Hi. It’s Mr. Andersen. And this is AP Physics essentials video 98. It is on the continuity
equation which is really an application of the conservation of mass in fluids. So if
we have a fire hose, the mass that goes in one side of the fire hose is equal to the
mass that comes out the other side. It seems pretty intuitive. But what happens if we change
the diameter of that fire hose. What if we put a nozzle on the front of it? The same
mass has to go through, but since we have restricted the area we are going to increase
the velocity. And that is the continuity equation, which is simply the conservation of mass in
fluids. And you will see it in one of two ways. You will see it represented as the mass
flow rate or the volume or volumetric flow rate. If we are looking at the mass flow rate,
this is what the equation looks like. And so it looks a little scary but it is really
not that bad. On the left side we are going to have somewhere in the fluid before, that
is 1. And then somewhere in the fluid after. And so the three parts are the density of
the fluid, the area, cross-sectional area of the fluid and then the velocity of the
fluid itself. Now if you think about water its density is not going to vary as it is
moving through a pipe. And so lots of times you will see it represented as the volume
flow rate. We have gotten rid of the density and so it is simply the cross-sectional area
times the velocity. And so an example of that. Let’s say we have a cross-sectional area
that is very, very, large right here and a low velocity. What happens as we decrease
the cross-sectional area? We should see an increase in the velocity. And this is a PHET
simulation that shows that. Again A1 times V1 or the cross-sectional area times the velocity
before equals the cross-sectional area times the velocity after. And so if we were to grab
a flux meter and just measure what the cross-sectional area is in this pipe, you will see that the
area is going to be 10 meters squared. If I measure the speed it is going to be 0.5
meters per second. And so A times V equals 5. What happens if I change the area where
the area is now 1 meter per second. We are going to see a velocity of 5, because 1 times
5 is 5. What would if I were to give you a totally different cross-sectional area? Let’s
say the cross-sectional area at this point is 5, then we should see a speed of 1. And
so once you know the cross-sectional area and the velocity in any part of that hose
you could calculate it anywhere else, as long as we have consistent density in the fluid
itself. And do you might see a continuity equation like this. Water flows through a
fire hose with a diameter of 7.5 centimeters. They are giving you the velocity at 4.0 meters
per second. And you need to calculate the velocity coming out if we restrict the size
of the nozzle to 2.3 centimeters. And so we are going to use the continuity equation,
A1 V1 equals A2 V2. What is going to be my area? Well I know my diameter here, but that
does not give me the area. I am going to have to use a little geometry to figure that out.
So I know the area. I know the velocity. I know the area coming out of the nozzle, but
we are going to solve for V2. So if I write out the formula we are going to use pi r squared
to calculate that cross-sectionals area where R1 is going to be the radius in the firehose
and R2 is going to be the radius in the nozzle itself. If you look at me working this out
I had to take this centimeters and convert it to meters. So that would be 0.75 meters.
And then since the radius is half of the diameter, that is where I am getting this radius here.
And this is the radius as we leave. And so I solve it and I get a velocity 2 of around
43 meters per second. And so it is going 4 meters per second around 10 miles per hour
in the hose but it is coming out around 100 miles per hour. And so did you learn to make
calculations of flow based on the continuity equation? I hope so. And I hope that was helpful.