Propane pressure?

[frag68]

I'm trying to calculate the pressure of a propane/air mixture after ignition, given a fixed volume. To be honest, I don't know where to start. I looked at using the constant volume leg of the Otto cycle as a reference, but I don't know the final temperature of the system. Assumptions would be that the system starts at atmospheric pressure and standard temperature (80F or 300K). Anyone have any thoughts that could point me in the right direction? Thank you!

 

[mbeychok]

frag68:

You really have not explained your problem in detail. I assume that you are talking about the pressure in an engine cylinder using propane fuel. Is that correct?

I also assume that by a "fixed volume", you meant the volume of the cylinder. Is that correct?

If both of my assumptions are correct:

--You must first know the amount of propane and of air that is admitted into the cylinder before ignition. Knowing those amounts, you could calculate the adiabatic flame temperature and the volume of combustion gases (i.e., carbon dioxide, water vapor, nitrogen, and oxygen). The nitrogen enters with the combustion air, and the combustion gases will include oxygen if your engine uses excess combustion air over and above the required stoichiometric amount.

--Knowing the temperature and the volume of combustion gases contained in the fixed cylinder volume, you could then use the ideal gas law (PV=nRT) to calculate a first estimate of the resultant pressure. If that pressure is fairly high (which I assume that it is), then you might want to calculate the compressibility factor (Z) of the combustion gas mixture and recalculate the pressure using PV=ZnRT.

I note that you are a mechanical engineer. The above methodology is fairly complicated and I would advise you to seek help from an experienced chemical engineer if you are unfamiliar with the above described calculations. Or perhaps an experienced automotive design engineer could help you.




Milton Beychok
(Visit me at www.air-dispersion.com)
.

 

[StoneCold]

frag68

If you are looking for a swag kind of answer you can go with seven times the absolute initial pressure.

I will probably get hammered for simplifying it down to that.

Regards
StoneCold

 

[frag68]

Actually, I'm using propane as a projectile propellant. But regardless, the ignition chamber volume is constant. I appreciate your replies. I'm not certain of the first methodology you described mbeychok. It seems like this should be a fairly straight forward calculation. For some reason I was thinking PV = m Cv (T2-T1) and solving for P. But I don't know T2.

 

[sailoday28]

See a good thermo book or a site suchas

http://dunx1.irt.drexel.edu/~hp37/L3.ppt#277,22,Slide

22

Regards

 

[mbeychok]

frag68:

Once the propane and air are ignited and burned, you no longer have propane and air in the rocket chamber ... you then have the combustion gases. (1) So you must first calculate the volume of the combustion gases. To do that you must know how much propane and air entered the rocket chamber. (2) Then you must calculate the temperature (i.e., the adiabatic flame temperature) of the combustion gases. (3) Then you can calculate the pressure using the gas law.

That will give you the first approximation of the pressure as I said before. Then you can make a better estimate by using the compressibility factor. Then you could go even further by taking into account the pressure drop between the external ambient air pressure and the internal rocket chamber pressure as it travels through the chamber outlet Venturi. Also keep in mind that external ambient pressure decreases as the rocket gains altitude.

It is definitely not a "fairly straight forward calculation"! Surely, NASA has probably got a computer program for such calculations. Why not contact them?


Milton Beychok
(Visit me at www.air-dispersion.com)
.

 

[25362]

The "theoretical" temperature of (perfect & complete) combustion for propane when burned with air (no xs air) would be 2115^oC.

However, taking into account the heat absorbed by the dissociation of water and carbon dioxide, at atmospheric pressure and above 1600^oC, I'd say the following "adiabatic" temperatures could be used as an approximation (taken from an old book of mine):

% xs air ^oC
0 1980
20 1800
40 1650
60 1500
80 1380
100 1270