Monroe L. Weber-Shirk S chool of Civil and Environmental Engineering External Flows CEE 331 June 25, 2015 CEE 331 June 25, 2015

Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

External FlowsExternal Flows

CEE 331

April 18, 2023

CEE 331

April 18, 2023

OverviewOverview

Non-Uniform Flow Boundary Layer Concepts Viscous Drag Pressure Gradients: Separation and Wakes Pressure Drag Shear and Pressure Forces Vortex Shedding

Non-Uniform Flow Boundary Layer Concepts Viscous Drag Pressure Gradients: Separation and Wakes Pressure Drag Shear and Pressure Forces Vortex Shedding

Non-Uniform FlowNon-Uniform Flow

In pipes and channels the velocity distribution was uniform (beyond a few pipe diameters or hydraulic radii from the entrance or any flow disturbance)

In external flows the boundary layer is always growing and the flow is non-uniform

In pipes and channels the velocity distribution was uniform (beyond a few pipe diameters or hydraulic radii from the entrance or any flow disturbance)

In external flows the boundary layer is always growing and the flow is non-uniform

Boundary Layer ConceptsBoundary Layer Concepts

Two flow regimes Laminar boundary layer Turbulent boundary layer

with laminar sub-layer

Calculations of boundary layer thickness Shear (as a function of location on the surface) Drag (by integrating the shear over the entire surface)

Two flow regimes Laminar boundary layer Turbulent boundary layer

with laminar sub-layer

Calculations of boundary layer thickness Shear (as a function of location on the surface) Drag (by integrating the shear over the entire surface)

Flat Plate: Parallel to FlowFlat Plate: Parallel to Flow

Ux

y

U U U

Why is shear maximum at the leading edge of the plate?

boundary layer thickness

shear

dudy

is maximum

Laminar Boundary Layer:Shear and Drag Force

Laminar Boundary Layer:Shear and Drag Force

5

Rexxd=

5

Rexxd= Rex

Uxn

=Rex

Uxn

=

Boundary Layer thickness increases with the _______ ______ of the distance from the leading edge of the plate

x

U 3

0 332.0 x

U 3

0 332.0

ll

d dxx

UwdxwF

0

3

0

0 332.0

ll

d dxx

UwdxwF

0

3

0

0 332.0

lUwFd3664.0 lUwFd3664.0

5x

Un

d =5x

Un

d =

On one side of the plate!On one side of the plate!

Based on momentum and mass conservation and assumed velocity distribution

Based on momentum and mass conservation and assumed velocity distribution

squareroot

Integrate along length of plate

Laminar Boundary Layer:Coefficient of Drag

Laminar Boundary Layer:Coefficient of Drag

2

2FC (Re)D

d fU Ar

= =2

2FC (Re)D

d fU Ar

= =lUwFd3664.0 lUwFd3664.0

lwU

lUwd

2

3)664.0(2C

lwU

lUwd

2

3)664.0(2C

lwU

lUwd

2

3328.1C

lwU

lUwd

2

3328.1C

Uld

328.1C

Uld

328.1C 1.328

CRe

d =1.328

CRe

d =

Rel

Uln

=Rel

Uln

=

Dimensional analysisDimensional analysis

Transition to TurbulenceTransition to Turbulence

The boundary layer becomes turbulent when the Reynolds number is approximately 500,000 (based on length of the plate)

The length scale that really controls the transition to turbulence is the _________________________

5

Rexxd=

5

Rexxd=Rex

Uxn

=Rex

Uxn

= ReU

d

dn

=ReU

d

dn

=ReRex x

d d=

ReRex x

d d= Re 5 Rexd =Re 5 Rexd =

boundary layer thickness

Re = 3500=

Transition to TurbulenceTransition to Turbulence

Ux

y

U U

U

turbulentturbulent

Viscous sublayerViscous sublayer

This slope (du/dy) controls 0.

Transition (analogy to pipe flow)

more rapidlymore rapidly

Turbulent Boundary Layer: (Smooth Plates)

Turbulent Boundary Layer: (Smooth Plates)

1/5

0.37Rexx

d= 1/5

0.37Rexx

d=

5/1

20 029.0

UxU

5/1

20 029.0

UxU

5/1

2

0

0 036.0

UlwlUdxwF

l

d

5/1

2

0

0 036.0

UlwlUdxwF

l

d

1/5C 0.072Red l-= 1/5C 0.072Red l-= 2

2FC Re,D

d fU A l

er

æ ö= =è ø2

2FC Re,D

d fU A l

er

æ ö= =è ø

Rex

Uxn

=Rex

Uxn

=

5/1

5/437.0

Ux

5/1

5/437.0

Ux

Derived from momentum conservation and assumed velocity distribution

Integrate shear over plateIntegrate shear over plate

Grows ____________ than laminar

5 x 105 < Rel < 107

x 5/4

Boundary Layer ThicknessBoundary Layer Thickness

Water flows over a flat plate at 1 m/s. Plot the thickness of the boundary layer. How long is the laminar region?

Water flows over a flat plate at 1 m/s. Plot the thickness of the boundary layer. How long is the laminar region?

Rex

Uxn

=Rex

Uxn

=

RexxU

n=

RexxU

n=

sm

smxx

/1

)000,500(/101 26

sm

smxx

/1

)000,500(/101 26

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 1 2 3 4 5 6

length along plate (m)

boun

dary

laye

r th

ickn

ess

(m)

.

-

1,000,000

2,000,000

3,000,000

4,000,000

5,000,000

6,000,000

7,000,000

8,000,000

9,000,000

10,000,000

Rey

nold

s N

umbe

r

laminarturbulentReynolds Number

Grand Coulee

x = 0.5 m

5/1

5/437.0

Ux

5/1

5/437.0

Ux

5x

Un

d =5x

Un

d =

Flat Plate Drag CoefficientsFlat Plate Drag Coefficients

0.001

0.01

1e+04

1e+05

1e+06

1e+07

1e+08

1e+09

1e+10

Rel

Uln

=Rel

Uln

=

lele

DfCDfC

1 x 10-3

5 x 10-4

2 x 10-4

1 x 10-4

5 x 10-5

2 x 10-5

1 x 10-5

5 x 10-6

2 x 10-6

1 x 10-6

( )[ ] 2.51.89 1.62log /DfC le

-= - ( )[ ] 2.5

1.89 1.62log /DfC le-

= -

( )0.5

1.328

ReDf

l

C =( )0.5

1.328

ReDf

l

C =( )[ ]2.58

0.455 1700Relog Re

Dfll

C = -( )[ ]2.58

0.455 1700Relog Re

Dfll

C = -

( )[ ]2.58

0.455

log ReDf

l

C =( )[ ]2.58

0.455

log ReDf

l

C =

0.20.072ReDf lC -= 0.20.072ReDf lC -=

Example: Solar Car Example: Solar Car

Solar cars need to be as efficient as possible. They also need a large surface area for the (smooth) solar array. Estimate the power required to counteract the viscous drag at 40 mph

Dimensions: L: 5.9 m W: 2 m H: 1 m Max. speed: 40 mph on solar power alone Solar Array: 1200 W peak

Solar cars need to be as efficient as possible. They also need a large surface area for the (smooth) solar array. Estimate the power required to counteract the viscous drag at 40 mph

Dimensions: L: 5.9 m W: 2 m H: 1 m Max. speed: 40 mph on solar power alone Solar Array: 1200 W peak

air = 14.6 x10-6 m2/s air = 1.22 kg/m3

Viscous Drag on ShipsViscous Drag on Ships

The viscous drag on ships can be calculated by assuming a flat plate with the wetted area and length of the ship

The viscous drag on ships can be calculated by assuming a flat plate with the wetted area and length of the ship

F F FD viscous wave

AU

dd

2

F2C

AU

dd

2

F2C

Rel

Uln

=Rel

Uln

=Lr3Lr3Fwave scales with ____

0 . 0 0 1

0 . 0 1

R e l

U ln

=R e l

U ln

=

lele

D fC D fC

1 x 1 0 - 3

5 x 1 0 - 4

2 x 1 0 - 4

1 x 1 0 - 4

5 x 1 0 - 5

2 x 1 0 - 5

1 x 1 0 - 5

5 x 1 0 - 6

2 x 1 0 - 6

1 x 1 0 - 6

( )[ ] 2 . 51 . 8 9 1 . 6 2 l o g /D fC le

-= -

( ) 0 . 5

1 . 3 2 8

R eD f

l

C =( )[ ]2 . 5 8

0 . 4 5 5 1 7 0 0

R el o g R eD f

ll

C = -

( )[ ]2 . 5 8

0 . 4 5 5

l o g R eD f

l

C =

0 . 20 . 0 7 2 R eD f lC -=

0 . 0 0 1

0 . 0 1

R e l

U ln

=R e l

U ln

=

lele

D fC D fC

1 x 1 0 - 3

5 x 1 0 - 4

2 x 1 0 - 4

1 x 1 0 - 4

5 x 1 0 - 5

2 x 1 0 - 5

1 x 1 0 - 5

5 x 1 0 - 6

2 x 1 0 - 6

1 x 1 0 - 6

( )[ ] 2 . 51 . 8 9 1 . 6 2 l o g /D fC le

-= -

( ) 0 . 5

1 . 3 2 8

R eD f

l

C =( )[ ]2 . 5 8

0 . 4 5 5 1 7 0 0

R el o g R eD f

ll

C = -

( )[ ]2 . 5 8

0 . 4 5 5

l o g R eD f

l

C =

0 . 20 . 0 7 2 R eD f lC -=

2CF

2d

viscous

U Ar=

2CF

2d

viscous

U Ar=

Separation and WakesSeparation and Wakes

Separation often occurs at sharp corners fluid can’t accelerate to go around a sharp

corner Velocities in the Wake are ______ (relative

to the free stream velocity) Pressure in the Wake is relatively ________

(determined by the pressure in the adjacent flow)

Separation often occurs at sharp corners fluid can’t accelerate to go around a sharp

corner Velocities in the Wake are ______ (relative

to the free stream velocity) Pressure in the Wake is relatively ________

(determined by the pressure in the adjacent flow)

small

constant

Flat Plate:StreamlinesFlat Plate:

Streamlines

U

0 1

2

3

4

2

0

2

2

21U

pp

U

vC p

2

0

2

2

21U

pp

U

vC p

Point v Cp p1234

0 1<U >0>U <0

>p0

>p0

<p0

<p0

Points outside boundary layer!

Application of Bernoulli Equation

Application of Bernoulli Equation

g

vp

g

Up

22

220

g

vp

g

Up

22

220

20

2

2

21U

pp

U

v

20

2

2

21U

pp

U

v

In air pressure change due to elevation is small

U = velocity of body relative to fluid

gv

hp

gv

hp

22

22

22

21

11

gv

hp

gv

hp

22

22

22

21

11

0

22

22pp

g

v

g

U

0

22

22pp

g

v

g

U

pC

Flat Plate:Pressure Distribution

Flat Plate:Pressure Distribution

2

0

2

2

21U

pp

U

vC p

2

0

2

2

21U

pp

U

vC p

1 0 -1 -1.2

rearfront ddd FFF rearfront ddd FFF

AppF rearfrontd AppF rearfrontd

AU

CCFrearfront ppd 2

2 A

UCCF

rearfront ppd 2

2

Cp

2

02U

ppC p

2

02U

ppC p

0

2

2pp

CU p

0

2

2pp

CU p

0.8

AU

Fd

22.18.0

2 A

UFd

22.18.0

2

Cd = 2

0

<U

>U

1

2

3

Bicycle page at Princeton

Drag of Blunt Bodies and Streamlined Bodies

Drag of Blunt Bodies and Streamlined Bodies

Drag dominated by viscous drag, the body is __________.

Drag dominated by pressure drag, the body is _______.

Whether the flow is viscous-drag dominated or pressure-drag dominated depends entirely on the shape of the body.

Drag dominated by viscous drag, the body is __________.

Drag dominated by pressure drag, the body is _______.

Whether the flow is viscous-drag dominated or pressure-drag dominated depends entirely on the shape of the body.

streamlined

bluff

Velocity and Drag: SpheresVelocity and Drag: Spheres

C ,Re, , ,d f shape orientationD

Meæ ö=

è øC ,Re, , ,d f shape orientation

DM

eæ ö=è ø

AU

Dd

2

F2C

AU

Dd

2

F2C

( )2

2FC ReD

d fU Ar

= = ( )2

2FC ReD

d fU Ar

= =2

CF

2AUdD

2

CF

2AUdD

Spheres only have one shape and orientation!Spheres only have one shape and orientation!

General relationship for submerged objectsGeneral relationship for submerged objects

Where Cd is a function of ReWhere Cd is a function of Re

How fast do particles fall in dilute suspensions?

How fast do particles fall in dilute suspensions?

What are the important parameters? Initial conditions After falling for some time...

What principle or law could help us?

What are the important parameters? Initial conditions After falling for some time...

What principle or law could help us?

Acceleration due to gravityAcceleration due to gravity

dragdrag

Newton’s Second Law...Newton’s Second Law...

Sedimentation:Particle Terminal Fall Velocity

maF maF

AU

Dd

2

F2C

AU

Dd

2

F2C

0 WFF bd 0 WFF bd

gW pp gW pp

2

2t

wPDd

VACF

2

2t

wPDd

VACF

3

3

4rp 3

3

4rp 2rAp 2rAp

WW

dFdF

bFbF

gF wpb gF wpb

velocity terminalparticle

tcoefficien drag

gravity todueon accelerati

densitywater

density particle

area sectional cross particle

volumeparticle

t

D

w

p

p

p

V

C

g

ρ

ρ

A

velocity terminalparticle

tcoefficien drag

gravity todueon accelerati

densitywater

density particle

area sectional cross particle

volumeparticle

t

D

w

p

p

p

V

C

g

ρ

ρ

A

Particle Terminal Fall Velocity (continued)

Particle Terminal Fall Velocity (continued)

bd FWF bd FWF

gV

AC wppt

wPD )(2

2

gV

AC wppt

wPD )(2

2

wPD

wppt

AC

gV

)(2 2

wPD

wppt

AC

gV

)(2 2

dAp

p

3

2

dAp

p

3

2

w

wp

D

t

C

gdV

3

4 2

w

wp

D

t

C

gdV

3

4 2

w

wp

D

t

C

gdV

3

4

w

wp

D

t

C

gdV

3

4

General equation for falling objectsGeneral equation for falling objects

Relationship valid for spheresRelationship valid for spheres

Drag Coefficient on a Sphere Drag Coefficient on a Sphere

0.10.1

11

1010

100100

10001000

0.10.1 11 1010 102102 103103 104104 105105 106106 107107

Reynolds NumberReynolds Number

Dra

g C

oeff

icie

ntD

rag

Coe

ffic

ient Stokes Law

24ReDC =24ReDC = Re=500000

Turbulent Boundary Layer

Drag Coefficient for a SphereEquations

Drag Coefficient for a SphereEquations

Laminar flow R < 1Laminar flow R < 1

Transitional flow 1 < R < 104Transitional flow 1 < R < 104

Fully turbulent flow R > 104Fully turbulent flow R > 104

1/ 2

24Re24 3

0.34Re Re0.4

D

D

D

C

C

C

=

@ + +

@

1/ 2

24Re24 3

0.34Re Re0.4

D

D

D

C

C

C

=

@ + +

@

Re tV d rm

=Re tV d rm

=

18

2wp

t

gdV

18

2wp

t

gdV

( )

0.3p w

tw

gdV

r r

r

-»

( )

0.3p w

tw

gdV

r r

r

-»

Example Calculation of Terminal Velocity

Example Calculation of Terminal Velocity

Determine the terminal settling velocity of a cryptosporidium oocyst having a diameter of 4 m and a density of 1.04 g/cm3 in water at 15°C.

Determine the terminal settling velocity of a cryptosporidium oocyst having a diameter of 4 m and a density of 1.04 g/cm3 in water at 15°C.

ms

kg1.14x1018

kg/m 999kg/m 1040m/s 189.m 4x10

3

33226

tV

ms

kg1.14x1018

kg/m 999kg/m 1040m/s 189.m 4x10

3

33226

tV

18

2wp

t

gdV

18

2wp

t

gdV

ms

kg1.14x10

m 4x10

m/s 189.

kg/m 999

kg/m 1040

3

6

2

3

3

d

g

ρ

ρ

w

p

ms

kg1.14x10

m 4x10

m/s 189.

kg/m 999

kg/m 1040

3

6

2

3

3

d

g

ρ

ρ

w

p

m/s1014.3 7 xVt m/s1014.3 7 xVt

cm/day 7.2 tV cm/day 7.2 tV Reynolds

Pressure Gradients: Separation and Wakes

Pressure Gradients: Separation and Wakes

Van Dyke, M. 1982. An Album of Fluid Motion. Stanford: Parabolic Press.

Diverging streamlinesDiverging streamlines

Adverse Pressure GradientsAdverse Pressure Gradients

Increasing pressure in direction of flow Fluid is being decelerated Fluid in boundary layer has less ______

than the main flow and may be completely stopped.

If boundary layer stops flowing then separation occurs

Increasing pressure in direction of flow Fluid is being decelerated Fluid in boundary layer has less ______

than the main flow and may be completely stopped.

If boundary layer stops flowing then separation occurs

inertiainertia

Streamlines diverge behind objectStreamlines diverge behind object

2

2p V

z Cgg

+ + =

Point of SeparationPoint of Separation

Predicting the point of separation on smooth bodies is beyond the scope of this course.

Expect separation to occur where streamlines are diverging (flow is slowing down)

Separation can be expected to occur around any sharp corners

Predicting the point of separation on smooth bodies is beyond the scope of this course.

Expect separation to occur where streamlines are diverging (flow is slowing down)

Separation can be expected to occur around any sharp corners(where streamlines diverge rapidly)

Drag on Immersed Bodies(more shapes)

Drag on Immersed Bodies(more shapes)

Figures 9.19-21 bodies with drag coefficients on p 392-394 in text. hemispherical shell 0.38 hemispherical shell 1.42 cube 1.1 parachute 1.4

Figures 9.19-21 bodies with drag coefficients on p 392-394 in text. hemispherical shell 0.38 hemispherical shell 1.42 cube 1.1 parachute 1.4

Why?

Vs ?

Velocity at separation point determines pressure in wake.

The same!!!

Shear and Pressure ForcesShear and Pressure Forces

Shear forces viscous drag, frictional drag, or skin friction caused by shear between the fluid and the solid

surface function of ___________and ______of object

Pressure forces pressure drag or form drag caused by _____________from the body function of area normal to the flow

Shear forces viscous drag, frictional drag, or skin friction caused by shear between the fluid and the solid

surface function of ___________and ______of object

Pressure forces pressure drag or form drag caused by _____________from the body function of area normal to the flow

surface area length

flow separation

Example: Matrix PowerExample: Matrix Power

Cd = 0.32

Height = 1.539 mWidth = 1.775 mLength = 4.351 mGround clearance = 15 cm100 kW at 6000 rpmMax speed is 124 mph

Calculate the power required to overcome drag at 60 mph and 120 mph.

Where does separation occur?

What is the projected area? ( )A H G W@ -

( ) 21.539 0.15 1.775 2.5A m m m m= - =

Electric VehiclesElectric 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.

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 windshieldSmooth connection to windshield

Drag on a Golf BallDrag 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 so that the point of separation moves back further on the ball. The golf ball's dimples increase the turbulence in the boundary layer, increase the _______ of the boundary layer, and delay the onset of separation. The effect is plotted in the chart, which shows that for Reynolds numbers achievable by hitting the ball with a club, the coefficient of drag is much lower for the dimpled ball.

DRAG ON A GOLF BALL comes mainly from pressure drag. The only practical way of reducing pressure drag is to design the ball so that the point of separation moves back further on the ball. The golf ball's dimples increase the turbulence in the boundary layer, increase the _______ of the boundary layer, and delay the onset of separation. The effect is plotted in the chart, which shows that for Reynolds numbers achievable by hitting the ball with a club, the coefficient of drag is much lower for the dimpled ball.

inertiainertia

Why not use this for aircraft or cars?

Effect of Turbulence Levels on Drag

Effect of Turbulence Levels on Drag

Flow over a sphere: (a) Reynolds number = 15,000; (b) Reynolds number = 30,000, with trip wire.

Flow over a sphere: (a) Reynolds number = 15,000; (b) Reynolds number = 30,000, with trip wire.

Point of separationPoint of separation

Causes boundary layer to become turbulentCauses boundary layer to become turbulent

Effect of Boundary Layer Transition

Effect of Boundary Layer Transition

Ideal (non viscous) fluid

Real (viscous) fluid: laminar

boundary layer

Real (viscous) fluid: turbulent boundary layer

No shear!No shear!Increased inertia in boundary layer

Spinning SpheresSpinning Spheres

What happens to the separation points if we start spinning the sphere?

What happens to the separation points if we start spinning the sphere?

LIFT!LIFT!

Vortex SheddingVortex Shedding

Vortices are shed alternately from each side of a cylinder

The separation point and thus the resultant drag force oscillates

Frequency of shedding (n) given by Strouhal number S

S is approximately 0.2 over a wide range of Reynolds numbers (100 - 1,000,000)

Vortices are shed alternately from each side of a cylinder

The separation point and thus the resultant drag force oscillates

Frequency of shedding (n) given by Strouhal number S

S is approximately 0.2 over a wide range of Reynolds numbers (100 - 1,000,000)

U

ndS

U

ndS

Summary: External FlowsSummary: External Flows

Spatially varying flows boundary layer growth Example: Spillways

Two sources of drag shear (surface area of object) pressure (projected area of object)

Separation and Wakes Interaction of viscous drag and adverse pressure

gradient

Spatially varying flows boundary layer growth Example: Spillways

Two sources of drag shear (surface area of object) pressure (projected area of object)

Separation and Wakes Interaction of viscous drag and adverse pressure

gradient

Solution: Solar CarSolution: Solar Car

AU

dd

2

F2C

AU

dd

2

F2C

UlRx

UlRx

2

CF

2AUdd

2

CF

2AUdd

( ) ( ) ( ) ( )23 3 23 10 1.22 / 17.88 / 11.88F 2

2d

x kg m m s m-

=( ) ( ) ( ) ( )23 3 23 10 1.22 / 17.88 / 11.88

F 22d

x kg m m s m-

=

U = 17.88 m/s

l = 5.9 m

air = 14.6 x 10-6 m2/s

Rel = 7.2 x 106

Cd = 3 x 10-3

air = 1.22 kg/m3

A = 5.9 m x 2 m = 11.8 m2 Fd =14 N

P =F*U=250 W

( )[ ]2.58

0.455

log ReDf

l

C =( )[ ]2.58

0.455

log ReDf

l

C =

0 . 0 0 1

0 . 0 1

R e l

U ln

=R e l

U ln

=

lele

D fC D fC

1 x 1 0 - 3

5 x 1 0 - 4

2 x 1 0 - 4

1 x 1 0 - 4

5 x 1 0 - 5

2 x 1 0 - 5

1 x 1 0 - 5

5 x 1 0 - 6

2 x 1 0 - 6

1 x 1 0 - 6

( )[ ] 2 . 51 . 8 9 1 . 6 2 l o g /D fC le

-= -

( ) 0 . 5

1 . 3 2 8

R eD f

l

C =( )[ ]2 . 5 8

0 . 4 5 5 1 7 0 0

R el o g R eD f

ll

C = -

( )[ ]2 . 5 8

0 . 4 5 5

l o g R eD f

l

C =

0 . 20 . 0 7 2 R eD f lC -=

0 . 0 0 1

0 . 0 1

R e l

U ln

=R e l

U ln

=

lele

D fC D fC

1 x 1 0 - 3

5 x 1 0 - 4

2 x 1 0 - 4

1 x 1 0 - 4

5 x 1 0 - 5

2 x 1 0 - 5

1 x 1 0 - 5

5 x 1 0 - 6

2 x 1 0 - 6

1 x 1 0 - 6

( )[ ] 2 . 51 . 8 9 1 . 6 2 l o g /D fC le

-= -

( ) 0 . 5

1 . 3 2 8

R eD f

l

C =( )[ ]2 . 5 8

0 . 4 5 5 1 7 0 0

R el o g R eD f

ll

C = -

( )[ ]2 . 5 8

0 . 4 5 5

l o g R eD f

l

C =

0 . 20 . 0 7 2 R eD f lC -=

Reynolds Number CheckReynolds Number Check

R<<1 and therefore in Stokes Law rangeR<<1 and therefore in Stokes Law range

VdR

VdR

ms

kg1.14x10

kg/m999m104m/s1014.3

3

367

xxR

ms

kg1.14x10

kg/m999m104m/s1014.3

3

367

xxR

R = 1.1 x 10-6

Solution: Power a Toyota Matrix at 60 or 120 mph

Solution: Power a Toyota Matrix at 60 or 120 mph

)(F2

C2

RfAU

Dd

)(

F2C

2Rf

AU

Dd

2

CF

2AUdD

2

CF

2AUdD

2

C 3AUP d

2

C 3AUP d

3 3 2(0.32)(1.2 / )(26.82 / ) (2.5 )2

kg m m s mP =

3 3 2(0.32)(1.2 / )(26.82 / ) (2.5 )2

kg m m s mP =

P = 9.3 kW at 60 mphP = 74 kW at 120 mph

Grand Coulee Dam Grand Coulee Dam

Turbulent boundary layer reaches surface!Turbulent boundary layer reaches surface!

Drexel SunDragon IVDrexel SunDragon IV

Vehicle ID: SunDragon IV (# 76)Dimensions: L: 19.2 ft. (5.9 m) W: 6.6 ft. (2 m) H: 3.3 ft. (1 m) Weight: 550 lbs. (249 kg)Solar Array: 1200 W peak; 8 square meters terrestrial grade solar cells; manf: ASE AmericasBatteries: 6.2 kW capacity lead-acid batteries; manf: US BatteryMotor: 10 hp (7.5 kW) brushless DC; manf: Unique MobilityRange: Approximately 200 miles (at 35 mph on batteries alone)Max. speed: 40 mph on solar power alone, 80 mph on solar and battery power.Chassis: Graphite monocoque (Carbon fiber, Kevlar, structural glass, Nomex)Wheels: Three 26 in (66 cm) mountain bike, custom hubsBrakes: Hydraulic disc brakes, regenerative braking (motor)

Vehicle ID: SunDragon IV (# 76)Dimensions: L: 19.2 ft. (5.9 m) W: 6.6 ft. (2 m) H: 3.3 ft. (1 m) Weight: 550 lbs. (249 kg)Solar Array: 1200 W peak; 8 square meters terrestrial grade solar cells; manf: ASE AmericasBatteries: 6.2 kW capacity lead-acid batteries; manf: US BatteryMotor: 10 hp (7.5 kW) brushless DC; manf: Unique MobilityRange: Approximately 200 miles (at 35 mph on batteries alone)Max. speed: 40 mph on solar power alone, 80 mph on solar and battery power.Chassis: Graphite monocoque (Carbon fiber, Kevlar, structural glass, Nomex)Wheels: Three 26 in (66 cm) mountain bike, custom hubsBrakes: Hydraulic disc brakes, regenerative braking (motor)

http://cbis.ece.drexel.edu/SunDragon/Cars.html

Pressure Coefficients on a WingPressure Coefficients on a Wing

NACA 63-1-412 Flowfield Cp's

2

0

2

2

21U

pp

U

vC p

2

0

2

2

21U

pp

U

vC p

Shear and Pressure Forces: Horizontal and Vertical Components

Shear and Pressure Forces: Horizontal and Vertical Components

dApD cossinF 0 dApD cossinF 0

dApL sincosF 0 dApL sincosF 0

lift

drag

U

Parallel to the approach velocity

Normal to the approach velocity

2

2UACF dd

2

2UACF dd

2

2UACF LL

2

2UACF LL

A defined as projected area _______ to force!normalnormal

drag

liftp < p0

negative pressure

p < p0

negative pressure

p > p0 positive pressurep > p0 positive pressure