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UT Arlington PHYS 1443 - Fluid Dymanics

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Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #21Wednesday, Nov. 19, 2003Dr. Mystery Lecturer1. Fluid Dymanics : Flow rate and Continuity Equation2. Bernoulli’s Equation3. Simple Harmonic Motion4. Simple Block-Spring System5. Energy of the Simple Harmonic OscillatorNext Wednesday’s class is cancelled but there will be homework!!Today’s Homework is #11 due on Wednesday, Nov. 26, 2003!!Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu2Flow 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 absorbs a lot of energyTwo main types of flowWater dynamics?? Hydro-dynamics Flow rate: the mass of fluid that passes a given point per unit time/mt∆∆since the total flow must be conserved1mt∆=∆11Vtρ∆=∆111Altρ∆=∆111Avρ111222AvAvρρ=12mmtt∆∆=∆∆Equation of ContinuityWednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3Example 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 300m3volume? Assume the air’s density remains constant.Using equation of continuity111222AvAvρρ=Since the air density is constant1122AvAv=Now let’s call the room as the large section of the duct2211AvAv==221/Altv=21Vvt=⋅23000.113.0900m=×Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu4Bernoulli’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 y1to y2is11Fl∆=111PAl∆Amount of work done on the other section of the fluid is2222WPAl=−∆()321Wmgyy=−−Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu5Bernoulli’s Equation cont’dThe net work done on the fluid is123WWWW=++11122221PAlPAlmgymgy=∆−∆−+From the work-energy principle22211122mvmv−=Since mass, m, is contained in the volume that flowed in the motion1122AlAl∆=∆and1122mAlAlρρ=∆=∆222112121122vvAlAlρρ−∆∆11122221PAlPAlmgymgy∆−∆−+112222111122AlAlAPPgylgyAlρρ=−−+∆∆∆∆Thus,Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu6Bernoulli’s Equation cont’dWe obtain221111222212221122111122AlAlAlAlAlvvPPgAlygyρρρρ−=−−+∆∆∆∆∆∆Re-organize211112Pvgyρρ++=Bernoulli’s Equation21111.2Pvgyconstρρ++=222212Pvgyρρ++222112211122vvPPgygyρρρρ−=−−+SinceThus, for any two points in the flowFor static fluid()21121PPgyyPghρρ=+−=+For the same heights( )22211212PPvvρ=+−The pressure at the faster section of the fluid is smaller than slower section.Pascal’s LawWednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Example 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 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( )( )2221121212PPvvgyyρρ=+−+−( )( )5322313.0101100.51.21109.852=×+×−+×××−522.510/Nm=×112AvA=21122rvrππ=20.0200.51.2/0.013ms×=Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu8kx−=FSimple Harmonic MotionWhat do you think a harmonic motion is?Motion that occurs by the force that depends on displacement, and the force is always directed toward the system’s equilibrium position.When a spring is stretched from its equilibrium position by a length x, the force acting on the mass is What is a system that has such characteristics?A system consists of a mass and a spring This is a second order differential equation that can be solved but it is beyond the scope of this class.It’s negative, because the force resists against the change of length, directed toward the equilibrium position.From Newton’s second lawFwe obtaina22dtxdWhat do you observe from this equation?Acceleration is proportional to displacement from the equilibriumAcceleration is opposite direction to displacementCondition for simple harmonic motionma=kx−=xmk−=xmk−=Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu9()cosAtωφ=+Equation of Simple Harmonic MotionThe solution for the 2ndorder differential equation What happens when t=0 and φ=0?Let’s think about the meaning of this equation of motion22dxkxdtm=−What are the maximum/minimum possible values of x?xAmplitudePhaseAngular FrequencyPhase constantxWhat is φ if x is not A at t=0?xA/-AAn oscillation is fully characterized by its:•Amplitude•Period or frequency•Phase constantGeneralized expression of a simple harmonic motion()φcosA= 'x=φ()'cos1x−=()00cos += AA=Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu10More on Equation of Simple Harmonic MotionLet’s now think about the object’s speed and acceleration.Since after a full cycle the position must be the sameSpeed at any given timeThe periodWhat is the time for full cycle of oscillation?xTOne of the properties of an oscillatory motionFrequencyHow many full cycles of oscillation does this undergo per unit time?fWhat is the unit?1/s=HzxvaMax speed maxvMax acceleration maxaAcceleration at any given timeWhat do we learn about acceleration?Acceleration is reverse direction to displacementAcceleration and speed are π/2 off phase:When v is maximum, a is at its minimum()()φω++=TtAcos()φπω++=2cos tAωπ2=T1=πω2=()φω += tAcosdtdx=()φωω +−= tAsinAω=dtdv=()φωω +−= tA cos2x2ω−=A2ω=Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu11Simple Harmonic Motion continuedLet’s determine phase constant and amplitudeBy taking the ratio, one can obtain the phase constantPhase constant determines the starting position of a simple harmonic motion.This constant is important when there are more than one harmonic oscillation involved in the motion and to determine the overall effect of the


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UT Arlington PHYS 1443 - Fluid Dymanics

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