Giant Fireballs and The Like

Sometimes the most random thoughts strike me on the commute to and from work. As a kind of geeky new years resolution I said to myself I'd follow through on these random musings and try to expand my horizons by working some of this stuff out. Queue musing number 1: propane bottle explosions.

Some prior knowledge, it's easy to lookup online:

• Propane in liquid form expands 270 times in volume as it converts from a liquid to a vapor.
• Cylinders are never full, the usually contain 80 percent propane liquid and 20 percent propane gas.
• An average cylinder contains 15 liters.
• 1 liter of propane, if completely vapourised, could create a fireball 9.6 cubic feet, or 0.27 cubic meters, in volume.

The problem

Driving to work the other morning, at one of the petrol stations I drive past, I noticed something I'd never noticed before. In a locked cage at one of the far ends of the forecourt is approximately 10 propane gas bottles. This got me wondering whether, if they all went up at the same time, if they were a sufficient distance from the pumps and people using the forecourt, and whether that's why they'd been placed in such a location.

Some quick Googling gives us the facts above. Everything we need to work this out.

The garage

A quick look on Google maps shows the cage, and a quick blast with the measuring tool tells me this cage is approximately 25m from the main forecourt.

The next question is, if one canister exploded how big would that explosion be? Doing the simple maths of 15 x 9.6 therefore shows us that one "full" tank of propane could create a fireball 144 cubic feet, or 4 cubic meters, in volume.

Ok, be we have 10 cylinders here. Lets make the sweeping generalisation that if one went up then they'd all go up. That gives us a resulting fireball of (15 x 10) x 9.6. Which is 1,440 cubic feet, or 40 cubic meters.

That sounds pretty big. But what does that translate to in terms of distance?

This can be done fairly simply using the following equation: V = ⁴⁄₃Πr³.

This takes some rearranging though and the assumption that V = (40m³ x 2) as I'm assuming that the blast radius will create a perfect half sphere (a big assumption). The equation to arrange looks like this then - 80m³ = ⁴⁄₃Πr³.

The re-arranged equation looks like: r³ = ^80m3⁄_⁴⁄3Π.

After working out the right hand side we get r³ = 19m³. I offended mathematicians everywhere with this by assuming Π was 3.142. Sorry. And the good thing here is the ³ cancels out on both sides leaving us with a blast radius of 19m. A whole 6m short of the nearest pump!

Conclusion

Although the ensuing fireball would fall short of the forecourt the shockwave wouldn't.

This was calculated as a bit of fun and is in no way an indication of forecourt safety. the actual chances of a propane bottle exploding are slim at best.

Have I got something wrong, or is my maths just crap? Please feel free to let me know in the comments or on Twitter.

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