[unclechimist]

Hi guys, I’m new in this forum.
I came from the world of modeling. I have some doubts about the chemical processes inherent in that environment and some questions for you.I hope you can help me.

The first thing to ask is to help resolve this apparent contradiction:
consider two basic mixtures used: methyl alcohol and nitro methane.
Data sheets shows  that the methyl alcohol has an ignition temperature to 12 ° C, a boiling point to 64.7 °C and an autoignition temperature to °464, density 0.79 (some data shet reports 400°C)
The nitro methane has an ignition temperature to 35° C, a boiling point at 100.8°C, and, though a temperature of a auto 418°C degrees, density 1,13
How do I explain this? From layman I would expect that the auto of nitro methane occurs at a temperature greater than that of alcohol-methyl

Prepare now two mixtures using as a base methyl alcohol: the first with a 5%, percentage of nitro methane, the second with a rate of 40%.
On top of the cylinder, and inside it there is a glow plug that, through electricity, bring to a temperature sufficient to trigger the less "Heat" and then leave at the same temperature for the subsequent mixture test-
Injected into a cylinder the first mixture so that the air inside the cylinder and the oxigen released by nitro methane are exactly the quantity used to burn the total mixture.
Through the rotation Axis, we climb the piston regular beginning  to collapse liquid and air. Let us assume the compression ratio of the engine is 6 / 1, which is the initial volume of the cylinder is 12 cc, at the end of the stroke volume will be reduced to 2cc.
At some point, the mixture reaches the  autoignition temperature.
We repeat the test with the mixture to 40% of nitro.

Which of the two mixtures, in the condition indicated (each with the proper amount of air and equal  temperature of plug) will go first?
Once lit, as will be different the two combustion in pressure, temperature and expansion speed of the flame?
Thanks.

[Borek]

From layman I would expect that the auto of nitro methane occurs at a temperature greater than that of alcohol-methyl

Why?

I am moving this thread to eneineering forum. While its is not strictly engineering, I have a feeling that Eugene can be able to add something about the temperature of autoignition of mixture. I can be wrong, though.

[unclechimist]

Cause nitromethane vs methanol has 35° > 12°, 100° > 64°, 400 < 418 ?
Can You give me more details?
Do You got answer to the other my question?
Thank You for Your availability.

[Borek]

These numbers are not comparable, they are measures of different properties.

Boiling point is irrelevant, as it is about intermolecular forces, while both ignition and autoignition are somehow related to the activation energy required to start the reaction - so these things are incomparable. That means you don't have three realationships 35° > 12°, 100° > 64°, 400 < 418, but only two 35° > 12°, 400 < 418. However, it is enough for both molecules to react in different way (ie different mechanism of oxidation, decomposition or whatever is the kinetically limiting step) for temperatures to differ. These molecules are very different, no wonder there is a difference.

Note: that's more of educated guess than knowledge.

[unclechimist]

Thanks first.
I understand that having differnt composition molecules have different temperatures of a flash of autoignition, but in mixtures, the molecules are mixed together, just the percentage. I was interested to know, to increase the percentage of nitromethane as varied pressure and temperature of the reaction of ignition.

[eugenedakin]

Borek is quite right about the question. The only way to know for sure about autoignition temperatures and pressures is by actual testing (anti-knock tests are still performed to this day because of this uncertainty).

In theory, the lowest boiling point fluid should combust first. In this case, methyl alcohol would combust because there are more vapors in the combustion chamber at equal temperatures and pressures. This assumes that the mixtures will not cause flooding or other real-world phenomenon. This is related to methyl alcohols lower boiling point.

Once combustion actually starts, the heat released by combustion can be stochiometrically calculated (I will leave you to determine the mathematics) - hotter gas expands more than cooler gas (relatively speaking). Also, there would probably be a greater increase in the volume of gases increases the pressure inside the combustion chamber (again, check the stoichiometry). I believe that nitromethane will probably have a greater release of energy due to these two factors. Plus, a higher piston level of nitromethanes increased pressure required for combustion, will also increase the speed at which a crankshaft turns due to a fulcrum-like effect.

I hope this helps. Feel free to provide suggestions 

Sincerely,

Eugene

[unclechimist]

Thanks Eugeene.
We assume mixtures will not cause flooding or other real-world phenomenon.
...the lowest boiling point fluid should combust first...
Why? I have to consider that the methanol (5% nitromethane) has more air to burn?
But if the Nitromethane evaporates first, in 40% mixture, it will not release more oxygen in the chamber than the other combustion?
Or the problem is related to the volume available, since, (assuming that there be oxygen for the complete combustion), the mixture 5% will have a volume of air much larger than the other.
For the pressure/temperature increasing (in correct stechiometric rate) I am completely in agreement.
How much should I drink nitro to get more oxygen to the brain to understand?
Hope U can help me again.

[eugenedakin]

Hi unclechimist,

Oops.. your right, I failed to mention the ignition point of Methyl Alchol which is at a lower temperature (thanks for pointing it out). With a lower ignition temperature, methyl alcolhol will ignite first - and is complemented by the lower boiling point of methyl alcohol.  I will explain these two aspects seperately. 

The lower boiling point of methyl alcohol 'should' have more gaseous methyl alcohol in the combustion chamber at a given temperature than nitromethane. As the ambient temperature approaches the boiling point, more methyl alcohol vapour is present than in liquid form. Ignition greatly prefers gasses (given everything is equal) than a liquid.

With a lower ignition point, the methyl alcohol will preferably ignite sooner. The ignition point  is the lowest temperature at which the chemical can form an ignitable mixture in air. Again, all things being equal, Methyl Alcohol requires a lower temperature to ignite  .

Yes, you are quite right. There is more air to burn in a 5% mixture than a  40% mixture. Performing stoichometry (with products of CO₂ and H₂O as products in your reaction. You will have to assume the concentration of available oxygen to be the oxygen content of atmospheric air. After performing the calcuations, you can determine the concentration of oxygen and if there is enough atmospheric air to completely combust all of the 'fuel'.

I hope this makes more sense. If I am ambiguous, feel free to ask.

Great question!

Eugene

[PeterPan]

Hi guys,

Another question regarding nitro methane and methanol mixtures.

Imagine this case:
- temperature is 20degC
- methanol mass mixture percentage 15%
- nitro methane mass mixture percentage 85%

What would be the mass mixture percentages at e.g.:
5degC, 10degC, 15degC, 25degC and 30degC

[Borek]

Mass percentages are not temperature dependent.

[PeterPan]

Hi Borek,

Actually maybe I should have started with the fact that I was using a hydrometer.
And never payed to much attention to how it really works (Just now finished my brush up course).

As this nice device is showing a percentage I sort of assumed it to be a mass mixture percentage. But actually it is a volume percentage at a specific reference temperature, which needs to be corrected if the temperature is different than the reference to take into account the density change. 

My final target was actually to see the effect on air fuel ratio which is based on air mass and fuel mass.

So thanks for waking me up (Mass = Volume * Density(rho) )

PeterPan