lecture 1 1 drag

8/18/2019 Lecture 1 1 Drag

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Flow over Bodies;Drag and Lift,Chapter 9

Dr. Hamdy A. Kandil

External Flow Bodies and vehicles in motion,

or with flow over them,experience fluid-dynamicforces and moments.

Examples include: aircraft,automobiles, buildings, ships,submarines, turbomachines.

These problems are oftenclassified as External Flows.

Fuel economy, speed,acceleration, maneuverability,stability, and control are

directly related to theaerodynamic/hydrodynamicforces and moments.


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External Flow

(b) Surface flow on amodel vehicle as indicatedby tufts attached to thesurface.

(a) Flow past a full-sizedstreamlined vehicle in the GMaerodynamics laboratory windtunnel, and 18-ft by 34-fttest section facility driven bya 4000-hp, 43-ft-diameterfan.

Two of NASA’s Wind Tunnels

Ames 80’ x 120’Langley


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Drag and Lift Concepts

Forces from thesurrounding fluid ona two-dimensionalobject:

(a) pressure force,(b) viscous force,(c) resultant force

(lift and drag).

Drag and Lift Fluid dynamic forces

are due to pressure(normal) and viscous

(shear) forces actingon the body surface. Drag: Force

component parallelto flow direction.

Lift: Force

component normal toflow direction.


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Drag Forces Drag forces can be found by

integrating pressure andwall-shear stresses.

Special Cases

= 90 = 0

is the angle between the normal vector and the direction of motion

nn

Drag due to friction only Drag due to pressure only

Drag Coefficient, CD. In addition to geometry, drag force F D is a function of

density and velocity V . drag coefficient, CD:

Area A is a reference area: can be frontal area (the areaprojected on a plane normal to the direction of flow) (dragapplications), plan-form area (wing aerodynamics), orwetted-surface area (ship hydrodynamics).

For applications such as tapered wings, C D may be afunction of span location. For these applications, a localC D,x is introduced and the total drag is determined by

integration over the span L


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Friction Drag and Pressure Drag Fluid dynamic forces are

comprised of pressure (form)and friction effects.

Often useful to decompose, FD = FD,friction + FD,pressure CD = CD,friction + CD,pressure

This forms the basis of modeltesting.

Friction drag

Pressure drag

Friction & pressure drag CD = CD,friction CD = CD,pressure

Streamlining Streamlining reduces drag by reducing FD,pressure, at

the cost of increasing wetted surface area andFD,friction.

Goal is to eliminate flow separation and minimize

total drag FD


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STREAMLINED

(a)

Flat

Plate

100%

Resistance (b)

Sphere

50%

Resistance

(c) Ovoid 15% Resistance (d) Streamlined 5% Resistance

Streamlining to reduce pressure drag

C D of Common Geometries For many geometries, total drag

C D is constant for Re > 104

C D can be very dependent upon

orientation of body. As a crude approximation,

superposition can be used to add

C D from various components of a

system to obtain overall drag.

222

42

1

2

1 DV C AV C F D D D


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C D of Common Geometries


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Some more shapes

Boundary Layer (BL) Approximation BL approximation bridges

the gap between the Euler(inviscid) and Navier-Stokes (NS) (viscous)equations, and between

the slip and no-slipBoundary Conditions (BC)at the wall.

Prandtl (1904) introducedthe BL approximation- BLis a thin region on thesurface of a body inwhich viscous effects are

very important andoutside of which the fluidbehaves as inviscid.


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Character of the steady, viscous flow past aflat plate parallel to the upstream velocity

(a)low Reynolds number flow,

(b)moderate Reynolds number flow,

(c)large Reynolds number flow.

ForceViscousForce InertiaVL

Re

Inertia force = ma =L3*V dV/dL= V2L2

Viscous Force = L2 dV/dL = V L

Boundary Layer on a Flat Plate

Not to scale

To scale

at y= : v = 0.99 V

at y= : v = 0.99 V


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Transition


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Turbulent boundary layer

Merging of turbulent spots and transition to turbulencein a natural flat plate boundary layer.

Top view

Side view

Grand Coulee Dam

Turbulent boundary layer reaches surface!


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Flat Plate Drag

Drag on flat plate is solely due to friction (FD=FDfriction) created by laminar, transitional, andturbulent boundary layers.

BL thickness, , is the distance from the plate atwhich v = 0.99 V

v = 0.99 V

Flat Plate Drag

Local friction coefficient

Laminar:

Turbulent: Average friction coefficient

For some cases, plate is long enough for turbulentflow, but not long enough to neglect laminar portion

CD, friction = Cf


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Flat Plate Drag

Transition takes place at a distance x given by:Rexcr=2X105 to 3X106 – We will use Rexcr=5X105

Drag coefficient may also be obtained from charts suchas those on the next slides

Friction dragcoefficient for aflat plate parallel tothe upstream flow.

Turbulent: CDf = f (Re, /L)Laminar: CDf = f (Re)


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Pressure Drag on a Flat Plate

Conditions depend onspecial features ofboundary layerdevelopment, including

onset at a stagnationpoint and separation, aswell as transition toturbulence.

Stagnation point: Location of zero velocity and maximumpressure. Followed by boundary layer development under afavorable pressure gradient and hence acceleration of the freestream flow

As the rear of the cylinder is approached, the pressure mustbegin to increase. Hence, there is a minimum in the pressure distribution, p(x),

after which boundary layer development occurs under theinfluence of an adverse pressure gradient

Cylinders in Cross Flow


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and is accompanied byflow reversal and adownstream wake.Location of separationdepends on boundarylayer transition.

– What features differentiate boundary development for the flatplate in parallel flow from that for flow over a cylinder? –Separation occurs when the velocity gradient reduces to zero.

Flow around a cylinder Inviscid flow past a

circular cylinder: (a)streamlines for theflow if there were noviscous effects. (b)pressure distributionon the cylinder’ssurface,(c) free-streamvelocity on thecylinder’s surface.


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Boundary layercharacteristicson a circularcylinder:

(a)boundary layerseparation location.

(b)typical boundarylayer velocityprofiles at variouslocations on thecylinder,

(c) surface pressuredistributions forinviscid flow andboundary layer flow.

Flow separation in a diffuser with a large angle


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Smooth Cylinder and Sphere Drag

VDRe

Drag coefficient as a function of Reynolds number for a smooth circularcylinder and a smooth sphere.

Effect of Surface Roughness

For blunt bodies an increase in may decrease CD bytripping the flow into turbulent at lower Re


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Sports Balls

Sports balls Many games involve balls

designed to use dragreduction brought aboutby surface roughness.

Many sports balls havesome type of surfaceroughness, such as theseams on baseballs orcricket balls and the fuzz

on tennis balls.

It is the Reynolds number (notthe speed) that determineswhether the boundary layer islaminar or turbulent.

Thus, the larger the ball, the

lower the speed at which arough surface can be of helpin reducing the drag.

VD

Re


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Drag on a Golf Ball Drag on a golf ball comes mainly from

pressure drag.

The only practical way of reducing

pressure drag is to design the ball sothat the point of separation moves backfurther on the ball.

The golf ball's dimples increase theturbulence in the boundary layer,increase the inertia of the boundarylayer, and delay the onset of separation.

The Reynolds number where theboundary layer begins to becometurbulent with a golf ball is 40,000

A non-dimpled golf ball would reallyhamper Tiger Woods’ long game

Why not use this for aircraft or cars?

GOLF BALL AERODYNAMICSLarge Wake of Separated Flow,High Pressure Drag

Laminar B.L. Separation Point

Reduced Size Wake of Separated

Flow, Lower Pressure Drag

Turbulent B.L. Separation Point


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Drag due to viscousboundary layer skin friction

D R A G






Pressure drag

(due to viscousflow separation, wake

D R A G


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D R A G

Total Drag

Baseball At the velocities of 50 to 130 mph dominant in baseball the air

passes over a smooth ball in a highly resistant flow. Turbulent flow does not occur until nearly 200 mph for a smooth

ball A rough ball (say one with raised stitches like a baseball) induces

turbulent flow A baseball batted 400 feet would only travel 300 feet if it was

smooth.


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Adverse pressure gradients (Tennis ball)

Separation of theboundary layers occurswhenever the flow tries

to decelerate quickly,that is whenever thepressure gradient ispositive (adversepressure gradient-pressure increases).

In the case of the tennisball, the flow initiallyaccelerates on theupstream side of the ball,while the local pressuredecreases in accord withBernoulli’s equation.

Near the top of the ball the localexternal pressure increases andthe flow should decelerate as thepressure field is converted tokinetic energy. However, because of viscouslosses, not all kinetic energy isrecovered and the flow reversesaround the separation point.

FAS

T ER


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History ofAutomobiles

History of Automobiles

Mercedes-Benz W125 in a wind tunnel

CD

= 0.170 1938

CD

= 0.365 1937


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History ofAutomobiles


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History of Cars

Automobile DragScion XB Porsche 911

CD = 1.0, A = 25 ft2, CD A = 25ft

2 CD = 0.28, A = 10 ft2, CD A = 2.8ft

2

• Drag force FD=1/2V2(CDA) will be ~ 10 times larger forScion XB

• Source is large CD and large projected area

• Power consumption P = FDV =1/2V3(CDA) for both scaleswith V3!


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Electric Vehicles Electric vehicles are designed to minimize drag.

Typical cars have a coefficient drag of 0.30-0.40.

The EV1 has a drag coefficient of 0.19.

Smooth connection to windshield

Automobile Drag


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Example-1

)2/()2/()2/(5.0)5.0*)2/(5.0(*2

)()(2

2222V AC V AV AV A

F F F

p D p p p

Drag D DragS Drag

Given: Pressure distribution isshown, flow is left to right.

Find: Find C D Solution: C D is based on the

projected area of the block from thedirection of flow. Force ondownstream face is:

The total force on each side face is:

The drag force on one face is:

The total drag force is:

Coefficient of Drag is: C D = 1

)2/(5.0)2/( 22 V A AV C F p p p Drag D

)2/(5.0)2/( 22 V A AV C F p p pS

5.0*)2/(5.0sin 2V AF F pS DragS

FS

2/2V

pC p

Ap

Example-2 Given: Flag pole, 35 m high, 10 cm diameter,

in 25-m/s wind, Patm = 100 kPa, T=20oC

Find: Moment at bottom of flag pole

Solution:

325 /20.1,/1051.1 mkgsm x

5

5 1066.1

1051.1

1.0*25Re x

x

VD

)219(Fig.1.1 DC

mkN

H V AC H F M o p D D

23

2

35*

2

25*2.1*35*10.0*1.1

2222

2

2

2V AC F p D D

Air properties at 20C

H=35

FD


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Example-3 Given: Spherical balloon 2-m diameter, filled with

helium at std conditions. Empty weight = 3 N.

Find: Velocity of ascent/descent.

Solution:

N

DF

DF D

W W F F F

air

Heair D

He Dair

He B D B y

422.1

26)077,2

287

1(3

6)(3

63

6

0

3

3

33

W B W He

F B

F D

D D p D

Do

o p D D

C C AC

F V

V AC F

739.0

225.1*2*)4/(

422.1*22

2

2

2

Iteration 1: Guess C D = 0.4 ??

smV o /36.14.0

739.0

Check Re

5

5 1086.1

1046.1

2*36.1Re x

x

VD

Iteration 2: Chart Re C D = 0.42

smV o /33.142.0

739.0

Vo

lecture 1 1 drag

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