2gGSX
15+ Year Contributor
- 1,956
- 30
- Feb 15, 2004
-
St. Louis,
Missouri
I propose that meth injection is not enough to significantly alter your octane rating as it's done by most people (e.g. M12 nozzle, delivering 750 cc/min). With dual M12's and 100% methanol, you'd get 2.61 lbs/min mass flow, or 1.137 lbs/min effective flow (flow in terms of gasoline, which is raw mass flow multiplied by 6.4/14.7). In a typical 450-500 lbs/min setup with 50 lbs/min airflow and an 11.5:1 AFR (4.34 lbs/min total fuel, assumed to be 93 octane), you'd end up with 98 octane [113*1.137/4.34 + 93*(4.34-1.137)/4.34]--and that's with 26% of your fuel coming pre-injector/intake manifold.
Stemming from this post: http://www.dsmtuners.com/forums/151401871-post32.html
Ignoring the fact that I just quoted myself...
The problem I have with that calculation is that it seems to decrease the effective octane of the original gas, which makes no sense. Furthermore, the octane properties of methanol should come from mass-flow, not the converted flow. When I try to do anything like that though, I end up with a total octane value of 110+ which makes even less sense. In the end, I think that it's just the nature of how the equation works.
THE BIG QUESTION IS: how much octane does injecting methanol actually add?
Stemming from this post: http://www.dsmtuners.com/forums/151401871-post32.html
While I agree that that is really rich (even more so with the 12 GPH, or 757 cc/min M12 nozzle spewing more fuel in), it seems to get rid of knock, which is always my first priority.
In my opinion you're just at the limits of your fuel, meth and all. Even after the meth injection, I'd put you between 92-94 octane.
Highlight this if you really want to see the fuzzy logic calculations
Assuming you *are* at 9.5:1, you'd be running about 4.63 lbs/min of fuel total. The 50/50 meth being injected gives ~380 cc/min, or .659 lbs/min of meth. Your effective octane rating would be 113*(.659*6.4/14.7[to factor in the lower energy content of methanol]/4.63)+91*(1-.659*6.4/14.7/4.63) = 92.36. That's not 100% accurate, but I'd put you right around there (within 1 or 2 points at most).
If you were to lean your AFR out, you'd have to either lower timing or drop boost. At this point, your only option would be to lower boost and thus some airflow. Your best way of figuring out which nets you more power would be on a dyno, although something like a 70-90 mph 3rd gear pull time comparison may work as well.
I'm actually just ranting now.
Ignoring the fact that I just quoted myself...
The problem I have with that calculation is that it seems to decrease the effective octane of the original gas, which makes no sense. Furthermore, the octane properties of methanol should come from mass-flow, not the converted flow. When I try to do anything like that though, I end up with a total octane value of 110+ which makes even less sense. In the end, I think that it's just the nature of how the equation works.
THE BIG QUESTION IS: how much octane does injecting methanol actually add?