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u/U = sin( /2);  = y/

u/U = sin( /2);  = y/. L A M I N A R. T A B L E 9.2. Given U and viscosity table 9.2 Sketch (x), *(x),  w (x). u/U = sin( /2);  = y/. Sketch (x), *(x),  w (x). u/U = sin( /2);  = y/. …. . W H I C H H A S M O R E F D ?. A. VS. B. U.

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u/U = sin( /2);  = y/

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  1. u/U = sin(/2);  = y/ L A M I N A R T A B L E 9.2 Given U and viscosity table 9.2 Sketch (x), *(x), w(x)

  2. u/U = sin(/2);  = y/ Sketch (x), *(x), w(x)

  3. u/U = sin(/2);  = y/ ….. 

  4. W H I C H H A S M O R E FD ? A VS B

  5. U Which plate experiences the most drag? LAMINAR / dp/dx = 0 Incompressible, steady, 2-D… A B

  6. L=L U Which plate experiences the most drag? A b=4L FD = wdA A = bL Cf = w/( ½ U2) = 0.730/Rex FD = 0.73b(LU3)1/2 FD(A) = 0.73(4L)(LU3)1/2 FD(B) = 0.73(L)(4LU3)1/2 FD(A)/FD(B) = 2 L=4L B b=L

  7. L U A b How Does Drag Change if 2U?

  8. L U How Does Drag Change if 2U? A b FD = wdA A = bL Cf = w/( ½ U2) = 0.730/Rex FD = 0.73b(LU3)1/2 If U goes to 2U, FD goes to (8)1/2

  9. (*/ = 0.344) Table 9.2

  10. (*/ = 0.344)

  11. 0.344

  12. y /  u / U Sinusoidal, parabolic, cubic look similar to Blasius solution.

  13. u/Umax = (y/)1/7 u/Ue = 2(y/) – (y/)2

  14. I give you Rexand x, how do you get w(x)? DL?

  15. Laminar boundary-layer over flat plate, zero pressure gradient U = 3.2 m/s 0.9 m 1.8 m u/U = sin[(/2)(/y)] Drag = ? using Table 9.2

  16. FD = wdA =  U2 d/dx bdx = U2 bL

  17. FD = wdA =  U2 d/dx bdx = U2 bL x=L

  18. u/y|Wall = /y U[(2y/) – (y/)2]y=0 u/y|Wall = U2/ PIPE: u/y|Wall = Ucl2/R Fully developed laminar flow

  19. u/y = /y {y/}1/7 = 1/7 {y/}-6/7 u/y|y=0 = 

  20. So how get wall for turbulent velocity profile u/U0 = (y/)1/7 ?

  21. u/Uo = (y/)1/7 w = 0.0332 (V)2[/(RV)]1/4 Eq(8.39) V/Uc/l = (2n2)/[(n+1)(2n+1)] Eq(8.24) w = 0.0233 (U0)2[/(Uo)]1/4

  22. If Re transition occurs at 500,000 what is the skin friction drag on fin?

  23. FD = CD (½  Uo2A) Laminar Flow – starting at x=0 CD = 1.33/ReL1/2 Turbulent Flow – starting at x=0 CD = 0.0742/Re1/5 {u/U = (y/)1/7} for 5x105<ReL<107 CD = 0.455 /(log ReL)2.58 For ReL < 109

  24. Calculating transition location, x: Assume xtr occurs at 500,000 Rex = Ux/ = 500,000 ReL = UL/ = 1.54 x 107 Rex/ReL = xtr/L = 500,000 / 1.54 x 107 xtr = 0.0325 L = 0.0325 x 1.65 = 0.0536 m

  25. ReL = UL/ = 1.54 x 107 Turbulent Flow – starting at x=0 CD = 0.455 /(log ReL)2.58 xtr = 0.054m L = 1.65m

  26. FLAT PLATE dp/dx = 0 LAMINAR: From Theory CD = 1.33 / ReL1/2 Eq. 9.33 TURBULENT: From Experiment 5x105< ReL<107 xtransition = 0 CD = 0.0742/ReL1/5 ReL<109 xtransition = 0 CD = 0.455/(logReL)2.58

  27. For Re transition of 500,000 and 5x105 < ReL < 109 CD = 0.455/(log ReL)2.58 – 1610/ReL (Eq. 9.37b) CD = 0.00281 – 0.000105 = 0.0027 ~ 4% less FD = CD2LH(1/2)U2 = 91.6 N 2 sides so 2 x Area

  28. CD = FD/( ½ Uo2); Cf(x) = w(x)/(1/2 Uo2) CD ~ 0.0028 Smooth flat plate, dp/dx = 0; flow parallel to plate

  29. CD = FD/( ½ Uo2); Cf(x) = w(x)/(1/2 Uo2) Where does transition occur for these data? These data include rough plates?

  30. CD = FD/( ½ Uo2); Cf(x) = w(x)/(1/2 Uo2) ReL

  31. breath

  32. Plastic plate falling in water, find terminal velocity -

  33. Plastic plate falling in water, find terminal velocity - + z DRAG BOUYANCY WEIGHT Given  of water and plastic - z Sum of forces = 0 FD = CD ½  U2A

  34. Plastic plate falling in water, find terminal velocity - + z DRAG BOUYANCY WEIGHT Sum of forces = 0 Drag Force = f (U) - z Assume: turbulent flow, transition at leading edge, doesn’t flutter, 5 x 105 < ReL < 107 Then CD = 0.0742/Reh1/5

  35. Net Force = 0 = FB + FD - W FB – W = (plastic - water) g (hLt) FD = CD ½  U2A = [0.0742/Reh1/5] ½ water U2A Reh = Uh/ Solving for U get 11.0 ft /sec Check Re number: 1.86 x 106 5 x 105 < ReL < 107

  36. THE END

  37. T H E E N D

  38. Extra Examples

  39. U2 = ? P2 – P1 = ? 2= 1.2 in. U1 = Uo = 80 ft/sec 1= 0.8 in.

  40. Continuity U1A1eff = U2A2eff A1eff = (h -2 *1) (h -2 *1) A2eff = (h -2 *2) (h -2 *2) *1 u/U = y/ =  * = 0.8 in *1 *1

  41. Continuity U1A1eff = U2A2eff A1eff = (h -2 *1) (h -2 *1) A2eff = (h -2 *2) (h -2 *2) *2 *2 = 1.2 in *2 *2

  42. First a bit of mathematic legerdemain before attempting Ex. 9.3 ~ = ?

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