Internal flow -- the HP equation

Prerequisites

As liquid flows through a pipe in a steady state, there is a tricky balance of forces, even when the fluid is flowing at a constant and steady rate. For simplicity, let's consider this case.

Because of adhesion between the liquid and the walls, the fluid sticks to the walls and doesn't move at the wall. But since it's moving in the middle, there is a variation in the velocity as a function of radius. This means that cylindrical shells of fluid are flowing over each other. This flow will be resisted by the liquid's internal viscosity. As a result, by Newton's 2nd law there has to be another force acting in the direction of flow to keep the fluid flowing as a constant velocity implies a balance of forces.

Since the velocity varies with distance from the wall, even in the steady flow, figuring out the balance of forces and the flow is quite tricky and involves some serious calculus. If you want to see the full detail, visit the page Internal flow -- the HP equation (advanced). For now, we'll work with a less accurate but simpler model that still gives the correct result.

Instead of worrying about the variation of flow with radius, let's assume a model in which the flow in the tube is uniform and doesn't vary in the tube. We know there is a velocity dependent drag, so let's include in our model a friction force between the wall of the tube and the fluid that's proportional to the velocity. We know that's not the actual source of resistance here, but since we know that a balance of forces is necessary and that the true resistive force is proportional to velocity, this should work. We'll just have to adjust the constants in the formula to match the more complex analysis, and this model will help us think about what's happening.

Consider a pipe with a liquid moving down it uniformly at a constant speed. Isolate a small cylinder of liquid (shown in blue in the figure below). We assume that there is a resistive force between the wall and the liquid that is proportional to the velocity.

Since the cylinder of liquid is moving at a constant speed, the resistive force must be balanced by some force pushing it down the pipe. The only thing touching the cylinder is the rest of the liquid on either side of the cylinder we have chosen. Each side exerts a pressure force on the disk of liquid.

Since there has to be a force balancing the resistive force, it must arise from a pressure difference. We conclude that if the fluid keeps moving at a constant speed, the pressure must drop as we go downstream. Here's the picture.

The resistive force is shown in the blue cylinder of fluid with arrows pointing to the left. The fluid upstream from the cylinder exerts a pressure force on it to the right, and the downstream fluid exerts a pressure force on the cylinder pushing to the left.

We can now derive a relation between the amount of fluid flowing ($Q$) and the pressure drop, $\Delta p$, if we make the following simple assumptions:

• The liquid is moving at a constant velocity, $v$.
• The liquid has a constant uniform density, so we can talk about the volumetric flow, $Q = Av$.
• The resistive force on the cylinder is proportional to its velocity and its length, $L$. We'll choose the proportionality constant so as to match the more sophisticated analysis:

$F_{resistive} = 8πμLv$

where $μ$ is the viscosity of the liquid. (The $8π$ comes from doing integrals.)

With these modeling assumptions, here's what the free-body diagram for the cylinder of fluid looks like.

where the $A_L$ and $A_R$ mean the areas on the left and the right. We take their directions from seeing which way the external fluid pushes on the stuff inside. They both have the same area, $A$. The upstream pressure must be greater than the downstream to keep the cylinder going. We'll define $p_{upstream} - p_{downstream} = Δp$.

Since these forces have to balance to maintain a constant velocity we get

$$F_{pressure\;difference} = F_{resistive}$$

$$\Delta p \;A = 8\pi \mu Lv$$

$$\Delta p = \bigg( \frac{8 \pi \mu L}{A} \bigg) v$$

But since we know that the volumetric flow can be given by $Q = Av$, we can replace $v$ by $Q/A$ and get a relationship between the amount of flow and the pressure drop across the cylinder:

$$\Delta p = \bigg( \frac{8 \pi \mu L}{A} \bigg) \bigg(\frac{Q}{A}\bigg)$$

$$\Delta p = \bigg( \frac{8 \pi \mu L}{A^2} \bigg) Q$$

For a pipe with a circular cross section, we can put in $A = \pi R^2$ to get

$$\Delta p = \bigg( \frac{8 \mu L}{\pi R^4} \bigg) Q$$

The factors in parentheses are all constants that have to do with the size of the pipe so we can give it a name and call it the resistance of the pipe, $Z$. In this form, the equation relates the volumetric flow to the pressure drop:

$$\Delta p = Z Q$$

$$Z = \frac{8 \mu L}{\pi R^4}$$

This is known as the Hagen-Poiseuille equation and it tells us that flow rate increases with pressure difference, and decreases with pipe length or fluid viscosity. These make sense. The more pressure that is applied to the fluid at the pipe opening, the faster the fluid should flow. The longer the pipe or the more sticky the fluid, the harder it is to get the fluid to flow.

One surprising result of the Hagen-Poiseuille equation is that flow rate increases with the fourth power of the tube radius. Therefore, pipe diameter has a huge effect on how easily fluid flows down a pipe. This has important biological implications.

There is a tradeoff in designing circulatory systems. There are two primary costs. The first is making the blood vessels and the blood. The smaller the total volume of the system, the less materials required and the lower the costs. However, the other cost in the system is making the pump to drive the flow through the blood vessels. The pump must be large enough to drive a given volumetric flow through the blood vessels. If you look at the Hagen-Poiseuille equation, as the blood vessels get smaller, the pump must work harder (create a bigger pressure) to get the same volume of fluid to flow. Apparently, this is a significant effort, as we use about 10% of our resting metabolic rate to power the circulatory system.*

It's also useful to consider the HP equation from the point of view of energy. When there are resistive forces, energy is extracted from macroscopic motion (mechanical energy) and is transferred to the microscopic motion of molecules (thermal energy). This is another way of understanding why the pressure must drop to maintain the fluid's speed against the resistive force of viscosity. The work done to maintain the pressure differential is what is eventually transferred to thermal energy. See the follow-on page on energy in fluids for details.

*Steven Vogel, Comparative Biomechanics: Life's Physical World (Princeton U. Press, 2013)

Joe Redish and Karen Carleton 10/26/11

Follow-ons

Article 426
Last Modified: February 28, 2019