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Ch. 15 Fluids

Ch. 15 Fluids. Fluids. Substances that flow readily, usually from high pressure to low pressure. They take the shape of the container rather than retain their shape. We refer both liquids and gases as fluids. Density. Is the mass of a substance divided by the volume of that substance.

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Ch. 15 Fluids

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  1. Ch. 15 Fluids

  2. Fluids • Substances that flow readily, usually from high pressure to low pressure. • They take the shape of the container rather than retain their shape. • We refer both liquids and gases as fluids.

  3. Density • Is the mass of a substance divided by the volume of that substance.

  4. Density Equation Mass Density = Volume

  5. m D V

  6. Density Units • Metric: kg / m3 or g / cm3 • English: lbm / ft3

  7. Sample Problem: What is the volume of my 2.16 g titanium wedding band if the density is 4.50g/ cm3?

  8. G: m = 2.16 g, D = 4.50 g/cm3 U: V = ? E: D = m / V  V = m/D S: V = 2.16 / 4.50 g/cm3 S: V = 0.48 cm3

  9. Pressure • is defined as force per unit of area.

  10. Each book exerts the same force, but which one exerts a greater pressure?

  11. Equation Force Pressure = Area

  12. F A P

  13. Units of Pressure • English: lb / ft2, lb / in2, or p.s.i. • Metric: N/m2 or Pascal (Pa)

  14. Sample Problem: If an airplane window that is 40 cm by 60 cm feels a pressure of 40,000 Pa. How much force is acting on the window?

  15. G: L = 40 cm = 0.4 m w = 60 cm = 0.6 m P = 40,000 Pa U: F = ? E: F = PA

  16. We don’t have area (A), A = Lw A = (.4)(.6) A = 0.24 m2

  17. F = (40,000)(.24) F = 9600 N

  18. Atmospheric Pressure The pressure of the earth’s atmosphere pushing down on you.

  19. Earth's atmosphere is pressing against each square inch of you with a force of 1 kilogram per square centimeter (14.7 pounds per square inch).

  20. Sample ProblemCalculate the net force on an airplane window if cabin pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

  21. Since pressure acts equally in all directions, gauge pressure is used to deal with this. • If you have a flat tire, is there a pressure in the tire or is the pressure zero? • Yes, it is equal to atmospheric pressure.

  22. Gauge Pressure (Pg) • Is the pressure inside an inflated object, that is above the atmospheric pressure. • Pg = P – Patm • If a tire is inflated to 35 psi, that means it is 35 psi above the atmospheric pressure. So the pressure inside the tire is actually 35 psi + 14.7 psi.

  23. Pressure increases with depth. The deeper the body of water or you go, the more water there is on top of you per square inch.

  24. Pressure depth relationship Phyd = rgh r – weight density of liquid

  25. The pressure of a liquid is the same at any given depth regardless the shape of the container.

  26. The pressure P is the same on the bottom of each vessel.

  27. Absolute Pressure • Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure. • pabs = patm + rgh • p2 = p1 + rgh

  28. Water seeks its own level

  29. Pascal’s Principle Pressure applied to a fluid in a closed container is transmitted equally

  30. to every point of the fluid and to the walls of the container.

  31. When the area of the 2 “pistons” are the same the input force is the same as the output force.

  32. What if you increased the area of the 2nd “piston” by 25? By Pascal’s Principle: P1 = P2

  33. By the definition of Pressure: F1/A1 = F2/A2 So if the 2nd area is increased by a factor of 25.

  34. In order to maintain equality, the force has to be increased by the same factor.

  35. Pascal Principle

  36. Distance each piston moved is different. F1d1 = F2d2

  37. Sample problem: The small piston on a hydraulic lift has an area of 0.20 m2. A car weighing 1.2 x 104 N sits on the rack with a large piston. If the area of the large piston is 0.9 m2, how much force must be applied to the small piston to support the car?

  38. G: A1 = 0.2 m2, F2 = 1.2 x 104 N, A2 =0.9 m2 U: F1 = ? E: F1/A1 = F2/A2 F1 = F2A1/A2

  39. S: F1=(1.2 x 104)(0.2)/(0.9) S: F1 = 2666.7 N

  40. How far will the 2nd piston rise if the 1st piston moves 1.2 m?

  41. The distance the 2nd piston moves is equal to the reciprocal of the factor of the ↑ or ↓ in area times the distance the 1st piston moves.

  42. Pascal’s Principle applies to all fluids, gases as well liquids.

  43. Buoyancy • The apparent loss of weight of an object that is submerged in a fluid.

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