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Ch. 9 Fluids

This text covers various problems and principles related to fluids, including pressure and Pascal's principle, buoyancy and Archimedes' principle. It also discusses compressibility, density, and common densities of different substances. The text is focused on sections 7-11 of Giambattista.

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Ch. 9 Fluids

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  1. Ch. 9 Fluids • Problems: 1, 9, 13, 21, 27, 35. • pressure & pascal’s principle • buoyancy & archimedes’ principle • (sections 7-11)

  2. Giambattista: • compressibility • pressure • /

  3. Density r = mass/volume • SI Unit: kg/m3 • common unit: 1 g/cm3 = 1,000 kg/m3 • Example densities: • Water: r = 1.00 g/cm3 • Aluminum: r = 2.70 g/cm3 • Iron: r = 7.87 g/cm3 • Lead: r = 11.3 g/cm3

  4. Pressure, p • p = force/area • Unit: pascal, Pa • 1 Pa = 1N/m2. • Example: 100N applied to 0.25m2. Pressure = 100/0.25 = 400N/m2.

  5. Gauge Pressure is Differential Pressure Pgauge = P – Patm

  6. Depth and Fluid Pressure • P = Po + rgd • P is the pressure at depth d. • Po is the pressure at the fluid surface • r is the fluid density • Example: increase in pressure at a depth of 1m in water = (1,000)(9.8)(1.0) Pa

  7. Barometer

  8. Pascal’s Principle

  9. Buoyancy and Archimedes’ Principle

  10. L=0.5m cube of steel in water • Vo = L3 = (0.5m)3 = 0.125 m3 • FB = (fl.density)(Vo)(g) • = (1,000kg/m3 )(0.125 m3 )(9.8N/kg) • = 1,225 N • W = mg = {(… kg/m3 )(… m3 )}(g) • = (7,800)(0.125)(9.8) • = 9,555 N • Weight under water: • W’ = W – FB = 8,330 N

  11. 1kg object is weighed in water • W’ = 9.2N. What is its density? • FB = W – W’ = 9.8N – 9.2N = 0.6N • FB = (fl.den.)(ob.vol.)(g) = 0.6N • FB = (1000)(ob.vol.)(g) = 0.6N • (ob.vol.) = 6.12x10-5cubic meter • (ob.den.) = mass/vol. = 1kg/(6.12x10-5) • 16,333 kg/m3. = 16.333 g/cm3. • This is not pure gold.

  12. S.G. Method Density is 16.33 x density of water, or 16.33 gram/cc.

  13. Summary • Pressure and its Variation with Depth • Pascal’s Principle • Archimedes Principle • (Continuity Equation (conservation of mass) • Bernoulli’s Equation (conservation of energy) • Viscosity and Viscous Drag • Surface Tension)

  14. Solid Deformation stress = force/area = pressure strain = fractional change in length Solid Moduli ~ stress/strain ~ stiffness Hooke’s Law is force/strain: k = F/x 14

  15. 15

  16. Solid Deformation Length Change: stress = F/A, strain = DL/Lo, Ex. Hooke’s Law F = kDx Young’s Modulus, Y = stress/strain. Volume Change: stress = F/A = p, strain = -DV/Vo, Bulk Modulus, B = -Dp/(DV/Vo) 16

  17. Example: Object mass 0.150kg density 2g/cm3 is weighed in water.

  18. Example: calculate speed water exits the hole in terms of the given parameters.

  19. 10cm Cube Submerged in Water • Bottom surface is 10cm deeper than top. • P2 = P1 + (fl.density)(g)(d) • Net force = (P2-P1)(Area) • = (fl.density)(g)(d)(0.1m)2. • = (1000)(9.8)(0.1m)(0.1m)2. • = 9.8N • Mass of (1cc = 1mL) of water = 1gram. • (10cm)3 = 1000cc = 1000grams = 1kg • Weight of 1kg is 9.8N

  20. Ex. Pascal principle • One piston moves d1= 5cm, another piston moves d2 = 0.10cm. What is the MA (mechanical advantage) of this hydraulic system? • MA = 5cm/0.10cm = 50 • e.g. if F1 = 2N, F2 = 2Nx50 = 100N

  21. Ex2. Pascal principle • One piston has diameter D1= 5cm, another piston D2 = 10cm. What is the MA (mechanical advantage) of this hydraulic system? • MA = (D2/D1)^2 = (10cm/5cm)^2 = 4

  22. Object weighs 5N in air and 3N in water. What is its density? • S.G. = (ob.den/wat.den) • =(mog/mwg) • =(5N/2N) • =2.5=2.5gram/cc

  23. Object has density 6gram/cc, mass 1kg. What is its apparent mass in water?

  24. Water density & Volume • 1cc = 1mL • Density of water: • 1gram/cc = 1gram/mL • Ex: 25 mL of water has mass 25grams • Mass = rhoxVolume • =1gram/mL x 25mL = 25grams

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