Fluid Properties and Units CVEN 311 Continuum Continuum All materials, solid or fluid, are composed of molecules discretely spread and in continuous motion. However, in dealing with fluid-flow flow relations on a mathematical basis, it is necessary to replace the actual molecular structure by a hypothetical continuous medium, called the continuum. In a continuum, the physical variable at a point in space is the averaged value of the variable in a small sphere. How good is the assumption? 10-3 cm 3x10 10 molecules of air
Dimensions and Units The dimensions have to be the same for each term in an equation (dimensionally homogeneous: check correctness!) Dimensions of mechanics are length L time mass T M force F ma MLT - temperature Dimensions and Units Quantity Symbol Dimensions i Velocity V LT -1 Acceleration a LT - Area A L Volume L 3 Discharge Q L 3 T -1 Pressure p ML -1 T - Gravity g LT - Temperature T Mass concentration C ML -3 Dimensions and Units Quantity Symbol Dimensions Density ML -3 Dynamic viscosity ML -1 T -1 Kinematic viscosity L T -1 Surface tension MT - Bulk mod of elasticity E ML -1 T - These are fluid properties! How many independent properties? 4 Systems of Units 1. British Gravitational (BG) Mass=slug slug, length=ft ft, force=lb g=3.17 ft/s^. International (SI) Mass=kg, length=m, force=n g=9 9.81 81m/s^
Density, Specific Weight Density ρ= mass/volume Water @60F = 194slugs/ft^3 1.94 =999kg/m^3 Func. of tem. (Fig 1.1) Max. @4 C Specific weight γ= weight/volume =density x g Density and Specific Weight 3 ) Dens sity (kg/m 3 1000 990 980 Density (mass/unit volume) Specific mass 970 960 density of water: 1000 kg/m 3 950 density of air at atmospheric pressure and 1000 15 C: 1. kg/m 3 999 Specific Weight (weight 998 per unit volume) ) 997 = g = 9806 N/m 3 Density (k kg/m 3 ) 0 50 100 0 10 0 Specific gravity Definition of a Fluid SG= density / density of water @ 4C No unit SG of mercury (Hg) =? Density=? a fluid, such as water or air, deforms continuously when acted on by shearing stresses of any magnitude. - Munson, Young, Okiishi Water Oil Air Why isn t steel a fluid?
Fluid Deformation between Parallel Plates Shear Stress F b Side view Force F causes the top plate to have velocity U. What other parameters control how much force is required to get a desired d velocity? Distance between plates (b) Area of plates (A) Viscosity! U AU F b F A U b du dy Fb AU dimension of Tangential force per unit area U b N s m Rate of angular deformation N m change in velocity with respect to distance rate of shear 1 s Fluid classification by response to shear stress Fluid Viscosity Newtonian Ideal Fluid Ideal plastic du tion dy deformat Rate of d Ideal Fluid 1 Shear stress Newtonian Ideal plastic du dy Examples of highly viscous fluids tar, 0w-50 oil Fundamental mechanisms (Fig.1.6) Gases - transfer of molecular momentum Viscosity increases as temperature increases. Viscosity increases as pressure increases. Liquids - cohesion and momentum transfer Viscosity decreases as temperature increases. Relatively independent of pressure (incompressible)
Role of Viscosity Statics Fluids at rest have no relative motion between layers of fluid and thus du/dy = 0 Therefore the shear stress is zero and is independent of the fluid viscosity Flows Fluid viscosity is very important when the fluid is moving Dynamic and Kinematic Viscosity Kinematic viscosity ( ) is a fluid property obtained by dividing the dynamic viscosity ( ) by the fluid density kg m s kg 3 m [m /s] Connection to Reynolds number! N s s k kg m N m s Re VD War in solving Fluid-Structure Interaction Example: Measure the viscosity of water Hydraulician (Hydraulics): Experimentalists Hydrodynamicist (Hydrodynamics): Applied mathematicians Prandtl s Contribution: Boundary Layer The inner cylinder is 10 cm in diameter and rotates at 10 rpm. The fluid layer is mm thick and 10 cm high. h The power required to turn the inner cylinder is 50x10-6 watts. What is the dynamic viscosity of the fluid? Outer cylinder Inner cylinder Thin layer of water
Viscosity Measurement: Solution Absolute/Gage Pressure AU F t r h F t P Fr r 3 h P t U r A rh Outer cylinder r = 5 cm Inner t = mm cylinder h = 10 cm P = 50 x 10-6 W 10 rpm Thin layer of water Pt 3 r h Absolute pressure=wrt to vacuum Gage pressure=wrt atmospheric p Pa = Pg + 14.7 (psi) (or 100kPa) -6 (50x10 W) (0.00 m) 1.16x10 3 (1.047/s) (0.05 m) (0.1m) -3 N s/m Perfect Gas Law Bulk Modulus of Elasticity PV = mrt => P= R R is the gas constant T is in Kelvin (K=C+73) (or R=F+460) Use absolute pressure for P and absolute temperature for T Relates the change in volume to a change in pressure changes in density at high pressure pressure waves sound water hammer a ty (GPa) lus of elasticit Bulk Modu E v.35.30.5.0.15.10.05.00 E v speed of sound - dp dv / V E v Water dp d / 0 0 40 60 80 100
Speed of Sound Assisting Grader In Water: 1500 m/s In Air: 340 m/s J. K. Kim kimjk007@neo.tamu.edu edu 979 446 150 Who can communicate faster, dolphin or bat? CHAP.1 HW#1 4, 7, 11, 3, 6, 45, 60, 6, 70, 95 Due 1/7 Fr. (5:00pm) liquid Vapor Pressure Vapo or pressure e (Pa) 8000 7000 6000 5000 4000 3000 000 1000 0 0 10 0 30 40 What is vapor pressure of water at 100 C? 101 kpa Connection forward to cavitation!
Cavitation: How can you boil the water at room temperature? Cavitation Damage Surface Tension Example: Surface Tension 0.080 0.075 0.070070 0.065 0.060 pr 0.055 0.050 Pressure increase in a spherical droplet R pr = R Surfac ce tensio on (N/m) p R 0 0 40 60 80 100 Surface molecules l Estimate the difference in pressure (in Pa) between the inside and outside of a bubble of air in 0ºC water. The air bubble is 0.3 mm in diameter. R 0 15 10 3 0.073 N/m p R p 970 Pa p 0.1510 3 m R = 0.15 x 10-3 m = 0.073 N/m
Outline the solution Restate the goal Identify the given parameters and represent the parameters using symbols Outline your solution including the equations describing the physical constraints and any simplifying assumptions -3 1.16x1016x10 N s/m