Note – throughout figures the boundary layer thickness is greatly exaggerated! CHAPTER 9: EXTERNAL...
Transcript of Note – throughout figures the boundary layer thickness is greatly exaggerated! CHAPTER 9: EXTERNAL...
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.
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”
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.
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
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
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?
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
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
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 =
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)
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
SIMPLIFYING ASSUMPTIONS OFTEN MADE FOR ENGINERING ANALYSIS OF BOUNDARY LAYER FLOWS
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).
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
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
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.
WHERE WE WANT TO GET…
Deriving MOMENTUM INTEGLAL EQso can calculate (x), w.
u(x,y)
Surface Mass Flux Through Side ab
Surface Mass Flux Through Side cd
Surface Mass Flux Through Side bc
Assumption : (1) steady (3) no body forces
Apply x-component of momentum eq. to differential control volume abcd
u
mf represents x-component of momentum flux;Fsx will be composed of shear force on boundaryand pressure forces on other sides of c.v.
X-momentumFlux =
cvuVdA
Surface Momentum Flux Through Side ab
u
X-momentumFlux =
cvuVdA
Surface Momentum Flux Through Side cd
u
U=Ue=U
X-momentumFlux =
cvuVdA
Surface Momentum Flux Through Side bc
u
a-b c-d
c-d b-c
X-Momentum Flux Through Control Surface
IN SUMMARYRHS X-Momentum Equation
X-Force on Control Surface
w is unit width into pagep(x)
Surface x-Force On Side ab
Note that p f(y)
w
w is unit width into pagep(x+dx)
Surface x-Force On Side cd
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
Why and not (bc)?w
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
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
-(w + ½ dw/dx]xdx)wdx
Surface x-Force On Side ad
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)
** ++
#
#
=
ab -cd bc
U
Divide by wdx
dp/dx = -UdU/dx for inviscid flow outside bdy layer
= from 0 to of dy
Integration by parts
Multiply by U2/U2 Multiply by U/U
If flow at B did notequal flow at C then could connect and make perpetual motionmachine.
C
C
HARDEST PROBLEM – WORTH NO POINTS
…BUT MAYBE PEACE OF MIND
(plate is 2% thick, Rex=L = 10,000; 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
= y/=0 when y=0=1 when y=
u/U
~ y/
= 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)
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)
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)
u/U = 2 -2
2 - 42 + 23 - 2 +23 - 4
Strategy: obtain an expression for w as a function of , and solve for (x)
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)
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)
THE END
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)
Re = 20,000Angle of attack = 6o
Symmetric Airfoil16% thick
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