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