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Fluid Mechanics. Chapter 10. Density. Recall that the density of an object is its mass per unit volume (SI unit is kg/m 3 ) The specific gravity of a substance is its density expressed in g/cm 3. Pressure in Fluids. Fluids exert a pressure in all directions
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Fluid Mechanics 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
Pressure in Fluids • Fluids exert a pressure in all directions • A fluid at rest exerts pressure perpendicular to any surface it contacts • The pressure at equal depths within a uniform fluid is the same SI Unit for Pressure is Pa 1 Pa= 1 N/m2 1 atm= 101.3 kPa=760 mm-Hg
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
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
Fb = Buoyant Force Fg = Gravity 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
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
70% Submerged 20% Submerged 100% Submerged Cubes floating in a fluid
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
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 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
Bernoulli’s Equation The kinetic energy of a fluid element is: The potential energy of a fluid element is:
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.
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
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
Sample Problem p.306 #40 • P2-P1=20318 Pa
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 • 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
Find pressure at the top 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
Sample Problem p.305 #37 Remember that Gauge Pressure is the pressure above atmospheric pressure. So to get gauge pressure, we need to subtract atmospheric Pressure from absolute pressure.