O2 sensor idea

[G. Scott Ponton]

Tom,
    Yes you can calculate the A/F ratio. Just not enough information in your
post. Need to know air density, fuel flow and specific gravity at known PW,
displacement of engine being tested (actual as it relates to VE).

answer to 3 is no.

    Back on my soap box for a moment. There seems to be a repeated mistake
here. Bruce was right with one of his posts. AFR means little or nothing if
the engine doesn't produce good power. Stoic is not about producing power.
It is about the proper balance of emissions which a given internal
combustion engine produces. Because of variences in the design of different
engines the amount of power produced is not tied to a specific AFR. The
whole point of trying to keep AFR as close to 14.7:1 is to minimize the
output of HC and CO. With a small deviation from stoic. one increases and
the other drops. (CO and HC).
    Lets make some assumtions to simplify the variables.
1. air is 20% oxygen and weights 1lb.
2. the fuel being burned is 45 % HC (the rest of the fuel is reasonably
unimportant for the sake of this example as it is mostly contaminants and
additives to modify its behavior under compression.) and weights 33.
3. VE =100%
4. eliminate all other variables for the moment. (I realize this isn't "REAL
WORLD")

    Ok now we can look at what happens when the chemical process of
combustion takes place. In a perfect situation, with the "facts" known above
what would be the perfect AFR for getting the maximum amount of heat from
the available fuel? (heat is what does the work).  By assuming (because I
can't remember the exact molecular weights of oxygen, carbon,and hydrogen
etc.) the weights given are accurate it would require 148.5 lbs. of air to
complete combustion. This would leave no free HC or O2 after combustion is
complete. So the AFR would be 4.5:1.

45%x33=14.85  =amount of HC per lb.
20%x1=.2   =amount of O2 per lb.
14.85x2=29.7  = amount of oxygen needed to complete combustion (not real
world as other parts of the
                             fuel  need to be oxydized also. 1 O2 per each
carbon and per each hydrogen)
29.7 / .2= 148.5    =amount of air needed
148.5 /33 = 4.5:1  =AFR

    Ok last part. Even in our perfect example what is wrong with this
scenario?

1.The amount of time it takes to oxydize the fuel. We only have a few ms to
complete our work when an engine is operating at 6,000 RPM (for example).

2. Whether we can completely mix the air and fuel so each O2 atom is
adjacent to each HC molecule.

    If we cannot accomplish both of the above we will have to increase the
amount of fuel until all the available oxygen is used in order to make the
most amount of heat from the available fuel. The design of the combustion
chamber and other such variables dictates the rest. But not all is lost.
>From this we can see that, up to a point, in a narrow range of reasonably
close AFRs, the oxygen content of the exhaust will remain fairly stable
until sufficient amounts of fuel are added at which point the oxygen content
will very sharply drop to some negotiable level.

    So is everyone near as confused as I am at this point?? :-)

Scott