Seems that you're enjoying the challenge of figuring this out too. I'm really happy to have someone so capable helping to figure this out with me. Thank you Chet. I'm still not up to speed yet and getting somewhat lost in some of the calculations. I'll post up some specific questions in the...
Thoughts... It's new territory for me. I'm having to look up a lot of the things you're mentioning to properly understand them, so I'm not quite eye to eye with you yet. Still a day or two worth of posts behind on the learning curve.
However, I have used equations for Adiabatic Compression...
You're ahead of me, which is fine. Once it's here, I can soak up the information at whatever rate.
I'm not sure that the final temperature is what I expected. Seems high, but if that's what it is, then that's what it is.
I'm guessing that for adiabatic compression, we'll need to average the...
Whatever's easier. As long as the results accurately indicate a trend. I pretty much don't care how precise the product is, as long as it's repeatable per individual air/fuel ratios.
The ultimate reason behind my trying to figure this out is, engines make more horsepower with richer than...
Based on this percent of total mass composition of air breakdown, in/at engineeringtoolbox.com
% of Total Mass, Air
Oxygen .... 23.20%
Nitrogen ... 75.47%
Carbon Dioxide .. 0.046 %
Hydrogen ... ~ 0%
Argon ... 1.28 %
Neon .... 0.0012%
Helium ..... 0.00007%
Krypton .... 0.0003%
I really appreciate your help with this Chestermiller
Imperial is easier for me to relate to. However, SI seems easier to calculate with.
Let's work in SI.
It seems that using the mass ratio to calculate nitrogen content is slightly more accurate than using the molar ratio. If there's not...
Since we're doing everything by weight, wouldn't you calculate the nitrogen weight from the oxygen weight and identify the number of moles N2 by it's weight per mole?
(12.5 * 32g/mol) = 400 g
400g / (79/21) = 1504 g
1504 / 28g/mol = 54 moles of N2
Simply basing it on the mole negates the...
Apparently the air composition of 79/21 is based on volume. By mass it's 23.2% O2 and 76.8% N2 and other inert gasses.
Adjusting for that brings the ratios quite a bit closer to what I expected to see.
15.09:1 for Octane
8.98:1 for Ethanol
6.46:1 for Methanol
The total mass now also balances between pre and post combustion
H2O ... 0.357452336 lb ... 8.03%
CO2 ... 0.7761953474 ... 17.44%
N2 ... 3.317310438 ... 74.53%
Total .. 4.4509581214 ... 100.00%
Had a thought. I think the numbers are off because the actual ratio is slightly lower than the 14.7 rounded figure I used to calculate fuel mass from air mass. One end based on rounded figure, the other end from the actual precise stoichiometric calculation. I'll math it out tomorrow and...
I only barely grasp what I've been able to deduce in a week of trying to figure this out on my own through google searches and YouTube lectures. I'll take that as a compliment. Thank you Chestermiller.
I believe I understand roughly how to write a balanced chemical reaction equation, however...
Hi guys, 42 year old engine hobbyist here, not a student. I've had great luck figuring out my questions in this portion of the forum in the past and look forward to your input. Keep in mind that I'm not formally educated, so if it's possible to dumb something down a bit I'd appreciate it...