Transcript Fluid Flow Concepts and Basic Control Volume Equations

External Flows
CEE 331
July 20, 2015
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Overview
 Non-Uniform
Flow
 Boundary Layer Concepts
 Viscous Drag
 Pressure Gradients: Separation and Wakes
 Pressure Drag
 Shear and Pressure Forces
 Vortex Shedding
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 is
always growing and the flow is nonuniform
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)
 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
Ux
Re x =
n
nx
d =5
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
rU A
2(0.664) w U 3l
Dimensional analysis
U 2lw
1.328w U 3l
U 2lw
Cd 
1.328 
Ul
Cd =
1.328
Re
Ul
Rel =
n
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
_________________________
Ux
Ud
Re x =
Red =
n
n
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
 
Ux
4/5 
d  0.37 x  
Re x =
n
U 
Grows ____________
more rapidly than laminar
d 0.37
= 1/ 5
x Re x
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
æ eö
Cd =
= f Re,
2
è
rU A
lø
Boundary 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?
0.1
10,000,000
laminar
turbulent
Reynolds Number
boundary layer thickness (m) .
0.09
0.08
Ux
Re x =
n
9,000,000
8,000,000
0.07
7,000,000
1/ 5
d  0.37 x
0.06
0.05
4/5
 
 
 
U 
6,000,000
5,000,000
0.04
4,000,000
nx
d =5
U
0.03
0.02
3,000,000
2,000,000
0.01
1,000,000
0
Reynolds Number

n Re x
x=
U
x
1x10 6 m 2 / s (500,000)
1m / s
x = 0.5 m
0
1
2
3
length along plate (m)
4
5
6
Grand Coulee
Flat Plate Drag Coefficients
0.01
CDf = [1.89 - 1.62log (e / l )]
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
- 2.5
CDf
CDf =
1.328
(Rel )0.5
CDf =
0.455
[log (Re )]
2.58
l
-
e
l
CDf = 0.072Rel- 0.2
1700
Rel
CDf =
0.001
0.455
[log (Re )]
Rel =
Ul
n
+1
0
1e
+0
9
1e
+0
8
1e
+0
7
1e
+0
6
1e
+0
5
1e
1e
+0
4
l
2.58
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

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
FD  Fviscous  Fwave
0.01
CDf = [1.89 - 1.62log (e / l )]
Fwave scales with ____
Lr3
CDf
CDf =
1.328
(Rel )0.5
C Df =
0.455
[ log (Re )]
2.58
-
l
CDf = 0.072 Rel- 0.2
1700
Rel
CDf =
0.001
0.455
[ log (Re )]
+0
9
+0
8
+1
0
1e
1e
+0
7
+0
6
+0
5
Rel =
1e
1e
1e
+0
4
l
1e
U A
2
Fviscous
Cd r U 2 A
=
2
1e
Cd 
2Fd
1 x 10-3
5 x 10-4
2 x 10-4
1 x 10-4e
5 x 10-5
l
2 x 10-5
1 x 10-5
5 x 10-6
2 x 10-6
1 x 10-6
- 2.5
Ul
n
Rel =
Ul
n
2.58
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)
Flat Plate:
Streamlines
3
U
0


p  p0 

Cp  1
 2
2
2 
U
 U 
v
Cp
p
>p0
0
1
<U >0
>p0
<p0
>U <0
<p0
v2
2
1
4
Point
1
2
3
4
Points outside boundary layer!
Application of Bernoulli
Equation
v12 p2
v22
 h1 
  h2 

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


p

p
0
C p  2
2 
 U 
U 2C p
 p  p0
2
Fd  Fd front  Fd rear
Fd   p front  prear A
<U 2

0 1
Fd  C p front  C prear
Fd  0.8  1.2

U 2
2
U 2
2
1 0.8
0
Cp
-1 -1.2
Cd = 2
A
A
Drag of Blunt Bodies and
Streamlined Bodies



Drag dominated by viscous
streamlined
drag, the body is __________.
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.
Bicycle page at Princeton
Velocity and Drag: Spheres
æe
ö
Cd = f
, Re, M, shape, orientation General relationship for
èD
ø submerged objects
Spheres only have one shape and orientation!
Cd 
2FD
U 2 A
2FD
= Cd = f (Re)
2
rU A
C d U 2 A Where C is a function of Re
FD 
d
2
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
 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
Fd  CD AP  w
Vt 2
Vt  particle terminal velocity
2
4
 p  r
3
3
Ap  r
Ap  particle cross sectional area
W
2
Cd 
2FD
U 2 A
Particle Terminal Fall Velocity
(continued)
Fd  W  Fb
CD AP  w
Vt 2 
Vt 2
  p (  p  w ) g
2
2 p (  p   w ) g
C D AP  w
p
2
 d
Ap 3
Vt 
2
Relationship valid for spheres
4 gd  p   w 
3 CD
w
General equation for falling objects
Vt 
4 gd  p   w 
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
Equations
24
d 2 g  p   w 
Laminar flow R < 1 CD =
Vt 
Re
18
24
3
4
Transitional flow 1 < R < 10 CD @ + 1/ 2 + 0.34
Re Re
r p - r w)
(
gd
Vt »
Fully turbulent flow R > 104 CD @0.4
0.3
Vt d r
Re =
m
rw
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
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
g
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)
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
Vs
?
0.38
Why?
1.42
Velocity at
1.1
separation point
determines pressure
1.4
in wake.
The same!!!
Shear and Pressure Forces
 Shear
forces
 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
 Pressure
forces
 pressure
drag or form drag
 caused by _____________from
the body
flow separation
 function of area normal to the flow
Example: Matrix Power
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
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
inertia of the
boundary layer, increase 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
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
shear (surface area of object)
 pressure (projected area of object)


Separation and Wakes

Interaction of viscous drag and adverse pressure
gradient
Solution: Solar Car
CDf =
0.455
Rx 
Ul
U = 17.88 m/s

l = 5.9 m
2Fd
Cd 
-6 m2/s
U 2 A

=
14.6
x
10
air
2
C d U A
Fd 
air = 1.22 kg/m3
2
6
Re
=
7.2
x
10
(3x10 )(1.22kg / m ) (17.88m / s ) (11.88m )
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
CDf = [1.89 - 1.62log (e / l )]
1 x 10-3
5 x 10-4
2 x 10-4
1 x 10-4e
5 x 10-5
l
2 x 10-5
1 x 10-5
5 x 10-6
2 x 10-6
1 x 10-6
- 2.5
CDf
CDf =
1.328
(Rel )0.5
C Df =
0.455
[ log (Re )]
2.58
l
-
CDf = 0.072 Rel- 0.2
1700
Rel
CDf =
0.001
0.455
[ log (Re )]
2.58
-3
Fd
3
P =F*U=250 W
+0
9
+0
8
+1
0
1e
1e
+0
7
+0
6
+0
5
Rel =
1e
1e
1e
1e
1e
+0
4
l
Ul
n
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
Cd 
FD 
2FD
U 2 A
 f (R )
C d U 2 A
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!
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


p

p
0
Cp  1
 2
2 
U2

U


v2
NACA 63-1-412 Flowfield Cp's
Shear and Pressure Forces:
Horizontal and Vertical Components
FD    p sin   t 0 cos dA
drag Parallel to the approach velocity
FL    p cos   t 0 sin  dA
lift Normal to the approach velocity
p < p0
U 2
negative pressure Fd  Cd A
2
A defined as projected
normal to force!
area _______
U 2
FL  CL A
2
lift

U
drag
p > p0 positive pressure