The work-energy theorem and the H-P equation

The work energy theorem in general is

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

and as we showed in the work-energy theorem in fluids, in a flowing incompressible fluid, it becomes

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

Let's consider the situation that we considered when we built the H-P equation. A uniform pipe containing an incompressible fluid (at a constant height) where we have to take into consideration a viscous drag force. We represented this model by the figure:

For this model, what happens to the work-energy theorem? Let's let the cross-sectional area of the pipe be $A$ and let's let the fluid move the width of the blue block of fluid, $L$.

Since the speed isn't changing, the term $Δ(½mv^2)$ is 0.

Since the height isn't changing, the term $Δ(mgh)$ is 0.

As we saw in our creation of Bernoulli's principle, if we watch the fluid move a distance $L$, the work done by the pressure term $(ΔP )AL$.

We now have to model the resistive force. From our discussion of the H-P equation, we'll choose $F^{viscous} = 8πμLv$ where $μ$ is the viscosity of the liquid. Since we are moving the fluid a distance $,$ and the force is constant, we can take it out of the integral. The distance integral is just $L$, so our work energy theorem becomes

$$0 = (ΔP )AL - 8πμL^2v$$

The resistive work is negative since it's direction is opposite the motion and so would reduce the kinetic energy in the work-energy theorem if the pressure forces weren't compensating.

Solving the resulting equation for $ΔP$ gives

$$ΔP = 8πμLv/A$$

Since our current (volume/s) is $Q = vA$, we can replace $v$ by $(Q/A)$ to get the H-P equation:

$$ΔP = (8πμL/A^2)Q$$

This approach gives a slightly different perspective on the H-P equation. Instead of seeing it as a balance of forces, we can see it as a balance of work done by the pressure and viscous forces. And in this form, it's pretty easy to see how we can refine our simple model to include the effects of changing height and changing vessel width!

Joe Redish 7/22/17