lecturenotes-March24

lecturenotes-March24 - Summary of Chapter 14 Fluids In this...

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Summary of Chapter 14 Fluids In this chapter we will explore the behavior of fluids. In particular we will study the following: Static fluids: Pressure exerted by a static fluid Methods of measuring pressure Pascal’s principle Archimedes’ principle, buoyancy
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Fluids: Here is what we will cover p at h = p 0 + ρ gh Pressure applied to a confned Fuid increases the pressure throughout by same amount Pascal’s Principle p out = p in F out = F in A out A in Buoyancy and Archimedes’ Principle Buoyant ±orce - is equal to the weight o² Fuid displaced by the object - is directed UPWARDs - Does not depend on the shape o² the object ONLY volume. - Applies to partially (Foat) or completely immersed object F B = m F g UPWARD
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The collisions of gas molecules on the wall of the tire keep it inflated Pressure P = F A “collisional force” Uniform force on flat area Units: pascal = Pa = N/m 2 1 atmosphere = 1 atm = 1.01 × 10 5 Pa = 760 torr = 14.7 lb/in 2 Hydrostatic Pressure: Water applies a force perpendicular to all of the surfaces in the pool, including the swimmer, the walls of the pool etc.
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Density density = ρ = m V Note: density solid > liquid > gas air = 1.21 kg/m 3 wood = 550 kg/m 3 water = 1000 kg/m 3 Al = 2700 kg/m 3 = 2.7 g/cm 3 Cu = 8960 kg/m 3 = 8.9 g/cm 3 “Vacuum” You can’t compress a liquid!!
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Fluids at Rest Static Equilibrium F = 0 a = 0 F = 0 F bottom F top mg = 0 p bottom A = p top A + mg p bottom = p top + mg A = p top + mg A h h p bottom = p top + gh m V p bottom = p top + ρ gh Pressure depends on depth NOT horizontal dimensions p at h = p atm + gh NOTE: p at h > p atm for h down p at h < p atm for h u p
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Fluids at rest: U‐Tube Consider the U-tube shown at the right. I contains two liquids in equilibrium: water with a density of 1000 kg/ m 3 and oil of unknown density. Measurements gives l=135 mm and d=12.3 mm. What is the density?
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Fluids at rest: U‐Tube Consider the U-tube shown at the right. I contains two liquids in equilibrium: water with a density of 1000 kg/ m 3 and oil of unknown density. Measurements gives l=135 mm and d=12.3 mm. What is the density? Both ends are open so the pressure above them is atmospheric Define the top of the oil column as the zero of height. The pressure at the oil/water interface on the left has to be identical to the pressure at the same height on the right. p Left = p 0 + ρ oil g l + d ( ) p right = p 0 + water g d ( ) oil g l + d ( ) = water g d ( ) oil = water d l + d ( )
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Examples 1. What is the pressure head at the bottom of a 98 ft (30 m) water tower? Δ p = p at h p atm = ρ gh = 1000 kg / m 3 ( ) 9.8 m / s 2 ( ) 30 m ( ) = 2.9 × 10 5 Pa = 42 psig 56.7 psi absolute ( ) Gauge pressure vs absolute pressure 2. What is the net force on Grand Coulee dam (width 1200 m - height 150 m)? p at h = p atm + gh dF = gy ( ) dA F = gy ( ) W dy 0 D = 1 2 gWD 2 = 1.3 × 10 11 N F = 2.9 × 10 10 lb
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More Examples 3. At what depth is the pressure two-times that of atmosphere?
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lecturenotes-March24 - Summary of Chapter 14 Fluids In this...

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