Re: Venturi question


"RP" <no_mail_no_spam@xxxxxxxxx> wrote in message
news:1158163964.500133.227550@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|
| Sorcerer wrote:
| > "RP" <no_mail_no_spam@xxxxxxxxx> wrote in message
| > news:1158151088.204766.146940@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > |
| > | Sorcerer wrote:
| > | > "RP" <no_mail_no_spam@xxxxxxxxx> wrote in message
| > | > news:1158122514.441410.277650@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | > | 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
| > | >
| > | > "Heat" is the mean kinetic energy of the molecules of the gas.
| > | > "Temperature" is the mean kinetic energy of the molecules of the gas
| > | > per unit volume.
| > | > Thus a glowing cigarette is red "hot", but only locally, it is not
going
| > | > to significantly raise the temperature of a room.
| > | >
| > | > In simple terms, "hot" gas pushes the piston simply by bumping into
it,
| > | > that increases the volume, the piston absorbs the mean kinetic
energy
| > | > of the gas molecules which it transfers to the wheels, the molecules
| > have
| > | > less kinetic energy and are "cooled".
| > | > When a diesel engine piston compresses air, it squeezes the gas
| > | > into a smaller volume and raises the temperature. Fuel is then
| > | > injected and the temperature is high enough to ignite the fuel.
| > | > As the piston passes TDC the burning process forces the piston
| > | > down to turn the flywheel and give the car kinetic energy.
| > | > The exhaust valve opens and the kinetic energy of the gas
| > | > is wasted as "heat", that heat is then distributed into a greater
| > | > volume and the temperate of the car's interior is raised to a
| > | > level of human comfort. When too high, the driver opens the
| > | > windows and thus increases the volume, lowering the quantity
| > | > of heat per unit volume.
| > | > Androcles
| > |
| > | Temperature is the KE per molecule.
| >
| > How awkward, my living room has rather a lot of molecules.
| > The temperature is somewhere between 0 Kelvin and 3000 Kelvin.
| > Adjust the thermostat for me, please.
| > Androcles
|
| :) You're much too predictable.

I would hope so, I like to be consistent. Better that than stupidly
claiming temperature is the KE per molecule.
Androcles.


.

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