SpudFiles

I would just like to know what percentage of the theoretical combustion pressure in the combustion chamber, in a hybrid, you have to calibrate the pressure behind the pilot?
I mean for example: If my pressure in the combustion chamber at the instant of the explosion, is 15 bar, the valve should open up to 50% of that pressure? 80% or much?
I'm rehearsing my first hybrid and although I have looked for this item here, I have not been lucky. Any advice will be very important to me. regards

You might do some searching on burst disk pressure. Back in the day, I know that there was some testing done when layered tin foil disks were the cats meow. But basically what was discovered is that if the opening pressure is too high, the pressure spike would already be starting to diminish before the projectile left the barrel. I do not think there is a hard and fast rule about having the valve open at XX% of max pressure, as the timing will change based on barrel length and total burn time. A shorter barrel will want a higher opening pressure, while a long barrel will want a lower pressure.

When I worked it out for my piston hybrid I did it by calculating the forces involved and then gave a bit of room for error.

I understand what you explain me. It will be a matter of trying gradually to draw valid conclusions for my gun.
I build a hybrid mini 3X, .22 caliber and a long 470mm cannon and noticed, on the evidence, when adjusting a little spring closes the valve of the combustion chamber, the power of the shot increases pretty ... that's why I have curiosity about this specific issue.
Thank you very much for the answers and explanations.

With respect to the maximum pressure (theoretical) of hybrid ... As this estimate?
Let's say my weapon test has a volume of the combustion chamber of 100 ml, and the mixture is compressed to 4X, 3.5% of butane gas. What would be the peak combustion pressure?
I'm interested in formulas normally used to estimate this.

MrCrowley wrote:http://www.thehalls-in-bfe.com/HGDT/

ok I'm reading it. Thank you

It is interesting the program but I need to understand how this pressure is calculated, albeit in a basic form, as influenced by the temperature, the air expands etc.
Say my combustor have 100ml air at atmospheric pressure. When I will compress 4 times 500 ml of air, 100 ml that had more 4 times this, 400ml), plus 17.5 ml of butane gas, 3.5%, and at room temperature, say 25 Cº.
When the mixture is ignited, at what temperature? I do not know, it expands, as volume up ?? Logically, as the combustion chamber still the same size (100 ml), the pressure will increase. The more heat more pressure, and the more fuel air mixture exists for those 100 ml, also, I have more pressure on heating this.

Why not just use HGDT and have it work that out for you? It provides estimates of peak combustion pressure

MrCrowley wrote:Why not just use HGDT and have it work that out for you? It provides estimates of peak combustion pressure

Simple curiosity only. I'm interested in how this issue occurs.

hectmarr wrote:With respect to the maximum pressure (theoretical) of hybrid ... As this estimate?
Let's say my weapon test has a volume of the combustion chamber of 100 ml, and the mixture is compressed to 4X, 3.5% of butane gas. What would be the peak combustion pressure?
I'm interested in formulas normally used to estimate this.

A "good enough" answer is that the peak pressure is just the hybrid number times the peak pressure in the 1X case. For propane in air (IIRC) the peak pressure is about 135 PSIG. For a 10X it would be 1350 PSIG. (someone else can chime in with this should be calculated using absolute pressure or gauge pressure.) The amount of energy in the chamber is linearly related to the amount of fuel (assuming a stoichiometric amount of air). The peak pressure in the chamber is linearly proportional to the energy (assuming no heat loss) and the number of molecules. For propane in air the change in the number of molecules on combustion is small (6 to 7 for the reactants and products but that is diluted by the 80% of the air that is nitrogen so the change in number of molecules is only about 3% and I would just ignore that factor).

Hectamarr-
If you are interested, there is a very (very very) old, but excellent, tool for doing gas law and combustion calculations called GasEq (http://www.gaseq.co.uk/).

jimmy101 wrote:Hectamarr-
If you are interested, there is a very (very very) old, but excellent, tool for doing gas law and combustion calculations called GasEq (http://www.gaseq.co.uk/).

I'm seeing gaseq. I try to download it and be curious with this, although old tool, it seems very interesting. Thanks for sharing the link.
I understand what you explain me.
I try to remember what is apprehended in school about the ideal gas law, to try to understand the basics these phenomena.
One question: What approximate temperature takes place in a combustion chamber of these weapons using butane or propane and air?
The point is that knowing this temperature can know the volume of a gas (air) to that temperature, and knowing the latter, could approximate the resulting pressure depending on the size of the camer combustion.
Logically these very common and known laws are for ideal gases but are also approximate.
  Gay-Lussac realaciona temperature to the volume of a gas and Boyle Mariotte, relates the volume with pressure.

Gaseq will calculate both the pressure and temperature of the combustion mix. The flame temp for propane in air is about 2000C.

Gay-Lussac's and Boyle's laws (and Charle's law) are all incorporated into the ideal gas equation. If you know PV=nRT you can derive the three other laws.

Have fun!

jimmy101 wrote:Gaseq will calculate both the pressure and temperature of the combustion mix. The flame temp for propane in air is about 2000C.

Gay-Lussac's and Boyle's laws (and Charle's law) are all incorporated into the ideal gas equation. If you know PV=nRT you can derive the three other laws.

Have fun!

Indeed, the results give very close to what you told me.
For possible hybrid 4x, and 100 ml volume combustion chamber, for example, and approximately, it would be as follows:
135 psi multiplied by 4x = 540 psi, as you have me.

Deputy what I figured with the general laws of ideal gases below.
My result is 561 psi. Enough "accuracy" for me with these results, even if a small gap between them.
Thank you very much for the clarification and information. Regards

http://picasion.com/i/2kYEx