jim rozen writes:

I suggest you read up an introductory thermodynamics book.

Puh-leeze.

If you do all the bookkeeping
all the numbers add up. Mechanical work, heat, kinetic
energy, stored potential energy, all sum to a constant.

You are just quibbling.

The fact is, energy (or the power, if you like, the time values are the
same on both sides, so it doesn't matter) goes into a regulator and comes
out. What comes out is less than what goes in. The difference is wasted,
ultimately as heat. Regulators are inherently WASTEFUL of ENERGY and
POWER, however you care to measure what the system is designed to deliver.
Regulators perform useful functions, but not efficiently.

In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.

Gary Hallenbeck writes:

I DO understand what CFM and SCFM mean. I worked with compressor
specification and air system analysis ,around the worl, for 25 years.
...
A CFM is just what it says it is. One Cubic Foot per
Minute at any temperature, pressure or humidity.

Your (former) employer disagrees. If you look at what Ingersoll-Rand
says in "Compressed Air Basics":

http://air.irco.com/asg/air_system_info/basic.asp

you'll see that CFM there refers to cubic feet of free air per minute.

I defy you to show me elsewhere on ingersoll-rand.com that says
otherwise.

Read the Grainger catalog section on "Guide to Air Compressors". You
will find this definition: "FREE AIR: The actual volume (CFM) of air
produced by the compressor pump at rated pressure(s)."

Peruse the Grainger catalog for Ingersoll-Rand (or any other brand)
compressors. In every case, you will see the "CFM" specified is for
cubic feet of FREE AIR per minute at delivery pressure.

I defy you to show me a label on an Ingersoll-Rand compressor at a
Sears, or a Home Depot, or a Harbor Freight sales floor, which makes any
sense at all without my definition. I have looked at many, and they all
specify their CFM ratings in terms of free air at delivery pressure.
Same for the other brands and distributors.

Perhaps you have simply misunderstood that CFM can be applied to any
delivery pressure, and in that sense is non-specific as to pressure.

Let me beg out of this quarrel. Either you believe that the letters
"CFM" (with variations such as "SCFM" denoting further special
conditions) in the context of air compressor performance refers to
volumes of free air per minute, or you don't. I don't deny that not
everybody understands this, or that it has been used (improperly) to
mean the volume of air at delivered pressure. But it is clear that
"free air" is the proper meaning, and the much more useful one, and the
one that must be understood when you read a specification label on an
air compressor.

In article , Richard J Kinch
says...

You are just quibbling.

Beggin your pardon but that's what this thread
is about.

The fact is, energy (or the power, if you like, the time values are the
same on both sides, so it doesn't matter) goes into a regulator and comes
out. What comes out is less than what goes in. The difference is wasted,
ultimately as heat.

No disagreement there, I was taking exception
to your use of certain terms, incorrectly.

You comment above about how gas regulators behave is
true for *all* devices that transform energy from
one form to another - whatever energy goes in, comes
out somewhere else, less some heat.

There's only one kind of machine that's exactly
100% efficient.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch wrote:
jim rozen writes:

I suggest you read up an introductory thermodynamics book.

Puh-leeze.

If you do all the bookkeeping
all the numbers add up. Mechanical work, heat, kinetic
energy, stored potential energy, all sum to a constant.

You are just quibbling.

The fact is, energy (or the power, if you like, the time values are the
same on both sides, so it doesn't matter) goes into a regulator and comes
out. What comes out is less than what goes in.

Power doesn't go anywhere. It is a *rate*, not an actual thing that moves.
Some amount of energy is stored in the tank, some lesser amount comes
out through the regulator valve as a mass flow. The rest remains in the
tank until used at some later time.

The amount of energy leaving the tank divided by the amount of time it
takes to leave is a rate, power, which we can calculate for any given
air draw from the tank. But there's only one power at any given draw,
ie energy is leaving the tank at only one rate.

In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the base
of a dam, the other 999,999,999,999,999 gallons in the reservoir are
wasted. Obviously not true at all. You use what you use, the rest stays
in the reservoir (with all of its potential energy intact) until you need it.

Gary

In article ,
says...
On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch wrote:


In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the base
of a dam, the other 999,999,999,999,999 gallons in the reservoir are
wasted. Obviously not true at all. You use what you use, the rest stays
in the reservoir (with all of its potential energy intact) until you need it.


Another faulty analogy. Since water is essentially
incompressible, the capacity of the water behind a dam to
do mechanical work is almost entirely due to its potential
energy -- its elevation in a gravitational field.

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

Explaining where the energy goes requires thermo.

Ned Simmons

In article , Richard J Kinch wrote:
Gary Hallenbeck writes:
I DO understand what CFM and SCFM mean. I worked with compressor
specification and air system analysis ,around the worl, for 25 years.
... A CFM is just what it says it is. One Cubic Foot per
Minute at any temperature, pressure or humidity.

Your (former) employer disagrees. If you look at what Ingersoll-Rand
says in "Compressed Air Basics":
http://air.irco.com/asg/air_system_info/basic.asp
you'll see that CFM there refers to cubic feet of free air per minute.
I defy you to show me elsewhere on ingersoll-rand.com that says
otherwise.

Richard is saying that CFM is at 1 atm, right? so an ideal compressor
would take in 'X' CFM/min and output that same amount, regardless of the
output pressure. does this make sense to anyone else? the discussions
of altitude and other losses seem to obscure this, imho (as does the
fact that no ideal compressor exists ;-).

also, i gather that SCFM is simply corrected for moisture content, and
some other factors that would not impact comparison of honest ratings?

now, the "power" required for a 90psi vs 175 psi output pressure makes
for real world examples, especially when you leave the "ideal" realm.
often said by the professor, ...an exercise for the student... --Loren

Read the Grainger catalog section on "Guide to Air Compressors". You
will find this definition: "FREE AIR: The actual volume (CFM) of air
produced by the compressor pump at rated pressure(s)."

On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote


It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more. Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight? Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment. Let's assume
adiabatic expansion (no heat exhange with the environment) to keep
things simple enough for me to understand.

I think the results would be about the same but I'm definitely open
to learning why not.

Gary Coffman writes:

Power doesn't go anywhere. It is a *rate*, not an actual thing that
moves.

You make a contemptibly false statement, ignorant of even high school
physics.

Consider an three-terminal electronic constant-voltage regulator as an
analogy to an air regulator. Say you input 7 volts DC at 1 amp to that
regulator, while drawing 5 volts at 1 amp out. The input is 7 watts of
POWER and the output is 5 watts of POWER. Yes, the POWER moves through
the device, an "actual thing that moves" as much as anything in physics.
The difference of 2 watts will be dissipated as waste heat in the
regulator device. In an air system, CFM is analgous to electric
current, and air pressure to electric potential (voltage).

Some amount of energy is stored in the tank, some lesser amount
comes out through the regulator valve as a mass flow. The rest remains
in the tank until used at some later time.

Absurd. If you run a given mass flow of air (CFM) at a higher pressure
into a regulator, and draw the same mass flow out the other side
(necessary by mass conversion), at a lower pressure, then the difference
in pressure times the flow rate (CFM) represents the the power being
lost in the regulator. The more the drop in pressure across the
regulator, the more the inefficiency.

In the extreme, you can make a device that will "regulate" compressed
air to output an arbitrarily large flow at an arbitrarily small
pressure, and dissipate all of the energy in the tank in the regulator
itself, such as diffuser device.

It is stupid to think that the difference across a regulator is somehow
"retained in the tank". You can continue the regulated flow until the
tank is empty, and then you have nothing in the tank. Depending on the
regulated pressure drop, you can do more or less work with the same flow
of air. The difference is wasted.

In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the base
of a dam, the other 999,999,999,999,999 gallons in the reservoir are
wasted. Obviously not true at all. You use what you use, the rest
stays in the reservoir (with all of its potential energy intact) until
you need it.

No, the analogy here would be regulating the head behind a hydraulic
dam, by allowing the excess to flow over a spillway. The spillway flow
performs the head regulation, but only at the expense of wasting the
power (or energy, if integrated over time) available from the water
"over the dam". Such wastes are inherent in any regulation scheme.

Loren Coe writes:

Richard is saying that CFM is at 1 atm, right?

CFM, so-called "free air". CFM with regard to compressed air means
cubic feet of free air per minute.

so an ideal compressor
would take in 'X' CFM/min and output that same amount, regardless of
the output pressure.

Any system without leaks must do this, not necessarily an "ideal" one,
since the air can't go anywhere else. Whatever goes in the inlet can
only exit via the delivery fitting, unless there are leaks. Of course,
this is not instantaneous equality, as the air may spend an arbitrary
amount of time in a reservoir tank before leaving for delivery.

also, i gather that SCFM is simply corrected for moisture content, and
some other factors that would not impact comparison of honest ratings?

Right, "S" in "SCFM" means CFM performance under conditions of a
standard temperature and humidity of the input air, namely 68 deg F and
36 percent relative humidity by my citations. There seems to be a
question about whether these actual values are standardized, as people
are citing a variety of purportedly standard values. This condition
applies to the *ambient* conditions, that is, the input air, *not* the
state of the *compressed* air, which unless treated by a cooler-drier
will typically be 100 percent humidity and 100s of deg F even when the
input air is moderately cool and dry.

Don Foreman writes:

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more.

Could be anything from near zero to 100 percent waste. Depends on the
difference in pressure across the regulator. No difference = no waste,
large difference = large waste.

Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight?

To the limits of travel (or it blows out the piston if there is no
limit). The cylinder hits the end of travel when the pressure reaches
50 psig in the expanded cylinder. The excess 150 psig are presumably
"wasted" if later vented, unless the system has some way to pass it on
to do other work.

Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment.

Same travel. You don't have the wasted excess 150 psig in the cylinder.
This is more efficient than the first case, but still inefficient
compared to using only a 50 psig source to start with, the latter
difference representing the significant losses in the regulator. So in
your example systems, the regulator is still inefficient, although it is
less inefficient than no regulator at all.

The practical lesson is that you typically want to set the pressure cut-
out on your compressor at just a bit more than the required pressure for
your application. Paying to store up excess pressure, and then
regulating it back down for the application, is a mechanically
(thermodynamically) inefficient round-trip, although sometimes usefully
so.

On Mon, 29 Dec 2003 06:17:25 GMT, Loren Coe wrote:
Richard is saying that CFM is at 1 atm, right? so an ideal compressor
would take in 'X' CFM/min and output that same amount, regardless of the
output pressure. does this make sense to anyone else? the discussions
of altitude and other losses seem to obscure this, imho (as does the
fact that no ideal compressor exists ;-).

Yes it makes sense.
Go he http://air.ingersoll-rand.com/CMP/sa...vertlevel3.htm
and then try to explain why the CFM ratings of these compressors don't
markedly change over a 75 to 175 PSI range.

For example, the IR 5 hp dual stage I have is rated at 14.9 CFM @75 PSI,
14.8 CFM @125 PSI, and 14.7 CFM @175 PSI. Same compressor, same
displacement, same speed, and nearly the same CFM over a pressure range
of more than 2:1. If we were to believe they're talking about CFM of
pressurized air, the gas law tells us we would see more than a 2:1 change
in CFM with a 2:1 change in pressure. But we don't. The only way we can
see the numbers they publish is if they're referring to CFM at 1 standard
pressure.

Gary

Gary Coffman writes:

For example, the IR 5 hp dual stage I have is rated at 14.9 CFM @75
PSI, 14.8 CFM @125 PSI, and 14.7 CFM @175 PSI. Same compressor, same
displacement, same speed, and nearly the same CFM over a pressure
range of more than 2:1. If we were to believe they're talking about
CFM of pressurized air, the gas law tells us we would see more than a
2:1 change in CFM with a 2:1 change in pressure. But we don't. The
only way we can see the numbers they publish is if they're referring
to CFM at 1 standard pressure.

Exactly.

On Mon, 29 Dec 2003 01:42:03 -0600, Richard J Kinch
wrote:

Don Foreman writes:

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more.

Could be anything from near zero to 100 percent waste. Depends on the
difference in pressure across the regulator. No difference = no waste,
large difference = large waste.

Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight?

To the limits of travel (or it blows out the piston if there is no
limit). The cylinder hits the end of travel when the pressure reaches
50 psig in the expanded cylinder. The excess 150 psig are presumably
"wasted" if later vented, unless the system has some way to pass it on
to do other work.

What excess 150 psig? We opened the cylinder to the reservoir so
both are at same pressure (50 PSIG) at the end of the experiment.

Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment.

Same travel.

So same amount of work done (50 lbf * travel distance) and same
residual pressure (50 PSIG) at the end. Sounds like a wash to me.

You don't have the wasted excess 150 psig in the cylinder.

What wasted excess 150 PSIG? Everything is at 50 PSIG at the end of
either experiment.

This is more efficient than the first case, but still inefficient
compared to using only a 50 psig source to start with, the latter
difference representing the significant losses in the regulator. So in
your example systems, the regulator is still inefficient, although it is
less inefficient than no regulator at all.

The practical lesson is that you typically want to set the pressure cut-
out on your compressor at just a bit more than the required pressure for
your application. Paying to store up excess pressure, and then
regulating it back down for the application, is a mechanically
(thermodynamically) inefficient round-trip, although sometimes usefully
so.

Gary Coffman wrote:

Yes it makes sense.
Go he http://air.ingersoll-rand.com/CMP/sa...vertlevel3.htm
and then try to explain why the CFM ratings of these compressors don't
markedly change over a 75 to 175 PSI range.

For example, the IR 5 hp dual stage I have is rated at 14.9 CFM @75 PSI,
14.8 CFM @125 PSI, and 14.7 CFM @175 PSI. Same compressor, same
displacement, same speed, and nearly the same CFM over a pressure range
of more than 2:1. If we were to believe they're talking about CFM of
pressurized air, the gas law tells us we would see more than a 2:1 change
in CFM with a 2:1 change in pressure. But we don't. The only way we can
see the numbers they publish is if they're referring to CFM at 1 standard
pressure.

I don't see it that way. If the spec isn't about CFM of pressurized air, then
why would the numbers differ at all for different output pressure? And why
quote CFM into different pressures? Why quote pressures at all?

Those numbers make sense to me as CFM of pressurized air. If you need more
volume than 14 cfm, you have to use a regulator to expand the air downstream
to get more flow. If you set your pressure switch to pump your air tank up
to 175 psi, then you have the headroom necessary to regulate it back down to
your desired end pressure but get quite a bit more volume.

Those specs relate as I see it to the pump and motor. If you run a pump at
a certain speed it is going to put out a fairly limited range of air flows
depending on its output pressure. A single stage pump's output flow will
drop much more markedly with output pressure than a two stage pump. An
ideal pump would have (to use an electrical analog) zero output impedance
so it would put out the same flow regardless of pressure.

It sounds like you have a pretty good compressor.

Grant Erwin

In article ,
says...
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote


It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more. Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight? Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment. Let's assume
adiabatic expansion (no heat exhange with the environment) to keep
things simple enough for me to understand.


Isothermal is easier. See if this helps:

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

I don't have time for a complete explanation right now, but
will try later if this isn't clear. The idea is to slowly
remove weight (W) while recording the position (D) of the
piston when the system is at both mechanical and thermal
equilibrium.

The area under the graphs represents the mechanical work
done. Though the end states are indistinguishable, the path
between the end states has to be accounted for.

Ned Simmons

On Mon, 29 Dec 2003 00:49:00 -0600, Don Foreman
wrote:

On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote


It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more. Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight? Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment. Let's assume
adiabatic expansion (no heat exhange with the environment) to keep
things simple enough for me to understand.

I think the results would be about the same but I'm definitely open
to learning why not.

Assuming that there is no friction between the piston and cylinder and
assuming that there was _just enough_ over 50 psi from the regulator to start
the weight moving then:-

with the regulator, the weight would move up at a constant , infinitesimal,
speed until it got where it was going.

Without the regulator, the weight would accelerate until a little below the
speed of sound and carry on at that speed until decelerating to a standstill a
_long_ way above the other weight, and then come back down etc.


the rest is left as an exercise for the student :-)


Mark Rand
RTFM

In article , Grant Erwin wrote:
Gary Coffman wrote:
Yes it makes sense.
Go he http://air.ingersoll-rand.com/CMP/sa...vertlevel3.htm
and then try to explain why the CFM ratings of these compressors don't
markedly change over a 75 to 175 PSI range.

For example, the IR 5 hp dual stage I have is rated at 14.9 CFM @75 PSI,
14.8 CFM @125 PSI, and 14.7 CFM @175 PSI. Same compressor, same
displacement, same speed, and nearly the same CFM over a pressure range
of more than 2:1. If we were to believe they're talking about CFM of
pressurized air, the gas law tells us we would see more than a 2:1 change
in CFM with a 2:1 change in pressure. But we don't. The only way we can

along with a 2:1 change in electrical current input to the motor, right?

see the numbers they publish is if they're referring to CFM at 1 standard
pressure.

I don't see it that way. If the spec isn't about CFM of pressurized air, then
why would the numbers differ at all for different output pressure? And why
quote CFM into different pressures? Why quote pressures at all?

well, the small differences are losses, more at higher pressure. Gary
is talking about IR specs for units rated "continuous duty" and that would
indicate to me that "tank losses" are not a large factor and the small
variation in the flows seems to indicate very nominal regulator loss.

....
Those specs relate as I see it to the pump and motor. If you run a pump at
a certain speed it is going to put out a fairly limited range of air flows
depending on its output pressure.

the only explanation for the additional pressures at similar volumes would
be additional power input to the system. perhaps that really is what goes
on? the motor _is_ taking more current as the output pressure goes up?
i have never measured the range of currents to a 220vac motor based on
the regulated output pressure, that would be a good afternoon project.

more, later, --Loren

I think I understand both examples. I don't think either refutes my
assertion that the weight ends up at the same place either way so work
delivered is the same. The "area under the path" argument works out
the same in both cases. Delivered work is the total available energy
(area of the whole triangle ( 1/2 * 30 * 2 = 30 in-lb ) minus the
remaining energy which is the area of the rectangle (15 * 1 = 15
in-lb) or 15 in-lb delivered.

Ned Simmons wrote:
In article ,
says...
On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch
wrote:


In general, regulation of *any power source* implies an inherent
waste versus the unregulated source: electrical power, compressed
air power, motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the
base of a dam, the other 999,999,999,999,999 gallons in the
reservoir are wasted. Obviously not true at all. You use what you
use, the rest stays in the reservoir (with all of its potential
energy intact) until you need it.


Another faulty analogy. Since water is essentially
incompressible, the capacity of the water behind a dam to
do mechanical work is almost entirely due to its potential
energy -- its elevation in a gravitational field.

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

Explaining where the energy goes requires thermo.

Ned Simmons

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?

The IR site makes it clear that it is inefficient to create higher pressure
than necessary and regulate it down. I believe this thread was started with
a reference to air for an HVLP gun. Using a high pressure compressor and an
HVLP conversion regulator is an extremely inefficient way to run an HVLP
gun. This can be confirmed by comparing the power use of low pressure
turbines and the compressor/regulator setup needed to duplicate the low
pressure, high volume needed.

Grant Erwin wrote:
Gary Coffman wrote:

Yes it makes sense.
Go he http://air.ingersoll-rand.com/CMP/sa...vertlevel3.htm
and then try to explain why the CFM ratings of these compressors
don't markedly change over a 75 to 175 PSI range.

For example, the IR 5 hp dual stage I have is rated at 14.9 CFM @75
PSI,
14.8 CFM @125 PSI, and 14.7 CFM @175 PSI. Same compressor, same
displacement, same speed, and nearly the same CFM over a pressure
range of more than 2:1. If we were to believe they're talking about
CFM of pressurized air, the gas law tells us we would see more than
a 2:1 change in CFM with a 2:1 change in pressure. But we don't. The
only way we can see the numbers they publish is if they're referring
to CFM at 1 standard pressure.

I don't see it that way. If the spec isn't about CFM of pressurized
air, then why would the numbers differ at all for different output
pressure? And why quote CFM into different pressures? Why quote
pressures at all?

Although the piston displacement and motor speed remain essentially
constant, slightly more air is going to flow into a lower pressure
reservoir.

Those numbers make sense to me as CFM of pressurized air.

Think about it. Do you really think a pump pumping into twice the pressure
is going to flow double the amount of input air? That's what you're claiming
if the specs are CFM of pressurized air.

Don Foreman writes:

What excess 150 psig? We opened the cylinder to the reservoir so
both are at same pressure (50 PSIG) at the end of the experiment

I misunderstood your postulated "reservoir" to be a regulated-pressure
source of 200 psig, like an active air compressor. I see now you must
have meant a closed reservoir with no make-up source, which is a
different problem. Your description still would seem incomplete, as you
don't say how long this cylinder is before it hits end-of-travel. But
let's assume it is very long, because, I think I see where you're going
with this.

In your experiment, if the cylinder is "long enough", the end condition
is the same whether a regulator (or other restriction) is interposed or
not, with both chambers (reservoir and cylinder) ending at 50 psig with
the same cylinder travel.

The answer to this seeming contradiction is this: depending on the
presence of a regulator (or other restriction) between the chambers, the
energy is wasted if different places. But it is still wasted, turned
ultimately to heat instead of doing work.

In one extreme case, if you assume the two chambers are instantly
connected by an imaginary "perfect" (no restriction) valve and piping,
then the waste heat ends up in the cylinder from the turbulence and
friction of the incoming air.

If you assume the two chambers are "slowly" connected, either through a
regulator, a slowly opening valve, very small orifice, very thin tube,
or other restriction, then part of the energy is wasted in the
restriction (through turbulence and friction) and part in the turbulence
appearing in the cylinder. In the extreme, you put all the wasted
energy into the restriction by opening it very slowly and making nearly
zero turbulence into the cylinder.

There is also friction in the air cylinder motion. If we postulate a
perfect cylinder with no friction, and an unrestricted connection, then
you will get a system with damped oscillation while the cylinder bounces
up and down, eventually settling into the same end condition, with the
waste heat in the gas itself.

These are all rather ordinary thermodynamic thought-experiments.
Nothing novel.

In article hp1Ib.1845$I07.1869@attbi_s53, Loren Coe wrote:
In article , Grant Erwin wrote:
Gary Coffman wrote:
Yes it makes sense.
Go he http://air.ingersoll-rand.com/CMP/sa...vertlevel3.htm
and then try to explain why the CFM ratings of these compressors don't
markedly change over a 75 to 175 PSI range.
...
Those specs relate as I see it to the pump and motor. If you run a pump at
a certain speed it is going to put out a fairly limited range of air flows
depending on its output pressure.

the only explanation for the additional pressures at similar volumes would
be additional power input to the system. perhaps that really is what goes
on? the motor _is_ taking more current as the output pressure goes up?
i have never measured the range of currents to a 220vac motor based on
the regulated output pressure, that would be a good afternoon project.
more, later, --Loren

well, it's not a dramatic increase in amps as pressure rises, not enough
to contradict Gary or Richard. my Sears 2hp (35+yrs old) starts on
an empty tank (0 pressure) at 11a and is at 12.7 amps at shutoff, 130 lbs.
that's about a 15% variation.

these are the (all approx) readings: 0lbs/11a, 30/11.5, 40-50/11.7, 60/12,
voltage230vac 90/12.5, 120-130lbs/12.7amps (shutoff)

unless i somehow flunked Physics and Chemistry it seems certain that IR
specs are stating 1atm in and out, at all pressures. my compressor tag
is mutilated for the 40 and 90psi ratings, but "10 cu.ft/min Displacement"
is listed.

what surprises me is that the current does not increase more. most system
losses must be occurring at 0 psi(?). the rpm does not perceptively
change from 0-130lbs, but you can hear it start "working". fwiw, with
max flow thru the regulator, the pressure kinda settles at 40-50/11.7a
with about 10 lbs drop accross the regulator (diaphram type). the higher
readings are @ zero flow.

In article ,
says...

I think I understand both examples. I don't think either refutes my
assertion that the weight ends up at the same place either way so work
delivered is the same. The "area under the path" argument works out
the same in both cases. Delivered work is the total available energy
(area of the whole triangle ( 1/2 * 30 * 2 = 30 in-lb ) minus the
remaining energy which is the area of the rectangle (15 * 1 = 15
in-lb) or 15 in-lb delivered.


The weight on the piston at the end of the expansion is
zero. As the graph shows, the force that is exerted varies
over the two paths.

It declines linearly from 30 to 0 in the unregulated case
(red line).

For the regulated case (the green line) the force is
constant at 15 lbs until the pressure in the vessel falls
below 2 atm, then declines to zero as expansion continues.

The work in the unregulated case is the entire area under
the red line = 30 in-lb. With the regulator in the circuit
the work performed is the area under the green line;
(15 X 1) + (15 X 1 x 1/2) = 22.5 in-lb.

Ned Simmons

Think about it. Do you really think a pump pumping into twice the pressure
is going to flow double the amount of input air? That's what you're claiming
if the specs are CFM of pressurized air.

I did think about it. I really do think that, yes. This is a 2-stage
compressor, so the first stage always "sees" a fairly constant load.
However, I don't want this thread to degrade to "will too .. will not .."
or some such unhelpful discourse. Let me ask you - if they are specifying
input air then why would that change with output pressure? And why say an
output pressure at all? Why not simply read "14.7 cfm @ 170 psi" at face
value? To me that reads "this compressor will deliver 14.7 cfm of compressed
air into a tank if that tank is at a pressure of 170 psi". Why shouldn't I
read this spec this way?

Grant Erwin

In article tF2Ib.683809$HS4.4843991@attbi_s01, Loren Coe says...

well, it's not a dramatic increase in amps as pressure rises, not enough
to contradict Gary or Richard. my Sears 2hp (35+yrs old) starts on
an empty tank (0 pressure) at 11a and is at 12.7 amps at shutoff, 130 lbs.
that's about a 15% variation.

If you are just measuring this with an amp-clamp
meter, you may be fooled by the phase angle of
the current. At start much of that 11 amps
may be reactive power, at finish it may
be mostly real.

Jim

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Grant Erwin wrote:
Think about it. Do you really think a pump pumping into twice the
pressure is going to flow double the amount of input air? That's
what you're claiming if the specs are CFM of pressurized air.

I did think about it. I really do think that, yes. This is a 2-stage
compressor, so the first stage always "sees" a fairly constant load.
However, I don't want this thread to degrade to "will too .. will not
.." or some such unhelpful discourse. Let me ask you - if they are
specifying input air then why would that change with output pressure?

I would expect it to change somewhat, the air is not going to flow as easily
into a fully pressurized tank as it would into 0 psi. As it turns out, it
doesn't change much. Why would it flow double the CFM at a higher pressure?


And why say an output pressure at all? Why not simply read "14.7 cfm
@ 170 psi" at face value? To me that reads "this compressor will
deliver 14.7 cfm of compressed air into a tank if that tank is at a
pressure of 170 psi". Why shouldn't I read this spec this way?

Grant Erwin

Well, that's what I always thought, too, but I think Richard Kinch has
enlightened us and the compressor manufacturer websites seem to support his
position. Measuring the inlet air makes sense from a practical point of view
as well, since it eliminates converting for pressure, which would make
comparisons more difficult. Ned Simmons has steered us in the right
direction regarding the potential energy of compressed air, which is
definitely lessened when you allow the volume to increase without recovering
any energy in the process. The transformer analogy is flawed, transformers
work both ways and with minimal loss. The same cannot be said for pressure
changes in compressed air.

Don Foreman wrote:
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote


It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more. Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight? Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment. Let's assume
adiabatic expansion (no heat exhange with the environment) to keep
things simple enough for me to understand.

I think the results would be about the same but I'm definitely open
to learning why not.

That's a good example. It exposes the fallacy in considering the regulated
air to be escaping into an open system. They would be close to being the
same in such a closed system, which is a somewhat similar system to some
practical compressed air uses. The air is going to flow until an equilibrium
is established at 50 psi. The inefficiency here is in compressing the air to
200 psi rather than 50 in the first place. In this case very little is
wasted. But in the extreme case of compressing air to 175 psi only to let it
flow through a spray gun at 5-10 psi, there is a great deal of waste, and
that's the way this thread started.

In article , jim rozen wrote:
In article tF2Ib.683809$HS4.4843991@attbi_s01, Loren Coe says...

well, it's not a dramatic increase in amps as pressure rises, not enough
to contradict Gary or Richard. my Sears 2hp (35+yrs old) starts on
an empty tank (0 pressure) at 11a and is at 12.7 amps at shutoff, 130 lbs.
that's about a 15% variation.

If you are just measuring this with an amp-clamp
meter, you may be fooled by the phase angle of
the current. At start much of that 11 amps
may be reactive power, at finish it may be mostly real.
Jim

gak! another ugly variable raised. yes, just an amp-clamp,
from RShack. maybe you do need a phd to argue in this thread(!).
in real power terms, what variation would be the most expected ?
--Loren

ps. just fyi, the start current was significantly higher at
35psi vs 0. just guessing, 20-25a vs 17-20. the branch
is on a 30a breaker. this comports w/my earlier experience
w/a similar vintage 1hp 120/240v Sears, it absolutely would
not start at 90psi when on a drop cord, but did just fine on
230v.

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

Grant Erwin writes:

I don't see it that way. If the spec isn't about CFM of pressurized
air, then why would the numbers differ at all for different output
pressure?

Because the pumping efficiency necessarily degrades as the output
pressure increases. The fact that the 5 hp example Coffman cites only
decreases slightly in rated CFM from 75 to 125 to 175 psi output
exhibits exactly this performance curve.

And why quote CFM into different pressures? Why quote
pressures at all?

Because different applications require different pressures, and
efficiency is higher when pressures are lower, and efficiency falls off
as pressures increase, so you want to set the tank cut-out as close to
the application pressure as you can.

Those numbers make sense to me as CFM of pressurized air. If you need
more volume than 14 cfm, you have to use a regulator to expand the air
downstream to get more flow. If you set your pressure switch to pump
your air tank up to 175 psi, then you have the headroom necessary to
regulate it back down to your desired end pressure but get quite a bit
more volume.

Absolutely impossible. You can't make a round-trip on pressure (pumping
in excess of the application pressure and then regulating it back down)
without it costing you in performance and efficiency, although other
goals may justify it. Doing what you propose will *decrease* the
output, not increase it.

Grant Erwin writes:

To me that reads "this compressor will deliver 14.7 cfm of compressed
air into a tank if that tank is at a pressure of 170 psi". Why
shouldn't I read this spec this way?

Because it would be wrong. It means 14.7 cubic feet of ambient air,
subsequently compressed to 170 psi.

I am about out of hope that you can understand why, since there have been
abundant explanations, examples, literature, and Web citations, all
consistently opposed to your notions.

ATP writes:

I would expect it to change somewhat, the air is not going to flow as
easily into a fully pressurized tank as it would into 0 psi. As it
turns out, it doesn't change much.

Yes, imagine a piston in a reciprocating compressor with leaf valves on the
output. Realize there is back pressure against those valves from the tank
pressure. The early portion of the piston stroke doesn't put anything into
the tank. Nothing goes into the tank until later in the stroke, when the
cylinder pressure first overcomes the tank pressure and forces the valve
open.

Surely we've all used a hand pump to inflate bicycle tires. Remember that
feeling that it didn't just get harder to pump against an almost-full tire,
but that your pumping strokes were of diminishing volume?

Why would it flow double the CFM at
a higher pressure?

Correct, the flow is strictly limited to the geometry of the piston
displacement, which is always the same, no matter what the pressure cut-off
is.

On Mon, 29 Dec 2003 09:56:07 -0800, Grant Erwin wrote:
Gary Coffman wrote:

Yes it makes sense.
Go he http://air.ingersoll-rand.com/CMP/sa...vertlevel3.htm
and then try to explain why the CFM ratings of these compressors don't
markedly change over a 75 to 175 PSI range.

For example, the IR 5 hp dual stage I have is rated at 14.9 CFM @75 PSI,
14.8 CFM @125 PSI, and 14.7 CFM @175 PSI. Same compressor, same
displacement, same speed, and nearly the same CFM over a pressure range
of more than 2:1. If we were to believe they're talking about CFM of
pressurized air, the gas law tells us we would see more than a 2:1 change
in CFM with a 2:1 change in pressure. But we don't. The only way we can
see the numbers they publish is if they're referring to CFM at 1 standard
pressure.

I don't see it that way. If the spec isn't about CFM of pressurized air, then
why would the numbers differ at all for different output pressure? And why
quote CFM into different pressures? Why quote pressures at all?

Volumetric efficiency of a piston pump (or an engine) is a function of the
pressure differential across it (among other things). As the back pressure
increases, volumetric efficiency drops. It doesn't drop a lot in the above
mentioned compressor, but it does drop.

If you look at the specs on my little Chinese portable compressor, you see
it drops a lot more with increasing output pressure. That's partially because
it is a single stage compressor, partially because it turns faster (volumetric
efficiency declines with speed), and partially because the valving isn't as
well designed as the IR compressor, so it doesn't breathe as well.

Gary

In article Ac6Ib.72824$VB2.143871@attbi_s51, Loren Coe says...

gak! another ugly variable raised. yes, just an amp-clamp,
from RShack. maybe you do need a phd to argue in this thread(!).
in real power terms, what variation would be the most expected ?
--Loren

A lot actually. For example, my unloaded phase conveter
draws about 8 amps at 240 volts, it would look like
a lot of power to an amp clamp. Measuring the phase angle
between current and voltage causes that number to drop
down to about 200 watts of real power consumed.

Sorry to muddy the waters - er, airflow!

Jim

==================================================
please reply to:
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==================================================

In article , jim rozen wrote:
In article Ac6Ib.72824$VB2.143871@attbi_s51, Loren Coe says...

gak! another ugly variable raised. yes, just an amp-clamp,
from RShack. maybe you do need a phd to argue in this thread(!).
in real power terms, what variation would be the most expected ? --Loren

A lot actually. For example, my unloaded phase conveter
draws about 8 amps at 240 volts, it would look like
a lot of power to an amp clamp. Measuring the phase angle
between current and voltage causes that number to drop
down to about 200 watts of real power consumed. Sorry
to muddy the waters - er, airflow! Jim

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.

11amps seemed like a lot for running free, basically, into an
empty tank. maybe i can just test at 0lbs and at cutoff using
the house meter? that sould not be too difficult, i can drop
the just few breakers to isolate the rest of the home, for a
few minutes. thanks! --Loren (in for a penny, in for...)

On Mon, 29 Dec 2003 18:33:09 -0500, Ned Simmons
wrote:

The weight on the piston at the end of the expansion is
zero. As the graph shows, the force that is exerted varies
over the two paths.

It declines linearly from 30 to 0 in the unregulated case
(red line).

The weight changes. Ooo-kay!

On Tue, 30 Dec 2003 02:55:01 GMT, "ATP"
wrote:

Don Foreman wrote:
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote


It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more. Consider
a reservoir of 1 cu ft volume containing 200 PSIG air, and a piston
& cylinder of 1 sq inch area lifting a 50 lb weight. First exhaust
air directly from the reservoir into the cylinder: how far does it
lift the weight? Then consider a regulator set to 50 PSI between
reservoir and cylinder and repeat the experiment. Let's assume
adiabatic expansion (no heat exhange with the environment) to keep
things simple enough for me to understand.

I think the results would be about the same but I'm definitely open
to learning why not.

That's a good example. It exposes the fallacy in considering the regulated
air to be escaping into an open system. They would be close to being the
same in such a closed system, which is a somewhat similar system to some
practical compressed air uses. The air is going to flow until an equilibrium
is established at 50 psi. The inefficiency here is in compressing the air to
200 psi rather than 50 in the first place. In this case very little is
wasted. But in the extreme case of compressing air to 175 psi only to let it
flow through a spray gun at 5-10 psi, there is a great deal of waste, and
that's the way this thread started.

Yes, that makes sense. 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. Thanks for the "get back on track" nudge!

In article _kpIb.705776$Fm2.607868@attbi_s04, Loren Coe says...

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.

Basically you need to be able to put current and
voltage waveforms on the scope at the same time,
so you can measure the phase angle between them.

I used a Fluke scope-meter, which is totally insulated
so I could simply wire the scope input across the
incoming 240 volt line, to show voltage. Then
to display current, I put a one ohm resistor in
series with the incoming line, and put the other
scope input at the other end of that resistor,
so the trace was an analog for current.

When the motor was just idling, the phase angle
was very, very close to 90 degrees, IIRC it was
about 86 degrees.

You could use two dual input plug-ins on a scope
like a Tek 7000 series, and use the 'difference'
option to show the same thing. Others have suggested
floating a scope chassis via isolation transformer.
Honestly I would strongly suggest NOT doing that.

Jim

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

Don Foreman wrote:
On Tue, 30 Dec 2003 02:55:01 GMT, "ATP"
wrote:

Don Foreman wrote:
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote


It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

If "some" means more than a few percent, please say more.
Consider a reservoir of 1 cu ft volume containing 200 PSIG air,
and a piston & cylinder of 1 sq inch area lifting a 50 lb weight.
First exhaust air directly from the reservoir into the cylinder:
how far does it lift the weight? Then consider a regulator set
to 50 PSI between reservoir and cylinder and repeat the
experiment. Let's assume adiabatic expansion (no heat exhange
with the environment) to keep things simple enough for me to
understand.

I think the results would be about the same but I'm definitely
open to learning why not.

That's a good example. It exposes the fallacy in considering the
regulated air to be escaping into an open system. They would be
close to being the same in such a closed system, which is a somewhat
similar system to some practical compressed air uses. The air is
going to flow until an equilibrium is established at 50 psi. The
inefficiency here is in compressing the air to 200 psi rather than
50 in the first place. In this case very little is wasted. But in
the extreme case of compressing air to 175 psi only to let it flow
through a spray gun at 5-10 psi, there is a great deal of waste, and
that's the way this thread started.

Yes, that makes sense. 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. Thanks for the "get back on track" nudge!

Wasn't trying to get you back on track, I like some of the twists and turns
this thread has taken, and your post corrected my thinking which had gone a
bit off track. I was initially thinking that if we allow a gas to expand, we
have lost potential energy since the larger volume of gas at a lower
pressure is capable of doing less work. This is not the case if the gas does
work in the process of that expansion, as in your example. The additional
turbulence/restriction created by the regulator must be accounted for, but
it may not have practical consequences in every application.

In any case, as a practical matter, we should strive to limit pressures,
allowing just enough headroom for variations in the volume needed, and use
turbines or vortex blowers for high volume/low pressure application. Venturi
jets are appropriate for some hi vol applications, more efficient than just
allowing the compressed air to expand.

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.







The IR site makes it clear that it is inefficient to create higher pressure
than necessary and regulate it down. I believe this thread was started with
a reference to air for an HVLP gun. Using a high pressure compressor and an
HVLP conversion regulator is an extremely inefficient way to run an HVLP
gun. This can be confirmed by comparing the power use of low pressure
turbines and the compressor/regulator setup needed to duplicate the low
pressure, high volume needed.

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


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.

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

Ned Simmons