The work-energy theorem in fluids


To begin our consideration of energy in fluid flow, let's start with a simplified model: an incompressible fluid flowing smoothly in a pipe. We seen in our discussion of the continuity equation that if the diameter of the pipe changes, the conservation of matter (amount in = amount out) implies that the speed of the fluid in the pipe has to change as the cross sectional area does ($A_1v_1 = A_2v_2$). Our result comes from a constraint condition: something that's true whatever details are going on. But if matter is speeding up or slowing down, Newton's 2nd law tells us that there have to be unbalanced forces. When our fluid changes speed, where do those forces come from? Analyzing those forces in the context of the work-energy theorem gives us a number of useful results.

As usual, we'll start with the simplest model that contains the effects we want to explore: change in pipe diameter, change in gravitational potential, and resistive energy loss.

Consider the flow of a fluid that can be approximated as incompressible (like water under typical conditions) in a pipe whose cross sectional area and height changes, as shown in the figure at the right. We know that the speed of the fluid has to change as it moves from the position marked $L_1A_1$ to that marked $L_2A_2$.

Let's look at the forces on a bit of the fluid in the context of the work-energy theorem so we can see how the pressure has to change along the flow in order to produce the necessary changes in speed. 

And just because it's easy to do in the work-energy context, we'll also include the effect of gravity, allowing the pipe to change its height as well as its diameter. (Of course this has real implications for animals that stand upright like humans and giraffes, and for tall plants, like trees.)

Assume the fluid is flowing steadily in this pipe. Consider a small cylinder of fluid of volume $V = A_1L_1$ entering the pipe. We know from the continuity equation that when it pushes out of the pipe at the top, it must be traveling at a higher speed in order to take out as much fluid as is put in. If this didn't happen, the fluid would have to squeeze more fluid into the same volume — something we are assuming doesn't happen. To increase the speed of the bit of fluid there have to be unbalanced forces. What are they? 

At each instant the cylinder feels the normal forces from the walls of the pipe, the force of gravity and two pressure forces: the one from behind pushing it forward, and the one from in front pushing it back. As it moves through the pipe, it also feels resistive forces (mostly dominated by viscosity).

By Newton 2, it will only change its velocity if the sum of those forces are non-zero. We could analyze this motion with Newton's second law and follow the fluid through the pipe. But we are only interested in the speed when it reaches a certain point in space — not in how it made the transition or how long it took to get there. That's the clue that work-energy might be an easier approach than force and acceleration.

The work-energy theorem states that the change of the kinetic energy of an object A  (here, the small cylinder of fluid) in moving from one point to another (here, from the entry point of the pipe to the exit point) is equal to the work done on that object by all the forces it feels added up along the path it travels. As a formal equation, this is

$$\Delta \big({1 \over 2} \;  m_A v_A^2 \big) = \sum_{j} \int  \overrightarrow{F}_{j \rightarrow A} \cdot \overrightarrow{\Delta r}_A$$

The summation (capital sigma) says add up for all the objects exerting forces on A, and the integral says add up the work each force does along the path a small step at a time.  Since we are only interested in speed, not direction, only forces along or against the direction of motion matter. (The dot product tells us to only include the force along the direction of motion.) This looks a mess, but it simplifies when we think about the physics.

  • The normal forces exerted by the walls on the fluid are perpendicular to the direction the fluid is moving, so they don't do any work on it. Only the forces in the direction of motion (along or opposite) change the object's speed. So we can ignore the walls.
  • We know that the work done by gravitational forces on an object that changes its height by an amount $Δh$ is $W = -mgΔh$ so we can move that work to the other side and treat it like a potential energy.
  • Of the pressure forces, only the outside pressure force pushing it in at the front end and the outside pressure force trying to hold it back at the outgoing end matter. The inner forces during the travel cancel. (This takes a bit of mapping physics to math to work out.)
  • Finally, any resistive drag forces decrease the mechanical energy (and convert it to thermal). 

Putting these all together, if the mass of fluid in the cylinder is $m$, the fancy equation turns into the result:

$$\Delta \big({1 \over 2} \;  m v^2 \big) + \Delta (mgh) = \int_1^2  P\overrightarrow{A} \cdot \overrightarrow{\Delta r} + \int_1^2  \overrightarrow{F}^{viscous} \cdot \overrightarrow{\Delta r}$$

(Are you surprised that the vector arrow in the work done by the pressure is on the area and not on the pressure? If you are, go back and read the pressure page.)

This still looks pretty messy. But remember, that at this stage, our equation is only expressing the relationships among the various terms that stand for various bits of physics. To get anything useful, we have to apply this equation to specific model systems, models in which we make different assumptions. Then, different terms drop and we can get some simpler but powerful results.

  • If we assume that we can neglect resistive forces, we can drop the integral on the right. We can then figure out the work done by the pressure forces. The result is Bernoulli's principle, a result that looks like a conservation of energy (density) in fluid flow.
  • If we assume that the fluid is stationary (or its speed is not changing), we'll wind up with the dependence of pressure on depth in a static fluid, a result we derived using force arguments in our derivation of Archimedes' principle
  • And it we look at a uniform flow at a fixed height but keep a viscous drag force, we'll wind up with the H-P Equation

Our work energy theorem not only unites these three results that look on the surface like different equations, they also give us ways to make corrections to the simple forms of the principles, each of which corresponds to highly simplified models.

Joe Redish 7/21/17


Article 458
Last Modified: February 24, 2019