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

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

4. 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/

5. = f()

6.  = 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)

7. 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 = 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/)2 Let y/ = 
u/U = 2 -2
Strategy: obtain an expression for w as a function of , and solve for (x)

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

9. 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)

10. Assuming  = 0 at x = 0, then c = 0
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)

11. 2/2 = 15x/(U) 2/x2 = 30/(Ux) = 30 Rex /x = 5.5 (Rex)-1/2
Strategy: obtain an expression for w as a function of , and solve for (x)

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

14. y

16. u/U = sin[(/2)()]

18. (*/ = 0.344)

19. 0.344

20. LAMINAR VELOCITY PROFILES: dp/dx = 0

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

23. 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 = U2[/(RU)]1/4
Replace Umax with Ue = U and R with  to get for flat plate:
w = U2[/( U)]1/4

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

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

26. FLAT PLATE; dp/dx=0; TURBULENT FLOW: u/U = (y/)1/7
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)

27. 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 = (/U)1/4dx
(4/5) 5/4 = (/U)1/4 x + c
Assume turbulent boundary layer begins at x=0
Then  = 0 at x = 0, so c = 0
= (/U)1/5 x4/5 (x/x)1/5
/x = (/[Ux])1/5 = 0.382/Rex1/5

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

30. 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

31. Unfavorable Pressure Gradient p/x > 0; U decreasing with x
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

32. 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