Momentum Integral Equation

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

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.

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

LAMINAR FLOW

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 =0 when y=0; =1 when y=  Not f(x)

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

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 = 0 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

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

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) 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)

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 *.

TURBULENT FLOW

BREATH

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 = 0.0332 (V)2[/(RV)]1/4

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

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 

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 

w = 0.0243  Uc/l2 [/(Uc/l )]1/4 w = 0.0332  V2 [/(RV)]1/4 V = 0.837 Uc/l; R  w = 0.0243  Uc/l2 [/(Uc/l )]1/4 u/Uo = (y/)1/8 = 1/8

u/Uo = (y/)1/8 = 1/8

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

 = 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 = 0.274 (/U)1/4x + c Assume  = 0 at x = 0, so c = 0 = {(5/4) 0.274 (/U)1/4x}4/5 = 0.424(/U)1/5x4/5  = 0.424 Rex-1/5x

 = 0.424 Rex-1/5x w = 0.0243  U2 (/(U))1/4 w = 0.0243  U2 (/(U 0.424 Rex-1/5x ))1/4 w = 0.0301  U2 Rex-1/5

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/

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

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

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

QUESTIONS

Which way is flow moving?

? 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

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

What is wrong with this figure?

What is wrong with this figure?

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 = 0.0594/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 = ?

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 = 0.03054 U = 0.288 m/s Rex ~ 3x108