George, not exactly
certain what you mean by “calculating Air/Fuel ratio”
.
I used the ratio of
the mass of the air to the fuel – which basically relies on the approximation
of 0.0765 lbs per cubic foot of air. So you calculate your air flow in
CFM then times 0.0765 to give you the air mass. The you can take
your desired A/F ratio – say 15:1 and divide 15 into your air mass and
that would give you the fuel mass required to achieve that ratio. Of if you
have any two of the three factors you can find the
third.
IF you mean from a
run time perspective - how do you know you Air/Fuel ratio - then
there are expensive testing instruments and I believe some fairly accurate
air/fuel ratio meters (based on the newer broad band O2 sensor) But
still several hundred dollars when I last
checked.
The narrow band O2
sensor is much cheaper and works just fine with 100 LL for OUR use. I
generally get closer to 200 hours using 100LL before the sensor appears to
lose too much sensitivity to continue to be useful. The common notion
that a few seconds running on leaded fuel will “Kill” an O2 sensor (at least the narrow band ) is
simply not true – at least not for our use. Doing that WILL apparently
degrade it for its intended use in an automobile fuel system (where it needs
to help the engine computer maintain a 14:1 A/F ratio), but if you just want a
general indication of whether you are lean, rich or in the middle, the cheap
O2 sensor works well.
You can even find
some narrow band units which will read out A/F in numeric values but if
using a narrow band O2 sensor, I question the accuracy of such units
myself. I suppose you could use a microprocessor and accurate
analog/digital converter and if you had the “Z” curve of your O2 sensor – you
might get close.
So basically if you
know the mass of air and mass of fuel, you have your Air/Fuel
ratio
So how to arrive at
those two factors, if you know the air pressure and temperature (at normal
atmospheric values) then you essentially know the air density from which
you can calculate air mass and then using your engine flow rates and with your
chosen Air/Fuel ratio - calculate your fuel flow, etc. But, frequently
it’s easier to use our fuel flow (which can be measured fairly
accurately)
So one r way to
approach the problem is as follows: You know the displacement of your
engine and assuming some Ve (volumetric efficiency) (85% - 110%) you can
calculate your air mass flow through the engine for any rpm. So how to
get an approximation of our volumetric efficiency (at least at WOT), its
fairly simple to get close.
Note the ambient
atmospheric pressure (manifold gauge pressure without engine running), fire up
your engine and when warmed up advance it to WOT and note the atmospheric
pressure inside your intake (i.e. your manifold pressure). Lets say
ambient pressure is 29.92 inches HG and lets say you are so lucky to read
29.92 “ Hg in manifold pressure - then theoretically your Ve is 100%.
But lets say your design is not perfect (few are) and your read 28.75”
then your Ve is 28.75/29.92 = 0.9608 or 96.08 % Ve –not bad, but not
perfect.
OK calculate your volumetric flow using our old
displacement formulas and as best I recall at 6000 rpm with a 13B at 100% VE =
277 CFM. Since our intake is not perfect we take our Ve of
0.96.08Ve*277 = 266 CFM actually going through your engine.
Recalling that a cubic foot of air approx = 0.0765 lbm/Cubic Foot, we have 266
* 0.0765 = 20.36 lbsm of air per minute.
Now we don’t
know our Air/Fuel ratio – but we do know our fuel flow at that rpm. Lets
say its 16 gallon/hour, turning that in to lbm/min we have 16 / 60 = 0.2666
gallon/min and we know mass of gasoline is approx 6 – 6.25 lbs/gallon.
So taking 6.0 lb/gallon we have
0.2666 * 6 = 1.6
lbm/min of fuel based on our fuel flow
indication.
Now taking both the
air mass 20.36 lbm/min and the fuel 1.6 lbm/min and we get 20.36/1.6 =
12.725 air/fuel ratio.
Which is very close
to the common “Best Power” ratio used by many.
So don’t know if this
answered any of your question – but, best I could come up with
{:>)
Ed
Ed
Anderson
Ed,
If 12.65: 1 gives
best power, what gives highest RPM is it best power or
14.7:1.
( I'm assuming
12,65:1)
Given we want to run
the coolest for take-off and climb, so do you run best power or
14.7:1.
(I'm assuming best
power) especially at elevation
What air fuel ratio
gives highest EGT, when leaning?
George (down
under)