Why does one need to LEAN OUT a CARB when climbing?

"Bertie the Bunyip" wrote in message
...
Tman wrote in news:-
:

wrote:
On Jan 19, 7:05 pm, "Todd W. Deckard" wrote:
It diverts low pressure air from the back of the venturi into the
fuel float bowl.
Never taken a close look at an airplane carb, but I understand that the
standard config (not this stromberg, but rather something like in a
basic 172N) has a fuel-bowl vent upstream of the venturi, making the air
pressure in the fuel bowl pretty much equal to the air intake
pressure.... Kind of like a standard auto carb from a few years ago when
they still had them. Let me know if that's not right.


They're basically a float bowl and a tube. not a lot to complicate them,
really. Probably the simplest carbs in operation today.

Bertie

You're an idiot.

Tman,

I've been pondering your question quite a bit. I believe I have it. In
deference to Dan my Camaro never idled correctly.

Ultimately your engine depends on the mass flow of air divided by the mass
flow of fuel. But the amount of fuel drawn up is a function of the pressure
difference in the carbureture venturi. So here goes:

The carb throat is a double venturi and a manometer between the opening and
the neck would show a theoretical pressure drop of:

p(opening) - p(neck) = .5 * density of air * { velocity(neck)^2 -
velocity(opening)^2 }
(Lets ignore carb ice for a second and say that the air is
incompressible).

Bernoulli got the idea from Newton thats why the 1/2 m v squared. Now
discouragingly this has the density
in front of it, which is why you posed the question. The difference in
pressures is directly proportional to the density.

Now the low pressure in the neck of the venturi is what is drawing the fuel
up (or properly the difference between the neck and ambient). Again we use
Bernoulli to describe the forces acting on a particle moving along a
streamline -- and this time it is properly incompressible.

{ Pressure / density } + .5 * { velocity ^ 2 } + gravity * change_in_height
= a constant

Again, Bernoulli copped it all from Newton and was just telling us that
kinetic energy + potential energy = a constant.

If we apply this to your fuel being drawn up we get:

(#) p (ambient) - p (neck) / density of fuel = .5 * (velocity of fuel) ^2 +
(gravity * vertical distance from bowl to jet)

However mass flow is the density * the size of the pipe * the velocity.

So the mass flow of air = density of air * carb barrel size * velocity
(opening)

But from (#) the mass flow of fuel is being determined by the pressure
difference (which also carries the air density)

so Air over Fuel cancels your density term.

An altitude compensating carburator puts a small vacuum on the fuel to
prevent the rho from dividing out.

Mr Wizard could have explained this better that as the air gets thinner it
sucks on the straw with less force but it takes less force to slurp up the
gas becuase of the reduced pressure. So the gas drawn up stays about the
same, however the mass flow of air drops off with density so the mixture
richens.

Q.E.D. Good question. If I ever become a physics teacher I am going to
put this one on the final!

I think you are making this waaaay to complicated. The volume of air going
through the venturi remains the same as you climb but the amount of oxygen
(the component needed to burn the fuel) decreases. The volume of air
remains the same so the fuel drawn out of the float bowl remains the same.
There is less oxygen so the mixture become rich.

That's my story and I am sticking to it.

--

*H. Allen Smith*
WACO - We are all here, because we are not all there.
"Todd W. Deckard" wrote in message
...
Tman,

I've been pondering your question quite a bit. I believe I have it. In
deference to Dan my Camaro never idled correctly.

Ultimately your engine depends on the mass flow of air divided by the mass
flow of fuel. But the amount of fuel drawn up is a function of the
pressure difference in the carbureture venturi. So here goes:

The carb throat is a double venturi and a manometer between the opening
and the neck would show a theoretical pressure drop of:

p(opening) - p(neck) = .5 * density of air * { velocity(neck)^2 -
velocity(opening)^2 }
(Lets ignore carb ice for a second and say that the air is
incompressible).

Bernoulli got the idea from Newton thats why the 1/2 m v squared. Now
discouragingly this has the density
in front of it, which is why you posed the question. The difference in
pressures is directly proportional to the density.

Now the low pressure in the neck of the venturi is what is drawing the
fuel up (or properly the difference between the neck and ambient). Again
we use Bernoulli to describe the forces acting on a particle moving along
a streamline -- and this time it is properly incompressible.

{ Pressure / density } + .5 * { velocity ^ 2 } + gravity *
change_in_height = a constant

Again, Bernoulli copped it all from Newton and was just telling us that
kinetic energy + potential energy = a constant.

If we apply this to your fuel being drawn up we get:

(#) p (ambient) - p (neck) / density of fuel = .5 * (velocity of fuel) ^2
+ (gravity * vertical distance from bowl to jet)

However mass flow is the density * the size of the pipe * the velocity.

So the mass flow of air = density of air * carb barrel size * velocity
(opening)

But from (#) the mass flow of fuel is being determined by the pressure
difference (which also carries the air density)

so Air over Fuel cancels your density term.

An altitude compensating carburator puts a small vacuum on the fuel to
prevent the rho from dividing out.

Mr Wizard could have explained this better that as the air gets thinner it
sucks on the straw with less force but it takes less force to slurp up
the gas becuase of the reduced pressure. So the gas drawn up stays about
the same, however the mass flow of air drops off with density so the
mixture richens.

Q.E.D. Good question. If I ever become a physics teacher I am going to
put this one on the final!

On Jan 21, 7:19 am, "Allen" wrote:
I think you are making this waaaay to complicated. The volume of air going
through the venturi remains the same as you climb but the amount of oxygen
(the component needed to burn the fuel) decreases. The volume of air
remains the same so the fuel drawn out of the float bowl remains the same.
There is less oxygen so the mixture become rich.


More properly, the weight of the air decreases. Oxygen still
makes up 21% of the air.

Dan

Allen wrote:
I think you are making this waaaay to complicated.
I know I am. I'm an engineer, it's what we do...
Anyways.

The volume of air going
through the venturi remains the same as you climb
I believe this is a good approximation -- volumetric efficiency should
not change (much) as an aircraft climbs into thinner air, and throttle
setting and RPM's approx constant.

but the amount of oxygen
(the component needed to burn the fuel) decreases.
The mass flow decreases of air (and oxygen), and therefore the amount of
fuel required decreases; I agree.

The volume of air
remains the same so the fuel drawn out of the float bowl remains the same.
This is the part I don't agree with. The mass rate of fuel drawn outta
the float bowl is driven by the pressure decrease in the venturi, which
depends on both the density and the velocity. Density goes down as you
climb. Amount of fuel drawn should not remain a constant.

If I can give a rough and poor analogy. As you climb into less dense
air, you don't achieve the same aerodynamic performance from the
airplane with a given airspeed. You will stall at a higher airspeed,
your airplanes gear and flaps can handle higher airspeeds, Vx and Vy are
higher airspeeds, and Vne is a higher airspeed. The air has to go
faster to have the same effect on the wings and other devices on the
bird. It is because of the decreased density... I'd propose that is the
same effect as is happening in the carb venturi. The air has to go
faster to have the same effect in pulling gas out.

Oh yeah, before somebody corrects me I'm particularly talking about TAS
above, I know these speeds stay roughly the same at IAS (just because
the ASI is also thus affected ).


That's my story and I am sticking to it.
And most people are satisfied with that, I'm wondering why it doesn't
stand up to a little deeper analysis.

Thanks, finally somebody on my nerdy wavelength, and a really thoughtful
reply, but...

Todd W. Deckard wrote:


The carb throat is a double venturi and a manometer between the opening and
the neck would show a theoretical pressure drop of:

p(opening) - p(neck) = .5 * density of air * { velocity(neck)^2 -
velocity(opening)^2 }
(Lets ignore carb ice for a second and say that the air is
incompressible).

agreed.


{ Pressure / density } + .5 * { velocity ^ 2 } + gravity * change_in_height
= a constant

I think you are speaking of the velocity of the gas in the orfice system
and the density of the gas, relative to the pressure differential that
is driving the fuel flow. If so, I agree (caveat below). Note that
this is equivalent to saying that the fuel flow is proportional to the
square root of the pressure differential, same assumption I made.


so Air over Fuel cancels your density term.

Check your math carefully, are you sure that you are not confusing the
density of the fuel (constant) and density of the air (decreasing)
terms. I gave this the quick and dirty back of the napkin verification,
and it seems I still had both density terms in the final equation
relating mass airflow to mass fuelflow. If you think you're right...
I'll do a little more rigorous playing with the terms.

My current hunch on this: The mass fuel flow is not proportional to the
square root of the pressure differential, but more or less directly
proportional to the differential. This is because of the viscous
friction effects of the avgas in going through the metering orfices. If
those effects predominate, (not surprising given the very small orfice
sizes), I'd say Bernoulli has little to say about the mass flow rate of
the avgas, and it is more linearly related to the pressure differential.


Q.E.D. Good question. If I ever become a physics teacher I am going to
put this one on the final!

I think I'm going to forward this to one of my old fluid dynamics profs

On Wed, 21 Jan 2009 12:29:21 -0500, Tman
wrote:

Thanks, finally somebody on my nerdy wavelength, and a really thoughtful
reply, but...

Todd W. Deckard wrote:


The carb throat is a double venturi and a manometer between the opening and
the neck would show a theoretical pressure drop of:

p(opening) - p(neck) = .5 * density of air * { velocity(neck)^2 -
velocity(opening)^2 }
(Lets ignore carb ice for a second and say that the air is
incompressible).

agreed.


{ Pressure / density } + .5 * { velocity ^ 2 } + gravity * change_in_height
= a constant

I think you are speaking of the velocity of the gas in the orfice system
and the density of the gas, relative to the pressure differential that
is driving the fuel flow. If so, I agree (caveat below). Note that
this is equivalent to saying that the fuel flow is proportional to the
square root of the pressure differential, same assumption I made.


so Air over Fuel cancels your density term.

Check your math carefully, are you sure that you are not confusing the
density of the fuel (constant) and density of the air (decreasing)
terms. I gave this the quick and dirty back of the napkin verification,
and it seems I still had both density terms in the final equation
relating mass airflow to mass fuelflow. If you think you're right...
I'll do a little more rigorous playing with the terms.

My current hunch on this: The mass fuel flow is not proportional to the
square root of the pressure differential, but more or less directly
proportional to the differential. This is because of the viscous
friction effects of the avgas in going through the metering orfices. If
those effects predominate, (not surprising given the very small orfice
sizes), I'd say Bernoulli has little to say about the mass flow rate of
the avgas, and it is more linearly related to the pressure differential.


Q.E.D. Good question. If I ever become a physics teacher I am going to
put this one on the final!

I think I'm going to forward this to one of my old fluid dynamics profs


You all are making this much too complicated. The reason you lean the
mixture as you climb is.... if you don't the check airman will flunk
you on your check ride.

Ron

:-)

On Jan 21, 10:29*am, Tman wrote:
Thanks, finally somebody on my nerdy wavelength, and a really thoughtful
reply, but...

Todd W. Deckard wrote:

The carb throat is a double venturi and a manometer between the opening and
the neck would show a theoretical pressure drop of:

p(opening) - p(neck) = .5 * density of air * { velocity(neck)^2 -
velocity(opening)^2 }
* * * * * * *(Lets ignore carb ice for a second and say that the air is
incompressible).

agreed.

{ Pressure / density } + .5 * { velocity ^ 2 } + gravity * change_in_height
= a constant

I think you are speaking of the velocity of the gas in the orfice system
and the density of the gas, relative to the pressure differential that
is driving the fuel flow. *If so, I agree (caveat below). *Note that
this is equivalent to saying that the fuel flow is proportional to the
square root of the pressure differential, same assumption I made.



so Air over Fuel cancels your density term.

Check your math carefully, are you sure that you are not confusing the
density of the fuel (constant) and density of the air (decreasing)
terms. *I gave this the quick and dirty back of the napkin verification,
and it seems I still had both density terms in the final equation
relating mass airflow to mass fuelflow. *If you think you're right...
I'll do a little more rigorous playing with the terms.

My current hunch on this: *The mass fuel flow is not proportional to the
* square root of the pressure differential, but more or less directly
proportional to the differential. *This is because of the viscous
friction effects of the avgas in going through the metering orfices. *If
those effects predominate, (not surprising given the very small orfice
sizes), I'd say Bernoulli has little to say about the mass flow rate of
the avgas, and it is more linearly related to the pressure differential.



Q.E.D. *Good question. * If I ever become a physics teacher I am going to
put this one on the final!

I think I'm going to forward this to one of my old fluid dynamics profs

I think your problem is the assumption that there's a linear
relationship between air density and pressure differential. The same
pressure differential is the force that lifts our airplanes off the
ground, and as they gain altitude the density decreases. The stall
speed will rise, but not linearly with the decrease in density; it's
the square root of the decrease in density that we're looking for. In
the carb, one-half the density should then cut the pressure
differential and therefore fuel flow by one quarter, which will give
us a mixture twice a rich as when we took off. We find half the
density at 18,000 feet, incidentally.

Dan

wrote:
In
the carb, one-half the density should then cut the pressure
differential and therefore fuel flow by one quarter, which will give
us a mixture twice a rich as when we took off.
At the risk of dragging on the subject ... Wouldn't that actually lean
it out, requiring one to compensate by richening the mixture at higher
altitudes? One-half the density i.e. one half the mass airflow at
constant velocity, fuel flow by quarter... sounds like that mixture is
leaner!

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?


like most arguments in aviation this one arises because a question is
asked with a misleading part in it.

"why do I lean my carb when I climb" is a bull**** question open to
much misinterpretation.

the fact is that most airfields are near sea level, say under 1,000ft
elevation anyway. there is actually no need to lean a carby at these
altitudes.
in fact the mixture is left rich so that as you climb the over rich
mixture aids in engine cooling.
continental's pilot notes will tell you this.

"why do I need to lean my carburettor at higher altitudes" is probably
a better wording of the question.

that is simple. the air is less dense so the amount of fuel it needs
to achieve full combustion is reduced. the density of the fuel doesnt
decrease so you need less of it.

learn to think in more precise terms and a lot of the confusion
vanishes.
Stealth Pilot