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Fluid Mechanics

Fluid Mechanics. Chapter 9. Defining a Fluid. A fluid is a nonsolid state of matter in which the atoms or molecules are free to move past each other, as in a gas or a liquid. Both liquids and gases are considered fluids because they can flow and change shape.

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Fluid Mechanics

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  1. Fluid Mechanics Chapter 9

  2. Defining a Fluid • A fluid is a nonsolid state of matter in which the atoms or molecules are free to move past each other, as in a gas or a liquid. • Both liquids and gases are considered fluids because they can flow and change shape.

  3. 9-1: Fluids and Buoyant Force • Liquids have a definite volume • Gasses do not • Liquids do not have a definite shape

  4. Density and Buoyant Force • In general, density is a measure of how much there is of a quantity in a given amount of space. • Quantity can be anything; # of trees or people, or amount of mass or energy

  5. Mass Density • The concentration of matter of an object is called the mass density. (Amount of mass per unit volume of a substance) • Represented with the Greek symbol, ρ(rho) Mass Density = mass/volume ρ = m/v S.I. Unit Kg / m3

  6. Common fluids and solids • Solids and liquids tend to be almost incompressible. i.e., change very little with changes in pressure.

  7. Buoyant Force • Buoyant Force – a force that acts upward on an object submerged in a liquid or floating on a liquid’s surface. • This force can keep objects afloat • Acts in direction OPPOSITE of gravity

  8. Buoyant Force

  9. Archimedes Principle • Law of Archimedes: The buoyant force is equal to the weight of the replaced liquid or gas.

  10. Buoyant Force and Archimedes’Principle • The Brick, when added will cause the water to be displaced and fill the smaller container. What will the volume be inside the smaller Container? The same volume as the brick!

  11. Buoyant Force and Archimedes’ Principle Archimedes’ principle describes the magnitudeof a buoyant force.Archimedes’ principle: Any object completely orpartially submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object.FB = Fg (displaced fluid) = mfgmagnitude of buoyant force = weight of fluid displaced

  12. Buoyant Force The raft and cargo are floating because their weight and buoyant force are balanced.

  13. Buoyant Force • Now imagine a small hole is put in the raft. • The raft and cargo sink because their density is greater than the density of the water. • As the volume of the raft decreases, the volume of the water displaced by the raft and cargo also decreases, as does the magnitude of the Buoyant force.

  14. Buoyant Force • For a floating object, the buoyant force equals the object’s weight. • The apparent weight of a submerged object depends on the density of the object. • For an object with density ρosubmerged in a fluid of density ρf, the buoyant force FBobeys the following ratio: Fg(object) = ρo FBρf

  15. Example A bargain hunter purchases a “gold” crown at a flea market. After she gets home, she hangs the crown from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, and the scale reads 6.86 N. Is the crown made of pure gold? Explain.

  16. Solution Choose your equations: Fg – FB = apparent weight Fg = ρo FB ρf Rearrange your equations: FB = Fg – (apparent weight) Ρo = Fg (ρf) FB

  17. Plug and Chug FB = 7.84 N – 6.86 N = 0.98 N Ρo = 7.84 N (1.00 10 kg/m) = 8.0 x 103 kg/m3 0.98 N • From the table in your book, the density of gold is 19.3 x 103 kg/m3. • Because 8.0 x 103 kg/m3 < 19.3 x 103 kg/m3, the crown cannot be pure gold.

  18. HOMEWORK • Page 324 • Practice 9A # 1 and 3 • Section Review # 1-3

  19. QOTD: Wed 1/13/2010 Get a calculator from the red bin and complete the following: 1: Calculate the actual weight, buoyant force, and apparent weight of a 5.0 X 10-5 m3 iron ball floating at rest in mercury. (density of mercury is 13.6 x 103 kg/m3) 2: How much of the ball’s volume is immersed in mercury?

  20. QOTD: Wed 1/13/2010 • Calculate the actual weight, buoyant force, and apparent weight of a 5.0 X 10-5 m3 iron ball floating at rest in mercury. (density of mercury is 13.6 x 103 kg/m3) 3.86N, 3.86 N, 0 N • How much of the ball’s volume is immersed in mercury? 2.89 x 10-5 m3

  21. Fluid Pressure and Temperature • Objectives: • Calculate pressure exerted by a fluid • Calculate how the pressure varies with depth in a fluid • Describe fluids in terms of temperature

  22. PRESSURE • THE MAGNITUDE OF THE FORCE ON A SURFACE PER UNIT AREA • PRESSURE = FORCE/AREA P = F A

  23. PRESSURE SI unit is the pascal (Pa) 1 pascal is equal to 1 N/m2 The pressure of the atmosphere at sea level is about 105 Pa * this amount of air pressure under normal conditions is the basis for another unit, the atmosphere (atm) The absolute air pressure inside a typical automobile tire is about 300,000 Pa, or about 3 atm.

  24. Applied Pressure • Transmitted equally throughout a fluid • In general, if the pressure in a fluid is increased at any point in a container, the pressure increases at all points inside the container by exactly the same amount. • This is noted by PASCAL • Which leads us to PASCAL’S PRINCIPLE

  25. 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.

  26. Pascal’s Principal • A hydraulic lift uses this principle. • A small force, F1, applied to a small piston of area A1 causes a pressure increase in a fluid, such as oil. • This increase in pressure, Pinc, is transmitted to a larger piston of area A2and the fluid exerts a force, F2, on this piston. See figure 9-7 in text

  27. Applying Pascal’s principle and the definition of pressure gives us the following equation : Pinc = F1 = F2 A1 A2 Rearranging the equation to solve for F2: From this equation, we see that F2 is greater than F1 by a factor equal to the ratio of the areas of the 2 pistons. F2 =F1A2 A1

  28. Pascal’s Principle Because the pressure is the same on both sides of the enclosed fluid in a hydraulic lift, a small force on the smaller piston (left) produces a much larger force on the larger piston (right).

  29. Pressure Pressure varies with depth in a fluid • Water pressure increases with depth because the water at a given depth must support the weight of the water above it.

  30. Fluid pressure as a function of depth • By using the symbol ρo for the atmospheric pressure at the surface, we can express the total pressure, or absolute pressure, at a given depth in a fluid of uniform density, ρ, as follows: ρ = ρo+ ρgh Absolute pressure = atmospheric pressure + (density x free fall accel x depth)

  31. Temperature • The temperature in a gas can be understood as what is happening to the atomic scale. • The higher the temperature of a gas, the fasterthe particles move. • The faster the particles move, the higher the rate of collisions against the walls of the container. • More momentum is transferred to container walls in given time interval, thus an increase in pressure results.

  32. SI units for Temperature:Kelvins and degrees CelsiusTo convert from Celsius to Kelvin, add 273.Ex: Room temp is about 20° Celsius or 293 k.

  33. HOMEWORK Pg 327 # 1-2 Pg 330 # 1 Pg 331 # 1, 3-4

  34. QOTD Friday: 1/15/2010

  35. FLUIDS IN MOTION • FLUID FLOW: • When a fluid is in motion, the flow can be characterized in one of two ways. The flow is said to be laminar if every particle that passes a particular point moves along the same smooth path traveled by the particles that passed that point earlier. • This path is called streamline. • Different streamlines CANNOT cross each other.

  36. Fluid Flow Cont. Flow of fluid becomes irregular, or turbulent, above a certain velocity or under conditions that can cause abrupt changes in velocity. • obstacles or sharp turns • Irregular motion of the fluid – eddy currents • Turbulent flow • Ex: water in wake of ship or in air currents of severe thunderstorm

  37. Ideal Fluid : fluid that has no internal friction or viscosity and is incompressible • Viscosity:amount of internal friction within a fluid. • Can occur when one layer of fluid slides past another.

  38. Principles of Fluid Flow • The mass flowing into the pipe MUST equal the mass flowing out of the pipe in the same time interval. Thus, m1 = m2 • Remember…m = pv • Volume of a cylinder: V = A (delta X) So, p1V1 = p2V2 p1A1 (Delta X 1) = p2A2 (Delta X 2)

  39. p1V1 = p2V2 p1A1 (Delta X 1) = p2A2 (Delta X 2) *The length of the cylinder is also the distance the fluid travels, which is equal to the speed of flow multiplied by the time interval (delta X = v delta t) • So, p1A1 v1 (Delta t) = p2A2 v2 (Delta t) ** for an ideal fluid, both the time interval and the density are the same on each side of the equation, so they cancel each other out. p1A1 v1 (Delta t) = p2A2 v2 (Delta t)

  40. Continuity Equation • After canceling each other out, we are left with the CONTINUITY EQUATION: A1v1 = A2v2 Simply state: area x speed in region 1 = area x speed in region 2

  41. Flow rate is CONSTANT throughout a pipe • Pressure in a fluid is related to the speed of flow. • BERNOULLI’S PRINCIPAL: The pressure in a fluid decreases as the fluid’s velocity increases

  42. Bernoulli’s Equation cont. • Explains the conservation of energy in fluids. Expressed as: P + ½ρv2 + ρgh = constant Pressure +KE per unit volume + PEgper unit volume = constant along a given streamline

  43. Homework • Pg 337: Practice #1 • Section Review # 1

  44. QOTD Tuesday 1/19/2010 • What is the continuity equation? • Using the equation, if a pipe narrows from a cross section of 2.0 m2 to 0.30 m2 and the speed through the wider area is 8.0 m/s, what is the speed of the water flowing through the narrow part?

  45. 9-4: Properties of Gases • Volume, Pressure, and Temperature are the three variables that completely describe the macroscopic state of an ideal gas.

  46. IDEAL GAS LAW • PV = NkBT Pressure x volume = # gas particles x Boltzmann’s constant x temperature KB = 1.38 x 10-23 J/K Temperature MUST be in Kelvins!!!!

  47. IDEAL GAS LAW • PV = nRT Pressure x volume = # moles of gas x molar gas constant x temperature n = # moles of gas Where one mole = 6.02 x 1023 particles R = 8.31 J/(mol * K)

  48. IDEAL GAS LAW • If the # of gas particles remains constant, then… P1V1 = P2V2 T 1 T2

  49. Sample • A sealed tank with a volume of 0.10 m3 contains air at 27 degrees Celsisus under pressure 18000 Pa. The valve can withstand pressures up to 50000. Will the valve hold if the air inside the tank is heated to 227 degrees C?

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