2 DensityRecall 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
3 Pressure in Fluids Fluids exert a pressure in all directions A fluid at rest exerts pressure perpendicular to any surface it contactsThe pressure at equal depths within a uniform fluid is the sameSI Unit for Pressure is Pa1 Pa= 1 N/m21 atm= kPa=760 mm-Hg
4 Pressure in FluidsGauge Pressure is a measure of the pressure over and above the atmospheric pressurei.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
5 Pressure in FluidsPascal’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
6 BuoyancyBuoyant force is the force acting on an object that is immersed in a fluidArchimedes Principle: The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the objectSince the buoyant force acts opposite of gravity, an object seems to weigh less in a fluidApparent Weight= Fg-FBFb = Buoyant ForceFg = Gravity
7 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 equilibriumFBIf Fb=Fg then the object will be in equilibrium and will FLOAT!Fg
8 Cubes floating in a fluid 70% Submerged20% Submerged100% Submerged
9 Density determines depth of submersion This equation gives the percent of the object’s volume that is submergedVf is the volume of fluid displacedVo is the total volume of the objectρo is the density of the objectρf is the density of the fluid
11 Continuity EquationContinuity 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 increasesThe quantity Av is the volume rate of flow
12 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
14 Bernoulli’s Equation The kinetic energy of a fluid element is: The potential energy of a fluid element is:
15 Bernoulli’s EquationThis 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.
16 Applications of Bernoulli’s Equation If a hole is punched in the side of an open container, the outside of the hole and the top of the fluid are both at atmospheric pressure.Since the fluid inside the container at the level of the hole is at higher pressure, the fluid has a horizontal velocity as it exits.
17 Applications of Bernoulli’s Equation If the fluid is directed upwards instead, it will reach the height of the surface level of the fluid in the container.
18 Sample Problem p. 306 #40What 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 wingThe height difference between the top of the wing and the bottom is negligible
19 Sample Problem p.306 #40The 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 upwardIf we know the difference in pressure we can use that to find the force
21 P.306 #43Water 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.
22 Find the flow velocity at the top A1 is area of first pipe= πr2 = 1.96x10-3 m2A2 is area of second pipe= πr2 = 5.31x10-4 m2V1= 0.6 m/s
23 Find pressure at the top P2= 1.86 x 105 Pa= 1.8 atm
24 Sample Problem p.305 #37What 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 0Point 2 is the top of the spray, so v=0 , P= atmospheric pressure, height = 12m
25 Sample Problem p.305 #37Remember that Gauge Pressure is the pressure above atmosphericpressure. So to get gauge pressure, we need to subtract atmosphericPressure from absolute pressure.