Slide 1Fluid PropertiesSysteme International (SI)Système International (SI)English (American) SystemDensityExample: Textbook Problem 2.8ExampleSpecific WeightSpecific GravitySlide 11Ideal Gas Law (equation of state)Slide 13Slide 14Example: Textbook Problem 2.4Slide 16Some Simple FlowsSlide 18Fluid DeformationSlide 20ViscosityShear in Different FluidsEffect of TemperatureSlide 24Typical Viscosity EquationsFlow between 2 platesSlide 27Slide 28Example: Textbook Problem 2.33Slide 30Example: Textbook Problem 2.34Slide 32Elasticity (Compressibility)Example: Textbook Problem 2.45Slide 35Surface TensionSlide 37Example: Capillary RiseExample: Textbook Problem 2.51Examples of Surface TensionExample: Textbook Problem 2.48Vapor Pressure (Pvp)Slide 43Vapor Pressure (Pvp) - continuedSlide 45Example: Relative HumiditySlide 47Saturation Vapor PressureSlide 491FLUID PROPERTIESFLUID PROPERTIESChapter 2CE319F: Elementary Mechanics of Fluids2Fluid Properties• Define “characteristics” of a specific fluid•Properties expressed by basic “dimensions”–length, mass (or force), time, temperature• Dimensions quantified by basic “units”We will consider systems of units, important fluid properties (not all), and the dimensions associated with those properties.3Systeme International (SI)•Length = meters (m)•Mass = kilograms (kg)•Time = second (s)•Force = Newton (N)–Force required to accelerate 1 kg @ 1 m/s2–Acceleration due to gravity (g) = 9.81 m/s2–Weight of 1 kg at earth’s surface = W = mg = 1 kg (9.81 m/s2) = 9.81 kg-m/s2 = 9.81 N • Temperature = Kelvin (oK)–273.15 oK = freezing point of water –oK = 273.15 + oC4Système International (SI)•Work and energy = Joule (J)J = N*m = kg-m/s2 * m = kg-m2/s2• Power = watt (W) = J/s• SI prefixes:G = giga = 109c = centi = 10-2 M = mega = 106m = milli = 10-3k = kilo = 103 = micro = 10-65English (American) System•Length = foot (ft) = 0.3048 m•Mass = slug or lbm (1 slug = 32.2 lbm = 14.59 kg)•Time = second (s)•Force = pound-force (lbf)–Force required to accelerate 1 slug @ 1 ft/s2• Temperature = (oF or oR)–oRankine = oR = 460 + oF•Work or energy = ft-lbf•Power = ft-lbf/s–1 horsepower = 1 hp = 550 ft-lbf/s = 746 WBanana SlugMascot of UC Santa Cruz6Density•Mass per unit volume (e.g., @ 20 oC, 1 atm)–Waterwater= 1,000 kg/m3 (62.4 lbm/ft3)–Mercury Hg= 13,500 kg/m3–Air air= 1.205 kg/m3•Densities of gases = strong f (T,p) = compressible•Densities of liquids are nearly constant (incompressible) for constant temperature•Specific volume = 1/density = volume/mass7Example: Textbook Problem 2.8•Estimate the mass of 1 mi3 of air in slugs and kgs. Assume air = 0.00237 slugs/ft3, the value at sea level for standard conditions8Example•A 5-L bottle of carbon tetrachloride is accidentally spilled onto a laboratory floor. What is the mass of carbon tetrachloride that was spilled in lbm?9Specific Weight•Weight per unit volume (e.g., @ 20 oC, 1 atm)water= (998 kg/m3)(9.807 m2/s)= 9,790 N/m3[= 62.4 lbf/ft3]air= (1.205 kg/m3)(9.807 m2/s)= 11.8 N/m3[= 0.0752 lbf/ft3]]/[]/[33ftlbformNg10Specific Gravity•Ratio of fluid density to density of water @ 4oC 3/1000 mkgSGliquidwaterliquidliquidWater SGwater = 1 Mercury SGHg = 13.55Note: SG is dimensionless and independent of system of units11Example•The specific gravity of a fresh gasoline is 0.80. If the gasoline fills an 8 m3 tank on a transport truck, what is the weight of the gasoline in the tank?12Ideal Gas Law (equation of state)TnRPVuTRVnPuRTRTVnMTMRVnMPuP = absolute (actual) pressure (Pa = N/m2)V = volume (m3)n = # molesRu = universal gas constant = 8.31 J/oK-molT = temperature (oK)R = gas-specific constant R(air) = 287 J/kg-oK (show)13Example•Calculate the volume occupied by 1 mol of any ideal gas at a pressure of 1 atm (101,000 Pa) and temperature of 20 oC.14Example•The molecular weight of air is approximately 29 g/mol. Use this information to calculate the density of air near the earth’s surface (pressure = 1 atm = 101,000 Pa) at 20 oC.15Example: Textbook Problem 2.4•Given: Natural gas stored in a spherical tank–Time 1: T1=10oC, p1=100 kPa–Time 2: T2=10oC, p2=200 kPa•Find: Ratio of mass at time 2 to that at time 1•Note: Ideal gas law (p is absolute pressure)16ViscosityViscosity17Some Simple Flows•Flow between a fixed and a moving plateFluid in contact with plate has same velocity as plate (no slip condition)u = x-direction component of velocityu=VMoving plateFixed plateyxVu=0ByBVyu )(Fluid18Some Simple Flows•Flow through a long, straight pipeFluid in contact with pipe wall has same velocity as wall (no slip condition)u = x-direction component of velocityrxR21)(RrVruVFluid19Fluid Deformation•Flow between a fixed and a moving plate•Force causes plate to move with velocity V and the fluid deforms continuously. u=VMoving plateFixed plateyxu=0Fluid t0t1t220Fluid Deformationu=V+VMoving plateFixed plateyxu=VFluid tt+txyLtFor viscous fluid, shear stress is proportional to deformation rate of the fluid (rate of strain)VLtyLyVtyV21Viscosity•Proportionality constant = dynamic (absolute) viscosity•Newton’s Law of Viscosity•Viscosity•Units•Water (@ 20oC):  = 1x10-3 N-s/m2•Air (@ 20oC):  = 1.8x10-5 N-s/m2•Kinematic viscositydydV /22///msNmsmmN V V+dv dydVKinematic viscosity: m2/s1 poise = 0.1 N-s/m21 centipoise = 10-2 poise = 10-3 N-s/m222Shear in Different Fluids•Shear-stress relations for different fluids•Newtonian fluids: linear relationship•Slope of line = coefficient of proportionality) = “viscosity”dydVdydVShear thinning fluids (ex): toothpaste, architectural coatings; Shear thickening fluids = water w/ a lot of particles, e.g., sewage sludge; Bingham fluid = like solid at small shear, then liquid at greater shear, e.g., flexible plastics23Effect of TemperatureGases: greater T = greater interaction between molecules = greater viscosity.Liquids: greater T = lower cohesive forces between molecules = viscosity down.2425Typical Viscosity EquationsSTSTTToo23Liquid:Gas:T = KelvinS = Sutherland’s constantAir = 111 oK+/- 2% for T = 170 – 1900 oKTbCeC and b = empirical constants26Flow between 2