Re: Aftermarket intake questions
- From: Alan Baker <alangbaker@xxxxxxxxx>
- Date: Sat, 07 Jul 2007 00:30:31 GMT
In article <4ult83t1nknkdjv7uk4gd6ub9lnremhi5e@xxxxxxx>,
rammm@xxxxxxxxxxxxxxxxxxxxxxxxxxx (Leon van Dommelen) wrote:
Alan Baker <alangbaker@xxxxxxxxx> wrote:
In article <6kvq835i3g3f6r9jd90dnj4krkkkcc0dmr@xxxxxxx>,
rammm@xxxxxxxxxxxxxxxxxxxxxxxxxxx (Leon van Dommelen) wrote:
"nosfatsug" <nosfatsug@xxxxxxx> wrote:
There is simply too much obfuscation so, I'm starting with a clean slate.
What I hear Leon saying is that at any given velocity of a car, lets say
a
Miata for interest of this group, the power output from the engine is the
same, assuming all other external factors are the same (e.g., engine
modifications don't change the profile and thus the drag of the car).
Also, assuming similar fuel mixture in the combustion chamber (which is
controlled by the ECU to meet exhaust and performance criteria),
Yes so far.
the amount
of air entering through the intake system stays the same (e.g., same
power,
same fuel to air ratio, and therefore same air volume).
No, this misses the exact point. It is the *amount of oxygen*, not the
*volume*
that stays the same. By design a Miata engine wants to take in 1.8 L air
by
volume every cycle. When it really needs only a fraction of that, during
highway cruise, say, the throttle keeps out the excessive air. That
means
that the engine fights the throttle, which costs power, hence fuel. Now
if
the
intake air goes down say 30 C in temperature, the volume that the engine
needs
becomes even smaller by about 10 percent. So the throttle must let the
engine
get even less volume, so they fight harder, so fuel consumption goes up.
No.
Because while resistance goes up, the speed of the flow goes down. And
power required to overcome the resistance to the flow is proportional to
the *square* of the speed of the flow, but only directly proportional to
the friction.
No. This is an internal flow, not flow around a car. Speed is irrelevant
here, it is total pressure that counts. The total pressure loss must be
greater for the smaller flow volume.
Internal flow follows the same laws of fluid dynamics as external flow
does.
And resistance is a meaningless term here also. The throttle is not trying
to move through the air. What counts is head loss, not force. (Although
the head loss of course finally shows up as a reduction in pressure on
the top piston surfaces.)
No, but the pistons are trying to draw air through the throttle. In
doing so, they must use power.
IOW, double the density of the air, you halve the required flow by
doubling the resistance. Thus you get an increase by a factor of two
from change in resistance, but a decrease by a factor of four from
change in flow, for an overall decrease in pumping losses by a factor of
two.
Think of taking the Miata to Denver. If it takes in the same number of
molecules of oxygen, in Denver the throttle might be wide open, and there
will be no real pumping losses. In Florida it will be partly closed at
exactly
the same produced power. Pumping losses, here we come.
In Denver, the resistance will be less (not none, BTW), but the air must
be moving much faster to make up for the change in density. Power
expended goes up in proportion to resistance (which has fallen) and up
in proportion to the square of the velocity (which has risen).
Then consider the fact that for *the air resistance of the car*, your
argument
above *does* apply. The air resistance in Denver will be much smaller too,
and
you are probably going at a good clip on this highway if your throttle is
wide
open in Denver. We in Florida suffer. :(
But then, we do not freeze to dead in winter as they do in most of the rest
of
the country. :) But then we freeze to dead in Summer. :( But you can
always
go outside to warm up some in your Miata. :))
Leon
--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling four feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect if you
sit in the bottom of that cupboard."
.
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