Characteristics of Fluid FlowsLaminar and Turbulent FlowFigure 2.2PowerPoint PresentationSlide 5Friction LossesExample 2.4 pg 24Friction loss in fittingsFriction loss in sudden enlargements in pipeFlow of air through particlesSlide 11Slide 12Fluid AirSlide 14Flow of air through floorsSlide 16Figure 10.1Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Characteristics of Fluid FlowsChapter 2 Henderson, Perry and YoungBAE 2023 Physical Properties 1Laminar and Turbulent Flow•Laminar: fluid flows in parallel elements, velocity remains constant but not always the same as the adjacent element•Turbulent: fluid moves in elemental swirls or eddies…velocity and direction of each element change with timeBAE 2023 Physical Properties 2Figure 2.2BAE 2023 Physical Properties 3•Velocity is highest at the center•At surface velocity is zero•Reynolds found 4 factors affecting velocity–D V ρ and μ so (pipe dia., average V, density and dynamic viscosity of fluid)–Re = D V ρ /μ–Re< 2130….laminar–Re>4000….turbulent–2130<Re<4000….impossible to predict exactlyBAE 2023 Physical Properties 4•Can be used for other shapes by calculating the “hydraulic diameter”•Dh = 4 (area of cross section)/(wetted perimeter)•For pipe Dh = 4(πD2/4) / (πD) = DBAE 2023 Physical Properties 5Friction Losses•Darcy F=f(L/D)V2/2g)•f is a function of the Re and relative roughness of the pipe. Table 2.2•And moody diagram Fig 2.6•Use either –For Re < 2130 f = 64/Re –For Re>2130 f = 0.25/(log10(ε/3.7D)+2.51/(Re f 0.5)))2BAE 2023 Physical Properties 6Example 2.4 pg 24BAE 2023 Physical Properties 7Friction loss in fittings•Follows Bernoullis equation…F = K(V2/2g) K = friction loss factorK is empirically determined Table 2-3BAE 2023 Physical Properties 8Friction loss in sudden enlargements in pipe•F = (V1 – V2)2 / 2g Fig 2.7•If expansion is large….velocity 2 becomes 0 and drops from the equation aboveBAE 2023 Physical Properties 9Flow of air through particles•Fixed bed of granular material….resistance is a function of –size, –shape, –surface configuration, –size distribution, –Method of placement (affects void space)BAE 2023 Physical Properties 10Flow of air through particles•Fixed bed of granular material….resistance is a function of –size, –shape, –surface configuration, –size distribution, –Method of placement (affects void space)BAE 2023 Physical Properties 11Flow of air through particles•Much empirical testing…–F = Δp/ = ϒ f(L/Dp)(V2/2g)•(specific weight of fluid or air)–When the particles are not spheres…•Dp = 6vp/sp –particle volume and particle surface area–f = ((1 – εp)/ε3p)(300(1-εp)/(Re + 3.5) •εp= bed voidageBAE 2023 Physical Properties 12Fluid AirBAE 2023 Physical Properties 13Flow of air through particles•Pressure loss–Δp/L = a V2/ln(1+bV) for clean loosely packed material–a and b are constants…see Table 2-5–Use 1.5 for packed or dirty material–After Δp is calculated…•Bernoullis F = Δp/ϒBAE 2023 Physical Properties 14Flow of air through floors•Perforated floors or walls that contain products add energy loss•Floors: F = (1.071)(V/Of)2 / (ρg) see pg 29-30•With product on the floor:–F = (1.071)(V/Ofεp)2 / (ρg) includes voidage fraction of the materialBAE 2023 Physical Properties 15BAE 2023 Physical Properties 16Dr. C. L. JonesBiosystems and Ag. Engineering•Air and water are used to remove foreign material from products•How much air required depends on the drag force FD ( sum of skin friction and pressure drag)•FD affected by density, abs. viscosity, area and velocity (equation 10.1)•Reference Figure 10.1Aero/Hydrodynamic PropertiesBAE 2023 Physical Properties 17Figure 10.1BAE 2023 Physical Properties 18Dr. C. L. JonesBiosystems and Ag. Engineering•FD depends on the drag coefficient CD which is quantified using the Reynolds number.•NRe = Vdpρf/η Where: »V = fluid velocity»dp = particle dimension»ρf= fluid density»η = absolute viscosity•NRe<1.0, Stokes flow, FD=3πdpμV (sphere)•NRe<1,000 Laminar flow• NRe >20,000 Turbulent flowAero/Hydrodynamic PropertiesBAE 2023 Physical Properties 19Dr. C. L. JonesBiosystems and Ag. Engineering•Terminal velocity: occurs when drag force balances gravitational force•See Table 10.1•For a sphere–Fdrag=CD(πd2/4)(ρfv2/2)•CD depends on the Reynold number of the particle: Rep= ρfvd/μ (restated from eqt. 10.1 in different terms)•If Rep<0.2, CD=24/Rep•If Rep>200,000, CD=0.44•If Repis between 500 and 200,000, CD=(24/Rep)(1.0 + 0.15(Rep)0.687)Aero/Hydrodynamic PropertiesBAE 2023 Physical Properties 20Dr. C. L. JonesBiosystems and Ag. EngineeringLecture 17 – Aero/Hydrodynamic Properties (Ch. 10)BAE 2023 Physical Properties 21Dr. C. L. JonesBiosystems and Ag. EngineeringRead Example Problem 10.1. You will need to be familiar with it. This examples shows how to find a Reynolds number for a particle, the drag coefficient and the terminal velocityAero/Hydrodynamic PropertiesBAE 2023 Physical Properties 22Dr. C. L. JonesBiosystems and Ag. EngineeringApplication example: Can corn stalks be separated from corn cobs pneumatically? What minimum air velocity can be used?How well will it be separated?How could we improve the separation?Aero/Hydrodynamic PropertiesBAE 2023 Physical Properties 23Dr. C. L. JonesBiosystems and Ag. EngineeringApplication example: A seed company would like to move soybeans through a pipe (5.25” inside diameter) pneumatically. What capacity should the air source (the fan) be rated for?Aero/Hydrodynamic
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