Given : Incompressible flow in a circular channel and Re = 1800, where D = 10 mm.

1. Given: Incompressible flow in a circular channel and Re = 1800, where D = 10 mm. Find: (a) Re = f (Q, D,) (b) Re = f(dm/dt, D,) (c) Re for same Q and D = 6 mm

2. Incompressible flow in a circular channel. Re = 1800, where D = 10mm Find: (a) Re = f (Q, D, ); (b) Re = f(dm/dt, D, ); • Equations: Re = DUavg/ = DUavg/ • Q = AUavgdm/dt = AUavg A = D2/4 • Re = DUavg/ = DQ/(A) = 4DQ/(D2) = 4Q/(D) • Re = DUavg/ = (dm/dt)D/(A) = (dm/dt)D4/(D2) = 4(dm/dt)/(D)

3. Incompressible flow in a circular channel. Re = 1800, where D = 10mm Find: (c) Re for same flow rate and D = 6 mm • Re = DUavg/ = DQ/(A) = 4DQ/(D2) = 4Q/(D) • Q = Re (D)/4 • (c) Q1 = Q2 • Re1D1/4 = Re2D2/4 • Re2 = Re1(D1/D2) • Re2 = 1800 (10 mm / 6 mm) = 3000

5. For most engineering pipe flow systems turbulence occurs around Re = 2300. On a log-log plot of volume flow rate, Q, versus tube diameter, plot lines that cor- respond to Re = 2300 for standard air and water at 15o. Q = ReD/4 Q = 2300 D/4 Air:  = / = 1.46 x 10-5 m2/s at 15oC Table A-10 Water:  = / = 1.14 x 10-6 m2/s at 15oC Table A-8

6. Re = Q/(A) = 4DQ/(D2) = 4Q/(D) Q = ReD/4 air ~ 13 water Why get higher Q’s for same Re and D in air than water? Re = 2300

8. What is the direction and magnitude (lbf/ft2) of shear stress on pipe wall ???

9. (P2-P1 = -P)

10. Pipe wall exerts a negative shear on the fluid. Consequently the fluid exerts a positive shear on the wall.

12. Given: Fully developed flow between two parallel plates, separated by h. Flow is from left to right. u(y) = [h2/(2)][dp/dx][(y/h)2 – ¼] y = h/2 y = 0 y = h/2 Net Pressure Force Plot:xy(y)

13. u(y) = [a2/(2)][dp/dx][(y/a)2 – ¼] Eq. 8.7 (y=0 at centerline & a = h) u(y) = [h2/(2)][dp/dx][(y/h)2 – ¼] xy =  du/dy xy=  [h2/(2)][dp/dx][2y/h2] =[dp/dx][y] xy=[dp/dx][y](for y=0 at centerline) h/2 Net Pressure Force 0 -h/2 xy = [dp/dx]{y–[a/2]}(for y=0 at bottom plate)

14. xy(y) = [dp/dx][y]dp/dx < 0 So xy is < 0 for y > 0 And xy is > 0 for y < 0 |Maximum xy| = y(+h/2) and y(-h/2) Shear stress forces u ??? Sign of shear stress and direction of shear stress forces “seem” contradictory??? y xy

15. xy y xx xz x z

16. sign convention for stress (pg 26): A stress component is positive when the direction of the stress component and the normal to the plane at which it acts are both positive or both negative. Stresses shown in figure are all positive

17. Shear sign convention Shear force + + Shear force y Plot:xy(y) xy

18. 1507 by Leonardo in connection with a hydraulic project in Milan

19. D = 6mm L = 25 mm P = 1.5MPa (gage) M = ?, SAE 30 oil at 20o Q = f(p,a), f =? Velocity of M = 1 mm/min a = ? Q

20. Fully Developed Laminar Flow Between Infinite Parallel Plates M = ? D = 6mm L = 25 mm P = 1.5MPa (gage) M = ? SAE 30 oil at 20o Velocity of M = 1 mm/min a = ?

21. Fully Developed Laminar Flow Between Infinite Parallel Plates Q= f(p,a) ? Q  p Q  a3 l l Q l= 2R =D

22. If V = 1 mm/min, what is a? Q = a3pl/(12L) Q = VA =UavgDa v D = 6mm L = 25 mm P = 1.5MPa (gage) M = ? SAE 30 oil at 20o Velocity of M = 1 mm/min a = ?

24. 1500 1500

25. 1500 Re = Va/ 1500

27. What is velocity profile, u(y), for a plate moving vertically at Uo through a liquid bath? Shear stress Uo x z B.C. y x y Shear Forces on Fluid Element

28. What is velocity profile, u(y), for a plate moving vertically at Uo through a liquid bath? ASSUME FULLY DEVELOPED: • -(zx+[dzx/dy][dy/2]dxdz • +(zx-[dzx/dy][dy/2]dxdz • gdxdydz = 0 • dzx/dy = - g • zx = gy + c1 Uo dy mg x y Shear Forces on Fluid Element

29. What is velocity profile, u(y), for a plate moving vertically at Uo through a liquid bath? ASSUME FULLY DEVELOPED: zx = gy + c1 zx = du/dy u(y) = gy2/[2] + c1y/ + c2 u(y=0) = U0 So c2 = Uo Uo dy mg x y Shear Forces on Fluid Element

30. What is velocity profile, u(y), for a plate moving vertically at Uo through a liquid bath? ASSUME FULLY DEVELOPED: zx = gy + c1 zx = du/dy u(y) = gy2/[2] + c1y/ + Uo du/dy (y=h) = 0 du/dy = zx / = 0 at y=h 0 = gh/ + c1/ c1 = -gh Uo dy mg x y Shear Forces on Fluid Element

31. What is velocity profile, u(y), for a plate moving vertically at Uo through a liquid bath? ASSUME FULLY DEVELOPED: zx = gy + c1 zx = du/dy u(y) = gy2/[2] + c1y/ + c2 Uo c2 = Uo c1 = -gh dy mg x u(y) = gy2/[2] + -ghy/ + Uo u(y) = g/{y2/2 –hy} + Uo y Shear Forces on Fluid Element

32. At y = 0 velocity at edge of film  0 but du/dy = 0

34. What is the maximum diameter of a vertical pipe so that water running down it remains laminar? D dx rz mg g(D2/4)(dx) ReD = VD/ = uavgD/ D = ? = 2300()/V

35. What is the maximum diameter of a vertical pipe so that water running down it remains laminar? D dx rz mg g(D2/4)(dx)

36. What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed D dx g(D2/4)(dx) + rz2rdx =0

37. What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed D dx g(D2/4)(dx) + rz2rdx =0 For fully developed, laminar, horizontal, pressure driven, Newtonian pipe flow: u(r) = -{(dp/dx)/4}{R2 – r2}

38. What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed  D dx g(D2/4)(dx) + rz2rdx =0

39. What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed D • Re = uacgD/ = VD/ • D = Re /uavg • Re = 2300 • = 1.0 x 10-6 m2/s at 20oC dx g(D2/4)(dx) + rz2rdx =0 D = 1.96 mm for water

42. uavg/umax = (y/R)1/n Not accurate at y=R and y near 0

46. u / Umax = (y/R)1/n

47. n = 1.85 log10ReUmax –1.96 from Hinze –Turbulence, McGraw Hill, 1975

48. Uavg/Umax ReUmax n = 1.85 log10(ReUmax) –1.96 Uavg/Umax = 2n2/((n+1)(2n+1)) For F.D. laminar flow uavg = ½ umax

49. Uavg/Umax ReUmax

50. Uavg/Umax = (1 – r/R)1/n = (y/R)1/2; n = f(Re) close to wall Power Law Velocity Profiles for Fully-Developed Flow in a Smooth Pipe u/U = Uavg/Umax