i make it fast..
on like 20x mix in a hybrid why does not the propane become liquid?
does there have to be 100% propane under pressure to become in liquid state?
noob question by spudfarm
The figure that matters is the partial pressure. This is basically the total pressure times the percentage of the gas mix that is element X. You should work in absolute pressure for these purposes.
For example, the partial pressure of nitrogen in air is ~11.5 psia.
Oxygen would be ~3.1 psia.
Argon and other trace elements make up the remaining ~0.2 psia.
The sum of the partial pressures in a gas will always be equal to the total pressure of the gas mixture.
Basically, in the case of propane, it needs to reach a given density to start to liquefy. In a hybrid, the density of propane is low, being interspersed with the elements of air, so propane wouldn't start to liquefy until around very high mixes.
For example, the partial pressure of nitrogen in air is ~11.5 psia.
Oxygen would be ~3.1 psia.
Argon and other trace elements make up the remaining ~0.2 psia.
The sum of the partial pressures in a gas will always be equal to the total pressure of the gas mixture.
Basically, in the case of propane, it needs to reach a given density to start to liquefy. In a hybrid, the density of propane is low, being interspersed with the elements of air, so propane wouldn't start to liquefy until around very high mixes.
Does that thing kinda look like a big cat to you?
No, he is saying that the partial pressure of the propane in the chamber must exceed the liquification pressure of propane at whatever temperature the chamber is at.
Using the old (inaccurate) system, mixes above 200x or so would cause the propane to liquify. In reality, it would liquify well before that if SB15's accurate fueling equations were used, but I wouldn't worry about it until a long way past 20x.
Using the old (inaccurate) system, mixes above 200x or so would cause the propane to liquify. In reality, it would liquify well before that if SB15's accurate fueling equations were used, but I wouldn't worry about it until a long way past 20x.
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.
To put it differently, there need to be X propane molecules per unit volume for it to liquefy.
In a hybrid, where only about 4% of the molecules are propane, the density of these molecules is low.
To get the propane to liquefy, you'd need to increase this density above this magic number (which changes with temperature).
Assuming the mix stayed stoichiometric, then the only way to get the propane to liquefy would be to increase the pre-ignition pressure.
This number would be around a 200x hybrid mix.
DYI... SB15's method won't actually change the ideal ratio of propane to air. Propane still needs be 4.03% of the gas mix by moles for a perfect mix. In a system where gas can be displaced, this is achieved by injecting 4.19% propane by volume.
What SB15's method does is correct for the volume differences that would arise in these molar quantities of gasses once they're pressurized in a system where gases can't be displaced from the chamber.
That means it's still about a 200x mix, SB15 just corrected fuelling errors made by people assuming the process to be a linear scaling of an atmospheric combustion. At no point was SB15 ever capable of changing how chemical reactions work.
You could also manage it with a rich mix, and slightly less pressure - or for that matter, any other way of boosting the density of the propane molecules high enough.
The normal way to achieve this effect is using 100% propane and about 120 psi - then shipping it out to stores, then having spudders buy the bottles.
In a hybrid, where only about 4% of the molecules are propane, the density of these molecules is low.
To get the propane to liquefy, you'd need to increase this density above this magic number (which changes with temperature).
Assuming the mix stayed stoichiometric, then the only way to get the propane to liquefy would be to increase the pre-ignition pressure.
This number would be around a 200x hybrid mix.
DYI... SB15's method won't actually change the ideal ratio of propane to air. Propane still needs be 4.03% of the gas mix by moles for a perfect mix. In a system where gas can be displaced, this is achieved by injecting 4.19% propane by volume.
What SB15's method does is correct for the volume differences that would arise in these molar quantities of gasses once they're pressurized in a system where gases can't be displaced from the chamber.
That means it's still about a 200x mix, SB15 just corrected fuelling errors made by people assuming the process to be a linear scaling of an atmospheric combustion. At no point was SB15 ever capable of changing how chemical reactions work.
You could also manage it with a rich mix, and slightly less pressure - or for that matter, any other way of boosting the density of the propane molecules high enough.
The normal way to achieve this effect is using 100% propane and about 120 psi - then shipping it out to stores, then having spudders buy the bottles.
Does that thing kinda look like a big cat to you?
Hmm... Assume a 10x mix - using SB15's calcs that is 10.64psi propane, and 10atm of air total (including air that was already there).SB15's method won't actually change the ideal ratio of propane to air
If we have a 100ci chamber, that is:
72.4ci of propane
1000ci of air.
For a total of 1072ci, of which 6.7% is propane. Significantly more than 4.03%, and this means that the partial pressure of the propane is higher than it would be at a 10x mix done in the old style (which likely wouldn't even ignite).
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.
I'm still rather curious as to how (or even if) the ratio of propane:air changes as pressure increases. No one has actually explained it completely yet, and I'm sure Rag has some sort of answer.
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.
And here is that answer:
In that case, there's a severe flaw in SB15's working (and a small one in mine, it's 4.02%, not 4.03% - slip of the finger.)
I'd largely glossed over reading the exact details of it, I'm not much of a fan of hybrids - I now need to look at it properly.
Now, it can be proved that propane reacts with oxygen like this:
C<sub>3</sub>H<sub>8</sub> + 5 O<sub>2</sub> ==> 4 H<sub>2</sub>O + 3 CO<sub>2</sub>
That won't change with pressure. Propane won't suddenly only need 4 molecules of oxygen (EDITED: oxygen, not air, silly Ragnarok) to react, or as 6.7% by moles suggests, the odd number of 3.12 molecules.
Try as you might, that won't ever change. If it did, engineers could have a field day by ramping up compression and being able to burn more fuel for the same oxygen. That just doesn't happen.
Nor does the percentage of oxygen in air change with compression - always 20.95%.
This means that propane will always need to be 4.02% of the mix by moles.
Where does 4.02% come from?
The percentage of oxygen in air is 20.95%. Divide that by 5 (the number of ratio of oxygen to propane) to get the well known 4.19%.
However, you've now got more moles of gas - 4.19% more. So to get the actual percentage of propane in the whole mix, you divide by 1.0419 to get 4.02%.
Thinking about it, it doesn't matter. Whether or not gas can be displaced from the chamber, you need to inject 4.19% propane by moles of air to leave you with that total 4.02% by total moles.
In a typical combustion, the extra 4.19% gas in the chamber will leak away, proportionally to the percentages of gasses in the mix.
That proportional loss still leaves the mix at 4.02% propane by moles.
Actually now I think, I begin to wonder what SB15's work actually needed to correct.
In that case, there's a severe flaw in SB15's working (and a small one in mine, it's 4.02%, not 4.03% - slip of the finger.)
I'd largely glossed over reading the exact details of it, I'm not much of a fan of hybrids - I now need to look at it properly.
Now, it can be proved that propane reacts with oxygen like this:
C<sub>3</sub>H<sub>8</sub> + 5 O<sub>2</sub> ==> 4 H<sub>2</sub>O + 3 CO<sub>2</sub>
That won't change with pressure. Propane won't suddenly only need 4 molecules of oxygen (EDITED: oxygen, not air, silly Ragnarok) to react, or as 6.7% by moles suggests, the odd number of 3.12 molecules.
Try as you might, that won't ever change. If it did, engineers could have a field day by ramping up compression and being able to burn more fuel for the same oxygen. That just doesn't happen.
Nor does the percentage of oxygen in air change with compression - always 20.95%.
This means that propane will always need to be 4.02% of the mix by moles.
Where does 4.02% come from?
The percentage of oxygen in air is 20.95%. Divide that by 5 (the number of ratio of oxygen to propane) to get the well known 4.19%.
However, you've now got more moles of gas - 4.19% more. So to get the actual percentage of propane in the whole mix, you divide by 1.0419 to get 4.02%.
Thinking about it, it doesn't matter. Whether or not gas can be displaced from the chamber, you need to inject 4.19% propane by moles of air to leave you with that total 4.02% by total moles.
In a typical combustion, the extra 4.19% gas in the chamber will leak away, proportionally to the percentages of gasses in the mix.
That proportional loss still leaves the mix at 4.02% propane by moles.
Actually now I think, I begin to wonder what SB15's work actually needed to correct.
Does that thing kinda look like a big cat to you?
3.12 molecules 
Well we're going to need to figure it out, aren't we?
I actually thought I knew what I was doing before Spudblaster read my post about the meter system for my hybrid and then posted the Hybrid Fueling Thread. I had assumed it was something more complex that I hadn't heard of yet, but with some further thought, it once again seems a bit fishy. The ratios shouldn't be able to change at all, unless you actually change the structure of the fuel or oxidiser.
Well Spudblaster, your turn
Well we're going to need to figure it out, aren't we?I actually thought I knew what I was doing before Spudblaster read my post about the meter system for my hybrid and then posted the Hybrid Fueling Thread. I had assumed it was something more complex that I hadn't heard of yet, but with some further thought, it once again seems a bit fishy. The ratios shouldn't be able to change at all, unless you actually change the structure of the fuel or oxidiser.
Well Spudblaster, your turn
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.
It's what I believe is called a fallacy. An apparently logical argument that is actually on closer examination at least in part, incorrect.DYI wrote:I had assumed it was something more complex that I hadn't heard of yet, but with some further thought, it once again seems a bit fishy.
As you put it, the ratios shouldn't change at all unless fuel or oxidiser is changed. However, how exactly you meet those ratios might require a little thought depending on the method - and in fact your definition of 10x hybrid... is it a 10 atm pre-ignition pressure; or 10 times the air originally in the chamber, then with propane added on top of that?
I've seen both.
Does that thing kinda look like a big cat to you?
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SpudBlaster15
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Last edited by SpudBlaster15 on Wed Jul 14, 2021 6:44 pm, edited 1 time in total.
And 100ci + 44ci = 144ci44
144ci / 100ci =1.44
1.44*14.7psia = 21.168psia
21.168psia - 14.7 = 6.468psig
So you're saying I was right all along?
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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SpudFarm
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this fueling stuff make me crazy... numbers numbers numbers all the way..
a question.. is it possible to just use a traditional fuel meter for 10x?
if i does like this (152/0.58-152)/meter volume? 152 is my chamber volume
SB15 told me that stuff on msn.
a question.. is it possible to just use a traditional fuel meter for 10x?
if i does like this (152/0.58-152)/meter volume? 152 is my chamber volume
SB15 told me that stuff on msn.
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