# Combustion in a closed system.

 Register Blogs Members List Search Today's Posts Mark Forums Read

 December 6, 2004, 15:10 Combustion in a closed system. #1 Jim Guest   Posts: n/a Dear all, If I have a completely closed system (a cube with all six sides walls) that contained a premixed fuel and air (at stoichiomteric ratio) and I 'burn' it, what happens to the density and the pressure? The volume is fixed and so is the mass. Therefore, if the density of the mixture drops due to combustion, then the volume has to increase unless the pressure increases and forces the density back to its original value. Is this correct? Does this occur instantaneously? What is the interaction (and time history) of the pressure and density.

 December 7, 2004, 06:29 Re: Combustion in a closed system. #2 matej Guest   Posts: n/a HI, It depends also on the heat transfer to the walls. Another important parameter is wall-area to vessel volume ratio, especially with small devices. When you remove the heat from the vessel during the combustion (or after) the pressure -- so the density will go down, but not exactly to the original state. with a simple search on sciencedirect.com I run into interesting article which might help you: A. K. Oppenheim and A. L. Kuhl, Dynamic features of closed combustion systems, Progress in Energy and Combustion Science, Volume 26, Issues 4-6, August 2000, Pages 533-564. (http://www.sciencedirect.com/science...56c0c032a5dbd1). I haven't gave you the answer, have I? )) matej

 December 7, 2004, 07:10 Re: Combustion in a closed system. #3 Jim Guest   Posts: n/a Thank you. It is an interesting paper. I am more interested in the case where the system is fully closed and there is no heat transfer through the walls. There is also no work done by any increase in pressure as the box is perfectly rigid. This may not be possible to achieve in reality, but I would still appreciate any observations on the likely behaviour of the density, pressure and temperature in such a box.

 December 7, 2004, 10:52 Re: Combustion in a closed system. #4 MLD Guest   Posts: n/a You would need to know about the thermodynamics of your reaction to calculate teh temperature. If you know this and you can guess your reaction products (or at least how many moles of reactant go to how many moles of gas), you can calculate the pressure from the ideal gas law. As for density, again you need to know the reaction products and calculate the mass. There has been work done by the University of Leeds chemistry group (really a definitive source for combustion researhc) on closed combustion behaviour. One of teh simplest systems being H2 + O2. I would suggest looking for authors: Griffiths, J; Gray, P; and Scott, S.K. An excellent monograph by Gray and Scott ("Chemical Oscillations and Instabilities") gives details of the extent of research at Leeds. All the best M

 December 7, 2004, 15:39 Re: Combustion in a closed system. #5 Jim_Park Guest   Posts: n/a If the system is closed (no mass enters or leaves), then it seems the density, which is mass / volume, can change only due to change in system volume. That is, the volume of the reaction chamber. A perfect chamber would be constant volume (one of your listed assumptions), so the density is constant through the process! This of course allows spatial variation of density within the volume - but the overall volume-averaged density has to be constant. What am I missing?

 December 9, 2004, 06:08 Re: Combustion in a closed system. #6 paul shaw Guest   Posts: n/a dear jim, does this restriction on the density prevent combustion from occuring though? if average denisty cannot change then the fuel can not burn - is this correct?

 December 9, 2004, 11:57 Re: Combustion in a closed system. #7 Jim_Park Guest   Posts: n/a I think it's pretty fundamental. If you have a fixed volume and a fixed mass (didn't you assume this?), the average density must be constant. There's a lot of mischief hidden in "average." Certainly, at the ignition point and time, the pressure and temperature will shoot up as energy is released. What does this imply for the density at the ignition point? Depends on your equation of state. For the density to change at the ignition point, mass must move within the volume. If the density goes up at the ignition point, it has to go down elsewhere. And vice versa. Otherwise, you're creating or destroying mass.

 December 9, 2004, 14:49 Re: Combustion in a closed system. #8 paul shaw Guest   Posts: n/a so if the density goes down somewhere (due to combustion) the density elsewhere must go up (it will be compressed?)? what if we assume incompressiblily (in that density is not a function of pressure). what would happen then. the highest density is likely to be the pure fuel, so how can it go up any further??? thankyou jim.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Mohan CFX 20 March 30, 2011 18:56 JB FLUENT 2 August 1, 2008 04:16 Martin Castillo FLUENT 3 June 20, 2007 21:38 Martin Castillo FLUENT 0 June 18, 2007 22:56 ypbanjare FLUENT 1 March 12, 2007 12:34

All times are GMT -4. The time now is 06:35.