At 2:45 AM, when most of the town slept soundly, Beehunter Ditch was anything but quiet. The monitoring station near 12th St NE and Highway 54 East recorded a surge—389 cubic feet per second (cfs) of water flowed through the ditch, a reminder of just how powerful even small waterways can be during peak rain conditions.
This got me to thinking — and oh how crazy my mind works at times! — what if that volume of water were diverted into the typical home? How big would the pipe need to be? And how long would it take to flood the house from floor-to-ceiling? Let’s geek out on the numbers, so we can put this all in perspective.
1. Pipe Sizing for Beehunter Ditch’s Peak Flow
The typical home is connected to the local water utility by a 3/4-inch diameter pipe, sufficient for showers, sinks, a dishwasher, ice maker, and washing machine. But to handle 389 cfs? That’s a whole different animal!
Let’s find out what diameter pipe would be needed to handle that flow rate using some simplified hydraulics.
Assumptions:
• Pipe runs full and under pressure (idealized for simplicity).
• Using the continuity equation:
Q = A \cdot v
Where:
• Q is flow (cubic feet per second)
• A is cross-sectional area (square feet)
• v is velocity (feet per second)
Let’s assume a velocity of 10 ft/s, reportedly a reasonable speed for pressurized flow in a large pipe.
Solving for area A:
A = \frac{Q}{v} = \frac{389}{10} = 38.9 \text{ ft}^2
Now solve for diameter using:
A = \frac{\pi d^2}{4} \Rightarrow d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 38.9}{\pi}} \approx \sqrt{49.5} \approx 7.04 \text{ ft}
Answer: You’d need a 7-foot diameter pipe to carry the peak flow of Beehunter Ditch at 10 ft/s. Wow! Just wow!
But let’s take it a step further, shall we?
Let’s say we’re hypothetically filling the typical house with said water flow—floor to ceiling. How long would it take at 389 cfs?
Assumptions:
• Average American home = 2,300 square feet
• Ceiling height = 8 feet
• Total volume to fill =
V = 2,300 \text{ ft}^2 \times 8 \text{ ft} = 18,400 \text{ ft}^3
Time to flood:
t = \frac{V}{Q} = \frac{18,400}{389} \approx 47.3 \text{ seconds}
Answer: It would take just under 48 seconds to fill the average house to the ceiling, if Beehunter Ditch’s peak flow were diverted straight into it.
So, the quiet trickle of Beehunter Ditch may seem harmless most days, but in moments like the one recorded at 2:45 AM on Saturday, it transforms into a force capable of overwhelming a home in less than a minute. To channel such power, you’d need a pipe larger than most storm sewers—far beyond that humble yet typical 3/4” line feeding water to your home.
That all said, the next time you pass by this local ditch, give it a little nod of respect. It’s not just a ditch; it’s a sleeping giant when it rains!
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And on the topic of rain, if you see a flooded roadway, officials remind you, “Turn Around, Don’t Drown!“