DRAG on “Blunt” Bodies. FLUID FLOW ABOUT IMMERSED BODIES U p4p4 p1p1 p2p2 p3p3 p6p6 p5p5 p7p7...
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Transcript of DRAG on “Blunt” Bodies. FLUID FLOW ABOUT IMMERSED BODIES U p4p4 p1p1 p2p2 p3p3 p6p6 p5p5 p7p7...
![Page 1: DRAG on “Blunt” Bodies. FLUID FLOW ABOUT IMMERSED BODIES U p4p4 p1p1 p2p2 p3p3 p6p6 p5p5 p7p7 p8p8 p9p9 p 10 p 11 p 13 p…p… p 12 10 99 88 77 66.](https://reader036.fdocuments.in/reader036/viewer/2022062423/56649f155503460f94c2aec7/html5/thumbnails/1.jpg)
DRAG on “Blunt” Bodies
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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.
DRAG
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“There is at present no satisfactory theory for the
forces on an arbitrary body immersed in a stream flowing at an arbitrary
Reynolds number.”
White – Fluid Mechanics
p = ?
Boundary layer theory can usually predict separation point but not pressure distribution in separated region
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DRAG Coefficient - CD
CD = FD/(1/2 U2A)
EXPERIMENTAL
EXPERIMENTAL
CD = f(shape, Re, Ma, Fr, /L)
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• FD = surface pdA + surface wall dA = surface pdA• pressure in wake essentially constant• pressure in wake can not be determined analytically
CD = FD/(1/2 U2A)• A = usually frontal area for “stubby” bodies• CD = 2 for Two-Dimensional and Re > 10,000
Flow over 2-D flat plate
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• b/h =1 square, CD = 1.18; (disk; CD = 1.17)
• CD independent of Re for Re > 1000 Question: CD = FD/(1/2 U2A)
What happens to CD if double area (b/h 2b/2h)?What happens to FD if double area (b/h 2b/2h)?
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CD = FD/(1/2 U2A)
The area A is usually one of three types:
Frontal area: the body area as seen from the stream – suitable for thick stubby bodies such as spheres, cylinders, cars, missiles, projectiles, torpedoes.Planform area: the body area as seen from above – suitable for wide flat bodies such as plates, wigs, hydrofoils
Wetted area: total area – customary for surface ships and barges.
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CD = FD/(1/2 U2A) = CD,pressure + CD,friction
t/c
~ 2-D
flat plate 100% friction drag
circular cylinder 3% friction drag
Rec = 106
x
½ drag due to friction for t/c = 0.25
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Cdmin = 0.06
0.25
tc
drag coefficient on a strut as a function of thickness/ chord
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Mostly pressure drag, separation point fixed
Frictiondrag
Character of CD vs Re curves for different 2-D shapes
press& fric
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True or False: Sharp-edged bodies are relatively insensitive to Re because separation points are fixed.
True or False: Smoothly-rounded bodies are relatively sensitive to Re because of laminar – turbulent transition /separation effects.
True or False: Horizontal flat plates are relatively sensitive to Re because of laminar – turbulent transition effects.
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sphere
Character of CD vs Re curves for 3-D shape - sphere
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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
CD = D/( ½ U2A)
?
??
?
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Drag Coefficient, CD, as a function of Re for a Smooth Sphere
SMOOTH SPHERE
CD = D/( ½ U2A)
~ 1/5th drag
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~82o
CD = 0.5
~120o
CD = 0.2
PRESSURE DRAGon SMOOTH SPHERE
inviscid theory
laminar bdy layer
turbulent bdy layer
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Drag Coefficient, CD, as a function of Re for a Smooth Sphere
SMOOTH SPHERE
FD = 3UD - theoryCD = 3UD/(1/2 U2A)
CD = 24/Re
FD U2
FD U
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Laminar boundary layerTurbulent flow in wakeSeparation point moving forward
Separation point fixed
At Re ~ 100095% of drag due to pressure
difference between front and back
Turbulentboundary
layer
LaminarFlow
* *
FD U2
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example
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A small grain of sand is stirred 10 m from the bottom by the passage of an oil tanker. Assume a constant current of 1 m/s. There is a coral reef 1km downstream from where the sand was stirred up. Is it possible for the current to deposit the sand on the coral reef?
Specific Gravity of sand = 2.3 Time for current to move sand 1 km = 1000m/(1m/s) = 1000s Time for sand to settle = 10m/Uterminal velocity
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weight
buoyancy
drag
Terminal velocity so F = 0
Fweight = Fbuoyancy + Fdrag
Drag Force = ½ SGH2OU2CD Assume Re < 1 so CD = 24/ReDrag Force = 3H2OUD
Gravity Force = SGH2OD3/6 H2O= g Buoyancy Force = H2OD3/6
U = (SG-1) H2OgD2/(18H2O) = 0.00632 m/s
Re = 0.564
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A small grain of sand is stirred 10 m from the bottom by the passage of an oil tanker. Assume a constant current of 1 m/s. There is a coral reef 1km downstream from where the sand was stirred up. Is it possible for the current to deposit the sand on the coral reef?
Specific Gravity of sand = 2.3 Time for current to move sand 1 km = 1000m/(1m/s) = 1000s Time for sand to settle = 10m/Uterminal velocity
= 10m/0.00632 m/s = 1582s
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Roughness
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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.
Recritical ~ 400,000Recritical ~ 50,000
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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* - Fox et al.From Van Dyke, Album of Fluid MotionParabolic Press, 1982; Original photographs By Werle, ONERA, 1980
Re = 15000
Re = 30000
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Picture of 8.5-in bowling ball entering water at 25 ft/sec.
Why different flow patterns?
![Page 27: DRAG on “Blunt” Bodies. FLUID FLOW ABOUT IMMERSED BODIES U p4p4 p1p1 p2p2 p3p3 p6p6 p5p5 p7p7 p8p8 p9p9 p 10 p 11 p 13 p…p… p 12 10 99 88 77 66.](https://reader036.fdocuments.in/reader036/viewer/2022062423/56649f155503460f94c2aec7/html5/thumbnails/27.jpg)
Dramatic differences in location of laminar and turbulent separation on an 8.5-in bowling ball entering water at 25 ft/sec.
smooth rough
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example
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UCSD Sr. design team asks you whether they should dimple ping pong balls in an effort to make them go faster. Assume that the figure below shows the total range of Re that roughness can affect Recritical. What would you suggest?
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UCSD Sr. design team asks you whether they should dimple ping pong balls in an effort to make them go faster. Assume that the figure below shows the total range of Re that roughness can affect Recritical. What would you suggest?
Determine Re numbers of interest – turns out that highest ping pong ball speeds correspond to Re < 400,000 so probably will not work.
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Sphere vs Cylinder
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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
CD = D / (1/2 U2Af)
D ~ U
CD = constantD ~ U2
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A
BC D
E
FLOW AROUND A SMOOTH CYLINDER
~82o ~120o
Smooth Cylinder
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vortex shedding
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for 250 < Re < 2 x 105 – smooth circular cylinder f = 0.198(U/d)(1-19.7/Re) [ G.I. Taylor (1886-1975)]
St = fD/U = 0.198 (1.19.7/Re) ~ 0.2 [Strouhal (1850-1922)]
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for 250 < Re < 2 x 105 smooth circular cylinder f = 0.198(U/d)(1-19.7/Re) [ G.I. Taylor (1886-1975)]
St = fD/U = 0.198 (1.19.7/Re) ~ 0.2 [Strouhal (1850-1922)]
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Re = 6.5 x 105
M = 0.61Schlicting
Boundary-Layer Theory
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Spiral blades used for break up of span wise coherence of vortex shedding from a cylindrical rod.
from Kundu & Cohen – FLUID MECHANICS
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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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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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“In general, we can not overstress the importance of streamlining to reduce drag at Re numbers above about 100.” – White, Fluid Mechanics
(b) has ? % of drag as (a) (c) has ? % of drag as (b)
2-D rectangular cylinder
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(b) has 1 – CD(b)/CD(a) x 100% = 45% (c) has 1 – CD(b)/CD(a) x 100% = 86%
2-D rectangular cylinder
CD = FD/(1/2 U2A)[FD(a)-FD(b)] /FD(a) x 100%
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• (d) has 1/8th the thickness and 1/300th the cross section
of (c), yet same drag
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STREAMLINING
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STREAMLINING
2 cm
~ same drag AND wake
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(a) Because the Re numbers of most familiar flows over blunt bodies are large, the per cent of drag caused directly by shear stresses (friction drag) is often quite small.
(b) For low Re flows most of the drag is due to shear stresses(friction drag).
(c) For streamlined bodies most of the drag may be due to Shear stresses (friction drag).
Which of the following statements is most true:(1) Only statements (a) and (b) are true(2) Only statements (a) and (c) are true(3) Only statements (b) and (c) are true(4) All statements are true
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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