Turbulence and Drag

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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(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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(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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(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

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Turbulence and Drag