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Fluids

Chapter 14 of Physics 210 covers the principles of fluids at rest and in motion, including definitions of density and pressure, Pascal's and Archimedes' principles, and the equation of continuity. It discusses the behavior of ideal fluids, buoyant forces, and provides examples to illustrate these concepts. Additionally, Bernoulli's equation is introduced to explain the relationship between pressure, velocity, and height in fluid dynamics.

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0% found this document useful (0 votes)
33 views48 pages

Fluids

Chapter 14 of Physics 210 covers the principles of fluids at rest and in motion, including definitions of density and pressure, Pascal's and Archimedes' principles, and the equation of continuity. It discusses the behavior of ideal fluids, buoyant forces, and provides examples to illustrate these concepts. Additionally, Bernoulli's equation is introduced to explain the relationship between pressure, velocity, and height in fluid dynamics.

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bns14
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Physics 210, Chapter 14 Fluids (Statics)

Recap and General Remarks on Chapters 5-8


Recap and General Remarks on Chapters 5-8
Recap and General Remarks on Chapters 5-8
Definitions

As opposed to solids fluids flow! They are controlled by the


boundaries of their containers as they cannot sustain a force that is
tangential to their surfaces
Definitions

Density:
∆m
ρ= ∆V , measured in [kg/m3 ]
Note: when ρ varies substatially with p then the fluid is
compressible (such are gases, while fluids are not).
Definitions

Pressure:
F
p= A, measured in pascal (Pa).
Here F is the magnitude of the normal force on area A. Note that
1atm = 1.01 × 105 Pa.
Fluids at rest

Let’s identify the forces and use both the above density and pressure
relations.
p2 = p1 + ρg(y1 − y2 )
Let p0 be the pressure of the atmosphere at the interface between
the liquid and air. Then we can write the pressure at depth h is
given by:
p = p0 + ρgh
Fluids at rest
Fluids at rest

p is said to be the absolute pressure while the difference p − p0 is


the gauge pressure.
Example

What is the density of the oil in the left arm of the tube? We are
given l, d and ρwater .
How does the barometer work?
Pascal’s Principle

A change in the pressure applied to an enclosed incompressible fluid


is transmitted undiminished to every portion of the fluid and to the
walls of its container.
Pascal’s Principle
Pascal’s Principle

Fi Fo
(
Ai = Ao
Ai di = Ao do

These together give us an equation for the work W :


W = Fo do = Fi di .
Pascal’s Principle

With a hydraulic lever, a given force applied over a given distance


can be transformed to a greater force applied over a smaller
distance.
Archemides’ Principle
Archemides’ Principle
Archemides’ Principle
Archemides’ Principle

When a body is fully or partially submerged in a fluid, a buoyant


~ b from the surrounding fluid acts on the body.The force is
force F
directed upward and has a magnitude equal to the weight mfg of
the fluid that has been displaced by the body.
Archemides’ Principle

Fb = mf g (buoyant force), where mf is the mass of the fluid that is


displaced by the body.
Floating

The block in static equilibrium is said to be floating. In general


When a body floats in a fluid, the magnitude Fb of the buoyant
force on the body is equal to the magnitude Fg of the gravitational
force on the body.
Fg = mf g
Floating

(apparent weight) = (actual weight ) - (magnitude of the buoyant


force)
Example

A block of density
ρ = 800kg/m3 floats face up in
a fluid of density
ρf = 1200kg/m3 . The block
has a height H = 6cm.
Example

By what depth h is the block


submerged?
Example

If the block is held fully


submerged and then released,
what is the magnitude of its
acceleration?
Ideal Fluids in Motion

We will be studying ideal fluid, which is simpler to handle


mathematically and yet provides useful results. Here are four
assumptions that we make about our ideal fluid; they all are
concerned with flow:
1- Steady flow: the velocity of the moving fluid at any fixed point
does not change with time.
2- Incompressible flow: its density has a constant, uniform value.
3- Nonviscous flow: an object moving through a nonviscous fluid
would experience no viscous drag force—that is, no resistive force
due to viscosity; it could move at constant speed through the fluid
4- Irrotational flow: in irrotational flow the test body will not rotate
about an axis through its own center of mass
The Equation of Continuity

You may have noticed that you can increase the speed of the water
emerging from a garden hose by partially closing the hose opening
with your thumb. Apparently the speed v of the water depends on
the cross-sectional area A through which the water flows.
The Equation of Continuity

The element’s speed is v , so


during a time interval ∆t, the
element moves along the tube a
distance ∆x = v ∆t. The
volume ∆V of fluid that has
passed through the dashed line
in that time interval ∆t is:
∆V = A∆x = Av ∆t.
The Equation of Continuity
The Equation of Continuity

A1 v1 = A2 v2 (equation of continuity).
RV = Av = a constant (volume flow rate, equation of continuity),
Rm = ρRV = ρAv = a constant (mass flow rate).
Example

A stream of water emerging from a faucet “necks down” as it falls.


This change in the horizontal cross-sectional area is characteristic of
any laminar (nonturbulant) falling stream because the gravitational
force increases the speed of the stream. Here the indicated
crosssectional areas are A0 = 1.2cm2 and A = 0.35cm2 . The two
levels are separated by a vertical distance h = 45mm. What is the
volume flow rate from the tap?
Example
Bernoulli’s equation
Bernoulli’s equation

p + 21 ρv 2 + ρgy = a constant
Proof of the Bernoulli’s equation
Example

In the old West, a desperado


fires a bullet into an open water
tank, creating a hole a distance
h below the water surface.What
is the speed v of the water
exiting the tank?
Example
Example
Example

The raft is made of solid square


pinewood. Determine whether
the raft floats in water and if so,
how much of the raft is beneath
the surface. ρpine = 530kg/m3
and ρwater = 1000kg/m3 .
Example
Example
Example
Example

Crown’s weight in air = 7.84 N


Weight in water (submerged) =
6.84 N
True ρgold = 19.33 kg/m3 .
Example
Example
Example
Example

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