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Chapter 9.1 Review Fluids

Chapter 9.1 Review Fluids. 1. What are the properties of a fluid? What states of matter are fluids?. No definite shape. The particles are free to flow around one another. Liquids and gases. 2. What is the volume flow rate in m 3 /s if a 2 m 3 barrel is filled in 2 minutes?.

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Chapter 9.1 Review Fluids

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  1. Chapter 9.1 ReviewFluids

  2. 1. What are the properties of a fluid? What states of matter are fluids?

  3. No definite shape. The particles are free to flow around one another.Liquids and gases.

  4. 2. What is the volume flow rate in m3/s if a 2 m3 barrel is filled in 2 minutes?

  5. Q = Av But Q = V/t, also.Q = 2 m3 / 120 sQ = 0.0167 m3 / s

  6. 3. At the flow rate found in question 2, with what speed is the water leaving a hose with an area of 2.95 x 10-3 m2?

  7. Q = Av0.0167 = 2.95 x 10-3 vv = 5.65 m/s

  8. 4. If the area of the hose in question 3 was increased to 7.35 x 10-3 m2, how does that affect the flow rate? How does it affect the speed of the water leaving the hose?

  9. It doesn’t affect the volume flow rate at all.Q = Av0.0167 = 7.35 x 10-3 vv = 2.27 m/s

  10. 5. What is Archimedes’ Principle?

  11. The buoyant force on a body in a fluid is equal to the weight of the fluid displaced.

  12. 6. A 10 N block of wood floats in water. What is the buoyant force?

  13. 10 N, the buoyant force on a FLOATING object is always just its weight.

  14. 7. A person has a volume of one cubic meter and a density of 925 kg/m3. If this person is in fresh water, will they float? If so, what volume of water will they displace? What volume of “person” is still above water?

  15. They will float. (Density is less that the water.)rVg person = rVg water rVg = rVg925(1)10 = 1000 V 10V = 0.925 m31.000 - 0.925 = 0.075 m3

  16. 8. If the person in question 7 were in salt water (density = 1025 kg/m3), how would the answers in question 7 change?

  17. They would still float.rVg person = rVg water rVg = rVg925(1)10 = 1025 V 10V = 0.902 m31.000 - 0.902 = 0.098 m3

  18. 6. What is the air pressure in Pascals at sea level? In bars? In atmospheres? In Torr?

  19. 1.013 x 105 Pa1 bar1 atmosphere760 Torr

  20. 7. What is the absolute pressure in Pascals 30 meters below the surface of fresh water? What is the gauge pressure at the same level?

  21. P2 = P1 + rghP2 = 105 + 1000(10)30P2 = 400000 Pagauge pressure is just rgh, 300000 Pa

  22. 8. If the pressure on a 0.025 m2 area is 3 x 105 Pa, what is the force on that surface?

  23. P = F/A, so P x A = FP x A = F3 x 105 Pa x 0.025 m2 =7500 N

  24. 9. What is Pascal’s Principle?

  25. Any change in pressure anywhere in a contained fluid is felt everywhere in the fluid.

  26. 10. A hydraulic system supports a car with a mass of 1500 kg on a cylinder with an area of 0.8 m2. What force must be applied to a cylinder with an area of 0.1 m2 to support the car? What is the “trade-off” in this situation?

  27. F/A = F/A1500 kg x 10 /0.8 m2 = F/ 0.1 m2F = 1875 NThe trade is distance to get force.

  28. 11. What is Bernoulli’s Principle? Give one example.

  29. A moving fluid applies less pressure. Shower curtain moving toward you in the shower.

  30. 12. A tube carrying water has an area of 0.05 m2. The water is moving at 0.1 m/s and is at a pressure of 2 x 105 Pa. If the tube reaches a constriction where the area becomes 0.01 m2, what is the new velocity and what is the new pressure?

  31. Av = Av0.05 x 0.1 = 0.01 x vv = 0.5 m/sP1 + 1/2 ρv12 + ρgh1= P2 + 1/2 ρv22 + ρgh22 x 105 + 1/2 (1000)0.12 = P2 + 1/2 (1000)0.52P2 = 199880 Pa

  32. 13. Pressure in a water pipe in an apartment is 1.1 x 105 Pa. If the pipe is vertical, what is the pressure 8 meters below the apartment?

  33. P1 + 1/2 ρv12+ρgh2 = P2 + 1/2 ρv22+ ρgh11.1 x 105 Pa + 1000(10)8 =P2 = 190000 Pa

  34. 14. What is the ideal gas law?

  35. PV = nRT

  36. 15. One mole of an ideal gas has a volume of 3 m3. If the temperature is 750 K, what is the pressure?

  37. PV = nRTP(3) = 1(8.31)750P = 2077.5 Pa

  38. 16. If the pressure in problem 15 is doubled at constant temperature, what is the resulting volume?

  39. PV/T =PV/TPV = PV2077.5 x 3 = 4155 x VV = 1.5 m3

  40. 17. The pressure of a gas is 1 atmosphere and the temperature is 600 K. If the temperature is reduced to 300 K without changing the volume, what is the resulting pressure?

  41. PV/T =PV/TP/T = P/T1/600 = P/300P = ½ atmospheres

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