[AR] Re: CO2
- From: Peter Fairbrother <zenadsl6186@xxxxxxxxx>
- To: arocket@xxxxxxxxxxxxx
- Date: Wed, 12 Aug 2015 06:43:16 +0100
On 10/08/15 05:53, David Gregory wrote:
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.
Hmmm, I rewrote the bit about enthalpy, and the method of calculation,
in the hope of making things clearer; which I reproduce below.
I think I will leave the above as an exercise for the reader though. I
don't have that much patience!
Hint: the enthalpy of sublimation is available under Phase Change Data
on the main NIST CO2 page.
-- Peter Fairbrother
Enthalpy is a measure of thermodynamic potential energy. It is a
universal quantity, possessed by all substances, solid, liquid, gas or
vapour. It is the amount of potential energy in a system, like a bucket
of water, a cubic meter of air, or the CO2 gas in the cartridge.
Enthalpy is defined as the internal energy of the system, plus PV, the
product of its pressure and volume. The product PV has dimensions of
energy. and both PV and the internal energy are measured or calculated
in units of energy.
In SI units enthalpy is measured in Joules; for substances in specified
conditions of pressure and volume, specific enthalpy is often given in
kiloJoules per mole of substance.
It can be tricky to measure (and hard to define) the absolute value of
the internal energy of a system, so enthalpy is commonly given as a
value compared against an arbitrarily chosen reference point; usually
the enthalpy the system or substance would have at 25C and 1 atm
pressure is set at zero, and the figure given for enthalpy under
different conditions is the amount of energy, in the form of heat and/or
work, required to change the system from 25C and 1 atm to those
different conditions.
For practical calculation purposes, where we are almost exclusively
interested in relative enthalpy or changes in enthalpy, this offset can
almost always be ignored - but if you are curious, it is why sometimes
the enthalpy of a system is given as a negative value. In these cases
the system does not have negative potential energy (!), just less energy
than it would have at 25C and 1 atm.
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.
If this happens then no energy is gained or lost, and by the principle
of conservation of energy. there can be no change in the potential
energy (=enthalpy) of the system.
The other way a process can be isoenthalpic is if a change in internal
energy is exactly balanced by an opposing change in PV energy.
The process of expansion in the case of the CO2 cartridge can be
considered to be isoenthalpic, 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 best known approximations to the equations
of state are verra complicated Captain involving something like 18
constants; then 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.
Actually, come to think of it, these days it is common to use NIST data
or software for almost everything. You can do it using the NIST REFPROP
software, but it costs (and it's pretty crappy). Otherwise, use the NIST
web pages.
The example in the question is a little unusual, as it involves three
phases - solid, liquid, vapour - so to begin with I will show you how to
work out a simpler process, expanding 100 ml of Nitrogen at 5 MPa and
20C into 1 litre total volume.
To begin with, we google NIST nitrogen, get
http://webbook.nist.gov/cgi/cbook.cgi?ID=7727-37-9, and click on Fluid
Properties. Choose isothermal, click to continue. Enter 293.15K for
temperature, 5Mpa for pressure (twice), press for data. We get
http://webbook.nist.gov/cgi/fluid.cgi?T=293.15K&PLow=5+Mpa&PHigh=5+MPa&PInc=&Applet=on&Digits=5&ID=C7727379&Action=Load&Type=IsoTherm&TUnit=K&PUnit=MPa&DUnit=mol%2Fl&HUnit=kJ%2Fmol&WUnit=m%2Fs&VisUnit=uPa*s&STUnit=N%2Fm&RefState=DEF
Change an axis to density, Change the other axis axis to enthalpy, click
on the single point shown, read off a density of 2.0638 mol/l and an
enthalpy of 8.2153 kJ/mol.
We started with 0.1 l, or 0.20638 mol of nitrogen. and we also end up
with 0.20638 mol. Our final volume is 1 liter, so our final density is
0.20638 mol/l.
Go back two pages to the thermophysical properties page, check the
isochoric properties box, click to continue. Enter density = 0.20638
mol/l and some sensible temperature ranges, eg 100K to 300 K, press for
data. We get:
http://webbook.nist.gov/cgi/fluid.cgi?D=0.20638+mol%2Fl&TLow=100&THigh=300&TInc=&Applet=on&Digits=5&ID=C7727379&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
The final state must be somewhere along that line, to find out where we
need to use the enthalpy.
To find out where, change the X axis to enthalpy, and find a point where
the enthalpy is close to 8.2153 kJ/mol, the initial value of enthalpy we
found 4 paragraphs above. The enthalpy has not changed, as the process
is isenthalpic. There is a point at 8.2144 kJ/mol, MPa - so the final
pressure is close to 0.48599 MPa.
Change the X axis back to temperature, find the point at 0.48599 MPa and
read off the final temperature of 283.67 K. These are the final conditions.
You can get an exact temperature from a pressure (and density) using the
NIST pages, but I don't know how to find an accurate pressure from an
enthalpy (and density). However, you can zoom in on the range to get
more points on the curve if more accuracy is needed.
So, back to the question. 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 CO2 at 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
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)
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