1. Momentum Integral Equation

2. Von Karman and Poulhausen derived momentum integral equation
(approximation) which can be
used for both laminar and
turbulent flow (with and
without pressure gradient)

3. Von Karman and Polhausen 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.

5. Want to solve for  in Laminar Flow
assume velocity profile,
u/U = f(y/=), similar profiles
u  Ue at y = ;
u/y  0 at y = 
u = 0 at y = 0
w = u/y = [U/]d(u/U)/d

6. LAMINAR FLOW

7. 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

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

9. Strategy: assume velocity profile: u/Uo = f(),
obtain an expression for w as a function of , and solve for 
= f ()

10. Assume velocity profile: u = a + by + cy2
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 = b + 2c so b = -2c
U = -2c2 + c2 = -c2 so c = -U/2 & b = 2U/
u = a + by + cy2 = 0 + 2Uy/ – Uy2/2
u/U = 2 -2

11. Assume velocity profile: u = a + by + cy2
Laminar Flow Over a Flat Plate, dp/dx = 0
Want to know w(x)
Assume velocity profile: u = a + by + cy2
u = a + by + cy2 = 0 + 2Uy/ – Uy2/2
u/U = 2(y/) – (y/)2
Let y/ = 
u/U = 2 -2

12. 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)

13. 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)

14. 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)
15dx = U(d)
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)

15. Exact Solution: /x = 5(Rex)-1/2
2/2 = 15x/(U)
2/x2 = 30/(Ux) = 30 Rex
/x = 5.48 (Rex)-1/2
Exact Solution: /x = 5(Rex)-1/2
Can also calculate drag on plate by integrating over w
~ since know w = 2U/
Since know  and u(x,y) can also calculate *.

16. TURBULENT FLOW

17. BREATH

18. Want to solve for  in Laminar Flow
u/Uo = (y/)1/n (from pipe)
u/Uo = 1/n similar profiles
2. w = u/y BLOWS UP at y = 0
w =  (V)2[/(RV)]1/4

19. Calculating drag on a flat plate,
zero pressure gradient – turbulent flow
u/Uo = (y/)1/8 *
Can’t use
wall = du/dy y=0*
u/y blows up at y = 0

20. u/Uo = (y/)1/8 Uo Uc/l; R  Calculating drag on a flat plate,
zero pressure gradient – turbulent flow
u/Uo = (y/)1/8
Uo Uc/l; R 

21. w = 0.0332  V2 [/(RV)]1/4 TO USE FOR FLAT PLATE
PIPE
TO USE FOR FLAT PLATE
need to Uavg to Uc/l; and R to 

22. w = 0.0243  Uc/l2 [/(Uc/l )]1/4
w =  V2 [/(RV)]1/4
V = Uc/l; R 
w =  Uc/l2 [/(Uc/l )]1/4
u/Uo = (y/)1/8 = 1/8

23. u/Uo = (y/)1/8 = 1/8

24. w = 0.0243  Uo2 [/(Uo )]1/4 w = (8/90)  Uo2d/dx
1/4d = 0.274(/U)1/4dx
(4/5)5/4 = (/U)1/4x + c

25.  = 0.424 Rex-1/5x (4/5)5/4 = 0.274 (/U)1/4x + c
Turbulent Flow
Assume tripped at leading edge so turbulent flow everywhere on plate
(4/5)5/4 = (/U)1/4x + c
Assume  = 0 at x = 0, so c = 0
= {(5/4) (/U)1/4x}4/5
= 0.424(/U)1/5x4/5
 = Rex-1/5x

26.  = Rex-1/5x
w =  U2 (/(U))1/4
w =  U2 (/(U Rex-1/5x ))1/4
w =  U2 Rex-1/5

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

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

30. Given: u/U = a + b + c2 + d 3
 = y/
Separating
u/y = 0
= 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
u/U = 3 2 -2 3

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

32. QUESTIONS

33. Which way is flow moving?

34. ? In laminar flow along a plate, (x), (x), *(x) andw(x):
Continually decreases
Continually increases
Stays the same
? In turbulent flow along a plate, (x), (x), *(x) andw(x):
?At transition from laminar to turbulent flow, (x), (x), *(x)
andw(x):
Abruptly decreases
Abruptly increases

35. Turbulent Laminar Turbulent Laminar wall 6:1 ellipsoid  natural
forced
natural
Turbulent
Laminar 

36. What is wrong with this figure?

37. What is wrong with this figure?

38. Are Antarctic Icebergs Towable Arctic News Record – Summer 1984; 36
80% of fresh water found in world – Antarctic ice
Are Antarctic Icebergs Towable Arctic News Record – Summer 1984; 36
Cf = 0.074/Re1/5 (Fox:Cf = /Re1/5);
Area = 1 km long x 0.5 km wide
sea water = 1030 kg/m3; sea water = 1.5 x 10-6 m2/sec
Power available = 10 kW; Maximum speed = ?

39. P = UD 104 = U Cf½  U2A = U [0.074 1/5/(U1/5L1/5)]( ½  U2A) 104 =
U14/5 [0.074(1.5x10-6)1/5/10001/5] x
[½ (1030)(1000)(500)]
U14/5 =
U = m/s
Rex ~ 3x108