Lecture 01 - Density, pressure and Pascal's principle

Lecture 01 - Density, pressure and Pascal's principle -...

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Unformatted text preview: What is a fluid? Fluids are “substances that flow”…. “substances that take the shape of the container” Lecture 1 Fluids: density, pressure, Pascal’s principle. Density, pressure Density: ρ= m V Pure water: 1000 kg/m3 Atoms and molecules must be free to move .. No long range correlation between positions (e.g., not a crystal). Gas or liquid… or granular materials (like sand) Atmospheric pressure The atmosphere of Earth is a fluid, so every object in air is subject to some pressure. At At the surface of the Earth, the pressure is Pressure: Units: p= F⊥ A patm ~ 1.013 x 105 Pa = 1 atm 10 Pascal (Pa) = 1 N/m2 psi (pounds per square inch) atmosphere bar 1 atm = 1.013 × 105 Pa Area of a hand ~ 200 cm2 = 0.02 m2 F = patmA ~ 2000 N on your hand due to air! DEMO: Piston and weight 1 bar = 105 Pa 1 DEMO: Plastic tube with cover Pressure vs. depth Imaginary box of fluid with density ρ with bases of area A and height h Ftop Fbottom = Ftop + mg m = ρAh Example: How deep under water is p = 2 atm? h= pbottom − ptop ρg = (10 1.01 × 105 Pa 3 kg/m 3 )(9.81 m/s ) 2 (i .e ., 1 atm is produced by a 10.3 m high column of water) Two liquids Y and G separated by a thin, light piston (so they cannot mix) are placed in a U-shaped container. What can you say about their densities? A. A. ρG < ρY If If liquid height was higher above A than above B pA > pB Net force → Net flow → • B This is not equilibrium! Water towers Water towers are a common sight in the Midwest… because it’s so flat! h1 Y h3 G h2 • A B. B . ρG = ρY • A = 10.3 m ACT: U tube • B C. ρG > ρY Pressure at A and B must be the same: Since h1 < h3 − h2 h phouse = patm + ρwaterhg ρ Y gh1 + ρG gh2 + patm = ρG gh3 + patm ρYh1 = ρG (h3 − h2 ) p(y) Fluid level is the same everywhere in a connected container (assuming no surface forces) A DEMO: U-tube with water and kerosene y Fbottom Fbottom/top pbottom = ptop + ρ gh Fluid in an open container Pressure is the same at a given depth, independently of the container. h mg Net force must be zero! Pbottom/top = DEMO: Pascal’s vases ⇒ ρ Y > ρG 2 So physics sucks, but how much? Your physics professor sucks on a long tube that rises out of a bucket of water. She can get the liquid to rise 5.5 m (vertically). What is the pressure in her mouth at this moment? Pascal’s principle xB h Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. A. 1 atm B. 0.67 atm xA C. 0.57 atm D. 0.46 atm Pascal’s Principle is most often applied to incompressible fluids (liquids): pmouth + ρwater gh = patm E. 0 atm pmouth = patm − ρwater gh = 10 Pa − (10 kg/m 5 DEMO: Sucking through a hose 3 3 )( 9.8 m/s ) (5.5 m) 2 Increasing p at any depth (including the surface) gives the same increase in p at any other depth = 46100 Pa = 0.46 atm Hydraulic chamber F1 F1 F2 = A1 A2 A F2 = 2 F1 A1 F2 can be very large… ACT: Hydraulic chambers F2 d2 d1 In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference di between the liquid levels. If A2 = 2A1, then: A. A. dA < dB dA A1 A1 ⎛ W = F1d1 = ⎜ F2 ⎜ ⎝ A1 ⎞⎛ A2 ⎞ ⎟⎜ d ⎟=Fd A2 ⎟⎜ 2 A1 ⎟ 2 2 ⎠⎝ ⎠ A10 B. dA = dB A2 C. dA > dB dB No energy is lost: Pressure depends only on the height of the water column above it. A2 M M A10 Incompressible fluid: A1d1 = A2d2 3 Measuring pressure with fluids Barometer Measures absolute pressure Top of tube evacuated (p = 0) Bottom of tube submerged into pool of mercury open to sample (p) p Pressure dependence on depth: h = g ρHg vacuum Vacuum p p=0 =0 Barometer atmosphere Sample p=p at p h h 0 Manometer Measures gauge pressure: pressure relative to atmospheric pressure. p − p0 Δh = Pressure dependence on depth: g ρHg p p patm p0 ∆h Δ A unit for pressure 760 mm Hg = 1 torr = 1 atm 4 ...
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This note was uploaded on 09/21/2011 for the course PHYS 222 taught by Professor Johnson during the Spring '07 term at Iowa State.

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