Hey guys, just wondering if someone could check over these calculations for manometric metering
Volume of the chamber = 19.6349541 * 28 = 549.778715 cm cubed
549.778715 cm cubed in inches cubed = 33.5495556 inches cubed
Calculation for 10X mix
(33.5495556 * 10)/0.958-(33.549556 * 10) = 14.7085694 inches cubed of propane
By how much has the chamber pressure risen?
14.7085694/33.5495556 = 0.438413241
0.438413241 * 14.7 = 6.44467464 psi for 10X mix
Add 9 atmospheres of air and fire
Also, is there an easier way to get the results of the amount of psi of propane I have to add from 1 - 15X mix?
Thanks Guys
Dave
Can someone check these metering calculations?
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Looks interesting to study.
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Got a side job fixing a completely broke toilet and plumbing...No plumbing experience besides spudding but I'm feelin' confident...
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BTW, (It's not poopy or anything, just a bunch of messed up internals and valve).
I have no time right now



BTW, (It's not poopy or anything, just a bunch of messed up internals and valve).
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Born To Be Alive!
I think the HGDT metering tool is flawed. I multiply mix by .61 to get propane pressure in psi.
For air, either of these work:
* psi= ((mix#-1)*14.7)+propane pressure
* psi= ((mix#-1)*14.7)+ (mix#*.61)
EDIT:
for 10x mix, I see 10*.61= 6.1psi propane followed by (14.7*9)+6.1= 138.4 psi air.
for an explanation of the .61 psi number, assume a 100cu in chamber.
the ideal mix is 4.2% of propane, so 100* .042= 4.2cu in propane
add this to the 100 cu in of air, and get 104.2.
divide this by 100 to get 1.042, which is in bar. (absolute). subtract the 1 to get relative pressure since atmosphere is nearly 1 bar, and convert .042 bar into psi to get .61 psi/x
the absolutely correct answer is 0.609158498 psi/x but .61 is close enough.
For air, either of these work:
* psi= ((mix#-1)*14.7)+propane pressure
* psi= ((mix#-1)*14.7)+ (mix#*.61)
EDIT:
for 10x mix, I see 10*.61= 6.1psi propane followed by (14.7*9)+6.1= 138.4 psi air.
for an explanation of the .61 psi number, assume a 100cu in chamber.
the ideal mix is 4.2% of propane, so 100* .042= 4.2cu in propane
add this to the 100 cu in of air, and get 104.2.
divide this by 100 to get 1.042, which is in bar. (absolute). subtract the 1 to get relative pressure since atmosphere is nearly 1 bar, and convert .042 bar into psi to get .61 psi/x
the absolutely correct answer is 0.609158498 psi/x but .61 is close enough.
POLAND_SPUD wrote:even if there was no link I'd know it's a bot because of female name
Okay, I seem to be pretty bad with these kind of calculation and just want to be clear about it
Could you check the number Psi of propane added to the chamber?
The chamber is 33.5495556 cubic inches
for 15x mix i get 9.66701196 psi of propane
and 5x mix i get 3.22233732 psi of propane
can someone check their results so i can confirm that the rest of my numbers are correct
Dave
Could you check the number Psi of propane added to the chamber?
The chamber is 33.5495556 cubic inches
for 15x mix i get 9.66701196 psi of propane
and 5x mix i get 3.22233732 psi of propane
can someone check their results so i can confirm that the rest of my numbers are correct
Dave
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Not to be self-promotional or anything, but I did a write-up a few days ago at http://www.spudstalker.net/hyfuel.html on hybrid fueling. It should answer your question about how much the propane affects final pressure, and what the final mixture pressure should be for x mix.