

U p

0 2   

C p

 2   

p



U p

0 2    

U

2

C p



p



p

0 2

F d F d

  

p F d front

 

F d rear p rear



A front

0 1 0

C p

-1 -1.2

F d

 

C p front



C p rear

 

U

2 2

A F d

  0 .

8  1 .

2  

U

2 2

A

C d = 2

Drag of Blunt Bodies and Streamlined Bodies  Drag dominated by viscous  Drag dominated by pressure  Whether the flow is viscous drag dominated or pressure drag dominated depends entirely on the shape of the body. Bicycle page at Princeton

Flow and Drag: Spheres C

d

=

f

ж e и

D

, Re,

M

,

shape orientation

ц ш General relationship for submerged objects Spheres only have one shape and orientation!

C

d

 2 F

D



U

2

A

r 2F

D

2

U A

= C

d

=

f

F

D

 C

d



U

2

A

2 Where

C d

is a function of Re

How fast do particles fall in dilute suspensions?

 What are the important parameters?

 Initial conditions Acceleration due to gravity  After falling for some time...

drag  What principle or law could help us?

Newton’s Second Law...

Sedimentation: Particle Terminal Fall Velocity 

F



ma F d



W F b

 

W



p

  0

p g F b

 

p



w g F d



C D A P



w V t

2 2 

p

 4 

r

3 3

A p

 

r

2

F d W F b



p

 particle volume

A p

 particle cross sectional area

ρ p

 particle density

ρ w g

  water density accelerati on due to gravity

C D V t

  drag coefficien t particle terminal velocity C

d

 2 F

D



U

2

A

Particle Terminal Fall Velocity (continued)

F d



W



F b C D V t A P

2  

w V t

2 2  2

p

( 

C D



p



A P



p

(  

w



w p

 )

g



w

)

g

General equation for falling objects 

p

 2

d

Relationship valid for spheres

A p V t

2  4

gd

3

C D

 3 

p

 

w



w



V t

 4

gd

3

C D

 

p

 

w



w



Drag Coefficient on a Sphere 1000 100 10 Stokes Law 1 0.1

0.1

1 24

C D

= Re 10 10 2 10 3 10 4 Reynolds Number 10 5 10 6 10 7 Re=500000 Turbulent Boundary Layer

Drag Coefficient: Equations Laminar flow R < 1

C D

= Transitional flow 1 < R < 10 4

C D

= Fully turbulent flow R > 10 4

C D

= 24 Re

V t

24 3 + Re Re 1/ 2 + 0.34



d

2

g

 

p

18   

w

 0.4

V t

»

gd

( r 0.3

p

r

w

r

w

) Re =

V d t

r m

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/cm 3

ρ p ρ w g d

 1040 kg/m 3  9 99 kg/m 3  9.

8 1 m/s 2  4x10  6 m

V t

in water at 15°C.

V t



d

2

g

  18

p

  

w

   4x10  6 m 2   9.

8 1 m/s 2  1040 kg/m 18   1.14x10

 3 s kg  m   3  9 99 kg/m 3    1.14x10

 3 kg s  m

V t

 3 .

14

x

10  7 m/s

V t

 2 .

7 cm/day

Reynolds

Pressure Gradients: Separation and Wakes Diverging streamlines Van Dyke, M. 1982. An Album of Fluid Motion. Stanford: Parabolic Press.

Adverse Pressure Gradients   Streamlines diverge behind object Increasing pressure in direction of flow

p

Fluid is being decelerated g  than the main flow and may be completely

V

2 2

g

=

C

stopped.

 If boundary layer stops flowing then separation occurs

Point 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 (where streamlines diverge rapidly)

Drag on Immersed Bodies (more shapes)  Figures 9.19-21 bodies with drag coefficients on p 392-394 in text.

 hemispherical shell  hemispherical shell  cube  parachute 0.38

1.42

1.1

1.4

Why?

Velocity at separation point determines pressure in wake Vs ?

Shear and Pressure Forces: Horizontal and Vertical Components F

D

F

L

U   

p

sin    

p

cos   t 0 cos   t 0 sin  

dA



dA

lift drag Parallel to the approach velocity lift p < p 0 Normal to the approach velocity negative pressure

F d



C d A



U

2

A

2 defined as projected 

F L



C L A



U

2 2 drag p > p 0 positive pressure

Shear and Pressure Forces  Shear forces  viscous drag, frictional drag, or skin friction  caused by shear between the fluid and the solid surface   Pressure forces  pressure drag or form drag   function of area normal to the flow

Example: Beetle Power

C d

= 0.38

Height = 1.511 m Width = 1.724 m Length = 4.089 m Ground clearance = 15 cm?

85 kW at 5200 rpm Where does separation occur?

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

Is the beetle streamlined?

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

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 so that the point of separation moves back further on the ball. The golf ball's dimples increase the turbulence in 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.

Why not use this for aircraft or cars?

Effect of Turbulence Levels on Drag  Flow over a sphere: (a) Reynolds number = 15,000; (b) Reynolds number = 30,000, with trip wire. Causes boundary layer to become turbulent Point of separation

Effect of Boundary Layer Transition Ideal (non viscous) fluid Real (viscous) fluid: laminar boundary layer Real (viscous) fluid: turbulent boundary layer No shear!

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

LIFT!

Vortex Shedding  Vortices are shed alternately from each side of a cylinder  The separation point and thus the resultant drag force oscillate  Dimensionless frequency of shedding given by Strouhal number S  S is approximately 0.2 over a wide range of Reynolds numbers (100 - 1,000,000)

S



nd U

Summary: 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

Solution: Solar Car F

d

= 0.455

C Df

= [

l

) ] 2.58

2 ( C F

d d

  C

d

2 F

d



U

2 

U A

2

A

2 3 )( 1.22

/ 3 ) ( 2

C Df

1e +04

F d

=14 N

( ( ) ) 0.5

[ [ [ [ ( ( ) ) ] ] 2.58

1e +05 1e +06 1e +07 Re

l l

= = 1e +08

R x

( ( e e / /

l l

) ) ] ] 1e +09 1e +10



Ul

1 x 10 5 x 10 -3 -4 2 x 10 -4 1 x 10 5 x 10 2 x 10 1 x 10 5 x 10 2 x 10 1 x 10 -4 -5 -5 -5 -6 -6 -6 e e e e



[ [ ( ( ) ) ] ] 2.58

U = 17.88 m/s

l

= 5.9 m  air  air = 14.6 x 10 -6 m 2 /s = 1.22 kg/m 3 ) 2 ( 11.88

m

2 )

Re l

= 7.2 x 10 6

A C d

= 3 x 10 -3 = 5.9 m x 2 m = 11.8 m 2

P

=F*U=250 W

Reynolds Number Check

R



Vd

 

R

  3 .

14

x

10  7 m/s  4

x

10  6 m  999 kg/m 3  1.14x10

 3 kg s  m R = 1.1 x 10 -6 R<<1 and therefore in Stokes Law range

Solution: Power a New Beetle at 60 mph C

d

 2 F

D



U

2

A

 F

D

 C

d



U

2

A

2

f

(

R

)

P

 C

d



U

3

A

2

P

 ( 0 .

38 )( 1 .

2

kg

/

m

3 )( 26 .

82

m

/

s

) 3 ( 2 .

0

m

2 ) 2 P = 8.8 kW at 60 mph P = 70 kW at 120 mph

Solution: Power a Plymouth Voyager at 60 mph C

d

 2 F

D



U

2

A

 F

D

 C

d



U

2

A

2

f

(

R

) Frontal Area 30.15 sq ft (2.80 sq m) Drag Coefficient 0.35

P

 C

d



U

3

A

2 How can

C d

larger object?

be smaller for

P

 ( 0 .

35 )( 1 .

2

kg

/

m

3 )( 26 .

82

m

/

s

) 3 ( 2 .

80

m

2 ) 2 P = 11.3 kW at 60 mph

Grand Coulee Dam Turbulent boundary layer reaches surface!

Drexel SunDragon IV  http://cbis.ece.drexel.edu/SunDragon/Cars.html

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 Americas Batteries: 6.2 kW capacity lead-acid batteries; manf: US Battery Motor: 10 hp (7.5 kW) brushless DC; manf: Unique Mobility Range: 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 hubs Brakes: Hydraulic disc brakes, regenerative braking (motor)

Pressure Coefficients on a Wing

C p

 1 

v

2

U

2  2   

p



U p

0 2   

NACA 63-1-412 Flowfield Cp's

Head Loss in Bends High pressure  Head loss is a function of the ratio of the bend radius to the pipe diameter (R/D) R n Possible separation from wall  Velocity distribution returns to normal several pipe diameters downstream D

K b

varies from 0.6 - 0.9

p

+ r у ф х

V

2 R

dn

+ g

z

= Low pressure

C h b



K b V

2 2

g