MOMENTUM INTEGRAL EQUATION
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 ~ 0   [u(x,y)/U e ] (1 – u(x,y)/U e )dy PowerPoint PPT Presentation


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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)/U e ] (1 – u(x,y)/U e )dy. (plate is 2% thick, Re x=L = 10,000; air bubbles in water).

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

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0 u x y u e 1 u x y u e 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


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


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

Not f(x)

  • = y/

  • =0 when y=0

  • =1 when y= 

u/U

 = y/


0 u x y u e 1 u x y u e dy

= f()


0 u x y u e 1 u x y u e dy

 = 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)


0 u x y u e 1 u x y u e dy

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 = Uso 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/)2Let y/ = 

u/U = 2 -2

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


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)


0 u x y u e 1 u x y u e dy

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)


0 u x y u e 1 u x y u e dy

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)


0 u x y u e 1 u x y u e dy

2/2 = 15x/(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)


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?


0 u x y u e 1 u x y u e dy

y


0 u x y u e 1 u x y u e dy

u/U = sin[(/2)()]


0 u x y u e 1 u x y u e dy

(*/ = 0.344)


0 u x y u e 1 u x y u e dy

0.344


0 u x y u e 1 u x y u e dy

LAMINAR VELOCITY PROFILES:dp/dx = 0


0 u x y u e 1 u x y u e dy

y / 

u / U

Sinusoidal, parabolic, cubic look similar to Blasius solution.


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 = 0.0233U2[/(RU)]1/4

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

w = 0.0233U2[/( U)]1/4


0 u x y u e 1 u x y u e dy

u/Umax = (y/)1/7

u/Ue = 2(y/) – (y/)2


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 = 0.0233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


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 = 0.0233U2[/( U)]1/4

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)


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

  • 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


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

  • /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


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 x

When velocity just above surface = 0,

then flow will separate; causes wake.

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


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 x

When velocity just above surface = 0, then flow will separate; causes wake.

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


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 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


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