Professor Ramamurti
Shankar: This is a
relatively simple topic.
In fact, any of you who took
any kind of high school physics,
you would have done this thing
with fluids.
So, this is the subject for
today.
Fluid dynamics and statics.
So, I'll start with the simple
static problem.
We are going to take--Whenever
I say fluid, you are free to
imagine water or oil.
That's the--that's a good
enough example of a fluid.
One important property of the
fluid, the density,
denoted by ρ;
the density of water I would
probably note by a subscript
w.
And you know that's mass per
unit volume.
For water that happens to be
1,000 kilograms per cubic meter.
So, let's not linger there.
I think that's a fairly simple
concept.
The more subtle concept is the
one of pressure.
So, you have the notion of a
pressure if you go in a swimming
pool and you dive down to the
bottom, you know,
the pressure is going up.
So, what's the formal
definition of pressure?
Is it a vector?
Is it a scale?
Or does it have magnitude?
Does it have a direction?
That's what I want to explain
to you.
So, you can pick a point on the
fluid, there,
and say the pressure there is
such and such.
But what do we mean by that?
We mean by that the following.
If you get into that fluid and
you want to carve out a little
space for yourself,
you know, make a little cube,
maybe a glass cube,
and you want to live inside
that cube.
So, I'm going to blow up the
cube like this.
The water is trying to push you
in from all sides and compress
this cube.
You therefore have to push out
on the two, on all the walls.
If the force you exert on this
wall is some F and the
area of that wall is A,
that ratio is called a
pressure.
So, pressure is an intensive
measure of how hard the water is
trying to push in.
If you don't put the cube,
that pressure is still there,
but one way to measure the
pressure is to try to go in
there and push the fluid out and
ask how hard does it push you.
And the unit for pressure
Newton per meters square--and we
use another name for that called
a Pascal.
One Pascal is one Newton per
meter squared.
Here's another example of
pressure.
You have a gas.
We'll do a lot of this after
the break.
There is a gas inside a
cylinder.
There's a piston.
Now, if the pressure of the gas
and the pressure of the outside
world are the same,
there's nothing you have to do,
assuming the piston is
massless.
But if you want to increase the
pressure in the gas,
you put some extra weight.
That mg will push down,
and mg divided by the
area of the piston will be extra
pressure you apply.
That's also the pressure of the
gas.
That pressure is the
atmospheric pressure,
plus the extra weight you
apply, divided by the area of
this piston.
The atmospheric pressure is
everywhere.
So, when you push down on the
piston here to compress the gas
further, you're adding to the
atmospheric pressure this extra
force divided by area.
Sometimes--I mean,
this is called the absolute
pressure.
And that's called the
atmospheric pressure,
and that's called the gauge
pressure.
So, the gauge pressure is the
pressure on top of atmospheric
pressure.
For example,
when your car has a flat,
the bright side of it is the
pressure inside the tire is in
fact equal to atmospheric
pressure.
It doesn't help you because
there's a pressure inside and
there's a pressure outside,
and if you want to keep your
car moving, you really have to
increase the pressure in the
tire.
And when you stick this gauge
in and you measure something,
32 pounds per square inch,
that's the gauge pressure.
That's in excess of the
atmospheric pressure.
So, you should understand that
the pressure is a condition in a
fluid, and one of the important
properties of pressure,
is that if you took a fluid and
you went to a certain height,
all points at that height have
the same pressure.
And we understand that as
follows.
So, I go to this fluid.
I imagine in my mind a little
cylindrical section of the same
fluid.
Just draw a dotted line around
that region and focus on that
little chunk of fluid and say,
"Can the pressure on the two
sides -- this face and that face
-- be different?"
The answer is "no."
Because if the pressure on the
left was bigger than the
pressure on the right,
I take a cylinder of some area
A, pressure times area on
the left will exceed pressure
time area on the right.
And therefore,
the fluid should move to the
right.
But it's not doing anything.
It's in equilibrium,
and the only way that can
happen is if it's pushed equally
from both sides.
So, notice that the pressure on
this chunk from the left points
this way, and then that chunk
points that way.
They're trying to push it in.
So, pressure cannot change at a
given depth.
But let's take a cylinder that
looks like this [draws on
board].
Remember this is not a real
cylinder.
This is the same water,
and mentally,
I have isolated a part of the
water that looks like a
cylinder;
maybe you want to draw it in
dotted lines.
But it's just a fluid of
cylindrical shape,
base area A and height
h.
I can ask the following
question.
Let's say h_1
is the measure of that one from
the surface, and
h_2 is the
depth from the surface to the
lower face of the cylinder.
We can now argue the pressure
in the top and bottom really
should not be equal.
You should think about it,
by the same argument I gave
earlier will tell me it cannot
be equal.
If they were equal,
since these two areas are
equal, the forces pushing up and
pushing down will cancel.
With no net force on the
cylinder, you can ask then,
"What's keeping the cylinder of
water from falling down?"
Well, there has to be a force
to equal the weight of that
cylinder.
Therefore, there has to be a
net upward force.
That means the pressure down
here, pushing up,
better be higher than the
pressure on the top.
We are now going to calculate
what the pressure difference is.
So, let's call the pressure
downstairs P_2
and upstairs
P_1.
So, the upward force is
P_1
[correction: should have said
P_2]
times A.
The downward force is
P_2
[correction: should have said
P_1]
times A.
That's the net upward force.
That's got to be equal to the
weight of that amount of water.
The weight is found by first
the mass, which is area times
h_2 -
h_1.
That's the volume of the
cylinder, times the density of
water or whatever fluid you
have.
That's the mass of the cylinder.
That's the weight of that
cylinder of liquid.
All I have done is balance the
gravitational force on this
cylinder with the net upward
force due to the different
pressures.
I think I made one mistake here.
Upward force,
in magnitude--I want to write
this P_2 times
A, and downward force I
want to write as
P_1.
How we keep track of signs is a
little subtle.
I'm balancing two magnitudes.
I'm balancing net upward force,
namely P_2 is
considered positive upward,
and this is the force of
gravity down.
I'm balancing the magnitudes.
So, the area cancels out.
Because this area -- see the
pressure between two points --
should not depend on the area of
this fictitious cylinder I took.
In fact, I find it is equal to
ρg times
h_2 -
h_1,
which is the difference in the
depth of these two points.
Yes?
Student: It's actually
like a [inaudible]
Professor Ramamurti
Shankar: Which one?
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah.
In fact, it really happens to
be normal, or perpendicular to
area because the pressure will
push straight up and straight
down.
How about forces on the sides
of cylinder?
They cancel at every height.
Because at every height the
push from the left and right are
equal;
I already showed you that.
So, for the mechanical
stability of this chunk,
I need all forces to add up to
zero.
Horizontally,
there is no gravity and the
pressure at every height is
equal.
So, that cancels out.
Vertically, there is the force
of gravity, and it's cancelled
by the difference in pressure.
So, we can see that
P_2 =
P_1 plus
ρg times the height
difference.
And people write this formula
as follows.
It's very standard to take
P_1 to be the
point right at the surface,
and P_2 to be
any point inside,
and to call the depth of that
simply as h.
Then the pressure at the point
P in the fluid is the
pressure at the top,
which is usually atmospheric
pressure, plus ρgh.
So, I don't have an
h_1 and an
h_2 because
h_1 I've chosen
to be zero.
And h_2 I'm
simply calling h.
This says something very simple.
If you go to a lake,
at the surface of the lake the
pressure is due to the
atmosphere.
You take a dive,
you go down some depth
h, the pressure goes up
by this amount.
Now, if you go to the bottom of
the ocean, it's going to be an
incredible amount of pressure.
That's why you and I cannot
survive in the bottom of the
ocean, because the outside
pressure -- inside is usual
atmospheric pressure -- you're
breathing the air into your
lungs,
you go down with that.
Outside is atmospheric plus
this, and that can kill you.
That's why when you build a
submarine, you've got to make
sure it can withstand the
pressure.
But fish don't have the
problem, because fish are
breathing the water.
The water is going into their
system and outside their system.
So, that's one way for you to
live.
If you're 20,000 feet under the
sea, start drinking the water.
But it's not a long-term
solution.
It'll work in the short time
because you'll be equalizing the
pressure.
Okay.
Now, how about the atmospheric
pressure?
What's the origin of that?
And the origin of the
atmospheric pressure is that we
are ourselves living in the
bottom of a pool,
but it's filled with air.
The air above our heads goes on
for maybe 100 miles,
but the density decreases,
eventually vanishes.
So, I cannot tell you precisely
where the atmosphere ends,
but I can say the following.
If I go far above where the
absolute pressure is zero,
in free space there is just
vacuum.
The pressure here is the
atmospheric pressure.
Atmospheric pressure at the
bottom of Earth is equal to zero
plus ρgh,
where ρ is the density
of air, g is g,
and h is the height of
the atmosphere.
And that happens to be 10^(5)
Pascals.
So, we are living in the bottom
of a pool where the pressure is
10^(5) Pascals,
relative to empty
space--interstellar space.
But as I said,
that pressure doesn't kills us
because the pressure can be felt
both from the outside pushing in
and inside going through your
nostrils and everything else,
pushing out.
But you have all seen the
dramatic experiment where you
take a can of something and you
heat it up so the air goes out;
then you seal it.
The air that's been driven out
reduces the pressure.
So, when you seal it and you
cool it, then the pressure
drops.
And even the drop in
pressure--it doesn't even drop
down to zero,
but it's big enough for the
while can to implode.
Now, let's see what you get
here.
Let's do the following.
This is ρgh,
and you can ask yourself the
following question.
If this was a swimming pool,
how high would the water be?
In other words,
what height of water above the
Earth would produce the same
pressure as our atmosphere does?
Then I say, if I want 10^(5)
Pascals, that's the density of
water at 10^(3),
g, let's pretend it's
ten.
And h is what I'm
looking for.
Cancel all the powers of 10;
h is like 10 meters.
And it turns out it's -- to the
best of my knowledge -- it's 32
feet.
These are the units we don't
use in the book,
but it's a very common way of
thinking, 32 feet.
So, we are at the bottom of a
pool.
If it was filled with water,
it would be 32 feet of water.
Alright.
So, now we are going to take
this formula P =
P_0 + ρgh and put
it to work.
Get some mileage out of that.
So, what are the things we can
do with that formula?
First thing you can do is to
build yourself a barometer.
Barometer you know is a way to
tell what the pressure is today.
Now, the atmospheric pressure,
when I said it's 10^(5),
that's the typical pressure.
It doesn't really stay locked
into the value.
Each day there are fluctuations.
That's why the weather person
tells you pressure is going up,
pressure is going down.
So, let's find a way to measure
the pressure,
and here is one way to do that.
You take a can of something,
fill it with some liquid,
take a test tube.
Evacuate it completely,
suck all the air out of it and
stick it into this.
When you do,
there is a complete vacuum here
and the atmosphere is pushing
down, so the fluid will rise up
to some height h.
And you can ask,
"How high will it go?
What's going to be the height?"
Well, it'll go to a height so
that zero pressure here,
plus this ρgh,
which is the pressure there,
will be the same as the
pressure here,
because they are two points at
the same height.
Pressure here is the
atmospheric pressure.
So pressure here,
atmospheric pressure,
is zero at the top of the tube,
plus ρgh.
So, if you build this gadget,
this barometer out of water,
the water column will rise to
the height of 32 feet.
But now, nobody wants a gadget
32-feet high,
so what you use instead is
mercury, because it's very
dense.
So, you want to get the same
atmospheric pressure,
but you want to have a bigger
ρ and get a smaller
h.
If you go look up a book and
find the density of mercury,
you're going to find the height
is something like,
I don't know,
750,780 millimeters.
That's why the weather guy says
the pressure today is so many
millimeters, and the mercury is
dropping.
Now, I'm not sure why they
bother to give the numbers,
because for most of us,
including me,
those numbers don't mean
anything.
Here's a number,
746 millimeters.
Does it speak to you?
Not to me.
So it--it speaks to one of you
guys?
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah.
Okay.
Some number.
It's not like saying today is
67 degree Fahrenheit.
I guess I know what that means,
but when I hear the mercury,
I think it's just a waste of
time.
Anyway, they're telling you how
much the mercury is.
Okay?
You can use any fluid you like,
but you've got to agree on
using mercury because if it's
760 millimeters,
and it's water the person is
talking about,
of course you are in serious
trouble.
So understood,
we're talking about mercury.
Also, mercury is used in
thermometers.
That's the subject for after
the break.
So there, the mercury falling
could stand for falling
temperatures and rising
temperatures,
but here it's for falling
pressure and rising pressure.
So, this is the first gadget
you can build with what I've
taught you.
The next gadget you can build
is--you can imagine this--you
trying to drink.
This is your ears and nose and
what not.
You've got a straw and you're
trying to drink something.
Now, remember the fluid is
water.
Okay?
It's not mercury now,
because you're doing different
experiment.
You're going to drink from a
straw.
So, how do you do that?
You know when you drink from a
straw you create a partial
vacuum in your mouth.
So, the pressure here--;So,
in the case of the fluid that
you want to drink,
this is your head.
The pressure here less than the
atmosphere.
If it's less than the
atmosphere by some amount,
then the fluid can start
climbing up to some height,
and climb to such a height so
that the lower pressure,
plus that ρgh will
equal atmospheric pressure.
So, you got to reduce the
pressure more and more in your
mouth until this fluid can climb
up.
If you just--if you want to
just make it to your mouth,
it depends on the length of the
straw, the height of the straw.
If you want to suck it up a
certain height,
then that height,
ρgh,
plus the pressure in your
mouth,
is the pressure here,
and that's the atmospheric
pressure.
So, if you want to drink water
from a well that is more than 32
feet deep, you are out of luck.
Even if your face and your head
are a complete vacuum,
you cannot get the water to
climb more than 32 feet.
So, that's another illustration
of this.
So, one more example of this
P = P_0 + ρgh,
is if I give you another fluid,
it doesn't mix with water and I
tell you find the density,
there are many ways.
One is to just find mass and
volume of that fluid and divide.
But here is another thing
people use.
You take what's called a U-tube.
Yeah, that's not--I know,
that's where they post all the
embarrassing videos,
but this was physics
contribution to pop culture long
before any of this happened.
This is the U-tube.
And the U-tube,
let's say you fill it up with
one fluid and this is the other
fluid.
So, this is oil,
and this is water.
If the two heights were equal,
then you know we're talking
about the same fluid.
But this is supposed to tell
you that oil is less dense than
water, and we can check that by
comparing two points in this
fluid at the same height,
and saying the pressure must be
the same at those points.
You understand how I get that?
That pressure and that pressure
are equal because you can draw a
cylinder there horizontally,
which cannot be pushed
sideways.
From here, to this pressure add
that ρgh and that
ρgh, and you come to
that point;
you conclude that pressure is
also equal.
You cannot jump into the new
fluid here because the ρ
for here and the ρ for
this are different.
But these two points have the
same pressure and they are in
the same fluid.
So, let me write that statement
that this pressure and this
pressure are equal by saying,
atmospheric pressure is on the
top for both of them.
That, plus ρ_1gh
_1 is equal to
atmospheric pressure plus
ρ_2,
gh_2,
where this is the second fluid
, height goes to
h_2,
first fluid,
the height is equal to
h_1,
the densities are
ρ_1 and
ρ_2.
Atmospheric pressure is
cancelled;
then you find
h_1/h_2 =
ρ_2/ρ
_1.
So, by comparing the heights
you can find the relative
density.
If one of them is water,
then, well, whatever it is.
If you know the density of one
fluid, you can find the density
of the other.
Okay, yet another application
of this law that the pressure is
equal at a given height is the
famous hydraulic press.
So, here is a--Here are two
pistons of different radii.
This has got cross section area
A_1;
this has got cross section area
A_2.
And here I want to put some
incompressible fluid.
Incompressible fluid is
something whose volume cannot be
changed no matter how much you
press it.
Now, water is pretty close to
incompressible.
It does have a compressibility,
but it's not going to change
very much for our purposes.
So now here,
I have a piston,
and I have a piston here.
And I push down here with a
force F_1,
and I ask, "What will I get at
the other side?"
You might think
F_1 =
F_2,
but since these fluids are at
the same height,
we only know P_1 =
P_2.
That means
F_1/A_1 =
F_2/A<
sub>2.
So, this fluid here will push
up here with a force
F_2,
which is equal to
F_1 times
A_2 over
A_1.
Did I get it right?
Ah.
Student: [inaudible]
Professor Ramamurti
Shankar: Pardon me.
Student: [inaudible]
Professor Ramamurti
Shankar: Did I make a
mistake here?
Student: [inaudible]
Professor Ramamurti
Shankar: But I want to
amplify the force.
So, what do we really want to
do?
Well, it's a matter of who you
want to emphasize.
So, what you want to emphasize
here--Maybe if I drew a picture
I will know exactly what I'm
trying to do.
This picture is fine.
In practice,
this is not what you do.
You want to push down here and
raise something there.
Because what you really imagine
is some elephant.
There's an elephant standing
here, and you want to lift the
elephant by applying a force
here.
So, let's still call your force
F_2,
and the force on the other side
is the force you apply times
A_1
/A_2.
So A_1
/A_2 could
be a hundred.
What that means is,
if you apply one Newton here,
you'll get a hundred Newtons on
the other side.
So, that's the way to max--to
convert a small force into a
large force.
But this is the oldest trick in
the book.
An even older one invented by
cave people, is that if you have
a support like this you can put
a weight here and,
uh, let me see,
that's right.
So, a tiny weight here can lift
a big weight here because that
mg times that distance
and this mg times this
distance will be equal.
But you must know even from
that example that you don't get
something for nothing.
In other words,
if you lift the elephant here
by pushing down here,
the fact that the forces don't
match is perfectly okay.
But the work you do here must
be the work delivered to the
other side.
The Law of Conservation of
Energy.
The work you do on this side is
the force multiplied by the
distance.
Now, force multiplied by
distance I'm going to write as
the pressure times the cross
section area,
times distance.
But what is A
times--A_2
times dX_2?
A_2 times
dX_2--if you
push the liquid here--if you
moved it a distance
ΔX_2,
area times
ΔX_2 is the
volume of fluid you push down
here.
That's the volume that'll come
up on the other side.
So, you're free to write that
as P_2 times
A_1
ΔX_1.
Okay, but P_2A
_1--I'm sorry,
yes, now, let's write
P_2 is the same
as P_1.
P_1 times
A_1 is
F_1.
So, it just says that the force
times distance on one side is
the force times distance on the
other side.
That means the work you
actually do is not amplified by
the process.
You cannot get more joules out
of one side by any device.
What you put in is what you'll
get out.
But still it's useful,
because in practice you may
have to move a whole meter here
to lift the elephant by one
centimeter.
But the point is,
you can lift elephants this
way.
That's the important thing.
Okay, that's why it's worth
doing.
And this is how you--the brake
in your car works.
You know, you pump the brake
pedal.
There's a little cylinder there
and there's a fluid there,
and the fluid is pushed by your
feet.
And you push it quite a bit,
several centimeters.
At the other end,
there's another cylinder whose
piston is right next to the drum
that's rotating,
and pushes on the drum.
And it exerts an enormous
amount of force,
but it moves a very tiny
amount.
The shoes that grab your
rotating drum move a very tiny
amount, whereas,
your feet move a large amount.
But that's the ratio of the
force that's transmitted.
The fluid has the same pressure.
The brake fluid has the same
pressure, but the force you
apply with your feet is much,
much smaller than the force
that the drum will exert on the
rotating--that the disc brakes
will exert on the rotating drum.
So, a lot of hydraulics is
based on this simple
amplification of force.
Alright.
Now, I move to the next topic
in this field,
which is the Archimedes'
principle.
So, we all know the conditions
under which this was discovered.
So, I will not go into that,
other than to say that Mr.
Archimedes noticed that if you
immerse something in a fluid it
seems to weigh less.
What I mean by that is that if
you attached it to some kind of
a spring balance,
and you weighed it so that the
kx of the spring was the
mg of the object,
and if you did the same thing
now you will find it seems to
weigh less.
And the question was,
"How much less?"
Archimedes' answer is very
simple.
The amount by which you lose
the weight, or the weight loss,
equals the weight of liquid
displaced.
Now, how do you show that?
Because we don't,
right now, so many years after
Archimedes, we will not accept
this on faith.
We want to be able to show this
is the case.
There are several ways to prove
this.
One way, which I like,
is to say if you,
if the thing you are hanging
here, was itself a chunk of
water shaped like that,
you don't have to do anything.
Because that chunk of water can
float at that height for free.
But now, if you took the chunk
of water and put a stone here of
the same shape,
the rest of the fluid doesn't
know what you're doing.
It applies the same force that
it would to its own colleagues.
Namely, if this is water,
the rest of the water is in a
configuration ready to support
that amount of water.
So, if you took that water out
and put something in,
the water will apply the same
amount of force and the rest of
it is your problem.
You apply the remaining force.
So, the water is ready to
support its own kind of any
volume.
That's the meaning of being in
equilibrium.
Any chunk of water is in
equilibrium;
therefore, it's getting an
upward force equal to the
weight.
What we are saying is,
if without disturbing the
environment, you take the water
out and put something in its
place,
the rest of the guys will apply
the same force.
Now, one formal way to prove
that--you can take various
geometries, but I prefer a
cylindrical geometry.
So this is, in this example,
not a piece of water,
but a new material you've
introduced inside the water.
And we want to find the
buoyancy force.
What is the net force of
buoyancy?
Well, it's the pressure at the
bottom times the area,
minus pressure at the top times
the area.
We have already seen the
difference in the pressure is
ρgh A,
but h is now this
height.
Well, h times A
is the volume of the water,
ρ times that is the
weight of the water,
the mass of the water.
That times g is the
weight of the liquid displaced.
You can prove this for a
cylinder;
you can prove this for all
oddball shapes,
by thinking a little harder.
But this is good enough.
So, basically the body weighs
less in water because the lower
part of the body is being pushed
up harder than the upper part of
the body is being pushed down,
because the pressure increases
with depth.
That's why you get this result
from Archimedes.
Now, you have to be a little
careful on writing the equation,
because if this was made of a
material like iron,
then the density of iron is
less than--is more than the
density of water.
So, the weight of that chunk of
iron will be more than the
weight of the water displaced.
So, you will have to provide a
net force, and you can support
it with the cable.
But suppose this was not made
of iron, but made of cork?
If it's cork,
it won't want to be there.
Right?
Because then,
the applied force by the water
is more than the weight it takes
to support it.
So, the cork will then bob up
to the surface.
It will look like this.
If you want to keep it down,
it's like a rubber ducky.
You want to keep the rubber
ducky inside the water level,
you've got to pull it down,
or you want to tie it to the--I
think one of the problems I gave
you--has somebody tied this
piece of whatever to the floor.
Then it'll stay.
But things will bob up to the
surface.
And the question is,
"How far up will it go?"
We know part of it's going to
be inside and part of it's going
to be outside.
And you can ask how much will
be outside and how much will be
inside?
You can already guess the
answer, but let's prove it.
Let f be the fraction
immersed, fractional volume
immersed.
Then, we can say the weight of
the liquid displaced is equal to
the fractional volume times
ρ,
times g,
times the total volume,
is the ρ of water.
You understand?
This is the weight of water
equal to the full volume;
this is the fraction of water
displaced.
So, this is simply the weight
of this shaded region here.
And that is going to be equal
to the weight of the thing
that's floating;
that is the ρg times
full volume.
If you cancel the g,
and you cancel the volume,
you find that the fraction that
is immersed is the density of
the material divided by density
of water.
In other words,
if this material is 90 percent
the density of water,
it will be immersed by 90
percent.
That's exactly what happens
with ice.
As you all know,
ice has a smaller density than
water.
One of the great mysteries.
Normally, when you cool
something it condenses and
reduces in volume and the
density will go up.
But ice actually expands when
you cool it.
That's why the density of ice
is less than the density of
water.
That's why ice floats on water.
That's the reason icebergs look
like this, because the big part
of the iceberg--;that's the very
peculiar property of ice.
That's why you have those
movies like the Titanic where
you've got a huge ice thing and
it's floating.
First of all,
if the density of ice was more
than the density of water,
these two actors would still be
alive.
Okay?
But what happens is it is
floating, and not only that;
what you see is maybe a small
fraction of the whole thing.
That's why those big ships went
down.
Okay, but it's all thanks to
this accident of nature that,
here is one substance which,
when cold, increases
its--decreases in density.
Alright.
Now, Archimedes' principle has
got hundreds of applications.
If you want to build a boat,
here's how we build a boat.
Here is the steel boat.
Now, you cannot come to me and
say, "How do you make a steel
boat float in water?"
Okay.
It's not a solid steel boat.
Okay?
If you are thinking about a
solid steel boat,
you should get in another line
of work.
This is a thing made out of
steel, but it's completely
hollow.
So, what we claim is,
this amount of water weighs the
same amount as that amount
of--see, the shaded region.
So, you can easily calculate
how deep this one should sink to
balance its weight.
Right?
That, I assume you know how to
do that.
Take this to be a rectangular
boat.
The cross section is
rectangular;
it's gone down to a depth
ρ, and to a depth
h.
Therefore, the volume of water
displaced, is h times the
area of the boat,
cross sectional of the area of
the floor of the boat.
That's the volume of water,
that's the mass of water,
that's the weight of water.
That's the weight of the boat.
If you tell me how many tons
the boat weighs and you give me
the area of the base and
g,
and density of water,
I'll tell you what height it'll
sink.
Then, of course,
you can load more and more
cargo in this,
you know, if you have a top
here, start putting more and
more cargo;
this will go down until the
boat looks like this.
And that's as far as you can
push it.
That's the kind of simple
calculation you will be asked.
How much cargo can the boat
take?
Well, that's very simple.
The weight of the boat plus
weight of the cargo is maximal
in this critical situation,
when it's just about to go
under.
Therefore, the total volume of
water displaced,
times the density,
times g,
will be the weight of your boat
plus cargo.
And these are all elementary
applications,
and they came from two
principles.
One is the pressure increases
with depth;
the second is that the weight
of the liquid displaced is the
weight of the--is the reduction
in weight.
So, you can imagine three
cases, one where the density of
the material is more than the
density of water,
in which case it'll start going
down and you will have to hold
it up with a cable,
but you won't have to apply the
same force as you would outside.
The second example is when the
density of the object is less
than the density of water,
in which case it will float,
with a certain fraction of it
immersed, and a certain fraction
of it outside.
The fraction immersed is simply
the same fraction as what you
get by dividing its density by
the density of water.
Also, the density of water
changes.
In the Dead Sea,
because of the salt
concentration,
density is a lot higher.
So, it's a lot easier for
people to float.
Okay.
Now, for the last and final
topic that's called Bernoulli's
Equation.
This is the first time I'm
going to consider fluids in
motion.
So far, my fluids were at rest.
And it's really very simple.
But notice one more time,
all I ever invoked was Newton's
Law of Motion.
You realize that?
I just took this fluid and took
that fluid, balanced the forces,
and said there should be no
acceleration.
If I can convince you of the
one thing, I have accomplished
something.
To realize that all the
mechanics we have done does not
appeal to any other law than
F = ma.
In all these problems of
equilibrium, a is zero,
F is zero.
Just from that fact,
and clever applications of F
= ma.
For example,
it's very clever to think of a
piece of the water and demand
that it be in equilibrium.
That's how we find how the
pressure varies with depth.
But there's no new principle so
far.
In fact, there's going to be no
new principles at all this term.
And relativity was different.
The old space-time was
modified, but non-relativistic
mechanics is all coming from
Newton's law,
as is this problem.
So, we're going to display for
the first time,
fluid motion.
We're going to take the most
general case of fluid motion,
where there's water in some
pipe.
This is the most famous picture
in all the textbooks.
We have not thought of a better
picture now.
We're all working on it,
but this is all we can come up
with after 300 years.
Water flowing in a pipe.
This point is going to be
called 1;
this point is going to be
called 2.
Everything here will have a
subscript 1.
That means, measured from some
ground, that's at a height
h_1,
that's at a height
h_2.
The velocity of the fluid here
is some V_1,
the velocity of the fluid there
is V_2.
And ρ is the density of
the fluid, and that's not
variable.
So, imagine now a steady flow
of some fluid through a pipe,
whose area of cross section is
changing, and the overall
altitude is also changing.
You can have a huge,
you know, pipe in the basement
where the whole supply to the
house comes, branches out into
little pipes maybe;
this could be the pipe in the
attic.
Small cross section possibly at
a bigger height,
it doesn't matter.
But we're not considering pipes
that break up into four or five
pipes.
This is just a single pipe.
So, here is the first law that
you have to satisfy.
If the fluid is incompressible,
there's a certain law.
And we are going to talk about
that law.
That's called the "equation of
continuity."
It relates the area here and
the velocity here,
to the area here and the
velocity here.
And the relationship should not
surprise you.
The basic premise is going to
be you're shoving in water from
the left, and the water cannot
pile up between here and here
because it's incompressible.
That means in that volume,
you can only pack in so much
water.
So, what comes in has to go out.
It follows that it's got to go
out much faster here because the
area is smaller.
And what's the relation?
You can almost guess it,
but let's prove that.
How much water do you think
comes in through this phase,
if you wait one second?
Can you visualize in your mind
that in one second a certain
amount of fluid comes in,
maybe until that point,
and the cross section of that
fluid is A,
the distance it travels is
V.
So, the volume of the fluid
coming in from the left is
really A_1V
_1.
That's called the flow rate
coming in.
And that's got to flow out,
and that's the flow rate
outside.
So, in an incompressible fluid,
if I tell you the flow rate at
one point, I've told you the
flow rate everywhere.
Think of cars going down a
freeway, and the freeway's
getting narrow,
but unlike in real life,
we don't allow the cars to pile
up.
We want the density of cars to
be the same.
That means if there's a narrow
road, they've got to go faster
to maintain the traffic.
That follows--Then,
it follows, that if I go to one
checkpoint and see how many cars
cross me per second here,
the same number will cross
anywhere else.
But the speeds will be in
inverse proportion to the area
so that the product remains the
same.
Now, let me show this to you in
another way.
It's going to be helpful.
Let me wait a small time
Δt.
In a small time Δt,
this front that was here would
advance there.
It'll go a distance
V_1 times the
Δt.
On the other end,
this front will advance the
distance, V_2
times Δt.
Now, the volume that got pushed
in the time Δt is
A_1V
_1Δt,
and that's the volume that came
out on the other side.
And you can cancel the
Δts and come up with the
result I gave you.
So wait a short time,
see what comes into the left
phase, and see what goes out of
the right phase and equate them.
Okay.
Now, we are going to find a
constraint between the state of
the fluid here and the state of
the fluid there.
What I'm going to do--Look,
think about what's going to
happen before you derive any
formula.
Suppose you're going uphill.
Does it make sense to you that
when the fluid climbs uphill
it's going to slow down?
It will slow down on the way to
the top, because it's got to
work against gravity.
Even if you threw a rock up
there, it's going to go up and
slow down.
Therefore, there's going to be
some connection between the gain
in height and the velocity of
the fluid.
Just from the Law of
Conservation of Energy.
That's all I'm going to do.
Here is what I'm going to do.
You remember if there are no
external forces on a system,
kinetic plus potential is
kinetic plus potential.
If there are external forces on
a system, kinetic plus potential
after minus kinetic plus
potential, before,
is the work done by external
forces.
That's what I'm going to use
here.
What are the external forces,
and what's the energy is what
I'm going to think about.
So, here's what I'm going to
focus on.
Take this region of fluid
trapped between these two cross
sections at t = 0.
Wait a short time Δt.
What happens to the body of
fluid?
It does what I told you.
In the front,
it advances to the new region
there.
Maybe I should draw better
pictures here.
This is the advance of the rear
guard, and in the front,
the fluid that used to
terminate here has gone up to
there.
In other words,
if I colored this fluid,
just this part of the fluid
between here and here,
a different color from the rest
of the fluid,
a short time later that colored
liquid will be occupying this
new volume.
It's the same numerical volume,
but it's a slightly different
volume.
I'm going to compare the energy
of this chunk of fluid before
and after.
You realize that in that
comparison this region in
between the shaded regions is
common to both the chunks of
fluid.
Pointwise, the fluid in the
same location is going at the
same velocity.
So, I don't have to worry about
that.
That's common to before and
after.
The only difference between
final minus initial is the
energy contained in that region
minus the energy contained in
this region.
The rest of it,
like here, it was part of the
old fluid;
it's part of the new fluid,
it's at the same height going
at the same speed.
So, you don't have to worry
about it.
So, what's the volume?
What's the energy of this
region?
The energy is going to be
ρ times
A_2 times
Δx_2.
Δx_2 is the
distance it advances here.
That's the volume of the fluid;
that's the mass of the fluid.
Sorry, that times g is
the mass of the fluid.
And the kinetic energy will be
mass times V^(2) over 2.
Potential energy will be--I'm
sorry, am I doing something
right?
I think I made a mistake.
Sorry about that.
There is no g here.
Let me do it more slowly.
What's the mass of the shaded
region?
It's got a base
A_2;
it's got a height
Δx_2.
That's a volume,
times density is the mass,
mv^(2) over 2 is the
kinetic energy.
Potential energy would be the
same mass times gh,
but I should call it
gh_2.
Yep?
Student: [inaudible]
Professor Ramamurti
Shankar: Oh,
you're worried about the height
varying over the tube.
Yeah, we are neglecting that
aspect.
That's an important point.
You can ask,
when you say
h_2 are you
talking about the middle or the
end and so on.
But we are going to be dealing
with problems where the height
differences are much bigger than
the cross section of the pipe.
So, we don't worry about that.
Okay.
Now, what is the same quantity
here?
It's going to be minus
ρA_1ΔX
_1,
times V_1^(2)
over 2 plus ρ times
gh_1.
That's the final energy minus
initial energy.
That's got to be equal to the
work done.
Now, who is doing the work on
this?
Can you guys understand?
Where is the work coming from?
Yep?
Student: [inaudible]
Professor Ramamurti
Shankar: Ah.
No.
Gravity you don't count as a
force when you have a potential
energy for gravity.
Gravity is included.
The minute you write
ρgh or mgh,
you don't count gravity again.
Student: [inaudible]
Professor Ramamurti
Shankar: Right.
But if I'm looking at the fluid
in question, who is acting on
that fluid?
Not what it's doing to others.
Who are the others doing
something to this fluid?
Yep.
Student: [inaudible]
Professor Ramamurti
Shankar: Now,
the walls of the pipe will
apply a normal force,
and that will not do any work.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah,
but that person is somewhere
here.
But remember,
in mechanics you don't have to
go to people in remote places.
I've told you,
when you apply Newton's law,
the only kinds of forces are
what?
Remember from day one?
Contact forces.
Except for gravity,
which reaches out and grabs
everything.
We've included gravity.
Who is this body of fluid in
contact with?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: This fluid?
In between?
No, but I'm talking about this
fluid.
I'm talking about all of this
fluid.
My taking the difference of
that shaded region and that
shaded region was a convenience,
but I'm looking at the energy
change in all of this fluid.
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah,
but the point is the fluid to
the left here is pushing it,
and it is pushing the fluid to
the right.
Okay?
That's the only way.
It is true there's a pump
somewhere pushing the fluid in
the beginning,
but you don't have to go all
the way to the pump.
In the end you only ask,
who is in contact with me?
Well, it's the guys to the left.
So, they will exert a force
P_1,
times A_1,
times Δx_1
because that's the force,
that's the distance,
that's the work done on the
fluid.
And the work done by the fluid
is P_2,
A_2,
Δx_2,
because if it moves to the
right it is doing work on the
left because it has work done on
it.
There you go.
Now you can see that
A_1Δx
_1 and
A_2Δx
_2 are all equal
because that is just the volume
of the fluid displaced in a
short time.
That is just the continuity
equation.
If you divide everything by
Δt, if you like,
then you will find that
Δx_2 over
Δt is
V_2.
I'm just using the fact
A_1V
_1 is
A_2V
_2.
So, if you cancelled those
factors, what do I get?
I'm going to just write it for
you.
Now, you guys can go home and
check it.
P_1 -
P_2 = ½
ρV_2^(2) +
ρgh_2,
minus ½ ρV_1^(2)
- ρgh_1.
Now, this is a derivation that
you can go home and study at
leisure in any book you like.
They're all the same;
the story is the same.
I can only tell you,
if you like to know,
what the trick behind the
derivation was.
I don't mind going over that
because maybe there are
different volumes in question
and you may have gotten
confused.
But first, let me write down
the result.
By writing everything with 1 on
the one side,
and 2 on the other side I get
this result: P_1 +
ρV_1^(2) over 2 +
ρgh_1 =
P_2 +
ρgh_2 + ½
ρV_2^(2).
This is nothing other than the
Law of Conservation of Energy
applied to unit volume.
If you take one meter cubed of
the fluid, its mass is ρ
times 1.
So, this is really a mass of
one cubic meter of the fluid.
This is its kinetic energy;
that's its potential energy.
That's the kinetic and that's
the potential.
You might say,
"Why aren't we just equating
kinetic plus potential to
kinetic plus potential?"
It's because this chunk of
fluid that I focused on is not
an isolated system.
It's getting pushed from people
behind it, and it's pushing
people ahead of it.
So, you've got to take the
difference between the work done
on it, and the work done by it.
Then, you will find they don't
quite cancel because the
pressures are not equal.
If the pressures are unequal,
you get an extra contribution.
But this is simply the Law of
Conservation of Energy.
I got it by writing the Law of
Conservation of Energy.
But it takes a while for you
guys to get used to applying the
Law of Conservation of Energy.
Maybe you are always used to
saying kinetic plus potential is
kinetic plus potential.
You got used to that,
or you also remember kinetic
plus potential may be less at
the end than the beginning,
because there is friction.
This is yet another thing.
Where there's no friction,
but there are external forces
acting on a body,
there is gravity,
which is included the minute
you write a potential energy
mgh.
There's the walls, too.
I think some of you brought up
a very good point.
The walls are exerting a force,
but that's perpendicular to the
fluid.
In a real fluid,
the walls will in fact exert a
force parallel to the fluid,
and it's called viscosity.
So, in a real fluid there will
be a drag on the fluid because
the fluid really doesn't like to
move right up against the walls.
It doesn't mind moving in the
middle of the tube.
So, as you go near the edges,
the speed of the fluid will
relax to zero.
There's a region where there's
a lot of dissipation.
So, we're ignoring what's
called viscosity.
We're ignoring all other losses.
Then, this is simply the Law of
Conservation of Energy.
And as to what I did with those
two volumes, maybe I'll repeat
this so you can all follow this.
I'm saying, take all that fluid
now, find the total energy in
it, and ask what happened to
that fluid a little later.
Well, fortunately,
here the pipe also had some
problems [laughter].
Okay.
Where is this fluid a short
time later?
I want you to think about it.
This end has moved here,
this end has moved there.
So, the fluid at the end of the
day is sitting there.
I want the energy of all of
these guys minus the energy of
all of this.
Then, you can see in the
comparison, it's only this part
that has gained,
and this part that has lost.
This part is common.
When I say common,
not only is the mass
common--the location--at every
location the height and velocity
are the same.
So, in the subtraction this
will cancel that.
That'll cancel that.
This will have nothing to
cancel and this will have
nothing to cancel.
So, just focus on the relative
increments, and that difference
is what gives you the net change
in energy, and you equate that
to the work done.
Okay.
So now, we are going to learn
how to use this Bernoulli.
The main thing to notice about
Bernoulli is,
for a minute,
if you focus on fluid at the
same height.
Just think about the same
height.
He tells you whenever the fluid
picks up speed,
it's going to lose pressure.
Because P + V^(2) or
something, is P + V^(2)
afterwards.
If you increase V,
you're going to decrease
P.
And that's the thrust of
Bernoulli on a qualitative
level.
I'll give you some quantitative
examples.
So, one example is a baseball.
Here's as baseball coursing
through air.
If you like,
you can sit and ride with the
baseball, and say the air going
backwards and the baseball is
going that way.
Now, suppose you spin the
baseball like this.
If you spin the baseball,
it carries some of its own air
due to friction between the
leather and the air.
So, that velocity is counter to
the velocity of the drift here
on the top and additive on the
bottom.
So, the actual air velocity on
the bottom will be more on the
bottom and less on the top
because you're subtracting this
vector from the top and adding
this vector on the bottom.
So, if the velocity is
different, this is the higher
velocity region,
this is the lower velocity
region.
So, pressure here will be less
than the pressure there.
So, the ball will sink down.
If you spun it the opposite
way, the ball will rise up.
If you spun the ball side to
side, the ball will curve from
left to right.
So, that's the spin on the ball.
If I spin the ball and release
it, then it produces extra
forces.
This rising and falling is on
top of the falling due to
gravity.
Even in a planet without
gravity you will have this extra
force.
It's coming simply because the
velocity of the top and bottom
have been modified.
Here's another example.
Here's an airplane wing.
The plane is going like this.
You'll go right with the plane,
in which case the air seems to
be doing this.
Far from the plane,
everything looks the same as if
the wing were not there.
But near the wing,
this is the flow of air past
the wing.
Notice that these guys above
the wing have to travel further
than the particles below the
wing so they can catch up here,
where everything is the same.
So, the velocity on top of the
airfoil is faster;
therefore, pressure is lower.
Pressure is lower than on the
bottom, then the difference in
pressure times the area of the
wing will push the wing up.
That's the lift you get when
you start a plane.
When you're on the runway,
this is why the plane goes up.
Once you're airborne,
you can tilt your wing.
When you tilt your wing to this
angle, then of course it's very
clear that you can get a
component of lift.
But I'm talking about when the
wing is not tilted but
horizontal.
That's what gives you the lift.
Now, what if you made a wing
that looks like this?
Okay.
So, you're all laughing.
But this has a use too.
Have you seen it anywhere?
Student: [inaudible]
Professor Ramamurti
Shankar: Yes?
NASCAR.
On the racetrack you want the
opposite effect,
because you want to push down
on the car, and therefore you
manufacture wings.
You go to the factory where by
mistake they built a wing that
looks like this,
and you can bring it to your
race car and attach it,
then it will keep the race car
down.
Of course, you also have the
other option of going to the
wing and tipping it over and
applying it to the airplane.
But if it did not occur to you,
if you thought once the wing is
made like I'm stuck with this
wing, well, go take it to your
race car, attach it.
That'll push it down.
Because you want to get
traction.
So, you can get the upward lift
or the downward lift.
Now, it turns out this lift
theory is actually somewhat
naive.
And I believed it for a long
time.
Now, I know the story is more
complicated.
There is a lot of truth to
this, but it's not the whole
story.
If you go into aeronautic
engineering, you will find out
that it's a little more
complicated.
The reason, the way to
calculate it precisely--But this
general notion that when the air
moves fast it loses pressure is
true.
So, here's another example.
If you have an atomizer,
you know, you have a perfume
here and you've got a pump.
And then, you have a tube here.
When you squeeze the pump,
the instant you squeeze the
pump, you're driving a lot of
air here at high velocity,
whereas the air here,
is at rest.
So, high velocity air has a
lower pressure than low velocity
air.
Therefore, it will suck the
perfume and spray it right on
your face [laughter].
Yes.
Okay.
Very good.
You know, I've taught you so
many things, planet and
supernovae, and galaxies,
but now I got a rise out of
this class.
They said, now we are telling
you something useful [laughter].
Alright.
So, now I'm going to do two
problems where I really start
putting numbers in.
This is very qualitative.
I'm going to do quantitative
problems.
And there are only two kinds.
So, when we are done,
we are done.
Here is a tank of water.
And I poke a hole in this
somewhere here,
at a depth, let's call it
h.
The question is,
"How fast will the water come
out of here?"
Turns out you can use this
Bernoulli's, whether it is
called Bernoulli's equation.
You can use Bernoulli to derive
this.
We just have to pick the points
one and two cleverly.
So, this is going to be point 1;
this is going to be point 2.
Point 2, in fact,
is just outside that hole.
So, what does Bernoulli say?
P_1,
which is atmospheric pressure,
plus ½ ρV_1^(2)
+ ρgh_1 =
P_2 + ½
ρV_2^(2) +
ρgh_2.
Let's modify this as follows:
let's pick h_2
to be zero, then
h_1 is just
this.
I mean, you can pick the zero
any way you like.
So, where the hole is I'm
calling zero.
So, I don't have to worry about
this one.
V_2 is what
I'm after.
What about P_1
and P_2?
P_1 is
atmospheric pressure there,
and right outside the hole it's
also the atmospheric pressure.
So, P_1 and
P_2 you cancel
because they're both equal to
atmospheric pressure.
How about the velocity here?
You know that if you punch a
hole in the tank and you drain
it, it's going to start moving
down.
But we're going to imagine
that's a huge tank,
ten meters diameter,
and this is a tiny pinprick.
Then, this velocity is
negligible.
You can put that back in if you
like.
I'm just going to ignore that
for now.
That tells me
ρgh_1 = ½
ρV_2^(2);
cancel the ρ,
and you find V^(2) =
2gh.
V_2^(2) =
2gh.
Now, you remember this formula
from day one.
This is the velocity a droplet
would have if it,
say, spilled over the top and
fell straight down.
Because that is just the
mgh being converted to ½
mv^(2).
Because by the Law of
Conversation of Energy,
the minute the fluid starts
coming out of here and draining
out of the top,
I have traded droplets of water
at the top for droplets of water
at the bottom.
Therefore, since they had
potential energy at the top,
they most likely have kinetic
energy at the bottom.
But be clear about one thing:
the drop coming here is not the
same drop moving at the top.
If I push it over the top,
that would be the same drop,
and that's very clear to all of
us.
The beauty of this is this
water is pushing down and
something's coming out here.
It's coming out with the right
speed so that drop by drop,
the kinetic energy of what
comes out here,
is the potential energy of
what's draining at the top.
So, you can punch the hole at
various places.
You know, once you punch a hole
here it's going to land there.
And you can imagine the fun you
can have with this kind of
problem.
Where do I punch the hole so
that if my dog is standing here
it's going to feed the dog?
Or don't feed the dog,
go up in front of the dog,
go past the dog.
Get the biggest range,
smallest range,
a whole bunch of problems.
They will combine this
Bernoulli who will tell you how
fast it's coming out with
chapter one of how to do
trajectories once the fluid is
coming out this way.
Okay.
Last topic.
It's again--it's not even a new
topic.
The last application of
Bernoulli has to do with
the--what you call the Venturi
meter.
By the way, you can imagine
this is just a cross section of
possible applications of fluid
dynamics.
You can take a whole course on
fluid dynamics.
So, let's take the following
problem.
So, you're going in an
airplane, and you want to find
out how fast the plane is going.
How do you think you find out?
I will show you know one device
people use to find the flow
rate, to find the speed of the
plane through the atmosphere.
So, you go to the plane and you
attach the following device to
the underside of the plane.
Well, this, I'm imagining in my
mind a symmetric thing.
This is a pipe with a
constriction.
Here the air is coming in at
the speed of the plane itself.
See, in real life the plane is
going through the air,
but go sit with the plane,
because the air is going
backwards at the speed of the
plane.
So, that's a cross section
A_1 here.
V_1 is what we
are trying to find.
I hope you understand.
That's our goal.
Then it comes to this region
where it has to speed up,
because A_1V
_1 is
A_2V
_2.
If it speeds up,
the pressure of the air is
going to be lower.
Let's first calculate the
pressure difference between this
point and this point.
Now they're both at the same
altitude.
Here's another example.
You're 5,000 feet above the
ground;
don't worry about altitude to
here or there or there.
That's, 5,000 is the big
height, and that ρgh
cancels on both sides.
We don't worry about the
variation in height over this
gadget.
It's a very,
very tiny thing you attach to
the underside of the plane.
So then, I can say
P_1 + ½
ρV_1^(2) +
ρgh_1,
I'm not going to write because
h_1 and
h_2 are going
be equal.
Then, it's P_2
½ ρV_2^(2).
Therefore, P_2 -
P_1 = ½
ρV_1^(2) -
V_2^(2).
Now, V_1 is
the speed of the plane.
Now, V_2 is
not the speed of the plane,
but we know what
V_2 is because
V_2 times
A_2 is equal to
V_1 times
A_1.
So, we can write here ½
ρV_1^(2) -
V_2^(2) =
V_1^(2) times
A_1 over
A_2^(2).
So, pull out the
V_1,
and you write it as ½
ρV_1^(2) times
one minus (A_1/A
_2)^(2).
So, let's digest this formula
for a second.
In this problem,
A_1 is bigger
than A_2.
So, you may worry that 1 -
A_1/A
_2^(2) is
negative,
but so is P_2 -
P_1 because
P_2 here is
certainly going to be lower than
P_1 here.
So, if you're happier,
you can flip it backwards and
write P_1 -
P_2 is something.
You see that?
You can flip it over,
provided you flip it over here.
But ask yourself,
"What do I need to know to find
the speed of the plane?"
The speed of the plane is here.
The ρ is the density of
what?
Can you tell me,
ρ is density of what?
Air;
ρ is the density of air.
A_1 and
A_2 are known,
because you designed the tube.
So, if you can find the
pressure difference,
somehow, you can read off the
speed of the plane.
So, what people do to find the
pressure difference is they go
and they take--they punch a hole
there and they put here a fluid,
like oil.
If the plane is not moving,
you can tell these two heights
must be equal because that's a
condition for the hydrostatic
equilibrium.
Because these two pressures are
equal.
But if the plane starts moving
and the pressure here is lower
than the pressure here.
Imagine this is a high pressure;
that's low pressure.
It'll push the fluid up and
it'll start looking like this,
with a little extra on the
other side.
And how much extra is it?
Well, let's think about it.
The pressure at these two
points is the same;
the pressure here is what I
called P_1;
the pressure here is
P_2 + ρg times
the height of this fluid.
This ρ is not the
density of air.
This ρ is the density
of, oil let's say.
So, the minute the plane begins
to move, the pressure here will
be higher than pressure here;
the fluid will be pushed up.
And the difference in these two
heights directly is a measure of
the pressure difference.
That, in turn,
is a measure of the velocity.
So, what you will try to do is
then, forget all about the
height difference,
and if you're clever enough you
can put marking so that by
counting the difference in the
markings you can translate to
the speed of the plane.
You can also use this to find
the rate at which oil is
flowing.
Suppose oil is flowing in a
pipe and you want to know the
rate at which oil is flowing.
Again, create a constriction in
the flow, then put some tubes.
This cannot be oil;
this has got to be some other
fluid.
Once again, there'll be
pressure difference and there'll
be height difference.
And from the height difference
you can find the rate at which
the fluid is flowing here.
So, what's the trick we use?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Oh,
for this one?
Student: [inaudible]
Professor Ramamurti
Shankar: Well,
I don't--I think all I'm--As
far as I can tell,
there's going to be a pressure
difference and that's going to
translate into height
difference.
And if the two do not mix,
I don't know the danger.
I think one danger you may have
is if the fluid comes in here,
and if it can penetrate this,
it will start mixing.
Right?
So, for that purpose it may be
better if this one is a higher
density than the one on top.
Now in practice,
I am not really sure what kind
of fluids people use in any
engineering thing.
My--I get very shaky when you
start going to the real world.
It's been my practice to avoid
it as much as possible,
which is why I chose this
career.
But people who really want to
build something have to worry
about what fluid to use,
you know, what starts mixing,
what doesn't mix,
what's the accuracy.
So, I don't.
If I thought hard about it,
I would try to reduce
everything to Newton's laws or
some laws of thermodynamics.
But it would take years and
years to go from there to
something practical.
So, I don't know in practice
what fluids people use.
So, this is another example of
Bernoulli's principle.
Final one, which is the kind of
a problem you sometimes do get.
I'm not going to write the
equations or do the numbers,
but I'll tell you how to think
about it.
Suppose you have a tank of
water.
And I make another tube here
and I cap it.
And the fluid here comes to
some height.
First question you can ask is,
"What's the height to which
fluid will rise here?"
Now, you know from your high
school days it will rise to the
same height, but what's the
argument that you will give
today for why that height has to
be the same?
You will use Bernoulli's
principle, and you will take two
points here, for example.
Luckily, there's no velocity to
worry about.
There is no height to worry
about.
So, P_1 =
P_2,
which is what we did use by
other considerations.
Those pressures are equal.
If that is atmospheric
pressure, that's atmospheric
pressure;
then, these columns have to be
equal.
They're starting with
atmosphere plus ρgh;
you want to hit the same number
here as there.
But now, if you open this pipe
and let the fluid really start
flowing, then the story is
different.
The minute fluid stops flowing
here--starts flowing--remember
from the Bernoulli,
if you've got a velocity,
you lose pressure.
Pressure here will be lower
than the atmosphere,
and then this will drop.
It will drop so that
atmosphere, plus that height
gives a pressure here,
that is atmosphere plus the
bigger height gives the pressure
here.
And the velocity here is
assumed to be negligible,
and the velocity here is some
velocity with which the fluid is
coming out.
Okay.
So guys, have a good holiday in
spite of what I've done to you
most reluctantly,
and I'll see you all after the
break.
 
