1 / 32

 ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

MOMENTUM INTEGRAL EQUATION. ASSUMPTIONS: steady, incompressible, two-dimensional no body forces, p = p(x) in boundary layer, d <<, d<<.  * ~ 0   (1 – u/U)dy.  ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy. (plate is 2% thick, Re x=L = 10,000; air bubbles in water).

megan-johnston
Download Presentation

 ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. MOMENTUM INTEGRAL EQUATION ASSUMPTIONS: steady, incompressible, two-dimensional no body forces, p = p(x) in boundary layer, d<<, d<< *~0(1 – u/U)dy  ~0[u(x,y)/Ue] (1 – u(x,y)/Ue)dy

  2. (plate is 2% thick, Rex=L = 10,000; air bubbles in water) For flat plate with dP/dx = 0, dU/dx = 0

  3. Realize (like Blasius) that u/U similar for all x when plotted as a function of y/ . Substitutions:  = y/; so dy/ = d Not f(x) • = y/ • =0 when y=0 • =1 when y=  u/U  = y/

  4. = f()

  5.  = 0.133 for Blasius exact solution, laminar, dp/dx = 0 u/U = f() Strategy: obtain an expression for w as a function of , and solve for (x)

  6. Laminar Flow Over a Flat Plate, dp/dx = 0 Want to know w(x) Assume velocity profile: u = a + by + cy2 B.C. at y = 0 u = 0 so a = 0 at y =  u = U so U = b + c2 at y =  u/y = 0 so 0 = b + 2c and b = -2c U = -2c2 + c2 = -c2 so c = -U/2 u = -2cy – (U/2) y2 = 2Uy/2 – (U/2) y2 u/U = 2(y/) – (y/)2Let y/ =  u/U = 2 -2 Strategy: obtain an expression for w as a function of , and solve for (x)

  7. Laminar Flow Over a Flat Plate, dp/dx = 0 u/U = 2 -2 Strategy: obtain an expression for w as a function of , and solve for (x)

  8. w = 2U/ u/U = 2 -2 2 - 42 + 23 - 2 +23 - 4 Strategy: obtain an expression for w as a function of , and solve for (x)

  9. 2U/(U2) = (d/dx) (2 – (5/3)3 + 4 – (1/5)5)|01 2U/(U2) = (d/dx) (1 – 5/3 + 1 – 1/5) = (d/dx) (2/15) Assuming  = 0 at x = 0, then c = 0 2/2 = 15x/(U) Strategy: obtain an expression for w as a function of , and solve for (x)

  10. 2/2 = 15x/(U) 2/x2 = 30/(Ux) = 30 Rex /x = 5.5 (Rex)-1/2   x1/2 Strategy: obtain an expression for w as a function of , and solve for (x)

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

  12. y

  13. u/U = sin[(/2)()]

  14. (*/ = 0.344)

  15. 0.344

  16. LAMINAR VELOCITY PROFILES:dp/dx = 0

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

  18. FLAT PLATE; dp/dx = 0; TURBULENT FLOW: u/U = (y/)1/7 {for pipe had u/U = (y/R)1/7 = 1/7} But du/dy = infinity, so use w from pipe for a u/U = (y/R)1/7 profile: w = 0.0233U2[/(RU)]1/4 Replace Umax with Ue = U and R with  to get for flat plate: w = 0.0233U2[/( U)]1/4

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

  20. FLAT PLATE; dp/dx = 0; TURBULENT FLOW: u/U = (y/)1/7 w = 0.0233U2[/( U)]1/4 Cf = skin friction coefficient = w/( ½ U2) Cf = 0.0466 [/( U)]1/4 CD = Drag coefficient = FD/(½U2A) = wdA/(½U2A) = (1/A)CfdA

  21. FLAT PLATE; dp/dx=0; TURBULENT FLOW: u/U = (y/)1/7 w = 0.0233U2[/( U)]1/4 w = U2 (d/dx)011/7(1- 1/7)d = U2 (d/dx) (1/(8/7) – 1/(9/7)) = U2 (d/dx) (63-56)/72 = U2 (d/dx) (7/72)

  22. FLAT PLATE; dp/dx=0; TURBULENT FLOW: u/U = (y/)1/7 • 0.0233U2[/( U)]1/4= U2 (d/dx) (7/72) • 1/4d = 0.240 (/U)1/4dx • (4/5) 5/4 = 0.240 (/U)1/4 x + c • Assume turbulent boundary layer begins at x=0 • Then  = 0 at x = 0, so c = 0 • = 0.382 (/U)1/5 x4/5(x/x)1/5 • /x = 0.382 (/[Ux])1/5 = 0.382/Rex1/5

  23. FLAT PLATE; dp/dx=0; TURBULENT FLOW: u/U = (y/)1/7 • /x = 0.382 (/[Ux])1/5 = 0.382/Rex1/5 Cf = skin friction coefficient = w/( ½ U2) Cf = 0.0466 [/( U)]1/4 Cf = 0.0594/Rex1/5

  24. 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

  25. 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

  26. Favorable Pressure Gradient (dp/dx<0), flow will never separate. Unfavorable Pressure Gradient (dp/dx>0), flow may separate. No Pressure Gradient (dp/dx = 0), flow will never separate. Logic ~ for flow to separate the velocity just above the wall must be equal to zero uy+dy = uo + u/yy=0 = u/yy=0= 0 for flow separation w = u/yy=0 Laminar Flow: w(x)/(U2) = constant/Re1/2; flat plate; dp/dx=0 Turbulent Flow: w(x)/(U2) = constant/Re1/5; flat plate; dp/dx=0 For both laminar and turbulent w is always positive so u/yy=0 is always greater than 0, so uy+dy is always greater than zero

More Related