How much dV is in fire extinguisher?

[Scotius]

I own couple of science-fiction antologies, and one particular short story made me curious. Storyline is fairly straightforward, but in the end the hero is forced to escape his orbital tug, because he overloaded the reactor pushing much bigger, damaged ship into a stable orbit. He does that using two fire extinguishers from the cockpit. Sadly, there is not much data - no orbital parameters, weight of the payload (pilot ), what kind of fire retardant was inside extinguishers and how much of the stuff two tanks held. Still - it had to be more than average sneeze, because it got the protagonist and data bank of his onboard AI out of nuclear explosion blast radius. So, would it be possible in real life? Does average handheld fire extinguisher hold any significant amount of deltaV - enough to push a grown man plus is space suit plus couple of kilos of electronics into safe distance from exploding nuclear reactor?

[m4v]

probably not a lot, since I suppose the tank/charge mass ratio of a fire extinguisher to be pretty bad, like the tank accounting half of the total mass.

[nyrath]

An engineer friend of mine did a few pointless calculations to figure that a hogs-head of beer would give about 0.0084 kN of thrust with a specific impulse of about 8.5 seconds.

[Mr Shifty]

Back of the napkin calcs:

Here are some CO2 fire extinguisher specs: http://www.aespl.com.sg/pdf/FIRE%20EXTINGUISHER-CO2.pdf

We'll assume your hero is using the smallest ones, which weigh 8.5kg full. They have 2kg of CO2 pressurized to 70 bars, which gives a total internal energy (assuming ideal gas) of 3*P*V = 3 * 70 bar * .54m * pi * (.114m/2)^2 = 116kJ. In vacuum, the internal energy of the gas (now solid) would be 0, so theoretically (neglecting state changes) all of the internal energy of the gas will be converted to kinetic energy upon leaving the nozzle.

If all of it gets converted to kinetic energy, you have 116kJ = 1/2 *m * v^2, divided up between a 100 kg dude and 2 kg of propellant so as to conserve momentum, which <math occurs here> gives you about 7 m/s for the dude (and 337 m/s for the propellant -- you can check my math!)

That's for one tank, so you could maybe get 14 m/s out of two extinguishers, which would be enough to raise your apoapsis on a 100 mile circular earth orbit by about 30 miles.

Edited July 23, 2013 by Mr Shifty

[Scotius]

That's not so bad. When the pilot returned from EVA, all alarms in the cockpit were in full swing and AI shut down already - so how much time he could have until reactor's meltdown (not counting the obvious "as much as author needed to get him safe") 10 minutes? 15? And he had to grab extinguishers, yank the datacore of his AI buddy out of its socket, then blast away from his ship. And did his ship really explode, or just turned into deadly source of radiation? Too bad author skimped on the important details at the end - instead we got teary reunions of rescued passengers with their families, and some philosophical waxing of ground crew *sigh*. Thank you, guys.

[metaphor]

If the exhaust velocity of the fire extinguisher is about 20 m/s (apparently it varies wildly between different types from 1 m/s to 100 m/s), and there is 2 kg of fuel in each fire extinguisher, and the total mass of astronaut + fire extinguisher is 100 kg, then they would only give about 1 m/s delta-v. You can get more than 1 m/s delta-v just by pushing against something with your legs and jumping off. So it doesn't seem plausible, unless he had an hour or more to get away from the explosion.

edit: Apparently some fire extinguishers can get up to 10 kg of fuel and 1000 atmospheres of pressure. That's almost 500 m/s exhaust velocity, which gives about 100 m/s delta-v. That's quite a lot. He would accelerate at about 0.5g and would be 1 km away from the ship within 20 seconds.

[Mr Shifty]

edit: Apparently some fire extinguishers can get up to 10 kg of fuel and 1000 atmospheres of pressure. That's almost 500 m/s exhaust velocity, which gives about 100 m/s delta-v. That's quite a lot. He would accelerate at about 0.5g and would be 1 km away from the ship within 20 seconds.

How are you calculating the exhaust velocity? I couldn't figure out how to get it, which is why I resorted to the energy balance.

[Moon Goddess]

(People are gonna notice just how weird I am here)

When I was a young teen I had this dream that I still remember to this day, I dreamed that I woke up on the moon, with basically the contents of my bedroom scattered across the lunar surface and for some reason i didn't need to breathe or any of that other pesky living stuff. So when I got board wandering about on the featureless moon I floated back to earth meeting the space shuttle along the way and rode back down with them (so as to not catch on fire)

Ever since then I've often tried to figure out, just what could have possibly been in my bedroom that would give me the lift to get escape velocity from the moon. I now see that a fire extinguisher (even if I had one) would not have done the trick.

[Kaleb]

Ever since then I've often tried to figure out, just what could have possibly been in my bedroom that would give me the lift to get escape velocity from the moon. I now see that a fire extinguisher (even if I had one) would not have done the trick.

Bedsprings.

[K^2]

How are you calculating the exhaust velocity? I couldn't figure out how to get it, which is why I resorted to the energy balance.

Under assumption of inviscid fluid, your best bet for an estimate is force balance. You know that the thrust is given by F = PA. But you can also derive thrust from exhaust velocity. F = vm', where m' is the mass flow rate equal to AvÃÂ. In other words, PA = F = ÃÂAv². For an ideal gas, you can also express density in terms of pressure. à= PM/(RT), where M is molar mass of the gas, 0.044kg/mol for CO2. So we have PA = PAMv²/(RT). Or v² = RT/M. Of course, you should keep in mind that temperature is going to drop as pressure does, with PV^γ remaining constant if the contents are just gas. At room temperature, the exhaust velocity will start at 238m/s and it's going to drop as the contents cool down.

Naturally, all of this is true for operation in vacuum. When operating in atmosphere, pressure differential will enter into all of this and things will be a little different.

[KOCOUR]

Duudes! It was future fireextinguishers with built in emergenci escape functions. :-)

[3_bit]

We could measure it with no attachments, and you should have no problem using those numbers to calculate what it would be with your weight/equipment.

[Mr Shifty]

At room temperature, the exhaust velocity will start at 238m/s and it's going to drop as the contents cool down.

Yeah, I knew a pure energy balance was going to have some unwarranted assumptions built in, but it's gratifying that I got well within an order of magnitude (337 m/s vice 238 m/s.) It does seem like the nozzle should have an impact; not sure how to calculate it since it's a simple tube to direct the , and not intended to create supersonic velocities. The video I linked, I thought, was a decent demonstration given that Dave gets to probably 3-5 mph using a single extinguisher, in air, with the friction of the skateboard wheels.

[K^2]

That's where inviscid assumption comes in. It eliminates any losses for gas going through the tube. For a more realistic estimate, you'd have to take into account the pressure differential across the tube. The thrust will then be equal to the pressure at the outer end of the tube times the cross section of the tube. And you compute pressure differential based on the flow rate. Unfortunately, things get a little complicated here. For incompressible fluid the solution is rather simple, but for a gas it's a bit more complicated. I would still expect the flow rate to be roughly proportional to the pressure differential, but any estimates I'd make on the proportionality constant are likely to be way off.

[Flixxbeatz]

What about soda?

[Epic DaVinci]

Forgive my ignorance of Physics, but would changing the nozzle diameter of the extinguisher give you any changes to the Dv output?

[Bunsen]

The pressure drop across the tube is going to disappear under the inaccuracies of ideal gas approximations when your working fluid is stored as a liquid and gets expelled as a solid/gas mixture. I think energy balancing is the easier way to get a decent approximation, but taking the phase changes (one of them incomplete) into account.