Note – throughout figures the boundary layer thickness is greatly exaggerated!

1. CHAPTER 9: EXTERNAL INCOMPRESSIBLE VISCOUS FLOWS • Can’t have fully developed flow • Velocity profile evolves, • flow is accelerating • Generally boundary layers • (Rex > 104) are very thin. Note– throughout figures the boundary layer thickness is greatly exaggerated!

2. Laminar Flow /x ~ 5.0/Rex1/2 THEORY Turbulent Flow (Rex > 106) /x ~ 0.16/Rex1/7 EXPERIMENTAL “At these Rex numbers bdy layers so thin that displacement effect on outer inviscid layer is small”

4. BOUNDARY LAYER THICKNESS:  is y where u(x,y) = 0.99 U  This definition for  is completely arbitrary, why not 98%, 95%, etc. Blasius showed theoretically that /x = 5/Rex (Rex = Ux/)

5. DISPLACEMENT THICKNESS: * = 0 (1 – u/U)dy Because of the velocity deficit, U-u, within the bdy layer, the mass flux through b-b is less than a-a. However if we displace the plate a distance *, the mass flux along each section will be identical.

6. The momentum thickness, , is defined as the thickness of a layer of fluid, with velocity Ue, for which the momentum flux is equal to the deficit of momentum flux through the boundary layer. [Flux of momentum deficit] U [(u)([U-u])] u U2 = total flux of momentum deficit Ue2=0 u(Ue – u)dy =0 [u/Ue] (1 – u/Ue)dy

7. Want to relate momentum thickness, , with drag, D, on plate Ue is constant so p/dx = 0; Re < 100,000 so laminar, flow is steady,  is small (so * is small) so p/y ~ 0 Conservation of Mass: -Uehw + 0huwdy + 0Lvwdx = 0 Ignoring the fact that because of *, Ue is not parallel to plate Jason Batin, what are forces on control volume?

8. p/dx = 0 p/dy = 0 Ue/y = 0 X-component of Momentum Equation a-b b-c c-d -D = - Ue2hw + 0hu2wdy + 0L Uevwdx v is bringing Ue out of control volume

9. p/dx = 0 Ue/y = 0 X-component of Momentum Equation -D = - Ue2hw + 0hu2wdy + 0L Uevwdx Conservation of Mass: -Uehw + 0huwdy + 0Lvwdx = 0 -Ue2hw + 0huUewdy + 0LvUewdx = 0

10. p/dx = 0 Ue/y = 0 X-component of Momentum Equation -D = - Ue2hw+0hu2wdy +Ue2hw-0huUewdy -D = 0hu2wdy +0huUewdy D/(Ue2w) = 0h (-u2/Ue2)dy +0h (u/Ue)wdy D/(Ue2w) = 0h(u/Ue)[1-u/Ue]dy ~ 0(u/Ue)[1-u/Ue]dy = 

11. p/dx = 0 Ue/y = 0 D/(Ue2w) =  D = Ue2w dD/dx = Ue2w(d/dx) D = 0L wallwdx (all skin friction) dD/dx = wallw =Ue2w(d/dx)

12. p/dx = 0 Ue/y = 0 D = Ue2w dD/dx = wallw =Ue2w(d/dx) • knowing u(x,y) then can calculate  and from  • can calculate drag and wall • the change in drag along x occurs at the expense on • an increase in  which represents a decrease in • the momentum of the fluid

13. SIMPLIFYING ASSUMPTIONS OFTEN MADE FOR ENGINERING ANALYSIS OF BOUNDARY LAYER FLOWS

14. Blasius developed an exact solution (but numerical integration was necessary) for laminar flow with no pressure variation. Blasius could theoretically predict: boundary layer thickness (x), velocity profile u(x,y)/U Moreover: u(x,y)/U vs y/ is self similar and wall shear stress w(x).

15. Dimensionless velocity profile for a laminar boundary layer: comparison with experiments by Liepmann, NACA Rept. 890, 1943. Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991

16. Blasius developed an exact solution (but numerical integration was necessary) for laminar flow with no pressure variation. Blasius could theoretically predict boundary layer thickness (x), velocity profile u(x,y)/U vs y/, and wall shear stress w(x). Von Karman and Poulhausen derived momentum integral equation (approximation) which can be used for both laminar (with and without pressure gradient) and turbulent flow

17. Von Karman and Polhausen method (MOMENTUM INTEGRAL EQ. Section 9-4) devised a simplified method by satisfying only the boundary conditions of the boundary layer flow rather than satisfying Prandtl’s differential equations for each and every particle within the boundary layer.

19. WHERE WE WANT TO GET…

20. Deriving MOMENTUM INTEGLAL EQ so can calculate (x), w.

21. u(x,y) Surface Mass Flux Through Side ab

22. Surface Mass Flux Through Side cd

23. Surface Mass Flux Through Side bc

24. Apply x-component of momentum eq. to differential control volume abcd Assumption : (1) steady (3) no body forces u

25. mf represents x-component of momentum flux; Fsx will be composed of shear force on boundary and pressure forces on other sides of c.v.

26. Surface Momentum Flux Through Side ab X-momentum Flux = cvuVdA u

27. Surface Momentum Flux Through Side cd X-momentum Flux = cvuVdA u

28. Surface Momentum Flux Through Side bc U=Ue=U X-momentum Flux = cvuVdA u

29. X-Momentum Flux Through Control Surface a-b c-d c-d b-c

30. IN SUMMARY RHS X-Momentum Equation

31. X-Force on Control Surface Surface x-Force On Side ab w is unit width into page Note that p  f(y) p(x) w

32. Surface x-Force On Side cd w is unit width into page p(x+dx)

33. Surface x-Force On Side bc p + ½ [dp/dx]dx is average pressure along bc w Force in x-direction: [p + ½ (dp/dx)] wd

34. Why  and not (bc)? w

35. Surface x-Force On Side bc psin  A p    w psin  in x-direction; (A)(psin) is force in x-direction Asin  = w So force in x-direction = p w

36. Surface x-Force On Side bc Note that since the velocity gradient goes to zero at the top of the boundary layer, then viscous shears go to zero.

37. Surface x-Force On Side ad -(w + ½ dw/dx]xdx)wdx

38. p(x) Fx = Fab + Fcd + Fbc + Fad Fx = pw-(p + [dp/dx]x dx) w( + d)+ (p + ½ [dp/dx]xdx)wd - (w + ½ (dw/dx)xdx)wdx Fx = pw-(p w + p wd + [dp/dx]x dx) w + [dp/dx]x dx w d) + (p wd + ½ [dp/dx]xdxwd) - (w + ½ (dw/dx)xdx)wdx # * * + + # - ½ (dw/dxxdx)dx = dw << w d  << 

39. =

40. U ab -cd bc

41. Divide by wdx dp/dx = -UdU/dx for inviscid flow outside bdy layer  =  from 0 to  of dy

42. Integration by parts Multiply by U2/U2 Multiply by U/U

46. C If flow at B did not equal flow at C then could connect and make perpetual motion machine. C

48. HARDEST PROBLEM – WORTH NO POINTS …BUT MAYBE PEACE OF MIND

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