In article , Gary Coffman says...

Note that this is also where energy is dissipated *if there is no
regulator present*. The presence or absence of a regulator
doesn't affect *where* the energy is dissipated, only the amount
that is dissipated at any given instant (more without the regulator
than with it, because the regulator is what limits mass flow).

I think I would have changed the emphaisis on the first part there,
to be:

That is where the energy is dissipated, even if there is
*NO* regulator present. You get the same effect if
a smaller hose is used, to passively limit the flow rate,
rather than a regulator.

But seeing as he also feels that the smaller hose would
be a 'loss element' I honestly don't think he would appreciate
the analogy.

Jim

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Gary Coffman writes:

If you have the same size pipe on the downstream side,
and the same mass of gas flowing in both the upstream and downstream
pipes, where did the pressure go?

Air expands in going through a regulator. The volume of the connecting
pipes has no relevance; this is just one example of your faulty analysis.
The volume of the *flow* is what counts.

So the energy not required to meet load demand at the set pressure
does indeed remain in the reservoir.

No. Some remains, some is lost.

1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

2. AIR EXPANDS GOING THROUGH A REGULATOR.

3. THUS, A REGULATOR ITSELF EXPENDS WORK.

You seem to deny these points and conclusions, and I can only assess that
you are hopelessly confused by your faulty analysis.

jim rozen writes:

But seeing as he also feels that the smaller hose would
be a 'loss element' I honestly don't think he would appreciate
the analogy.

Are you saying hoses do not themselves lose (waste, consume, dissipate,
whatever) power?

Richard J Kinch wrote:

1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

This is key to my misunderstanding. In everything I've read on the previous
(original) thread, this wasn't explained, only asserted. So, I have to ask:

Why does any expansion of compressed air necessarily do work?

Grant Erwin

In article ,
says...

Air is driven through the aperture of a valve by a force which is the
product of the pressure differential existing across the aperture times
the area of the aperture. A small aperture will necessarily have less
force pushing air through it, and will push less air in a given time. But
that doesn't affect the amount of work the air can do on the load side.

Work isn't a time function. Work is force times distance. How long it
takes is irrelevant. Less work is done pushing air through a small
aperture than through a large one of the same length. In other words,
open the valve fully, no regulation, and more work is required to move
air through the length of the valve than would be required if the valve
were nearly closed.

Of course you move more air per unit time in the former case, but
work isn't a time function, so that's not relevant when calculating
dissipation.


This is a proud day indeed for RCM. Gary has just abolished
both the second law of thermodynamics and entropy with a
rhetorical flourish. All the more miraculous is this was
all done without resorting to pesky physics, mathematics or
experimental evidence, and in the face of numerous
counterexamples.

All that's left is construction of a device to move air
from a low pressure to high pressure vessel without doing
any work, beyond those annoying "parasitic losses."

If you listen carefully you can hear Messrs. Rumford,
Joule, Carnot, Boltzmann and Helmholtz applauding from
beyond the grave.

Ned Simmons

In article , Richard J Kinch
says...

Are you saying hoses do not themselves lose (waste, consume, dissipate,
whatever) power?

That's the gist of it. If you have a small hose, you cannot
support a large flow to run, say, a large impact wrench.
The same as trying to fill your washing machine through
a piece of 1/8 inch diamteter copper line. It will just
take a long time.

But the small piece of copper line does not dissipate
energy when doing so.

Jim

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In article , Richard J Kinch
says...

2. AIR EXPANDS GOING THROUGH A REGULATOR.

And yet, oddly enough, it will expand even
*without* the regulator. I don't understand
how you would run your impact wrench _without_
expanding the air.

Jim

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In article ,
says...
Richard J Kinch wrote:

1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

This is key to my misunderstanding. In everything I've read on the previous
(original) thread, this wasn't explained, only asserted. So, I have to ask:

Why does any expansion of compressed air necessarily do work?


Your intuition is right, expansion of a gas doesn't
necessarily do work, but the entropy of the gas increases
as it expands.

As I've said before, the "lossy" character of regulators is
easy to demonstrate without resorting to thermodynamics.
However, understanding the energy balance of the loss is
impossible without thermo, and entropy is central to the
explanation.

Ned Simmons

Ned Simmons wrote:

Richard J Kinch wrote:
1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

This is key to my misunderstanding. In everything I've read on the previous
(original) thread, this wasn't explained, only asserted. So, I have to ask:

Why does any expansion of compressed air necessarily do work?
Grant

Your intuition is right, expansion of a gas doesn't
necessarily do work, but the entropy of the gas increases
as it expands.

As I've said before, the "lossy" character of regulators is
easy to demonstrate without resorting to thermodynamics.
However, understanding the energy balance of the loss is
impossible without thermo, and entropy is central to the
explanation.

I'm not looking for a demonstration, Ned. A simple explanation would do.
If you find it "impossible" to explain then that's one thing, but could
you at least try? I'm very frustrated with people who simply assert complex
facts without trying to explain them. They do teach thermodynamics to college
freshmen, and I don't usually fail to understand concepts from freshman physics.
Grant

Grant Erwin wrote:
Ned Simmons wrote:

Richard J Kinch wrote:
1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

This is key to my misunderstanding. In everything I've read on the
previous (original) thread, this wasn't explained, only asserted.
So, I have to ask:

Why does any expansion of compressed air necessarily do work?
Grant

Your intuition is right, expansion of a gas doesn't
necessarily do work, but the entropy of the gas increases
as it expands.

As I've said before, the "lossy" character of regulators is
easy to demonstrate without resorting to thermodynamics.
However, understanding the energy balance of the loss is
impossible without thermo, and entropy is central to the
explanation.

I'm not looking for a demonstration, Ned. A simple explanation would
do. If you find it "impossible" to explain then that's one thing, but
could you at least try? I'm very frustrated with people who simply
assert complex facts without trying to explain them. They do teach
thermodynamics to college freshmen, and I don't usually fail to
understand concepts from freshman physics. Grant

Maybe it's a little beyond the scope of a newsgroup discussion. I don't
understand why there seems to be some sort of onus here placed on Ned and
Richard to prove or explain the basics of thermodynamics. Just because they
"teach" thermo to college freshmen doesn't mean they fully understand the
concept of entropy.

In article , Grant Erwin says...

Why does any expansion of compressed air necessarily do work?

Start with an empty air tank, and hook your compressor up to it.
The compressor runs, and increases the pressure in the tank.
The electrical energy that ran the compressor is now stored,
in a way, inside the now-compressed air in the tank.

So now there is some energy in the tank that one can use for
'doing stuff.'

To use the energy, one must make the compressed gas expand.
You could:

1) hook the tank up to another, empty tank. Now you have a process
(thermodynamic lingo) where some of the gas at one pressure
transfers over to the empty tank, so you now have a larger
volume at a lower pressure. The first tank cools off (if this
is 'adiabatic', or thermally insulated) and the second one heats
up. If you've ever seen a scuba tank filled you will see that
the second tank heats up. This is an outward sign that there
is work going on - work being done to compress the gas in the
second tank.

2) you could hook the tank up to an air piston, and have the
piston move up and lift a weight or do some other mechanical
work, maybe undo a bolt if there is an impact wrench involved.
Like gary says, the total amount of *energy* used in lifting
the weight is the same if it goes up fast, or slow. That is,
if you open the valve to the piston a lot, or a little. Of
course the time rate of energy transformation ("power") is
larger if you whack the valve wide open, rather than crack it
just a bit.

Think of the tank as a storage battery - to get out the
work the compressor did, you need to allow the air to
perform work by expanding. To avoid the (bad) [1] electrical analogy
you should look at the gas law. If you want to add a small
volume of gas to a reservoir at some pressure, you need
to do some work. The process works backwards, you extract
that work out of the system if the same volume of gas
is expanded out.

What if you just take the tank and don't hook it up to
*anything*? That is, simply open the valve and vent it to
the atmosphere? Then you are pretty much at example (1) above,
but the second 'tank' is really, really large. So when the
surrounding environs heat up, the temperature rise is really
tiny - so small you don't notice it.

I hope this provides some insight.

Jim

[1] and do we ever know it's bad.

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In article ,
says...
In article , Richard J Kinch
says...

Are you saying hoses do not themselves lose (waste, consume, dissipate,
whatever) power?

That's the gist of it. If you have a small hose, you cannot
support a large flow to run, say, a large impact wrench.
The same as trying to fill your washing machine through
a piece of 1/8 inch diamteter copper line. It will just
take a long time.

But the small piece of copper line does not dissipate
energy when doing so.


Actually it does; the temperature of the water will rise as
a result of the pressure drop as it flows thru the copper
tube. Some of the heat will be dissipated (in the sense
that a resistor dissipates energy) by the tube.

I also can't see any problem, in the context of this
discussion, with saying a hose, or any other restriction,
dissipates energy when a compressible gas flows thru it.
The energy may not be dissipated by the regulator in the
same sense as a resistor, but that use of "dissipate" is an
example of jargon, and presumably we've rejected electrical
analogies.

The use of "dissipation" to describe the loss of available
energy in a compressible fluid may confuse folks used to
the way it's applied to an electrical resistance, but that
doesn't make using the word in another sense incorrect. I
think it's pretty descriptive of the increase in entropy of
an expanding gas, and consistent with the ordinary
dictionary definition of dissipate-arguably more consistent
than the electrical usage.

Ned Simmons

In article , Ned Simmons
says...

The use of "dissipation" to describe the loss of available
energy in a compressible fluid may confuse folks used to
the way it's applied to an electrical resistance, but that
doesn't make using the word in another sense incorrect. I
think it's pretty descriptive of the increase in entropy of
an expanding gas, and consistent with the ordinary
dictionary definition of dissipate-arguably more consistent
than the electrical usage.

The term is pretty specific as a rule. Basically it
means a process that converts any kind of energy into
thermal energy. Definitely means, 'increase in
entropy' as you say.

In the electrical sense, it implies that somewhere
there is a resistor burning the power, with associated
johnson noise.

Jim

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In article ,
says...
Ned Simmons wrote:

Richard J Kinch wrote:
1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

This is key to my misunderstanding. In everything I've read on the previous
(original) thread, this wasn't explained, only asserted. So, I have to ask:

Why does any expansion of compressed air necessarily do work?
Grant

Your intuition is right, expansion of a gas doesn't
necessarily do work, but the entropy of the gas increases
as it expands.

As I've said before, the "lossy" character of regulators is
easy to demonstrate without resorting to thermodynamics.
However, understanding the energy balance of the loss is
impossible without thermo, and entropy is central to the
explanation.

I'm not looking for a demonstration, Ned. A simple explanation would do.
If you find it "impossible" to explain then that's one thing, but could
you at least try? I'm very frustrated with people who simply assert complex
facts without trying to explain them. They do teach thermodynamics to college
freshmen, and I don't usually fail to understand concepts from freshman physics.
Grant



Unfortunately the demonstration is much easier than the
explanation. I just looked at Resnick and Halliday (my
freshman physics text) and it devotes about 40 pages to a
good quick introduction to thermo. I took a full semester
thermo course as a sophomore or junior MechE student around
1972 and haven't thought about it much since. This is my
long winded way of saying that even if I felt qualified,
I'd be reluctant to try to explain one of the most abstract
concepts in the classical ME curriculum here.

That said, if you're willing to spend some time digging for
material you find readable, here are a few of the concepts
that are relevant. Please don't take any of this as a
rigorous presentation, and don't waste time with web sites
that use entropy to prove the existence (or non-existence)
of God. g

The first law is the conservation of energy, pretty
straightforward and intuitive to most of us. However, many
processes can be imagined that, although they do not
violate the first law, do not proceed naturally. For
example a hot chunk of metal will not jump up in the air
and drop in temperature to compensate for its increased
potential and kinetic energy.

The second law predicts which processes *will* occur
naturally, and entropy is the property that quantifies the
2nd law. Any process that occurs on its own will move in a
direction such that net entropy increases. Entropy is *not*
conserved, and to my mind, is not at all intutitive.

Entropy can also be thought of as an indicator of how much
of the energy in a system is available to do work. As
entropy rises, energy available for work falls. Relevant to
this thread, the entropy of a gas increases as it expands,
reducing available work.

That's as far as I'm sticking my neck out-hopefully it'll
give you a place to start.

Ned Simmons

On Fri, 16 Jan 2004 12:56:25 -0800, Grant Erwin

I'm not looking for a demonstration, Ned. A simple explanation would do.
If you find it "impossible" to explain then that's one thing, but could
you at least try? I'm very frustrated with people who simply assert complex
facts without trying to explain them. They do teach thermodynamics to college
freshmen, and I don't usually fail to understand concepts from freshman physics.

No doubt, and I don't doubt that you'd understand thermo as well with
some effort, if you cared to invest it, but you may have to hurt your
head more than you want to or think it's worth. . Nobody
understands it without some effort. It's a mess.

Unfortuantely, that's *exactly* what thermo profs do: assert complex
facts with equations without explaining. I never did get a
satisfactory explanation of entropy. "What is entropy"? "It's S".
Mmph.
"Yessir, but what is the meaning of it?" "It's the integral of Q/ T
over a path of integration." Oh, right, that clears it right up and
thankyouverymuch you old jerk. Passed course (got a B) by
regurgitating equations with no idea what they meant in any intuitive
or sensible context, as did the rest of the herd.

I've cracked a few books over the past few weeks and I think I sort of
understand what's going on now, but not nearly well enough to explain
it clearly and succinctly even to myself. It's not magic, but it
isn't a simple matter lending itself to a simple intuitive
understanding with simple analogies. There are too many different
and interdependent things going on: change of pressure, temperature
and volume, and energy content in the form of pressure time actual
volume (cu ft of gas at given absolute pressure) vs energy content as
heat which is not the same as temperature except perhaps in terms of
mass. I'm not trying to be obscure, that's as clearly as I can
state it given my admittedly limited present state of understanding.
This cat has a lot of tails on it.

The only "simple" observation I can offer from what I've learned from
my recent Excedrin time with the books is that energy in the form of
heat * in excess of that manifesting itself as an increase in
pressure of given volume or volume at given pressure by rasing
temperature* may not be recoverable as work. It has been
asserted several times that when gas is expanded then work is
necessarily done. I think that's an incorrect inference though I may
be accused of splitting nits for saying so. There's little or no
dispute that the available energy in the expanded gas is less, so the
energy hadda go somewhere -- but not necessarily as work. Some of it
may have been converted to heat content -- which (again) is not the
same as temperature because temperature also depends upon pressure and
volume. It doesn't leave the regulator to ambient because the
regulator doesn't run hot, so it must remain in the downstream gas --
but at least some of it may no longer be available to do work.

We've seen many places that PV = nT but T is temperature, not heat.
The recoverable portion of the heat conveyed to the downstream gas is
that which increases V (or effectively reduces V drawn from the
reservoir to Gary's point) in the regulated P pipe downstream.

Maybe the difference could be found by calculating the V and T of the
downstream gas (P is set by the regulator) from the equations in
Machinery's handbook, calculate available energy from that, and then
calculate the heat energy at that T from the relationship Jim Rozen
noted using the Stephen-Boltzmann contant for energy (per molecule, I
think) as 3/2 power of temperature. The arcane math and terminology
of termogoddammics was derived from first principles in the first
place, so maybe it'd be easer to get ahold of by doing that for one's
self.

I'll admit to not being that curious. But that may well be how a
termogoddammics course should be taught, at least for openers.
Students then might get some feel for where the hell the math
conventions, definitions and terms came from.

The exhaust gas is usually colder than ambient but not as cold as it
might be were the regulator not there because it's expanding from a
lower pressure. I think the unaccounted-for "where did it go?"
energy is heat content lost out the exhaust even though the exhaust
is still colder than ambient.

One of the laws of thermodynamics asserts that no useful work can be
extracted from the heat energy in a gas that's already at or below the
temperature of the "sink" -- in this case, ambient. That energy is
still there, but it's not available to do work. If you send that
exhaust air in a perfectly-insulated container and bring it to
Minnesota during January, I'll recover some of that energy back as
work!

At that point I called a halt to my Excedrin time in favor of Miller
time, quietly bowed out from fracas and the hell with it.
But I'm still following along!

On Fri, 16 Jan 2004 02:19:30 -0500, Gary Coffman
wrote:

Work is not a time function. You can conduct air from a tank to a piston
though either a large pipe, or a restricted pipe. The resulting work done
by the piston is the same, only the rate of doing the work changes.
When doing an energy accounting (change in potential energy vs
work done), rates don't matter.

Air is driven through the aperture of a valve by a force which is the
product of the pressure differential existing across the aperture times
the area of the aperture. A small aperture will necessarily have less
force pushing air through it, and will push less air in a given time. But
that doesn't affect the amount of work the air can do on the load side.

Would you agree that downstream pressure affects the amount of work
the air can do on the load side, and that pressure drop in an
aperture depends on flow rate?

jim rozen writes:

Are you saying hoses do not themselves lose (waste, consume, dissipate,
whatever) power?

That's the gist of it.

Very well then, we'll leave it at that.

Grant Erwin writes:

1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

Why does any expansion of compressed air necessarily do work?

Because it necessarily applies a force (pressure times some area being
expanded) along some distance (the expansion), which is the *definition* of
work.

jim rozen writes:

The first tank cools off (if this
is 'adiabatic', or thermally insulated) and the second one heats
up. If you've ever seen a scuba tank filled you will see that
the second tank heats up. This is an outward sign that there
is work going on - work being done to compress the gas in the
second tank.

The elevated temperature of compressed air is due to the concentration of
heat which pre-existed in the input air, not from newly added heat (aside
from a slight bit of friction). The compressed air holds the work of
compression as a recoverable potential, not as heat.

Scuba tanks filled from a bulk cylinder bank (cylinders previously cooled
to room temp) do not heat up like scuba tanks filled on-the-fly from a
running compressor drawing free air.

jim rozen writes:

What if you just take the tank and don't hook it up to
*anything*? That is, simply open the valve and vent it to
the atmosphere? Then you are pretty much at example (1) above,
but the second 'tank' is really, really large. So when the
surrounding environs heat up, the temperature rise is really
tiny - so small you don't notice it.

No. Compression and expansion do not create heat. If you vent a compressed
tank to the atmosphere, the energy is spent in three ways: (1) work
expanding the planetary atmosphere, which is not heat, and which is
theoretically recoverable, (2) friction heating at the orifice, and (3)
kinetic energy in the motion of the exiting flow, which kinetic energy
eventually turns to heat as the moving air settles.

If you were to take a tank of compressed air above the atmosphere and vent
it in the vacuum of space, then (1) is no longer present, (2) is about the
same, and (3) does not turn to heat, but the gas retains its kinetic
energy, moving outward and expanding forever.

In article , Richard J Kinch
says...

If you were to take a tank of compressed air above the atmosphere and vent
it in the vacuum of space, then (1) is no longer present, (2) is about the
same, and (3) does not turn to heat, but the gas retains its kinetic
energy, moving outward and expanding forever.

The case of venting it into vacuum is the same as venting it to
atmosphere. Basically you are now connecting two reservoirs,
the second of which is at zero pressure to start rather than
15psi absolute.

The second reservoir heats up. If your world view does not
allow this then you have never seen a scuba tank filled. A
theoretical treatment that fails to predict real-world experimental
outcomes is called 'wrong.'

Jim

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In article , Richard J Kinch
says...

Scuba tanks filled from a bulk cylinder bank (cylinders previously cooled
to room temp) do not heat up like scuba tanks filled on-the-fly from a
running compressor drawing free air.

Again, if your 'theory' that you ascribe to fails to predict what
actually happens in real life, it is simply plain flat out
wrong. I and about a dozen other ng members have personally
performed both examples (filling from a compressor, and filling
a tank from another tank) and can assure you that the reservoir
being filled does not care one single iota what is at the
other end of the fill fitting.

It warms up in both cases. If you want do the same experiment
yourself, or read up on the details of the gas law.

Jim

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In article , Don Foreman says...

We've seen many places that PV = nT but T is temperature, not heat.

Another nit if you will to pick:

PV = nkT

Universal gas law - this is applicable to 'ideal' gasses,
and as such, the temperature is indeed an exact measure of
the internal energy, or heat, of the gas.

Temperature in general is not the same thing as heat, but
for ideal gasses, the relation is true; knowing the temperature
allows one to calculate the exact average kinetic energy of
the gas atoms in the volume.

Jim

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jim rozen writes:

The case of venting it into vacuum is the same as venting it to
atmosphere.

I disagree. Nothing to push against, no force, no work, no heat. Just
eternal expansion.

jim rozen writes:

It warms up in both cases.

Differently so, and not from the reason you state.

jim rozen writes:

If your world view does not
allow this then you have never seen a scuba tank filled.

Ones "world view" should not take a casual observation of elevated scuba
tank temperatures to deduce nonsense like your "compressing a gas creates
heat".

In article , Richard J Kinch
says...

The case of venting it into vacuum is the same as venting it to
atmosphere.

I disagree. Nothing to push against, no force, no work, no heat. Just
eternal expansion.

Look, you seem to be a very thoughtful individual, who maybe has not
had a great deal of formal training on the subject. I've had a little,
and there are others here who have had a lot. So making bald
statements like the one above simply won't fly. An introductory
book on thermodynamics will show you that the two cases (connecting
a charged reservoir of compressed gas to two different second
reservoirs - one at zero pressure absolute, and the other at 15psi
absolute - are handled exactly the same way mathmatically.

And that the two cases where the second reservoir is about the
same size as the first, or very much larger, are also part and
parcel of the same behavior.

The rules don't require any piston to push against, when connecting
two compressed gas bottles, each at different pressures or sizes.
The gas law explains what happens when the valve is opened, and
a number of posters here have pointed this out. Replying
"but it's just not *so*" may be emotionally satisfying but really
does not have a great deal of predictive power, nor does it
enlighten you about the subject at hand.

Thermodynamics does indeed explain how things like ideal gasses
behave, what entropy really is (and not just the popular
definition one finds in the media) and how one can predict exactly
how efficient a given process or engine is going to be, and what
the source of the various losses are. It arose partly because
at the time, there were real world engines being manufactured
and folks wanted a theoretical basis for their behavior - which
had been wanting. It also arose to explain certain non-classical
behaviors and was a glimpse into the idea that the world really
was not explainable with newtonian physics alone - that one
could not simply model things with little levers and gears and
so on, and *still*get*the*right*answer*.

Your attempts to explain the behavior of ideal gasses without
using thermodynamics is leading you to the same point that our
predecessors reached: the wrong answer.

Charging a compressed gas bottle from a compressor, or from
another, higher pressure bottle, will yeild the same result: the
bottle that is filling will warm up. Thermodynamics says this
happens in both cases, and further that it happens for the
exact same reason. And the experiement has been done time and
time again over the years since the theory was written down,
and each and every time the experiment has agreed with the
theory.

Jim

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Years ago I worked for the 3M Co. in their complex in St. Paul MN. It covered
about 3/4 sq. mile and had about 18 buildings. I had just completed a 2 yr.
company funded AC&R course. Even before that I was an evangelical supporter of
the law of conservation of energy (and matter). The company came out with a
program saying that whoever could come up with a program that would save the
company money, they would reward the person with a generous amount of the
savings. I went over to the boilerhouse and saw 3 - 5000 HP steam turbine
driven R-12 compressors which were delivering a huge amount of liquid
refrigerant at 175 PSI through an orifice to almost atmospheric. It seemed to
me that there must be a lot of energy disappearing. I considered the idea of a
metorite entering the earth's atmosphere at high speed. At first it develops
heat by friction but the relative motion between the gas and solid produces
cooling. Later in the process different factors take over and the meteorite
burns up. I surmise that the same thing is happening here but at a lower
velocity level and the heat is carried away by the refrigerant which is cool
because it has evaporated.My idea was to install some sort of turbine-generator
in place of the orfice to extract the energy that would otherwise be wasted as
heat. My research told me this is commonly done in air and gas liquefaction
plants. I made some calculations and concluded that this could result in a
savings of 8 to 10 % in the cost of air conditioning. I submitted my idea but
the head honchos said that there were easier ways to get that kind of savings
in the system. Too bad They didn't tell me what they were as I would have
submitted them. In the end I came away empty-handed.
Engineman1

On 17 Jan 2004 09:39:39 -0800, jim rozen
wrote:

In article , Don Foreman says...

We've seen many places that PV = nT but T is temperature, not heat.

Another nit if you will to pick:

PV = nkT

Universal gas law - this is applicable to 'ideal' gasses,
and as such, the temperature is indeed an exact measure of
the internal energy, or heat, of the gas.

Your V is not the actual volume of the gas, Jim. . A 1 cu ft
container can be made to have 100 PSIA pressure at any temperature
you like.

Machinery's Handbook 23d Edition, p 2369 says,

"P x V/ T = 53.3 where P is absolute pressure in lbf/ft^2, V is
volume in cubic feet *of one pound of air at the given pressure and
temperature* and T is absolute temperature in deg R. "

Ratiometric relationships (with appropriate exponents) between
P1,V1,T1, P2, V2 and T2 still hold because the constant cancels.

jim rozen writes:

Charging a compressed gas bottle from a compressor, or from
another, higher pressure bottle, will yeild the same result: the
bottle that is filling will warm up.

One of these two is approximately isothermal, the other approximately
adiabatic. One is a net compression, the other a net expansion. One pumps
in heat from the ambient air, the other moves it from one part of the
system to another. The results are not the same.

Look, you seem to be a very thoughtful individual, who maybe has not
had a great deal of formal training on the subject.

Wrong. Again. I laugh and wave goodbye.

In article , Don Foreman says...

PV = nkT

Your V is not the actual volume of the gas, Jim. . A 1 cu ft
container can be made to have 100 PSIA pressure at any temperature
you like.

The n above gives the number of atoms, basically.

In that equation, the V really is the actual volume. In your
example, if you have a cubic foot of a gas at 100 psia,
at say, 300K, then I can tell you how many moles of gas
there are.

Jim

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In article , Richard J Kinch
says...

Charging a compressed gas bottle from a compressor, or from
another, higher pressure bottle, will yeild the same result: the
bottle that is filling will warm up.

One of these two is approximately isothermal, the other approximately
adiabatic.

From my former post:

(quote)
================================================== =====================
1) hook the tank up to another, empty tank. Now you have a process
(thermodynamic lingo) where some of the gas at one pressure
transfers over to the empty tank, so you now have a larger
volume at a lower pressure. The first tank cools off (if this
is 'adiabatic', or thermally insulated) and the second one heats
up. If you've ever seen a scuba tank filled you will see that
the second tank heats up. This is an outward sign that there
is work going on - work being done to compress the gas in the
second tank.
================================================== =====================

I was saying that in both cases (tank charge, or compressor charge)
the process was defined as adiabatic. I apologize for the confusion.

Jim

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Tom Quackenbush writes:

You lost me. If you crack the valve on an O2 bottle, even in a
vacuum, the gas will go in one direction and the bottle in the
opposite direction. Doesn't that require force?

If we assume the geometry of, oh, a conventional scuba tank, part of the
kinetic energy will be imparted to the bottle, which will acquire both spin
and translational motion, But since the "system" is simply the bottle and
contents, all it can push on, is itself. The net effect is to propel the
expelled gas and the bottle, converting the potential energy into kinetic
energy. How much of this kinesis is in the flying bottle versus the gas
depends on geometry of the exit orifice.

Consider the case where the gas output is vented out in two orifices
pointed out in opposite directions from the center of the bottle. The
thrusts are exactly opposite, so the bottle won't move. All the kinetic
energy goes into the motion of the expanding gas.

Consider the case where you can make a massless bottle dissolve instantly,
leaving the compressed gas. You just have a glob of gas that expands
forever. The work potential is all converted into kinetic energy of the
expanding gas.

On 17 Jan 2004 20:56:41 GMT, (Engineman1) wrote:

Years ago I worked for the 3M Co. in their complex in St. Paul MN. It covered
about 3/4 sq. mile and had about 18 buildings. I had just completed a 2 yr.
company funded AC&R course. Even before that I was an evangelical supporter of
the law of conservation of energy (and matter). The company came out with a
program saying that whoever could come up with a program that would save the
company money, they would reward the person with a generous amount of the
savings. I went over to the boilerhouse and saw 3 - 5000 HP steam turbine
driven R-12 compressors which were delivering a huge amount of liquid
refrigerant at 175 PSI through an orifice to almost atmospheric.
snip

Reality check: R-11 or R-12? They use R-11 in huge centrifugal
compressor chiller systems like that, and R-12 in automotive systems.

I read that far and went "what's wrong with that statement?..."

-- Bruce --
--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
5737 Kanan Rd. #359, Agoura CA 91301 (818) 889-9545
Spamtrapped address: Remove the python and the invalid, and use a net.

In article ,
Grant Erwin wrote:
Ned Simmons wrote:

Richard J Kinch wrote:
1. ANY EXPANSION OF COMPRESSED AIR NECESSARILY DOES WORK.

This is key to my misunderstanding. In everything I've read on the previous
(original) thread, this wasn't explained, only asserted. So, I have to ask:

Why does any expansion of compressed air necessarily do work?
Grant

Your intuition is right, expansion of a gas doesn't
necessarily do work, but the entropy of the gas increases
as it expands.

As I've said before, the "lossy" character of regulators is
easy to demonstrate without resorting to thermodynamics.
However, understanding the energy balance of the loss is
impossible without thermo, and entropy is central to the
explanation.

I'm not looking for a demonstration, Ned. A simple explanation would do.
If you find it "impossible" to explain then that's one thing, but could
you at least try? I'm very frustrated with people who simply assert complex
facts without trying to explain them. They do teach thermodynamics to college
freshmen, and I don't usually fail to understand concepts from freshman
physics.

To a good approximation, the energy of a gas depends solely on its
temperature, not on its volume. A gas consists of lots of gas molecules,
bouncing around; their speed determines the temperature. So suppose you
have two containers, one filled with gas, and another empty (filled with
vacuum). You connect them, and open the valve between them. The gas
molecules bounce around into the newly connected container, so now you
have two containers each with half the gas. But the gas temperature is
the same (or very close). No work was done on the outside world; the gas
molecules did a bit of work on each other (the ones at the leading edge
of the front going through the valve being accelerated by the ones behind
them), but given time, the distribution of energy inside the gas will
equalize itself out again. Maybe a bit of energy went into making an
audible hiss, which could be heard outside the device, but not much: a
little heat is a lot of energy, and even a loud sound has little energy.
Otherwise no energy escaped.

Nevertheless something was lost. You could have expanded that gas
against a piston, and done useful work. If you had, the gas would have
lost heat as it did the work. It then would have sucked in heat from its
surroundings, to regain its original temperature. In net terms, heat
from the outside would have been converted into useful work. But how to
describe what the gas lost, as some of it went through the valve? It's
not energy. It's something else, and that something else has been nailed
down exactly, and given the name of entropy.


--
Norman Yarvin

In article ,
Ned Simmons wrote:
In article ,
says...

I'm not looking for a demonstration, Ned. A simple explanation would
do. If you find it "impossible" to explain then that's one thing, but
could you at least try? I'm very frustrated with people who simply
assert complex facts without trying to explain them. They do teach
thermodynamics to college freshmen, and I don't usually fail to
understand concepts from freshman physics.
Grant

Unfortunately the demonstration is much easier than the
explanation. I just looked at Resnick and Halliday (my
freshman physics text) and it devotes about 40 pages to a
good quick introduction to thermo.

But that book is pretty pathetic; it drones on and on, not saying things
as accurately and concisely as they should be said. Fermi's book on
thermodynamics is much better (as befits one of the greatest physicists
of the last century): it spends about the same number of pages on a
really good explanation of everything up to and including entropy. It's
not easy reading, but it isn't difficult for stupid reasons; it is as
easy as the subject permits.

Please don't take any of this as a
rigorous presentation, and don't waste time with web sites
that use entropy to prove the existence (or non-existence)
of God. g

Also don't waste time listening to people who try to say that entropy is
a way of measuring the amount of chaos in a system, as opposed to the
amount of order. Entropy doesn't correspond to everyday notions of order
and chaos. What these people say is that increasing entropy means
increased chaos; but if, for instance, you pour one jar of paint into
another, resulting in a chaotic-looking mess of colors, that system has
less entropy than it does after you mix the colors thoroughly, creating a
nice, even, orderly, uniform color.

The first law is the conservation of energy, pretty
straightforward and intuitive to most of us. However, many
processes can be imagined that, although they do not
violate the first law, do not proceed naturally. For
example a hot chunk of metal will not jump up in the air
and drop in temperature to compensate for its increased
potential and kinetic energy.

The second law predicts which processes *will* occur
naturally, and entropy is the property that quantifies the
2nd law. Any process that occurs on its own will move in a
direction such that net entropy increases. Entropy is *not*
conserved, and to my mind, is not at all intutitive.

Entropy can also be thought of as an indicator of how much
of the energy in a system is available to do work. As
entropy rises, energy available for work falls. Relevant to
this thread, the entropy of a gas increases as it expands,
reducing available work.

That's as far as I'm sticking my neck out-hopefully it'll
give you a place to start.

Your neck is safe.


--
Norman Yarvin

Richard J Kinch wrote:

jim rozen writes:

The case of venting it into vacuum is the same as venting it to
atmosphere.

I disagree. Nothing to push against, no force, no work, no heat. Just
eternal expansion.

I think I see another like my dad. He always said a rocket would not
work
in space since there was nothing to "push against". :-)
...lew...

"Lewis Hartswick" wrote in message
...
I think I see another like my dad. He always said a rocket would not
work
in space since there was nothing to "push against". :-)
...lew...

I had the same argument with my high school physics teacher back in 1953. He
swore that rockets wouldn't work in the vacuum of space with nothing to push
against. I wasn't totally correct in my argument, I was arguing that the
main force driving the rocket
was the unbalanced high pressure in the rocket engine (zero force on the
open nozzle end and thousands of lbs force on the closed other end of the
engine). I was, I think, more correct than he was, though. I didn't know
enough then to even consider the ejection
velocity of the exhaust, etc.

Harold Burton

In article , Norman Yarvin says...

But how to
describe what the gas lost, as some of it went through the valve? It's
not energy. It's something else, and that something else has been nailed
down exactly, and given the name of entropy.

A poster previously stated that he felt that a discussion of
thermodynamics on this ng was a bit to involved. I think
your comments say otherwise.



Jim

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In article , Norman Yarvin says...

... Fermi's book on
thermodynamics is much better (as befits one of the greatest physicists
of the last century): it spends about the same number of pages on a
really good explanation of everything up to and including entropy. It's
not easy reading, but it isn't difficult for stupid reasons; it is as
easy as the subject permits.

Small world. We went out on a treck to a Borders bookstore in
a nearby town, and they had that small volume on the shelf. It
did not however, have any section on ideal gasses. I probably
should have bought it anyway.

Jim

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