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class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> AHL Fluid dynamics <a href="#" class="mark-page-favorite pull-right" data-pid="1249" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../physics.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../849/engineering.html">Engineering</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">AHL Fluid dynamics</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 60 minutes"><i class="fa fa-clock-o"></i> 60'</span> </ol> <article id="main-article"> <p><img alt="" src="../../ahl-engineering/rapids-355737_640.jpg" style="float: left; width: 250px; height: 180px;">Fluid dynamics is the study of moving fluids. As with moving bodies, mass and energy are conserved leading to equations that can be used to determine unknown quantities. An understanding of fluid properties enables us to calculate drag forces and the categorization of fluid flow.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Streamlines</p> </div> </div> <div class="panel-body"> <div> <p>Flowlines show the movement of particles in a fluid.</p> <p>Streamlines follow the velocity vectors of particles in a fluid. As with all vectors, streamlines will never intersect as they show the resultant velocity. The closer the streamlines, the higher the velocity.</p> <p>For steady flow, flowlines and streamlines will be indistinguishable.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388880033"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Continuity</p> </div> </div> <div class="panel-body"> <div> <h4>Conservation of mass</h4> <p>Mass is conserved in fluid flow. This means that mass flowrate is constant: the mass of fluid arriving at any point must equal the mass leaving the point in any given period of time. This is a statement of continuity. Mass flowrate is the mass passing a point per unit time.</p> <h4>Continuity equation</h4> <p>Since density is constant for incompressible (ideal) fluids, if mass flowrate is constant, volumetric flowrate must be constant:</p> <p style="text-align: center;"><span class="math-tex">\(Av=\text{constant}\)</span></p> <ul> <li><span class="math-tex">\(A\)</span> is cross-sectional area of the container (m<sup>2</sup>)</li> <li><span class="math-tex">\(v\)</span> is velocity of the fluid (ms<sup>-1</sup>)</li> </ul> <p>The velocity of a fluid increases as the radius of a pipe decreases.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388879865"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Bernoulli's principle</p> </div> </div> <div class="panel-body"> <div> <p>Mass is not the only conserved quantity in a moving fluid; energy is also conserved. In steady flow, the sum of kinetic energy, potential energy and internal energy remains constant. One consequence is Bernoulli's principle.</p> <p>Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. The resulting equation is:</p> <p style="text-align: center;"><span class="math-tex">\({1\over 2}\rho v^2+\rho gz+p=\text{constant}\)</span></p> <ul> <li><span class="math-tex">\(\rho\)</span> is fluid denisty (kg m<sup>-3</sup>)</li> <li><span class="math-tex">\(v\)</span> is velocity (ms<sup>-1</sup>)</li> <li><span class="math-tex">\(g\)</span> is gravitational field strength (on Earth, 9.81 N kg<sup>-1</sup>)</li> <li><span class="math-tex">\(z\)</span> is height (m)</li> <li><span class="math-tex">\(p\)</span> is pressure (Pa)</li> </ul> <p>There are many applications of the Bernoulli equation including flow out of a container, determining the speed of a plane and venturi tubes.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388879824"></iframe></div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388883016"></iframe></div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388879853"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Measuring velocity</p> </div> </div> <div class="panel-body"> <div> <h4>Pitot tubes</h4> <p>The speed of a plane can be measured using a pitot tube. As the plane flies, there is relative motion between the plane and the air.</p> <p style="text-align: center;"><img alt="" src="../../ahl-engineering/airspeed_p1230157.jpg" style="width: 250px; height: 237px;"></p> <p>A tube points directly forwards to collect the air and is connected to a manometer to measure the dynamic pressure, the pressure caused by the relative motion of the plane and the air:</p> <p style="text-align: center;"><span class="math-tex">\(\Delta p=\rho_f g\Delta h\)</span></p> <ul> <li><span class="math-tex">\(\Delta p\)</span> is the dynamic pressure (Pa)</li> <li><span class="math-tex">\(\rho_f\)</span> is the density of the fluid in the manometer tube (kg m<sup>-3</sup>)</li> <li><span class="math-tex">\(g\)</span> is gravitational field strength (on Earth, 9.81 N kg<sup>-1</sup>)</li> <li><span class="math-tex">\(\Delta h\)</span> is the height difference in the U-tube (m)</li> </ul> <p style="text-align: center;"><img alt="" src="../../ahl-engineering/pitot_tube_manometer.svg" style="width: 188px; height: 150px;"></p> <h6 style="text-align: center;">By Cmglee - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=74391265</h6> <p>If we assume that the air is incompressible, Bernoulli's equation applies. The pitot tube is horizontal, so there is no height difference and the air is stopped as it reaches the manometer:</p> <p style="text-align: center;"><span class="math-tex">\({1\over 2}\rho_a u^2+\Delta p=0\)</span></p> <p style="text-align: center;"><span class="math-tex">\(\Rightarrow {1\over 2}\rho_a u^2=-\Delta p\)</span></p> <ul> <li><span class="math-tex">\(\rho_a\)</span> is the density of the air (kg m<sup>-3</sup>)</li> <li><span class="math-tex">\(u\)</span> is the relative velocity of the air and the plane (ms<sup>-1</sup>)</li> <li><span class="math-tex">\(\Delta p\)</span> is the dynamic pressure (Pa)</li> </ul> <h4>Venturi meter</h4> <p>A Venturi meter is a device for measuring the pressure of a fluid. A constriction is made in a pipe to reduce cross-sectional area, which, according to continuity, increases velocity:</p> <p style="text-align: center;"><span class="math-tex">\(A_1v_1=A_2v_2\)</span></p> <ul> <li><span class="math-tex">\(A\)</span> is cross-sectional area (m<sup>2</sup>)</li> <li><span class="math-tex">\(v\)</span> is velocity (ms<sup>-1</sup>)</li> </ul> <p style="text-align: center;"><img alt="" src="../../ahl-engineering/venturi5.svg" style="width: 400px; height: 225px;"></p> <p>If the pipe is horizontal, the pressure must decrease to satisfy Bernoulli's principle. Applying Bernoulli's equation:</p> <p style="text-align: center;"><span class="math-tex">\({1\over 2}\rho {v_1}^2+p_1={1\over 2}\rho {v_2}^2+p_2\)</span></p> <ul> <li><span class="math-tex">\(\rho\)</span> is fluid density (kg m<sup>-3</sup>)</li> <li><span class="math-tex">\(v\)</span> is velocity (ms<sup>-1</sup>)</li> <li><span class="math-tex">\(p\)</span> is pressure (Pa)</li> </ul> <p>Vertical 'Venturi' tubes above the initial section and the constriction are open to the atmosphere. The pressure difference can be determined from the difference in the heights the the fluid reaches:</p> <p style="text-align: center;"><span class="math-tex">\(p_1-p_2=\rho g \Delta h\)</span></p> <ul> <li><span class="math-tex">\(p\)</span> is pressure (Pa)</li> <li><span class="math-tex">\(\rho\)</span> is fluid denisty (kg m<sup>-3</sup>)</li> <li><span class="math-tex">\(g\)</span> is gravitational field strength (on Earth, 9.81 N kg<sup>-1</sup>)</li> <li><span class="math-tex">\(\Delta h\)</span> is the height difference in the Venturi tubes (m)</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388880050"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Stokes' law</p> </div> </div> <div class="panel-body"> <div> <p>The drag force on a spherical object in a fluid can be calculated using Stokes’ law:</p> <p style="text-align: center;"><span class="math-tex">\(F_D=6\pi \eta rv\)</span></p> <ul> <li><span class="math-tex">\(F_D\)</span> is drag force (N)</li> <li><span class="math-tex">\(\eta\)</span> is viscosity (Pa s)</li> <li><span class="math-tex">\(r\)</span> is the radius of the sphere (m)</li> <li><span class="math-tex">\(v\)</span> is velocity (ms<sup>-1</sup>)</li> </ul> <p style="text-align: center;"><img alt="" src="../../ahl-engineering/stokes_sphere.svg" style="width: 200px; height: 288px;"></p> <h6 style="text-align: center;">By Kraaiennest - self-madeBased on: G.K. Batchelor (1967) An introduction to fluid dynamics, Cambridge University Press. Pages 230–235., CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=4115921</h6> <p>This equation applies only for low Reynolds numbers when flow is laminar. Note that the drag force increases with velocity.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388880026"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Viscosity</p> </div> </div> <div class="panel-body"> <div> <p>Viscosity of a fluid is its resistance to deformation. For example, treacle has a higher viscosity than water.</p> <p style="text-align: center;"><img alt="" src="../../ahl-engineering/flower-honey-1911749_640.jpg" style="width: 250px; height: 129px;"></p> <p>It emerges from the friction between adjacent layers of fluid in relative motion, such as when fluid flows faster in the centre of a pipe than at the walls. The higher the viscosity, the higher the pressure difference required to sustain fluid flow throughout a pipe.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/388883603"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Flow</p> </div> </div> <div class="panel-body"> <div> <p>Fluid flow can be categorised as laminar (a) or turbulent (b).</p> <p style="text-align: center;"><img alt="" src="../../ahl-engineering/laminar_and_turbulent_flows.svg" style="width: 400px; height: 245px;"></p> <h4>Laminar</h4> <p>In laminar flow, particles in the fluid follow smooth paths in layers with no mixing. Motion is orderly with particles close to the walls of a pipe moving in straight lines parallel to the walls. Laminar flow occurs at low velocities in fluids of high viscosity, below a Reynolds number threshold at which the flow becomes turbulent. Stokes' law applies and drag forces are low.</p> <h4>Turbulent</h4> <p>In turbulent flow, fluid particles experience changes in pressure and flow velocity due to excessive kinetic energy in a fluid of low viscosity. Vortices and eddy currents develop causing mixing of the layers of fluid. The Reynolds number exceeds the threshold for laminar flow and Stokes' law does not apply.</p> <p>Examples include storm clouds, smoke from a chimney and fast-flowing rivers. An advantage of turbulent flow is a reduction of lift, as required in Formula 1 racing.</p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Reynolds number</p> </div> </div> <div class="panel-body"> <div> <p>Laminar and turbulent flow can be distinguished by observation or by calculation. The latter requires determination of the Reynolds number, a dimensionless value for fluid flow:</p> <p style="text-align: center;"><span class="math-tex">\(R={vr\rho \over \eta}\)</span></p> <ul> <li><span class="math-tex">\(R\)</span> is Reynolds number (dimensionless)</li> <li><span class="math-tex">\(v\)</span> is velocity (ms<sup>-1</sup>)</li> <li><span class="math-tex">\(r\)</span> is the radius of the pipe (m)</li> <li><span class="math-tex">\(\rho\)</span> is fluid density (kg m<sup>-3</sup>)</li> <li><span class="math-tex">\(\eta\)</span> is viscosity (Pa s)</li> </ul> <p>Laminar flow conditions prevail for <span class="math-tex">\(R<1000\)</span>.</p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use flashcards to practise your recall.</em></p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="3" data-mode="Normal" data-topics="1033" data-subject-id="6" data-n-flashcards="16" style="text-align:center">Show flashcards</a></div><hr> <p><em>Use quizzes to practise application of theory.</em></p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#54523790"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="54523790"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> AHL fluid dynamics <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-384-1249" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>This image represents an ideal fluid flowing in a pipe:</p><p style="text-align: center;"><img alt="" src="https://www.thinkib.net/files/physics/activities/fluids3.png" style="width: 485px; height: 166px;"></p><p>Calculate <span class="math-tex">\(v_1\over v_2\)</span> if <span class="math-tex">\({A_1\over A_2} = 4\)</span>.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>16</span></label> </p><p><label class="radio"> <input type="radio"> <span>2</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1/4</span></label> </p><p><label class="radio"> <input type="radio"> <span>4</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(Av=\text{ constant}\)</span></p><p><span class="math-tex">\({v_1\over v_2}={A_2\over A_1}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This image represents an ideal fluid flowing in a pipe:</p><p style="text-align: center;"><img alt="" src="https://www.thinkib.net/files/physics/activities/fluids3.png" style="width: 485px; height: 166px;"></p>If 4 m<sup>3</sup> flows through the wide section per second, calculate the volume flowing through the narrow section per second.</div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2 m<sup>3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>16 m<sup>3</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>4 m<sup>3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>8 m<sup>3</sup></span></label> </p></div><div class="q-explanation"><p>Continuity tells us that mass flowrate is constant. This fluid is incompressible (ideal) so volumetric flowrate is constant.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This image represents water flowing in a pipe:</p><p style="text-align: center;"><img alt="" src="https://www.thinkib.net/files/physics/activities/fluids3.png" style="width: 485px; height: 166px;"></p><p>If <span class="math-tex">\(v_1\)</span> is 1 ms<sup>-1</sup> and <span class="math-tex">\(v_2\)</span> is 4 ms<sup>-1</sup>, calculate the difference in pressure.</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>7.5 kPa</span></label> </p><p><label class="radio"> <input type="radio"> <span>15 Pa</span></label> </p><p><label class="radio"> <input type="radio"> <span>15 kPa</span></label> </p><p><label class="radio"> <input type="radio"> <span>7.5 Pa</span></label> </p></div><div class="q-explanation"><p>Using Bernoulli's equation for horizontal fluid flow:</p><p><span class="math-tex">\(p_1 + {1\over 2}ρ{v_1}^2 = p_2 + {1\over2}ρ{v_2}^2\)</span></p><p><span class="math-tex">\(p_1 - p_2 = {1\over2} ρ{v_2}^2 - {1\over 2}ρ{v_1}^2=500(4^2-1^2)=7500\text{ Pa}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This image represents water flowing in a pipe:</p><p style="text-align: center;"><img alt="" src="https://www.thinkib.net/files/physics/activities/fluids3.png" style="width: 485px; height: 166px;"></p>If the pressure difference between the two sides is 4.5 kPa and <span class="math-tex">\(v_1\)</span> = 4 ms<sup>-1</sup>, calculate <span class="math-tex">\(v_2\)</span>.</div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>5 ms<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>6 ms<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>3 ms<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>25 ms<sup>-1</sup></span></label> </p></div><div class="q-explanation"><p>Using Bernoulli's equation for horizontal fluid flow:</p><p><span class="math-tex">\(p_1 + {1\over 2}ρ{v_1}^2 = p_2 + {1\over2}ρ{v_2}^2\)</span></p><p><span class="math-tex">\({1\over2} ρ{v_2}^2 - {1\over 2}ρ{v_1}^2=p_1 - p_2\)</span></p><p><span class="math-tex">\(500({v_2}^2-4^2)=4500\)</span></p><p><span class="math-tex">\(v_2 = 5\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This image is from the PhET Fluid Flow simulation:</p><p style="text-align: center;"><img alt="" src="../../screenshot-2020-01-03-at-06.30.11(1).png" style="width: 500px; height: 244px;"></p><p>If the fluid is water and the difference in height is 2 m, determine the pressure in the higher section of pipe.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>138 kPa</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>98 kPa</span></label> </p><p><label class="radio"> <input type="radio"> <span>116 kPa</span></label> </p><p><label class="radio"> <input type="radio"> <span>120 kPa</span></label> </p></div><div class="q-explanation"><p>Using Bernoulli's equation for constant speed:</p><p><span class="math-tex">\(p_1+\rho g z_1=p_2+\rho gz_2\)</span></p><p><span class="math-tex">\(p_1-p_2=10\text{ }000(z_2-z_1)=20\text{ kPa}\)</span></p><p><span class="math-tex">\(p_2 =98\text{ kPa}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This diagram represents water flowing out of a small hole in a bucket:</p><p style="text-align: center;"><img alt="" src="https://www.thinkib.net/files/physics/activities/bernie2.png" style="width: 242px; height: 286px;"></p><p>If <span class="math-tex">\(h\)</span> = 10 cm, calculate the velocity of water coming out of the hole.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>4 ms<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>It depends on the size of the hole</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>√2 ms<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>2 ms<sup>-1</sup></span></label> </p></div><div class="q-explanation"><p>Using conservation of energy or Bernoulli's equation:</p><p><span class="math-tex">\(v = \sqrt{2}gh\)</span></p><p>Note: the velocity does not depend on the diameter of the hole. However, flow rate would.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This image is from the PhET Fluid Flow simulation:</p><p style="text-align: center;"><img alt="" src="../../screenshot-2020-01-03-at-07.21.44.png" style="width: 500px; height: 340px;"></p><p>We can see that <span class="math-tex">\(p_1\)</span> = 105.132 kPa and <span class="math-tex">\(p_2\)</span> = 87.437 kPa. If the flow rate is increased...</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span><em><span style="font-size: 15px;">p</span></em></span></label><sub>1 </sub>and <em>p</em><sub>2</sub> decrease but <em>p</em><sub>1</sub> - <em>p</em><sub>2</sub> stays the same. </p><p><label class="radio"> <input type="radio"> <span><em>p</em><sub>1 </sub>and <em>p</em><sub>2</sub> increase but <em>p</em><sub>1</sub> - <em>p</em><sub>2</sub> stays the same.</span></label> </p><p><label class="radio"> <input type="radio"> <span><em><span style="font-size: 15px;">p</span></em></span></label><sub>1 </sub>and <em>p</em><sub>2</sub> increase but <em>p</em><sub>1</sub> - <em>p</em><sub>2</sub> decreases. </p><p><label class="radio"> <input type="radio"> <span><em>p</em><sub>1 </sub>and <em>p</em><sub>2</sub> increase and <em>p</em><sub>1</sub> - <em>p</em><sub>2</sub> increases.</span></label> </p></div><div class="q-explanation"><p>If flow rate is increased for the same cross-sectional area, then velocity is increased. The pressure is reduced.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This image is from the PhET Fluid Flow simulation:</p><p style="text-align: center;"><img alt="" src="../../screenshot-2020-01-03-at-07.21.44.png" style="width: 500px; height: 340px;"></p><p>We can see that <span class="math-tex">\(p_1\)</span> = 105.132 kPa and <span class="math-tex">\(p_2\)</span> = 87.437 kPa. If the left side was lowered...</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>The value for <em>p</em><sub>1</sub> would increase and <em>p</em><sub>2</sub> would remain the same</span></label> </p><p><label class="radio"> <input type="radio"> <span>The values for <em>p</em><sub>1</sub> and <em>p</em><sub>2</sub> would increase</span></label> </p><p><label class="radio"> <input type="radio"> <span>The value for <em>p</em><sub>1</sub> would increase and <em>p</em><sub>2</sub> would decrease</span></label> </p><p><label class="radio"> <input type="radio"> <span>The value for <em>p</em><sub>2</sub> would increase and <em>p</em><sub>1</sub> would remain the same</span></label> </p></div><div class="q-explanation"><p>The left side would be deeper and therefore at higher pressure.</p><p>The right hand side remains at the same height and so pressure is the same. While one might think that the greater height difference would cause a reduction in the right hand pressure, this is compensated for by the increased pressure on the left.</p><p><em>Still unconvinced?</em> Imagine that the right hand side was exposed to the atmosphere. Then there would clearly be no effect from simply lowering another part of the pipe.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram shows a pitot static tube.</p><p style="text-align: center;"><img alt="" height="206" src="../../screenshot-2020-01-06-at-08.22.25.png" width="312"></p><p>If the flow rate was increased...</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>h<sub>1</sub> would increase and h<sub>2</sub> would increase</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>h<sub>1</sub> would decrease and h<sub>2</sub> would stay the same</span></label> </p><p><label class="radio"> <input type="radio"> <span>h<sub>1</sub> would stay the same and h<sub>2</sub> would increase</span></label> </p><p><label class="radio"> <input type="radio"> <span>h<sub>1</sub> would increase and h<sub>2</sub> would stay the same</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(p_1\)</span> would decrease due to the increase in <span class="math-tex">\(v\)</span>.</p><p><span class="math-tex">\(p_2\)</span> remains the same as velocity here is still zero.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two 25 m diameter spheres are released above the Earth.</p><p style="text-align: center;"><img alt="" height="236" src="../../screenshot-2020-01-06-at-09.05.07.png" width="235"></p><p>The mass of the blue sphere is 40 kg and the red 30 kg. Compared to the red sphere, the blue will have...</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The same initial acceleration and higher terminal velocity</span></label> </p><p><label class="radio"> <input type="radio"> <span>The same initial acceleration and terminal velocity</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Higher initial acceleration and higher terminal velocity</span></label> </p><p><label class="radio"> <input type="radio"> <span>Lower initial acceleration and higher terminal velocity</span></label> </p></div><div class="q-explanation"><p>The buoyancy on each is the same. Resolving forces at the start:</p><p><span class="math-tex">\(ma=mg-B\)</span></p><p><span class="math-tex">\(a={mg-B\over m}=g-{B\over m}\)</span></p><p>The red ball has a larger mass and therefore a larger acceleration as the fractional term is smaller.</p><p>Resolving forces at terminal velocity, drag + buoyant force = weight. 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