Ch 9: Part B – Fluid Flow About Immersed Bodies Flow Stream U Drag = pressure + friction.
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Transcript of Ch 9: Part B – Fluid Flow About Immersed Bodies Flow Stream U Drag = pressure + friction.
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Ch 9: Part B – Fluid Flow About Immersed Bodies
Flow StreamU
Drag =pressure+ friction
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Summary of Paradoxes
(1) In the first experiment we found that sometimes an increase of speed actually produces a decrease of drag.
(2) Sometime roughening increases drag and sometime it decreases drag.
(3) Sometime streamlining increases drag and sometime it decreases drag.
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FLUID FLOW ABOUT IMMERSED BODIES
Up4
p1
p2
p3
p6
p5
p7 p8p9
p10
p11
p13p…
p1210
9
8
7
65
4
3
2
1 ……
Drag due to surface stresses composed of normal (pressure) and tangential (viscous) stresses.
All we need to know is p and on body to calculate drag. Could dofor flat plate with zero pressure gradient because U and p, which were constant, we knew everywhere. If = 0 then pressure distributionis symmetric, so no net pressure force (D’Alembert’s Paradox - 1744)
DRAG
LIFT
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LOW
ReD
HIGH
ReD
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DRAG Coefficient - CD
FD = f(d,V, , )*
CD = FD/(1/2 U2A) = f(Re)* ignored roughness
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CD on flat plate (no pressure gradient) in laminar and turbulent flow
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DRAG COEFFICIENT - CD
CD = FD / (1/2 U2A)
Flow over a flat plate: FD = plate surface wdA
CD = PSwdA / (1/2 U2A)
Cf = w/(1/2 U2) {Cf = shear stress or skin friction coef.}
CD = (1/A)PSCf dA (good for laminar and turbulent flow)
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Flow over a flat plate with zero pressure gradient: CD = (1/A) PS CfdA
Cf = 0.664/Re1/2 for laminar flow (Blasius solution – flat plate laminar flow
& no pressure gradient)
CD = (1/A)A (0.664/Re1/2) dA = (bL)-1 0
L (0.664 U-1/2x-1/21/2) bdx = (0.664/L) (/U)1/2 (2)x1/2o
L = 1.33 ( / LU)1/2
CD = 1.33 (ReL) -1/2 for laminar flow over a flat plate, with no pressure gradient
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Flow over a flat plate with zero pressure gradient: CD = (1/A) PS CfdA
Cf = 0.0594/Re1/5 for turbulent flow (u/U = [y/]1/7) (Blasius correlation: f = 0.316/Re1/4; Re 105)
CD = (1/A)A (0.0594/Re0.2) dA = (bL)-1 0
L (0.0594 (U/)-0.2x-0.2 bdx
= (0.0594/L) (/U)0.2 [x0.8/0.8]oL
= 0.0742(/UL)0.2
CD = 0.0742 (ReL) –0.2 for turbulent flow over a flat plate, with no pressure gradient - 5x105 <ReL<107
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CD = 1.33 (ReL) -1/2 for laminar flow over a flat plate, with no pressure gradient ~ Re < 5x105
CD = 0.0742 (ReL) –0.2* for turbulent flow over flatplate, with no pressure gradient ~ 5x105 <ReL<107
CD = 0.455/ log (ReL)2.58* for turbulent flow over flatplate, with no pressure gradient ~ ReL<109
* Assumes turbulent boundary layer begins at x=o
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CD correction term for partly laminar / partly turbulent
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CD correction term for partly laminar / partly turbulent
? ADD
ORSUBTRACT
CORRECTION TERM ???
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Must account for fact that turbulence does not start at x = 0-must subtract B/ReL
CD correction term = B/ReL = Retr(CDturb – Cdlam)/ReL
Retr
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CD correction term = B/ReL = Retr(CDturb – CDlam)/ReL
For Retr = 5 x 105
CD = 0.0742/ReL1/5 – Retr(CDturb – CDlam)/ReL
CD = 0.0742/ReL1/5
– 5x105[0.0742/ (5x105)1/5–1.33/(5x105)1/2]/ReL
CD = 0.0742/ReL1/5 – 1748/ReL
Retr
5 x 105 < ReL < 107
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CD correction term = B/ReL = Retr(CDturb – CDlam)/ReL
For Retr = 5 x 105
CD = 0.0742/ReL1/5 – Retr(CDturb – CDlam)
CD=0.0742/ReL1/5–5x105[0.455/ (log[5x105])1/5–1.33/(5x105)1/2]
CD = 0.455/(logReL)2.58 – 1600/ReL
Retr
5 x 105 < ReL < 109
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SMOOTH FLAT PLATE NO PRESSURE GRADIENT
CD = 0.0742 (ReL) –0.2
CD = 0.455/ log (ReL)2.58
CD = 1.33 (ReL) -1/2
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Rough Flat Plate
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FLAT PLATE
CD = D/( ½ U2A)
ReL
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PIPE
FLAT PLATE
CD = D/( ½ U2A)
f = (dp/dx)D/( ½ U2)
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Flat Plate Perpendicular to Flow Direction
CD = FD/(1/2U2bh)
for Reh > 1000, CD very weak function
of Re.
CD ~ 2 Newton “guessed”
Separation points fixed
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Drag Force = p/t = (mv)/tm ~ UAf = mass per second passing through area
v ~ U-0 = UCD = D/(1/2 U2Af) ~ UAfU/(1/2U2Af)
CD ~ 2 Newton
Value is right order of magnitude,& Re insensitivity predicted correctly.
(fixed)
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Mostly pressure drag, separation point fixed
Frictiondrag
Character of CD vs Re curves for different shapes
press& fric
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• Flow parallel to plate – viscous forces important and Re dependence
• Flow perpendicular to plate –pressure forces important and no strong Re dependence
What about Re dependence for flow around sphere?
Re
CD ?
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Drag Coefficient, CD, as a function of Re for a Smooth Sphere
SMOOTH SPHERE
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Drag Coefficient, CD, as a function of Re for a Smooth Sphere
SMOOTH SPHERE
FD = 3VDCD = ?
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CD = FD/(½ U2R2) = 6UR/(½ U2R2) = 24/Re
Laminar boundary layerTurbulent flow in wakeSeparation point moving forward
Separation point fixed
95% of drag due to pressure difference between front and back
Turbulentboundary
layer
LaminarFlow
* *
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IDEAL FLOW* LAMINAR FLOW TURBULENT FLOW
S e p a r a t i o n
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~82o
~120o
PRESSURE DRAG
DRAG
IF NO VISCOSITYWHAT WOULD BE
TOTAL DRAG ?
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Smooth
Trip By roughening surface can “trip” boundary layer so turbulent which resultsin a favorable momentumexchange, pushing separation point furtherdownstream, resultingin a smaller wake andreduced drag.
125 yd drive with smooth golf ball becomes 215 ydsfor dimpled*From Van Dyke, Album of Fluid MotionParabolic Press, 1982; Original photographs By Werle, ONERA, 1980
Re = 15000
Re = 30000
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Drag coefficient as a function of Reynolds number for smooth circularcylinders and smooth spheres. From Munson, Young, & Okiishi,
Fundamentals of Fluid Mechanics, John Wiley & Sons, 1998
ASIDE: At low very low Reynolds numbers Drag UL
CD = D / (1/2 U2Af) D ~ U
CD = constantD ~ U2
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Drag coefficient as a function of Reynolds number for smooth circularcylinders and smooth spheres. From Munson, Young, & Okiishi,
Fundamentals of Fluid Mechanics, John Wiley & Sons, 1998
ASIDE: At low very low Reynolds numbers Drag UL
CD = D / (1/2 U2Af) D ~ U
CD = constantD ~ U2
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Drag coefficient as a function of Re for a smooth cylinder and smooth sphere.
ReDcrit ~ 3 x 1053-D relieving effectCdcylinder>CDsphere
Is ReDcritical constant?
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Effect of surface roughness on the drag coefficient of a sphere in theReynolds number range where laminar boundary layer becomes turbulent.
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vortex shedding
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Theodore Von Karman
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A
B
C
D
E
FLOW AROUND A SMOOTH CYLINDER
~82o ~120o
Smooth Sphere
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Vortex Shedding St = UD/f =0.21
for 102 < Re < 107
PICTURE OF SHEDDING
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PICTURE OF SHEDDING
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Flow Separation
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FLOW SEPARATION
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Fig. 9.6
Uupstream = 3 cm/sec; divergent angle = 20o; Re= 900; hydrogen bubbles
Unfavorable pressure gradient necessary for flow separation to be “possible” but separation
not guaranteed.
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Water, velocity = 2 cm/s, cylinder diameter = 7 cm, Re = 1200Photographed 2 s after start of motion; hydrogen bubble technique
Back flow
0 velocity at y = dy
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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
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Viscous flow around
streamlined body
streamlines divergevelocity decreases
adverse pressure gradient
streamlines convergesvelocity increases
adverse pressure gradient
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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
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Streamlining
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STREAMLINING
First employed by Leonardo da Vinci –First coined by d’Arcy Thompson – On Growth and Form (1917)
CD ~ 0.06CD ~ 2 for flat plate
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STREAMLINING
(a)
(b)
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CD = FD /(1/2 U2A) FD = CD (1/2 U2A)
CD = 2.0
CD = 1.2
CD = 0.12
CD = 1.2
CD = 0.6
d =
d/10
d =
d =
d = As CD decreases,what is happening
to wake?
Is there a wakeassociated with
pipe flow?
If CD decreases does that necessarily imply that the drag decreases?
2 - D
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(note that frictional force increased from (b) to (c) but net force decreased)
(note that although CD decreased from(d) to (e) that the Drag force did not.
CD = 2.0
CD = 1.2
CD = 0.12
CD = 1.2
CD = 0.6
*
*
*
*
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First flight of a powered aircraft 12/17/03 120ft in 12 secondsOrville Wright at the controls
Same drag at 210 mph
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The End