Monday, November 9, 1998

Chapter 9: Archimedes’ principle compressibility bulk modulus fluids & Bernoulli’s equation

Archimedes had this whole process figuredout some 2000 years ago! He said,

A body wholly or partially submerged ina fluid is buoyed up by a force equal to theweight of the displaced fluid.

So, the cork naturally float with just theright portion of its volume under the water’ssurface so that the buoyant force upwardfrom the water equals the gravitational force.

If we have a cork with density of0.8 g/cc, what fraction of its volumewill be below the surface in a pool ofwater when it reaches equilibrium?

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

Fg

FBF FB g

The gravitational force:

F mg V gg cork tot ( )

F Vg tot( )7840 2 kg

m2s

Fg

FB

The buoyant force is given bythe weight of the displaced water.

If we have a cork with density of0.8 g/cc, what fraction of its volumewill be below the surface in a pool ofwater when it reaches equilibrium?

F m g V gB H O H O sub 2 2

( )

F VB sub( )9800 2 kg

m2s

Now, set this equal tothe gravitational force...

If we have a cork with density of0.8 g/cc, what fraction of its volumewill be below the surface in a pool ofwater when it reaches equilibrium?

Fg

FB

F V V Fg tot sub B ( ) ( )7840 98002 2 kg

m2s

kg

m2s

V

Vsub

tot

( )

( )

7840

9800

2

2

kg

m2skg

m2s

V

Vsub

tot

0 8.Remarkable!

What else happens to our cork while it’scompletely submerged? After all, there’s a pressure on all its surfaces, acting inward,from the water which surrounds it...

P P

P

P

The water exerts a pressure force that triesto compress the cork (i.e., reduce the volumeof the cork).

P P

P

PWe call this type ofstress a volume stressand define it simply as

volume stressF

AP

This type of stress iscaused by a change in pressure.

The response of an object to an increase inpressure around its side is to…

DECREASE ITS VOLUME!

P P

P

PYou can now probablyguess how we’ll definethe volume strain

volume strainV

V

The way an object responds to a volume stressis again simply a property of the material fromwhich the object is made. We call this propertythe bulk modulus:

Bvolume stress

volume strain

P

V V

/

P P

P

POften this characteristicof materials is tabulatedas the inverse of thebulk modulus, and knownas the compressibility:

kB

V V

P

1

/

[ ][ ]

[ ] / [ ]B

P

V V

Pa

[ ][ ]

kB

1 1

Pa

Seawater has a bulk modulus of23 X 1010 N/m2. Find the change inthe volume which 1024 kg occupiesat a depth where the pressure is800 atm. Use a surface density ofseawater of 1024 kg/m3.

BP

V V

/

23 10799 1013 1010

3

Pa atm Pa

atm( )( . )

/V V

V V/ .

8.1 10 Pa

Pa

7

23 1035 10

104

We’re now going to spend some time examiningthe behavior of liquids as they flow or movethrough pipes, the atmosphere, the ocean,...

Let’s trace out the motion of a given pieceor parcel of water as if flows through achannel.

These lines, which tell uswhere a parcel has beenand in which direction itis going, are calledtrajectories.

If the flow is in a conditionknown as steady state (notvarying) then the trajectoriesare the same as thestreamlines.

The streamlines tell us the instantaneousdirection of motion of a parcel in a flow,whereas the trajectories trace out exactlywhere the parcel has been.

Real flows often result in turbulence --a condition in which the flow becomesirregular.

turbulent region

Real flows are also often viscous. Viscositydescribes the internal “friction” of a fluid, orhow well one layer of fluid slips past another.

To simplify our problems, we’re going tostudy the behavior of a class of fluids knownas “ideal” fluids. These fluids have thefollowing 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, densityand pressuer at each point remain constant)4) The fluid moves without turbulence.

This is really just a conservation of massargument. It says that if I put in 5 g of watereach second at the left end of my hose, thenunder steady-state flow conditions, I mustget out 5 g of water each second at the rightend of the hose.

m

t

m

tin out

5 g/s 5 g/s

For ideal fluids in steady-state (unchanging)flows, this must be true regardless of theshape of the hose. For instance, I could havea hose that’s narrower at the left end wherethe fluid enters the hose than it is at the rightend where fluid leaves.

5 g/s 5 g/s

Nevertheless, the mass entering at theleft each second must equal the massexiting at the right.

v1 v2A1 A2

The mass entering at the left side is given by

m

t

V

t

A x

tA vin 1 1 1

1 1

And similarly, the mass leaving at right is

m

t

V

t

A x

tA vout 2 2 2

2 2

v1 v2A1 A2

These two quantities must be equal, leavingus with the relationship

A v A v1 1 2 2

A v A v1 1 2 2

An ideal fluid flows through a pipeof cross-sectional area A. Suddenly,the pipe narrows to half it’s originalwidth. What happens to the speedof the flow in the pipe?

A r 2 So, the cross-sectionalarea goes down by afactor of 4. That meansthat the velocity mustgo up by a factor of 4!A v A v1 1 2 2

In examining flows through pipes in theEarth’s gravitational field, Bernoulli founda relationship between the pressure in thefluid, the speed of the fluid, and the heightoff the ground of the fluid.

The sum of the pressure (P), the kineticenergy per unit volume (0.5v2), and thepotential energy per unit volume (gy) hasthe same value at all points along astreamline.

Pm

Vv

m

Vgy

1

22 constant

P v gy 1

22 constant

There’s a nice derivation of this in thebook…so I won’t derive it here…but let’sapply this to a problem.

A large water tower is drained by apipe of cross section A through avalve a distance 15 m below thesurface of the water in the tower.If the velocity of the fluid in the bottompipe is 16 m/s and the pressure atthe surface of the water is 1 atm, whatis the pressure of the fluid in the pipeat the bottom? Assume that thevelocity of the fluid in the tank isapproximately 0 m/s downward.

A

15 m

v = 16 m/s

v = 0 m/s

P v gy 1

22 constant

A

15 m

v = 16 m/s

v = 0 m/sLet’s look at the conditionsat the top of the tower:

P v gy 1

22

1013 101

210 0 10 9 8 153 3

32 3

3. ( )( ) ( ( . )( )) Pa m / s m / s mkg

m

kg

m

2

= 248 kPa

This must match the conditions for the pipeat the bottom of the tower...

A

15 m

v = 16 m/s

v = 0 m/s

P 1

210 16 10 9 8 03

32 3

3( )( ) ( ( . )( ))kg

m

kg

m

2 m / s m / s m

P + 128 kPa = 248 kPa

P = 120 kPa

Let’s look at the conditionsin the pipe at the bottom ofthe tower:

P v gy 1

22