1. Ch 9: Part B – Fluid Flow About Immersed Bodies Flow Stream U Drag = pressure + friction

2. Summary of Paradoxes (1) In the first experiment we found that sometimes an increase of speed actually produces a decrease of drag. (2) Sometime roughening increases drag and sometime it decreases drag. (3) Sometime streamlining increases drag and sometime it decreases drag.

3. FLUID FLOW ABOUT IMMERSED BODIES U p4p4 p1p1 p2p2 p3p3 p6p6 p5p5 p7p7 p8p8 p9p9 p 10 p 11 p 13 p…p… p 12  10 99 88 77 66 55 44 33 22 11 ………… Drag due to surface stresses composed of normal (pressure) and tangential (viscous) stresses. All we need to know is p and  on body to calculate drag. Could do for flat plate with zero pressure gradient because U and p, which were constant, we knew everywhere. If  = 0 then pressure distribution is symmetric, so no net pressure force (D’Alembert’s Paradox - 1744) DRAG LIFT

4. L O W Re D H I G H Re D

5. DRAG Coefficient - C D F D = f(d,V, ,  )* C D = F D /( 1 / 2  U 2 A) = f(Re) * ignored roughness

6. C D on flat plate (no pressure gradient) in laminar and turbulent flow

7. DRAG COEFFICIENT - C D C D = F D / ( 1 / 2  U 2 A) Flow over a flat plate: F D =  plate surface  w dA C D =  PS  w dA / ( 1 / 2  U 2 A) C f =  w /( 1 / 2  U 2 ) {C f = shear stress or skin friction coef.} C D = (1/A)  PS C f dA (good for laminar and turbulent flow)

8. Flow over a flat plate with zero pressure gradient: C D = (1/A)  PS C f dA C f = 0.664/Re 1/2 for laminar flow (Blasius solution – flat plate laminar flow & no pressure gradient) C D = (1/A)  A (0.664/Re 1/2 ) dA = (bL) -1  0 L (0.664 U -1/2 x -1/2 1/2 ) bdx = (0.664/L) ( /U) 1/2 (2)x 1/2  o L = 1.33 ( / LU) 1/2 C D = 1.33 (Re L ) -1/2 for laminar flow over a flat plate, with no pressure gradient

9. Flow over a flat plate with zero pressure gradient: C D = (1/A)  PS C f dA C f = 0.0594/Re 1/5 for turbulent flow (u/U = [y/  ] 1/7 ) (Blasius correlation: f = 0.316/Re 1/4 ; Re  10 5 ) C D = (1/A)  A (0.0594/Re 0.2 ) dA = (bL) -1  0 L (0.0594 (U/ ) -0.2 x -0.2 bdx = (0.0594/L) ( /U) 0.2 [x 0.8 /0.8]  o L = 0.0742( /UL) 0.2 C D = 0.0742 (Re L ) –0.2 for turbulent flow over a flat plate, with no pressure gradient - 5x10 5 <Re L <10 7

10. C D = 1.33 (Re L ) -1/2 for laminar flow over a flat plate, with no pressure gradient ~ Re < 5x10 5 C D = 0.0742 (Re L ) –0.2* for turbulent flow over flat plate, with no pressure gradient ~ 5x10 5 <Re L <10 7 C D = 0.455/ log (Re L ) 2.58* for turbulent flow over flat plate, with no pressure gradient ~ Re L <10 9 * Assumes turbulent boundary layer begins at x=o

11. C D correction term for partly laminar / partly turbulent

12. C D correction term for partly laminar / partly turbulent ? ADD OR SUBTRACT CORRECTION TERM ???

13. Must account for fact that turbulence does not start at x = 0 -must subtract B/Re L C D correction term = B/Re L = Re tr (C Dturb – C dlam )/Re L Re tr

14. C D correction term = B/Re L = Re tr (C Dturb – C Dlam )/Re L For Re tr = 5 x 10 5 C D = 0.0742/Re L 1/5 – Re tr (C Dturb – C Dlam )/Re L C D = 0.0742/Re L 1/5 – 5x10 5 [0.0742/ (5x10 5 ) 1/5 –1.33/(5x10 5 ) 1/2 ]/Re L C D = 0.0742/Re L 1/5 – 1748/Re L Re tr 5 x 10 5 < Re L < 10 7

15. C D correction term = B/Re L = Re tr (C Dturb – C Dlam )/Re L For Re tr = 5 x 10 5 C D = 0.0742/Re L 1/5 – Re tr (C Dturb – C Dlam ) C D =0.0742/Re L 1/5 –5x10 5 [0.455/ (log[5x10 5 ]) 1/5 –1.33/(5x10 5 ) 1/2 ] C D = 0.455/(logRe L ) 2.58 – 1600/Re L Re tr 5 x 10 5 < Re L < 10 9

16. SMOOTH FLAT PLATE NO PRESSURE GRADIENT C D = 0.0742 (Re L ) –0.2 C D = 0.455/ log (Re L ) 2.58 C D = 1.33 (Re L ) -1/2

17. Rough Flat Plate

18. FLAT PLATE C D = D/( ½  U 2 A) Re L

19. PIPE FLAT PLATE C D = D/( ½  U 2 A) f = (dp/dx)D/( ½  U 2 )

21. Flat Plate Perpendicular to Flow Direction C D = F D /( 1 / 2  U 2 bh) for Re h > 1000, C D very weak function of Re. C D ~ 2 Newton “guessed” Separation points fixed

22. Drag Force =  p/  t =  (mv)/  t m ~  UA f = mass per second passing through area  v ~ U-0 = U C D = D/( 1 / 2  U 2 A f ) ~  UA f U/( 1 / 2  U 2 A f ) C D ~ 2 Newton Value is right order of magnitude, & Re insensitivity predicted correctly. (fixed)

23. Mostly pressure drag, separation point fixed Friction drag Character of C D vs Re curves for different shapes press & fric

25. Flow parallel to plate – viscous forces important and Re dependence Flow perpendicular to plate – pressure forces important and no strong Re dependence What about Re dependence for flow around sphere? Re CDCD ?

26. Drag Coefficient, C D, as a function of Re for a Smooth Sphere SMOOTH SPHERE

27. Drag Coefficient, C D, as a function of Re for a Smooth Sphere SMOOTH SPHERE F D = 3  VD C D = ?

28. C D = F D /(½  U 2  R 2 ) = 6  UR/(½  U 2  R 2 ) = 24/Re Laminar boundary layer Turbulent flow in wake Separation point moving forward Separation point fixed 95% of drag due to pressure difference between front and back Turbulent boundary layer Laminar Flow * *

29. IDEAL FLOW* LAMINAR  FLOW TURBULENT  FLOW S e p a r a t i o n

30. ~82 o ~120 o PRESSURE DRAG DRAG IF NO VISCOSITY WHAT WOULD BE TOTAL DRAG ?

31. Smooth Trip By roughening surface can “trip” boundary layer so turbulent which results in a favorable momentum exchange, pushing separation point further downstream, resulting in a smaller wake and reduced drag. 125 yd drive with smooth golf ball becomes 215 yds for dimpled* From Van Dyke, Album of Fluid Motion Parabolic Press, 1982; Original photographs By Werle, ONERA, 1980 Re = 15000 Re = 30000

32. Drag coefficient as a function of Reynolds number for smooth circular cylinders and smooth spheres. From Munson, Young, & Okiishi, Fundamentals of Fluid Mechanics, John Wiley & Sons, 1998 ASIDE: At low very low Reynolds numbers Drag   UL C D = D / ( 1 / 2  U 2 A f ) D ~ U C D = constant D ~ U 2

33. Drag coefficient as a function of Reynolds number for smooth circular cylinders and smooth spheres. From Munson, Young, & Okiishi, Fundamentals of Fluid Mechanics, John Wiley & Sons, 1998 ASIDE: At low very low Reynolds numbers Drag   UL C D = D / ( 1 / 2  U 2 A f ) D ~ U C D = constant D ~ U 2

34. Drag coefficient as a function of Re for a smooth cylinder and smooth sphere. Re Dcrit ~ 3 x 10 5 3-D relieving effect C dcylinder >C Dsphere Is Re Dcritical constant?

35. Effect of surface roughness on the drag coefficient of a sphere in the Reynolds number range where laminar boundary layer becomes turbulent.

36. vortex shedding

37. Theodore Von Karman

38. A B C D E FLOW AROUND A SMOOTH CYLINDER ~82 o ~120 o Smooth Sphere

39. Vortex Shedding St = UD/f =0.21 for 10 2 < Re < 10 7 PICTURE OF SHEDDING

41. Flow Separation

42. FLOW SEPARATION

43. Fig. 9.6 U upstream = 3 cm/sec; divergent angle = 20 o ; Re= 900; hydrogen bubbles Unfavorable pressure gradient necessary for flow separation to be “possible” but separation not guaranteed.

44. Water, velocity = 2 cm/s, cylinder diameter = 7 cm, Re = 1200 Photographed 2 s after start of motion; hydrogen bubble technique Back flow 0 velocity at y = dy

45. Favorable Pressure Gradient  p/  x < 0; U increasing with x Unfavorable Pressure Gradient  p/  x > 0; U decreasing with x When velocity just above surface = 0, then flow will separate; causes wake. Gravity “working”against friction Gravity “working” with friction

46. Viscous flow around streamlined body streamlines diverge velocity decreases adverse pressure gradient streamlines converges velocity increases adverse pressure gradient

47. Favorable Pressure Gradient  p/  x < 0; U increasing with x Unfavorable Pressure Gradient  p/  x > 0; U decreasing with x When velocity just above surface = 0, then flow will separate; causes wake. Gravity “working”against friction Gravity “working” with friction

48. Streamlining

49. STREAMLINING First employed by Leonardo da Vinci – First coined by d’Arcy Thompson – On Growth and Form (1917) C D ~ 0.06C D ~ 2 for flat plate

50. STREAMLINING (a) (b)

51. C D = F D /( 1 / 2  U 2 A) F D = C D ( 1 / 2  U 2 A) C D = 2.0 C D = 1.2 C D = 0.12 C D = 1.2 C D = 0.6 d = d/10 d = As C D decreases, what is happening to wake? Is there a wake associated with pipe flow? If C D decreases does that necessarily imply that the drag decreases? 2 - D

52. (note that frictional force increased from (b) to (c) but net force decreased) (note that although C D decreased from (d) to (e) that the Drag force did not. C D = 2.0 C D = 1.2 C D = 0.12 C D = 1.2 C D = 0.6 * * * *

53. First flight of a powered aircraft 12/17/03 120ft in 12 seconds Orville Wright at the controls Same drag at 210 mph

54. The End