Chapter 9: Archimedes’ principle
compressibility
bulk modulus
fluids & Bernoulli’s equation
Monday, November 9, 1998
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.
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.
FB  Fg
FB
The gravitational force:
Fg  mg  (  corkVtot ) g
Fg  (7840
kg
m2 s2
)Vtot
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.
FB  mH 2 O g  (  H 2 OVsub ) g
FB  (9800
kg
m2 s2
FB
)Vsub
Now, set this equal to
the gravitational force...
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?
Fg  (7840
kg
m2 s2
Vsub (7840

Vtot (9800
Vsub
 0.8
Vtot
)Vtot  (9800
kg
m2 s2
)
kg
m2 s2
)
kg
m2 s2
)Vsub  FB
FB
Fg
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...
P
P
P
P
The water exerts a pressure force that tries
to compress the cork (i.e., reduce the volume
of the cork).
We call this type of
stress a volume stress
and define it simply as
F
volume stress 
 P
A
P
P
P
P
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!
You can now probably
guess how we’ll define
the volume strain
V
volume strain 
V
P
P
P
P
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:
volume stress
P
B

volume strain
V / V
Often this characteristic
of materials is tabulated
as the inverse of the
bulk modulus, and known
as the compressibility:
P
P
P
1
V / V
k 
B
P
[ P]
[ B]  
 Pa
[ V ] / [V ]
1
1
[k ] 

[ B] Pa
P
Seawater has a bulk modulus of
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.
P
B
V / V
3 Pa
(799 atm)(1013
.  10 atm )
10
23  10 Pa  
V / V
7
8.1  10 Pa
4
V / V  
 35
.  10
10
23  10 Pa
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.
If the flow is in a condition
known as steady state (not
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.
turbulent region
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.
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
5 g/s
min mout

t
t
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.
5 g/s
5 g/s
Nevertheless, the mass entering at the
left each second must equal the mass
exiting at the right.
v1
A1
A2
v2
The mass entering at the left side is given by
min
V1
A1x1


 A1v1
t
t
t
And similarly, the mass leaving at right is
mout
V2
A2 x2


 A2 v2
t
t
t
v1
A1
A2
v2
These two quantities must be equal, leaving
us with the relationship
A1v1  A2 v2
A1v1  A2 v2
An ideal fluid flows through a pipe
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?
Ar
2
A1v1  A2 v2
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 (0.5v2), and the
potential energy per unit volume (gy) has
the same value at all points along a
streamline.
1m 2 m
P
v  gy  constant
2V
V
1 2
P  v  gy  constant
2
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.
15 m
v = 0 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
v = 16 m/s
1 2
P  v  gy  constant
2
Let’s look at the conditions
at the top of the tower:
15 m
v = 0 m/s
1 2
P  v  gy 
2
A
v = 16 m/s
1
1013
.  10 Pa  (103
2
3
kg
m3
)(0 m / s)  (10
2
(9.8 m / s2 )(15 m) 
3 kg
)
m3
= 248 kPa
This must match the conditions for the pipe
at the bottom of the tower...
Let’s look at the conditions
in the pipe at the bottom of
the tower:
15 m
v = 0 m/s
1 2
P  v  gy 
2
A
v = 16 m/s
1
P  (103
2
kg
m3
)(16 m / s)  (10
2
(9.8 m / s2 )(0 m) 
3 kg
)
m3
P + 128 kPa = 248 kPa
P = 120 kPa