Let me try a simpler explanation.

14.7 to one is the normal fuel air ratio in gasoline engines.

As you climb the air (14.7) gets less and mixture goes rich so you
have to lean (reduce) the fuel flow to keep the14.7 to one ratio that
engine likes.

Big John

************************************************** **************************

On Mon, 19 Jan 2009 14:23:57 -0500, Tman
wrote:

Somebody posed that seemingly simple question to me, but kept coming
back to the point that they stumped me.... And I am stumped. What do
you see wrong with the logic in this dialog?

Q: Why do I need to lean out my carb when I climb?

A: Ahem, seems you forgot your PPL ground school. The air is less
dense. Fewer air molecules per unit volume. Therefore, you need less
gas, so you lean it out!

Q': Um, ok. Well I looked at my ground school text, and it shows how a
carby operates. Apparently, avgas is kept at a constant level in a
float bowl, which is vented upstream of a venturi. Air flows through
the venturi, and creates a lower pressure, the resulting differential
pressure forcing the avgas across an orfice and into the airstream,
where it mixes it all up in a nice and precise ratio.

A': OK, go on.

Q'': Well, as you climb, I understand the air gets less dense. Let's
assume for simplicity that the volumetric efficiency of the engine
remains fixed, therefore the velocity in the venturi remains the same.
Now the air is less dense, and from the previous chapter in ground
school 101, the differential pressure "p" is related to the density "r"
given a certain velocity "v" like this:
p = 1/2 r v^2
So given a constant velocity, and a decreasing density, won't the
differential pressure decrease, effectively metering less avgas across
that orfice?

A'': OK; I'm sure you're simplifying assumptions are too simple, you
missed something there.

Q''': OK, let's get a little more precise. The mass airflow rate, m,
through a carby is m = c v , where c is a constant for a certain
throttle setting, v is the velocity. Substituting that into the eq's
above , we see that p = 1/2 r m^2 / c^2. Now we know that the mass flow
rate for a liquid across an orfice is very close to proportional to the
square root of the pressure drop [ I actually had to check up on this
one, but it appears to be so:
http://www.efunda.com/formulae/fluid..._flowmeter.cfm ] --
and of course the density of the avgas doesn't change appreciably[!].
Therefore, the avgas flow rate is proportional to "r^(1/2) m". From
this point of view the carby at a constant air density can be viewed as
a device that meters a constant mass proportion mixture of avgas and
air, across a range of mass airflows -- ignoring the effects of
accelerator pumps, full-throttle enrichers, idle circuits and all that.
But note that as the density decreases, the fuel proportion to air
decreases -- suggesting that one would need to ENRICH the mixture when
climbing into less dense air. Assuming that the desired mass proportion
of fuel/air is approximately the same across varying densities (which
seems very reasonable to both of us).

A''': OK, I do follow that (after some work)... and I'm stumped.

Granted, some simplifying assumptions here, but no convincing
explanation of why you would need to lean that red know when climbing...
(and I don't question that you in fact do)....

Anyone see what is amiss?

T