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Note – throughout figures theboundary layer thickness is
greatly exaggerated!
CHAPTER 9: EXTERNAL INCOMPRESSIBLE
VISCOUS FLOWS
•Can’t have fully developed flow•Velocity profile evolves,
flow is accelerating• Generally boundary layers(Rex > 104) are very thin.
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Laminar Flow/x ~ 5.0/Rex
1/2
THEORY
Turbulent Flow (Rex > 106)/x ~ 0.16/Rex
1/7
EXPERIMENTAL
“At these Rex numbersbdy layers so thin that displacement effect onouter inviscid layer is small”
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Blasius showed theoretically that /x = 5/Rex (Rex = Ux/)
BOUNDARY LAYER THICKNESS: is y where u(x,y) = 0.99 U
This definition for is completely arbitrary, why not 98%, 95%, etc.
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Because of the velocity deficit, U-u, within the bdy layer, the mass flux through b-b is less than a-a. However if we displace the plate a distance *, the mass flux along each section will be identical.
DISPLACEMENT THICKNESS:
* = 0 (1 – u/U)dy
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u
U[(u)([U-u])]
[Flux of momentum deficit]
U2 = total flux of momentum
deficit
The momentum thickness, , is defined as the thickness of a layer of fluid, with velocity Ue, for which the momentum flux is equal to the deficit of momentum flux through the boundary layer.
Ue2
= 0 u(Ue – u)dy = 0 �[u/Ue] (1 – u/Ue)dy
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Want to relate momentum thickness, , with drag, D, on plate Ue is constant so p/dx = 0; Re < 100,000 so laminar,
flow is steady, is small (so * is small) so p/y ~ 0
Conservation of Mass: -Uehw + 0
huwdy + 0Lvwdx = 0
Ignoring the fact that because of *, Ue is not parallel to plate
Jason Batin, what are forces on control volume?
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X-component of Momentum Equation
p/dx = 0p/dy = 0
Ue/y = 0
-D = - Ue2hw + 0
hu2wdy + 0L Uevwdx
v is bringing Ue out of control volume
a-b c-d b-c
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X-component of Momentum Equation
p/dx = 0Ue/y = 0
-D = - Ue2hw + 0
hu2wdy + 0L Uevwdx
Conservation of Mass: -Uehw + 0
huwdy + 0Lvwdx = 0
-Ue2hw + 0
huUewdy + 0LvUewdx = 0
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X-component of Momentum Equation
p/dx = 0Ue/y = 0
-D = - Ue2hw+0
hu2wdy +Ue2hw-0
huUewdy -D = 0
hu2wdy +0huUewdy
D/(Ue2w) = 0
h (-u2/Ue2) dy + 0
h (u/Ue)wdyD/(Ue
2w) = 0h (u/Ue)[1-u/Ue] dy
~ 0 (u/Ue)[1-u/Ue] dy =
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p/dx = 0Ue/y = 0
D/(Ue2w) =
D = Ue2w
dD/dx = Ue2w(d/dx)
D = 0L wallwdx (all skin friction)
dD/dx = wallw =Ue2w(d/dx)
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p/dx = 0Ue/y = 0
D = Ue2w
dD/dx = wallw =Ue2w(d/dx)
• knowing u(x,y) then can calculate and from can calculate drag and wall
• the change in drag along x occurs at the expense on an increase in which represents a decrease in the momentum of the fluid
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SIMPLIFYING ASSUMPTIONS OFTEN MADE FOR ENGINERING ANALYSIS OF BOUNDARY LAYER FLOWS
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Blasius developed an exact solution (but numerical integration was necessary) for
laminar flow with no pressure variation. Blasius could theoretically predict:
boundary layer thickness (x), velocity profile u(x,y)/U
Moreover: u(x,y)/U vs y/ is self similarand wall shear stress w(x).
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Dimensionless velocity profile for a laminar boundary layer:comparison with experiments by Liepmann, NACA Rept. 890,1943. Adapted from F.M. White, Viscous Flow, McGraw-Hill, 1991
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Blasius developed an exact solution (but numerical integration was necessary) for laminar flow with no pressure variation.
Blasius could theoretically predict boundary layer thickness (x), velocity profile u(x,y)/U vs y/, and wall shear stress w(x).
Von Karman and Poulhausen derived momentum integral equation (approximation) which can be
used for both laminar (with and without pressure gradient) and
turbulent flow
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Von Karman and Polhausen method (MOMENTUM INTEGRAL EQ. Section 9-4)
devised a simplified method by satisfying only the boundary
conditions of the boundary layer flow rather than satisfying Prandtl’s
differential equations for each andevery particle within the boundary layer.
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WHERE WE WANT TO GET…
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Deriving MOMENTUM INTEGLAL EQso can calculate (x), w.
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u(x,y)
Surface Mass Flux Through Side ab
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Surface Mass Flux Through Side cd
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Surface Mass Flux Through Side bc
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Assumption : (1) steady (3) no body forces
Apply x-component of momentum eq. to differential control volume abcd
u
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mf represents x-component of momentum flux;Fsx will be composed of shear force on boundaryand pressure forces on other sides of c.v.
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X-momentumFlux =
cvuVdA
Surface Momentum Flux Through Side ab
u
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X-momentumFlux =
cvuVdA
Surface Momentum Flux Through Side cd
u
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U=Ue=U
X-momentumFlux =
cvuVdA
Surface Momentum Flux Through Side bc
u
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a-b c-d
c-d b-c
X-Momentum Flux Through Control Surface
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IN SUMMARYRHS X-Momentum Equation
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X-Force on Control Surface
w is unit width into pagep(x)
Surface x-Force On Side ab
Note that p f(y)
w
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w is unit width into pagep(x+dx)
Surface x-Force On Side cd
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Surface x-Force On Side bc
p + ½ [dp/dx]dx is average pressure along bc
Force in x-direction: [p + ½ (dp/dx)] wd
w
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Why and not (bc)?w
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Surface x-Force On Side bc
psin in x-direction; (A)(psin) is force in x-direction
Asin = wSo force in x-direction = p w
psin
w
A
p
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Note that since the velocity gradient goes to zero at the top of the boundary layer,
then viscous shears go to zero.
Surface x-Force On Side bc
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-(w + ½ dw/dx]xdx)wdx
Surface x-Force On Side ad
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Fx = Fab + Fcd + Fbc + Fad
Fx = pw -(p + [dp/dx]x dx) w( + d) + (p + ½ [dp/dx]xdx)wd - (w + ½ (dw/dx)xdx)wdx
Fx = pw -(p w + p wd + [dp/dx]x dx) w + [dp/dx]x dx w d) + (p wd + ½ [dp/dx]xdxwd) - (w + ½ (dw/dx)xdx)wdx
d <<
=
- ½ (dw/dxxdx)dx
dw << w
p(x)
** ++
#
#
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=
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ab -cd bc
U
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Divide by wdx
dp/dx = -UdU/dx for inviscid flow outside bdy layer
= from 0 to of dy
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Integration by parts
Multiply by U2/U2 Multiply by U/U
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If flow at B did notequal flow at C then could connect and make perpetual motionmachine.
C
C
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HARDEST PROBLEM – WORTH NO POINTS
…BUT MAYBE PEACE OF MIND
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(plate is 2% thick, Rex=L = 10,000; air bubbles in water)
For flat plate with dP/dx = 0, dU/dx = 0
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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/
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= 0 u/Ue(1 – u/Ue)dy = y/; d = 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)
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Laminar Flow Over a Flat Plate, dp/dx = 0
Assume velocity profile: u = a + by + cy2
B.C. at y = 0u = 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)
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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)
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u/U = 2 -2
2 - 42 + 23 - 2 +23 - 4
Strategy: obtain an expression for w as a function of , and solve for (x)
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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)
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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)
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THE END
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Illustration of strong interaction between viscid and inviscid regions in the rear of a blunt body.
Re = 15,000
Re = 17.9(separation at Re = 24)
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Re = 20,000Angle of attack = 6o
Symmetric Airfoil16% thick
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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
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