Re: Venturi question


matt271829-news@xxxxxxxxxxx wrote:
RP wrote:
Sorcerer wrote:
<matt271829-news@xxxxxxxxxxx> wrote in message
news:1158097941.649861.101980@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| Sorcerer wrote:
| > <matt271829-news@xxxxxxxxxxx> wrote in message
| > news:1158004973.423968.104990@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | Hi
| > |
| > | It seems to be a well-established fact that the temperature drops
| > | inside a venturi tube (e.g. causing icing in carburettors). The only
| > | explanation I've been able to find is that "the expansion of fluid as
| > | it passes the throat causes a temperature decrease". I'm not sure what
| > | "passes" means here. The pressure inside the throat is lower than the
| > | pressure either side, right?
| >
| > No, not right. A venturi is narrower in the middle.
| > The gas is squeezed down, that RAISES its pressure and temperature,
| > it cools by losing heat to the tube, but then it cools again as it
leaves
| > the exit upon expansion.
|
| The standard explanations (e.g.
| http://en.wikipedia.org/wiki/Venturi_tube) say that pressure is lower
| in the narrow part of the tube.

"Wikipedia does not have an article with this exact name. Please search for
Venturi tube) in Wikipedia to check for alternative titles or spellings."

Wackypedia is a disaster, any idiot can write it, and they do.

Let's put it this way... As a piston falls in the engine, the lowest
pressure is in the manifold, the highest is atmospheric in the air filter.

Venturi tubes are also used to measure water flow and the introduction
of chlorine gas to kill bacteria; the lowest pressure is in the reservoir,
the highest at the faucet. Likewise air pressure is lowest at the top of the
atmosphere.

Thus in the case of a carburettor the flow is from high pressure to low,
but in the water example it is from low pressure to high. If you say
the pressure is lower in the narrow part of the tube, lower than what,
inlet or exhaust?

| If that really means that the density
| of the gas (in terms of molecules per litre, say) is lower in the
| narrow part, then the gas would be expanding as it *entered* the narrow
| part, and compressing as it *exited* the narrow part. This is what I
| can't get my head round. Is it really true that the gas is being
| compressed as it *exits* the narrow part??

The gas is decompressed as it leaves. If you let the air out of a truck
tyre the valve will frost up. The pressure in the valve stem is
slightly lower than the pressure in the tyre (100 PSI), but is much higher
than atmospheric (15 PSI).
So if you have a restriction the gas or water squeezed down at
the start of the tube and then drops pressure as it leaves, but
in the water example the exit pressure is greater than the entry
pressure.
In numbers, 100 ft head of water at the faucet, 0 feet head at the
reservoir which is on a hill.
Faucet closed :
At 50 feet, there is 50 feet head of water on each side the venturi.
Faucet open:
Water flows, there is then 50 feet head on the inlet side of the
venturi, 49 feet head on the other, and 99 at the faucet.

In other words the faucet is supporting the weight of 99%
of the water, the venturi 1%.

Look up Boyle's Law and Charles's Law.
Androcles


I think the question is "where did the heat go?" So far we've
outlined three distinct cooling processes between us. Actually four,
because you and tadchem mentioned the vapor compression cycle
(refrigerator). This cooling cycle is not however due, or even remotely
related to the venturi effect. This would fall under the category of
plain old evaporative cooling instead. On the other hand evaporative
cooling can occur at the venturi outlet providing that the mixture
exiting the outlet contains liquid in some form. But this cooling
effect wouldn't be related to the venturi effect, it would only be
incidental.

The process that you outlined isn't adiabatic. It requires a heat
sink. This was dubbed the "gas cycle" when it was first invented. The
restricted portion of the venturi would have to be maintained at a
temperature lower than the gas within it. If no heat sink is present,
then the cooling effect will only occur temporarily, i.e. until the
venturi and gas temps equalize.

The adiabatic process that tadchem outlined causes cooling of the
entire volume of compressed gas, not just the gas at the outlet. This
is evident in his example of the aerosol can. It is the entire can that
cools rather than just the nozzle. The throttling effect (Joule-Thomson
process) that I mentioned causes a temperature drop that is
proportional to the drop in pressure, or in other words a cooling of
the gas exiting any orifice, be it a venturi or otherwise. Though there
would be some cooling within the connecting tubing and in the venturi,
it would be a much smaller effect than that occurring at the outlet
where the pressure drop is greatest. Evaporative cooling would occur
in similar fashion, since the throttling effect is essentially just a
further evaporation of the gas, so to speak. In the throttling effect,
even though the gas is already in vapor form, it still has internal
"cohesive" PE, and depending upon conditions, that PE may be either net
positive or net negative. The gas may cool down, or it may heat up,
depending upon its initial density and temperature, and of course it's
composition. This is one of the reasons that real gases aren't ideal.

So take your pick, or mix and match.

Matt, can you describe in more detail the actual system that you have
in mind? I'm no expert on this subject, but I might be able to describe
the process in terms other than this or that "effect". Basically, any
time a gas cools, it's because it has lost heat energy. That's a
bit of a tautology, but at least it tells us that the heat had to go
somewhere. I *think* your question is two-fold: How did the heat
transfer occur, and where did the heat go. Based upon your questions
so far, I'm really not sure what you're looking for other than a
better understanding.

But now that you've brought the subject up, I'd like to know a little
more about it too. Maybe someone else can help us all out.

Richard Perry

Thanks for your reply Richard. The question just arose in conversation
- the phenomenon of Carburetor icing was brought up, and it was
suggested that the cooling was due to the venturi effect, but we
weren't sure exactly how it happened. I've just come across another
Wikipedia article actually -
http://en.wikipedia.org/wiki/Carburetor_icing - which says that "Carb
icing occurs when there is humid air, and the temperature drop in the
venturi causes the water vapour to freeze." Unfortunately neither this
nor any other reference I've found explains why this temperature drop
should occur. That's really all I was interested in understanding.

As you say, various different potential cooling processes have been
mentioned. If expansion of the gas is in some way involved then I'm not
at all sure where this is supposed to occur - you say that the pressure
drop is greatest "at the outlet", but as I understand it the pressure
is lowest in the narrowest part of the tube, which seems to imply that
"expansion" would occur as the gas passes from the inlet into the
narrow part. Is that wrong?

Hmmmm!


Maybe this will help. Pressure is the force per unit area acting on a
surface. Since force is a vector, then pressure is also technically a
vector. The pressure on the walls of the venturi is the pressure
exerted in a direction transverse to the direction of flow of the gas.
However, the pressure of the gas along the line of motion is higher
when looking upstream. The former is called the static pressure, and
the latter is called the velocity pressure. By comparing the static
pressure to the velocity pressure you can actually determine the
velocity of the gas through any part of the tube. If there weren't a
continual drop in velocity pressure along the line of motion, then
there would be no flow.

In order to put you on an eye-to-eye level with the moving volume of
gas let's place you right in the gas stream, moving along with it. From
this new perspective the temperature of the gas is much lower because
the molecules have less KE wrt you. Since the temperature is lower,
and since the volume and density remain unchanged through this
transformation, then the pressure must also be lower according to the
Ideal Gas Law. You are now experiencing the static pressure of the gas,
the same pressure experienced by the walls of the venturi tube. The
tube, however, oddly experiences a higher temperature than you do. This
happens for the following reason: The KE of the molecules of gas is
higher wrt the stationary walls, and temperature is a measure of
average KE per molecule of an ideal gas, thus the temperature of the
gas wrt the walls must be higher than it is wrt you. The change in
momentum per second (force) is however the same as that experienced by
you per unit area, thus you and the wall are experiencing the same
static pressure. It is this trick that the venturi plays on the walls
of the tube that produces the venturi effect.

Now if you suddenly come to a stop within the tubing, the gas molecules
will begin striking you with a much greater force (greater change in
momentum) because wrt this stationary frame the molecules have a
greater energy, which translates to greater speed, and thus to a
greater momentum in that direction. The temperature of the gas also
seems to be quite a bit higher, so that what was room temperature gas
entering the stream may even scald you as it passes by and around you.
You've seen pictures and/or footage of the Shuttle during reentry where
the heat shield is glowing hot.

Conversely on the back side of you the velocity pressure is lower than
static pressure. The temperature will thus seem to be a bit lower in
this vicinity than room temperature, if there is laminar flow. With a
5 degree nozzle at the outlet, the surface of the venturi will
experience this lower velocity pressure to some extent since it faces
upstream to an extent, and thus it will experience the lower
temperature associated with the reduced velocity pressure.

Any water vapor present in the moving stream can easily drop in
temperate enough to condense into droplets and even freeze. With a
little turbulence these may even accumulate on the walls.

I think that this is probably the particular cooling effect that you
had in mind.

I've caught some flak over this description in the past by those who
only understand the process in terms of equations, but unless I
acknowledge any technical errors made within, then just ignore any
conspiracy theories that these people generate. :)

Richard Perry

.

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