In this screencast we will explore the concept of hydrostatic pressure and specifically how
pressure varies throughout a fluid at rest.
First let's start with our general form of the equation of motion in a fluid in which
there are no shearing stresses.
You can see here where del of the pressure is the pressure gradient.
What this del operator implies is that we are looking at the pressure on all sides.
In this case it is a Cartesian system that is applying to this fluid element.
We would be looking at the y, x, and z directions and looking at how the pressure varies across
this element.
Gamma here is our specific weight for the fluid and we are applying this only in the
k vector which is also our z direction so this is our component that accounts for the
weight of the fluid particle.
Lastly we have this equal to rho times the acceleration vector.
This is density.
This is our acceleration vector.
Again with a vector we are looking at all of our Cartesian coordinates.
You can see that this is just a force balance.
We have our mass times our acceleration divided by a volume.
If we multiply both sides by volume we would force on the right side.
For our specific weight which is weight times volume we would just have weight.
Our pressure times a volume gives us a force.
You can check the units on that to verify.
This basically means that if we apply some kind of pressure gradient to a fluid element
based on its weight and that pressure we are going to cause that element to move.
If we take this fluid element and say it is at rest so our acceleration term here is 0.
Therefore we could rewrite this and show that our pressure is going to be only a function
of this z component.
What does this mean?
It means we would have the differential form of pressure as a function of z equal to the
specific weight of the fluid.
We could rewrite this in the form of an ordinary differential equation.
This basically becomes the fundamental equation for fluids at rest.
Let's take an example where you can imagine yourself in a swimming pool.
That swimming pool is going to have some depth that we will call z.
We will say that the surface is z=0.
If we wanted to plot what the pressure is on our body as a function of depth, such as
the pressure on your ears as you go deeper, we can see that the pressure is going to increase
as we go deeper into the pool.
Hence that is why we have this negative sign here which indicates that our pressure decreases
as we go up in elevation or increases as we go down.
The same thing happens right now on your body.
You have the force of all of this air weighing down on you causing some kind of pressure
specifically on your body.
As you go higher in elevation there is less air above you to put pressure on your body.
The question becomes how does this specific weight vary as a function of z and can it.
Since specific weight is a weight per volume this is going to be a function of the density
of the fluid.
If the density changes with elevation then the specific weight will change with elevation.
For this video let us just focus on incompressible fluids in which the density of the fluid is
constant.
In other words it is not a function of pressure.
For most liquids this is true but for most gases it is not hence why we can compress
a gas so easily and change its density.
If we integrate our equation above.
Going back to our fundamental equation for a fluid at rest.
We can separate variables and integrate.
If we integrate from z1 to z2 and P1 to P2 we get P2 minus P1 is equal to the specific
weight of the fluid times the difference in elevation.
I have taken out the negative sign and distributed it.
Another way of writing this is in this form.
The reason this might be more common is typically our z2 is going to be a higher elevation so
we will keep the right side positive.
The reason we do this again thinking of some location on our ground as z1 and some higher
elevation as z2 we can think about this change in elevation as the height of our fluid and
we can write that as h.
Plugging that in we have a new equation where we can say P1, the pressure on the ground,
is equal to P2 plus the specific weight of the fluid times the height of that fluid.
Let's think of a tube of fluid that we have.
We will say water.
We take a little bit of water and we put it at the bottom.
It is going to have some pressure associated with that much water.
As we add more water that pressure is going to increase because the weight of this water
is now acting downwards on the water below it.
Since the fluid is at rest that force must counterbalance the weight above it.
You can imagine that if we keep adding water this force is going to get larger towards
the bottom and therefore the pressure at the bottom is going to increase.
All of the fluid particles stacked on top of each other are going to exert some kind
of weight and therefore pressure on the fluid particles on the bottom and since they are
not accelerating and they are not moving the force has to counterbalance it so the pressure
at the bottom is going to be much greater than the pressure at the top.
If we think of this height of fluid as "h" as we did before we could refer to it as a
pressure head or a head pressure.
That is the height of the fluid at some specific weight that results in the pressure difference
P1 minus P2.
That is just rewriting this equation in that form.
Hopefully this helps you understand why a unit conversion of 1 atm can be converted
to a head pressure of feet of water which is roughly 33.9 feet of water.
That means that is we have a column of water that is 33.9 feet high then the pressure at
the bottom the pressure being exerted would be 1 atm or 33.9 feet of water.
With that in mind let's take a look at an example where we have 3 containers.
All having a fluid of height h and they all have the same fluid that has a specific weight
designated gamma fl.
If we took this particle, this particle, and this particle which particle is experiencing
the highest pressure?
Since our fluid is at rest it is not moving in the x direction, the y direction, or the
z direction pressure is equal on all sides.
So pressure here is at P1 is the same pressure because it has the same fluid height above
it.
Interestingly enough if we look at this third container if we drew this element in the middle
then you can see that there is very little argument that the pressure at this height
would be the same as the other two containers because it has the same height of fluid above
it but if you look at this particle to the left that is below this curvature of the container
it must also have the same pressure because since there is no acceleration the force,
the pressure acting on the left and right side of this element, must be the same and
therefore it is the same every where along that z plane.
The pressure on the left side closest to the container wall is still the same pressure.
Hopefully that gives you an idea of the concept of hydrostatic pressure of that of a fluid
at rest and we will use this as a basis for a lot of the work that we are going to do
in our fluid dynamic.
Take a look at some of the other screencasts on how we can utilize hydrostatic pressure
as a mechanical advantage for things such as hydrolic jacks.