Presentation on theme: "Chapter 11 Fluids. Density and Specific Gravity The density ρ of an object is its mass per unit volume: The SI unit for density is kg/m 3. Density is."— Presentation transcript:
Density and Specific Gravity The density ρ of an object is its mass per unit volume: The SI unit for density is kg/m 3. Density is also sometimes given in g/cm 3 ; to convert g/cm 3 to kg/m 3, multiply by 1000. Water at 4°C has a density of 1 g/cm 3 = 1000 kg/m 3. The specific gravity of a substance is the ratio of its density to that of water. (10-1)
Pressure in Fluids Pressure is defined as the force per unit area. Pressure is a scalar; the units of pressure in the SI system are pascals: 1 Pa = 1 N/m 2 Pressure is the same in every direction in a fluid at a given depth; if it were not, the fluid would flow.
Atmospheric Pressure and Gauge Pressure At sea level the atmospheric pressure is about this is called one atmosphere (atm). 1 atm = 14.7psi = 760 torr = 760mm Hg Another unit of pressure is the bar: Standard atmospheric pressure is just over 1 bar. This pressure does not crush us, as our cells maintain an internal pressure that balances it.
Pressure at a Depth P = F = mg = ρVg = ρAhg = ρhg A A A A 330 m Find the pressure on the bottom of the submarine due to the water above it. Find the pressure on the bottom of the submarine due to the water And air above it.
10-4 Atmospheric Pressure and Gauge Pressure Most pressure gauges measure the pressure above the atmospheric pressure – this is called the gauge pressure. The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
Archimedes’s Principle Any object completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the fluid displaced by the object.
Buoyant Force The upward force is called the buoyant force The physical cause of the buoyant force is the pressure difference between the top and the bottom of the object
Buoyant Force, cont. The magnitude of the buoyant force always equals the weight of the displaced fluid For a floating object, the buoyancy force must equal the weight of the object. B = W Some objects may float high or low
Example: A balloon having a volume of 1.5 cubic meters is filled with ethyl alcohol and is tethered to the bottom of a swimming pool. Calculate the tension in the cord tethering it to the bottom of the swimming pool. Fy = B – T – W = 0 Therefore, T = B – W T = water Vg – alcohol Vg T = ( water – alcohol )Vg T = (1000kg/m 3 – 806kg/m 3 ) (1.5m 3 )(9.8m/s 2 ) T = 2851.8 N B TW FBD
11.6 Archimedes’ Principle Example 9 A Swimming Raft The raft is made of solid square pinewood. Determine whether the raft floats in water and if so, how much of the raft is beneath the surface.
Example 7 A Car Lift The input piston has a radius of 0.0120 m and the output plunger has a radius of 0.150 m. The combined weight of the car and the plunger is 20500 N. Suppose that the input piston has a negligible weight and the bottom surfaces of the piston and plunger are at the same level. What is the required input force?
EQUATION OF CONTINUITY The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow. SI Unit of Mass Flow Rate: kg/s
11.8 The Equation of Continuity Incompressible fluid: Volume flow rate Q:
11.8 The Equation of Continuity Example 12 A Garden Hose A garden hose has an unobstructed opening with a cross sectional area of 2.85x10 -4 m 2. It fills a bucket with a volume of 8.00x10 -3 m 3 in 30 seconds. Find the speed of the water that leaves the hose through (a) the unobstructed opening and (b) an obstructed opening with half as much area.
Bernoulli’s Equation Relates pressure to fluid speed and elevation Bernoulli’s equation is a consequence of Conservation of Energy applied to an ideal fluid Assumes the fluid is incompressible and nonviscous, and flows in a nonturbulent, steady-state manner
Bernoulli’s Equation, cont. States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline
Example: Water is contained in a tank. The water level is 5 meters above a hole in the tank where water exits through a hole to the atmosphere as shown below. The diameter of the hole is 0.01 meters, and the diameter of the tank is 3 meters. The tank is open to atmosphere. Calculate the velocity of the water exiting the tank. The pressure at the top of the tank is equal to the pressure at the exit since they are both open to atmosphere. Therefore P t and P e cancel out. Since the tank diameter is so much larger than the exit hole, the velocity of the water level drop can be approximated as zero. The exit height is taken as zero, therefore y e = 0 The equation above reduces to: Where y t =h v e = 9.9m/s