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Weve done fluid statics . Now, F luid Dynamics

(fluid flow), which is much more interesting!COURSE THEME: NEWTONS LAWS OF MOTION !

NOW Sects. 14.5 - 14.7: Methods to analyze the dynamics of

fluids in motion . First, we need to discuss F LUI D L ANGUAGE.

Weve introduced a lot of this language while talkingabout fluid statics. But, there is some other terminology we need to discuss before we discuss Newtons Laws (Especially Newtons 2 nd Law!) inFluid Language!

Section 14.5: Fluid Dynamics

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Two main types of fluid flow:

1. Laminar Flow (or Streamline Flow ) Steady flow Each particle of the fluid follows a smooth path The paths of the different particles never cross each other Every fluid particle arriving at a given point has the same velocity

The path taken by the particles is called a streamline

Types of Fluid Flow

Paths of the particles look qualitatively like this!

Well assume this type of flow

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Two main types of fluid flow:

2. Turbulent Flow Irregular flow which has small whirlpool-like regions Its turbulent flow when the particles go above some critical speed

Streamlines can cross each other

Types of Fluid Flow

Paths of the particles canLook qualitatively like this!

Well not discuss this type.

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Viscosity is a measure of the amount of

internal friction in the fluid.

This internal friction or VI SCOUS F ORCE ,

comes from the resistance that two adjacentlayers of fluid have to moving relative to eachother.

Viscosity causes part of the kinetic energy of afluid to be converted to internal energy.

Viscosity

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We make four simplifying assumptions in our treatment of fluid flow to make the analysis easier:

1. The f luid is nonviscous Internal friction is neglected

2. The f low is steady The velocity of each point remains constant

3. The f luid is incompressible

The density remains constant4. The f low is ir rotational

The fluid has no angular momentum about any point

Ideal Fluid Flow

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Streamlines The path a particle takes in steady flow is

a streamline

The velocity of each particle

is tangent to a streamline

A set of streamlines

is called aTUBE OF F L OW

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Consider a fluid moving through a pipe of

nonuniform diameter. The particles movealong the streamlines in steady flow.

The mass m 1 in the small portionof pipe of length x1, crossingarea A1 in some time t , must beexactly the same as the mass m 2 inlength x2, crossing area A2 in thesame time t .

Why? Because no fluid particles

leak out of the pipe! The fluid has

Conservation of Mass!

Equation of Continuity

m 1 = mass of fluidin this volume

m 2 = mass of fluidin this volume

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Conservation of Mass: m 1 = m 2 (1)

For point 1 & point 2, the definition of density in terms of mass m & volume V gives: m = V.

For points1 & 2, use V = Ax (1) gives

r 1A1v1 = r 2A2v2 (2) Fluid is incompressible so, r = constant

(2) gives: A1v1 = A 2v2 (3) (3) is called the EQUATION OF

CONTINUITY FOR FLUIDS The product of the area and the fluid

speed at all points along a pipe isconstant for an incompressible fluid

Av mass f low rate Units: mass per time interval

or kg/s

Av volume f low rate Units: volume per time interval

or m 3/s

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Mass flow rate (mass of fluid passing a point per second) is constant: 1A1v1 = 2A2v2

Equation of Continuity PHYSICS : Conservation of Mass!!

For an incompressible fluid (1 = 2 = )

Then A1v1 = A 2v2 Or: Av = constant

Where cross sectional area A is large, velocityv is small, where A is small, v is large.

Volume flow rate: (V/ t) = A( x/ t) = Av

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Implications of Equation of ContinuityA1v1 = A 2v2

The fluid speed v is low wherethe pipe is wide (large A) The fluid speed v is high wherethe pipe is constricted (small A)

The product, Av , is called thevolume f low rate or f lux .Av = constant says that the

volume that enters one end of the pipe in a given time equals thevolume leaving the other end in thesame time (If no leaks are present!)

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PHYSICS : Conservation of Mass!!

A1v1 = A 2v2 Or Av = constant

Smal l pipe cross section larger v

L arge pipe cross section smaller v

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Example: Estimate Blood Flowr cap = 4 10 -4 cm, r aorta = 1.2 cm

v1 = 40 cm/s, v 2 = 5 10 -4 cm/s Number of capillaries N = ?A2 = N (r cap )2, A 1 = (r aorta )2

A1v1 = A 2v2N = (v

1/v

2)[(r

aorta)2/(r

cap)2]

N 7 10 9

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Example: Heating DuctSpeed in duct:

v1 = 3 m/sRoom volume:

V2 = 300 m 3

Fills room everyt =15 min = 900 sA1 = ?

A1v1 = Volume flow rate = ( V/ t) = V 2/t

A1 = 0.11 m 2