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~ 0 [u(x,y)/U e ] (1 – u(x,y)/U e * ~ 0 (1 – u/U)dy MOMENTUM INTEGRAL EQUATION UMPTIONS: steady, incompressible, two-dimensional body forces, p = p(x) in boundary layer, d<<, d<<

~ 0 [u(x,y)/U e ] (1 – u(x,y)/U e )dy

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MOMENTUM INTEGRAL EQUATION. ASSUMPTIONS: steady, incompressible, two-dimensional no body forces, p = p(x) in boundary layer, d 

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Page 1: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

~ 0 [u(x,y)/Ue] (1 – u(x,y)/Ue)dy

* ~ 0 (1 – u/U)dy

MOMENTUM INTEGRAL EQUATION

ASSUMPTIONS: steady, incompressible, two-dimensionalno body forces, p = p(x) in boundary layer, d<<, d<<

Page 2: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy
Page 3: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

(plate is 2% thick, Rex=L = 10,000; air bubbles in water)

For flat plate with dP/dx = 0, dU/dx = 0

Page 4: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

Realize (like Blasius) that u/U similar for all x when plotted as a function of y/ . Substitutions: = y/; so dy/ = d

= y/=0 when y=0=1 when y=

u/U

= y/

Not f(x)

Page 5: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

= f()

Page 6: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

Strategy: obtain an expression for w as a function of , and solve for (x)

= 0.133 for Blasius exact solution, laminar, dp/dx = 0

u/U = f()

Page 7: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

Laminar Flow Over a Flat Plate, dp/dx = 0

Assume velocity profile: u = a + by + cy2

B.C. at y = 0 u = 0 so a = 0 at y = u = U so U = b + c2

at y = u/y = 0 so 0 = b + 2c and b = -2c U = -2c2 + c2 = -c2 so c = -U/2

u = -2cy – (U/2) y2 = 2Uy/2 – (U/2) y2 u/U = 2(y/) – (y/)2 Let y/ =

u/U = 2 -2

Want to know w(x)

Strategy: obtain an expression for w as a function of , and solve for (x)

Page 8: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

Laminar Flow Over a Flat Plate, dp/dx = 0

u/U = 2 -2

Strategy: obtain an expression for w as a function of , and solve for (x)

Page 9: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

u/U = 2 -2

2 - 42 + 23 - 2 +23 - 4

Strategy: obtain an expression for w as a function of , and solve for (x)

w = 2U/

Page 10: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

2U/(U2) = (d/dx) (2 – (5/3)3 + 4 – (1/5)5)|01

2U/(U2) = (d/dx) (1 – 5/3 + 1 – 1/5) = (d/dx) (2/15)

Assuming = 0 at x = 0, then c = 0

2/2 = 15x/(U)

Strategy: obtain an expression for w as a function of , and solve for (x)

Page 11: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

2/2 = 15x/(U)

2/x2 = 30/(Ux) = 30 Rex

/x = 5.5 (Rex)-1/2

x1/2

Strategy: obtain an expression for w as a function of , and solve for (x)

Page 12: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy
Page 13: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

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

Page 14: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

y

Page 15: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy
Page 16: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

u/U = sin[(/2)()]

Page 17: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy
Page 18: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

(*/ = 0.344)

Page 19: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

0.344

Page 20: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

LAMINAR VELOCITY PROFILES: dp/dx = 0

Page 21: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

y /

u / U

Sinusoidal, parabolic, cubic look similar to Blasius solution.

Page 22: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy
Page 23: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

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 = 0.0233U2[/(RU)]1/4

Replace Umax with Ue = U and R with to get for flat plate:

w = 0.0233U2[/( U)]1/4

Page 24: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

u/Ue = 2(y/) – (y/)2

u/Umax = (y/)1/7

Page 25: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

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

w = 0.0233U2[/( U)]1/4

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

Cf = 0.0466 [/( U)]1/4

CD = Drag coefficient = FD/(½U2A) = wdA/(½U2A) = (1/A)CfdA

Page 26: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

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

w = 0.0233U2[/( U)]1/4 w = U2 (d/dx) 0

11/7(1- 1/7)d = U2 (d/dx) (1/(8/7) – 1/(9/7))

= U2 (d/dx) (63-56)/72 = U2 (d/dx) (7/72)

Page 27: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

0.0233U2[/( 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 + cAssume turbulent boundary layer begins at x=0Then = 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/Rex

1/5

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

Page 28: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

/x = 0.382 (/[Ux])1/5 = 0.382/Rex1/5

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

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

Cf = 0.0466 [/( U)]1/4

Cf = 0.0594/Rex1/5

Page 29: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy
Page 30: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

Favorable Pressure Gradientp/x < 0; U increasing with x

Unfavorable Pressure Gradientp/x > 0; U decreasing with xWhen velocity just above surface = 0,then flow will separate; causes wake.

Gravity “working”against friction Gravity “working” with friction

Page 31: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

Favorable Pressure Gradient p/x < 0; U increasing with x

Unfavorable Pressure Gradient p/x > 0; U decreasing with xWhen velocity just above surface = 0, then flow will separate; causes wake.

Gravity “working”against friction Gravity “working” with friction

Page 32: ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy

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 zerouy+dy = uo + u/yy=0 = u/yy=0 = 0 for flow separation

w = u/yy=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/yy=0 is always greater than 0, so uy+dy is always greater than zero