# Fluid Flow Concepts and Basic Control Volume Equations by 4ZtQyQU

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```									                  External Flows

CEE 331
September 11, 2012
School of Civil and
Monroe L. Weber-Shirk    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 non-
uniform
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          U          U
to                                      boundary
layer
y                                             thickness
U                                      d
x
shear

Why is shear maximum at the leading edge of
the plate? du
is maximum
dy
Laminar Boundary Layer:
Shear and Drag Force
d   5                            Ux             nx
=                       Re x =           d =5
x   Re x                         n              U
square
Boundary Layer thickness increases with the _______
root
______ of the distance from the leading edge of the plate
Based on momentum and mass
U 3
t 0  0.332                       conservation and assumed
x                  velocity distribution
U 3
l         l
Fd  w t 0 dx  w 0.332           dx      Integrate along length of plate
0         0            x

Fd  0.664w U 3l            On one side of the plate!
Laminar Boundary Layer:
Coefficient of Drag
2FD
Fd  0.664w U l    3
Cd =    2
= f (Re)
rU A

2(0.664) w U 3l          Dimensional analysis
Cd 
U 2lw

1.328w U 3l              1.328                 1.328
Cd                        Cd                   Cd =
U 2lw                   Ul                    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   Red d  d  5      Red = 5 Re x
Re x =    Red =          = = =
n        n    Re x x x  Re x
Red = 3500
Transition to Turbulence
U            U                  U
d

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                        Ux                      
4/5 
= 1/ 5                Re x =           d  0.37 x  
 
x Re x                        n                      U 
Grows ____________ than laminar
more rapidly
1/ 5
                Derived from momentum conservation
t 0  0.029 U  
 
2

 Ux              and assumed velocity distribution
x 5/4
1/ 5
l
 
Fd  w t 0 dx  0.036 U wl  
2
 
Integrate shear over plate
0                       Ul 
2FD    æ eö
Cd = 0.072Rel- 1/ 5      5 x 105 < Rel < 107       Cd =      = f Re,
2
rU A    è    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                                                                                             Ux
0.09
turbulent                                               9,000,000
Re x =
0.08                Reynolds Number                                         8,000,000                                   n
boundary layer thickness (m) .

0.07                                                                        7,000,000
 
1/ 5
n Re x

Reynolds Number
0.06       d  0.37 x   4/5
 
 
6,000,000                             x=
0.05
U 
5,000,000
U
0.04                                                                        4,000,000                           1x10 6 m 2 / s (500,000)
x
0.03
nx       3,000,000                                    1m / s
0.02                                                          d =5          2,000,000
U                                             x = 0.5 m
0.01                                                                        1,000,000

0                                                                         -
0        1               2              3              4   5     6
length along plate (m)                                                        Grand Coulee
Flat Plate Drag Coefficients
0.01
- 2.5             1 x 10-3
CDf = [1.89 - 1.62log (e / l )]                                                5 x 10-4
2 x 10-4
1 x 10-4   e
5 x 10-5
2 x 10-5   l
CDf                                                                                                  1 x 10-5
5 x 10-6
2 x 10-6
1 x 10-6
CDf = 0.072Rel- 0.2
1.328                    0.455               1700
CDf =                  CDf =                 2.58   -
(Rel )0.5              [log (Re )]
l
Rel                                     0.455
0.001
CDf =                 2.58
[log (Re )]
l
4

5

6

7

8

9

0
+0

+0

+0

+0

+0

+0

+1
1e

1e

1e

1e

1e

1e

1e

Ul
Rel =
n
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
- 2.5           1 x 10-3
CDf = [1.89 - 1.62log (e / l )]                                           5 x 10-4
2 x 10-4
1 x 10-4 e

2Fd                         Cd r U 2 A
5 x 10-5
2 x 10-5
l

Cd 
CDf                                                                                           1 x 10-5
5 x 10-6
2 x 10-6

Fviscous   =
1 x 10-6
C Df = 0.072 Re l- 0.2
1.328                   0.455                 1700
CDf =                CDf =                  2.58   -
(Rel )0.5            [ log (Re )]               Rel

U A
l                                                      0.455
C Df =

2
2.58

2                                           0.001                                                                                            [ log (Re )]
l

4

5

6

7

8

9

0
+0

+0

+0

+0

+0

+0

+1
1e

1e

1e

1e

1e

1e

1e
Ul
Rel =
n

Ul
Lr3
Fwave scales with ____                                     Rel =
n
Separation and Wakes

 Separation   often occurs at sharp corners
 fluidcan’t accelerate to go around a sharp
corner
 Velocities  in the Wake are ______ (relative
small
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                        Cp  1
v2    p  p0 
 2
2                                       2 
 U 
2
4                  U
0         1        Point   v   Cp      p
1       0    1     >p0
2       <U >0      >p0
3       >U <0      <p0
4                  <p0
Points outside boundary layer!
Application of Bernoulli
Equation
p1       v12 p2       2
v2                  In air pressure change due to
 h1       h2 
        2g         2g                  elevation is small
2      2
p0 U    p v
                           U = velocity of body relative to fluid
 2g  2g
U2        v2       p       p0
               
2g        2g              

v2     p  p0 
1  2  2
 U  2 
  Cp
U             
Flat Plate:
Pressure Distribution
                                   U 2C p
p  p0 
p  p0    C p  2
2

 2
v
Cp  1                                                                p  p0
     2                  2 
 U                U 
2
U                                                 2
>U 3
Fd  Fd front  Fd rear

<U 2                                       Fd   p front  prear A

U 2
0 1                                       
Fd  C p    front
 Cp   rear
    2
A

U 2
Fd  0.8  1.2                        A
2
1 0.8             0    Cp           -1 -1.2                 Cd = 2
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 viscous-
drag dominated or pressure-
drag 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!

2FD
Cd 
U 2 A

2FD                        C d U 2 A Where C is a function of Re
2
= Cd = f (Re)     FD                    d
rU A                             2
How fast do particles fall in
dilute suspensions?
 Whatare the important
parameters?
 Initialconditions 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   Ap  particle cross sectional area
Fd  Fb  W  0                                  ρ p  particle density
Fd
W  ppg                                     ρw  water density
Fb   p  w g                                g  acceleration due to gravity
2                       C D  drag coefficient
Vt
Fd  CD AP  w                                  Vt  particle terminal velocity
2
W              2FD
4                                          Cd 
 p  r   3
Ap  r   2

3                                                  U 2 A
Particle Terminal Fall Velocity
(continued)
Fd  W  Fb

Vt 2
CD AP  w            p (  p  w ) g
2
2 p (  p   w ) g
Vt 2                             General equation for falling objects
C D AP  w
p       2
 d             Relationship valid for spheres
Ap 3
4 gd  p   w                       4 gd  p   w 
Vt 
2                                      Vt 
3 CD         w                        3 CD     w
Drag Coefficient on a Sphere
1000
Drag Coefficient

100                  Stokes Law

10

1

0.1
0.1   1   10   102    103   104   105   106   107
24                            Re=500000
CD =           Reynolds Number
Re
Turbulent Boundary Layer
Drag Coefficient for a Sphere
Equations
24                  d 2 g  p   w 
Laminar flow R < 1 CD =                Vt 
Re                        18
4 C @
24    3
Transitional flow 1 < R < 10 D        + 1/ 2 + 0.34
Re Re
gd ( r p - r w )
Fully turbulent flow R > 104 CD @0.4           Vt »
0.3    rw

Vt d r
Re =
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/cm3 in water at 15°C.
d 2 g  p   w 
ρ p  1040 kg/m    3
Vt 
ρw  999 kg/m 3                                           18

g  9.81 m/s   2
Vt   
4x10   6
m  9.81 m/s 2 1040 kg/m 3  999 kg/m 3 
2

             kg 
1.14x10 3      
d  4x10 6 m                                       18
            sm


kg
  1.14x10 3                               Vt  3.14 x107 m/s
sm
Vt  2.7 cm/day
Reynolds
and Wakes

Diverging streamlines

Van Dyke, M. 1982. An Album of Fluid Motion. Stanford:
Parabolic Press.
Streamlines diverge behind object
 Increasing  pressure in direction of flow
p     V2
 Fluid is being decelerated                 +z+    =C
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)
 Figures9.19-21 bodies with drag
coefficients on p 392-394 in text.
 hemispherical shell       0.38
Why?
 hemispherical shell       1.42
Velocity at
 cube                      1.1  separation point
 parachute                 1.4  determines pressure
in wake.

Vs             ?        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
 function of ___________and ______of object
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
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.
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     Real (viscous)     Real (viscous)
viscous) fluid    fluid: laminar    fluid: turbulent
boundary layer     boundary layer

Increased inertia in
No shear!
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          nd
S
 Frequency of shedding (n) given          U
by Strouhal number S
 S is approximately 0.2 over a
wide range of Reynolds numbers
(100 - 1,000,000)
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
Solution: Solar Car
0.455                                                                                                             Ul
CDf =                  2.58                     U = 17.88 m/s                     Rx 
[log (Rel )]                       
2Fd
0.01

l = 5.9 m
CDf = [1.89 - 1.62log (e / l )]
- 2.5           1 x 10-3
5 x 10-4

Cd 
2 x 10-4
1 x 10-4 e
5 x 10-5
2 x 10-5
l
CDf                                                                                           1 x 10-5
5 x 10-6

air = 14.6 x 10-6 m2/s
2 x 10-6

U 2 A
1 x 10-6
C Df = 0.072 Re l- 0.2
1.328                   0.455                 1700
CDf =                CDf =                  2.58   -
(Rel )0.5            [ log (Re )]
l
Rel
C Df =
0.455
2.58
0.001                                                                                            [ log (Re )]
l
4

5

6

7

8

9

0
+0

+0

+0

+0

+0

+0

+1
C d U 2 A
1e

1e

1e

1e

1e

1e

1e
Ul
Rel =

air = 1.22 kg/m3
n

Fd 
2
(3x10 )(1.22kg / m ) (17.88m / s ) (11.88m )
-3          3                             Rel = 7.2 x 106             2                                                                 2

Fd   =2
2
Cd = 3 x 10-3
Fd =14 N                                A = 5.9 m x 2 m = 11.8 m2
P =F*U=250 W
Reynolds Number Check
Vd
R


R
3.14 x10     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 (R )
U 2 A
C d U 2 A
FD 
2
C d U 3 A
P
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

        
v2      p  p0 
Cp  1     2
           NACA 63-1-412 Flowfield Cp's
U2      U 2 
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
area _______ to force!
U 2
U                                                        FL  CL A
lift                                2

drag
p > p0 positive pressure

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