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.

DRAG on “Blunt” Bodies

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

“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

DRAG Coefficient - CD

CD = FD/(1/2 U2A)

EXPERIMENTAL

EXPERIMENTAL

CD = f(shape, Re, Ma, Fr, /L)

• 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

• 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)?

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.

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

Cdmin = 0.06

0.25

tc

drag coefficient on a strut as a function of thickness/ chord

Mostly pressure drag, separation point fixed

Frictiondrag

Character of CD vs Re curves for different 2-D shapes

press& fric

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.

sphere

Character of CD vs Re curves for 3-D shape - sphere

• 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 ?

Drag Coefficient, CD, as a function of Re for a Smooth Sphere

SMOOTH SPHERE

CD = D/( ½ U2A)

?

??

?

Drag Coefficient, CD, as a function of Re for a Smooth Sphere

SMOOTH SPHERE

CD = D/( ½ U2A)

~ 1/5th drag

~82o

CD = 0.5

~120o

CD = 0.2

PRESSURE DRAGon SMOOTH SPHERE

inviscid theory

laminar bdy layer

turbulent bdy layer

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

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

example

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

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

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

Roughness

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

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

Picture of 8.5-in bowling ball entering water at 25 ft/sec.

Why different flow patterns?

Dramatic differences in location of laminar and turbulent separation on an 8.5-in bowling ball entering water at 25 ft/sec.

smooth rough

example

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?

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.

Sphere vs Cylinder

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

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

A

BC D

E

FLOW AROUND A SMOOTH CYLINDER

~82o ~120o

Smooth Cylinder

vortex shedding

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)]

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)]

Re = 6.5 x 105

M = 0.61Schlicting

Boundary-Layer Theory

Spiral blades used for break up of span wise coherence of vortex shedding from a cylindrical rod.

from Kundu & Cohen – FLUID MECHANICS

Flow Separation

FLOW SEPARATION

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.

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

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

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

Streamlining

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

“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

(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%

• (d) has 1/8th the thickness and 1/300th the cross section

of (c), yet same drag

STREAMLINING

STREAMLINING

2 cm

~ same drag AND wake

(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

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

(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

*

*

*

*

First flight of a powered aircraft 12/17/03 120ft in 12 secondsOrville Wright at the controls

Same drag at 210 mph

The End