That was my original hypothesis (not that precise ratio--but some ratio of those two hydrocarbons). Yet I can't get the chemistry to work. Can you show me the stoichiometry calculations that demonstrate that ratio (or any ratio) of those two hydrocarbons being stoich at 14.7? Or provide some...
Using data from Bruce Hamilton's superbly cited http://faqs.org/faqs/autos/gasoline-faq/part1/", I'm getting oxygen as a 23.14% constituent of air, by weight. So using 23.14% instead of the 21% I used before:
1 mole C8H18 and 12.5 moles of O2.
12.5 moles O2 = 400 units mass.
3.32 times as...
I understand what your saying. But my understanding is the 14.7 a/f ratio is an ideal number (i.e. it assumes perfect combustion). If that's not the case then what defines the variables used to arrive at 14.7 a/f ratio for gasoline? I mean literally--what gasoline, engine, "air", etc. where...
I found an SAE paper: Stoichiometric Air-Fuel Ratios of Automotive Fuels
http://www.sae.org/technical/standards/J1829_200210"
Anyone have access to SAE docs? I can't pay $50 for that. Plus, I'm suspicious it doesn't actually define a "standard" gasoline but rather guides one through...
Thanks. I think I've already done that though:
1 mole C8H18 and 12.5 moles of O2.
12.5 moles O2 = 400 units mass.
3.76 times as much nitrogen as oxygen in air (79%/21%): (3.76)(12.5 moles) = 47 moles N2. 47 moles N2 = 1316 units mass.
1316 + 400 = 1716
1716.599 / 114.232 = 15.03:1...
I did the math again with less rounding and I'm getting:
iso-octane: 15.03:1
n-heptane: 15.08:1
I'm a little concerned about using 3.76x as much nitrogen in the air as oxygen (21% oxygen, 79% nitrogen--I think by volume?). Could that simplistic representation of the air account for the...
I'm trying to figure out where the often cited 14.7:1 stoich a/f ratio for gasoline comes from. I've always thought it was calculated from a primary reference fuel containing some mixture of iso-octane and n-heptane. But I've done the calculations for those hydrocarbons and my numbers don't...
I worked out stoich for ethanol also and came up with 8.95:1. 9:1 is the figure I've seen for ethanol so I think I'm doing the calculations correctly. But if that's so and my iso-octane and n-heptane calculations are without error, that leads me to believe that the 14.7:1 figure is NOT based...
Ok...
iso-octane: 2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O
and
n-heptane: C7H16 + 11 O2 ----> 7 CO2 + 8 H2O
So then I need to account for the mass of these equations and factor in the non-reactive nitrogen component somehow? Hopefully I'm not getting ahead of myself...
Molecular...
This is what I've got so far:
iso-octane = C8H18
n-heptane = C7H16
air = 21% O2, 79% N2 (not exactly, but good enough for my purposes)
combustion products = CO2 + H2O
Where do I go from here?
I'm looking for some help with the stoichiometry of gasoline. 14.7:1 is commonly cited as the stoich ratio for gasoline and air. I believe this ratio is calculated using a primary reference fuel consisting of a mixture of n-heptane and iso-octane. Precisely what mixture is unknown to me...
I see. "Light" and "heavy" ends of the distillation process, not "light" and "heavy" components of individual hydrocarbons.
Is it accurate that heavy hydrocarbons burn less easily? If so, does that hold true even when both varieties have been fully vaporized? In other words, do heavy...
Thanks for the reply. However, carburetors are relevant to me. Nearly all off-road motorcycles still use carburetors. All of the engines in which I am making specific fuel choices use carburetors.