I'm missing something...

[Brian Dessent]

> I am missing something.  How can that NOT be a Mass flow device.  If an
> engine was running @ 20% throttle position on a flat road (constant load) @
> sea level it would be pulling X amount of air though the flow meter.  @ sea
> level there are going to be X' amount of molecules in the air & the flapper
> is going to move to some value.
> 
> @ 5000 feet using the same 20% throttle position on the "same" flat road is
> going to simulate the "same" load (ignoring the air resistance issue, say 20
> mph, so it is negligible), the air is only going to be 83.3% as dense
> (24.92/29.92), so it is going to have 83.3% fewer molecules in the air to
> "push" the flow meter open.
> 
> would this not "compensate" for the altitude change?  I would expect that
> depending on how "smart" the ECM was it would compensate right or wrong, but
> it ought to be doing some compensation.

I don't know the textbook answer either but here's my reasoning...

In order for the flapper door to open at all, there must be a pressure
differential across it to create a force that acts againt the spring. 
If the pressure was the same on both sides, there would be no net force
and it would stay closed.  The way the pressure difference is created is
from the velocity of air moving across it.  (I guess technically there
could be a pressure difference with zero velocity, but that would be a
transient state and not a steady-state since as soon as the pressure is
not equal on both sides the flapper would begin to open and there would
be flow, hence velocity.)  Hence the amount of force on the flapper is
proportional to the velocity of the flow.  Conveniently, the amount of
force on the flapper directly controls the cross-sectional area through
which the flow passes, so now we know the volumetric flow rate once we
have velocity and area (it's the product of the two.)

In order to get from a volumetric flow rate to actual mass flow rate you
would need to know the temperature and ambient (atmospheric) pressure. 
(Just differentiate both sides of P*V = m*R*T with P, T, and R constant
, and you get m-dot = V-dot * P / T / R, m-dot is mass flow rate, V-dot
is volumetric flow rate [V is volume here not velocity.])

I know I could easily have made a serious error in my own understanding,
can anyone add their approval or disapproval?

Brian

PS - I'm not sure this is relevant or not but in your example, wouldn't
one need more than 20% throttle opening in the second case (at
elevation) for "all things to be equal", i.e. engine operating at the
same load?
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