2. A model of a river boat is to be tested at 1:13.5 scale. The boat is designed to travel at 9mph in fresh water at 10 o C. (1)Estimate the distance from the bow on the full-scale ship where transition occurs. (2) Where should transition be stimulated on model? What equations are relevant to this problem? What approach should we take?

3. A model of a river boat is to be tested at 1:13.5 scale. The boat is designed to travel at 9mph in fresh water at 10 o C. Estimate the distance from the bow where transition occurs. Where should transition be stimulated on model? Things we know: Re = UL/ Re L transitions ~ 5 x 10 5 (experimentally determined for flat plate and dP/dx = 0) Want same fraction of L to be turbulent for both cases: [x tr /L]model = [x tr /L]boat so ration of x tr = ratio of boat lengths = 13.5

4. ?

6. Three unknowns, A, B, and C- will need three boundary conditions. What are they?

11. (  */  = 0.344)

12. 0.344

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

15. u/U e = 2(y/  ) – (y/  ) 2 u/U max = (y/  ) 1/7

19. R

20. R (interesting) u = (-R 2 /4  )(dp/dx)(1-[r/R]) 2 eq. 8.12  wall = (R/2)(dp/dx) U max = (-R 2 /4  )(dp/dx) du/dr = -2U max /R du/dy = 2U max /R

21. * *

24. 1 m U = 2.7 m/s cd

25. (5) Two-Dimensional (6) Incompressible assumed assumptions

27. cd u(x,y)/U = y/  =  (y) 1 0

28. cd

29. F Drag cd 

30. d c

32. U 1 = U o = 80 ft/sec  1 = 0.8 in. U 2 = ? P 2 – P 1 = ?  2 = 1.2 in.

33. Continuity U 1 A 1eff = U 2 A 2eff A 1eff = (h -2  * 1 ) (h -2  * 1 ) A 2eff = (h -2  * 2 ) (h -2  * 2 ) *1*1 *1*1  * = 0.8 in *1*1 u/U = y/  = 

34. Continuity U 1 A 1eff = U 2 A 2eff A 1eff = (h -2  * 1 ) (h -2  * 1 ) A 2eff = (h -2  * 2 ) (h -2  * 2 ) *2*2 *2*2  * 2 = 1.2 in *2*2

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

38. L = 1.8 m U = 3.2 m/s  = ?;  * = ?;  w = ? b = 0.9m Laminar and assume u/U = sin(  /2);  = y/   */  =  1 o (1-u/U) d   /  =  1 o u/U(1-u/U) d   w =  U 2 d  /dx =  u/  y  0

39.  w =  u/  y  0 =  (U/  ) d(u/U)/d(y/  )  0  w =  (U/  ) dsin(  /2) /d   0  w =  (U/  )(  /2) cos(  /2)  0  w =  (U/  )(  /2) =   U/(2  ) u/U = sin(  /2);  = y/ 

40.  */  =  1 o (1-u/U) d  u/U = sin(  /2);  = y/   */  =  1 o (1 – sin(  /2) )d   */  =  + (2/  )cos (  /2)  from 1 to 0  */  = 1 - 2/  = 0.363  * = 0.363 

41.  /  =  1 o u/U(1-u/U) d   = 0.137 

42. Finding   */  =  1 o (1-u/U) d   /  =  1 o u/U(1-u/U) d   w =  U 2 d  /dx

43.  w =  u/  y  0

44. ….. u/U = sin(  /2);  = y/ 

47. F drag =  0 L  w bdx =  0 L  U 2 (d  /dx)bdx =  0 L  U 2 bd  =  U 2 b  0 L F drag =  U 2 b  L (whew!)

49. Calculating drag on a flat plate, zero pressure gradient – turbulent flow * *  u/  y blows up at y = 0 Can’t use  wall =  du/dy  y=0 *

51. Turbulent Flow Tripped at leading edge so turbulent flow everywhere on plate  = 0.424 Re x -1/5 x

52.  w = 0.0243  U 2 ( /(U  )) 1/4  w = 0.0243  U 2 ( /(U 0.424 Re x -1/5 x )) 1/4  = 0.424 Re x -1/5 x  w = 0.0301  U 2 Re x -1/5

54. u/U = (y /  ) 1/6 u/U = (y /  ) 1/7 u/U = (y /  ) 1/8 u/U y/  Re increases, n increases, wall shear stress increases, boundary layer increases, viscous sublayer decrease

56. LAMINAR BOUNDARY LAYER AT SEPARATION Given: u/U = a + b  + c  2 + d  3 What are boundary conditions?

57. Given: u/U = a + b  + c  2 + d  3  = y/   = 0; u = 0; a = 0  = 0;  u/  y = 0; b = 0  =  ; u =U; 1 = c + d  =  ;  u/  y = 0; 2c + 3d = 0 2(1-d) + 3d = 2 + d = 0 so d = -2 and c = 3 Separating  u/  y = 0 u/U = 3  2 -2  3

58. dp/dx = 0 Separating Flow u/U = 3  2 -2  3 dp/dx > 0

59. U

60. If Re transition occurs at 500,000 evaluate Re L, x tr, and power to overcome skin friction drag on fin

61. Re L = UL/ = 1.54 x 10 7 Calculating transition location, x: Assume x tr occurs at 500,000 Re x = Ux/ = 500,000 Re x /Re L = x tr /L = 500,000 / 1.54 x 10 7 x tr = 0.0325 L = 0.0325 x 1.65 = 0.0536 m

62. FLAT PLATE dp/dx = 0 LAMINAR: From Theory C D = 1.33 / Re L 1/2 Eq. 9.33 TURBULENT: From Experiment 5x10 5 < Re L <10 7 x transition = 0 C D = 0.0742/Re L 1/5 Re L <10 9 x transition = 0 C D = 0.455/(logRe L ) 2.58

63. For Re transition of 500,000 and 5x10 5 < Re L < 10 9 C D = 0.455/(log Re L ) 2.58 – 1610/Re L (Eq. 9.37b) C D = 0.00281 – 0.000105 = 0.0027 ~ 4% less F D = C D 2LH(1/2)  U 2 = 91.6 N 2 sides so 2 x Area

64. Smooth flat plate, dp/dx = 0; flow parallel to plate C D ~ 0.0028

66. Re h = Uh/, don’t know U Assume: turbulent flow, transition at leading edge, doesn’t flutter, 5 x 10 5 < Re L < 10 7 Then C D = 0.0742/Re L 1/5 Plastic plate falling in water, find terminal velocity - Sum of forces = 0 Drag Force = f (U) DRAG BOUYANCY WEIGHT + z - z

67. Net Force = 0 = F B + F D - W F B – W = (  plastic -  water ) g (hLt) F D = C D ½  U 2 A = [0.0742/Re L 1/5 ] ½  water U 2 A Re h = Uh/ Solving for U get 11.0 ft /sec Check Re number: 1.86 x 10 6 5 x 10 5 < Re L < 10 7

69. mg = 120 kg U = 6 m/s F D = C D ½  U 2 A Sum of forces = 0, pick A so U = 6 m/s

70. mg = 120 kg U = 6 m/s F D = C D ½  U 2 A Sum of forces = 0, pick A so U = 6 m/s C D = 1.42 for open hemisphere F D = C D ½  U 2 A = W = mg A =  d 2 /4 d 2 = [mg]/[C D  U 2  /8] d = [(8/  )(120kg)(9.8 m/s 2 ) (1/1.42)(1/1.23 kg/m 3 )(1/6m/s) 2 d = 6.9 m

72. F D = C D ½  U 2 A F D = ma = m(dU/dt) = m(dU/dx)(dx/dt) C D ½  U 2 A = m (dU/dx) U

73. dU/U = C D  A dx /(2m) Integrating from x=0 to x=  x gives: ln U f – lnU i = C D  A  x /(2m) C D = 2m ln {U f/ U i } /[  A  x] C D = 0.299

76. Find POWER = F U for this condition And then see if that is enough to win bet.

77. air

78. 8.33 m/s 2.78 m/s

81. F D = C D ½  U 2 A A = 23.4 ft 2 C D = 0.5 F R = 0.015 x 4500 lbf F D = F R to find U where aerodynamic force = frictional force Total Power = F D U + F R U

82. F D [ C D ½  U 2 A] = F R [.015xW] (0.5)½ 0.00238slug/ft 3 U 2 (23.4ft 2 ) = 0.015 x 4500 lbf U 2 = 67.5/0.0139 U = 69.7 ft/s = 47.5 mph Where does aerodynamic force = frictional force ?

86. Average of 44 sport cars is 0 43, not much better than the 0.47 of ’37 the Lincoln Zephyr

90. (245 km/hr)