Registered users: Bing [Bot], Exabot [Bot], Google [Bot], Google Adsense [Bot], Yahoo [Bot]
 
User Information


Site Menu


Sponsored


Who is online
In total there are 82 users online :: 5 registered, 1 hidden and 76 guests Most users ever online was 218 on Wed Dec 07, 2016 6:58 pm Registered users: Bing [Bot], Exabot [Bot], Google [Bot], Google Adsense [Bot], Yahoo [Bot] based on users active over the past 5 minutes 

The Team
Administrators
Global Moderators


Sponsored


Propane Injection Math  An Exception to Like Terms?For a while, it has bugged me that propane injection math supposedly did not follow a very important rule in math and science: the units were not in like terms.
The simplest version of an equation for propane injection governing is simply a rearranged and slightly modified version of Boyle's Law, that is, P<sub>1</sub>V<sub>1</sub> = P<sub>2</sub>V<sub>2</sub>. In our context, rearranged, it would look like this: <div align="center"></div> Where P<sub>m</sub> is in the form of atmospheres, an absolute measurement of pressure. Since P<sub>2</sub> is equal to one atmosphere, the numerator of the fraction is simply 4.2% of the chamber volume. For so long, it has grieved me supremely that this equation does not work for spudding purposes. For some reason, the popular, and correct, I might add, equation has put the left hand side of the equation in gauge terms. This means, that if you come up with 1 atmosphere as the answer, you would read that on a gauge, as 14.7 PSI. This is 2 atmospheres. Thinking about it further, I think I've developed a hypothesis: When injecting propane, the average spudder closes his injection valve right after opening it. This leaves one atmosphere of propane left inside of the meter. So, if you were to simple use the correct absolute answer from the equation above, and adjust it for gauge terms, (subtract an atmosphere), you would be short in propane volume proportional to the meter volume. Rearranging Boyle's law again into the rearranged form and setting V<sub>2</sub> = V<sub>1</sub>, the terms cancel out leaving P<sub>2</sub> = P<sub>1</sub>. <div align="center"></div> So, to add another meter volume's worth of propane into the chamber to make up for this, we simply use the equation: <div align="center"></div> Where P<sub>m</sub> is in atmospheres. This equations yields the correct absolute pressure that the meter pipe needs to be at in order to obtain a correct fuel mixture. But, since we spudders rarely use absolute pressure, we can simple use the first equation and say that P<sub>m</sub> is in gauge terms. Thanks for listening to my hypothesis. While I'm here, I might as well say it now: this equation will have a percent error. This is because if you inject into a closed chamber, (which is what the value of 4.2% was chosen for, which is for injection into a closed system), the pressure will increase slightly, throwing off of equation by the slightest bit. This little error is not enough to bug me, and a few minutes of simple algebra could compensate for it, but by simply replacing 4.2% with 4.03%, and changing your system to an open one which displaces air as you inject propane, you can obtain the exact value you may be after. Any comments or theories of your own are welcome, just make sure you at least prove what you're saying before yelling at me that I'm wrong.
What would be the typical error using the existing math?
Sorry if you have answered this in your Hypothisis, my brain starts to swim after i read the words "simple algebra"
America, the greatest gangster of all time. With 200 million odd foot soldiers at it's whim and call.
When you fill your car with refined oil remember that it has been paid for with blood and guts, some from your own countrymen, most not.
That's the thing: there wouldn't be an error using the existing "math", because, like I said, people simply switch up the lefthand side of the equation to gauge pressure from absolute pressure.
This thread was to create a mathematical explanation for why this worked. As you can see, if you remove the 1 from the final derived equation, you can determine that P<sub>m</sub> would then be in gauge terms. The final equation in the post gives the meter pressure in absolute terms. All of this would be helpful to people building "fueltools" for themselves or others, so that they do not hit the same misconception that I did, that P<sub>m</sub> would be in absolute terms simply by using Boyle's law, because this is incorrect.
(Since I cannot read the formulas in the first post)
I use the equation P<sub>m</sub> = (((V<sub>c</sub>*0.042)/V<sub>m</sub>)+P<sub>a</sub>)*PSI<sub>a</sub>, which can be derived from simple logic. You need 0.042% of the chamber's volume in propane. Using a set meter volume, the pressure required (assuming no air is present in the meter) will equal the required volume of fuel divided by the meter volume, plus an additional atmosphere to compensate for the propane left within the meter. This will yield the results in (absolute) atmospheres, which is multiplied by the atmospheric pressure in PSIA, which will yield results in PSIA. Edited due to a mental lapse
Last edited by SpudBlaster15 on Sat Oct 27, 2007 3:19 pm, edited 1 time in total.
People should not be afraid of their governments. Governments should be afraid of their people.
Hmmm, seems to me that the first error is the assumptions the this equation is correct;
P<sub>meter</sub>V<sub>meter</sub> = (0.042)V<sub>chamber</sub> It is close to being correct but shouldn't it really be; P<sub>meter</sub>V<sub>meter</sub> = (0.042)(V<sub>chamber</sub> + V<sub>meter</sub>)
Now you're thinking like I was. That would be assuming the average spudgunner left his meter ball valve open long enough for the gases to diffuse throughout both the chamber and the meter, which would take quite a bit without some sort of fan. Simply put, since you are short a meter's volume worth of propane, you simply tack on another atmosphere. Another meter's volume worth of propane is injected into the chamber, and the mixture is corrected.
EDIT: wait, SpudBlaster.
I get the equation, but this would give you the answer in PSIA, not PSIG. The result of the equation: <div align="center">P<sub>m</sub> = (((V<sub>c</sub>*0.042)/V<sub>m</sub>)+P<sub>a</sub>)</div> Would not result in gauge atmospheres. Without adding an additional atmosphere to the original equation given by Boyle's Law, you would get your required pressure in gauge atmospheres. Adding the atmosphere gives you your result in absolute pressure, which is the point of my work. So, the output of your equation would give you results in absolute PSI, not gauge PSI. The whole point of my work was to create an expression which gives you your required meter pressure in absolute pressure. If you were to use simply Boyle's Law, you would obtain a value that is one atmosphere too low, due to propane remaining in the meter, i.e. : you would obtain the gauge reading. So, to restate that, using Boyle's Law alone will give you the value of gauge atmospheres, even though your input on the right side of the equation was in absolute atmospheres. By adding a +1, (atmosphere), you can correct this to obtain the correct absolute reading. If you desire to obtain the answer in gauge pressure, simply don't tack on the extra atmosphere. And why can't you read the equations in the first post? EDIT II: (Last one I swear!) Here's a little about what I'm talking about. A simple example to show you what I mean. <div align="center"><table><tr><td></td></tr></table></div> A simple Excel file I created with the equation I constructed, (which simply adds another atmosphere), and Boyle's law. The equation which adds an atmosphere gives absolute PSI, Boyle's law by itself yields the correct gauge pressure. Notice that your equation is identical to the equation which yields absolute pressure, not gauge pressure. And, of course, here's a fueltool reading to check against. <div align="center"></div> Remember, as well, this is not a groundbreaking development here. I was simply annoyed by the whole unlike terms thing which arises from propane remaining in the meter. Don't think I'm running around screaming that we've been doing it all wrong for years, because the way we do it actually works, by switching up your LH terms.
Let me get this straight. In simple words: You are saying that the formula we use for propane metering doesn't account for the air that is already in the propane meter when before we add propane in the meter? Is this what you are stating?
Ecclesiastes 1:9  What has been will be again, what has been done will be done again; there is nothing new under the sun.
Mark:
I seem to be having a terrible logicalthinking day. I wrote an excel spreadsheet for personal use using the equation P<sub>m</sub> = ((V<sub>c</sub>*0.042)/V<sub>m</sub>)*PSI<sub>a</sub>, as this results in gauge pressure. This is also the equation I use when demonstrating the validity of my reasoning on the forums. While replying to this thread, I figured I would write the equation to calculate absolute pressure. I then wrote gauge and PSIG for some reason unknown to me at this time. You should know better than to assume that I do not understand propane injection Math.
People should not be afraid of their governments. Governments should be afraid of their people.
SpudBlaster:
Yeah, I figured something was goin' on there. Only reason I rambled on there for a time was because I'm having a really hard time putting things into words today. Origin: No, that's not what I'm stating. What I'm stating is that I'm finding an equation for absolute pressure to satisfy the mathnazi part of my brain. Jimmy: I see where you're coming from. The thing is, we are thinking of two different processes for injecting propane. Your process is to calculate 4.2% of the combined volume of meter and chamber and allow the gas to diffuse across the entire volume. I don't think too many spudders do this, though. It would require bleeding the meter pipe after every shot and a long wait. Most people inject propane and then close the valve. Tacking on another atmosphere to your pressure displaces the desired quantity of propane into the chamber and an atmosphere remains in the meter.
Umm ouch, im brain dead This is only for combustions right, not hybrids?
4SPC, My 4" piston 3" porting cannon
Memo: Fix up copper cannon Fix up 4SPC Start Stirrup pump Start Toolies piston bazooka
Mark:
Regardless of whether you leave the valve open long enough for the air in the chamber to fully equilibrate with the fuel in the meter pipe you still need to take into account the volume of fuel that is left in the meter pipe. Perhaps like this; PmeterVmeter = (0.042)(Vchamber)  Vmeter That is, PmeterVmeter is proportional to the amount of fuel you start with in the meter. The "Vmeter" on the right side of the equation takes into account the volume of fuel left in the meter (and assumes no air migrates into the meter from the chamber). Vmeter probably should be (Vmeter)*(Pmeter,final) but "Pmeter,final" is very close to 1 ATM (absolute). So if Pmeter is in atmospheres (absolute) then "Pmeter,final" is close enough to 1 to be called 1.
 
Who is onlineRegistered users: Bing [Bot], Exabot [Bot], Google [Bot], Google Adsense [Bot], Yahoo [Bot] 
