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UT Arlington PHYS 1443 - Fluid Dymanics

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PHYS 1443 – Section 501 Lecture #24AnnouncementsPascal’s Law and HydraulicsBuoyant Forces and Archimedes’ PrincipleFlow Rate and the Equation of ContinuityExample for Equation of ContinuityBernoulli’s EquationBernoulli’s Equation cont’dSlide 9Example for Bernoulli’s EquationSimple Harmonic MotionEquation of Simple Harmonic MotionMore on Equation of Simple Harmonic MotionSimple Harmonic Motion continuedExample for Simple Harmonic MotionMonday April 26, 2004 1 PHYS 1443-501, Spring 2004Dr. Andrew BrandtPHYS 1443 – Section 501Lecture #24Monday, April 26, 2004Dr. Andrew Brandt1. Fluid Dymanics : Flow rate and Continuity Equation2. Bernoulli’s Equation3. Simple Harmonic Motion4. Simple Block-Spring SystemMonday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt2Announcements•The last HW on Ch. 12+13 is due Weds. 4/28 (decided not to have HW during last week)•An optional take-home quiz to replace your lowest quiz is due tonight•No optional sections (*) in Ch 11, 12, 13 Explicitly, covered 11.1-5,7 12.1-6 13.1-8 14.1-5•We’ll stop class a little early for teaching evaluationsMonday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt3Pascal’s Law and HydraulicsA change in the pressure applied to a fluid is transmitted undiminished to every point of the fluid and to the walls of the container.The resultant pressure P at any given depth h increases as much as the change in P0. This is the principle behind hydraulic pressure. Therefore, the resultant force F2 isWhat happens if P0is changed?PSince the pressure change caused by the the force F1 applied on to the area A1 is transmitted to the F2 on an area A2.ghPP0d1d2F1A1A2F22FIn other words, the initial force multiplied by the ratio of the areas A2/A1 is transmitted to F2 on the surface.Note the actual displaced volume of the fluid is the same and the work done by the forces are still the same.2F11AF22AF121Fdd112FAAAFP 2/11 mNPa VMMonday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt4Buoyant Forces and Archimedes’ PrincipleWhy is it so hard to put a beach ball under water while a piece of small steel sinks in the water?The water exerts a force on an object immersed in the water. This force is called Buoyant force.How does the Buoyant force work?Let’s consider a cube whose height is h and is filled with fluid and at its equilibrium. Then the weight Mg is balanced by the buoyant force B.This is called Archimedes’ principle.The magnitude of the buoyant force always equals the weight of the volume of fluid displaced by the submerged object.BBMghAnd the pressure at the bottom of the cube is larger than the top by gh.PTherefore,Where Mg is the weight of the fluid.gFMgAB /ghB PAghAVgBgFVgMgMonday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt5Flow Rate and the Equation of ContinuityStudy of fluid in motion: Fluid DynamicsIf the fluid is water: •Streamline or Laminar flow: Each particle of the fluid follows a smooth path, a streamline•Turbulent flow: Erratic, small, whirlpool-like circles called eddy current or eddies which absorb a lot of energyTwo main types of flowHydro-dynamics Flow rate: the mass of fluid that passes a given point per unit time/m tD Dsince the total flow must be conserved1mtD=D1 1Vtr D=D1 1 1A ltr D=D1 1 1A vr1 1 1 2 2 2A v A vr r=1 2m mt tD D=D DEquation of ContinuityMonday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt6Example for Equation of ContinuityHow large must a heating duct be if air moving at 3.0m/s along it can replenish the air every 15 minutes in a room of 300m3 volume? Assume the air’s density remains constant.Using equation of continuity1 1 1 2 2 2A v A vr r=Since the air density is constant1 1 2 2A v A v=Now let’s view the room as a large section of the duct2 211A vAv= =2 21/A l tv=21Vv t=�23000.113.0 900m=�Monday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt7Bernoulli’s EquationBernoulli’s Principle: Where the velocity of fluid is high, the pressure is low, and where the velocity is low, the pressure is high. Amount of work done by the force, F1, that exerts pressure, P1, at point 11W =Work done by the gravitational force to move the fluid mass, m, from y1 to y2 is1 1F lD =1 1 1P A lDAmount of work done on the other section of the fluid is2 2 2 2W P A l=- D( )3 2 1W mg y y=- -Monday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt8Bernoulli’s Equation cont’dThe net work done on the fluid is1 2 3W W W W= + +1 1 1 2 2 2 2 1P A l P A l mgy mgy= D - D - +From the work-energy principle2 22 11 12 2mv mv- =Since mass, m, is contained in the volume that flowed in the motion1 1 2 2A l A lD = Dand1 1 2 2m A l A lr r= D = D22 2 112121 12 2v vA l A lr r-D D1 1 1 2 2 2 2 1P A l P A l mgy mgyD - D - +1 1 2 2 2 2 1 111 2 2A l A l AP P gyl gyA lr r= - - +D D D DThus,Monday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt9Bernoulli’s Equation cont’dWe obtain2 2 1 1 1 1 2 2 2 2 12 22 1 1 2 2 111 12 2A l A l A l A l A lv v P P g A ly gyr r r r- = - - +D D D D D DRe-organize21 1 112P v gyr r+ + =Bernoulli’s Equation21 1 11.2P v gy constr r+ + =22 2 212P v gyr r+ +2 22 1 1 2 2 11 12 2v v P P gy gyr r r r- = - - +SinceThus, for any two points in the flowFor static fluid( )2 1 1 2 1P P g y y P ghr r= + - = +For the same heights( )2 22 1 1 212P P v vr= + -The pressure at the faster section of the fluid is smaller than slower section.Pascal’s LawMonday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt10Example for Bernoulli’s EquationWater circulates throughout a house in a hot-water heating system. If the water is pumped at a speed of 0.5m/s through a 4.0cm diameter pipe in the basement under a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter pipe on the second floor 5.0m above? Assume the pipes do not divide into branches.Using the equation of continuity, flow speed on the second floor is2v =Using Bernoulli’s equation, the pressure in the pipe on the second floor is( )( )2 22 1 1 2 1 212P P v v g y yr r= + - + -( )( )5 3 2 2 313.0 10 1 10 0.5 1.2 1 10 9.8 52= � + � - + � � �-5 22.5 10 /N m= �1 12A vA=21 122r vrpp=20.0200.5 1.2 /0.013m s� �� =� �� �Monday April 26, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt11kxFSimple Harmonic MotionHarmonic Motion is motion that occurs due to a force that depends on


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UT Arlington PHYS 1443 - Fluid Dymanics

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