, 9 min, 1651 words
Tags: fire-stuff physics
Yet another post on pump operations, sorry y'all. This one, like a previous post, is me coming to terms with a statement about water flow in a textbook. This time around, I'm looking at how to estimate additional hydrant capacity if you're already flowing some amount of water from your hydrant.
Big picture, I'll do some calculations based on Bernoulli's equation to show how much additional water flow a hydrant can provide, then I'll briefly compare that to two of the three methods discussed in Sykes and Sturtevant's 1999 Fire Apparatus Operator: Pumper textbook.
For the below, I'll be assuming we know the static pressure of the hydrant (note the intake pressure before you start flowing water), along with the current flow rate and current residual hydrant pressure (look at the intake pressure now).1
We'll start with Bernoulli's equation, which says the total energy at a point along the flowstream will be constant, both over time and along the flow, and writes out the energy in terms of three "pressures": static pressure (what you can measure on a gauge), hydrostatic pressure (think gravitational potential energy), and hydrodynamic pressure (think kinetic energy):
\begin{align} p_{tot} &= p_{s} + p_{hs} + p_{dyn}\\ &= p_s + \rho g h + \frac12 \rho v^2 \end{align}
In the above, $p_s$ is the static pressure, $\rho$ is the water's density, $g$ is a gravitational constant in Earth's gravitational field, $h$ is the water's height over some arbitrarily defined starting point, and $v$ is the water's velocity. For this calculation, we'll imagine that we don't have to worry about elevation changes, so we can drop the $\rho g h$ term for the rest of our analysis.
Now we're going to define three points in time that we're interested in:
The beauty of energy conservation is that for all three of these times, the total energy will have to be the same.
For any further calculations, we'll need to be able to convert between flow rate (e.g. in gallons per minute) and water velocity (e.g. in feet per second). The former is much easier to estimate in the field, while the latter is what we actually need to calculate anything with Bernoulli's equation.
If we take the cross-section of whatever supply line(s) we're using to be a constant, and call it $A$, then the total flow rate will just be the product of the water's velocity and that cross-sectional area:
\begin{align} Q &= A * v\\ v &= \frac{Q}{A} \end{align}
Phew, that wasn't so scary, was it?
Alrighty, now let's take a look at how our intake's static pressure will change with a given flow rate. To do this we'll compare time 0 (no flow) and time 1 (arbitrary flow) using Bernoulli's equation:
\begin{align} p_0 + \frac12\rho {v_0}^2 &= p_1 + \frac12\rho {v_1}^2\\ p_0 &= p_1 + \frac12\rho\left(\frac{Q_1}{A}\right)^2 \end{align}
Rearranging a bit, we can see the pressure drop is:
\begin{align} (p_0 - p_1) &= \frac12\rho\left(\frac{Q_1}{A}\right)^2 \end{align}
Or, if we want to flip this on its head, flow rate as a function of pressure drop:
\begin{align} \left(\frac{Q_1}{A}\right)^2 &= \frac{2(p_0 - p_1)}{\rho}\\ \frac{Q_1}{A} &= \sqrt{\frac{2(p_0-p_1)}{\rho}}\\ Q_1 &= A\sqrt{\frac{2(p_0-p_1)}{\rho}} \end{align}
Now, this isn't particularly useful at this moment – after all, we already know the pressures and flow rates at times zero and one! Where it starts to get useful, though, is when we start comparing the flow rates at different times. (See also Physics factor problems for a discussion of factor problems in general.)
Okay, if we're already successfully flowing one line's worth of water from the hydrant, what we really care about is how many more lines we can hook up without endangering our water supply. That is, we care about max amount of water we can flow compared to however much we're flowing now. In math terms, we care about $Q_{max}/Q_1$. To be a bit more general, I'll say that we care about the ratio of two flow rates $Q_2/Q_1$. Plugging in our equation above, we can cancel a bunch of stuff:
\begin{align} \frac{Q_2}{Q_1} &= \frac{A\sqrt{\frac{2(p_0-p_2)}{\rho}}}{A\sqrt{\frac{2(p_0-p_1)}{\rho}}}\\ &= \sqrt{\frac{p_0-p_2}{p_0-p_1}}\\ \left(\frac{Q_2}{Q_1}\right)^2 &= \frac{p_0-p_2}{p_0-p_1} \end{align}
Now we're in business! This equation relates only to pressures (which we can read from gauges) and flow rates (which we can estimate based on what lines are connected and what stage of fire operations we're in).
With that physics-y stuff out of the way, let's take a look at the estimation methods provided by Sykes and Sturtevant: the percentage method and the squaring-the-lines method.
In their words:
- Determine the percentage drop using the formula $(\text{static pressure} - \text{residual pressure})\times \left(\frac{100}{\text{static pressure}}\right)$.
- Determine the amount of additional hydrant capability using the percentage drop and the following table:
Percentage drop $\ \ \ $ Extra water available 0-10% 3 times original flow 11-15% 2 times 16-25% 1 time Over 25% less than original flow
Let's translate this approach into the terms we were using in our calculations above. This is all about the maximum possible flow rate $Q_{max}$, which will correspond to a minimum acceptable residual hydrant pressure $p_{min}$.2 With that in mind, we get:
\begin{align} \left(\frac{Q_1}{Q_{max}}\right)^2 &= \frac{p_0 - p_1}{p_0 - p_{min}} \end{align}
If we make the assumption that $p_{min}$ is much smaller than the initial static pressure of the hydrant,3 then the right side of this equation is pretty much exactly the percentage drop their method refers to, just without the factor of 100. Note: since $p_{min}$ isn't exactly zero, our percentage drop will slightly underestimate the true ratio of pressure drops.
So, if we want to double the water flow rate ($Q_1/Q_{max}=\frac{1}{2}$), we'll need our percentage drop to be approximately $\left(\frac12\right)^2=0.25=25\%$ With a bit of wiggle room for $p_{min}$ not actually being zero, 16-25% seems reasonable here.
What if we want to triple our flow rate (that is, have 2 times the original flow available as extra water)? Then we get $Q_1/Q_{max}=\frac13$, so the percentage drop should be $1/9=0.11=11\%$. Again, with a bit of fudgign for $p_{min}$, 11-15% seems reasonable.
Finally, if we want to get three times the original flow, we have $Q_1/Q_{max}=\frac14$, so the percentage drop is in the range of 6%. With fudge factor...sure, 0-10% seems legit.
For this one I'm going to paraphrase, because I think the original textbook makes things unnecessarily complicated and omits some important notes. Basically:
The new pressure drop is the ratio of new lines to those already flowing water, squared, multiplied by the original pressure drop.
If we switch out line count for flow rate (assuming all these lines are the same diameter/general flow rate), then we can translate that into math:
$$ (p_0-p_2) = \left(\frac{Q_2}{Q_1}\right)^2(p_0-p_1) $$
That's exactly what we found up above! No fudging or room for error necessary.
As an example of how to use this, let's say that you're operating one of our particularly energetic hydrants, which has a static pressure of 100 psi. You're already flowing one inch and three quarter line off it, at 125 gallons per minute, and you see a pressure drop of 15 psi. If you hook up another inch and three quarter line at the same gallonage, what will be the resulting residual hydrant pressure?
\begin{align} p_0 - p_2 &= \left(\frac{Q_2}{Q_1}\right)^2(p_0-p_1)\\ &= \left(\frac{250\ \text{gpm}}{125\ \text{gpm}}\right)^2\left( (100\ \text{psi}) - (85\ \text{psi})\right)\\ &= \left(2\right)^2\left(15\ \text{psi}\right)\\ &= 60\ \text{psi} \end{align}
So the resulting residual pressure in this case would be 40 psi, so still safe for the water system. Hooray!
At least for me, this was fairly helpful in understanding where these rules come from – it's all about pressure and flow rates, and Bernoulli's equation continues to help out with converting between the two. Plus it's nice to confirm that the standard wisdom really does work with the underlying physics.
In practice, the trickiest of these on our engines is estimating flow rates, as long as you remember to check the hydrant's static pressure when you first get it hooked up.
The textbook suggests 20 psi here. Our department SOPs say 30.
This assumption probably holds up better for big city departments. Some of our less enthusiastic hydrants have static pressures of around 40 psi, according to the city's dataset (discussed here), which is not noticeably much larger than the 30 psi minimum residual pressure we operate with...