 ~ 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

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 = 10000 (air bubbles in water) For flat plate with dP/dx = 0, dU/dx = 0

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/

= f()

 = 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)

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

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)

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)

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)

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)

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

y

u/U = sin[(/2)()]

(*/ = 0.344)

0.344

LAMINAR VELOCITY PROFILES: dp/dx = 0

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

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

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

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 = 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

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)

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

Cf = skin friction coefficient = w/( ½ U2) 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

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

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

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