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Physics 103: Lecture 18 Fluids

Physics 103: Lecture 18 Fluids. Today’s lecture will cover Pascal’s Principle Archimedes’ Principle Fluids in motion: Continuity & Bernoulli’s equation. Midterm II. Midterm Exam II, Thur Nov 5 th , 5:45 - 7 PM Arrange alternate times today before/after lecture

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Physics 103: Lecture 18 Fluids

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  1. Physics 103: Lecture 18Fluids • Today’s lecture will cover • Pascal’s Principle • Archimedes’ Principle • Fluids in motion: Continuity & Bernoulli’s equation Physics 103, Fall 2009, U.Wisconsin

  2. Midterm II • Midterm Exam II,Thur Nov 5th, 5:45 - 7 PM • Arrange alternate times today before/after lecture • Material from Chapters 5-8 inclusive • One page of notes, both sides, (8.5” x 11”) allowed • You can re-use your equation sheet from exam 1 if you only wrote on one side. • 20 multiple choice questions • Scantron will be used - bring #2 HB pencils + calculator • Know your student Id number and section number (starts with 3) • All relevant physics constants will be provided as with exam 1 • Exam Rooms • 165 Bascom: 302, 303, 304, 306, 312, 318, 320, 324 B-10 Ingraham: 305, 313, 317, 321, 322, 327, 328, 329, 330 • 3650 Humanities: 307, 308, 309, 310, 311, 314, 315, 319, 323, 326 • Alternative exam room: Chamberlin 4320 (lab room) • (all the same as Exam I) Physics 103, Fall 2009, U.Wisconsin

  3. Pressure in a fluid or gas book v Fx v Air molecule • Impulse to book: • Fxt = px = (Mvx) • Fx = (Mvx)/ t • Force is perpendicular to surface • Force proportional to area of surface • pressure (p) • p = Force/area [N/m2] • 1 N/m2 = 1 Pascal (Pa) Physics 103, Fall 2009, U.Wisconsin

  4. Pressure = Force per Unit Area Which will hurt more? • If you are pricked by a nail with a force equal to your weight • If your entire weight is supportedby a bed of similar nails • Both will hurt the same Physics 103, Fall 2009, U.Wisconsin

  5. Pressure y Atmospheric Pressure • Even when there is no breeze air molecules are continuously • bombarding everything around - results in pressure • normal atmospheric pressure = 1.013 x 105 Pa Physics 103, Fall 2009, U.Wisconsin

  6. Pressure and Depth • Examine the darker region, assumed to be a fluid  • It has a cross-sectional area A • Extends to a depth h below the surface • Three external forces act on the region • -P1A + P2A - Mg = 0P2 = P1 + Mg/A = P1 + Mgh/V = P1 + gh • At the surface compared to at depth h • Po is normal atmospheric pressure • 1.013 x 105 Pa = 14.7 lb/in2 Physics 103, Fall 2009, U.Wisconsin

  7. Fluids Summary Atmospheric Pressure • Even when there is no breeze air molecules are continuously • bombarding everything around - results in pressure • normal atmospheric pressure = 1.013 x 105 Pa • Pressure (P) • P = Force/Area [N/m2] • 1 N/m2 = 1 Pascal (Pa) Density = Mass/Volume  = M / V units = kg/m3 Pressure variation with depth ∆ P = r g h Physics 103, Fall 2009, U.Wisconsin

  8. Pressure and DepthBarometer: a way to measure atmospheric pressure p1=0 h p2=patm • p2 = p1 + gh • patm = gh • Measure h, determine patm • example--Mercury •  = 13,600 kg/m3 • patm = 1.013 x 105 Pa •  h = 0.760 m = 760 mm(for 1 atm) Physics 103, Fall 2009, U.Wisconsin

  9. Pascal’s Principle F1 A1 A1 F2 A2 • A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. • This principle is used in hydraulic system • P1 = P2 • (F1 / A1) = (F2 / A2) • Can be used to derive large gain by making A2 much larger than A1 • F2 = F1 (A2 / A1) • Work done is the same: height by which the surface A2 rises is smaller than the change in the height of surface with area A1. Physics 103, Fall 2009, U.Wisconsin

  10. Archimedes Principle Buoyant Force (B) • weight of fluid displaced • B = fluid gVdisplaced • W = object gVobject • object sinks if object > fluid • object floats if object < fluid If object floats…. • B=W • Therefore fluid gVdisplaced = object gVobject • Therefore Vdisplaced/Vobject = object / fluid Physics 103, Fall 2009, U.Wisconsin

  11. CORRECT Archimedes Principle Weight of a glass filled to the brim with water is Wb. A cube of ice is placed in it, causing some water to spill. After the splilled water is cleaned up, the weight of the glass with ice cube is Wa. How do the weights compare: 1. Wb > Wa. 2. Wb < Wa. 3. Wb = Wa. Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice Physics 103, Fall 2009, U.Wisconsin

  12. CORRECT Preflight 7 Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up causing the water to spill. 2. Go down. 3. Stay the same. Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice When ice melts it will turn into water of same volume Physics 103, Fall 2009, U.Wisconsin

  13. Fluid Flow Fluid flow without friction • Volume flow rate: DV/Dt = ADd/Dt = Av (m3/s) • Continuity: A1 v1 = A2 v2 • i.e., flow rate the same everywhere • e.g., flow of river or a garden hose • Water through a narrow hose moves faster Physics 103, Fall 2009, U.Wisconsin

  14. Faucet A1 V1 A2 V2 A stream of water gets narrower as it falls from a faucet (try it & see). Explanation: the equation of continuity The velocity of the liquid increases as the water falls due to gravity. If the volume flow rate is conserved, them the cross-sectional area must decrease in order to compensate The density of the water is the same no matter where it is in space and time, so as it falls down and accelerates because of gravity,the water is in a sense stretched, so it thins out at the end. Physics 103, Fall 2009, U.Wisconsin

  15. Bernoulli’s Equation • Pressure drops in a rapidly moving fluid • whether or not the fluid is confined to a tube • For incompressible, frictionless fluid: Physics 103, Fall 2009, U.Wisconsin

  16. Applications of Bernoulli’s Principle • Wings and sails • Higher velocity on one side of sail or wing versus the other results in a pressure difference that can even allow the boat to sail into the wind • Water pressure/velocity at your house • Velocity measurement Physics 103, Fall 2009, U.Wisconsin

  17. Sailing Against the Wind • v2>v1 • Therefore: • P1>P2 • Pressure difference causes a force Physics 103, Fall 2008, U. Wisconsin

  18. Problem 1 (a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6 m/s parallel to its front surface and 3.5 m/s along its back surface. Take the density of air to be 1.29 kg/m3. (b) Discuss whether this force is great enough to be effective for propelling a sail boat. Physics 103, Fall 2009, U.Wisconsin

  19. Problem 2 (a) What is the pressure drop due to Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9 cm diameter fire hose while carrying a flow of 40 L/s? (b) To what maximum height above the nozzle can this water rise neglecting air resistance. Physics 103, Fall 2009, U.Wisconsin

  20. Extra Physics 103, Fall 2009, U.Wisconsin

  21. CORRECT Heavy Ice Two identical glasses are filled to the same level with water. One glass has a cube of regular ice floating in it and the other has a cube of special ice-9, heavier than water, which sinks to the bottom. Which of the two glasses weighs more? 1. The glass with the regular ice 2. The glass with the ice-9 3. Both glasses weigh the same The ice-9 sinks. The buoyant force equal to the weight of the displaced water is not sufficient to counter the weight of the ice-9. Physics 103, Fall 2009, U.Wisconsin

  22. CORRECT Preflight 8 Which weighs more: 1. A large bath tub filled to the brim with water. 2. A large bath tub filled to the brim with water with a toy battle ship floating on it. 3. Both weigh the same. Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of the ship in water Physics 103, Fall 2009, U.Wisconsin

  23. CORRECT Preflight 9 An oil tanker is floating in a port. The oil tanker is loaded with oil. The density of oil is less than that of water. The waterline (i.e., a line marked on the outside of a ship) of a loaded tanker compared to that of the empty tanker is: 1. lower. 2. the same. 3. higher. Physics 103, Fall 2009, U.Wisconsin

  24. CORRECT Preflight 10 Marbles dropped into a partially filled bathtub sink to the bottom. The downward force on the bottom of the tub with the marbles, compared to the situation without the marbles is: 1. lower. 2. the same. 3. higher. The downwards force with the marbles is the sum of the weights of the marbles and the water column above it. Marbles sank. Therefore, their weight is higher compared to the volume occupied by water which originally filled their place. Therefore, the total force is higher. Physics 103, Fall 2009, U.Wisconsin

  25. Melting Ice and Volume CORRECT Two identical glasses are filled to the same level with water. One of the two glasses has ice cubes floating in it. When the ice cubes melt, in which glass is the level of the water higher? 1. The glass without ice cubes 2. The glass with ice cubes originally 3. The level is the same in both. The weight of water from molten ice cubes is equal to the water originally displaced by the cubes. Physics 103, Fall 2009, U.Wisconsin

  26. Floating Balls CORRECT Two identical glasses are filled to the same level with water. One of the two glasses has plastic balls floating in it. If the density of the plastic balls is less than that of ice, which of the two glasses weighs more? 1. The glass without plastic balls 2. The glass with plastic balls 3. Both glasses weigh the same The plastic balls displace exactly their own weight in water, so the two glasses weigh the same amount. Physics 103, Fall 2009, U.Wisconsin

  27. Sunken Balls CORRECT Two identical glasses are filled to the same level with water. Solid steel balls are at the bottom in one of the glasses. Which of the two glasses weighs more? 1. The glass without steel balls 2. The glass with steel balls 3. Both glasses weigh the same The steel balls sink. The buoyant force equal to the weight of the displaced water is not sufficient to counter the weight of the steel balls. Therefore, the glass with steel balls weighs more. Physics 103, Fall 2009, U.Wisconsin

  28. CORRECT Preflight 4 Two hoses, one of 20-mm diameter, the other of 15-mm diameter are connected one behind the other to a faucet. At the open end of the hose, the flow of water measures 10 liters per minute. Through which pipe does the water flow faster? 1. The 20-mm hose 2. The 15-mm hose 3. Water flows at the same speed in both cases 4. The answer depends on which of the two hoses comes first in the flow When a tube narrows, the same volume occupies a greater length. For the same volume to pass through points 1 and 2 in a given time, the velocity must be greater at point 2. The process is reversible. Physics 103, Fall 2009, U.Wisconsin

  29. Preflight 5 CORRECT A large bucket full of water has two equal diameter drains. The water level in the bucket is kept constant by constantly refilling it. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which is bent downward such that the bottom of this pipe even with the other hole, like in the picture below: Though which drain is the water spraying out with the highest speed? 1. The hole 2. The pipe 3. Same Since the pressures at the two drains are the same and the liquid leaves the pipe at the same height, their speeds are the same. Physics 103, Fall 2009, U.Wisconsin

  30. Preflight 6 An artery with cross sectional area of 1 cm2 branches into 20 smaller arteries each with 0.5 cm2 cross sectional area. If the velocity of blood in thicker artery is v, what is the velocity of the blood in the thinner arteries? 1. 0.1 v 2. 0.2 v 3. 0.5 v 4. v 5. 2 v Physics 103, Fall 2009, U.Wisconsin

  31. Problem (a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6 m/s parallel to its front surface and 3.5 m/s along its back surface. Take the density of air to be 1.29 kg/m3. (b) Discuss whether this force is great enough to be effective for propelling a sail boat. Physics 103, Fall 2008, U. Wisconsin

  32. Torricelli’s Theorem P1, v1, h1 h P2=P1 , v2 , h2 Physics 103, Fall 2009, U.Wisconsin

  33. Velocity Measurement: Pitot tube Physics 103, Fall 2009, U.Wisconsin

  34. Notes on Moduli • Solids have Young’s, Shear, and Bulk moduli • Liquids have only bulk moduli Physics 103, Fall 2009, U.Wisconsin

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