How do you derive Bernoulli's equation?
Consider the following diagram where water flows from left to right in a pipe that changes both area and height. As before, water will speed up and gain kinetic energy
at constrictions in the pipe, since the volume flow rate must be maintained for an incompressible fluid even if those constricted sections move upward. But now since the constriction also causes the fluid to move upward, the water will be gaining gravitational potential energy
as well as kinetic energy
. We will derive Bernoulli's equation by setting the energy gained by the fluid equal to the external work done on the fluid.
Let's assume the energy system we're considering is composed of the volumes of water 1 and 2 as well as all the fluid in between those volumes. If we assume the fluid flow is streamline, non-viscous, and there are no dissipative forces affecting the flow of the fluid, then any extra energy
added to the system will be caused by the external work
done on the fluid from pressure forces surrounding it.
We can express this mathematically as,
First we'll try to find the external work done on the water. None of the water between points 1 and 2 can do external work since that water is all part of our energy system. The only pressures that can directly do external work on our system are and as shown in the diagram. The water at to the left of volume 1 will do positive work since the force points in the same direction as the motion of the fluid. The water at to the right of volume 2 will do negative work on our system since it pushes in the opposite direction as the motion of the fluid.
For simplicity's sake we'll consider the case where the force from water pressure to the left of volume 1 pushes volume 1 through its entire width . Assuming the fluid is incompressible, this must displace an equal volume of water everywhere in the system, causing volume 2 to be displaced through its length a distance .
Work can be found with
. We can plug in the formula for the force from pressure
into the formula from work to get
. So, the positive work done on our system by the water near point 1 will be
and the work done by the water near point 2 will be
Plugging these expressions for work into the left side of our work energy formula we get,
But the terms and have to be equal since they represent the volumes of the fluid displaced near point 1 and point 2. If we assume the fluid is incompressible, an equal volume of fluid must be displaced everywhere in the fluid, including near the top. So, . We can just write the volume term simply as since the volumes are equal. This simplifies the left side of the work energy formula to,
That takes care of the left hand side. Now we have to address the right hand side of this equation. This is a crucial and subtle part of the derivation. Remember that our system includes not only the shaded portions of water near point 1 and 2, but also all the water in between those two points. How will we ever be able to account for all the change in kinetic energy and gravitational potential energy of all parts of that large and winding system?
Well, we have to make one more assumption to finish the derivation. We're going to assume that the flow of the fluid is steady. By "steady flow" we mean that the speed of the fluid passing by a particular point in the pipe doesn't change. In other words, if you stood and stared at any one particular section of the transparent pipe, you would see new water moving past you every moment, but if there's steady flow, then all the water would have the same speed when it moves past that particular point.
So how does the idea of steady flow help us figure out the change in energy of the big winding system of fluid? Consider the diagram below. Our energy system consists of the greyed out fluid (volume 1, volume 2, and all fluid in between). In the first image, the system has some amount of total energy . In the second image the entire system had work done on it, gains energy, shifts to the right, and now has some different total energy . But notice that the energy of the fluid between the dashed lines will be the same as it was before the work was done assuming a steady flow. Water changed position and speed in the region between the dashed lines, but it did so in such a way that it will be moving with the exact same speed (e.g. and ) and have the same height as the previous water had in that location. The only thing that's different about our system is that volume 2 now extends into a section of the pipe it wasn't in previously, and now nothing in our system is occupying the old position behind volume 1.
Overall this means that the total change in the energy of the system can be found by simply considering the energies of the end points. Namely, we can take the kinetic and potential energy
that now exists in volume 2 after the work was done and subtract the kinetic and potential energy
that no longer exists behind volume 1 after the work was done. In other words,
Plugging this into the right hand side of the work energy formula we get,
Now we'll substitute in the formulas for kinetic energy and gravitational potential energy to get,
In this equation and represent the pressures of the fluid in volumes 1 and 2 respectively. The variables and represent the speeds of the fluid in volumes 1 and 2 respectively. And and represent the height of the fluid in volumes 1 and 2 respectively.
But since we are assuming the fluid is incompressible, the displaced masses of volumes 1 and 2 must be the same . So getting rid of the subscript on the 's we get,
We can divide both sides by and drop the parenthesis to get,
We can simplify this equation by noting that the mass of the displaced fluid divided by volume of the displaced fluid is the density of the fluid . Replacing with we get,
Now, we're just going to rearrange the formula using algebra to put all the terms that refer to the same point in space on the same side of the equation to get,