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External Flows CEE 331 September 11, 2012 School of Civil and Monroe L. Weber-Shirk Environmental Engineering Overview Non-Uniform Flow Boundary Layer Concepts Viscous Drag Pressure Gradients: Separation and Wakes Pressure Drag Shear and Pressure Forces Vortex Shedding Non-Uniform Flow In pipes and channels the velocity distribution was uniform (beyond a few pipe diameters or hydraulic radii from the entrance or any flow disturbance) In external flows the boundary layer is always growing and the flow is non- uniform Boundary Layer Concepts Two flow regimes Laminar boundary layer Turbulent boundary layer with laminar sub-layer Calculations of boundary layer thickness Shear (as a function of location on the surface) Drag (by integrating the shear over the entire surface) Flat Plate: Parallel to Flow U U U to boundary layer y thickness U d x shear Why is shear maximum at the leading edge of the plate? du is maximum dy Laminar Boundary Layer: Shear and Drag Force d 5 Ux nx = Re x = d =5 x Re x n U square Boundary Layer thickness increases with the _______ root ______ of the distance from the leading edge of the plate Based on momentum and mass U 3 t 0 0.332 conservation and assumed x velocity distribution U 3 l l Fd w t 0 dx w 0.332 dx Integrate along length of plate 0 0 x Fd 0.664w U 3l On one side of the plate! Laminar Boundary Layer: Coefficient of Drag 2FD Fd 0.664w U l 3 Cd = 2 = f (Re) rU A 2(0.664) w U 3l Dimensional analysis Cd U 2lw 1.328w U 3l 1.328 1.328 Cd Cd Cd = U 2lw Ul Re Ul Rel = n Transition to Turbulence The boundary layer becomes turbulent when the Reynolds number is approximately 500,000 (based on length of the plate) The length scale that really controls the transition to turbulence is the boundary layer thickness _________________________ Ux Ud Red d d 5 Red = 5 Re x Re x = Red = = = = n n Re x x x Re x Red = 3500 Transition to Turbulence U U U d y turbulent U x Viscous sublayer to This slope (du/dy) controls t0. Transition (analogy to pipe flow) Turbulent Boundary Layer: (Smooth Plates) 1/ 5 d 0.37 Ux 4/5 = 1/ 5 Re x = d 0.37 x x Re x n U Grows ____________ than laminar more rapidly 1/ 5 Derived from momentum conservation t 0 0.029 U 2 Ux and assumed velocity distribution x 5/4 1/ 5 l Fd w t 0 dx 0.036 U wl 2 Integrate shear over plate 0 Ul 2FD æ eö Cd = 0.072Rel- 1/ 5 5 x 105 < Rel < 107 Cd = = f Re, 2 rU A è lø Boundary Layer Thickness Water flows over a flat plate at 1 m/s. Plot the thickness of the boundary layer. How long is the laminar region? 0.1 10,000,000 laminar Ux 0.09 turbulent 9,000,000 Re x = 0.08 Reynolds Number 8,000,000 n boundary layer thickness (m) . 0.07 7,000,000 1/ 5 n Re x Reynolds Number 0.06 d 0.37 x 4/5 6,000,000 x= 0.05 U 5,000,000 U 0.04 4,000,000 1x10 6 m 2 / s (500,000) x 0.03 nx 3,000,000 1m / s 0.02 d =5 2,000,000 U x = 0.5 m 0.01 1,000,000 0 - 0 1 2 3 4 5 6 length along plate (m) Grand Coulee Flat Plate Drag Coefficients 0.01 - 2.5 1 x 10-3 CDf = [1.89 - 1.62log (e / l )] 5 x 10-4 2 x 10-4 1 x 10-4 e 5 x 10-5 2 x 10-5 l CDf 1 x 10-5 5 x 10-6 2 x 10-6 1 x 10-6 CDf = 0.072Rel- 0.2 1.328 0.455 1700 CDf = CDf = 2.58 - (Rel )0.5 [log (Re )] l Rel 0.455 0.001 CDf = 2.58 [log (Re )] l 4 5 6 7 8 9 0 +0 +0 +0 +0 +0 +0 +1 1e 1e 1e 1e 1e 1e 1e Ul Rel = n Example: Solar Car Solar cars need to be as efficient as possible. They also need a large surface area for the (smooth) solar array. Estimate the power required to counteract the viscous drag at 40 mph Dimensions: L: 5.9 m W: 2 m H: 1 m Max. speed: 40 mph on solar power alone Solar Array: 1200 W peak air = 14.6 x10-6 m2/s air = 1.22 kg/m3 Viscous Drag on Ships The viscous drag on ships can be calculated by assuming a flat plate with the wetted area and length of the ship FD Fviscous Fwave 0.01 - 2.5 1 x 10-3 CDf = [1.89 - 1.62log (e / l )] 5 x 10-4 2 x 10-4 1 x 10-4 e 2Fd Cd r U 2 A 5 x 10-5 2 x 10-5 l Cd CDf 1 x 10-5 5 x 10-6 2 x 10-6 Fviscous = 1 x 10-6 C Df = 0.072 Re l- 0.2 1.328 0.455 1700 CDf = CDf = 2.58 - (Rel )0.5 [ log (Re )] Rel U A l 0.455 C Df = 2 2.58 2 0.001 [ log (Re )] l 4 5 6 7 8 9 0 +0 +0 +0 +0 +0 +0 +1 1e 1e 1e 1e 1e 1e 1e Ul Rel = n Ul Lr3 Fwave scales with ____ Rel = n Separation and Wakes Separation often occurs at sharp corners fluidcan’t accelerate to go around a sharp corner Velocities in the Wake are ______ (relative small to the free stream velocity) Pressure in the Wake is relatively ________ constant (determined by the pressure in the adjacent flow) Flat Plate: Streamlines 3 U Cp 1 v2 p p0 2 2 2 U 2 4 U 0 1 Point v Cp p 1 0 1 >p0 2 <U >0 >p0 3 >U <0 <p0 4 <p0 Points outside boundary layer! Application of Bernoulli Equation p1 v12 p2 2 v2 In air pressure change due to h1 h2 2g 2g elevation is small 2 2 p0 U p v U = velocity of body relative to fluid 2g 2g U2 v2 p p0 2g 2g v2 p p0 1 2 2 U 2 Cp U Flat Plate: Pressure Distribution U 2C p p p0 p p0 C p 2 2 2 v Cp 1 p p0 2 2 U U 2 U 2 >U 3 Fd Fd front Fd rear <U 2 Fd p front prear A U 2 0 1 Fd C p front Cp rear 2 A U 2 Fd 0.8 1.2 A 2 1 0.8 0 Cp -1 -1.2 Cd = 2 Drag of Blunt Bodies and Streamlined Bodies Drag dominated by viscous streamlined drag, the body is __________. Drag dominated by pressure drag, the body is _______. bluff Whether the flow is viscous- drag dominated or pressure- drag dominated depends entirely on the shape of the body. Bicycle page at Princeton Velocity and Drag: Spheres æe ö Cd = f , Re, M, shape, orientation General relationship for èD ø submerged objects Spheres only have one shape and orientation! 2FD Cd U 2 A 2FD C d U 2 A Where C is a function of Re 2 = Cd = f (Re) FD d rU A 2 How fast do particles fall in dilute suspensions? Whatare the important parameters? Initialconditions Acceleration due to gravity After falling for some time... drag What principle or law could help us? Newton’s Second Law... Sedimentation: Particle Terminal Fall Velocity p particle volume F ma Fb Ap particle cross sectional area Fd Fb W 0 ρ p particle density Fd W ppg ρw water density Fb p w g g acceleration due to gravity 2 C D drag coefficient Vt Fd CD AP w Vt particle terminal velocity 2 W 2FD 4 Cd p r 3 Ap r 2 3 U 2 A Particle Terminal Fall Velocity (continued) Fd W Fb Vt 2 CD AP w p ( p w ) g 2 2 p ( p w ) g Vt 2 General equation for falling objects C D AP w p 2 d Relationship valid for spheres Ap 3 4 gd p w 4 gd p w Vt 2 Vt 3 CD w 3 CD w Drag Coefficient on a Sphere 1000 Drag Coefficient 100 Stokes Law 10 1 0.1 0.1 1 10 102 103 104 105 106 107 24 Re=500000 CD = Reynolds Number Re Turbulent Boundary Layer Drag Coefficient for a Sphere Equations 24 d 2 g p w Laminar flow R < 1 CD = Vt Re 18 4 C @ 24 3 Transitional flow 1 < R < 10 D + 1/ 2 + 0.34 Re Re gd ( r p - r w ) Fully turbulent flow R > 104 CD @0.4 Vt » 0.3 rw Vt d r Re = m Example Calculation of Terminal Velocity Determine the terminal settling velocity of a cryptosporidium oocyst having a diameter of 4 m and a density of 1.04 g/cm3 in water at 15°C. d 2 g p w ρ p 1040 kg/m 3 Vt ρw 999 kg/m 3 18 g 9.81 m/s 2 Vt 4x10 6 m 9.81 m/s 2 1040 kg/m 3 999 kg/m 3 2 kg 1.14x10 3 d 4x10 6 m 18 sm kg 1.14x10 3 Vt 3.14 x107 m/s sm Vt 2.7 cm/day Reynolds Pressure Gradients: Separation and Wakes Diverging streamlines Van Dyke, M. 1982. An Album of Fluid Motion. Stanford: Parabolic Press. Adverse Pressure Gradients Streamlines diverge behind object Increasing pressure in direction of flow p V2 Fluid is being decelerated +z+ =C g 2g Fluid in boundary layer has less ______ inertia than the main flow and may be completely stopped. If boundary layer stops flowing then separation occurs Point of Separation Predicting the point of separation on smooth bodies is beyond the scope of this course. Expect separation to occur where streamlines are diverging (flow is slowing down) Separation can be expected to occur around any sharp corners (where streamlines diverge rapidly) Drag on Immersed Bodies (more shapes) Figures9.19-21 bodies with drag coefficients on p 392-394 in text. hemispherical shell 0.38 Why? hemispherical shell 1.42 Velocity at cube 1.1 separation point parachute 1.4 determines pressure in wake. Vs ? The same!!! Shear and Pressure Forces Shear forces viscous drag, frictional drag, or skin friction caused by shear between the fluid and the solid surface length function of ___________and ______of object surface area Pressure forces pressure drag or form drag caused by _____________from the body flow separation function of area normal to the flow Example: Matrix Power Cd = 0.32 Height = 1.539 m Width = 1.775 m Length = 4.351 m Ground clearance = 15 cm 100 kW at 6000 rpm Max speed is 124 mph Where does separation occur? Calculate the power required to overcome drag at 60 mph and 120 mph. What is the projected area? ( A @ H - G )W A = (1.539m - 0.15m)1.775m = 2.5m2 Electric Vehicles Electric vehicles are designed to minimize drag. Typical cars have a coefficient drag of 0.30-0.40. The EV1 has a drag coefficient of 0.19. Smooth connection to windshield Drag on a Golf Ball DRAG ON A GOLF BALL comes mainly from pressure drag. The only practical way of reducing pressure drag is to design the ball so that the point of separation moves back further on the ball. The golf ball's dimples increase the turbulence in the inertia boundary layer, increase the _______ of the boundary layer, and delay the onset of separation. The effect is plotted in the chart, which shows that for Reynolds numbers achievable by hitting the ball with a club, the coefficient of drag is much lower for the dimpled ball. Why not use this for aircraft or cars? Effect of Turbulence Levels on Drag Flow over a sphere: (a) Reynolds number = 15,000; (b) Reynolds number = 30,000, with trip wire. Causes boundary layer to become turbulent Point of separation Effect of Boundary Layer Transition Ideal (non Real (viscous) Real (viscous) viscous) fluid fluid: laminar fluid: turbulent boundary layer boundary layer Increased inertia in No shear! boundary layer Spinning Spheres What happens to the separation points if we start spinning the sphere? LIFT! Vortex Shedding Vortices are shed alternately from each side of a cylinder The separation point and thus the resultant drag force oscillates nd S Frequency of shedding (n) given U by Strouhal number S S is approximately 0.2 over a wide range of Reynolds numbers (100 - 1,000,000) Summary: External Flows Spatially varying flows boundary layer growth Example: Spillways Two sources of drag shear (surface area of object) pressure (projected area of object) Separation and Wakes Interaction of viscous drag and adverse pressure gradient Solution: Solar Car 0.455 Ul CDf = 2.58 U = 17.88 m/s Rx [log (Rel )] 2Fd 0.01 l = 5.9 m CDf = [1.89 - 1.62log (e / l )] - 2.5 1 x 10-3 5 x 10-4 Cd 2 x 10-4 1 x 10-4 e 5 x 10-5 2 x 10-5 l CDf 1 x 10-5 5 x 10-6 air = 14.6 x 10-6 m2/s 2 x 10-6 U 2 A 1 x 10-6 C Df = 0.072 Re l- 0.2 1.328 0.455 1700 CDf = CDf = 2.58 - (Rel )0.5 [ log (Re )] l Rel C Df = 0.455 2.58 0.001 [ log (Re )] l 4 5 6 7 8 9 0 +0 +0 +0 +0 +0 +0 +1 C d U 2 A 1e 1e 1e 1e 1e 1e 1e Ul Rel = air = 1.22 kg/m3 n Fd 2 (3x10 )(1.22kg / m ) (17.88m / s ) (11.88m ) -3 3 Rel = 7.2 x 106 2 2 Fd =2 2 Cd = 3 x 10-3 Fd =14 N A = 5.9 m x 2 m = 11.8 m2 P =F*U=250 W Reynolds Number Check Vd R R 3.14 x10 7 m/s 4 x10 6 m 999kg/m 3 3 kg 1.14x10 sm R = 1.1 x 10-6 R<<1 and therefore in Stokes Law range Solution: Power a Toyota Matrix at 60 or 120 mph 2FD Cd f (R ) U 2 A C d U 2 A FD 2 C d U 3 A P 2 (0.32)(1.2kg / m3 )(26.82m / s)3 (2.5m2 ) P= 2 P = 9.3 kW at 60 mph P = 74 kW at 120 mph Grand Coulee Dam Turbulent boundary layer reaches surface! Drexel SunDragon IV http://cbis.ece.drexel.edu/SunDragon/Cars.html Vehicle ID: SunDragon IV (# 76) Dimensions: L: 19.2 ft. (5.9 m) W: 6.6 ft. (2 m) H: 3.3 ft. (1 m) Weight: 550 lbs. (249 kg) Solar Array: 1200 W peak; 8 square meters terrestrial grade solar cells; manf: ASE Americas Batteries: 6.2 kW capacity lead-acid batteries; manf: US Battery Motor: 10 hp (7.5 kW) brushless DC; manf: Unique Mobility Range: Approximately 200 miles (at 35 mph on batteries alone) Max. speed: 40 mph on solar power alone, 80 mph on solar and battery power. Chassis: Graphite monocoque (Carbon fiber, Kevlar, structural glass, Nomex) Wheels: Three 26 in (66 cm) mountain bike, custom hubs Brakes: Hydraulic disc brakes, regenerative braking (motor) Pressure Coefficients on a Wing v2 p p0 Cp 1 2 NACA 63-1-412 Flowfield Cp's U2 U 2 Shear and Pressure Forces: Horizontal and Vertical Components FD p sin t 0 cos dA drag Parallel to the approach velocity FL p cos t 0 sin dA lift Normal to the approach velocity p < p0 U 2 negative pressure Fd Cd A 2 A defined as projected normal area _______ to force! U 2 U FL CL A lift 2 drag p > p0 positive pressure