Course Content I > Energy: The Quantity of Motion >Energy in fluid flow
6.5.2
Prerequisites
To begin our In the pre-requisite, we say how the work-energy theorem yields a description of fluid speed in a pipe. Now let's simplify our model even further. We'll not only assume an incompressible fluid, but we'll also assume that we can ignore resistive forces.
Consider the work done by the pressure forces on the block of fluid that is shown in the figure below. We DON'T mean on the little blue pieces. We are looking at the whole fluid that begins at the left of the lower blue piece (L1A1) and ends at the right of the upper blue piece on the right (L2A2). The blue pieces are only there to illustrate that, from the continuity equation, that as the left end moves a distance L1, the right end moves a distance L2.
The work energy theorem is
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Let's first drop the resistive forces. We know the work done by gravity just leads to a PE term, Δ(mgh).We then only have to calculate the work done by the pressure forces.
We know the signs of the work done by the pressures at the beginning and the end since the pressure at the beginning creates a force (P1A1) that is in the same direction as the motion (the dot product is +1), tending to increase the KE, and the pressure at the end creates a force (P2A2) that is in the opposite direction as the motion (the dot product is -1), tending to decrease the KE.
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We also know from continuity, that as the surface on the left of the block of fluid moves a distance L1, the surface on the right moves a distance L2. So the total work done by these forces is P1A1L1 - P2A2L2. This gives the simplified result
Since the fluid is incompressible, the incoming volume is the same as the outgoing so we can write
Since the fluid is incompressible the density won't change in traveling through the pipe, so we can also write the mass as equal to the density times the volume
Putting both of these into our work-energy equation, we get
Since the volume and density don't change, we can take them out of the delta. (This is like Δ(mgh) = (mgh2‑mgh1) = mg(h2‑h1)=mgΔh.) There is a V in every term so we can cancel it giving Bernoulli's principle:
(We get a negative sign on the pressure term since our delta always means "final minus initial" and our final here is 2 and our initial is 1. So ΔP = P2 - P1 and our equation has the negative of that.) Putting the pressure term on the left and opening up the deltas, we get two potentially useful forms of Bernoulli's principle:
Bernoulli's principle provides insight into lots of properties of fluid flow. Airplanes use it in the Pitot tubes they use to measure external air pressure. It has biological implications in situations ranging from the flow of water in sponges and caddisfly larvae to the air flow in prairie dog burrows.* Of course, just like mechanical energy, it's only a starting point since it ignores the loss of mechanical energy to thermal due to resistive forces. To see what happens there, check out the follow-on page on the work-energy theorem and the H-P equation.
Follow-ons:
* S. Vogel, Comparative Biomechanics: Life's Physical World, 2nd ed. (Princeton U. Press, 2013) p. 117 ff.
Joe Redish 7/21/17
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