Transcript Concept of Drag

Concept of Drag
Viscous Drag and 1D motion
Concept of Drag
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Drag is the retarding force exerted on a
moving body in a fluid medium
It does not attempt to turn the object,
simply to slow it down
It is a function of the speed of the body,
the size (and shape) of the body, and
the fluid through which it is moving
Drag Force Due to Air

The drag force due to wind (air) acting on an
object can be found by:
FD = ½ ρ CDV2A
where:
FD = drag force (N)
CD = drag coefficient (no units)
V = velocity of object (m/s)
A = projected area (m2)
ρ = density of air (kg/m3) {1.2 kg/m3}
Drag Coefficient: CD
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The drag coefficient is a function of the shape of
the object and the Reynolds Number (Re)
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Re = (velocity * length-scale) / (kinematic viscosity)
For a spherical shape the drag coefficient ranges
from 0.1 to 300, as shown on the next slide.
Drag Coefficient for Spheres
Dropping a Ping Pong Ball
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If you dropped a ping pong ball down
the stairwell in this building (height 50
feet), and the stairwell had a vacuum in
it, how long would it take for the ping
pong ball to hit the floor?
If you left the air in the stairwell would
it take longer, shorter, or the same time
to hit the bottom?
Looking at the ball in detail
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Drawing a Free Body
Diagram (FBD) of the ball is
shown to the right
Since all the drag force is
doing is slowing the ball
down, it is directly vertical
and upwards
FD=f(v)
FG=mg
Numerical Analysis
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If you have two data points (time,
position), then you can approximate the
velocity of the body.
Given the points (2 s, -15m) and (2.1 s,
-17m), what is the approximate velocity
at 2.1 seconds?
If the next data point is (2.2 s, 19.05m), what is the velocity at 2.2
seconds?
Solution


 15m    17m 
m
V [@ 2.1] 
 20
(2s  2.1s)
s


 17m    19.05m 
m
V [@ 2.2] 
 20.5
(2.1s  2.2s)
s
Now Find Acceleration
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Given the velocities at 2.1s and 2.2s,
what is the acceleration at 2.2s?
Data points are (time, velocity):
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(2.1s, -20 m/s)
(2.2s, -20.5 m/s)
Acceleration Solution

m 
m 
  20 s     20.5 s 
m

 


A[@ 2.2] 
 5 2
2.1s  2.2s
s
Continuing the process
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The ultimate goal of this numerical
analysis is to find the drag force on the
body
Now that we have the acceleration, we
can find the total force acting on the
body (F=ma), the force of gravity
(Fg=mg), and Drag Force (FD)
Implementing this in Matlab
(Section 8.3)
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Suppose you have two arrays defined,
one with the elevation of the ping pong
ball and the other with the time
intervals
You can use numerical methods to
determine the drag coefficient by
estimating the acceleration of the ball at
each time step
Central Differencing
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To find the average velocity over a time
interval, you divide the change in
position by the change in time.
This is called forward differencing.
A better estimate is central differencing,
where you estimate the velocity at a
point by considering the points on
either side of it.
Sample Data
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Type the sample
data into Matlab and
plot Height versus
time
Use central
differencing to find
the velocity at the
points 0.1 through
0.9
T (s)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
h (m)
0
-0.05
-0.2
-0.44
-0.76
-1.17
-1.65
-2.19
-2.78
-3.41
-4.08
The plotted data
Central Differencing for Velocity
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To find the velocity at time 0.1 seconds,
use the formula:
h0.2  h0  0.20m  0m
m
v0.1 

 1.0
t0.2  t0
0.2s  0s
s
>> t1=[0:.1:1];
>> h1=[0 -.05 -.2 -.44 -.76 -1.17 -1.65 -2.19 -2.78 -3.41 -4.08];
>> v1=(h1(3:10)-h1(1:8))./(t1(3:10)-t1(1:8))
v1 =
-1.0000 -1.9500 -2.8000 -3.6500 -4.4500 -5.1000 -5.6500 -6.1000
Central Differencing for Acceleration
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To find the acceleration at time 0.2
seconds, use the formula:
a0.2 

v0.3  v0.1
t0.3  t0.1
m
m
 2.80  1.0
s
s  9.0 m

0.3s  0.1s
s2
Notice you must be careful with the indices
>> a1=(v1(3:8)-v1(1:6))./(t1(4:9)-t1(2:7))
a1 =
-9.0000 -8.5000 -8.2500 -7.2500 -6.0000 -5.0000
Finding the Drag Force
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With the net acceleration known at each time step,
the acceleration due to drag can be found be
subtracting the acceleration due to gravity.
Then the drag force, at each time step, can be found
from Newton’s second law (for a mass of 0.0025 kg).
>> a2=a1-(-9.8)
a2 =
0.8000
1.3000
1.5500
2.5500
3.8000
4.8000
0.0039
0.0064
0.0095
0.0120
>> f2=a2*.0025
f2 =
0.0020
0.0032
Finding the Drag Coefficient
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If you know the velocity at each time
step and the drag force at each time
step, you can plot F versus v2 and fit a
straight line to the data.
Care must be taken to make sure you
line up the correct force (a 1x6 array)
with the correct velocity (a 1x8 array).
Force 1 matches velocity 2, etc.
Using Matlab
>> v2=v1(2:7)
v2 =
-1.9500 -2.8000 -3.6500
-4.4500 -5.1000 -5.6500
>> v3=v2.^2
v3 =
3.8025 7.8400 13.3225
19.8025 26.0100 31.9225
>> plot(v3,f2,v3,f2,'o')
Using Least Squares
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Cast the data points
into array form using
the base equation and
then solve using matrix
math.
Note that currently V2
and FD are row vectors
and need to be
transposed.
>> k=v3'\f2'
k=
3.5928e-004
v 2 * k  FD
 3.8025 
0.0020
 7.8400 
0.0032




13.3225
0.0039
19.8025 * k  0.0064




26.0100
0.0095
31.9225
0.0120




For this problem, \ is the
least squares operator!
Final Form of Drag Equation
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The final form of the drag equation can be
written, which will allow the drag force to
be calculated at any velocity.
FD  0.000359v
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where:
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FD is in Newtons
V is in meters per second
2