow being intercepted the energy contained in a 19 km\/h water flow is equivalent to that contained in an air mass moving at about 180 km\/h.\u00a0 Thus, ocean currents represent a potentially significant currently untapped reservoir of energy.<\/p>\n<p>The total worldwide power in ocean currents has been estimated to be about 5000 GW with power densities of up to 15 kW\/m2.\u00a0 The relatively constant extractable energy density near the surface of the Florida Straits Current is about 1 kW\/m2 of flow area.\u00a0 It has been estimated that capturing just 1\/1000th of the available energy from the Gulf Stream, which has 21 000 times more energy than Niagara Falls in a flow of water that is 50 times the total flow of all the world\u2019s freshwater river, would supply Florida with 35% of its electrical needs.<\/p>\n<p>Countries that are interested in and pursuing the application of ocean current energy technologies include the European Union, Japan and China.<\/p>\n<p>Equations Introduced or Used for this Section:<\/p>\n<p style=\"text-align: center;\">[latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/p>\n<p>Where&#8230;<\/p>\n<ul>\n<li>[latex]P[\/latex] is power in watts<\/li>\n<li>[latex]\\varrho[\/latex] is density of water (fresh water is roughly 1000 kg\/m<sup>3<\/sup>).<\/li>\n<li>[latex]A[\/latex] is rotor swept area exposed to the water (m<sup>2<\/sup>)<\/li>\n<li>[latex]v[\/latex] is water speed (m\/s)<\/li>\n<\/ul>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.5.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>The Amazon River can reach an average current speed of 7 km\/h at the height of the wet season.\u00a0 What is the expected power that can be produced by an underwater turbine of diameter 2.0 m, assuming an efficiency of 42% for the power generating turbine? (Density of fresh water is around 1000 kg\/m<sup>3<\/sup>)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-976 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1.png\" alt=\"\" width=\"311\" height=\"311\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1.png 1200w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-300x300.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-1024x1024.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-150x150.png 150w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-768x768.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-65x65.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-225x225.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/example-12.5.1-350x350.png 350w\" sizes=\"auto, (max-width: 311px) 100vw, 311px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]P = \\frac{1}{2} \\varrho A v^3[\/latex]<\/li>\n<li>[latex]P = \\frac{1}{2}(1000 \\text{kg\/m}^3) \\pi (2.0 \\text{ m})^2 (1.94 \\text{ m\/s})^3[\/latex]<\/li>\n<li>[latex]P = 12\\;200 \\text{ W or } 12.2 \\text{ KW } (\\approx 12 \\text{ KW})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 12.5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>The Fraser River has an average current speed ranging from 0.25 m\/s to 1.25 m\/s. What is the expected range of power that can be produced by an underwater turbine of diameter 1.5 m?\u00a0 (Density of fresh water is around 1000 kg\/m<sup>3<\/sup>)<\/li>\n<li>One of the strongest ocean currents is the Gulf Stream, which moves at speeds greater than 1.5 m\/s. What is the maximum power that can be extracted by a 3.0 m diameter underwater turbine in this current?\u00a0 How many of these turbines would be needed to replace a 50 MW coal-powered electric generator? (Density of ocean water is around 1003 kg\/m<sup>3<\/sup>)<\/li>\n<li>What maximum power could be extracted from a 1.0 m diameter turbine in water moving at 20 km\/h? (This is the speed of water falling 1.57 m.) How does this compare to what a similar sized turbine could extract from a Category 2 Hurricane where the wind (20\u00b0 C) is moving at 160 km\/h?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox\">Article to Read: <a href=\"https:\/\/oilprice.com\/Latest-Energy-News\/World-News\/Scientists-Develop-Wave-Energy-Device-That-Generates-Low-Cost-Electricity.html\">Scientists Develop Wave Energy Device That Generates Low-Cost Electricity<\/a><\/div>\n<h1>Exercise Answers<\/h1>\n<h2>12.1 Power<\/h2>\n<ol class=\"twocolumn\">\n<li>10<span style=\"margin-left: 0.25em;\">000<\/span> W<\/li>\n<li>5.4 MJ<\/li>\n<li>30 s<\/li>\n<li>89<span style=\"margin-left: 0.25em;\">500<\/span> W or 89.5 kW<\/li>\n<\/ol>\n<h2>12.2 Power &amp; Change in Energy<\/h2>\n<ol class=\"twocolumn\">\n<li>1300 W<\/li>\n<li>368 W<\/li>\n<li>7840 J, 1120 W<\/li>\n<li>23<span style=\"margin-left: 0.25em;\">700<\/span> W<\/li>\n<li>62<span style=\"margin-left: 0.25em;\">000<\/span> W<\/li>\n<li>0.947 MW<\/li>\n<li>\n<ol type=\"i\">\n<li>50.5 MW<\/li>\n<li>22.8 MW<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2>12.3 Power &amp; Accelerated Motion<\/h2>\n<ol class=\"twocolumn\">\n<li>0.0371 m\/s<\/li>\n<li>2.28 m\/s<\/li>\n<li>13<span style=\"margin-left: 0.25em;\">700<\/span> W or 13.7 kW\n<ol type=\"i\">\n<li>11<span style=\"margin-left: 0.25em;\">400<\/span> W<\/li>\n<li>45<span style=\"margin-left: 0.25em;\">700<\/span> W<\/li>\n<\/ol>\n<ol type=\"i\">\n<li>8500 N<\/li>\n<li>[latex]2.28 \\times 10^5[\/latex] W = 228 kW<\/li>\n<\/ol>\n<\/li>\n<li>27.4 m\/s<\/li>\n<li>7.7 m\/s<\/li>\n<li>152<span style=\"margin-left: 0.25em;\">000<\/span> W = 152 kW<\/li>\n<li>9.3\u00b0<\/li>\n<\/ol>\n<h2>12.4 Large Tree Evaporation, Human\u2019s Radiated Heat, Power Output of the Sun<\/h2>\n<ol class=\"twocolumn\">\n<li>\n<ol type=\"i\">\n<li>98<span style=\"margin-left: 0.25em;\">000<\/span> J<\/li>\n<li>2.27 W<\/li>\n<\/ol>\n<\/li>\n<li>[latex]2.2 \\times 10^6[\/latex] J<\/li>\n<li>[latex]3.8 \\times 10^{26}[\/latex] W<\/li>\n<\/ol>\n<h2>12.4.1 The Wind Resource<\/h2>\n<ol>\n<li>\n<ol class=\"twocolumn\" type=\"i\">\n<li>0.785 m<sup>2<\/sup><\/li>\n<li>9.46 m\/s<\/li>\n<li>421 Rotors<\/li>\n<li>9.18 m\/s<\/li>\n<li>45.4 kW<\/li>\n<\/ol>\n<ol type=\"i\">\n<li>The curve should flatten out due to decreased air density<\/li>\n<li>The curve should become steeper climb due to increased air density<\/li>\n<li>The ratio of 600 to 300 watts is roughly the same as the ratio of (10 m\/s)<sup>3<\/sup> to (8 m\/s)<sup>3<\/sup>. Alternatively, on can look at the ratio of [latex]\\dfrac{P}{v^3}[\/latex] which equals 0.6 for both speeds.<\/li>\n<li>0.997 m<sup>2<\/sup> or 1 m<sup>2<\/sup><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2>12.5 Power from Ocean and River Currents<\/h2>\n<ol>\n<li>13.8 W at 0.25 m\/s to 1.73 kW at 1.25 m\/s<\/li>\n<li>120 kW maximum power, approximately 4200 turbines to replace 50 MW<\/li>\n<li>67 kW for water compared to 41 kW from a hurricane<\/li>\n<\/ol>\n<h3>Media Attributions<\/h3>\n<ul>\n<li>&#8220;The Horse Mill&#8221; by John F. Herring (1795 &#8211; 1865) is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Horse_Gin_in_use.jpg\">A Horse Gin In Use<\/a>&#8221; by Roger Griffith is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/en.wikipedia.org\/wiki\/File:Four_pulleys.svg\">Pulley diagram with 4 pulleys<\/a>&#8221; by <a title=\"User:Prolineserver\" href=\"https:\/\/commons.wikimedia.org\/wiki\/User:Prolineserver\">Prolineserver<\/a> and <a title=\"User:Tomia\" href=\"https:\/\/commons.wikimedia.org\/wiki\/User:Tomia\">Tomia<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/3.0\/deed.en\">CC BY-SA 3.0 Unported<\/a> licence.<\/li>\n<li>&#8220;<a href=\"https:\/\/en.wikipedia.org\/wiki\/File:Amazonrivermap.svg\">Amazonrivermap<\/a>&#8221; by <a class=\"external text\" href=\"https:\/\/en.wikipedia.org\/wiki\/User:Kmusser?rdfrom=commons:User:Kmusser\">Kmusser<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/3.0\/deed.en\">CC BY-SA 3.0 Unported<\/a> licence.<\/li>\n<\/ul>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-169-1\">Watt\u2019s improvements on the steam engine efficiency are credited as the start of the expansion of the industrial revolution. The unit of watts to measure power was named in honor of James Watt for his steam engine improvements in 1889. <a href=\"#return-footnote-169-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-169-2\">Wikipedia has an extensive list of the various form and measures of horsepower: <a href=\"https:\/\/en.wikipedia.org\/wiki\/Horsepower\">https:\/\/en.wikipedia.org\/wiki\/Horsepower<\/a> <a href=\"#return-footnote-169-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-169-3\"><a href=\"http:\/\/nyfalls.com\/niagara-falls\/faq5\/\">Niagara Falls FAQ<\/a> <a href=\"#return-footnote-169-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><li id=\"footnote-169-4\"><a href=\"https:\/\/en.wikipedia.org\/wiki\/Thorntonbank_Wind_Farm\">Thorntonbank Wind Farm<\/a> <a href=\"#return-footnote-169-4\" class=\"return-footnote\" aria-label=\"Return to footnote 4\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":125,"menu_order":12,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-169","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/169","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/users\/125"}],"version-history":[{"count":25,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/169\/revisions"}],"predecessor-version":[{"id":1244,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/169\/revisions\/1244"}],"part":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/169\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/media?parent=169"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=169"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/contributor?post=169"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/license?post=169"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}