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Wednesday, November 18, 2009 46 The Final Exam Schedule is Final! Friday, December 18, 1:30 - 4:30 pm Frank Kennedy Brown & Gold Gyms The whole course 30 multiple choice questions Formula sheet provided Seating (this info is on Aurora) Brown Gym: A - S Gold Gym: T - Z Wednesday, November 18, 2009 47 WileyPLUS Assignment 4 Chapters 8, 9, 10 Due: Friday, November 27 at 11 pm Next Week: Tutorial & Test 4 Wednesday, November 18, 2009 48 Fluids, continued... Pascal’s Principle: a change in pressure is transmitted equally throughout an enclosed fluid. Archimedes’ Principle: the buoyant force acting on an object is equal to the weight of fluid displaced by the object. Wednesday, November 18, 2009 49 Clicker Question: Focus on Concepts, Question 8 Two objects, A and B, have the same volume and are completely submerged in a liquid, although A is deeper than B. Which object, if either, experiences the greater buoyant force? A) Object B, because the closer an object is to the surface, the greater is the buoyant force. B) Both objects experience the same buoyant force. C) Object A, because, being at a greater depth, it experiences a greater pressure. B) The buoyant force is equal to the weight of fluid displaced by the object and the objects have the same volume. Wednesday, November 18, 2009 50 Clicker Question: Focus on Concepts, Question 11 Two beakers are filled to the same level with water. One of them has a plastic ball floating in the water. If the beakers are placed on a scale, one at a time, which weighs more? A) The beaker without the plastic ball weighs more. B) The beaker with the plastic ball weighs more. C) The beakers have the same weight. C) Archimedes: the weight of the ball is the same as the weight of the “missing” water displaced by the ball. So the ball could be replaced by the missing water without changing the total weight and then the picture would be as on the left. Wednesday, November 18, 2009 51 Fluids in Motion Streamline (orderly, or steady) flow – the velocity of a fluid at some point does not change with time. Streamlines show the path of fluid particles. Streamlines do not cross. Wednesday, November 18, 2009 52 Fluids in Motion Unsteady flow – the velocity at a point in the fluid changes with time. Turbulent flow is an extreme case of unsteady flow for a fastmoving fluid – the velocity changes erratically from moment to moment, as at sharp obstacles or bends. Viscous flow – a type of friction impeding the relative motion of layers of a fluid, as in molasses. Bernoulli’s equation, to follow, applies to streamline flow. Wednesday, November 18, 2009 53 Freely Flowing Fluids: The Equation of Continuity !m2 !m1 Volume = A1v1!t Volume = A2v2!t (flowing out) (flowing in) What goes in must come out That is: Mass flowing in = mass flowing out (Δm2 = Δm1) So, "2 A2v2Δt = "1 A1v1Δt and "2 A2v2 = "1 A1v1 Equation of continuity Wednesday, November 18, 2009 54 The Equation of Continuity ρ1 A1 v1 = ρ2 A2 v2 !m = " A v = mass flow rate !t If the fluid is incompressible, "1 = "2, and the volume flow rate (volume per second) is Q = v1A1 = v2A2 V = vAΔt A v vΔt Wednesday, November 18, 2009 55 11.-/50: Oil is flowing at a speed of 1.22 m/s through a pipeline with a radius of 0.305 m. How much oil flows in 1 day? V = vAΔt A v vΔt The volume rate of flow (volume per second) is Q = vA A = πr2 = π ! 0.3052 = 0.2923 m2. So, Q = (1.22 m/s)(0.2923 m2) = 0.3565 m3/s. In 24 hours, the flow is 3.08!104 m3. Wednesday, November 18, 2009 56 11.58/52: The volume rate of flow in an artery supplying the brain is 3.6!10-6 m3/s. If the radius of the artery is 5.2 mm, determine the average blood speed. Q 3.6 × 10−6 m3 /s v= = = 0.0424 m/s A π × 0.00522 Find the average blood speed if a constriction reduces the radius of the artery by a factor of 3 (without reducing the flow rate). v = Q/A, and r is reduced to r/3, so the speed is increased by a factor of 32 = 9. So, v = 9 ! 0.0424 = 0.381 m/s Wednesday, November 18, 2009 57 Bernoulli’s Equation Q = vA = constant F2 = P2A2 F1 = P1A1 Fluid of constant density The fluid speeds up when it gets to the constriction. What is the force that causes this acceleration? There must be a drop in pressure force that accelerates the fluid to the right. F2 > F1, that is, P2A2 > P1A1 Wednesday, November 18, 2009 58 Bernoulli’s Equation Q = vA = constant F2 = P2A F1 = P1A Pipe of constant cross-sectional area The fluid flows up hill at constant speed through a pipe of constant area. Where does the force come from to push the fluid up the hill at constant speed? There must be a decrease in pressure that pushes the fluid up the hill. P2 > P1 Wednesday, November 18, 2009 59 Bernoulli’s Equation m F1 = P1A1 F2 = P2A2 m Follow a fluid element of mass m through the pipe from region 2 at the left to region 1 at the right. Work Wnc done by the pressure forces increases the mechanical energy of the fluid element. Wednesday, November 18, 2009 60 Bernoulli’s Equation m m Work-energy equation: Wnc = !PE + !KE 1 1 = (mgy1 − mgy2) + ( mv21 − mv22) 2 2 What is the work done by non-conservative forces? Wednesday, November 18, 2009 61 Bernoulli’s Equation Volume V m The pressure difference between the two ends of the fluid element of mass m exerts a net force on the mass – (F + !F) − F = (P + !P)A − PA So that, !F = !P × A When the fluid element moves through its own length, s, the net force does work on it: !Wnc = !Fs = !P(A × s) = !P ×V The work done by the whole pressure difference between the ends of the pipe, P2 – P1 should then be: Wnc = (P2 − P1) ×V Wednesday, November 18, 2009 62 Bernoulli’s Equation Back to the work-energy equation: Wnc = !PE + !KE 1 1 Wnc = (P2 − P1) ×V = (mgy1 − mgy2) + ( mv21 − mv22) 2 2 Divide by V and use density " = m/V 1 1 P2 − P1 = (!gy1 − !gy2) + ( !v21 − !v22) 2 2 1 1 P1 + !gy1 + !v21 = P2 + !gy2 + !v22 2 2 Bernoulli’s Equation • For streamline flow • Streamlines form “virtual pipes” Wednesday, November 18, 2009 63 Bernoulli’s Equation For streamline flow: 1 P + ρgy + ρv 2 = constant 2 If the speed of flow increases, then the pressure decreases If the height increases, then the pressure decreases When the speed of flow is constant, Bernoulli’s equation gives same variation of pressure with depth as for a static fluid: If v is constant, P2 = P1 + "g(y1 - y2) Wednesday, November 18, 2009 64 Bernoulli’s Equation Q = vA = constant F2 = P2A2 F1 = P1A1 Fluid of constant density v increases, y is constant, so P must decrease Wednesday, November 18, 2009 65 Clicker Question: Focus on Concepts, Question 16 Blood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. A hemoglobin molecule moves from the narrow region into the wider region. What happens to the pressure acting on the molecule as it moves from the narrow to the wider region? 1 P + ρgy + ρv 2 = constant 2 A) The pressure decreases. B) There is no change in the pressure. C) The pressure increases. C) The pressure increases because the speed decreases: P + "v2/2 = constant Wednesday, November 18, 2009 66 Clickers! vA vC vB Fluid is flowing from left to right through the pipe. Points A and B are at the same height, but the cross-sectional areas of the pipe differ. Points B and C are at different heights, but the crosssectional areas are the same. Rank the pressures at the three points, from highest to lowest. A) A and B (a tie), C B) C, A and B (a tie) C) B, C, A D) C, B, A E) A, B, C Wednesday, November 18, 2009 vB > vA, what force speeds up the fluid? C is higher than B, what force pushes the fluid uphill? E) PA > PB > PC 67 Applications of Bernoulli’s Equation Because of the enlargement of a blood vessel, the cross-sectional area A1 of an aorta increases to A2 = 1.7A1. The speed of the blood (" = 1060 kg/m3) through a normal portion of the aorta is v1 = 0.4 m/s. Assuming that the aorta is horizontal, find the amount by which the pressure P2 in the enlarged region exceeds P1 in the normal region. First, what is the speed of the blood in the enlarged region? Equation of continuity: v1 A1 = v2 A2 so, v2 = v1 A1 0.4 = = 0.235 m/s A2 1.7 A1 v2 v1 = 0.4 m/s P1 Bernoulli: 1 1 P1 + !v21 = P2 + !v22 2 2 A2 = 1.7A1 P2 (no change of height) Wednesday, November 18, 2009 68 1 1 P1 + !v21 = P2 + !v22 2 2 A1 A2 = 1.7A1 v1 = 0.4 m/s 1 So, P2 − P1 = ![v21 − v22] 2 v2 = 0.235 m/s P1 P2 1 P2 − P1 = (1060 kg/m3)[0.42 − 0.2352] = 55 Pa 2 That is, the pressure is greater in the already weakened enlarged section, putting greater stress on it. The pressure must be larger because there has to be a force that slows down the blood as it enters the enlarged section. Wednesday, November 18, 2009 69 11.67/60: Water is circulating through a closed system of pipes in a two-floor building. On the first floor, the water has a gauge pressure of 3.4!105 Pa and a speed of 2.1 m/s. On the second floor, which is 4 m higher, the speed of the water is 3.7 m/s. The speeds are different because the pipe diameters are different. What is the gauge pressure on the second floor? Wednesday, November 18, 2009 70 Bernoulli’s Equation For streamline flow of an incompressible fluid (or close to incompressible in some situations such as airflow over an airplane wing) 1 P + !gy + !v2 = constant 2 Equation of continuity for fluid of constant density: vA = constant Wednesday, November 18, 2009 71 Lift force on an airplane wing Streamlines Front of wing Streamlines become compressed over curved top of wing. Compare: Wednesday, November 18, 2009 72 11.63/59: An airplane wing is designed so that the speed of the air across the top of the wing is 251 m/s when the speed of the air below the wing is 225 m/s. The density of the air is 1.29 kg/m3. Find the lift on a wing of area 24 m2. Imagine streamlines with uniform air conditions in front of the plane and that the streamlines divide and pass above and below the wing. 1 1 1 P0 + !gy + !v20 = P1 + !gy + !v21 = P2 + !gy + !v22 2 2 2 In front of plane Above wing Below wing 1 1 So, P2 − P1 = !(v21 − v22) = × (1.29 kg/m3) × (2512 − 2252) = 7983 Pa 2 2 The net upward force is then: 7983xA = 7983!(24 m2) = 191,600 N Wednesday, November 18, 2009 73 View from above Ball is moving to the right Streamlines of air moving relative to the ball Wednesday, November 18, 2009 74 11.74/66: A liquid flows through a horizontal pipe of radius 0.02 m. The pipe bends straight upward through a height of 10 m and joins another horizontal pipe with radius 0.04 m. What flow rate is needed to keep the pressures in the two horizontal pipes the same? h = 10 m v1 v2 r2 = 0.04 m r1 = 0.02 m Wednesday, November 18, 2009 75