Let’s talk about pressure and at the end, we will go over some example problems
to see how to apply what we learned. Pressure is the normal force exerted by a fluid per unit
area. We usually use pressure when it comes to dealing with gases or liquids. Since pressure is
force per unit area, the units would be newtons per square meter. This is also called a pascal,
represented with the letters Pa. There are also other units of pressure that
might come in handy, such as bar, standard atmosphere, psi and more.
Usually, you will hear something called the absolute pressure. That’s the actual pressure at
a given position. We also use pressure measuring devices and they have a little gauge on them
that shows the difference between the absolute pressure and the local atmospheric pressure.
That difference is called gage pressure. We can write it as an equation like this.
When we have pressures below the atmospheric pressure, it’s called vacuum pressures.
These are measured by vacuum gages, and it tells the difference between atmospheric pressure and
absolute pressure. The equation for it is this. To understand this better, imagine you have a
tire. You plug your pressure gauge in to see the tire pressure and you see it shows 33 PSI.
What this is showing is that the air pressure is 33 psi
above the atmospheric pressure. If we wanted to figure out the absolute pressure in the tire,
we’d have to add the pressure shown on the gage and the atmospheric pressure together.
You probably know that pressure changes with depth. So in simple terms, the deeper we go,
the higher the pressure and vice versa. Imagine you want to figure out the pressure a certain
distance below the ocean surface. How can we do that? We can use this equation. The first term is
the atmospheric pressure. This is the pressure at the surface from where we are calculating
the depth. This surface is also called the free surface. The rho symbol represents the density of
the substance. So here, that would be the density of sea water. This term is the acceleration due
to gravity, which is 9.81 m/s^2 or 32.2 ft/s^2. Lastly, we have h, which represents the depth
below the free surface. In other words, where we’re trying to find pressure at.
If we use a gage, then our equation is simply this.
Sometimes, we will have to figure out the density using something called the specific gravity value.
It’s simply the density of a substance that is given
relative to the density of something we usually know. Most of the time,
it’s given relative to water, which has a density of 1000 kg/m^3. So for example, if we’re told that
the specific gravity of seawater is 1.025, how do we calculate the density from that?
To do so, we can use this equation. Here, SG is the specific gravity, rho
is the density, and the bottom is the density of water. All you need to do is plug in 1.025 for SG,
and 1000 for the density of water and you get the density of sea water, giving us 1025 kg/m^3.
Now let’s talk about pressure measuring devices. One of the most common instruments
we use to measure atmospheric pressure called a barometer. This is why atmospheric pressure is
usually called barometric pressure. We can figure out the atmospheric pressure using this equation.
Here, we use the density of mercury since most barometers use mercury,
g is the acceleration due to gravity, and h is the height of the mercury column above the free
surface. Most of the time, this height is given with these units, and it’s called the Torr.
So here, this 700 mm is the height of mercury, and you will usually convert this to meters.
Lastly, let’s discuss the Manometer. It’s a device that can be used to measure pressure
differences using fluid columns. Here, we have a simple manometer.
The pressure at point 1 is equal to point 2 since pressure doesn’t vary with horizontal direction.
Remember, if we go down, pressure increases, but going horizontally doesn’t make a difference.
To figure out the pressure, we can use the same equation we talked about before.
If we have a monometer with multiple liquids, like the one here, then the process is pretty
similar. Keep In mind that pressure increases downwards, and decreases when going up and to
use the proper density for different liquids. In other words, you add the pressure when you go down
and subtract when you go up. All of this will make more sense through examples, so let’s get started.
In this question, we have a vacuum gage connected to a chamber and it reads 35
kilo pascals. The atmospheric pressure is 92 kilo pascals.
We need to find the absolute pressure in the chamber. So we can use the equation we
talked about earlier where all we need to do is subtract the pressure in the vacuum chamber
from the atmospheric pressure. So the absolute pressure in the chamber is 57 kilopascals.
In this question, we need to determine the atmospheric pressure when the barometric
reading is 750 mmHg. To solve this problem, we can use this equation, where the pressure is equal to
the density times the acceleration due to gravity times height of the mercury column above the free
surface. We know the density of mercury which is given to us. The acceleration due to gravity is
9.81 m/s^2. We also have the height, which is 750 mm, and in meters, that’s 0.75m. So let’s write
our equation down. Solving gives us our answer in pascals, or we can write in kilopascals. If
you don’t know how we got pascals, please check the video where I cover unity conversion ratios.
In this problem, we need to figure out the pressure exerted on a diver 45 m below the surface
of the sea. To find the pressure, we first need to figure out the pressure exerted by the ocean,
and then add that to the atmospheric pressure. So our equation will look like this and this
part is the pressure exerted by the ocean. We know the atmospheric pressure since it’s given
to us. We also know the acceleration due to gravity and we know the depth, which is 45 m.
Now notice that we don’t have the density of seawater, instead, we’re given the specific
gravity of seawater. So we first need to figure out the density. To do that,
we can use this equation. We know the density of water,
and we have the specific gravity of sea water. Let’s plug them in and solve.
Now we can plug values into our equation. Let’s solve and that is our answer.
Let’s take a look at one last example involving a manometer.
We need to figure out the pressure difference between the two pipelines.
We know the density of seawater, the density of mercury, and the density of water.
Now what we will do is start going through the pipe. In simple terms, we are going to
calculate the pressure at each segment, and add or subtract them using our usual equation.
Remember that we add the pressure when we go down, and subtract when we go up.
So first, we have P1, that’s the pressure in the freshwater pipe. Then we go down so we
have the density of water times gravity times the height of the water column.
Next we have the mercury area. That’s negative since its going up.
Now comes the air. This is also negative since we’re going up. One thing to note here
is that the density of air is 1.2 kg/m^3, and the height is only 0.7m. The pressure
difference is going to be very tiny, so we can actually just forget about it.
Next the sea water area. This is positive since we’re going back down.
Lastly, we have p2 which is the sea water. The difference between the two pipelines is
the subtraction of P2 from P1. So we if isolate for those terms, we can write our equation like
this. We can now plug our values in, I’ll do each part separately. Now let’s plug the
values into our equation. Solving gives us the difference in pressure between the two pipelines.
If you calculate the pressure created by the air column, it’ll be 0.0082 kPa
and like we talked about before, it’s negligible.
That should cover the types of problem you will face when it comes to pressure and tools used to
measure pressure. I hope this video helped. Thanks for watching and best of luck with your studies.