Lecture 03 archimedes. fluid dynamics.

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Lecture 03 archimedes. fluid dynamics.

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Lecture 03 archimedes. fluid dynamics.

  1. 1. Lecture 3 Archimedes principle. Fluid dynamics.
  2. 2. ACT: Side tube A sort of barometer is set up with a tube that has a side tube with a tight fitting stopper. What happens when the stopper is removed? A. Water spurts out of the side tube. vacuum stopper B. Air flows in through the side tube. C. Nothing, the system was in equilibrium and remains in equilibrium. DEMO: Side tube
  3. 3. Buoyancy and the Archimedes’ principle A box of base A and height h is submerged in a liquid of density ρ. ytop Net force by liquid: ΣF = Fbottom − Ftop = Apbottom − Aptop A ybottom ( = A ( patm + ρ gybottom ) − A patm + ρ gy top = A ρhg = ρVg Ftop ) h Fbottom direction up Archimedes’s principle: The liquid exerts a net force upward called buoyant force whose magnitude is equal to the weight of the displaced liquid.
  4. 4. In-class example: Hollow sphere A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the minimum diameter necessary for this sphere not to sink in water? (ρwater = 1000 kg/m3) A. It will always sink. FB B. 0.11 m C. 0.21 m mg D. 0.42 m E. It will always float. The sphere sinks if mg > FB mg > ρwater R < 3 4 πR 3g 3 3m = 0.106 m 4πρwater Minimum diameter = 2R ≥ 0.21 m
  5. 5. Density rule DEMO: Frozen helium balloon A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the minimum diameter necessary for this sphere not to sink in water ( ρwater = 1000 kg/m3) ? Answer: R = 0.106 m. And what is the average density of this sphere? 5 kg m ρsphere = = = 1000 kg/m3 = ρwater 3 4 4 πR 3 π ( 0.106 m ) 3 3 An object of density ρobject placed in a fluid of density ρfluid • sinks if ρobject > ρfluid • is in equilibrium anywhere in the fluid if ρobject = ρfluid • floats if ρobject ρfluid This is why you float on the sea (1025 kg/m3) but not on a pool (1000 kg/m3) …
  6. 6. ACT: Styrofoam and lead A piece of lead is glued to a slab of Styrofoam. When placed in water, they float as shown. Pb styrofoam What happens if you turn the system upside down? styrofoam styrofoam Pb Pb A C. It sinks. B The displaced volume in both cases must be the same (volume of water whose weight is equal to the weight of the lead+Styrofoam system)
  7. 7. ACT: Floating wood Two cups are filled to the same level with water. One of the two cups has a wooden block floating in it. Which cup weighs more? A. Cup 1 B. Cup 2 C. They weigh the same. 1 2 Cup 2 has less water than cup 1. The weight of the wood is equal to the weight of the missing liquid (= “displaced liquid”) in 2. DEMO: Bucket of water with wooden block
  8. 8. Attraction between molecules Wood floats on water because it is less dense than water. But a paper clip (metal, denser than water!) also floats in water… (?) . Molecules in liquid attract each other (cohesive forces that keep liquid as such!) Very small attraction by air molecules. On the surface: Net force on a molecule is inward. In the bulk: Net force on a molecule is zero. …And this force is compensated by the incompressibility of the liquid.
  9. 9. Surface tension Overall, the liquid doesn’t “like” surface molecules because they try to compress it. Liquid adopts the shape that minimizes the surface area. Any attempt to increase this area is opposed by a restoring force. The surface of a liquid behaves like an elastic membrane. The weight of the paper clip is small enough to be balanced by the elastic forces due to surface tension.
  10. 10. Drops and bubbles Water drops are spherical (shape with minimum area for a given volume) Adding soap to water decreases surface tension. This is useful to: • Force water through the small spaces between cloth fibers • Make bubbles! (Large area and small bulk)
  11. 11. How wet is water? Molecules in a liquid are also attracted to the medium it is in contact with, like the walls of the container (adhesive forces). Water in a glass Fadhesive > Fcohesive Water in wax- or teflon-coated glass Fadhesive < Fcohesive Or: surface tension in air/liquid interface is larger/smaller than surface tension in wall/liquid interface
  12. 12. Fluid flow Laminar flow: no mixing between layers Turbulent flow: a mess…
  13. 13. Dry water, wet water Within the case of laminar flow: Slower near the walls Faster in the center Same speed everywhere Real (wet) fluid: friction with walls and between layers (viscosity) Ideal (dry) fluid: no friction (no viscosity)
  14. 14. Flow rate Consider a laminar, steady flow of an ideal, incompressible fluid at speed v though a tube of cross-sectional area A v dt A dV = Avdt Volume flow rate dV = Av dt Mass flow rate dm = ρAv dt
  15. 15. Continuity equation The mass flow rate must be the same at any point along the tube (otherwise, fluid would be accumulating or disappearing somewhere) ρ1Av1 = ρ2Av2 1 2 v2 A2 v1 A1 ρ1 If fluid is incompressible (constant density): ρ2 Av1 = Av2 1 2
  16. 16. Example: Garden hose When you use your garden faucet to fill your 3 gallon watering can, it takes 15 seconds. You then attach your 1.5 cm thick garden hose fitted with a nozzle with 10 holes at the end. You turn on the water, and 4 seconds later water spurts through the nozzle. When you hold the nozzle horizontally at waist level (1 m from the ground), you can water plants that are 5 m away. a) How long is the hose? b) How big are the openings in the nozzle? Volume flow rate dV = Ahosevhose dt dV 3 gallons 3.785 liters 1 m3 = = 7.6 × 10 −4 m3 /s dt 15 s 1 gallon 1000 liter vhose dV −4 3 dt = 7.6 × 10 m /s = 1.1 m/s = Ahose π 1.5 × 10 −2 m 2 ( Length of hose = vhoset = ( 1.1 m/s ) ( 4 s ) = 4.3 m )
  17. 17. When you use your garden faucet to fill your 3 gallon watering can, it takes 15 seconds. You then attach your 1.5 cm thick garden hose fitted with a nozzle with 10 holes at the end. You turn on the water, and 4 seconds later water spurts through the nozzle. When you hold the nozzle horizontally at waist level (1 m from the ground), you can water plants that are 5 m away. a) How long is the hose? b) How big are the openings in the nozzle? Ahosevhose = Anozzlevnozzle 2 ∅2 vhose = 10∅nozzlevnozzle hose We use kinematics to determine vnozzle: x = 0 + vnozzlet t = g 2 t 2 2  x  g 0 =h +x −  ÷ 2  vnozzle ÷   x h vnozzle x 0 = h + vnozzlet − ∅nozzle → ∅hose vhose ( 1.5 cm ) = = 10 vnozzle 10 vnozzle = gx = 2( h + x ) 2 ( 9.8 m/s ) ( 5 m) 2 2 ( 6 m) 1.1 m/s = 0.073 cm = 0.73 mm 4.5 m/s 2 = 4.5 m/s

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