Login    Register
User Information
Username:
Password:
We are a free and open
community, all are welcome.
Click here to Register
Sponsored
Who is online

In total there are 67 users online :: 3 registered, 0 hidden and 64 guests


Most users ever online was 155 on Mon Aug 15, 2016 1:40 am

Registered users: Bing [Bot], Google [Bot], Yahoo [Bot] based on users active over the past 5 minutes

The Team
Administrators
Global Moderators
global_moderators.png CS

Liquid C02 Cartidge?

A place to ask general spud cannon related questions.
Sponsored 
  • Author
    Message

Liquid C02 Cartidge?

Unread postAuthor: kjjohn » Fri Dec 17, 2010 8:07 pm

Would it be possible to have a cartridge filled with liquid C02, and use perhaps an electrothermal-type effect to vaporize almost all of the liquid at once? If so, this could have a much higher power density than other compressed-gas cartridges. The cartridge could be as simple as a small metal cylinder with a 800+psi burst disk.
  • 0

User avatar
kjjohn
1st Lieutenant
1st Lieutenant
 
Posts: 270
Joined: Sat Jul 11, 2009 3:54 pm
Reputation: 0

Unread postAuthor: Labtecpower » Sat Dec 18, 2010 12:30 pm

First of all, CO2 turns into a solid rather than into a liquid when compressed. (the well-known "dry ice")
If heated, it may have a higher vapor-pressure than 800 PSI.

But, I think it would be possible.
(or I don't see why it wouldn't be possible :roll:)
  • 0

User avatar
Labtecpower
Major General
Major General
 
Posts: 1267
Joined: Sat Feb 20, 2010 6:38 am
Location: Herb Island
Reputation: 7

Unread postAuthor: Fnord » Sat Dec 18, 2010 1:13 pm

It's possible, though maybe a little complicated to implement.

I don't think CO2 actually is liquid at room temp. It's supercritical around that point, meaning the gas and liquid are basically the same density.

But whatever. Commencing Science! :

For simplicity lets assume a 1 cubic inch cartridge.
1 cubic inch of 800 psi CO2 would weigh about 1.75 grams,
.839 joules to raise 1 gram +1 degree C.

Lets say you want to raise it +30C. That would put the pressure at... ~2500psi? (I am no wiz with pv=nrt) This should allow a burst disk with a decent safety factor.

You'll need to dump about 44 joules assuming no losses.
Certainly possible though probably not good for a rapid-fire solution.

edit: gas phase diagram for carbon dioxide

room temp is roughly 300 K
  • 0

Attachments
630px-Carbon_dioxide_pressure-temperature_phase_diagram_svg.png
630px-Carbon_dioxide_pressure-temperature_phase_diagram_svg.png (36.25 KiB) Viewed 1303 times
Image
User avatar
Fnord
Major General
Major General
 
Posts: 2244
Joined: Tue Feb 13, 2007 9:20 pm
Location: Pripyat
Reputation: 7

Unread postAuthor: Technician1002 » Sat Dec 18, 2010 3:39 pm

CO2 can be a liquid at room temperature, but only under high pressure. The critical point is a little above room temperature, but not much.
The critical temperature for carbon dioxide is 31.1°C, and the critical pressure is 73 atm. Above the critical temperature, the fluid is called super-critical fluid.

At atmospheric pressure, it is either a solid at very cold temperature, or a gas at warmer temperatures. Dry ice sublimates directly into gas with no liquid phase at atmospheric pressure.

Simply compressing CO2 does not create a solid. It liquefies. When the liquid is then dropped in pressure, the phase change to a gas causes enough cooling to make solid form. Only under extreme pressure can you make a solid by compression, Study the phase diagram for room temperature. There is a lot of liquid before you reach solid.

Quote from http://www.science.uwaterloo.ca/~cchieh/cact/c123/phasesdgm.html
  • 0

User avatar
Technician1002
Chief of Staff
Chief of Staff
 
Posts: 5190
Joined: Sat Apr 04, 2009 11:10 am
Reputation: 14

Unread postAuthor: DYI » Sat Dec 18, 2010 4:06 pm

Firstly, thanks go to Fnord for posting that diagram here, because I was going to link to it in my response otherwise. I'll try to go against the pattern set by two of the three previous posts, and give you some possibly useful information that you might NOT already know from a cursory glance at Wikipedia.

Your idea is achievable, if not particularly helpful. This stems from the fact that, as I tend to state incessantly, CO<sub>2</sub> is a terrible propellant gas, and electrothermal tech is bulky and expensive. If you're going to spend several thousand dollars building a launcher that you can't hold or move, it should at least be made energetic and fast.

Loading the cartridges would be the easy part. Open them up, cool to roughly the sublimation point of CO<sub>2</sub>, and pack them about half full with crushed dry ice (this based on the density of the solid and the liquid). Designing and building the cartridges is the hard part. Having designed (but not implemented) ET and ETC cartridges, I can say that they're very complicated compared to traditional firearm cartridges. You obviously have your own lathe if you're asking this question, but it's prudent to warn you that you'll be looking at possibly over an hour of machining per shot, assuming that you reload used cartridges with new seals, insulators and capillary tubes.

Now, on to the science. CO<sub>2</sub>'s critical point occurs at 304 K, roughly ten degrees above room temperature. Heated above that temperature it becomes a supercritical fluid. Practically speaking, this means that you can't slosh it around in the container any more, as it has become filled to constant density. For most of our applications, we can treat supercritical fluids as gases. In fact, in some of our modern high pressure launchers (over 500psi), nitrogen is actually a supercritical fluid.

At its critical point, the enthalpy of vaporization of a substance is zero. As such, to "vaporize" the liquid CO<sub>2</sub>, all you need to do is warm it by about ten degrees, which by itself is very easy to achieve (based on cubic fit I have for liquid CO<sub>2</sub>'s heat capacity, it takes about 34J/g). The real issue is with it converting back to a liquid in the barrel. This is where GGDT comes in. Model this hypothetical launcher, and take a look at the pressure-temperature. If the temperature is maintained above 304 K at all times, there's no problem. If it drops below that value, the pressure should be below the line on the phase diagram indicating a liquid state. If it isn't, then you need to make the CO<sub>2</sub> hotter.

This is where the tricky bit comes in - converting the temperature you want to the energy you should input. Take a look at pages two and three of this for an idea of why. Regrettably, I've been unable to find a polynomial fit, and I have neither the dataset nor the motivation to construct my own. You, however, might. As you can see, CO<sub>2</sub> is a particularly nasty gas to do these calculations for. You're looking for a constant volume heat capacity, not constant pressure, but that's easily enough obtained by looking at the ratio of specific heats, which is much easier to come across data for.

Looking at the 3-D graph in my link, you can imagine the process you're working with taking some diagonal path across from low pressure and low temperature to increased values of each. I'm not you, but I'd settle for approximations and ballpark figures at this point (or I could write code to model it for you, but I'm afraid I'd have to start asking for hourly pay at that point :lol: ). So, here's my approximation: imagine a plane corresponding to a roughly "average" of that surface in 3-space. I'll assume a C<sub>p</sub> of 3 kJ/(kg * K). This converts to a C<sub>V</sub> of roughly 2.4 J/(g*K). For example, if you want to heat 1g of CO<sub>2</sub> to 600K (from its critical point of 304), you need to add roughly 710J.

How much energy do you need to actually pump into the cartridge? Well, a good capillary tube plasma generator might be 70% efficient at actually creating the plasma. The mixing process is tremendously complicated (turbulent flow, non-equilibrium chemistry and thermodynamics all jammed together with a healthy dollop of plasma physics on top) and depends on the geometry of the chamber and capillary tube. Some parts of the fluid will be heat to thousands of degrees, some will be almost unaffected, and it's all up to the firing process to determine what happens from there, and how this translates to barrel temperature. One thing that's certain is you'll need a good deal more than (710/0.7) J to make it work the way you hope. Remember though, that's just an example. Fooling with GGDT will tell you more about the temperature you actually need.

Also remember that you can't treat the CO<sub>2</sub> as a perfect gas in your calculations. It's far too dense for that in this case, and you'll need a better equation of state. The only constant here (before the disk ruptures) is the density - i.e., you know the number of entities, the volume they occupy, and the fact that you're dealing with CO<sub>2</sub>. Go find yourself a suitable equation of state for the temperatures and pressures you're looking to work with, because you'll need that to calculate the GGDT inputs.

Finally, if you do decide to try this, good luck, and tell us all about it if you succeed.
  • 0

Spudfiles' resident expert on all things that sail through the air at improbable speeds, trailing an incandescent wake of ionized air, dissociated polymers and metal oxides.
User avatar
DYI
Lieutenant General
Lieutenant General
 
Posts: 2861
Joined: Sat Jul 07, 2007 8:18 pm
Location: The People's Republic of Canuckistan
Country: Turks and Caicos Islands (tc)
Reputation: 9

Return to General Spud Cannon Related

Who is online

Registered users: Bing [Bot], Google [Bot], Yahoo [Bot]

Reputation System ©'