[AR] Re: CO2
- From: "John Dom" <johndom@xxxxxxxxx>
- To: "\(AR\) ocket List" <arocket@xxxxxxxxxxxxx>
- Date: Wed, 12 Aug 2015 09:59:46 +0200
Complex indeed.
I would assume that the entirety of the CO2 cartridge is saturated liquid.
The network54 paper says their 25 g cartridges are charged to 75 vol.-% with
liquid CO2, after cooling them that is: see page 3. Unfortunately the CO2
temperature in the cartridge right after filling is not mentioned so it is not
possible to find the pressure of the 75% liquid 25% gas mix ratio on the
liquid-gas line of the PT graph to check the ratio for the ratio at say, 20°C .
But maybe there is a way I do not know of. To fill the cartridges up completely
might have them exceed the burst pressure (>483 bar) when exposed to heat like
sunshine I read somewhere. The article compares the CO2 cartridge energy with a
.38 shell: impressive potential.
<
http://www.network54.com/Realm/Iden/Co2.pdf>
http://www.network54.com/Realm/Iden/Co2.pdf
The clip below I discovered shows puncture tests of different CO2 cartridge
sizes launched from a pipe. The guy would have had a stable trajectory with a
stick or with fins attached. OK as a pneumatic alternative to a tiny model
rocket combustion motor. The ephemeral white smoky cartridge exhaust means dry
ice presence, demonstrating the gas/solid temperature must be briefly below the
triple point of minus 56,6°C before it de-sublimates. Observation is easier &
more reliable than an enthalpy balance.
Since the punctured cartridge flies about wildly, the CO2 vapor pressure does
perform a lot of work. Also, during the flight, a water-ice coat forms on it
(isenthalpic?). Could freeze the hand. Since this is a rocketry list, a
contraption with a 10 L pressure CO2 chamber aboard, filled with a heavy
cylinder & manometer on the pad could fly quite high! The question is if the
CO2 line to the exhaust should not better run from the top of such chamber
instead of from the bottom whereby the liquid phase would be (prematurely?)
expelled. Next, I wonder if compressed air chambers are not to be preferred;
But there are no such tiny cheap cartridges for air I guess.
<
https://www.youtube.com/watch?v=VgC9YsODubo>
https://www.youtube.com/watch?v=VgC9YsODubo
You have 0.1818 moles of CO2, but without knowing the volume of the cartridge
you cannot calculate the mass of liquid and/or vapor - you don't know how
much of the CO2 is in liquid form, how much is vapor - and so you cannot
calculate the initial enthalpy.
I weighed the 8 g cartridge volume by filling an empty one it with water. It is
11 mL. An empty cartridge weighs 20 g. So the average density of the phase mix
inside the cartridge is 0,008 kg/0,011 L = 0,73 kg/L. Less dense than water.
But even knowing the initial enthalpy, the final one requires % solid in the
expanded gas and heats of evaporation and sublimation. This has probably all
been evaluated by cartridge manufacturers. I’m not going to.
jd
From: arocket-bounce@xxxxxxxxxxxxx [
mailto:arocket-bounce@xxxxxxxxxxxxx] On
Behalf Of David Gregory
Sent: maandag 10 augustus 2015 6:54
To: arocket@xxxxxxxxxxxxx
Subject: [AR] Re: CO2
Peter has more patience than I do, hats off. For bonus points, you can actually
calculate how much (by mass) of the CO2 will be solid and how much is still
vapor. I would assume that the entirety of the CO2 cartridge is saturated
liquid.
On Sun, Aug 9, 2015 at 9:02 PM, Peter Fairbrother <
<
mailto:zenadsl6186@xxxxxxxxx> zenadsl6186@xxxxxxxxx> wrote:
On 08/08/15 23:55, John Dom wrote:
I would like to be able to see the equations which are involved.
Enthalpy balance and heat of evaporation, gas law etc..
hokay. will be long and complex ...
Enthalpy is a "thermodynamic potential", a quantity which can be measured
and/or calculated for a particular system, eg a bucket of water, a cubic meter
of air, or the CO2 gas in the cartridge. It is defined as the internal energy
of the system, plus PV, the product of pressure and volume. It is a universal
quantity, possessed by all substances - eg the unleashed explosive power of TNT
would be considered part of the internal energy and thus part of the enthalpy
of a block of TNT: the PV and heat of the hot gas left immediately after a
detonation would also be parts of the enthalpy.
In SI enthalpy is measured in Joules, or for substances in particular
conditions of pressure and volume it is often measured in Joules per gram (or
per mole) of substance. An isoenthalpic (isenthalpic) process is a process in
which the enthalpy does not change during the process. Generally, this means no
mechanical work is done to the surroundings, and no energy is transferred to or
from the system in the form of heat.
The process of expansion in this case is isenthalpic, because no work is done
to the surroundings, only to the gas itself; and we assume it happens quickly
enough that no significant heat transfer takes place. The process is
isenthalpic, so you know the final enthalpy and density. Look up P and T on
the NIST webbook.
In cases like this, where a substance does not behave like an ideal or even a
non-ideal gas, and especially where it is close to its critical and/or triple
points, and the actual equations of state are verra complicated Captain
involving something like 18 constants; it is very much easier to use NIST or
similar data, and the results will be far more accurate than using the more
normal theoretical simplifications.
So. Final density is 8g/l (ignoring the cartridge volume, see below), or 0.1818
moles/litre (a mole of CO2 is 44.01 grams).
A constant-density graph for 0.1818 moles/litre is available at
<
http://webbook.nist.gov/cgi/fluid.cgi?D=0.181776869&TLow=216.592+K&THigh=300&TInc=1&Applet=on&Digits=5&ID=C124389&Action=Load&Type=IsoChor&TUnit=K&PUnit=MPa&DUnit=mol%2Fl&HUnit=kJ%2Fmol&WUnit=m%2Fs&VisUnit=uPa*s&STUnit=N%2Fm&RefState=DEF>
http://webbook.nist.gov/cgi/fluid.cgi?D=0.181776869&TLow=216.592+K&THigh=300&TInc=1&Applet=on&Digits=5&ID=C124389&Action=Load&Type=IsoChor&TUnit=K&PUnit=MPa&DUnit=mol%2Fl&HUnit=kJ%2Fmol&WUnit=m%2Fs&VisUnit=uPa*s&STUnit=N%2Fm&RefState=DEF
But what is the initial (= final) enthalpy? In eg kJ/mol? At 293.15K, with both
liquid and vapour present in equilibrium, initial pressure is 5.73 MPa. liquid
density is 17.373 mol/l, vapour density is 4.413 mol/l. Enthalpy of the vapour
is 17.950 kJ/mol, enthalpy of the liquid is 11.261 kJ/mol. All those figures
can be found from the graphs on the NIST pages.
You have 0.1818 moles of CO2, but without knowing the volume of the cartridge
you cannot calculate the mass of liquid and/or vapour - you don't know how much
of the CO2 is in liquid form, how much is vapour - and so you cannot calculate
the initial enthalpy.
However, expanding to the final density given above, and with an overall
enthalpy somewhere in between those values, final conditions are off to the
left of the constant-density graph linked to above.
Try setting the X axis to enthalpy to see why.
The temperature will be below the triple point, and you _will_ get CO2 snow.
(that's why the constant-density graph won't go any lower - density becomes a
somewhat meaningless quantity. A volume could contain lots of snow, or only a
little, but still have the same pressure and temperature)
-- Peter Fairbrother
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