External Flows
CEE 331
March 24, 2016
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Overview
The Fss connection to Drag
Boundary Layer Concepts
Drag
Shear Drag
Pressure Drag
Pressure Gradients: Separation and Wakes
Drag coefficients
Vortex Shedding
Fss: Shear and Pressure Forces
U
Shear forces:
Major losses in pipes
viscous drag, frictional drag, or skin friction
caused by shear between the fluid and the solid
surface
length object
function of ___________and
______of
surface area
Flow
Pressure forces
expansion
pressure drag or form drag
losses
caused by _____________from
the body
flow separation
function of area normal to the flow Projected area
U
Non-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 (the flow
influenced by the solid object) is always growing
and the flow is non-uniform
 We need to calculate shear in this non-uniform
flow!
Boundary 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)
 Viscous Drag (by integrating the shear over the entire
surface)
Flat Plate: Parallel to Flow
U
to
y
U
U
d
U
boundary
layer
thickness
x
shear
Why is shear maximum at the leading edge of
the plate? du
dy
is maximum
Laminar Boundary Layer:
Shear and Drag Force
d
5

x
Re x
Re x 
Ux

d 5
x
U
square
Boundary Layer thickness increases with the _______
root of the distance from the leading edge of the plate
______
Based on momentum and mass
3
U
conservation and assumed
t 0  0.332
x
velocity distribution
l
l
Fd  w t 0 dx  w 0.332
0
0
Fd  0.664w U 3l
U 3
dx
Integrate along length of plate
x
On one side of the plate!
Laminar Boundary Layer:
Coefficient of Drag
Fd  0.664w U l
3
Cd 
Cd 
2Fd
Cd 
 f (Re)
2
U A
2(0.664) w U 3l
Dimensional analysis
U 2lw
1.328w U 3l
U 2lw
Cd 
1.328 
Ul
Cd 
1.328
Rel
Rel 
Ul

Transition 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
boundary layer thickness
_________________________
Re x 
Ux

Red 
Ud

d
5
Red d
 = 
x
Re x
Re x x
Red  5 Re x
Red = 3500
Transition to Turbulence
U
U
d
U
y
turbulent
U
x
Viscous
sublayer
to
This slope (du/dy) controls t0.
Transition (analogy to pipe flow)
Turbulent Boundary Layer:
(Smooth Plates)
1/ 5
 
d  0.37 x  
Re x 

U 
Grows ____________
more rapidly than laminar
d
0.37
 1/ 5
x Re x
Ux
4/5
1/ 5
 
t 0  0.029 U  
 Ux 
Derived from momentum conservation
and assumed velocity distribution
2
x 5/4
1/ 5
 
Fd  w t 0 dx  0.036 U wl  
 Ul 
0
l
2
Cd  0.072Rel1/ 5
5 x 105 < Rel < 107
Integrate shear over plate
2Fd


Cd 
 f  Re, 
2
U A
l

Boundary Layer Thickness
 Water flows over a flat plate at 1 m/s. How long is the laminar region?
0.02
2,000,000
d  0.37 x
boundary layer thickness (m) .
0.018
0.016
4/5
 
 
 
U 
laminar
turbulent
Reynolds Number
1,800,000
0.014
1,400,000
0.012
1,200,000
0.01
d 5
0.008
Re x 
1,600,000
x
1,000,000
800,000
U
0.006
600,000
0.004
400,000
0.002
200,000
0
Reynolds Number
1/ 5
x
x
Ux

 Re x
U
1x10 6 m 2 / s(500,000)
1m / s
x = 0.5 m
0
0.5
1
1.5
length along plate (m)
2
Grand Coulee
Flat Plate Drag Coefficients
Cd f  1.89  1.62log   / l  
0.01
2.5
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
1.328
0.5
 Rel 
Cd f 
laminar
Cd f
1700
Rel

00
00
0
Cd f  0.072Rel0.2
00
00
10
00
00
00
00
00
00
00
10
Rel 
Ul
e
l
Cd f 
0.455
log  Rel  
2.58
10

0
2.58
00
00
10
00
00
00
10
00
0
00
log  Rel  
0
0.001
0.455
00
Cd f 
10
transitional
10
rough
Turbulent boundary
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 on the solar panel 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 Ships
The viscous drag on ships can be calculated
by assuming a flat plate with the wetted
area and length of the ship
0.001
10
00
00
0
10
00
00
00
10
00
00
00
0
10
00
00
00
00
10
00
00
00
00
0
U 2 A
Cd U 2 A
Fd f 
2
10
00
00
Cd 
2Fd
0.01
10
00
0
Fd  Fd f  Fwave
Rel 
Ul

Fwave scales with ____
Lr3 (based on _______
Froude similarity)
U
Separation and Wakes
Separation often occurs at sharp corners
fluid can’t accelerate to go around a sharp
corner
Velocities in the Wake are ______
small (relative
to the free stream velocity)
Pressure in the Wake is relatively ________
constant
(determined by the pressure in the adjacent
flow)
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
V2
z
C
Fluid is being decelerated

2g
Fluid in boundary layer has less ______
inertia
than the main flow and may be completely
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)
Flat Plate:
Streamlines
3
U
0
p in wake is uniform


p  p0 

Cp  1
 2
2
2 
U
4
 U 
Point v
Cp
p
Cp = 1
1
______
________
____
>p0
0
<U ________
2 ______
>p0
0 < Cp < 1 ____
3 ______
____
<p0
>U ________
Cp < 0
<U ________
<p0
4 ______
____
Cp < 0
Points outside boundary layer!
v2
2
1
Application of Bernoulli
Equation
v12
p2
v22
 z1 

 z2 

2g 
2g
p1
2
2
p0 U
p v

 
 2g  2g
U2
2g

v2
2g

p


In air pressure change due to
elevation is small
U = velocity of body relative to fluid
p0

 p  p0 
  Cp
1  2  2
2 
U
 U 
v2
Flat Plate:
Pressure Distribution


v
p

p
0
Cp  1
 2
2
2 
U
U 

>U 3
2
<U 2


p

p
0
C p  2
2 
 U 
Front of plate
0 1
Back of plate
U 2C p
 p  p0
2
Fd  Fd front  Fd rear
Fd   p front  prear A

Fd  C p
front
 Cp
rear
Fd  0.8  1.2

U 2
2
U 2
2
1 0.8
0
Cp
-1 -1.2
Cd = 2
A
A
Drag Coefficient of Blunt and
Streamlined Bodies
 Drag dominated by viscous
drag, the body is __________.
streamlined
 Drag dominated by pressure
drag, the body is _______.
bluff
 Whether the flow is viscousdrag dominated or pressuredrag dominated depends
entirely on the shape of the
body.
 This drag coefficient is
calculated from a measured
value of ____
Fss
Flat plate
Cd 
2Fd
U 2 A
Bicycle page at Princeton
Drag Coefficient at High
Reynolds Numbers
Figures 9.28-9.30 bodies with drag
coefficients on p 593-595 in text.
hemispherical shell
hemispherical shell
cube
parachute
Vs
?
0.38
Why?
1.42
Velocity at
1.1
separation point
determines pressure
1.4
in wake.
The same!!!
SUVs have got Drag…
Ford Explorer 2002 Cd = 0.41
2 Drag
Cd 
U 2 A
Cd U 2 A
Drag 
2
Automobile Drag Coefficients
(High Reynolds Number)
Cd = 0.32
Height = 1.539 m
Width = 1.775 m
Length = 4.351 m
Ground clearance = 15 cm
100 kW at 6000 rpm
Max speed is 124 mph
Where does separation occur?
Calculate the power required to overcome drag at 60 mph and 120 mph.
What is the projected area?
A   H  G W
A  1.539m  0.15m 1.775m  2.5m2
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
Plan view of car?
Velocity and Drag: Spheres


Cd  f  , Re, M, shape, orientation  General relationship for
D
 submerged objects
Spheres only have one shape and orientation!
2Fd
Cd 
U 2 A
2Fd
 Cd  f  Re 
2
U A
Cd U 2 A
Fd 
2
Where Cd is a function of Re
Sphere Terminal Fall Velocity
 p  particle volume
 F  ma
Fb
Fd  Fb  W  0
ρ p  particle density
Fd
W  ppg
ρw  water density
g  acceleration due to gravity
C D  drag coefficient
Fb   p  w g
Vt 2
Fd  Cd AP  w
2
4
 p  r
3
3
Ap  r
Ap  particle cross sectional area
Vt  particle terminal velocity
W
2
2Fd
Cd 
U 2 A
Sphere Terminal Fall Velocity
(continued)
Fd  W  Fb
Vt 2
Cd AP  w
  p (  p  w ) g
2
Vt 
2
2 p (  p   w ) g
Cd AP  w
p
2
 d
Ap 3
Relationship valid for spheres
4 gd   p   w 
Vt 
3 Cd
w
2
General equation for falling objects
4 gd   p   w 
Vt 
3 Cd
w
Drag Coefficient on a Sphere
Drag Coefficient
1000
100
Stokes Law
10
1
0.1
0.1
1
24
Cd 
Re
10
102
103
104
Reynolds Number
105
106
107
Re=500000
Turbulent Boundary Layer
Drag Coefficient for a Sphere:
Terminal Velocity Equations
Valid for laminar and turbulent
24
Laminar flow R < 1 Cd 
Re
Transitional flow 1 < R < 104
Fully turbulent flow R >
Re 
Vt d 

104
Cd  0.4
4 gd   p   w 
Vt 
3 Cd
w
Vt 
d 2 g  p   w 
18
gd   p   w 
Vt 
0.3
w
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.
ρ p  1040 kg/m
3
Vt 
ρw  999 kg/m 3
g  9.81 m/s
2
Vt
d  4x10 6 m
  1.14x10 3
kg
sm
4x10

6
d 2 g  p   w 
18
m  9.81 m/s 2 1040 kg/m 3  999 kg/m 3 


3 kg


181.14x10


sm
2
Vt  3.14 x107 m/s
Vt  2.7 cm/day
Reynolds
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
inertia of
in the boundary layer, increase the _______
the boundary layer, and delay the onset of
separation.
 What is the Reynolds number where the
boundary layer begins to become turbulent with
a golf ball? _________
40,000
 Why not use this for aircraft or cars?
Boundary layer is already turbulent
At what velocity is the boundary
layer laminar for an automobile?
Rel 
U
Ul

 air
m2
 1.5 x10
s
 Rel
5
l  4m
l
Rel  500,000
2

5 m 
1.5 x10 s   500000 

U
 1.9m / s  6.8km / hr
 4m 
Effect of Turbulence Levels on
Drag
Flow over a sphere with a trip wire.
Causes boundary layer to become turbulent
Re=15,000
Re=30,000
Point of separation
Effect of Boundary Layer
Transition
Ideal (non
viscous) fluid
No shear!
Real (viscous)
fluid: laminar
boundary layer
Real (viscous)
fluid: turbulent
boundary layer
Increased inertia in
boundary layer
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 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)
S
nd
U
Summary: External Flows
 Spatially varying flows
 boundary layer growth
 Example: Spillways
 Two sources of drag (Fss)
 shear (surface area of object)
 pressure (projected area of object)
 Separation and Wakes
 Interaction of viscous drag and adverse pressure
gradient
Challenge
Cd U 2 A
Drag 
2
I’m going on vacation and I can’t back all
of our luggage in my Matrix. Should I put it
on the roof rack or on the hitch?
Challenges
Cd U 2 A
Drag 
2
How long would L have to be to double the
drag of a sphere?
L
V=30 m/s
D=3m
Cd U 2 A
Drag 
2
Challenges
2
6m
  14.6 10
s
How long would L have to be
to double the drag of a sphere?
V=30 m/s
L
V D
Re 

6
Re  6.164  10
D=3m
0.01
00
00
10
10
00
00
00
00
00
00
0
00
00
10
10
00
00
00
00
0
10
00
00
00
10
00
0
00
10
0
Find drag of sphere
Guess at Re for plate
Find drag coefficient for plate
(note different area)
0.001
Solve for L
Elongated sphere
V=30 m/s
A 
L
D=3m
2 Drag
2
C.dplate   V
C.dsphere  0.2
2 Drag
L 
2
C.dplate   V   D
2
2  C.dsphere   V   D

L 
2
2

Drag 
2
8C.dplate   V   D
L 
C.dsphere D
4C.dplate
L  50 m
2
C.dsphere   V   D
8
C.dplate  0.003
Solution: Solar Car
Cd f 
0.455
Rx 
Ul
U = 17.88 m/s

l = 5.9 m
2Fd
Cd 
U 2 A
air = 14.6 x 10-6 m2/s
C d U 2 A
Fd 
air = 1.22 kg/m3
2
6
Re
=
7.2
x
10
3
x
10
1.22
kg
/
m
17.88
m
/
s
11.88
m





l
2
2
Cd = 3 x 10-3
Fd =14 N
A = 5.9 m x 2 m = 11.8 m2
log  Rel 
2.58
0.01
3
Fd
3
P =F*U=250 W
10
00
00
0
10
00
00
00
10
00
00
00
0
10
00
00
00
00
10
00
00
00
00
0
10
00
00
10
00
0
0.001
2
2
Reynolds Number Check
R
Vd


3.14 x10
R
7
m/s 4 x10 6 m 999kg/m 3 
3 kg
1.14x10
sm
R = 1.1 x 10-6
R<<1 and therefore in Stokes Law range
Solution: Power a Toyota Matrix
at 60 or 120 mph
2Fd
Cd 
 f (Re)
2
U A
Cd U 2 A
Fd 
2
P
C d U 3 A
2
(0.32)(1.2kg / m3 )(26.82m / s)3 (2.5m2 )
P
2
P = 9.3 kW at 60 mph
P = 74 kW at 120 mph
Grand Coulee Dam
Turbulent boundary layer reaches surface!
Reflections on Drag
0.01
 What are 3 similarities with Moody
diagram?
 Fully turbulent boundary layer
 Transition between laminar and turbulent on
the plate
 Why more detail in transition region
here than in Moody diagram?
 Are any lines missing on the graph?
0
00
00
10
10
00
00
00
00
00
00
0
00
00
00
00
00
00
0
10
00
00
00
00
00
0
10
10
10
 Why 2 curves for smooth (red and
green)
0.001
10
 Laminar
 Smooth
 Rough
Function of conditions
at leading edge
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)