Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 9: Fluids •Introduction to Fluids •Pressure •Pascal’s Principle •Gravity and Fluid Pressure •Measurement of Pressure •Archimedes’ Principle •Continuity Equation •Bernoulli’s Equation •Viscosity and Viscous Drag •Surface S rface Tension 1 Fluids A liquid will flow to take the shape of the container that holds it A gas will completely fill its container it. container. Fluids are easily deformable by external forces. A liquid is incompressible. Its volume is fixed and is impossible to change change. 2 FLUIDS (both liquids and gases)Best quantity to characterize a fluid is the density ρ ρ = m/V [kg/m³] m=Mass, M V= V Volume V l ρwater= 1000 kg/m³ Mass of 1L H2O (L = liter = 1dm³) m1L Water = 1000 kg/m³ 1dm³ = 1000 kg/m³ 10-³m³ = 1kg 1 gallon of water = 3.79 L = 3.79kg Substance Density (kg/m3) Substance Density (kg/m3) Gold 19,300 Olive oil 920 Mercury 13,600 Ice 917 Lead 11 300 11,300 Ethyl alcohol 806 Silver 10,500 Cherry (wood) 800 Iron 7860 Balsa wood 120 Aluminum 2700 Styrofoam 100 Ebony (wood) 1220 Oxygen 1.43 Whole blood (37 ºC) 1060 Air 1.29 Sea ater Seawater 1025 Heli m Helium 0 179 0.179 Fresh water 1000 3 3 Volumerefr = 1 ⋅ 0.6 ⋅ 0.75m = 0.45m 3 G G Weight Air , refr = (0.45m ⋅1.29 3 ) ⋅ g = (580.5 g ) ⋅ g m G G Weighteggs , refr = (12 ⋅ 44 g ) ⋅ g = (528 g ) ⋅ g 3 kg 4 Static Dynamic 5 Pressure Pressure arises from the collisions between the particles of a fluid with another object (container walls for example). There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall. 6 By Newton’s 3rd Law, there is a force on the wall due to the particle. Pressure is defined as F P= . A The units of pressure are N/m2 and are called Pascals (Pa). Note: 1 atmosphere (atm) = 101.3 kPa (older unit: 1 bar = 105 Pa) Are we under pressure for our entire life? - YES Air above us has a weight. weight Atmospheric Pressure Pat= 1.01 x 105 N/m² = 101 kPa ≈ 1 bar 7 (average value on earth´s surface, the higher you go the lower the pressure) pascal (Pa) bar (bar) technical atmosphere (at) atmosphere (atm) torr (Torr) pound-force per square inch (psi) 1 Pa ≡ 1 N/m2 10−5 1.0197×10−5 9.8692×10−6 7.5006×10−3 145.04×10−6 1 bar 100,000 ≡ 106 dyn/cm2 1.0197 0.98692 750.06 14.5037744 1 at 98,066.5 0.980665 ≡ 1 kgf/cm2 0.96784 735.56 14.223 1 atm 101,325 1.01325 1.0332 ≡ 1 atm 760 14.696 1 torr 133.322 1.3332×10−3 1.3595×10−3 1.3158×10−3 ≡ 1 Torr; ≈ 1 mmHg 19.337×10−3 1 psi 6,894.76 68.948×10−3 70.307×10−3 68.046×10−3 51.715 ≡ 1 lbf/in2 8 9 The pressure in a fluid (gas, liquid) acts equally in all directions and acts at right angles to any surface 10 Example (text problem 9.1): Someone steps on your toe, exerting a force of 500 N on an area of 1 1.0 0 cm2. What is the average pressure on that area in atmospheres? 2 ⎞ 2⎛ 1m −4 2 1.0 cm ⎜ ⎟ = 1.0 ×10 m ⎝ 100 cm ⎠ F 500 N Pav = = A 1.0 ×10 -4 m 2 1 atm ⎞ 6 2 ⎛ 1 Pa ⎞⎛ = 5.0 × 10 N/m ⎜ ⎟ 2 ⎟⎜ 5 ⎝ 1 N/m ⎠⎝ 1.013 × 10 Pa ⎠ = 49 atm 11 Gravity’s Gravity s Effect on Fluid Pressure (What you feel at a start/landing of an airplane or when you go scuba diving) FBD for the fluid cylinder P1A A cylinder of fluid fl id w P2A 12 Apply A l N Newton’s t ’ 2ndd Law L to the fluid cylinder: ∑F = P A− PA− w = 0 2 1 P2 A − P1 A − (ρAd )g = 0 P2 − P1 − ρgd = 0 ∴ P2 − P1 = ρgd or P2 = P1 + ρgd If P1 (the pressure at the top of the cylinder) is known, then th above the b expression i can b be used d tto fifind d th the variation i ti off pressure with depth in a fluid. 13 If the top of the fluid column is placed at the surface of the fluid then P1=P fluid, Patm open. t if the container is open P = Patm + ρgd Example (text problem 9.15): At the surface of a freshwater lake lake, the pressure is 105 kPa. kPa (a) What is the pressure increase in going 35.0 m below the surface? P = Patm + ρgd d ΔP = P − Patm = ρgd 35m ( )( ) = 1000 kg/m 9.8 m/s (35 m ) 3 2 = 343 kPa = 3.4 3 4 atm 14 Example: The surface pressure on the planet Venus is 95 atm. t How H f below far b l th the surface f off the th ocean on Earth E th do d you need to be to experience the same pressure? The densityy of seawater is 1025 kg/m g 3. P = Patm + ρgd 95 atm = 1 atm + ρgd ρgd = 94 atm = 9.5 × 10 N/m 6 (1025 kg/m )(9.8 m/s )d = 9.5 9 5 × 10 3 2 6 N/m 2 2 d = 950 m ≈ 0.600miles 15 16 17 Measuring Pressure A manometer i aU is U-shaped h d tube that is partially p y filled with liquid. Both ends of the tube are open to the atmosphere. P 1 = P2 18 A container of gas is connected to one end of the U-tube If there is a pressure difference between the gas and the p a force will be exerted on the fluid in the U-tube. atmosphere, This changes the equilibrium position of the fluid in the tube. 19 From the figure: At point C Also Pc = Patm PB = PB' The pressure at point B is the pressure of the gas. PB = PB ' = PC + ρgd PB − Patm = ρgd Pgauge = ρgd Gauge pressure is the pressure relative to the local atmospheric or ambient pressure. 20 A Barometer P2: pressure in glass tube P1 = P2 + ρhg = 0 + ρhg Therefore: Pat= ρhg 1 atmosphere = Pat = 760 mm Hg The atmosphere pushes on the container of mercury which forces mercury up the closed, inverted tube. The distance d is called the barometric pressure. 21 Example (text problem 9.22): An IV is connected to a patient’s patient s vein. vein The blood pressure in the vein has a gauge pressure of 12 mm of mercury. At least how far above the vein must the IV bag be placed in order for fluid to flow into the vein? Assume that the density of the IV fluid is the same as blood. Blood: Pgauge = ρ Hg gh1 h1 = 12 mm IV: The pressure is Th i equivalent i l t to raising a column of mercuryy 12 mm tall. Pgauge = ρ blood gh2 22 At a minimum, the gauge pressures must be equal. When h2 is large enough, fluid will flow from high pressure to low pressure. Pgauge = ρ Hg gh1 = ρ blood gh2 ρ Hgg gh1 h2 = ρ blood g ⎛ ρ Hg ⎞ ⎛ 13,600 kg/m 3 ⎞ H ⎟⎟h1 = ⎜⎜ ⎟(12 mm ) = ⎜⎜ 3 ⎟ ⎝ 1060 kg/m ⎠ ⎝ ρ blood ⎠ = 154 mm 23 Pascal’s Pascal s Principle A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid fluid. (This is useful in making a hydraulic lift.) The applied force is transmitted to the piston of crosssectional area A2 here. here Apply a force F1 here to a piston of crosssectional area A1. 24 Mathematically, ΔP at point 1 = ΔP at point 2 F1 F2 = A1 A 2 ⎛ A2 ⎞ ⎟⎟ F1 F2 = ⎜⎜ ⎝ A1 ⎠ 25 Example: Assume that a force of 500 N (about 110 lbs) is applied li d tto th the smaller ll piston i t iin th the previous i fifigure. F For each h case, compute the force on the larger piston if the ratio of the piston areas ((A2/A1) are 1, 10, and 100. p Using g Pascal’s Principle: A2 A1 1 10 100 F2 500 N 5000 N 50,000 N 26 The work done pressing the smaller piston (#1) equals the work done by the larger piston (#2). W1 = W2 F1d1 = F2 d 2 F1 d2 = d1 F2 27 Example: In the previous example, for the case A2/A1 = 10, it was ffound d th thatt F2/F1 = 10. 10 If the th larger l piston i t needs d tto rise by 1 m, how far must the smaller piston be depressed? Using the result on the previous slide, F2 d1 = d 2 = 10 m F1 28 Archimedes’ Principle An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object. F1 An FBD for an object floating submerged in a fluid. w F2 The total force on the block due to the fluid is called the buoyant force. FB = F2 − F1 where F2 > F1 29 The magnitude of the buoyant force is: FB = F2 − F1 = P2 A − P1 A = (P2 − P1 )A From before: P2 − P1 = ρg gd The result is FB = ρgdA = ρ fluid gV fluid ,displaced 30 Archimedes’ Principle: p A fluid exerts an upward p buoyant y force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object. FB = ρ fluid gV fluid ,displaced 31 32 33 Example (text problem 9.30): A flat-bottomed barge loaded with coal has a mass of 3 3.0×10 0×105 kg. kg The barge is 20 20.0 0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline? ∑F = F B FB FBD for the barge −w=0 FB = w mw g = (ρ wVw )g = mb g w ρ wVw = mb ρ w ( Ad ) = mb mb 3.0 ×105 kg d= = = 1.5 m 3 ρ w A 1000 kg/m (20.0 m *10.0 m ) ( ) 34 35 Example (text problem 9.34): A piece of metal is released under water. water The volume of the metal is 50 50.0 0 cm3 and its specific gravity is 5.0. What is its initial acceleration? (Note: when v=0, there is no drag force.) FB FBD for the metal w ∑F = F B − w = ma The buoyant force is the weight g of the fluid displaced p by the object FB = ρ waterVgg ⎛ ρ waterV ⎞ ρ waterVg FB −g = − g = g⎜ − 1⎟ Solve for a: a = ⎜ρ V ⎟ m ρ objectVobject ⎝ object object ⎠ 36 Example continued: Since the object is completely submerged V=Vobject. specific ifi gravity it = ρ ρ water 3 is the where ρwater = 1000 kg/m t density of water at 4 °C. ρ object = 5.0 Given specific gravity = ρ water ⎛ ρ waterV ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ a=g −1 = g⎜ − 1⎟ = g ⎜ − 1⎟ = −7.8 m/s 2 ⎜ρ V ⎟ ⎝ S .G. ⎠ ⎝ 5.0 ⎠ object object ⎝ ⎠ 37 38