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CHAPTER 13 FLUIDS • Density ! Bulk Modulus ! Compressibility

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CHAPTER 13 FLUIDS • Density ! Bulk Modulus ! Compressibility Liquids CHAPTER 13 FLUIDS FLUIDS Gases • Density To begin with ... some important definitions ... ! Bulk modulus ! Compressibility Mass m DENSITY: , i.e., ρ = • Pressure in a fluid Volume V ! Hydraulic lift [M] ! Hydrostatic paradox Dimension: ⇒ 3 [ L] • Measurement of pressure −3 Units: ⇒ kg ⋅ m ! Manometers and barometers • Buoyancy and Archimedes Principle ! Upthrust Force F ! Apparent weight PRESSURE: , i.e., P = Area A • Fluids in motion [M] ! Continuity Dimension: ⇒ [L][T]2 ! Bernoulli’s equation Units: ⇒ N ⋅ m−2 ⇒ Pascals (Pa) Mass = Volume × Density Question 13.1: What, approximately, is the mass of air Volume of room ≈ 15 m ×10 m × 3 m −3 in this room if the density of air is 1 .29 kg ⋅ m ? 3 −3 ∴ M ≈ 450 m ×1.29kg ⋅ m = 580 kg (over half-a-ton!) The weight of a medium apple is ~ 1 N, so the mass of a medium apple is ~ 0.1 kg. 3 A typical refrigerator has a capacity of ~ 18 ft . 3 3 3 ∴ 18 ft ⇒ 18 × (0.305 m) = 0.51 m . 3 But 1 m of air has a mass of 1 .3 kg, so the mass of air in the refrigerator is Question 13.2: How does the mass of a medium sized 0 .51×1.3 kg = 0.66 kg, apples compare with the mass of air in a typical i.e., approximately the mass of 6 apples! 3 refrigerator? the density of cold air is ~ 1.3 kg/m . We don’t notice the weight of air because we are immersed in air ... you wouldn’t notice the weight of a bag of water if it was handed to you underwater would you? definitions (continued) ... 5 Atmospheric pressure: P o = 1×10 Pa ∴ Force on the ceiling from floor of room above is ΔP BULK MODULUS: B = pressure area 1 105 (15 m 10 m) ΔV × ≈ × × × ( V) ≈ 1.2 ×107 N. [M] Dimension: ⇒ 2 [L][T] −2 Units: ⇒ N ⋅ m −1 Compressibility ⇒ B Δ P Why doesn’t it collapse under that weight ... ? Δ V Because pressure operates equally in all directions! ! Why? F Δ P Δ P When air molecules “bounce” off V the walls they produce an impulse: F Δt = Δp ⇒ pressure Δ P Since the molecules are traveling Gases are easily compressed (B ⇒ very small). with equal speeds in all directions Liquids and solids much less compressible. ... the pressure is the same ! Pressure at a depth in a fluid ... P ! Area = A P ! h w = mg P Imagine a cylindrical body of the fluid with its top face at Question 13.3: A balloon has a radius of 10 cm. By how the surface of the fluid. At equilibrium there is no net much does the radius change if the balloon is pushed force acting on the surfaces of the cylinder. down to a depth of 10 m in a large tank of water? (The bulk modulus of air is 2 ×105N⋅ m−2.) ∴ ∑Fy = 0 at the lower face, i.e., P × A = P! × A + mg . But the mass of fluid in the cylinder is m = ρV = ρAh. ∴ P = P! + ρgh . What’s the pressure in water at a depth of 10m, say? 3 3 2 5 ρ gh = 1×10 kg/m × 9.81 m/s ×10 m ≈10 Pa. 5 P ! = Pat ≈1.01×10 Pa. ∴ P ≈ 2Pat. The pressure difference is What’s the difference in the air pressure from the r Δ P = hρg ceiling to the floor in this room ... ? ΔP But B = . ΔV h ( V) The room height is ~ 3m so the pressure difference is: ΔP = hρg = 3 ×1.29 × 9.81 ≈ 38Pa Δ r ΔV ΔP hρg ∴ = = . ΔP 38 V B B ∴ ≈ ≈ 3.8 ×10−4 (0.038%) P 1 105 For an air-filled balloon at a ! × depth of 10m, we have i.e., negligible. ΔV 10 m ×1×103 kg/m3 × 9.81 m/s2 = ≈ 0.5 (50%). V 2 ×105 N/m2 Pascal’s principle ... Assuming the balloon is spherical, P ! + ΔP 4 If additional pressure ( Δ P) is V = πr3 so Δ V = 4πr2 × Δr. 3 applied, it is transmitted ΔV 3Δr through the whole fluid: ∴ = ≈ 0.5, h 1 h V r 2 ∴ P1 = P! + h1ρg + ΔP P 1 so, for an initial radius r = 0.1 m, and P 0.5r 2 P = P + h ρg + ΔP. Δr ≈ = 0.017m (1.7 cm). 2 ! 2 3 Blaise Pascal (1623-1662) Hydraulic lift: F 1 Area A 1 Area A 2 Δ x2 F1 Δx Area A Area A 1 1 2 F 2 F h 2 Get a larger force OUT than you put IN? Too good to be true? same pressure No, not really, because, to do work (like lift something If a force F 1 is applied to the left hand piston, the heavy) the force F is applied through a distance x . F1 2 Δ 2 additional pressure, P1 = , is transmitted through the A1 But by conservation of energy A2 whole fluid. Therefore, on the surface of the right hand F2Δx2 = F1Δx1 ⇒ Δx1 = Δx2. A1 F2 piston, P2 = = P1. A2 So, although F 1 < F2, it is applied through a greater distance Δx1 > Δx2. F1 A1 ⎛ A2⎞ ∴ = , i.e., F2 = ⎜ ⎟ F1. Examples: F2 A2 ⎝ A1⎠ • lifts Wow ... the force is amplified!! • dentist’s chair Mechanical advantage • hydraulic brake systems We can use these ideas to measure pressure: Here’s something surprising ... no matter the shape of a vessel, the pressure depends only on the P ! P = 0 vertical depth. It is known as the hydrostatic paradox. h h y y 2 P 2 P ! P ! y 1 y 1 P ! P ! P ! P ! Manometer Barometer h P + ρgy1 = Po + ρgy 2 0 + ρgy 2 = Po + ρgy1 This is absolute pressure P ! + ρgh i.e., P − P! = ρgh . i.e., P ! = ρgh . This is the Gauge pressure Atmospheric pressure DISCUSSION PROBLEM 13.1: Barometer #2 The drawing shows two P P = 0 ! pumps, #1 and #2 to be used P ! = ρgh , i.e., h = ρg for pumping water from a very 5 P ! ≈ 1.01×10 Pa deep well (~30 m deep) to Using water: ground level. Pump #1 is h 3 −3 submerged in the water at the ρ = 1×10 kg ⋅ m #1 5 bottom of the well; the other P P 1.01×10 ! ! ∴h = ~ 10m. pump, #2, is located at ground 1×103 × 9.81 level. Which pump, if either, can be used to pump water Using mercury: to ground level? 3 −3 ρ = 13.6 ×10 kg ⋅ m 1.01×105 A: Both pumps #1 and #2. ∴h = ~ 0.76m 13.6 ×103 × 9.81 B: Pump #1. ‘Standard pressure’ is 760mm of Hg. C: Pump #2. D: Neither pump #1 nor pump #2. If the object is floating then ... w = B. Buoyancy and the concept of upthrust B B V s V L w = mg V s w = mg Weight of object is w = mg = ρsVs g. If V L is the volume submerged, then the weight of liquid displaced is Archimedes Principle : when an object is partially or ρ LVL g. But according to Archimedes principle, this is wholly immersed in a fluid, the fluid exerts an upward equal to the upthrust (B). force ... upthrust ... (or buoyant force, B) on the object, ρs ∴ρsVs g = ρLVL g ⇒ VL = Vs. which is equal to the weight of fluid displaced. ρL Example: what volume of an iceberg is submerged? Submerged: w > B 3 −3 ρ s = 0.92 ×10 kg ⋅ m . 3 −3 ρ L = 1.03×10 kg ⋅ m . Weight of object w = mg = ρsVs g 3 Weight of liquid displaced = ρ V g VL ρs 0.92 ×10 L s ∴ = = 3 = 0.89, Vs ρL 1.03×10 If ρ s > ρL the object will sink. i.e., 89% of an iceberg is submerged! What ... you don’t believe me ... Question 13.4: On Earth, an ice cube floats in a glass of water with 90% of its volume below the level of the water. If we poured ourselves a glass of water on the Moon, where the acceleration due to gravity is about 1 th 6 of that on Earth, and dropped in an ice cube, how much of the ice cube would be below the level of the water? When an ice cube floats, the weight of the ice cube ( = ρsVsg) equals the weight of the water displaced ( = ρLVLg), which is proportional to V L, the volume of the ice cube below the surface. ∴ ρLVLg = ρsVsg, ρs i.e., VL = Vs, ρL which is independent of g. So, the volume submerged Question 13.5: A piece of copper of mass 0.50 kg is would remain the same! suspended from a spring scale. If it is fully submerged in water, what is the reading (in N) on the spring scale? Since both the weight of an object, which is floating, and 3 −3 (The density of copper is 9 .0 ×10 kg ⋅ m .) the weight of a fluid it displaces are proportional to the local value of g, the submerged volume does not depend on g. T (apparent weight) T Question 13.6: A beaker containing water is placed on a B w = mg = Vsρsg balance; its combined mass reading on the balance is 1.200 kg. In (a) below, a copper block is hanging freely from a spring scale, which has a mass reading of 0.20 kg. Identify the forces acting on the block. If the copper block is totally immersed in the water, as At equilibrium ∑Fy = T + B + (−w) = 0.
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