Turbulent flow in hoses

11 September, 2022, 6 min, 1103 words

Tags: fire-stuff physics

Yet another pump operations/water flow post, sorry folks! If only it weren't so darn interesting...

This is another "Zeph convinces themself that the book is right" post, this time on a statement about laminar versus turbulent flow in hoses:

The pressures in hose are typically high enough to cause turbulent flow.

If you're interested in a demonstration of this, buckle in and come along! If not, might I suggest different reading material?

Turbulence

Okay, first, some definitions:

• Laminar flow: when water flows in an orderly way from one point to another. Generally this means that following a hypothetical water molecule will give you a straight line. On a small scale, it means that even when something introduces a little turbulence – maybe a bit of roughness in a pipe or a change of direction – the fluid's flow is stable enough that it damps out the turbulence and returns to smooth flow.
• Turbulent flow: the opposite of laminar flow! Basically, even if you have the smoothest, straightest pipe in the world, past a certain point the flow will be topsy turvy no matter what you do.

Two major things affect the boundary between laminar and turbulent flow: viscosity (how "thick" a liquid is – envision honey versus water), which damps out turbulence; and velocity, which encourages it by making everything move faster. The balance between these two components is expressed in a unitless value called the Reynolds number (no apostrophe). It's defined like so:

$$ \text{Re} = \frac{v L}{\nu} $$

where $v$ is the fluid's velocity, $L$ is a characteristic length (for round pipes or hoses, it's the inner diameter), and $\nu$ is the kinematic viscosity of the fluid.¹

The cool thing about the Reynold's number is that we can say pretty reliably whether the flow of a fluid will be turbulent just by looking at this one value. In general, if $\text{Re}$ is less than 2300, flow will be laminar, and if $\text{Re}$ is more than 2900, it'll be turbulent. The in between area is a bit of a transitional space, with properties of both laminar and turbulent flow present at the same time.

Turbulence in a hose

At this point, I'm interested in understanding how much water flow we can have through a hose before the flow becomes particularly turbulent. We'll say that $Q_1$ is the flow rate at which $\text{Re}$ hits 2300 and the flow starts to get turbulent, and we'll call $Q_2$ the flow rate where the flow is conclusively turbulent at $\text{Re}=2900$.

Now, if we want to express the Reynolds number in terms of total volumetric flow $Q$,² we just have to remember that the total volume flowing is the water's speed times the hose's cross sectional area $A$, or $Q=vA$:

$$ \text{Re} = \frac{Q L}{\nu A} $$

Let's say the inner diameter of the hose is $D$, and substitute in the hose's cross sectional area $A=\frac{\pi D^2}{4}$:

\begin{align} \text{Re} &= \frac{4 Q D}{\pi\nu D^2}\\ &= \frac{4 Q}{\pi \nu D} \end{align}

Of the values in here, the only two that can vary are flow rate $Q$ – by increasing the gallonage on a combination nozzle, for instance – and the hose diameter $D$. If we want to solve for $Q$, we can rearrange a bit:

\begin{align} Q &= \frac{\text{Re}\ \pi\nu D}{4}\\ &= \text{Re}\frac{\pi\nu}{4} D \end{align}

Now let's stick in some values for our constants $\pi$ and $\nu$

\begin{align} Q &= \text{Re}\frac{\pi\nu}{4} D\\ &= \text{Re}\frac{(3.14)\cdot\left(1.0\cdot 10^{-6}\frac{\text{m}^2}{\text{s}}\right)}{4} D\\ &= 7.85\cdot10^{-7}\cdot\text{Re}\cdot D \cdot\frac{\text{m}^2}{\text{s}}\\ \end{align}

Now for my favorite part of all this: more unit conversion.

\begin{align} Q &= 7.85\cdot10^{-7}\cdot\text{Re}\cdot D \cdot\frac{\text{m}^2}{\text{s}}\\ &= 7.85\cdot10^{-7}\cdot\text{Re}\cdot D \cdot\frac{\text{m}^2}{\text{s}} \left(\frac{\text{m}}{\text{m}}\right) \left(\frac{60\ \text{s}}{\text{min}}\right)\\ &= 4.7\cdot10^{-5}\cdot\text{Re}\cdot\frac{D}{\text{m}}\frac{\text{m}^3}{\text{min}}\\ &= 4.7\cdot10^{-5}\cdot\text{Re}\cdot\frac{D}{\text{m}}\frac{\text{m}^3}{\text{min}} \left(\frac{264\ \text{gal}}{\text{m}^3}\right)\\ &= 0.012\cdot\text{Re}\cdot\frac{D}{\text{m}}\frac{\text{gal}}{\text{min}} \end{align}

In words: the flow rate (in gallons per minute) that produces a Reynolds number of $\text{Re}$ in a pipe or hose of diameter $D$ is the diameter in meters multiplied by the Reynolds number multiplied by 0.012. What that means for some standard hose diameters:

In other words, with a 5-inch supply hose, if you're flowing fewer than 3.5 gallons per minute, you're likely in laminar flow. Fun fact: you're never going to do that (think more like 500-1000 gpm). So the textbook is right! We can pretty safely assume that all water flow through fire hose is fully turbulent.³

Wrapping up

In summary:

• The Reynolds number is a dimensionless value that describes the balance of viscosity (which tends to make a fluid flow smoothly) and velocity (which tends to lead to turbulence).
• Reynolds numbers above 2900 mean fully turbulent flow.
• In fire hose at any reasonable flow rate (more than 5 gpm, say), we can safely assume that all flow is fully turbulent.
• SI units are still superior to whatever the heck gpm and inches are.
• If you want to do this more simply, consider an online Reynolds number calculator like this one.

Footnotes