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CHAPTER 9 Solids and Fluids Fluid Mechanics (only) pages 261-277

CHAPTER 9 Solids and Fluids Fluid Mechanics (only) pages 261-277.  = M V. M = kg V = m 3.  = kg/m 3. F = N A = m 2. P = N/m 2 = Pascal (Pa). P = F/A. Fluids and Fluid Mechanics. Density: mass of substance divided by its volume.

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CHAPTER 9 Solids and Fluids Fluid Mechanics (only) pages 261-277

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  1. CHAPTER 9Solids and FluidsFluid Mechanics (only)pages 261-277

  2.  = M V M = kg V = m3  = kg/m3 F = N A = m2 P = N/m2 = Pascal (Pa) P = F/A Fluids and Fluid Mechanics Density: mass of substance divided by its volume The ratio of a substances density to the density of water at 4°C (which is 1.0x103 kg/m3) The specific gravity of a substance is: 1. Dimensionless 2. Magnitude differs from the substances density by a factor of 103 Specific Gravity: (sp.gr.) Pressure: A force applied to an area

  3. M = 177 kg Example Problem (Density/Sp.gr./Pressure) A lead cube with a side of .250m rests on the floor. Calculate the mass of the cube, the force exerted on the floor, and the pressure on the floor beneath the cube (lead = 11.3x103 kg/m3)  = M/V M = v M = (11.3 x 103 kg/m3)(.250m x .250m x .250m) F = mg F = (177 kg)(9.8m/s2) F = 1730 N P = F/A P = (1730 N) / (.250m x .250m) NOTE: 1 atmosphere of pressure = 101,300 pascals P = 27,700 N/m2 P = 27,700 pascals

  4. Mg + PoA = PA M = V Vg + PoA = PA Vg + Po = P A Static head pressure Pressure In/Under Fluids Po = atmospheric pressure above the water P = pressure in the water at a point under the shaded block of water If the shaded block of water is not falling or rising, then FNet,y = 0 hg + Po = P P = Po + gh

  5. o 10.m A 50.m To water users B Example Problem (Head Pressure) A water tower sits on top of a 50.m platform in the middle of the city (which is on level ground). The tower’s water tank is 10.m high. Assuming the tank is full of water, what is the pressure of the bottom of the tank and what is the pressure in the water pipe at grade level below the tower. Sketch PA = Po + gh PA = 101.3kpa + (1.0x103kg/m3)(9.8m/s2)(10.m) PA = 101.3kpa + 98,000N/m2 1 N/m2 = 1 pa PA = 101.3kpa + 98.0kpa PA = 199.3kpa PB = PA + (1.0x103kg/m3)(9.8m/s2)(50.m) PB = 689.32kpa NOTE: 101.3kpa = 1.00atm = 14.7psi = 760mmHg

  6. 760mmHg 1 cm 1 inch 10mm 2.54cm Example Problem (Barometer) How high will liquid Mercury rise in a barometer tube when the air pressure is 1.00atm? [note: Hg = 13.6x103kg/m3] Analysis Pressure at bottom of rising mercury column is equal to Po = 1.00atm. Po is forcing the liquid up. When in equilibrium, the static head pressure of the column plus the pressure above the column (0.00atm) equals Po. Po = gh +P 1.00atm = 101,300N/m2 101,300N/m2 = (13.6x103kg/m3)(9.80m/s2)h h = .760mHg h = 760mmHg h = 29.9 inches Hg As in “the barometric pressure is at 29.2inches and rising.”

  7. A 1 h 2 Fluid FBuoyant = Mg M = mass of displaced fluid Buoyant Forces Archimedes’s Principle: Any body completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the displaced fluid. (What else did Archimedes say 2200 years ago that is still repeated today?) P1 A = F1 (Downward Force) P2 A = F2 (Upward Force) FNet,up = FBuoyant = F2 – F1 = P2A – P1A = A (P2 – P1) = A (gh2 – gh1) = A g (h2 – h1) = A gh = Vg = Mg

  8. Example Problem (Floating Boat) A particular boat could displace 4.50m3 of water before water would pour over the sides of the boat. The boat itself weighs 35,000N when filled with gasoline and other provisions. If each fisherman wishing to go out in the boat has a mass of 100.kg, how many fisherman can the boat hold? Analysis At maximum load the FBuoyant = WBoat+ WMen FBuoyant = Vg = (1.0x103kg/m3)(4.50m3)(9.80m/s2) = 44,100N 44,100N = 35,000N + N(100.kg)(9.80m/s2) N = 9.29 Fisherman N = 9 Fisherman

  9. water = 1.0x103kg/m3 FT FBuoyant V =.250m3 gold = 19.3x103kg/m3 Fg Example Problem (Sunken Treasure) A box containing .250m3 of gold bars has been located at the bottom of the sea. How much force will it require to lift the box upward through the water? (Neglect the mass of the box.) Analysis If lifting at constant velocity, Fup = Fdown FT + FBuoyant = Fg FT + (water)(Vgold)(g) = (gold)(Vgold)(g) FT = 44,800 N Notice: FT = (gold – water) Vg FT = Vg

  10. 1 2 v1 v2 A1 A2 v1A1 = v2A2 Continuity Equation Continuity Equations for Flowing Fluids Analysis Mass passing point 1 each second must equal mass passing point 2 each second (at steady-state). v1A1 = v2A2

  11. P1V + ½ Mv12 + Mgy1 = P2V+ ½ Mv22 + Mgy2 V V Energy in a flowing fluid can be transferred pressure velocity elevation Bernoulli’s Equation for Flowing Fluids Bernoulli’s Equation is really a conservation of energy equation and pressure x volume has units of energy (N/m2· m3 = N · m = Joule). Kinetic and Potential Energy are also conserved. P1 + ½ (M/V) v12 + (M/V) gy1 = P2 + ½ (M/V) v22 + (M/V) gy2 P1 + ½ v12 + gy1 = P2 + ½ v22 + gy2 P + ½ v2 + gy = Constant

  12. 3 1 2 v1 A1 A2 A1 v2 v3 Bernoulli’s Equation w/Closed Systems At constant elevation: P1 + ½ v12 = P2 + ½ v22 = P3 + ½ v32 = constant v2> v1 v3< v1 P2 < P1 P3 > P1

  13. v1 v v2 P2 > P1 Flight v1> v2 p = ½ v12 – ½ v22 p · Area (wing) = F F = FNet FNet is in upward direction

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