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7/30/2019 Turbulence and Drag
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The drag force
The magnitude of the drag force moving through
some resistive medium can be roughly described by
D =1
2
C D Aρv 2.
A: cross-sectional area normal to object’s path
ρ: density of resistive medium v : object’s speed
C D : drag coefficient
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Viscosities: How resistant to shear a fluid is
The “dynamic viscosity” (η) of a fluid measures
how “non-shearable” the fluid is.
Substanceη
N · s/m2Hydrogen 9.6×10
−6
Air 1.8×10−5
Gasoline 2.9×10−4
Water 9.6×10−4
Mercury 1.5×10−3Glycerine 9.5×10−1
chocolate syrup 10−25molten glass 10−1000
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How to tell which C D to use?
Calculate the Reynolds number of your situation:
Re =Lρv
η
where L is some “characteristic dimension” of theobject (like the diameter of a sphere).
A small Re means slow motion through a stiff
medium (ball bearing falling through glycerine, forexample, or bacterium swimming in the body);
large Re means fast motion through a slippery
medium (bullet through air).
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Smooth spheres vs. rough ones
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Changes of “flow regime”
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Effects of increasing the velocity
Click to start
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Small Re and large Re
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Small Re
For small Re (Re < 1), C D = 24/Re , and so the
drag force becomes
D = 6πηrv ,
with r the radius of the sphere. This is known as
“Stokes’ Law”. It’s valid for slow motion throughviscous fluids.
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... and large Re
For spheres moving with Reynolds number
between 1000 and 300000, we have C D ≈ 0.5.
This is known as the “quadratic model” of fluidresistance. For macroscopic objects moving
through air (bullets, skydivers, baseballs), the
quadratic model is pretty good.
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Example: water balloon
A spherical water balloon of radius 10.0 cm is
dropped from the top of a tall building. (a)
Calculate the balloon’s terminal velocity. (b)Calculate the velocity of the balloon after 2.00
seconds. (c) compare the answer in (b) to that
obtained in the case of no air resistance.
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Example: water balloon
(a) Calculate the balloon’s terminal velocity.
Solution:
For a sphere we take C D = 0.5. The density of dry
air is ρ ≈ 1.20 kg/m3 and the balloon’s mass is
calculable from the density of water and its
volume: m ≈ 4.19 kg. Then
v T =
2mg
C D Aρ≈ 66.0 m/s.
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Example: water balloon
(b) Calculate the velocity of the balloon after 2.00
seconds.Solution:
The velocity at any time can be calculated directly
from our solution of the differential equation:
v =−v T tanh
gt
v T =− (66.0 m/s) tanh
9.80 m/s2 ·2.00 s
66.0 m/s
≈−19.0 m/s.
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Example: water balloon
(c) compare the answer in (b) to that obtained inthe case of no air resistance.
Solution:
The constant-acceleration (no resistance) modelpredicts that after 2.00 s the velocity is
v = v 0−g · t =
= 0 m/s−9.80 m/s2·2.00 s ≈−19.6 m/s.
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Example: water balloon plot of v vs. t