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FuelingI'm starting to get into hybrids and I wanted to know if I'm actually understanding how to properly fuel it.
I probably wouldn't be able to go past 8x mix with the one I'm starting to plan out, so (as I've heard) a metering pipe won't really be worth making; that being said I'll have to measure the fuel directly into the chamber right? And then when measuring out the fuel I'd use something like ((Chamber volume)x(Mix number))/.958*((Chamber volume)x(Mix Number))=Cubic inches of fuel to find how many cubic inches of fuel I need to put into my chamber, now I need to find out what percentage of my chamber the fuel is taking up so I'd do something like (Cubic inches of fuel)/(Chamber volume)=% taken up by fuel to find what percentage of my chamber is taken up by fuel. Then I'd take the number I got from that ^ and multiply it by the pressure of one atmosphere (about 14.7 PSI) to determine how much pressure of fuel I need to add into my chamber (% taken up by fuel)x(14.7 PSI)=Pressure created in chamber by fuel *If I'm remembering correctly the .958 comes from the stoichiometric ratio of 95.8:4.2 So just to make sure I'm understanding this, lets say I have a 100CI chamber and want to go to 8x mix; so using the formulas above, I'd first inject about 5 PSI worth of fuel, and then pressurize it up to 7 atmospheres (about 103 PSI) with air (so the total pressure is now about 108 PSI)? Thanks in advanced!
Whatever the mind can conceive and believe, it can achieve
Re: FuelingThis is your goto thread: hybridcannons/topic13602.html
Alternatively, have you considered syringe fuelling? Much simpler than a meter pipe! howtodatabase/topic22301.html
Re: FuelingThank you! I've read through that a couple of times, and I'll probably go through it a couple more (the 101 thread, I think with my first hybrid I'd go with manometric fueling)
Whatever the mind can conceive and believe, it can achieve
Re: FuelingFor manometric, you don't need to work out the chamber volume. You'd need to work out the PSI required for the specific fuel (propane/MAPP/butane) but thankfully this has been done by other people so you just need to do basic math.
I believe you use 0.644PSI of propane per mix. So a 10X mix would measure out 6.44PSI of propane in the chamber. (14.7 x 9) = 132.3PSI of air would also be added. For MAPP, the mix number is something like 0.72 but I can't recall exactly so you may have to look around for that number. Not sure if someone has done the calculations for butane. Anyway, that number (0.644PSI or 0.72PSI) is the same regardless of chamber size when using manometric metering.
Re: FuelingOoh okay thank you! For some reason I took the second part of the Hybrid Fueling 101 thread as being manometric (instead of using a meter pipe), this makes perfect sense now. I feel stupid now xD
Whatever the mind can conceive and believe, it can achieve
Re: FuelingNo, you're right. The second part is for manometric meters but it is the entire process for calculating the pressure required. It is the formulas required to arrive at the fuel pressure required for a given mix. However, because that fuel pressure for a given mix is the same in every manometric hybrid, regardless of chamber size, you don't need to do the calculations unless you're using a fuel that's not propane/MAPP/butane (as these fuels have already been calculated). This is the advantage of manometric metering; I can use the same manometric fuel meter on both my hybrids (one has a chamber of 1.4L or so, the other only 0.35L or so). So if you're using propane, you know that each mix requires 0.644PSI of propane in the chamber. That's all you need to know. 1X = 0.644PSI propane 2x = 1.288PSI propane + 14.7PSI air 5x = 3.22PSI propane + 58.8PSI air 10x = 6.44PSI propane + 132.3PSI air You may want to check what atmospheric pressure is at your location. Mine is closer to 14.5PSI (sealevel) so I use that instead of 14.7PSI. Won't make a huge difference but it could help if you're having trouble igniting mixes.
Re: FuelingThank you, this has helped a lot!
Whatever the mind can conceive and believe, it can achieve
Re: Fueling
Re: FuelingThere are two slightly different definitions of the X number in wide usage:
1. amount of fuel for 1X is equal to the amount which could be burned at the desired mix ratio if one started with a SEALED chamber full of air at 1atm and injected fuel until the desired mix ratio was reached. Under this definition, a 1X mix is actually at slightly higher than atmospheric pressure. 2. amount of fuel for 1X is the amount required to achieve the desired mix ratio with a charge pressure of exactly 1 atmosphere. The whole "X" thing is a bit ambiguous. For example, if you use oxygen rather than air as an oxidizer, both definitions cause deflagration pressure much lower than would be seen if the oxidizer was air. It's much less confusing to just specify a mix ratio and a starting pressure. Here's how I do hybrid fueling now, using the example of a propane/air mix: Using GasEq, I'll enter the reactants. In this case oxygen, nitrogen, and propane. Start off with 21 oxygen, 79 nitrogen to approximate air. Enter a number for propane, and you'll see the "mole fraction" displayed on the right (calculated as f=n_C3H8/(n_O2+n_N2+n_C3H8)). Choose the "Adiabatic T and composition at constant V" option to model the case of a closed combustion chamber. Choose a product set from the dropdown menu (HC/O2/N2 extended works well for most applications, but you may need to add some, especially if you have something like helium in your reactant list which isn't in the product list), enter a starting temperature and pressure, and click calculate. This will display the postcombustion information on the right side. Using this method you can find the mix ratio you want, based on the desired properties of the combustion products. Then alter the starting pressure until you find the firing pressure you're looking for. You now have a starting pressure, and a mole fraction of fuel. Assuming that the fuel and oxidiser obey an ideal gas law (a good assumption in your case), and that your chamber starts off full of air at 1 atmosphere of pressure: P_fuel=mole_fraction_fuel*P_charge (multiply by 14.7 for pressure in psi) P_final=P_charge Simple, eh? An example: we start with a chamber full of air, and have chosen a mix ratio of 0.05:0.95 (fuel:air). We want to reach a maximum theoretical pressure of 100atm (absolute). Using GasEq, we discover that we need a charge pressure of 10.34atm (absolute). So, using the fuel pressure equation, we find that we need to add fuel until the chamber pressure is 0.517atm (gauge). After the fuel addition, we simply raise the chamber pressure to 9.34atm (gauge) with air, and we have our perfect 0.05:0.95 mix. This happens to be in the neighbourhood of 10X, but we never needed to know that. Incidentally, you will actually find that, in most cases, the "stoichiometric" mix is not the one that will provide optimal performance. Slightly rich mixes tend to provide higher peak pressure for a given charge pressure as well as yielding slightly lower molecular weights in the combustion products. You can investigate this effect for yourself using GasEq.
Spudfiles' resident expert on all things that sail through the air at improbable speeds, trailing an incandescent wake of ionized air, dissociated polymers and metal oxides.
Re: FuelingThank you! I'll just have to figure out how to use GasEq now xD
Whatever the mind can conceive and believe, it can achieve
 
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