1 / 21

Fluid Mechanics

Fluid Mechanics. Ellen Akers. Fluids. A fluid is a substance that has the ability to flow and change its shape. Gases and liquids are both fluids. Liquids have a definite volume and are considered to be incompressible. Gases do not have a definite volume. Density.

gisela-hutchinson
Télécharger la présentation

Fluid Mechanics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Fluid Mechanics Ellen Akers

  2. Fluids • A fluid is a substance that has the ability to flow and change its shape. • Gases and liquids are both fluids. • Liquids have a definite volume and are considered to be incompressible. • Gases do not have a definite volume.

  3. Density • The amount of mass per unit volume of a substance. • Equation • ρ : (greek letter “rho”) density • m : mass (in kilograms) • V : volume (in cubic meters, m3) • 1 m3 = 1,000 Liters • SI Unit for mass density is the kilogram per meter cubed, or kg/m3.

  4. Pressure • The amount of force applied over a certain area. • Equation • P : Pressure • F : Force (in Newtons) • A : Area (in meters squared, m2) • The SI unit for pressure is the Pascal (Pa) • 1 Pa = 1 N/m2 • The amount of air pressure at sea level is called one atmosphere (1 atm) • 1 atm = 1.013 x 105 Pa = 760 torr= 760 mmHg = 14.7 psi

  5. Pressure • How much a liquid weighs and how much pressure it exerts depends on its density • Liquids are practically incompressible so pressure only dependent on its density and depth, NOT the shape of the container or size of the bottom of the container • Pressure of a liquid does not depend on volume, weight or total amount • Pressure of water at 1 m is the same in a pool, pond or lake

  6. Pressure • Since weight is 1000 kg x 9.8 N = 9800 N • Density of water is 9800 N/m3 • Water pressure in a pool or lake is 9800 N/m3 x depth • In SI pressure = pascals • 1Pa = 1 N/m2 result would be 9800 Pa • Seawater has a density of 10,000 N/m3

  7. Example 1 The area of the sole of Henry’s foot is 0.024 m2. If he has a mass of 85 kg, how much pressure does Henry’s foot exert on the ground when he is standing on one leg? Solution: P = F/A The force applied is Henry’s weight, Fg mg Fg = (85)(9.8) = 833 N P = 833 / 0.024 P = 3.47 x 104 Pa

  8. Pressure and Net Force • If there is a pressure difference between two sides of an object, it will experience a net force. • Examples • The pressure on the bottom of the red box is greater than at the top. It will feel an upward net force. • When the hot soda can is placed in ice water the gas inside compresses so there is less pressure inside the can. The surrounding air crushes the can.

  9. Pressure vs. Depth • In a fluid of constant density, the pressure increases with the depth below the surface of the fluid. • For example, as a submarine dives deeper underwater, the pressure against the side of the hull increases, and the side of the hull must be strong enough to withstand high pressures. • The absolute pressure in a fluid at a given depth is given by the equation • P : absolute pressure (in Pascals) • P0 : atmospheric pressure (at the surface of the fluid, in Pascals) • ρ : density of the fluid (in kg/m3) • g : gravitational acceleration (g = 9.8 m/s2) • h : depth below the surface of the liquid (in meters)

  10. Example 2 Calculate the absolute pressure at an ocean depth of 1.0 kilometer. Assume the density of the water is 1.025 x 103 kg/m3 and that the atmospheric pressure is 1.01 x 105 Pa. Solution: P = P0 + ρgh P = (1.01 x 105) + (1.025 x 103)(9.8)(1000) P = 1.015 x 107 Pa

  11. Buoyancy Ellen Akers

  12. Volume Displacement • The volume of an irregular object is equal to the volume that is displaced when it is placed in a liquid. • Archimedes is said to have discovered this principle when commissioned by the king of Syracuse to determine whether or not his crown was made of pure gold.

  13. Archimedes’ Principle • The upward force exerted on an object placed in a fluid is equal to the weight of the fluid displaced by the object. • This upward force is called the Buoyant Force, or Buoyancy. • The maximum amount of buoyancy that can be exerted on an object is given by the equation: • FB : buoyant force (in Newtons) • ρ : density of the fluid (in kg/m3) • V : volume of fluid displaced (in m3) • g : gravitational acceleration (g = 9.8m/s2)

  14. Buoyancy • The apparent loss of weight of objects when submerged in a liquid • Water exerts and upward force opposite to gravity • Greater forces at great depths • Forces of sides are equal • Weight > buoyant force – sinks • Weight = buoyant force – remain at any level like a fish • Weight < buoyant force – rise to top and float

  15. Buoyancy • What determines whether an object will sink or float? • If the density of the object is greater or less than the density of the fluid. • How does a large metal ship float on water? • It is filled mostly with air, which makes the overall density less than that of water. • It displaces enough water so that the buoyant force balances the weight of the ship. • If an object is floating on the surface of the fluid, then the buoyant force is equal to the weight of the object. • Fg = FB • If an object has sunk and is resting at the bottom of the fluid, there is still an upward buoyant force and the object will feel “lighter”. • The apparent weight of an object in a fluid can be found by subtracting the buoyant force from the gravitational force. • Apparent weight = Fg – FB

  16. Principle of Floatation • A floating object displaces a weight of fluid = to its own weight • Every ship must be designed to displace a weight of water equal to its own weight

  17. Example 3 A steel ship has a total mass of 3,000,000 kg. What minimum volume of sea water must the ship displace in order to float?(the density of sea water is 1,024 kg/m3) Solution: Fg = FB because the boat is floating Fg = mg mg = ρVg FB = ρVg V = mg / ρg V = m / ρ V = (3x106)/(1024) V = 2,930 m3

  18. Example 4 What is the apparent weight of a 60 kg solid iron cannonball with a radius of 12 cm that is resting on the bottom of a freshwater pond at a depth of 18 meters?(The density of fresh water is 1,000 kg/m3 and the density of iron is 7.86 x 103 kg/m3) Solution:Apparent Weight = Fg – FB V = 4/3πr3 = 4/3π(.12)3 = mg – ρVg V = 7.238x10-3 m3 = (60)(9.8) – (1000)(7.238x10-3)(9.8) Apparent Weight = 517 N

  19. Motion of Fluids Ellen Akers

  20. Fluid Flow • Laminar Flow • Each particle of a fluid follows the path of the one before it. • The path is called a “streamline”. • Turbulent Flow • Particles of a fluid exhibit random motion. • Irregular motions of the fluid are called “eddy currents”.

  21. Bernoulli’s Principle • The pressure in a fluid is indirectly related tothe speed of the fluid. • Where the speed of a fluid is low the pressure is high, and where the speed of a fluid is high the pressure is low. • The wing of an airplane (called an “airfoil”) uses this principle to generate lift. • The air flowing over the top of the wing moves faster than the air underneath. • This causes there to be a higher pressure under the wing which causes an upward force.

More Related