# Pascal's Principle Archimede's Principle Fluid Flows Continuity - PDF - PDF

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```					     Physics 111
Thursday,
November 11, 2004

• Ch 15:   Fluids
Pascal’s Principle
Archimede’s Principle
Fluid Flows
Continuity Equation
Bernoulli’s Equation
Toricelli’s Theorem
Thurs
Phys
Nov
Announcements               111
.11.

• Wednesday, 8 - 9 pm in NSC 118/119
• Sunday, 6:30 - 8 pm in CCLIR 468
Thurs
Phys
Nov
Announcements                   111
.11.

This week’s lab will be another physics
workshop - on fluids this time. No quiz
this week. Bring PHYSLETS book & CD.
Thurs
Phys
Nov
Announcements                        111
.11.

Friday 3:20 pm in Neils 224
Free cookies in Neils 231 at 3 pm

5 class participation points

Tom Morris on forensic accident reconstruction.
See how the physics you’ve learned this
semester contributes to determining how
accidents occur and who’s responsible.
Thurs
Phys
Nov
Announcements                111
.11.

Thursday, Nov. 18, 2004
11:50 am – 1:05 pm
Ch. 7, 8, 9.5 - 9.6, 10 - 13

• Hint: Be able to do the homework
you’ll do fine on the exam!
• You may bring one 3”X5” index card (hand-
written on both sides), a pencil or pen, and a
scientific calculator with you.
• I will put any constants and mathematical
formulas that you might need on a single page
attached to the back of the exam.
Thurs
Phys
Nov
Announcements                 111
.11.

Thursday, Nov. 18, 2004
11:50 am – 1:05 pm
Ch. 7, 8, 9.5 - 9.6, 10 - 13
Format:
• Three sections (like homework).
Do 2 of 3.
Topics: Energy, Collisions, SHM, Rotation
• One essay question. Required.
Topics: Work, Energy
• One section of multiple choice. Required.
Topics: Gravity/Orbits, Torque/Rotation, SHM,
Collisions, Energy/Work
Thurs
Phys
Nov
Ch 15: Fluid Mechanics        111
.11.

We found last time that in
hydrostatic equilibrium…

P2 = P + ρ g(Δh)
1

ΔP = ρg ( Δh)
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                    111
.11.

NOTE: our derivation here assumes a uniform
density of molecules at a given layer in the
atmosphere. In the real atmosphere, density
Decreases with altitude. Nevertheless, our
pressure and force balance diagram applies
so long as our layer is sufficiently thin so that
within it, the density is approximately constant.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                 111
.11.

If the atmosphere is in equilibrium (which would
imply a uniform temperature and no winds
blowing), the pressure at a given height above
the surface would be the same around the Earth.

The same arguments can be made for
pressure under water. All other things being
equal, the pressure at a given depth below
the surface is the same.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                       111
.11.

A scuba diver explores a reef 10 m below the
surface. The density of water is 1 g/cc. What is
the external water pressure on the diver?

Worksheet #1
We know the pressure at the surface of the water is
1 atm = 101.3 kPa. The change in pressure as the
diver drops under a 10 m column of water is given by

ΔP = ρ g(Δh) = (10 kg/m )(9.8 m/s )(10 m)
3        3             2

ΔP = 9.8 × 10 kg/m/s = 9.8 × 10 N/m
4         2             4       2

P = P0 + ΔP = (101.3 + 98) kPa = 199.3 kPa
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                 111
.11.

In solving the last problem, we applied a
principle that we haven’t even defined yet, but
that probably made good sense to us.

We said that the surface pressure at the
bottom of the atmosphere equaled the
pressure in the surface layer of water.

If this weren’t true, the water would fly out of the
oceans or sink rapidly toward the ocean floor!
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                   111
.11.

In fact, Pascal’s Principle guarantees this will
be true. It states:
The pressure applied to an enclosed liquid is
transmitted undiminished to every point in
the fluid and to the walls of the container.

Which means, that the pressure below the
surface of the water is equal to the surface
pressure + the pressure due to the column
of water above a given level.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics             111
.11.

Worksheet #2
≥
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                       111
.11.

Worksheet #2

A container is filled with oil and fitted on both ends with
pistons. The area of the left piston is 10 mm2; that of the right
piston 10,000 mm2.What force must be exerted on the left
piston to keep the 10,000-N car on the right at the same height?

1. 10 N
2. 100 N
3. 10,000 N
4. 106 N
5. 108 N
6. insufficient information
PI, Mazur (1997)
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                111
.11.

What happens to a cork when we try to
submerge it in water?

It shoots right back up to the surface.

What’s responsible for the motion of the cork?

There must be a force acting upward on
the cork greater in magnitude than gravity.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics             111
.11.

But what happens to the
cork when it gets to the     It floats!
surface?

So what must be the net force on the
cork as it’s floating on the surface?

ZERO!         What’s changed?
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                  111
.11.

What have we noticed about our strange
underwater force on the cork?

• It’s greater than gravity when the cork
is completely submerged.

• It’s equal to gravity when the cork floats
on the surface, only partially submerged.
• Our new force relates to the volume of
the cork that’s underwater!
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                   111
.11.

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.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                  111
.11.

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    Worksheet #3
body diagram for                         B
our cork.
The gravitational force:

W = mg = ( ρcorkVtot )g                 W

W = (7840
kg
)Vtot   B =W
m2s2
Thurs
Phys
Nov
Ch 15: Fluid Mechanics               111
.11.

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
B
the weight of the displaced water.

B = mH 2 O g = ( ρ H 2OVsub )g
kg                  W
B = (9800           )Vsub
m2s2
Now, set this equal to           B =W
the gravitational force...
Thurs
Phys
Nov
Ch 15: Fluid Mechanics               111
.11.

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?
kg
W = (7840                )Vtot              B
m2s2
kg
= (9800           )Vsub = B
m2s2

kg                     W     le!
Vsub (7840             ) V                   kab
=             m2s2    sub
= 0.8        ar
em
B =W
kg
Vtot (9800             ) V              R
2 2
m s        tot
Thurs
Phys
Nov
Ch 15: Fluid Mechanics             111
.11.

Worksheet #4
≥
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                                111
.11.

Worksheet #4

A lead weight is fastened on top of a large solid piece of
Styrofoam that floats in a container of water. Because of the
weight of the lead, the water line is flush with the top surface
of the Styrofoam. If the piece of Styrofoam is turned upside
down so that the weight is now suspended underneath it,

1. the arrangement sinks.
2. the water line is below the top surface of the Styrofoam.
3. the water line is still flush with the top surface of the Styrofoam.

PI, Mazur (1997)
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                 111
.11.

We’re now going to examine 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.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                111
.11.

If the flow is in a condition
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 actually been.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                     111
.11.

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.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                 111
.11.

To simplify our problems, we’re going to
study the behavior of a class of fluids known
as “ideal” fluids.

1) The fluid is nonviscous (no internal friction)
2) The fluid is incompressible (constant density)
3) The fluid motion is steady (velocity, density
and pressure at each point remain constant)
4) The fluid moves without turbulence.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                  111
.11.

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.

Δmin Δmout
5 g/s                            5 g/s       =
Δt   Δt
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                     111
.11.

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.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics               111
.11.

v1    A1                             A2   v2
The mass / time entering at the left side is
Δmin    ΔV1    A1Δx1
=ρ     =ρ       = ρ A1v1
Δt     Δt      Δt
And similarly, the mass / time leaving at right is
Δmout    ΔV2    A2 Δx2
=ρ     =ρ        = ρ A2 v2
Δt       Δt     Δt
Thurs
Phys
Nov
Ch 15: Fluid Mechanics               111
.11.

v1   A1                            A2   v2

These two quantities must be equal, leaving
us with the relationship

ρ A1v1 = ρ A2v2       A1v1 = A2 v2
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                    111
.11.

An ideal fluid flows through a pipe
of cross-sectional area A. Suddenly,
the pipe narrows to half it’s original
width. What is the ratio of the final
to the initial speed of the fluid flow?

1)   4:1           Worksheet #5
2)   2:1       2 Ar∝
3)   1:1
4)   1:2
5)   1:4
A1v1 = A2v2
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                  111
.11.

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 (0.5ρv2), and the potential
energy per unit volume (ρgy) has the same
value at all points along a streamline.
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                111
.11.

1m 2 m
P+    v + gy = constant
2V    V
1 2
P + ρv + ρgy = constant
2

We can derive Bernoulli’s Equation using
a conservation of energy argument.
Skip
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                    111
.11.
Δx2
A2 P2
v2
Δx1
The change in kinetic            P1 A1               h2   g
energy of the fluid
v1    h1
between the two ends
must equal the net                The hatched regions have
work done on the fluid.           the same mass of fluid.
1        1
ΔK = mv − mv
2
2
2   2
2
1      Wg = −ΔU
And the work done by          = −(mgh2 − mgh1 )
gravity on the fluid is
given by                      = mg(h1 − h2 )
Thurs
Phys
Nov
Ch 15: Fluid Mechanics               111
.11.
Δx2
A2 P2
v2
Δx1
Next, we consider the   P1 A1                h2 g
work done by the
v1   h1
pressure forces at each
end of the pipe:          The hatched regions have
W1 = P A1Δx1 = P ΔV       the same mass of fluid.
1         1

W2 = −P2 A2 Δx2 = − P2 ΔV

Giving a net     Wnet = W1 + W2 + Wg
work on
the fluid of       = P ΔV − P2 ΔV + mg(h1 − h2 )
1
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                          111
.11.
Δx2
A2 P2
v2
Now, putting it all               Δx1
P1 A1                   h2      g
together, we have
v1    h1
Wnet = ΔK
1
P ΔV − P2 ΔV + mg(h1 − h2 ) = m(v − v )
1                                      2
2
2
2
1

Rearranging we get…
1                                 1
P ΔV + mgh1 + mv = P2 ΔV + mgh2 + mv
1                  2
2
1                           2
2
2
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                         111
.11.
Δx2
A2 P2
v2
Δx1
P1 A1               h2    g
Dividing by ΔV…                  v1    h1

2                           2
mgh1 mv          mgh2 mv
P+     +     = P2 +
1
+                   2
1
ΔV    2ΔV        ΔV    2ΔV
Finally, identifying density…

P1 + ρgh1 + ρv = P2 + ρgh2 + ρv
1
2
2
1
1
2
2
2
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                       111
.11.

Airplane wings

Bernoulli’s Principle:            P + ρv = P2 + ρv
1    2         1   2
1   2    1         2   2

If v1 > v2, then P1 < P2. That creates a pressure
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                              111
.11.

Curveballs
muttley.ucdavis.edu/Book/Sports/instructor/curveball-01.html
P2
v2                              Bernoulli’s Principle:

P + ρv = P2 + ρv
1
1
2
2
1
1
2
2
2

v1
P1

If v1 > v2, then P1 < P2. That creates a pressure
gradient force that causes the “curveball.”
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                        111
.11.

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 cross-sectional area of the tank is
much bigger than that of the drain pipe.

Worksheet #6

1 2
P + ρv + ρ gy = constant
2
Thurs
Phys
Nov
Ch 15: Fluid Mechanics               111
.11.

Pictorial Representation:
Knowns:
vtop = 0
vtop, Ptop                     Ptop = 101.5 kPa
g
vpipe = 16 m/s
H = 15.0 m
A          y
H

ρ = 1000 kg/m3

x   g = 9.81 sm
2

vpipe, Ppipe
Unknowns:
Ppipe
Thurs
Phys
Nov
Ch 15: Fluid Mechanics              111
.11.

v = 0 m/s             Mathematical Representation:

Let’s look at the conditions
15 m

A       at the top of the tower:
1 2
v = 16 m/s            Ptop + ρvtop + ρ gytop =
2
1 3 kg
101.3 × 10 Pa + (10 3 )(0 m/s) 2 + (103 kg )(9.8 m/s2 )(15 m) =
3

2    m                   m3

= 248 kPa
This must match the conditions for the pipe
at the bottom of the tower...
Thurs
Phys
Nov
Ch 15: Fluid Mechanics                     111
.11.

v = 0 m/s         Mathematical Representation:

Let’s look at the conditions in the
15 m

A       pipe at the bottom of the tower:
1 2
v = 16 m/s        Ppipe + ρv pipe + ρ gy pipe =
2
1 3 kg                3 kg
Ppipe + (10 3 )(16 m/s) + (10 3 )(9.8 m/s2 )(0 m) =
2

2    m                  m

Ppipe + 128 kPa = 248 kPa

Ppipe = 120 kPa

```
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