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Injection PsiI understand most of the main principles behind a hybrid but i still have a few questions that I have yet to figure out. First of all, let me get this straight (please tell me if im wrong about anything, i don't really wana blow myself up)
For a standard 4x hybrid launch: 1) Fill propane meter (meter is 4.2% of chamber volume) then inject into chamber 2) Fill propane meter again then inject 3) Fill propane meter again then inject 4) Fill propane meter again then inject 5) 4.2 X 4 = 16.8% propane in chamber 6) To reduce the propane percentage back down to 4.2: pump in 3 atmospheres of air (since one already exists? im not sure; also using 14.7 as the accepted pressure of the atmosphere) making the chamber pressure 44.1 psi 7) Fire! Just wondering if i have the process cemented into my head enough to write it down from memory (plz critque my procedure) Questions: 1. What pressure would i want to inject the propane at. In other words, what pressure is the propane meter when injected? 2. Propane Regulation? would this be a good idea, eliminating the need for a pressure gage in the propane meter?kinda somewhat the same as the first question 3. Ignition: I DON'T want to stress my chamber and increase the risk of explosion, so what is the best way to ignite the mix? Through a pipe nipple between the meter and chamber (in the front? middle? or end? of the chamber)? Also, i believe im going to attempt the homemade schradervalvesparkplug i've seen on here, I would need two of them, a close distance apart, completing the circuit and creating a spark in the chamber to ignite...right? 4) Final question at last; I've heard about DDT although almost always a side thought, i know it creates extreme pressure/heat and will destroy your cannon and possible kill you. I've heard the best preventative measure is multiple spark gaps? Is this true? if so, where do they go and are they in one circuit or two seperate ones? I seriously apologize for long rantim bored (It's raining here and I can't go outside to shoot or nothing) and YES! i have searched some and came up with this. Im not an unexpreienced noob spudbuilder, ive built many launchers: copper and pvc, piston, ball valve, sprinkler valve, and QEV. Bu I must admit that i have never experimented with combustions or hybrids so please help me hehe. Again...sorryanyone who bothered to read this and help me out, i seriously admire you
Hey Isomer, what a first post! To start with you are correct with your loading procedure. To figure out the pressure needed in your propane meter you need to know the capacities of your chamber and meter. I am not the best member to offer answers in regards to calculating this pressure as i do not use meters on my Hybrids.
The best idea would be to give us an idea of the size of launcher you wish to build. Normally i would suggest that you build a few pneumatics and combustions before undertaking a Hybrid build, however judging by your post you seem like the type of person that could handle it. As long as the construction techniques and materials used are ok a Hybrid is as safe as any other launcher. Good luck with your spudding future and welcome to Spudfiles.
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Nice first post Isomer.
The injection pressure doesn't really matter, so long as it is above atmospheric pressure (0 psig or 14.7psia), and below the pressure of the propane tank (usually between 90 and 120 psi, although it can drop to below 60 on a really cold day, and climb over 150 in a hot car). Your meter should be designed accordingly. Propane regulators, while expensive, difficult to obtain, and complete overkill, are a nice bonus to a hybrid, adding to its overall ease of use. While DDT used to be a major concern in hybrid construction, further research and testing by our members has indicated that it is a minor worry. Exceptions to this rule include: Cannons with exceptionally long and thin chambers. A possible DDT was created with a 6x propane/air mix in a 1/2"x60" chamber. Cannons with exceptionally high energy ignition systems (think kW+ sparks, unobtainable with normal equipment like stunguns) Cannons using very brissant fuels like hydrogen, or even worse, acetylene. Acetylene can DDT very easily Cannons using very high pressure, pure oxy/fuel mixes. Still difficult to DDT It is a common misconception that DDT will destroy cannons completely. It places extreme wear on all components of a launcher, and will cause steel pipes to fatigue and rupture after a few shots. Good luck with a hybrid, and if you value your limbs, don't use plastic pipe (especially not PVC).
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.
thanks you for responding:
@Novacastrian: I am planning for the chamber to be a 12" piece of schedule 4020.5" steel pipe The volume (not including the reducers, space from chamber to unionvery small as barrell and union will be half inch, or space from chamber to ball valvegonna be used as injection for fuel/air) would be: 1.25*1.25*3.14*12=58.88 square inches + 23 square inches to account for extra spaceusing 62 square inches making the total fuel meter volume roughly 2.6 square inches? im not sure though, got distracted in the middle of calculating thats a rough calculation of my size of launcher And i have made many, many pneumatics just not combustions @DYI thanks a ton for the pressure injection. I was thinking of just using a basic air regulator from lowe's? I thought i saw someone else do something similar or would i have to get a propane specific one ***If someone could point out some reliable, easy ignition systems capable of up to 6x and how to instal for consistant ignition including level of power and maybe even a circuit diagram but if not, it's ok i can do somemore research, i get bored quickly lol ****Another injection pressure question: since the presure of the propane would directedly affect the amount of propane squished into the chamber would i add my "X" air pressure ontop of the injection pressure of the propane? and how would preexisting atmostpheric pressure fit into that equation, lol **Simplified** A=atmostpheric pressure P=pressure at which the propane is injected C=fireready chamber pressure X=amount of air pressure added for certain level of "X" So, does C=P+X+A? im not sure EDIT: What im really saying is if i should add my desired level of air pressure to the pressure of the injected propane and put in that amount?
For each "x", the fuel injected will add ~0.5psig to the chamber. Thus, a 6x mix will be at a total of (14.7x6)+(0.5x6)= 91.2psia, or 76.5psig (what will show up on a normal gauge, which is calibrated to read 0 at the ambient pressure of ~14.7psia or 1atm). Unfortunately, a bit of thought will reveal that this isn't exactly accurate, since the fuel added is 3.5% of the initial volume, not 3.5% of the (volume+fuel).
I just figured out using logic and grade 9 math how to solve this. There may be a more advanced an easier way if you know more math. My method: Chamber volume=10ci. After propane injection, chamber should be, according to my knowledge of stoichiometry, 3.5% propane for best combustion, and 96.5% air. So, 10/96.5=0.1036..., x 100 = 10.36... 10.36 is what the total ci of gas would be at atmospheric pressure. Compressing 10.36ci into 10 ci of space yields a pressure 1.036... times greater than atmospheric, which means 14.7 x 1.036, which = 15.233 psia, or 0.533 psig. So instead of .5 psi per "x", you need .533 per "x". This means, for a 6x mix, you will actually have a final pressure of (14.7x6)+(.533x6)=~91.399psia, or ~76.699 psig for a 6x mix. Unfortunately, you have no idea how propane meters work. I believe I mentioned earlier that the injection pressure doesn't matter, as long as it will flow, and isn't above the critical point of propane. But basically, you should have 0.533 psi of propane per "x", and 14.7 psi more air per "x". For your chamber size (assuming 60cubic inches, not square inches), this translates to: Per "x", you need 2.176 ci of propane (at 1 atm). This means that if your meter was 2.6 ci, it wouldn't work, because its pressure would be lower than that of the surrounding atmosphere to get the right amount of propane. If you used a 0.5 ci meter, it would need to be filled to ~62 psig for every "x". As long as the air was filled through the same port as the propane, you wouldn't need to subtract the meter volume from your calculations.
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.
OK, DYI, I apologize for my ignorance and everything but that last post just threw me off quite a bit.
1) Could you reexplain the equation (including reasons) for the .533 psi of propane per "x". I understand the math involved, but I don't understand why, most likely a problem in the understanding of basic principles. 2) I don't understand how i "have no idea how propane meters work" because my fault with the system is that if i inject varying pressures of propane, it's the same volume but different raw amounts, isnt it? EDIT: I think I could solve this by making a meter smaller than the required ci of propane per "X" and calculating the psi for the correct amount of propane? 3) Could you diagram the equation to arrive at 62 psi, i can't quite grasp it. Also, im assuming .5 ci meter was used for ease of math and to come out with a psi that would be greater than atmospheric and less than the propane bottle itself? EDIT #2: I figured out the equation for the 62 psi: [2.176/.5]14.7 reason: amount of propane/meter volume*atmospheric pressure=psi About your last paragraph: I came about 2.6 ci because i used 4.2% as the optimal propane percentage for combustion instead of 3.5 4) Would it be a better idea to increase my chamber volume to arrive at a practical meter volume with a good psi? **To make answering easier, I've numbered Q's so if you could answer that way it would be awesome** Thank you SO much for answering and helping me out, I really appreciate it!
1. Assuming 21% O2 in air (which may be a bit off), the correct percentage of propane in air for a complete burn would be 3.5%.
Assuming a 10ci chamber for the purpose of illustration, if you added 3.5% of 10 ci, you would have a lean mix, because then you would have 10.35ci total, but only enough propane for 10 ci total. So, you have to find what 10ci (the amount of air) is 96.5% of. You do this by dividing 10 by 96.5, and multiplying the answer by 100. After you have that number, you calculate how much less space it is being forced into. It is being forced into about 1.04 times less space. You multiply 14.7 (ambient pressure) by 1.04, and subtract 14.7. The answer is .533, which is the psi of propane needed for every "x". 2. Your assumption in how to fix your design is correct. Which is exactly what I explained. If you need 2.1 ci of propane, you need a meter with less volume than 2.1ci. 0.5 ci would be convenient, so I used it in my example. 3. 0.5 is 4.2 times smaller than 2.1. You are forcing 2.1ci of gas into 0.5ci of space. Multiply ambient pressure (14.7psia) by 4.2 to get 61.74 psia, or 47.04 psig. About the 4.2% number. After some basic calculations, it can be seen to be flawed. My reasoning: Stoichiometric propane combustion: C3H8 + 5O2 > 3CO2 + 4H2O Therefore, one C3H8 for every 5 O2. Since that equates to 1/6th (or 16.66...%), then propane would combust perfectly at 16.66... percent in pure oxygen. Divide that by ~5, and you get 3.333, which you can substitute into the equations if you like. I used 3.5 to simplify things, which would equate to 21 percent O2 in air rather than the more likely 20%. 4. A 0.5 ci meter is quite doable, with precise measurement. Unless you're building something the size of FEAR (don't ask, search), you'll need some measure of accuracy anyway, and traditional meters have been built for hybrids 1/4 the size of the one you have planned. For a 1/4" ID meter pipe (and 1/4" pipe isn't really 1/4", although some 3/8" and 5/16" tube is), it would need to be about 10" long, which isn't exactly miniscule. I have no idea where the 4.2% propane came from, but exceedingly simple calculations seem to prove it to be completely wrong.
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.
<a href="http://www.spudfiles.com/forums/accuratelymeasuringyourchamberandmetert9899.html">Read this.</a>
You can make the propane meter any size you want, just follow the directions in that topic and ask if you have any other questions. When you find the correct meter pressure for a 1x mixture (no compressed air), the rest is extremely easy. Correct mixtures: 1x: 1 shot of propane 2x: 2 shots of propane + 14.7 (15) psi 3x: 3 shots of propane + 29.4 (30) psi 4x: 4 shots of propane + 44.1 (4445) psi 5x: 5 shots of propane + 58.8 (5860) psi 6x: 6 shots of propane + 73.5 (7374) psi
doesn't the propane meter have to be equal to or smaller than the volume of propane going into the chamber because it would be lower than atmospheric pressure if it wasn't? not to start of on a bad note
Thanks everyone for the great posts, helped me not destroy my future spudding career, although i still haven't sold the 'rents on my designs Another question: Hybrids can funtion as very calculated combustions in a 1x mix correct? Would all the calculations for a hybrid really shine when compared to a "X amount of sprays and ignite gun? TYTY AGAIN im probably buying all my parts tomorrow, YAY! EDIT: noname, mcr had the #1 vid in 2007 for famous last wordsjust thought i'd say and i too think they're nonmetal but they have some catchchy toons. Sorry for goin OT just thought i'd say
Wow, I didn't realize you could have a total volume greater than 100%! Hmmm, 100% oxygen and 16.6...% propane, this is so... wait a second... I think you mean divide by 4 to get the required proportion of propane in atmospheric air. This yields a result of 4.15%, which people have decided to round up to 4.2%.
Based on the actual stoichiometric ratio (95.85% air, 4.15% propane), the chamber pressure will actually rise a quantity closer to (14.7*0.0415) = ~0.61PSI per "shot" of propane.
Not necessarily. My hybrid launcher's meter is ~30% of my chamber's volume. However, my filling/venting method leaves an atmosphere of air in the meter rather than an atmosphere of propane, which enables me to use the scale on my pressure gauge as if it read in absolute pressure.
People should not be afraid of their governments. Governments should be afraid of their people.
SB15, propane would make up 16.6%, and oxygen would make up the rest. 1/6 + 5/6 = 6/6 = 100%. 100% is not >100% as far as I know.
Air is about 1/5 O2, so it would seem logical to divide the figure by 5. If propane/O2 was 16.6% propane, wouldn't propane air be 3.3%? I know you're probably right, but how exactly did you arrive at that number? You have calculated two different burst pressures for the same pipe, differing by ~100%, so I'm kind of inclined to wonder. That, and the fact that your calculated ratio actually give 0.636 psi of propane per "x", assuming an airtight chamber. 0.62 psi would only work if the propane displaced the air from the chamber, which it can't do in a hybrid.
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.
Well, I should have looked over your calculation more closely, as more was incorrect than I originally observed.
As you said, the correct ratio of oxygen to propane is (5/6)% : (1/6)%. Atmospheric air is ~21% oxygen. (1/6)/((5/6)/21) = 4.2% When did I calculate 2 different burst pressures for the same pipe? I have a feeling we are not discussing the same issue in terms of the pressure rise caused by the injected propane, but I am too tired to read the thread again.
People should not be afraid of their governments. Governments should be afraid of their people.
Ah, thanks for clearing that up.
If you injected the amount of propane needed for 10ci into a 10ci air filled, airtight chamber, you would have a lean mix, because you would have enough propane for 10ci total, not for the new total which is caused by adding the propane in the first place. Therefore, you need to figure out what 10ci is 95.8% of. Then you take that number, subtract 10, and you get the final amount of propane in ci that you really need. Although this problem is unnoticeable with smaller mixes in large chambers, it could become very problematic with higher mixes.
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.
 
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