Internal flow -- the HP equation (advanced)

To consider the detailed solution to the problem of laminar flow in a pipe, one has to permit the velocity to vary as a function of radius. Because of the no-slip condition, the flow velocity will be zero at the pipe wall and highest in the center.  By symmetry, the velocity will only be a function of the distance from the center of the pipe ($r$). Since we are assuming there is only flow along the pipe, the pressure has to be independent of the radius and only depend on the distance along the pipe (otherwise pressure differences would result in unbalanced forces that would accelerate the fluid towards the walls instead of along the pipe).

For an incompressible fluid in steady state flow, the equation that determines the motion of the parts of the fluid is the Navier-Stokes equation (essentially Newton's 2nd law applied to a small bit of fluid). The solution in this case is a “parabolic” flow that varies as distance squared according to the equation:

$$v(r) = \frac{\Delta p}{L} \frac{(a^2 - r^2)}{4 \mu}$$

Here, $v(r)$ is the velocity as a function of $r$, the radial distance from the pipe's centerline, $Δp$ is the difference in pressure between the two ends of the pipe, $L$ is the pipe length, a is the pipe radius and $μ$ is the fluid viscosity.  

One of the things we’d like to know is how fast a volume of fluid can flow through the pipe.  This volumetric flow rate will be the area of the pipe times the linear velocity of fluid down the pipe.  Since for each annulus (ring of fluid) of the pipe, the radius is different, both the area and the linear velocity vary as a function of distance from the pipe center ($r$). To deal with this, we have to integrate with respect to pipe's radius to get an actual value for the total volumetric flow rate. If we consider an annulus (ring) that has a radius $r$ and a smallthickness $dr$, the area of this annulus is $2πr dr$.  If we multiply this area by the corresponding flow velocity at that radius, $v(r)$, and integrate with respect to pipe radius, we get the total flow through the pipe:

$$Q = \int^a_0 v(r) 2 \pi r dr = \int^a_0 \frac{\Delta p}{L} \frac{(a^2 - r^2)}{4 \mu} 2\pi r dr$$

We can pull everything that doesn’t depend on r out of the integral, rearrange and simplify:

$$Q = \frac{\pi \Delta p}{2 \mu L} \int_0^a (a^2r - r^3) dr =  \frac{\pi \Delta p}{2 \mu L}  \bigg[ \frac{1}{2} a^2 r^2 - \frac{1}{4}r^4\bigg]^{r = a}_{r = 0}$$

If we evaluate and simplify this we get:

$$Q = \frac{\pi \Delta p}{2 \mu L}\frac{a^4}{4} = \frac{\pi a^4 \Delta p}{8 \mu L}$$

or, solving for $\Delta p$,

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

Since we

This is known as the Hagen-Poiseuille (HP) 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.

The HP equation is an example of gradient driven flow — a transport of matter or energy that arises from a driving force represented as a change in space of something we typically refer to as a potential pushing through a resistive force to create a terminal velocity. Examples of this also include Ohm's law for the flow of electric current, Joule's law for the transport of heat, and Fick's law for the transport of chemical concentrations. (In the last case, the mechanism is somewhat different, but we can rely on the idea, "If the equations are the same, the analogy is good." Fick's Law equation has the same structure as the others: a spatial variation of a scalar field leading to a transport.)

Karen Carleton 10/26/11 and Joe Redish 2/15/19