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3/18/2014 What is a fluid A substance that can flow • Liquids (difficult to compress) Gases (easy to compress) • • Chapter 9 Basic properties: • Fluids The field of science that describes the fluid motion is called “Fluid Mechanics” • Density • • • • • • • • • ρ= ∆m ∆V Mass per unit volume. Unit kg/m3. The density can be different at each point of the fluid. An object is uniform if the density is constant at any point of the body A fluid is incompressible if the density does not change as a result of an applied pressure. Pressure Fluids take the form of the container. When shear stress is applied they flow. The molecules in fluids are in constant random motion (Brownian motion). • • • P= p= Some examples Interstellar space Laboratory vacuum Air (1 atm 20C) Ice Water Seawater Earth Neutron star • • • • • • • • ∆F ∆A Force per unit area (the force is perpendicular to the surface) Scalar: the pressure has the same value in all directions! Units: 1Pa=1N/1m2. Other popular units 1 atm = 1 bar = 1.01x105 Pa = 760 torr = 14.7 lb/in2 10-20 kg/m3 10-17 kg/m3 1.21 kg/m3 0.917x103 kg/m3 0.998x103 kg/m3 1.024x103 kg/m3 5.5x103 kg/m3 1018 kg/m3 Some Examples • • • • Sea level atm. pressure Bottom of the ocean Center of the Earth Center of the Sun 105 Pa 1.1x108 Pa 4x1011 Pa 2x1016 Pa 1 3/18/2014 Fluid Pressure Pascal’s Principle Pressure arises from the collisions between the particles of a fluid with another object (container walls for example). A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container. • There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall. ∆p = ∆pext 7 Hydraulic Lever • • An input force applied over a distance is transformed into bigger force over a shorter distance ∆p = • Gravity’s Effect on Fluid Pressure Fi Fo = Ai Ao Fo = Fi F2 = F1 + mg F1 = p1 A F2 = p2 A p2 A = p1 A + ρVg Ao > Fi Ai V = A(y1 − y 2 ) p2 = p1 + ρg(y1 − y 2 ) The work done by the input force and the output force is the same! V = Ai di = Ao do W = Fo do = do Fi Static equilibrium If the top of the fluid column is placed at the surface of the fluid, then P1 = Patm =P0 if the container is open. Ao d = d o Fi i = Fi di Ai do p = patm + ρgh Example (text problem 9.19): At the surface of a freshwater lake, the pressure is 105 kPa. (a) What is the pressure increase in going 35.0 m below the surface? Hydrostatic pressure Example: The surface pressure on the planet Venus is 95 atm. How far below the surface of the ocean on Earth do you need to be to experience the same pressure? The density of seawater is 1025 kg/m3. P = Patm + ρgd ∆P = P − Patm = ρgd ( )( P = Patm + ρgd 95 atm = 1 atm + ρgd ) = 1000 kg/m 3 9.8 m/s 2 (35 m ) ρgd = 94 atm = 9.5 ×10 6 N/m 2 (1025 kg/m )(9.8 m/s )d = 9.5 ×10 = 343 kPa = 3.4 atm 3 2 6 N/m 2 d = 950 m 11 12 2 3/18/2014 Measuring pressure Gauge pressure • Absolute pressure = total pressure Gauge pressure = total - atmospheric The atmospheric pressure on the free surface of a liquid is often neglected. • • • Mercury barometer: measures the atmospheric pressure. p0 = ρgh • pgauge = ρgh Open tube manometer: measures the gauge pressure of a gas. pg = p − p0 = ρgh Fluids at Rest: Examples Sample problem • • The closed U-tube shown in the figure contains a liquid of density ρ. What is the pressure difference ∆P = PA - PB? What gauge pressure must a machine produce to suck mud of density 1800 kg/m3 up a tube by a height of 1.5 m? PA = PB + ρgh p g = p − patm h Manometer • The (open) U-tube shown in the figure contains two liquids in static equilibrium: Water of density ρW is in the right arm, and an oil of unknown density ρX is in the left arm. What is the density of the unknown oil in terms of d and L? patm = p + ρgh p g = − ρgh = −2.6×10 4 Pa pa PB = P0 + ρ x g ( L + d ) PB = P0 + ρW gL ρX = mud University of Florida Archimedes Principle-Buoyancy Force • r Fb When a body is partially or fully submerged in a fluid a buoyant force from the surrounding fluid acts on the body. • • • The buoyancy force is upward* It is applied at CM (displaced fluid) The magnitude is: Fb = F2 − F1 Fb = ρ f gh2 A − ρ f gh1 A = ρ f gV Fb = m f g where mf is the mass of the displaced fluid! L B B L ρW L+d Page 16 Example (text problem 9.28): A flat-bottomed barge loaded with coal has a mass of 3.0×105 kg. The barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline? FBD for the barge Apply Newton’s 2nd Law to the barge: ∑F = F FB B h1 −w=0 FB = w h2 F2 d PHY 2053 y F1 h PA = PB + ρgh ∆P = PA − PB = ρgh p B A mw g = (ρ wVw )g = mb g ρ wVw = mb x w ρ w ( Ad ) = mb d= mb ρw A = 3.0 ×105 kg (1000 kg/m )(20.0 m *10.0 m ) = 1.5 m 3 18 3 3/18/2014 Example (text problem 9.40): A piece of metal is released under water. The volume of the metal is 50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration? (Note: when v = 0, there is no drag force.) y FBD for the metal Example continued: Since the object is completely submerged V=Vobject. specific gravity = Apply Newton’s 2nd Law to the piece of metal: FB ∑F = F B where ρwater = 1000 kg/m3 is the density of water at 4 °C. − w = ma x w The magnitude of the buoyant force equals the weight of the fluid displaced by the metal. FB = ρ waterVg Solve for a: ρ V F ρ Vg − g = g water − 1 a = B − g = water ρ objectVobject m ρ objectVobject ρ ρ water Given specific gravity = ρ object = 5 .0 ρ water ρ V 1 1 a = g water − 1 = g − 1 = g − 1 = −7.8 m/s 2 ρ V S .G. 5.0 object object 19 20 4