On Mon, 29 Dec 2003 23:04:25 GMT, ATP wrote:

I don't understand why this is not clear to everyone here. Which has more
energy-one liter of 200 psi gas or two liters of 100 psi gas? Can you do
any work going from one state to the other? In one direction you can,

True. There is one catch in this example (and a lot of similar ones in this
thread) though... If you start with one liter of 200 psi gas, and let it
expand through an air motor into a receiving tank until you have two liters
of 100 psi gas, you can extract some work, but you don't end up with those
two liters in the receiving tank.

To accomplish this task you must use a one-liter receiving tank. At the end
of the demonstration, one liter is still in the original source tank,
because the flow stopped when the receiving tank reached 100 psi. Putting
that second liter of air into the source tank cost you more than putting
the first liter in, and more than putting a constant 100 psi against the
motor until the receiving tank was at 100 psi.

Loren

Also e-mailed due an ISP suffering Altsheimer's:

Loren Coe wrote:

does anyone have a suggestion as to how i measure this with
just "normal" equipment, including an O'scope? running back
and forth to the utility co.power meter doesn't seem like a
very good approach.

If your scope is dual trace, you can do what you want with a current
transformer. You need a 120V to 12V transformer rated for a dozen or so
amps. Connect a 10 Ohm, 10 Watt resistor across the 120V winding.
Connect the 12V winding in series with a motor lead. The "backwards"
transformer will have 11A flowing in the "12V" winding which will become
1.1A in the "120V" winding. 1.1A in a 10 ohm resistor is 11 volts so
the other winding (the "12V" winding) will drop only 1.1V. This will
have negligable effect on the motor operation. If you now pick up the
hot motor lead in one channel of the scope and the secondary of the
current transformer in the other channel, you can measure the time
difference between the waveforms and calculate the phase shift.

Ted

Ned Simmons wrote:
In article ,
says...
On Mon, 29 Dec 2003 23:04:25 GMT, "ATP"
wrote:



I don't understand why this is not clear to everyone here. Which
has more energy- one liter of 200 psi gas or two liters of 100 psi
gas? Can you do any work going from one state to the other? In one
direction you can, as long as you don't just waste it with a
regulator.

Putting it another way, as you compress a given volume of gas, does
it have more potential energy the more you compress it? Of course
it does, so isn't there a loss as it expands through a regulator?

Good question! I think the answer is "no" in an isothermal
process.

Power flow is pressure * volume flow, volume here being actual
volume at that pressure. To avoid confusion with what "CFM"
means, let's define volume flow as actual cu in / sec at given
pressure. If I have 1 cu in/ sec at 200 PSIA on the high side I
have 2 cu in/sec at 100 PSIA on the low side (if temp is the same).
Power out = power in ergo no power loss in the regulator.

No, speaking loosely thermodynamically, the power is
proportional to the pressure *squared* times flow. Pressure
* flow is proportional to power for a non-compressible
fluid, for instance in a hydraulic system, but not for a
compressible gas.

Look at the example I posted again, except now imagine a
case where the initial pressure in the reservoir is 5
atmospheres absolute (twice the gage pressure of the
original case), and no regulator. The initial force on the
piston will be twice the unregulated 3 atm case, and the
total travel of the piston will also be doubled. The total
work done by the piston will be 4x.

http://www.suscom-maine.net/~nsimmons/expansion.jpg


Are you assuming the piston is doing the maximum work theoretically possible
by pushing against the greatest possible resistance that would still allow
some minimal motion? Otherwise it would seem that only twice the work would
be done, since the potential energy of a fixed weight would only be doubled
at equilibrium. Now I think I know what you meant by what happens along the
path. This brings up another interesting point- we can design grossly
inefficient mechanisms for using air by not taking advantage of the
available force- the air is going to eventually expand and exhaust to
atmospheric pressure, even if the resistance along the way is sub-optimal.

All this talk about gas flow rates brings up a few questions about a project
I'm working on.
I built a boiler and steam engine to power a bicycle and I'm using a propane
burner to heat the boiler. It was hard to tell when I had the gas flow set
properly so I put a flowmeter in the line. It is the type of flowmeter used to
regulate medical oxygen. It has a calibrated plastic tube in which a small
steel ball floats and there is a needle valve on the outlet to adjust the gas
flow to the burner jet. I've operated it with a direct connection to the
propane tank at about 120 PSI and using a regulator putting out 12 PSI. Am I
correct in assuming that if the flowmeter is set to read say, 5 liters per
second in each case, the actual gas output will be about 10 times as much with
the direct connection?
Engineman1

Don Foreman wrote:
On Mon, 29 Dec 2003 23:04:25 GMT, "ATP"
wrote:



I don't understand why this is not clear to everyone here. Which has
more energy- one liter of 200 psi gas or two liters of 100 psi gas?
Can you do any work going from one state to the other? In one
direction you can, as long as you don't just waste it with a
regulator.

Putting it another way, as you compress a given volume of gas, does
it have more potential energy the more you compress it? Of course it
does, so isn't there a loss as it expands through a regulator?

Good question! I think the answer is "no" in an isothermal process.

Power flow is pressure * volume flow, volume here being actual volume
at that pressure. To avoid confusion with what "CFM" means, let's
define volume flow as actual cu in / sec at given pressure. If I
have 1 cu in/ sec at 200 PSIA on the high side I have 2 cu in/sec at
100 PSIA on the low side (if temp is the same). Power out = power in
ergo no power loss in the regulator.

But you observed that it takes more power to squeeze a given volume
of atmospheric air to a higher pressure, so where does that excess
power go?

There's a checkvalve in the compressor. The pump squeezes the air
until the pressure equals that in the tank; there's no flow until that
happens. The power up to that point has gone into potential energy in
the form of cylinder pressure. When the checkvalve opens, flow
occurs at essentially constant pressure if the reservoir is large
compared to the cylinder, and energy is transfered to the reservoir.
When the cylinder pressure drops below reservoir pressure, the
checkvalve closes, flow stops, and the potential energy in the
cylinder pressure is returned to the flywheel until the next
compression stroke.

At 100 PSI the pump pushes twice as much volume at half the pressure
per cycle as it does at 200 PSI but the energy transfer is the same.
It just pushes with less force for a longer time *while moving air*
at lower pressure. In the higher pressure case there's more
alternating exchange of energy between cylinder and flywheel.

Thermodynamics enters in if heat is lost during compression as is
generally the case. Thus Ingersoll's generally-true note that
compressing to pressure higher than needed is inefficient. The loss
is at the compressor, not at the regulator. The compressor gets hot
but the regulator does not, right?

However, if thermal energy is preserved by being removed from the
compressor and used to warm expanding air, then there is essentially
no loss in efficiency.

Take a look at the example Ned posted and consider that the piston is used
to push some mechanical device of variable resistance- the resistance is
infinitesimally less than the available force from the air side along the
entire path. At maximum extension the resistance approaches 0. I think these
conditions woud extract the maximum amount of work out of our fixed volume
of pressurized air. Under these conditions a doubling of pressure would
result in a doubling of force and distance, a quadrupling of work
accomplished, as Ned stated. In your earlier example with a fixed weight,
much of the potential of the air is squandered- consider an airtight piston
of negligible weight- it would also reach an equilibrium with 1 atm on each
side, but there would be almost no work accomplished. The amount of work
that is accomplished is dependent on the design of our device- to say that
at the end a certain weight was only lifted x amount in each case does not
prove that the unregulated case is not capable of doing more work. My
apologies to Ned if I have misunderstood his example.

Don Foreman writes:

Real compressors aren't isothermal so the
compressed air is hot. Some of the work done by the compressor goes
into heating the air, and that heat is lost to the environment so the
process is inefficient.

Compressed air is hot (has a high temperature), not so much from the
addition of heat, as from the compression.

Richard J Kinch wrote:
Don Foreman writes:

Real compressors aren't isothermal so the
compressed air is hot. Some of the work done by the compressor goes
into heating the air, and that heat is lost to the environment so the
process is inefficient.

Compressed air is hot (has a high temperature), not so much from the
addition of heat, as from the compression.

Yes, that's the way a heat pump works.

Don Foreman writes

Power flow is pressure * volume flow, volume here being actual volume
at that pressure.

No, power = pressure * mass flow, which is why that definition of
"actual volume" instead of "free air volume" isn't useful for such
calculations.

Thermodynamics enters in if heat is lost during compression as is
generally the case.
Thus Ingersoll's generally-true note that
compressing to pressure higher than needed is inefficient.

Losing heat isn't the reason for inefficiency of excess pressure. If it
were, insulation would cure it. Regulating down is inherently
inefficient.

The loss
is at the compressor, not at the regulator. The compressor gets hot
but the regulator does not, right?

The regulator can get downright chilly. Temperature is not heat or a
measure of heat. Something can get colder and gain heat at the same
time.

Loren Amelang writes:

There is one catch in this example (and a lot of similar ones in this
thread) though... If you start with one liter of 200 psi gas, and let
it expand through an air motor into a receiving tank until you have
two liters of 100 psi gas, you can extract some work, but you don't
end up with those two liters in the receiving tank.

A valid point.

However the application to the original problem is that you can regulate
the pressure down via a regulator (essentially a feedback-controlled
variable restriction) or you could conceivably accomplish the same
regulation down via an air motor. The latter case produces work output
that somehow must be accounted for in the former, namely by waste heat.

In article , Ted Edwards says...

If your scope is dual trace, you can do what you want with a current
transformer. You need a 120V to 12V transformer rated for a dozen or so
amps. Connect a 10 Ohm, 10 Watt resistor across the 120V winding.
Connect the 12V winding in series with a motor lead. The "backwards"
transformer will have 11A flowing in the "12V" winding which will become
1.1A in the "120V" winding. 1.1A in a 10 ohm resistor is 11 volts so
the other winding (the "12V" winding) will drop only 1.1V. This will
have negligable effect on the motor operation. If you now pick up the
hot motor lead in one channel of the scope and the secondary of the
current transformer in the other channel, you can measure the time
difference between the waveforms and calculate the phase shift.

I don't understand how he is going to display voltage
with this setup, unless he ties one side of the
incoming line (240, yes?) to the scope common rail,
which is typically by definition power ground. That
will cause smoke to leak out someplace.

Jim

==================================================
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==================================================

Engineman1 writes:

Am I
correct in assuming that if the flowmeter is set to read say, 5 liters
per second in each case, the actual gas output will be about 10 times
as much with the direct connection?

If you mean a rotameter (what it sounds like you are describing) these read
volumetric flowrate at standard conditions (volumes of gas at 14.7 psia and
70 deg F). So no, you would not be at all correct with that factor of 10.

Also, your meter is calibrated for oxygen, and you must also correct
proportionately for the different molecular weight of propane. (Assuming
you have pure propane--I hear it sometimes has a butane fraction.)

On Wed, 31 Dec 2003 19:37:33 -0500, Ned Simmons
wrote:

In article ,
says...
On Mon, 29 Dec 2003 23:04:25 GMT, "ATP"
wrote:



I don't understand why this is not clear to everyone here. Which has more
energy- one liter of 200 psi gas or two liters of 100 psi gas? Can you do
any work going from one state to the other? In one direction you can, as
long as you don't just waste it with a regulator.

Putting it another way, as you compress a given volume of gas, does it have
more potential energy the more you compress it? Of course it does, so isn't
there a loss as it expands through a regulator?

Good question! I think the answer is "no" in an isothermal process.

Power flow is pressure * volume flow, volume here being actual volume
at that pressure. To avoid confusion with what "CFM" means, let's
define volume flow as actual cu in / sec at given pressure. If I
have 1 cu in/ sec at 200 PSIA on the high side I have 2 cu in/sec at
100 PSIA on the low side (if temp is the same). Power out = power in
ergo no power loss in the regulator.

No, speaking loosely thermodynamically, the power is
proportional to the pressure *squared* times flow. Pressure
* flow is proportional to power for a non-compressible
fluid, for instance in a hydraulic system, but not for a
compressible gas.

Thermodynamics are not at issue when the checkvalve opens because
pressure is essentially constant. Power flow is pressure * volume
flow -- look at the units. ( lbf/in^2) * (in^3/sec) = in * lbf /
sec which is power.

Look at the example I posted again, except now imagine a
case where the initial pressure in the reservoir is 5
atmospheres absolute (twice the gage pressure of the
original case), and no regulator. The initial force on the
piston will be twice the unregulated 3 atm case, and the
total travel of the piston will also be doubled. The total
work done by the piston will be 4x.

The piston analogy was a bad idea, apologies for that. It obscures
the issue of transferred power vs store and release of potential
energy. Work is done in raising the weighted piston (assuming the
weight doesn't become zero in the process) that can be recovered when
the piston returns to it's initial position.


But you observed that it takes more power to squeeze a given volume
of atmospheric air to a higher pressure, so where does that excess
power go?

There's a checkvalve in the compressor. The pump squeezes the air
until the pressure equals that in the tank; there's no flow until that
happens. The power up to that point has gone into potential energy in
the form of cylinder pressure. When the checkvalve opens, flow
occurs at essentially constant pressure if the reservoir is large
compared to the cylinder, and energy is transfered to the reservoir.
When the cylinder pressure drops below reservoir pressure, the
checkvalve closes, flow stops, and the potential energy in the
cylinder pressure is returned to the flywheel until the next
compression stroke.

It seems you're neglecting the fact that the compressor
needs to draw in a fresh charge of air on the down stroke.
The dead volume above the piston is very small at TDC, so
there may be a small "kick" as the crank goes over the top
and the small volume of remaining air expands, but unless
the pressure in the cylinder falls below atmospheric (or
the pressure of the previous stage) pretty quickly, the
cylinder will not refill.

Said fact agreed to and not neglected. The fact remains that the
energy transferred to the reservoir is volume flow * pressure while
checkvalve is open, while energy expended compressing the air that
did not flow is recovered (less heat loss -- I addressed that) until
the pressure in the cylinder is back to atmospheric.

Marks Handbook has a discussion of single and multistage
air compression and the thermo behind the relative
efficiencies.

I'm familiar with it. That treatment assumes that heat of
compression is lost. If it is (as it usually is) then compressing
beyond end-use pressure is indeed inefficient. My point was and is
that the loss isn't in the regulator but in the heat loss at the
compressor. If that heat energy is preserved then there is no
significant penalty to overcompressing and post regulation. There are
also some significant economies in pipe sizes to minimize flow loss
(less at higher pressure and lower volume) and reservoir sizes
required if reservoirs are used.

The original question in this thread was about loss in a regulator.
I continue to assert that there is little power or energy loss in a
regulator. I'll further note that there is value in a small shop
in minimizing reservoir size by operating at higher pressure with a
2-stage compressor and regulating down at the point of use.

I once had a similar discussion with Scott Foss (a leading expert
in the economics of compressed air usage) and a VP of Engrg at
Ingersoll in Paducah KY. We were in complete agreement both with
our analyses and with the realization that it's very difficult to get
plant engineers to think beyond rote recitation of that which is
printed in Marks with underlying assumptions unquestioningly accepted
though not fully understood. Too bad, since some plants spend
over $1mil/year in energy cost to compress air, so even minor
percentage savings can be significant.

Current practice is still to reduce leaks and stage pumps to demand.
Good ideas to be sure, but far short of what could be done with
advanced controls and energy management at the enterprise level.

Don Foreman writes:

I continue to assert that there is little power or energy loss in a
regulator.

Astonishingly, amazingly stubborn. Equivalent to belief in perpetual
motion, and denial of energy conservation.

Compressing air requires work.

Decompressing air spends this work, either usefully as through a tool, or
wastefully as in a restriction (aka, a regulator).

A regulator can be had to regulate arbitrarily close to zero output
pressure and any desired flow. The original uncompressed volume of air is
then released at near-zero pressure. Near-zero pressure times a finite
volume, yields near-zero work. Where did the energy go?

In article , Richard J Kinch
says...

I continue to assert that there is little power or energy loss in a
regulator. (don)

Astonishingly, amazingly stubborn. Equivalent to belief in perpetual
motion, and denial of energy conservation.

And yet, Don is correct. The energy did *not*
show up in the body of the regulator.

Take the example where you start with one volume
of gas at some large pressure. Then allow the
gas to expand, through a regulator, to twice the
volume and a lower pressure.

Now compare the energy of the original and final states:
One tank of gas at a large pressure, compared with
*two* tanks (represening three times the original
volume) at the new, lower pressure. Remember there's
still gas left in the supply tank, at the outlet
tank's pressure, you have to account for that
energy.

Sure, there's less potential energy in the
final, stored gas of that two-tank system, than
in the original, one-tank configuration.
Don's point: energy does *not* appear as
heat "in the regulator" during the gas transfer.

- He's correct.

The gas regulator does not get hot.

That's not perpetual motion - it's just physics.

Doing the conservation of energy math properly
requires you to consider a bunch of things, like
the thermal (heat) energy *in* the gas, the kinetic
energy during the flow, the amount of heat
that is given up to, or from, the surroundings
to the volumes of gas, and any actual mechanical
work done during the expansion.

This subject is actually pretty well understood
and was, even a long time ago. Basically thermodynamics
came about because folks wanted to build steam engines.
Granted the physicists would say that the theory came
first, but we all know better.

It's a tolerably complicated subject; a proper
treatment has all kinds of equations and funny symbols
littering the discussion. Some here have complained
that putting stuff like that in a post is needlessly
complicated. I agree. But to *really* understand what
goes on when somthing as blastedly simple as 'one
tank of compressed air gets hooked up to an empty
tank' the rigorous treatment is the ONLY way to get
the correct answer.

Jim

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==================================================

On Wed, 31 Dec 2003 23:45:21 -0600, Richard J Kinch
wrote:

Don Foreman writes

Power flow is pressure * volume flow, volume here being actual volume
at that pressure.

No, power = pressure * mass flow, which is why that definition of
"actual volume" instead of "free air volume" isn't useful for such
calculations.

Power is integral of force(t) * distance (t) dt in my physics book.
At constant pressure, that's pressure * volume flow rate.

Thermodynamics enters in if heat is lost during compression as is
generally the case.
Thus Ingersoll's generally-true note that
compressing to pressure higher than needed is inefficient.

Losing heat isn't the reason for inefficiency of excess pressure. If it
were, insulation would cure it. Regulating down is inherently
inefficient.

The loss
is at the compressor, not at the regulator. The compressor gets hot
but the regulator does not, right?

The regulator can get downright chilly. Temperature is not heat or a
measure of heat. Something can get colder and gain heat at the same
time.

There's a law of thermodynamics that says otherwise.

On Wed, 31 Dec 2003 16:51:21 -0800, Loren Amelang
wrote:

On Mon, 29 Dec 2003 23:04:25 GMT, ATP wrote:

I don't understand why this is not clear to everyone here. Which has more
energy-one liter of 200 psi gas or two liters of 100 psi gas? Can you do
any work going from one state to the other? In one direction you can,

True. There is one catch in this example (and a lot of similar ones in this
thread) though... If you start with one liter of 200 psi gas, and let it
expand through an air motor into a receiving tank until you have two liters
of 100 psi gas, you can extract some work, but you don't end up with those
two liters in the receiving tank.

To accomplish this task you must use a one-liter receiving tank. At the end
of the demonstration, one liter is still in the original source tank,
because the flow stopped when the receiving tank reached 100 psi. Putting
that second liter of air into the source tank cost you more than putting
the first liter in, and more than putting a constant 100 psi against the
motor until the receiving tank was at 100 psi.

But the additional energy invested in the first case isn't lost. At
the end of the first experiment you still have 2 liters of 100 PSI
air remaining at the end of the experiment. In the second case you
have only 1 liter of 100 PSI air remaining. Further, if you
discharged thru the motor from the 200 PSI tank to the receiving tank
you got twice as much work because the intial delta-P was 200 PSI
rather than 100 since the receiver starts at 0. Net result: you
get 4 times as much work: twice as much while discharging to 100 PSI
and twice much afterwards as you discharge 2 liters (rather than 1)
from 100 PSI to 0.

jim rozen wrote:

In article , Ted Edwards says...

If your scope is dual trace, you can do what you want with a current
transformer. You need a 120V to 12V transformer rated for a dozen or so
...

I don't understand how he is going to display voltage
with this setup, unless he ties one side of the
incoming line (240, yes?) to the scope common rail,
which is typically by definition power ground. That
will cause smoke to leak out someplace.

The transformer provides isolation so the "12V" current sensing winding
can go in either motor lead as dictated by convenience. The transformer
and resistor converts the current to a proportional voltage. One probe
and ground goes across the resistor which is across the "120V" winding
and isolated from the other winding.

The other channel of the scope is sensing voltage. The _probe_ goes to
the hot lead, the scope ground goes to ground. You may have to allow
for the small difference between neutral and ground but this can be
measured separately.

If you still have doubts, I could post a circuit to the dropbox.

BTW, this method is commonly used to measure currents in lines kilovolts
above ground. The common clamp on ammeter also works on this scheme.
The clamp contains the core and the wire which it encircles forms a one
turn winding.

Ted

Engineman1 wrote:

properly so I put a flowmeter in the line. It is the type of flowmeter used to
regulate medical oxygen. It has a calibrated plastic tube in which a small
steel ball floats and there is a needle valve on the outlet to adjust the gas
flow to the burner jet.

That sounds like the same type of flow regulator that is used for Argon
on a TIG outfit. All the ones I've seen are calibrated for an input
pressure of 50psi. The ones on my TIG have a regulator with no external
adjustment to drop bottle pressure to 50psi. The tubes actually say
that they read cfm flow at 50 psi input. Also, they have scales on all
four sides with calibrations for different gasses.

Ted

In article , Ted Edwards says...

The other channel of the scope is sensing voltage. The _probe_ goes to
the hot lead, the scope ground goes to ground.

But the compressor runs off of 240 volts. Don't you
need to put the scope across both hot leads?

Jim

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In article wmNIb.11726$Do6.4460932
@news4.srv.hcvlny.cv.net,
says...
The amount of work
that is accomplished is dependent on the design of our device- to say that
at the end a certain weight was only lifted x amount in each case does not
prove that the unregulated case is not capable of doing more work. My
apologies to Ned if I have misunderstood his example.


Exactly. And in this case the regulator is the only feature
of the system that can be blamed for the reduced work
output.

Ned Simmons

In article ,
says...
On Wed, 31 Dec 2003 19:37:33 -0500, Ned Simmons
wrote:

In article ,
says...
On Mon, 29 Dec 2003 23:04:25 GMT, "ATP"
wrote:



I don't understand why this is not clear to everyone here. Which has more
energy- one liter of 200 psi gas or two liters of 100 psi gas? Can you do
any work going from one state to the other? In one direction you can, as
long as you don't just waste it with a regulator.

Putting it another way, as you compress a given volume of gas, does it have
more potential energy the more you compress it? Of course it does, so isn't
there a loss as it expands through a regulator?

Good question! I think the answer is "no" in an isothermal process.

Power flow is pressure * volume flow, volume here being actual volume
at that pressure. To avoid confusion with what "CFM" means, let's
define volume flow as actual cu in / sec at given pressure. If I
have 1 cu in/ sec at 200 PSIA on the high side I have 2 cu in/sec at
100 PSIA on the low side (if temp is the same). Power out = power in
ergo no power loss in the regulator.

No, speaking loosely thermodynamically, the power is
proportional to the pressure *squared* times flow. Pressure
* flow is proportional to power for a non-compressible
fluid, for instance in a hydraulic system, but not for a
compressible gas.

Thermodynamics are not at issue when the checkvalve opens because
pressure is essentially constant. Power flow is pressure * volume
flow -- look at the units. ( lbf/in^2) * (in^3/sec) = in * lbf /
sec which is power.

You're still neglecting the compressibility of the medium.
Units consistency does not guarantee a correct analysis;
ft*lb work vs. lb*ft torque, for example.

As I said, in a hydraulic system the mechanical power
available is equal to pressure * flow. That power is used
(or wasted and turned into heat, as occurs in a hydraulic
regulator or relief) when there is a pressure drop across a
device.

The difference in the case of a compressible fluid is that
there is additional energy available, either as heat or
work, when the fluid expands as a result of the delta P.

Consider the difference between the amount of work that can
be done by a tank of compressed air vs. a tank of
pressurized water.


Look at the example I posted again, except now imagine a
case where the initial pressure in the reservoir is 5
atmospheres absolute (twice the gage pressure of the
original case), and no regulator. The initial force on the
piston will be twice the unregulated 3 atm case, and the
total travel of the piston will also be doubled. The total
work done by the piston will be 4x.

The piston analogy was a bad idea, apologies for that. It obscures
the issue of transferred power vs store and release of potential
energy. Work is done in raising the weighted piston (assuming the
weight doesn't become zero in the process) that can be recovered when
the piston returns to it's initial position.

The piston analogy is exactly on point. It illustrates the
loss thru a regulator, which is the whole point of this
thread, in as simple a manner as I can imagine.

Let me try one more time. You have the system I illustrated
before. Initially the reservoir is charged to 3 atm abs.
There is a 30 lb weight on the piston balancing the
pressure. You reduce the weight in small increments and
when the system comes to mechanical and thermal equilibrium
after each reduction, record the remaining weight and
position of the piston. Since the system is free to
exchange heat with the environment and we wait for
equilibrium before taking our measurements, the process is
isothermal. The red line represents the resulting data. The
area under the line equals the mechanical work done by the
expanding air.

Now repeat the experiment, except put a regulator set to 2
atm between the reservoir and the cylinder and balance the
reduced pressure with a 15 lb weight. The green line
represents the recorded values. Obviously the work done by
the expanding gas is less.

The gas was at the same pressure, temperature, and volume
at the beginning of the two processes, and the end states
are identical as well, so clearly there had to be a loss.
If it wasn't the gas expanding thru the regulator before
doing any mechanical work, then what?

Additionally, without the regulator the process is
reversible merely by replacing the weights, i.e., reversing
the direction of the work. Not so if the regulator is in
the system.

Ned Simmons

In article ,
says...

Sure, there's less potential energy in the
final, stored gas of that two-tank system, than
in the original, one-tank configuration.
Don's point: energy does *not* appear as
heat "in the regulator" during the gas transfer.

- He's correct.

The gas regulator does not get hot.

Well, he's correct in the observation that the regulator
does not get hot, not in the conclusion he draws from that
fact--that there is thus no loss in the regulator.




It's a tolerably complicated subject; a proper
treatment has all kinds of equations and funny symbols
littering the discussion. Some here have complained
that putting stuff like that in a post is needlessly
complicated. I agree. But to *really* understand what
goes on when somthing as blastedly simple as 'one
tank of compressed air gets hooked up to an empty
tank' the rigorous treatment is the ONLY way to get
the correct answer.


Yes, without the thermo the best you can do is illustrate a
point with a narrowly defined ideal case. That's what I was
trying to do with the reservoir/regulator/piston example,
which if treated isothermally can be solved with Boyle's
Law. But a thorough analysis and complete understanding
requires thermo with all its ugly abstract concepts.

Ned Simmons

In article , Ned Simmons
says...

The gas regulator does not get hot.

Well, he's correct in the observation that the regulator
does not get hot, not in the conclusion he draws from that
fact--that there is thus no loss in the regulator.

The 'loss' is not in the regulator. It's inherent
in the process of expanding the gas into the second
tank. The same thing happens with no regulator at
all, if you simply connect the two tanks together.

Jim

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In article , Ned Simmons
says...

Additionally, without the regulator the process is
reversible ...

But the example I used, of simply connecting two
tanks and allowing the pressure to equilibrate between
the two of them, is not. Regulator or not, you cannot
put the gas back into the first, smaller tank, without
doing work on it. The entropy of the system goes up
when you connect the two tanks.

Jim

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In article ,
says...
In article , Ned Simmons
says...

Additionally, without the regulator the process is
reversible ...

But the example I used, of simply connecting two
tanks and allowing the pressure to equilibrate between
the two of them, is not. Regulator or not, you cannot
put the gas back into the first, smaller tank, without
doing work on it. The entropy of the system goes up
when you connect the two tanks.


Sure. Just an extreme example of expansion without doing
*any* reversible work. The reservoir/piston example is at
the opposite extreme, the reservoir/regulator/piston falls
between.

Ned Simmons

In article ,
says...
In article , Ned Simmons
says...

The gas regulator does not get hot.

Well, he's correct in the observation that the regulator
does not get hot, not in the conclusion he draws from that
fact--that there is thus no loss in the regulator.

The 'loss' is not in the regulator. It's inherent
in the process of expanding the gas into the second
tank. The same thing happens with no regulator at
all, if you simply connect the two tanks together.


I'm not sure what you're getting at here. Saying the loss
is "in the regulator" isn't any sloppier than the other
jargon we've all been tossing about. How about "due to a
regulator", or "due to the expansion of the gas as a result
of the delta P across the orifice of the regulator",
or....?

Ned Simmons

jim rozen wrote:

But the compressor runs off of 240 volts. Don't you
need to put the scope across both hot leads?

You could use another transformer - any handy 240 to whatever and put a
light load on it. Ground one side of the secondary and hook the other
side to the probe.

If it really is split 240 and not two lines of a three phase service,
looking at only one hot from ground should have very little error in
phase.

Ted

Richard J Kinch wrote:

A regulator can be had to regulate arbitrarily close to zero output
pressure and any desired flow. The original uncompressed volume of air is
then released at near-zero pressure. Near-zero pressure times a finite
volume, yields near-zero work. Where did the energy go?

Near-zero isn't zero. Near-zero pressure at finite volume is obtained
from near-zero volume at finite prssure. Where's the problem?
E = Integral P dV

Back a ways you attempted to use a series regulator as an analogy. This
doesn't work since, in a series regulator, Iout ~= Iin and Vout Vin.
A switching regulator might be a better analogy: Vin*Iin ~= Vout*Iout.

Ted

In article , Ted Edwards says...

If it really is split 240 and not two lines of a three phase service,
looking at only one hot from ground should have very little error in
phase.

This is probably true, from the standpoint of
determining phase angle between voltage and current.
The phase of the L1 to common should look pretty
much the same as L1 to L2. Of course he would
have to use the correct voltage when figuring
I x V x cos(angle).

Jim

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In article , Ned Simmons
says...

I'm not sure what you're getting at here. Saying the loss
is "in the regulator" isn't any sloppier than the other
jargon we've all been tossing about. How about "due to a
regulator", or "due to the expansion of the gas as a result
of the delta P across the orifice of the regulator",
or....?

There doesn't have to *be* a diaphram/spring/orifice
regulator in the connection between the two tanks in
my example. You still see the same thermodynamics
for a plain, large valve that connects the two tanks,
and is opened to permit gas to flow from one to the
other.

Jim

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In article , Ted Edwards wrote:
Also e-mailed due an ISP suffering Altsheimer's:
Loren Coe wrote:
does anyone have a suggestion as to how i measure this with
just "normal" equipment, including an O'scope? running back
and forth to the utility co.power meter doesn't seem like a
very good approach.

If your scope is dual trace, you can do what you want with a current
transformer. You need a 120V to 12V transformer rated for a dozen or so
amps. Connect a 10 Ohm, 10 Watt resistor across the 120V winding.
Connect the 12V winding in series with a motor lead. The "backwards"
transformer will have 11A flowing in the "12V" winding which will become
1.1A in the "120V" winding. 1.1A in a 10 ohm resistor is 11 volts so
the other winding (the "12V" winding) will drop only 1.1V. This will
have negligable effect on the motor operation. If you now pick up the
hot motor lead in one channel of the scope and the secondary of the
current transformer in the other channel, you can measure the time
difference between the waveforms and calculate the phase shift. Ted

i have been following you guys, haven't had the incentive to continue but
that will come, i have a spare bell xmfr and i know how to float a scope,
it has been awhile, tho. does anyone think the power meter would be of
any use (the one on the side of the home)?

Good Luck in the New Year to all rcm'rs, --Loren

In article ,
says...
In article , Ned Simmons
says...

I'm not sure what you're getting at here. Saying the loss
is "in the regulator" isn't any sloppier than the other
jargon we've all been tossing about. How about "due to a
regulator", or "due to the expansion of the gas as a result
of the delta P across the orifice of the regulator",
or....?

There doesn't have to *be* a diaphram/spring/orifice
regulator in the connection between the two tanks in
my example. You still see the same thermodynamics
for a plain, large valve that connects the two tanks,
and is opened to permit gas to flow from one to the
other.


That's obvious to you and me.

I never intended to give the impression that a regulator is
the only example of lossy expansion, but that was the
subject of the original, and most of the subsequent, posts.
I certainly hope everyone following this accepts that
allowing the air to expand freely, as in your example,
lowers the potential to do mechanical work. Paradoxically,
not everyone seems to believe that if the expansion occurs
across a regulator there is also a loss.

Ned Simmons

On Thu, 1 Jan 2004 19:59:34 -0500, Ned Simmons
wrote:

In article ,
says...

Sure, there's less potential energy in the
final, stored gas of that two-tank system, than
in the original, one-tank configuration.
Don's point: energy does *not* appear as
heat "in the regulator" during the gas transfer.

- He's correct.

The gas regulator does not get hot.

Well, he's correct in the observation that the regulator
does not get hot, not in the conclusion he draws from that
fact--that there is thus no loss in the regulator.

I did not conclude that there is no loss because it doesn't get hot.
My assertion was based on power in vs power out, which are the same.

Ted Edwards writes:

A regulator can be had to regulate arbitrarily close to zero output
pressure and any desired flow. The original uncompressed volume of
air is then released at near-zero pressure. Near-zero pressure times
a finite volume, yields near-zero work. Where did the energy go?

Near-zero isn't zero. Near-zero pressure at finite volume is obtained
from near-zero volume at finite prssure. Where's the problem?

Eh? The final volume of exhaust at ambient pressure is always the same,
namely the volume of free air taken in by the compressor (intook?). You
can regulate the output to as close to zero as you like. Which is to
say, you can take any tank of compressed air and vent it to the
atmosphere, with as close to all of the energy dissipated by the
regulator as you care to choose. In the limit, all of it.

E = Integral P dV

And the limit of E is thus zero as P approaches zero, since V is
independent of P for small P (the compressed volume can only expand to
the original free air volume).

Remember that every expanding container does work against the
atmosphere.

Back a ways you attempted to use a series regulator as an analogy.
This doesn't work since, in a series regulator, Iout ~= Iin and Vout
Vin.

No, Iout = Iin in a 3-terminal regulator, except for a small overhead
that runs the chip itself. Or, model it as a magic variable resistor if
you like.

A switching regulator might be a better analogy: Vin*Iin ~=
Vout*Iout.

Hardly. A switching regulator is just a transformer, with an DC-AC
converter on the input, and AC-DC converter on the output. This would
be analogous to my imaginary "regulator" consisting of an air motor
regenerating the reservoir, not a conventional variable-restriction
regulator.

Don Foreman writes:

Something can get colder and gain heat at the same
time.

There's a law of thermodynamics that says otherwise.

Compressed air exiting an orifice gets colder (temperature decreases) from
expansion, while gaining some new heat from the friction and turbulence of
passing through the orifice. Which law of thermodynamics "says otherwise"?

jim rozen writes:

But the compressor runs off of 240 volts. Don't you
need to put the scope across both hot leads?

No, you really just want to know the phase angle between the current versus
voltage. So any picture of the utility AC voltage will do (such as the
internal scope test signal!). You don't need to look directly at the
voltage supplied to the motor.

Ned Simmons writes:

The difference in the case of a compressible fluid is that
there is additional energy available, either as heat or
work, when the fluid expands as a result of the delta P.

True. A container of pressurized gas expanded isothermally into an ambient
vacuum has theoretically *infinite* attainable work, since the pressure
decreases inversely with the expanded volume, the integral of this pressure
with the change in volume is logarithmically increasing with the expansion,
and the expansion (into a vacuum) can be infinite.

In article , Richard J Kinch
says...

But the compressor runs off of 240 volts. Don't you
need to put the scope across both hot leads?

No, you really just want to know the phase angle between the current versus
voltage. So any picture of the utility AC voltage will do (such as the
internal scope test signal!). You don't need to look directly at the
voltage supplied to the motor.

Not *just* any picture. The signal has to have exactly
the same phase as the incoming 240 volt line - because
when one is doing this, the power is mostly reactive,
so one is measuring a pretty small difference between
two large values. It would be worth checking to see
that the line phase reference really is good.

Jim

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In article , Richard J Kinch
says...

Compressed air exiting an orifice gets colder (temperature decreases) from
expansion, while gaining some new heat from the friction and turbulence of
passing through the orifice.

In that case, it's a simple accounting job to see that
the warming from friction is much smaller than the
cooling from expansion. The gas is not 'cooling as
it gains heat.' The NET change in internal heat is
negative.

It's simply cooling. The resulting container of colder
gas contains less internal heat after the expansion.

The internal energy of a gas is a simply proportional
to 3/2 kT. Hotter gas has always has more internal
heat energy.

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In article , Ned Simmons
says...

I never intended to give the impression that a regulator is
the only example of lossy expansion, but that was the
subject of the original, and most of the subsequent, posts.
I certainly hope everyone following this accepts that
allowing the air to expand freely, as in your example,
lowers the potential to do mechanical work. Paradoxically,
not everyone seems to believe that if the expansion occurs
across a regulator there is also a loss.

If I might take a moment to dither about the semantics
here.

"lowering the potential to do mechanical work" is
a bit different in some sense, than "loss." Because
of the history of electrical analogy here, this may
be confusing.

Most folks think of the term 'loss' as a point loss,
ie a lumped circuit element that is dissipative. The
regulator, or the connection between the two tanks
in the case of no regulator, or even the expansion
orifice in the case where gas is expanding into
vacuum, are not analogous to electrical lumped circuit
elements like resistors. No heat appears in them
during the process. [1]

(electrical analog discussion really are
doomed to fail here...)

Sure the end result having two tanks with three times
the volume, filled at a lower pressure, has less stored
potential. But the difference does not appear in the
regulator. The energy was not lost in the regulator.

Jim

[1] aside from the turbulent flow in the regulator
as has been noted. This is a small effect.

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