Video Transcript
In this video, we’re going to learn
about Bernoulli’s equation. We’ll learn what this equation
says, what its limitations are, and how to work with it practically.
To get started, imagine that, in
preparation for the upcoming summer vacation and time spent at the neighborhood
pool, you’re developing a human-powered water cannon. The canon consists of a large
cylinder holding water, sealed in by a movable platform that people can stand on top
of.
The pressure this creates pushes
water through a tube attached to the side of the cylinder. You want to design the canon so
it’s able to shoot water all the way across the pool and onto unsuspecting
sunbathers on the other side. To learn how to design such a
canon, it will be helpful to learn a bit about Bernoulli’s equation.
Several hundred years ago, well
before space travel or highway systems or hydraulic construction equipment, the
Swiss scientist Daniel Bernoulli set out to understand fluid flow better. He found that when you combine
various terms having to do with fluids, the terms pressure, density, fluid speed,
and fluid elevation, that they formed a constant value.
Specifically, he found that
pressure 𝑝 plus one-half fluid density 𝜌 times its speed squared plus its density
times the acceleration due to gravity multiplied by its elevation all is equal to a
constant value.
What this means is that if we have
a pipe of varying thickness with a fluid that flows through it, then if we picked a
location in that pipe, say we pick this spot, and at that point measure pressure,
density, speed, and elevation, then the value we would get when we compute this sum
is equal to the value we would get if we picked a totally different point in the
pipe. The values of these individual
parameters would change. But this combination would stay
constant.
If we call the first point that we
chose point one and the second point point two, then we can equally write this
expression, that the combination of those terms at point one is equal to the
combination of those terms at point two. This equation is known as
Bernoulli’s equation. It’s an expression of fluid
properties in a steady flow state.
To understand what this equation
means a bit better, let’s look at the units involved in each term. For the pressure term involved in
each side of this equation, we know that the SI unit of pressure is the pascal or in
base units newtons per meter squared. A newton per meter squared is
equivalent on a units basis to a joule per meter cubed. These are the same units. This means it’s possible to express
pressure in units of energy per unit volume.
Moving on to the second term,
one-half the density 𝜌 times 𝑣 squared, the units of this term are kilograms per
meter cubed for density 𝜌 and meters per second quantity squared for the speed
squared. Looking at these units, we realize
that a joule is equal to a kilogram meter squared per second squared, which means
this second term as well can be expressed as an amount of energy in joules per unit
volume.
With the third term, 𝜌 times 𝑔
times ℎ, the units of 𝜌 are kilograms per cubic meter. The units of gravitational
acceleration are meters per second squared. And the units of ℎ are meters. We again see a kilograms meter
squared per second squared term, which we can write as joules.
What we’re seeing is that, across
the board, Bernoulli’s equation involves terms that represent energy per unit
volume. In other words, we can say that
Bernoulli’s equation is a statement of energy conservation. Here, our initial state is a state
we’ve called one and our final state is a state we’ve called two.
Just as the conservation of energy
principle depends on an assumption, namely, of working with an isolated system,
Bernoulli’s equation has some assumptions as well. First, we assume that the fluid
flowing is flowing steadily. There are no air bubbles or
turbulent eddies or vortices involved. Second, we assume that the density
of the fluid involved 𝜌 is constant.
Notice that, in the statement of
Bernoulli’s equation, our density symbols don’t have a subscript for one and
two. We assume they’re the same
everywhere. And finally, this equation assumes
that negligible energy is lost to friction in the process of this fluid flowing. When these assumptions hold, we
know that Bernoulli’s equation does as well.
When Bernoulli developed this
equation, there was a particular connection between pressure and fluid speed that he
highlighted. Say that we shift our points one
and two so that one is now at the widest part of our flow and two is near the
narrowest part. By our intuition, we might expect
correctly that the speed of the fluid at point two is greater than the speed at
point one.
What Bernoulli saw is that that
also means that the pressure at point two is less than the pressure at point
one. In other words, there’s a
connection between higher speed and lower pressure. This idea, that higher fluid speed
means lower fluid pressure, is sometimes called the Bernoulli effect.
We experience this effect in
practical ways, including when we’re riding airplanes. The wing on an airplane is designed
so that as the wing moves through the air, the air on the top of the wing has to
move a longer path than the air on the bottom to get to the back end of the
wing. If we call a point just on the top
of the wing point one and a point just below the wing point two, then we can say
that the speed of the air on the top of the wing, 𝑣 one, is greater than the speed
of the air under the wing, 𝑣 two.
The Bernoulli effect implies then
that the pressure on the wing’s topside is less than the pressure on the wing’s
underside. Because of this pressure imbalance,
a net lift force is created on the wing. The Bernoulli effect isn’t the only
reason a plane stays airborne. But it is one of the main
contributors to the lift on the plane.
As we move toward applying
Bernoulli’s equation practically, here are three different things to keep in
mind. First, when using Bernoulli’s
equation, we’ll want to explicitly identify and probably draw in on a diagram we
have the location of points one and two, the spots we choose. It’s important to keep these points
in mind because all of our variables are defined with respect to these
locations.
Second, we want to be careful to
use consistent units all throughout our solution. The different terms involved in
Bernoulli’s equation may be given to us in different unit sets. But we’ll want to put them all on a
common basis. And finally, and this may seem
silly, but be careful with calculator order of operations. Bernoulli’s equation involves so
many terms that it can be confusing which terms we’re multiplying, dividing, adding,
or subtracting when we solve on our calculator.
Now let’s get some practice with an
example involving Bernoulli’s equation.
A container of water has a
cross-sectional area of 0.100 meters squared. A piston sits on top of the
water, as shown. There is a spout located 0.150
meters from the bottom of the tank, open to the atmosphere. And a stream of water exits the
spout. The cross-sectional area of the
spout is 7.00 times 10 to the negative fourth meters squared. What is the speed of the water
as it leaves the spout? How far from the spout does the
water hit the floor? Ignore all friction and
dissipative forces.
We can call the speed of the
water as it leaves the spout 𝑣 sub two. And the distance from the spout
the water hits the floor we’ll label 𝑑. Since this example involves a
flowing fluid and all friction and dissipative forces are ignored, that means
this is an opportunity to apply Bernoulli’s equation.
This relationship tells us that
the pressure plus half the fluid density times its speed squared plus its
density times 𝑔 times its altitude ℎ is constant throughout any point in our
fluid system.
Looking at our diagram, we’ll
want to pick two points over which to apply Bernoulli’s equation. We’ll pick point number one to
be at the location right beneath our piston. And point number two is right
at the spout of our outlet.
Applying Bernoulli’s equation
to our scenario, we write that 𝑝 one plus one-half 𝜌 𝑣 one squared plus 𝜌𝑔
times ℎ one is equal to 𝑝 two plus one-half 𝜌 𝑣 two squared plus 𝜌𝑔 ℎ
two. We’ll treat 𝑔, the
acceleration due to gravity, as exactly 9.8 meters per second squared.
Looking at this long
expression, let’s see if there’re any terms we can eliminate. At point one, the speed of our
fluid is zero. It won’t be in motion. So one-half 𝜌 times 𝑣 one
squared goes to zero. We can also condense a bit by
simplifying our 𝜌𝑔 ℎ one and 𝜌𝑔 ℎ two terms. If we subtract 𝜌𝑔 ℎ two from
each side, then we can combine our 𝜌 times 𝑔 terms.
Since we want to solve for 𝑣
two, the speed of the water as it leaves the spout, we can do that algebraically
now. When we do, we find that 𝑣 two
is the square root of two over 𝜌 times the quantity 𝑝 one minus 𝑝 two plus 𝜌
times 𝑔 times the quantity ℎ one minus ℎ two.
In the problem statement, we’re
told that 𝜌 is equal to 1000 kilograms per cubic meter. But we’ll want to solve for 𝑝
one minus 𝑝 two and ℎ one minus ℎ two. When we consider the pressure
term, we know that 𝑝 one consists of two pressures. The pressure due to the
atmosphere plus the pressure due to the piston comprise 𝑝 one.
On the other hand, 𝑝 two,
which we’re told is exposed to the atmosphere, consists solely of atmospheric
pressure. This means that 𝑝 one minus 𝑝
two is equal to atmospheric pressure plus pressure due to the piston minus the
same amount of atmospheric pressure. So that term cancels out. So then what is the pressure
due to the piston?
If we recall that pressure is
equal to force over area, we know that the pressure due to the piston will be
the pressure created by the force of gravity divided by the cross-sectional area
of our cylinder. That area, which we’re told in
the problem statement is 0.100 meters squared, we can call 𝐴 one. So the pressure due to the
piston and therefore 𝑝 one minus 𝑝 two is equal to the piston’s mass, given as
20.0 kilograms, multiplied by 𝑔 divided by 𝐴 one. We insert this expression for
𝑝 one minus 𝑝 two in our expression for 𝑣 two.
And now we look to solving for
ℎ one minus ℎ two. Because we’re given the heights
at these two points we’ve chosen, 0.500 meters and 0.150 meters, ℎ one minus ℎ
two is simply the difference between them, or 0.350 meters. Plugging that into our
expression for 𝑣 two, we’re now ready to insert values for our other
variables.
When we plug in for 𝜌, 𝑚, 𝑔,
𝐴 one, and ℎ one minus two and enter these values on our calculator, we find
that 𝑣 two is 3.28 meters per second. That’s the speed at which water
leaves the spout. Then if we call 𝑑 the
horizontal distance that the water travels before it hits the ground, that’s
what we want to solve for next.
We knew that as the water
falls, its speed in the horizontal direction, which is 𝑣 two, will be constant
over its descent. That means if we can solve for
the time it takes for the water to fall from the spout down to the ground, when
we multiply that time by 𝑣 two, that will give us 𝑑. We’re not able to solve for
that time using motion in the horizontal direction. But we can solve for it using
motion in the vertical.
Because the acceleration of the
water once it leaves the spout is constant, that means the kinematic equations
of motion describe how it moves. In particular, we can use the
kinematic equation, which says that distance is equal to initial speed times
time plus one half 𝑎 times time squared.
In our case, we can write that
ℎ two, the height of the spout, is equal to one-half 𝑔 times 𝑡 squared. We don’t have an initial speed
term because, initially, our speed in the vertical direction is zero.
Rearranging to solve for 𝑡, we
see it’s equal to the square root of two ℎ two over 𝑔. Plugging in for these values
and entering this expression on our calculator, the time 𝑡 it takes for the
water to fall from the spout to the ground is approximately 0.175 seconds. Recalling that, for a constant
speed, distance traveled equals speed times time, we can write that 𝑑 is equal
to 𝑣 two times 𝑡. Plugging in for these values,
when we calculate 𝑑, we find that, to three significant figures, it’s 0.574
meters. That’s how far from the end of
the spout the water reaches the ground.
Let’s summarize what we’ve learned
about Bernoulli’s equation. We’ve seen that Bernoulli’s
equation describes steady incompressible fluid flow. When we looked at this equation, we
saw that it was essentially a statement of conservation of energy. We also saw, as part of this
equation, the Bernoulli effect, that higher speeds create lower pressure. And we saw how this has to do with
creating lift on the wing of an airplane.
And finally, we learned three
things to keep in mind when solving Bernoulli’s equation examples. First, we saw to choose two points:
one for point one and one for point two. Second, we want to be careful with
units with all the terms in Bernoulli’s equation and, finally, to be careful
entering expressions on our calculator when they’re long and extended. Bernoulli’s equation is a powerful
tool for understanding fluid flow. And now we know how to use it.
