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# Physics 111 - PowerPoint PPT Presentation

Lecture 30. Chapter 9: Archimedes’ principle compressibility bulk modulus fluids & Bernoulli’s equation. Monday, November 9, 1998. Physics 111. The Physics 111 Help Session. Today!. Mondays 5:00 - 6:30 pm 8:00 - 9:00 pm. NSC Room 118. Achimede's Principle.

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Chapter 9: Archimedes’ principle

compressibility

bulk modulus

fluids & Bernoulli’s equation

Monday, November 9, 1998

Physics 111

Help Session

Today!

Mondays

5:00 - 6:30 pm

8:00 - 9:00 pm

NSC Room 118

Archimedes had this whole process figured

out some 2000 years ago! He said,

A body wholly or partially submerged in

a fluid is buoyed up by a force equal to the

weight of the displaced fluid.

So, the cork naturally float with just the

right portion of its volume under the water’s

surface so that the buoyant force upward

from the water equals the gravitational force.

FB

Fg

If we have a cork with density of

0.8 g/cc, what fraction of its volume

will be below the surface in a pool of

water when it reaches equilibrium?

?

Let’s look at the free body diagram for our cork.

The gravitational force:

FB

Fg

If we have a cork with density of

0.8 g/cc, what fraction of its volume

will be below the surface in a pool of

water when it reaches equilibrium?

?

The buoyant force is given by

the weight of the displaced water.

Now, set this equal to

the gravitational force...

FB

Fg

If we have a cork with density of

0.8 g/cc, what fraction of its volume

will be below the surface in a pool of

water when it reaches equilibrium?

?

Remarkable!

P

P

P

What else happens to our cork while it’s

completely submerged? After all, there’s a pressure on all its surfaces, acting inward,

from the water which surrounds it...

The water exerts a pressure force that tries

to compress the cork (i.e., reduce the volume

of the cork).

P

P

P

We call this type of

stress a volume stress

and define it simply as

This type of stress is

caused by a change in pressure.

The response of an object to an increase in

pressure around its side is to…

DECREASE ITS VOLUME!

P

P

P

You can now probably

guess how we’ll define

the volume strain

The way an object responds to a volume stress

is again simply a property of the material from

which the object is made. We call this property

the bulk modulus:

P

P

P

P

Often this characteristic

of materials is tabulated

as the inverse of the

bulk modulus, and known

as the compressibility:

23 X 1010 N/m2. Find the change in

the volume which 1024 kg occupies

at a depth where the pressure is

800 atm. Use a surface density of

seawater of 1024 kg/m3.

?

We’re now going to spend some time examining

the behavior of liquids as they flow or move

through pipes, the atmosphere, the ocean,...

Let’s trace out the motion of a given piece

or parcel of water as if flows through a

channel.

These lines, which tell us

where a parcel has been

and in which direction it

is going, are called

trajectories.

varying) then the trajectories

are the same as the

streamlines.

The streamlines tell us the instantaneous

direction of motion of a parcel in a flow,

whereas the trajectories trace out exactly

where the parcel has been.

Real flows often result in turbulence --

a condition in which the flow becomes

irregular.

Real flows are also often viscous. Viscosity

describes the internal “friction” of a fluid, or

how well one layer of fluid slips past another.

To simplify our problems, we’re going to

study the behavior of a class of fluids known

as “ideal” fluids. These fluids have the

following set of properties:

1) The fluid is nonviscous (no internal friction)

2) The fluid is incompressible (constant density)

3) The fluid motion is steady (velocity, density

and pressuer at each point remain constant)

4) The fluid moves without turbulence.

5 g/s

Continuity Equation

This is really just a conservation of mass

argument. It says that if I put in 5 g of water

each second at the left end of my hose, then

under steady-state flow conditions, I must

get out 5 g of water each second at the right

end of the hose.

5 g/s

Continuity Equation

For ideal fluids in steady-state (unchanging)

flows, this must be true regardless of the

shape of the hose. For instance, I could have

a hose that’s narrower at the left end where

the fluid enters the hose than it is at the right

end where fluid leaves.

Nevertheless, the mass entering at the

left each second must equal the mass

exiting at the right.

v1

v2

A1

A2

Continuity Equation

The mass entering at the left side is given by

And similarly, the mass leaving at right is

v1

v2

A1

A2

Continuity Equation

These two quantities must be equal, leaving

us with the relationship

of cross-sectional area A. Suddenly,

the pipe narrows to half it’s original

width. What happens to the speed

of the flow in the pipe?

?

So, the cross-sectional

area goes down by a

factor of 4. That means

that the velocity must

go up by a factor of 4!

In examining flows through pipes in the

Earth’s gravitational field, Bernoulli found

a relationship between the pressure in the

fluid, the speed of the fluid, and the height

off the ground of the fluid.

The sum of the pressure (P), the kinetic

energy per unit volume (0.5rv2), and the

potential energy per unit volume (rgy) has

the same value at all points along a

streamline.

There’s a nice derivation of this in the

book…so I won’t derive it here…but let’s

apply this to a problem.

A

15 m

v = 16 m/s

A large water tower is drained by a

pipe of cross section A through a

valve a distance 15 m below the

surface of the water in the tower.

If the velocity of the fluid in the bottom

pipe is 16 m/s and the pressure at

the surface of the water is 1 atm, what

is the pressure of the fluid in the pipe

at the bottom? Assume that the

velocity of the fluid in the tank is

approximately 0 m/s downward.

?

A

15 m

v = 16 m/s

Let’s look at the conditions

at the top of the tower:

= 248 kPa

This must match the conditions for the pipe

at the bottom of the tower...

A

15 m

v = 16 m/s

Let’s look at the conditions

in the pipe at the bottom of

the tower:

P + 128 kPa = 248 kPa

P = 120 kPa