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Fluids and Elasticity

Fluids and Elasticity. Chapter 15. Density ( r ). r = mass/volume Rho ( r ) – Greek letter for density Units - kg/m 3 Specific Gravity = Density of substance Density of water (4 o C) Unitless ratio Ex: Lead has a sp. Gravity of 11.3 (11.3 times denser than water. Ex: 1.

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Fluids and Elasticity

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  1. Fluids and Elasticity Chapter 15

  2. Density (r) r = mass/volume • Rho ( r) – Greek letter for density • Units - kg/m3 • Specific Gravity = Density of substance Density of water (4oC) • Unitless ratio • Ex: Lead has a sp. Gravity of 11.3 (11.3 times denser than water

  3. Ex: 1 Estimate the mass of air in this classroom

  4. Pressure • Force per unit area • P = F/A • Unit - N/m2 (Pascal) • The larger the area, the less the pressure • Shoeshoes • Elephant feet • Bed of nails

  5. Fluid Pressure • A fluid exerts the same pressure in all directions at a given depth • P = Po + rgh • The atmosphere is a fluid • Po often 1 atm (101.3 kPa)

  6. Pressure: Example 1 A water storage tank is 30 m above the water faucet in a house. Calculate the pressure at the faucet: We will neglect the atmospheric pressure since it is the same at the tank and at the surface DP = rgh = (1000 kg/m3)(9.8 m/s2)(30 m) DP = 29,000 kgm2/m3s2 = 29,000 kg m/s2m2 DP = 29,000 N/m2

  7. Pressure: Example 2 The Kraken can live at a depth of 200 m. Calculate the pressure the creature can withstand (neglect atmospheric pressure)

  8. Pressure: Example 2 DP = rgh = (1000 kg/m3)(g)(200 m) DP = 1.96 X 106 N/m2

  9. Atmospheric Pressure • 1 atm = 1.013 X 105 N/m2 = 101.3 kPa • 1 bar = 1 X 105 N/m2 (used by meteorologists) • Gauge pressure P = Patm + PG Absolute pressure atmospheric pressure Gauge pressure

  10. We usually measure gauge pressure • Ex: A tire gauge reads 220 kPa, what is the absolute pressure? P = Patm + PG P = 101.3 kPa + 220 kPa = 321 ka

  11. Straw Example You can pick up soda in a straw using your finger. Why doesn’t the soda fall out?

  12. Another Straw Example What pushes soda up a straw when you drink through it?

  13. Torrecilli’s Work P = Patm + PG P = Patm + rgh What is the highest column of water that the atmosphere can support?

  14. P = Patm + rgh 0 = 1.013 X 105 N/m2 + (1000kg/m3)(9.8m/s2)(h) h = 10.3 m • No vacuum pump can pump more than ~30 feet

  15. Try the same calculation with mercury P = Patm + rgh 0 = 1.013 X 105 N/m2 + (13,600kg/m3)(9.8m/s2)(h) h = 0.760 m (760 mm) 1 atm = 760 mm Hg (760 torr)

  16. Can an astronaut attach suction cups to the boots of his spacesuit to help him climb around the space shuttle while in space?

  17. Pascal’s Principle • Pressure applied to a confined fluid increases the pressure the same throughout Pin = Pout Fin = Fout Ain Aout for pistons: DF = rg(A1 + A2)d2

  18. Pascal’s Principle: Ex 1 A hydraulic lift can produce 200 lb of force. How heavy a car can be lifted if the area of the lift is 20 times larger that the input of the Force? Fin = Fout Ain Aout Fout = Fin Aout = (200 lb) (20) = 4000 lbs Ain 1

  19. Pascal’s Principle: Ex 1 A hydraulic lift has the car rest ona 25 cm pipe. The lift the car, compressed air pushes on a 6.0 cm pipe. • Calculate the force needed to lift a 1300 kg car. • Calculate how much the air pressure force must be increased to lift the car 2.0 m.

  20. Buoyancy • Buoyancy • The “lift” provided by water • Objects weight less in water than out • Caused by pressure differential between top and bottom of an object. Fbouyant = rgV

  21. Derivation of the Buoyancy Formula Fb = F2 – F1 P = F/A F = PA F = rghA Fb = rgh2A – rgh1A Fb = rgA(h2 -h1) Fb = rgV

  22. Archimedes Principle “The bouyant force on an object equals the weight of fluid displaced by the object” w’ = weight of an object in water (or any liquid) w’ = mg - Fb

  23. Buoyancy: Example 1 A 7000-kg ancient statue lies at the bottom of the sea. Its volume is 3.0 m3. How much force is needed to lift it? Fb = rgV Fb = (1000 kg/m3)(9.8 m/s2)(3.0m3) Fb = 2.94 X 104 kg-m/s2 Fb = 2.94 X 104 N Fb mg

  24. w’ = mg - Fb w’ = (7000 kg)(9.8m/s2) - 2.94 X 104 N w’ = 3.92 X 104 N Say, isn’t w’ just the sum of the forces? Yep. SF = w’ Fb mg

  25. Buoyancy: Example 2 Archimedes tested a crown for the king. Out of water, it masses 14.7 kg. In water, it massed 13.4 kg. Was the crown gold? w’ = mcrg – Fb w’ = mcrg – rgVcr (13.4 kg)(g) = (14.7 kg)(g) – (1000 kg/m3)(g)(Vcr) 131 N = 144 N – (9800 kg/ms2)(Vcr) Vcr = 0.00133 m3

  26. Now we can calculate the density of the crown: • = m/V = 14.7 kg/0.00133 m3 • = 11,053 kg/m3 Gold’s density is about 19,000kg/m3. This is much closer to lead.

  27. Example 3 A cube of wood that is 10 cm on a side is held underwater by tying a string to the cube and the bottom on the container. The wood has a density of 700 kg/m3. • Draw a free body diagram showing all the forces on the block. • Calculate the force of bouyancy • Calculate the tension in the string.

  28. Floating • Objects that are less dense than water will float • Part of the object will be above the water line • A case of static equilibrium SF = 0 Fb mg

  29. Floating: Example 1 A 1200 kg log is floating in water. What volume of the log is under water? SF = 0 SF = 0 = mg – Fb mg = Fb mg = rgVlog Vlog = mg rg Fb mg

  30. Vlog = m (Hey, the g’s cancel!) r Vlog = 1200 kg = 1.2 m3 1000 kg/m3

  31. Floating: Example 1 A wooden raft has a density of 600 kg/m3, an area of 5.7 m2, and a volume of 0.60 m3. How much of the raft is below water in a freshwater lake?

  32. Let’s first calculate the mass of the raft: r = m/V m = rV = (600 kg/m3)(0.60 m3) = 360 kg Now we can worry about the raft. SF = 0 SF = 0 = mg – Fb mg = Fb mg = rgVsubmerged mg = rghsubmergedA

  33. mg = rghsubmergedA m = rhsubmergedA (Hey, the g’s cancelled!) hsubmerged = m/rA hsubmerged = 360 kg = 0.063 m (1000 kg/m3)(5.7 m2)

  34. Floating: Example 3 Suppose a continent is floating on the mantle rock. Estimate the height of the continent above the mantle (assume the continent is 35 km thick).

  35. SF = 0 = mg – Fb 0 = mcg – rmangVc(submerged) mcg = rmangVc(submerged) mc = rmanVc(submerged) We don’t know the mass of the continent rc = mc/Vc(total) mc = rcVc(total) mc = rmanVc(submerged)

  36. mc = rmanVc(submerged) mc = rcVc(total) rmanVc(submerged) = rcVc(total) Vc(submerged) = rc = (2800 kg/m3) = 0.85 Vc(total) rman (3300 kg/m3) This means that 85% of the continent is submerged, and only 15% is above: (0.15)(35 km) = 5.25 km

  37. Floating: Ex 4 A block is placed in water and 5.8 cm is submerged. The same block is placed in an unknown liquid and 4.6 cm is submerged. Calculate the density of the unknown liquid. Assume the same face of the block pointed downward in both cases (A).

  38. Fluid Flow

  39. Laminar Flow – Smooth, streamline flow (laminar means “in layers”) • Turbulent Flow – erratic flow with “eddies” • Viscosity – Internal friction of a liquid • High viscosity = slow flow • Viscosity is NOT the same as density

  40. Equation of Continuity A1v1 = A2v2 A = Area of a pipe v = velocity of the liquid

  41. Equation of Continuity v1A1 = v2A2 • Fluid will flow faster through a smaller opening • Placing your finger over a hose opening.

  42. The term “vA” is the “volume rate of flow” A = m2 v = m/s vA = m3/s Q = vA

  43. Eqn. Of Continuity: Example 1 A garden hose has a radius of 1.00 cm and the water flows at a speed of 0.80 m/s. What will be the velocity if you place your finger over the hose and narrow the radius to 0.10 cm? A1 = pr2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2 A2 = pr2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2

  44. A1v1 = A2v2 v2 = A1v1 A2 v2 = (3.14 X 10-4 m2)(0.80 m/s) = 80 m/s (3.14 X 10-6 m2)

  45. Eqn. Of Continuity: Example 2 A water hose 1.00 cm in radius fills a 20.0-liter bucket in one minute. What is the speed of water in the hose? A1 = pr2 = (3.14)(1 cm)2 = 3.14 cm2 Remember that Av is volume rate of flow. A2v2=20.0 L 1 min 1000 cm3= 333 cm3/s 1 min 60 s 1 L

  46. A1v1 = A2v2 v1 = A2v2/A1 v1 = 333 cm3/s = 160 cm/s or 1.60 m/s 3.14 cm2

  47. Eqn. Of Continuity: Example 3 A sink has an area of about 0.25 m2. The drain has a diameter of 5 cm. If the sink drains at 0.03 m/s, how fast is water flowing down the drain? Ad = pr2 = (p)(0.025 m)2 = 1.96 X 10-3 m3 Advd = Asvs vd = Asvs/Ad=[(0.25 m2)(0.03 m/s)]/(1.96 X 10-3 m3) vd = 3.82 m/s

  48. Eqn. Of Continuity: Example 4 The radius of the aorta is about 1.0 cm and blood passes through it at a speed of 30 cm/s. A typical capillary has a radius of about 4 X 10-4 cm and blood flows through it at a rate of 5 X 10-4 m/s. Estimate how many capillaries there are in the human body.

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