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Chapter 10 Density Recall that the density of an object is its mass per unit volume (SI unit is kg/m3) The specific gravity of a substance is its density expressed in g/cm3 m V Pressure in Fluids Fluids exert a pressure in all directions F Force P A Area SI Unit for Pressure is Pa 1 Pa= 1 N/m2 1 atm= 101.3 kPa=760 mm-Hg A fluid at rest exerts pressure perpendicular to any surface it contacts The pressure at equal depths within a uniform fluid is the same P gh (density )( g )( depth) Pressure in Fluids Gauge Pressure is a measure of the pressure over and above the atmospheric pressure i.e. the pressure measured by a tire gauge is gauge pressure. If the tire gauge registers 220 kPa then the absolute pressure is 321 kPa because you have to add the atmosphere pressure (101 kPa) If you want the absolute pressure at some depth in a fluid then you have to add atmosphere pressure P Po gh Pressure in Fluids Pascal’s Principle: Pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and to the walls of the container F1 F2 A1 A2 Buoyancy Buoyant force is the force acting on an object that is immersed in a fluid Archimedes Principle: The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the object Since the buoyant force acts opposite of gravity, an object seems to weigh less in a fluid Apparent Weight= Fg-FB Fb = Buoyant Force Fg = Gravity Sinking vs Floating Think back to free body diagrams If the net external force acting on an object is zero then it will be in equilibrium FB If Fb=Fg then the object will be in equilibrium and will FLOAT! Fg Cubes floating in a fluid 70% Submerged 20% Submerged 100% Submerged Density determines depth of submersion This equation gives the percent of the object’s volume that is submerged Vf is the volume of fluid displaced Vo is the total volume of the object ρo is the density of the object ρf is the density of the fluid o Vo f Vf Summary of Floating For an object to float FB= Fg ρo ≤ ρf If ρo = ρf then the Object will be Completely submerged But not sinking. If ρo is less than ρf Then the amount submerged can be found with Vf Vo o f Continuity Equation •Continuity tells us that whatever the volume of fluid in a pipe passing a particular point per second, the same volume must pass every other point in a second. •If the cross-sectional area decreases, then velocity increases Continuity Equation A1v1 A2 v2 The quantity Av is the volume rate of flow Bernoulli’s Principle The pressure in a fluid decreases as the fluid’s velocity increases. Fluids in motion have kinetic energy, potential energy and pressure How do planes fly? Bernoulli’s Equation The kinetic energy of a fluid element is: The potential energy of a fluid element is: PE mgh ( V ) gh Bernoulli’s Equation This equation is essentially a statement of conservation of energy in a fluid. Notice that volume is missing. This is because this equation is for energy per unit volume. Bernoulli' s Equation 1 2 1 2 P1 v1 gh1 P2 v2 gh2 2 2 Sample Problem p. 306 #40 What is the lift (in newtons) due to Bernoulli’s principle on a wing of area 80 m2. If the air passes over the top and bottom surfaces at speeds of 350 m/s and 290 m/s, respectively. Let’s make point 1 the top of the wing and point 2 the bottom of the wing The height difference between the top of the wing and the bottom is negligible 1 2 1 2 P1 v1 gh1 P2 v2 gh2 2 2 Sample Problem p.306 #40 The net force on the wing is a result of the difference in pressure between the top and the bottom. P1 is exerted downward, P2 is exerted upward If we know the difference in pressure we can use that to find the force 1 2 1 2 P1 v1 P2 v2 2 2 1 2 1 2 1 P2 P1 v1 v2 (1.29kg / m3 )(340 2 290 2 ) 20318 Pa upward 2 2 2 Sample Problem p.306 #40 P2-P1=20318 Pa F ( P2 P1 ) A (20318Pa)(80m2 ) 1.63x106 N upward P.306 #43 Water at a pressure of 3.8 atm at street level flows into an office building at a speed of 0.60 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6 cm in diameter by the top floor, 20 m above street level. Calculate the flow velocity and the pressure in such a pipe on the top floor. Ignore viscosity. Pressures are gauge pressures. Find the flow velocity at the top Continuity Equation A1v1 A2 v2 A1 is area of first pipe= πr2 = 1.96x10-3 m2 A2 is area of second pipe= πr2 = 5.31x10-4 m2 V1= 0.6 m/s A1v1 (1.96 x103 m 2 )(0.6m / s) m v2 2.2 4 2 A2 5.31x10 m s Find pressure at the top 1 2 1 2 P1 v1 gh1 P2 v2 gh2 2 2 1 2 1 2 P2 P1 v1 v2 gh2 2 2 N N N P2 3.84 x10 Pa 180 2 2420 2 196200 2 m m m 5 P2= 1.86 x 105 Pa= 1.8 atm Sample Problem p.305 #37 What gauge pressure in the water mains is necessary if a fire hose is to spray water to a height of 12.0 m? Let’s make point 1 as a place in the water main where the water is not moving and the height is 0 Point 2 is the top of the spray, so v=0 , P= atmospheric pressure, height = 12m 1 2 1 2 P1 v1 gh1 P2 v2 gh2 2 2 Sample Problem p.305 #37 P1 P 2 gh Patm (1000kg/m )(9.81m/s )(12m) 3 2 Remember that Gauge Pressure is the pressure above atmospheric pressure. So to get gauge pressure, we need to subtract atmospheric Pressure from absolute pressure. Gauge Pressure P1 P atm 1.2 x105 Pa