HYBRID FUELING 101
Okay, so just simplifying the calculations on the first page for calculating for chamber fueling, I get the equation mix/(1-fuel%)-mix = Fuel pressure
Then fill with 1-mix# atm of oxidizer
Correct?
Then fill with 1-mix# atm of oxidizer
Correct?
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if I use butane I'd need 4.54639248psi for a 10X mix, right ? (assuming that the ideal mix is 3%, though I am not sure if that's correct)
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- jackssmirkingrevenge
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It is.POLAND_SPUD wrote:(assuming that the ideal mix is 3%, though I am not sure if that's correct)
Sound advice.doyourselfit wrote:@POLAND: Although manometric meters are nice, they're not the most practical with lower mixes, especially if you're using a small chamber.
I'm chucking this here too:
[youtube][/youtube]
For a hybrid mix, just multiply the resulting fuel volume by whatever mix number you want to achieve.
It's important that you consider the dead space after the check valve of your pump to be part of the chamber volume when making your calculations, especially for small chambers.
The calculation is as follows:
(ideal fuel % / 100) x (chamber + pump dead volume) x mix number
In the case of this Beto shock pump, the dead volume is 1.75mL. For a 10mL chamber using butane to say 5x, the calculation is as follows:
3/100 x (10+1.75) x 5 = 1.76mL of butane
Had we not considered the pump dead volume, the result would have been 1.5mL of fuel, an error of 15% which would most likely prevent ignition. If it were a 100mL chamber on the other hand, the error would have been of 1.5%, and likely the mix would have still ignited.
After the fuel is injected, all you have to do is pressurise with your pump to the following pressure (assuming you're using psi):
(mix number - 1) x 14.7
In the case of our 5x mix, the calculation is as follows:
(5-1) x 14.7 = 59 psi
If your gauge can read bar, simply pressurise to (mix number - 1) bar, so 4 bar in this case.
Last edited by jackssmirkingrevenge on Tue Feb 27, 2018 7:36 pm, edited 1 time in total.
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Yeah, I got that 3% from your vidI'm chucking this here too:
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
yeah it might be. But is 10x mix a low mix ?Sound advice
AFAIK 4.54639248psi = 235mmhg, which is in range of my gauge... So guess I might as well give it a try
W8 a second... so I am supposed to pressurise the chamber to 9bar for a 10X mix ? (note that I still want to use manometric metering)If your gauge can read bar, simply pressurise to (mix number - 1) bar, so 4 bar in this case
Lol if jsr is posting fuelling calculations then it means that the apocalypse is near
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
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Hence your "though I am not sure if that's correct" WPOLAND_SPUD wrote:Yeah, I got that 3% from your vid
![Image](http://www.cityu.edu.hk/webimage/symbol/html/anchor.gif)
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
You're more likely to believe Jimmy, it's actually 3.23%, but I wanted to keep things simple for your average youtube viewer
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
jimmy101 wrote:Complete combustion of butane (C4H10) in air (~21% O2, 29% nitrogen) produces carbon dioxide (CO2) and water as described by the equation:
2 C4H10 + 13 O2 = 8 CO2 + 10 H2O + heat
The equation tells us that for each 13 molecules of oxygen we need 2 molecules of butane. Since air is only ~21% oxygen we need (21%)(2/13) = 3.23% by volume butane in the combustion chamber.
Luckily, we have a bit of leeway:
jimmy101 also wrote:Butane in air will only burn if the percentage of butane is in the range of 1.9% to 8.4%. Outside that range the mixture can not be ignited.
... so 3% is just fine
![Wink ;)](./images/smilies/icon_e_wink.png)
Perhaps, if you already have the equipmentSo guess I might as well give it a try
For a 10x mix, you want 10 bar in the chamber. You already have 1 bar in there (assuming you're not working in a vacuum) and when your gauge reads zero, it's zero relative to atmospheric pressure, it should be reading 1 bar.W8 a second... so I am supposed to pressurise the chamber to 9bar for a 10X mix ? (note that I still want to use manometric metering)
Yep, 2012 is fast approaching and my army of space monkeys isn't nearly as big as it needs to be!Lol if jsr is posting fuelling calculations then it means that the apocalypse is near
hectmarr wrote:You have to make many weapons, because this field is long and short life
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I've just realized something... to fuel a hybrid one has to fill a closed chamber (already filled with air at atmospheric pressure) with propane/butane/mapp. When the pressure in the chamber reaches a set value regulated air has to be added
What if I just purge the chamber with propane to make sure that it contains only propane (not air) and then add air at X bar ? Wouldn't that be a better and simpler idea?
so if the ideal ratio for butane is 3% then fill the chamber completely, close the vent and fill with air to 33bar
EDIT
lol either it is a pretty clever idea or I am just to tired to think.
Anyway this is what I think would work fine:
1.butane is feed into the chamebr via a check valve
2. on the other end of the chamber is a 3 way valve
3. In normal position the 3 way acts as a vent, when actuated it blocks off vent port and fills the chamber with regulated air
4. When the valve is actuated the check valve blocks the flow (since the pressure in the chamber increases)
What if I just purge the chamber with propane to make sure that it contains only propane (not air) and then add air at X bar ? Wouldn't that be a better and simpler idea?
so if the ideal ratio for butane is 3% then fill the chamber completely, close the vent and fill with air to 33bar
EDIT
lol either it is a pretty clever idea or I am just to tired to think.
Anyway this is what I think would work fine:
1.butane is feed into the chamebr via a check valve
2. on the other end of the chamber is a 3 way valve
3. In normal position the 3 way acts as a vent, when actuated it blocks off vent port and fills the chamber with regulated air
4. When the valve is actuated the check valve blocks the flow (since the pressure in the chamber increases)
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how many 3way valves do you own?
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lol a lot... but it isn't important - this thing would need just one
maybe I should have explained it better. The idea is simple instead of metering regulated fuel and then air I purge the chamebr with butane. If the ideal mix for butane is ~3% and I've got 100% of it in the chamber so I need to add 33 times the volume of the chamber of air.
To do so I just fill it up with air to 33 bar
So is it a good idea or am I retarded?
maybe I should have explained it better. The idea is simple instead of metering regulated fuel and then air I purge the chamebr with butane. If the ideal mix for butane is ~3% and I've got 100% of it in the chamber so I need to add 33 times the volume of the chamber of air.
To do so I just fill it up with air to 33 bar
So is it a good idea or am I retarded?
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Well, your idea works so long as you don't mind being stuck to a certain mix number.
Achieving a full purge of the chamber could be an issue in terms of fuel wasted if you're doing it using the method you described above. Although it may not be quite so convenient, depending on implementation, filling the chamber with a liquid and then displacing it with fuel would be less wasteful, and more likely to result in the chamber being completely purged.
The purging method would also make it easier to attain the maximum ~220X achievable with gaseous propane at roughly room temperature (although it wouldn't alter the maximum, only bring the required fueling pressure down to the same value as the tank pressure).
Also, I'd like to add that in the case where pure O<sub>2</sub> is being used as the oxidiser, this method isn't the greatest idea - detonations are even more likely when that comparatively small amount of nitrogen is removed completely from the mix, as Larda's experience testing HyGaC20 indicates.
Achieving a full purge of the chamber could be an issue in terms of fuel wasted if you're doing it using the method you described above. Although it may not be quite so convenient, depending on implementation, filling the chamber with a liquid and then displacing it with fuel would be less wasteful, and more likely to result in the chamber being completely purged.
The purging method would also make it easier to attain the maximum ~220X achievable with gaseous propane at roughly room temperature (although it wouldn't alter the maximum, only bring the required fueling pressure down to the same value as the tank pressure).
Also, I'd like to add that in the case where pure O<sub>2</sub> is being used as the oxidiser, this method isn't the greatest idea - detonations are even more likely when that comparatively small amount of nitrogen is removed completely from the mix, as Larda's experience testing HyGaC20 indicates.
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I'm having some fueling trouble thats just making me feel like an idiot....
I have a 3/4"x12" chamber with taser ignition and 1/2" cam-lock union, but for some reason I'm having lots of trouble getting my mix to ignite. I put in 13Kpa of propane and 2bars of air for a 3x mix, but nothing's happening. Any tips?
I have a 3/4"x12" chamber with taser ignition and 1/2" cam-lock union, but for some reason I'm having lots of trouble getting my mix to ignite. I put in 13Kpa of propane and 2bars of air for a 3x mix, but nothing's happening. Any tips?
Just use this. If you can figure out the error with column G while you're at it, kudos and thanks.
And a note to the user: Put in the units that you wish it to return.
And a note to the user: Put in the units that you wish it to return.
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Taking the pump's dead space in the calculationsis only needed when using a fill valve wich has an open connection with the pump , right?
I'm planning on using a dunlop valve when filling it, because it doesn't leak when disconnected, like a schrader valve.
I'm planning on using a dunlop valve when filling it, because it doesn't leak when disconnected, like a schrader valve.
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If the fill valve is pressure activated then technically it is acting as a check valve and in this case the pump dead volume need not be taken into account.
Note that my shock pump has a separate knob on the attachment head that allows you to independently depress the schrader valve stem, and therefore the chamber does not leak when the head is disconnected.
Note that my shock pump has a separate knob on the attachment head that allows you to independently depress the schrader valve stem, and therefore the chamber does not leak when the head is disconnected.
hectmarr wrote:You have to make many weapons, because this field is long and short life
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I simplified the equation for Manometric fueling a bit, hope it helps some people.
(X/R-X)*P
Where X is your mix number, R is fuel ratio (for propane it's .958, butane .9677), and P is ambient pressure (usually 14.7 psi or 1 bar). For example, for propane at sea level with PSI:
(X/.958-X)*14.7
For a 10X mix:
(10/.958-10)*14.7=6.44 psi, the same as SB15 got for a 10X mix.
For measuring in bar, it gets even easier, since atmospheric pressure is 1 bar (well 1.01325, but close enough)
X/.958-X
10/.958-10 = 0.438413361 bar
The math gets much easier when you take out chamber volume, and now you don't have to measure it either![Very Happy :D](./images/smilies/icon_e_biggrin.png)
Edit:
I think you can simplify the equation much further. Take the equation from above for propane, and calculate for a 1X Mix
(1/.958-1)*14.7 = .64447 psi
And then you can multiply that by the mix number to get the amount of psi.
Easy, huh?
Edit: Just read one of Crowley's posts where he said the exact same thing.
Oh well.
(X/R-X)*P
Where X is your mix number, R is fuel ratio (for propane it's .958, butane .9677), and P is ambient pressure (usually 14.7 psi or 1 bar). For example, for propane at sea level with PSI:
(X/.958-X)*14.7
For a 10X mix:
(10/.958-10)*14.7=6.44 psi, the same as SB15 got for a 10X mix.
For measuring in bar, it gets even easier, since atmospheric pressure is 1 bar (well 1.01325, but close enough)
X/.958-X
10/.958-10 = 0.438413361 bar
The math gets much easier when you take out chamber volume, and now you don't have to measure it either
![Very Happy :D](./images/smilies/icon_e_biggrin.png)
Edit:
I think you can simplify the equation much further. Take the equation from above for propane, and calculate for a 1X Mix
(1/.958-1)*14.7 = .64447 psi
And then you can multiply that by the mix number to get the amount of psi.
Easy, huh?
Edit: Just read one of Crowley's posts where he said the exact same thing.
![Rolling Eyes :roll:](./images/smilies/icon_rolleyes.png)
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So I'm sorry that this is a really dumb question, but I was looking at the load mix calculator on HGDT and it said that the chamber pressure after the fuel injection (using propane at a 10x mix) was only 5.89PSI whereas it's been stated throughout the thread that 6.44PSI's the proper pressure, am I missing something?
Also, I know it was mentioned somewhere here but I don't believe it was answered, you'd add 132PSI of air in on top of the fuel pressure right? (so for a 10x mix with 6.44PSI of propane the chamber pressure after air injection would be 138.44PSI?).
Also, I know it was mentioned somewhere here but I don't believe it was answered, you'd add 132PSI of air in on top of the fuel pressure right? (so for a 10x mix with 6.44PSI of propane the chamber pressure after air injection would be 138.44PSI?).
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