I have been thinking about the issue of model equivalency between an air
regulator and a transformer again. Recall that I had postulated that the
equations of air flow and pressure with respect to an air regulator might
be similar to the equations of electricity flow and pressure (amps/volts)
with respect to a transformer. For an ideal transformer, of course, the
product of amps and volts on either side of the transformer is constant.
Thus I had postulated that if air were flowing e.g. at 10 cfm at 180 psi
and it were regulated down to 90 psi through an (unachievable) ideal
regulator, the output would be 20 cfm at 90 psi.

That discussion generated much heat but little light some months ago (GTA).
Many people felt that if you have an air compressor which can generate
e.g. 10 cfm into 90 psi that you cannot ever get more than 10 cfm out of it
no matter what. (It is possible that they didn't feel this way, but that is
what I perceived, but as usual I may have been wrong.)

As an interesting corollary to this discussion I just found an interesting
equation which I had not known, which is the mathematical relationship between
SCFM and CFM when the air pressure is expressed in pounds per square inch (psi):

SCFM = CFM * SQRT[(Pg + 14.7)/14.7] ;; Pg expressed in psi

For example if a compressor is rated to deliver 10 cfm into 90 psi then it
could equivalently be rated to deliver approximately 26 SCFM. So beware of
SCFM ratings unless you have the above equation handy!

(In case anyone is curious, I got the above equation from a Sylvania Web page:
http://www.sylvania.com/pmc/heaters/air/using.htm)

I propose an experiment: an air regulator with an airflow meter on either
side of it, plumbed in series with it. If I attached my compressor to the
input of this and a big air grinder (running unloaded) or some other fairly
constant load, let the tank charge up fully and then open the tank valve
to energize the system, let the system stabilize and then pull the trigger
on the air grinder and take readings off the airflow meters for several
different regulator settings, that should produce data which would either
support or refute my postulated equations of flow/pressure on either side
of the regulator.

So. Does my proposed experiment make sense? Does anyone have a couple of
suitable airflow meters they would like to loan me or sell me real cheap?

Grant Erwin
Kirkland, Washington

SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi SCFM
on either side of a regulator is the same. SCFM on either side of
a compressor is the same.

All the comressors are rated in SCFM even if the the label only
says CFM. Think of the standard piston compressor: it runs at
some speed, has a certain displacement. At low pressures it
essentially delivers the CFM of the discplacement times the
speed. As the pressure goes up, there are some losses so the CFM
at higher pressures drops somewhat, typically 10% to 20% less
going from 40 to 90 psi.

Grant Erwin wrote:

I have been thinking about the issue of model equivalency between an air
regulator and a transformer again. Recall that I had postulated that the
equations of air flow and pressure with respect to an air regulator might
be similar to the equations of electricity flow and pressure (amps/volts)
with respect to a transformer. For an ideal transformer, of course, the
product of amps and volts on either side of the transformer is constant.
Thus I had postulated that if air were flowing e.g. at 10 cfm at 180 psi
and it were regulated down to 90 psi through an (unachievable) ideal
regulator, the output would be 20 cfm at 90 psi.

That discussion generated much heat but little light some months ago (GTA).
Many people felt that if you have an air compressor which can generate
e.g. 10 cfm into 90 psi that you cannot ever get more than 10 cfm out of it
no matter what. (It is possible that they didn't feel this way, but that is
what I perceived, but as usual I may have been wrong.)

As an interesting corollary to this discussion I just found an interesting
equation which I had not known, which is the mathematical relationship
between
SCFM and CFM when the air pressure is expressed in pounds per square
inch (psi):

SCFM = CFM * SQRT[(Pg + 14.7)/14.7] ;; Pg expressed in psi

For example if a compressor is rated to deliver 10 cfm into 90 psi then it
could equivalently be rated to deliver approximately 26 SCFM. So beware of
SCFM ratings unless you have the above equation handy!

(In case anyone is curious, I got the above equation from a Sylvania Web
page:
http://www.sylvania.com/pmc/heaters/air/using.htm)

I propose an experiment: an air regulator with an airflow meter on either
side of it, plumbed in series with it. If I attached my compressor to the
input of this and a big air grinder (running unloaded) or some other fairly
constant load, let the tank charge up fully and then open the tank valve
to energize the system, let the system stabilize and then pull the trigger
on the air grinder and take readings off the airflow meters for several
different regulator settings, that should produce data which would either
support or refute my postulated equations of flow/pressure on either side
of the regulator.

So. Does my proposed experiment make sense? Does anyone have a couple of
suitable airflow meters they would like to loan me or sell me real cheap?

Grant Erwin
Kirkland, Washington

Roy J wrote:
SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi

Perhaps you meant to write, "one cubic foot of air @ 14.7 psi per minute"?

SCFM on either side of a regulator is the same.

That's interesting. How did you come to that conclusion? Applying that to
my hypothetical situation of an airflow of 10 cfm @ 180 psi regulated down
to 90 psi, that would yield about 13.7 cfm at 90 psi. I suppose we're assuming
constant temperature throughout.

Grant Erwin
Kirkland, Washington

The one thing we must agree on is that the MASS flow rate thru a regulator
or compressor is conserved. (Assuming no leaks)

One cubic foot per minute of air at 14.7 psia (a for absolute) is equal to
some mass flow rate kg/minute, grams/second. I don't care what units.
So an SCFM is equivalent to a mass flow rate.
The convervation of mass principle says there will be the same mass flow
rate on the downstream side of the regulator. Hence Roy J's assertion that
SCFM is the same on either side of regulator.

Erich


"Grant Erwin" wrote in message
...
Roy J wrote:
SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi

Perhaps you meant to write, "one cubic foot of air @ 14.7 psi per minute"?

SCFM on either side of a regulator is the same.

That's interesting. How did you come to that conclusion? Applying that to
my hypothetical situation of an airflow of 10 cfm @ 180 psi regulated down
to 90 psi, that would yield about 13.7 cfm at 90 psi. I suppose we're
assuming
constant temperature throughout.

Grant Erwin
Kirkland, Washington

Yes PER MINUTE, adding coffee helps one add all the proper terms
at the end!

And I was assuming constant temp, minimal moisture, etc etc to
calm down the calculations. In the case of a shop compressor
running a lower duty cycle, this is reasonable. Youcan get some
pretty weird numbers when you locate the compressor outside,
sucking outside air, then pipeing it inside.

The conclusion is the STANDARD cfm part. That implies the number
of air molecules (composed of O2, N2, Co2 and whatever else) is
conserved on both sides of the regulator.

Grant Erwin wrote:
Roy J wrote:

SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi


Perhaps you meant to write, "one cubic foot of air @ 14.7 psi per minute"?

SCFM on either side of a regulator is the same.


That's interesting. How did you come to that conclusion? Applying that to
my hypothetical situation of an airflow of 10 cfm @ 180 psi regulated down
to 90 psi, that would yield about 13.7 cfm at 90 psi. I suppose we're
assuming
constant temperature throughout.

Grant Erwin
Kirkland, Washington

Grant Erwin writes:

For example if a compressor is rated to deliver 10 cfm into 90 psi
then it could equivalently be rated to deliver approximately 26 SCFM.
So beware of SCFM ratings unless you have the above equation handy!

(In case anyone is curious, I got the above equation from a Sylvania
Web page: http://www.sylvania.com/pmc/heaters/air/using.htm)

I believe this is incorrect. The "S" means standard temperature and
humidity of the free air. See Machinery's Handbook.

http://www.truetex.com/aircompressors.htm

Grant Erwin writes:

That's interesting. How did you come to that conclusion? Applying that
to my hypothetical situation of an airflow of 10 cfm @ 180 psi
regulated down to 90 psi, that would yield about 13.7 cfm at 90 psi. I
suppose we're assuming constant temperature throughout.

No, you fundamentally misunderstand.

CFM (or SCFM) has nothing to do with pressure. It is a measure of the
flow rate of air, expressed at atmospheric pressure.

The CFM (or SCFM) going into your tool is the same as the CFM (or SCFM)
exhausted, even though the power has been spent. Think of the CFM as
the amount of exhaust (free air) that comes out of the tool.

An air regulator is not at all analogous to an electric transformer.
The proper analogy is to a three-terminal voltage regulator. Energy is
lost; that is how the regulator works. The CFM (or SCFM) on either side
of the air regulator is necessarily equal. The pressure drops. Power
is lost and turned into heat.

SCFM is just CFM measured with a standard temperature and humidity in
the free air. Hotter or wetter input air will degrade the compressor
performance. Cooler or drier input air (than the standard) will improve
it.

An "SCFM" (standard CFM) is a CFM produced with input air at 68 deg F
and 36 percent relative humidity.

I have a 600 CFM compressor (!) in my garage that uses only 1/3 HP!
Read how:

http://www.truetex.com/aircompressors.htm

On Tue, 23 Dec 2003 02:19:00 -0600, Richard J Kinch wrote:
An air regulator is not at all analogous to an electric transformer.
The proper analogy is to a three-terminal voltage regulator. Energy is
lost; that is how the regulator works. The CFM (or SCFM) on either side
of the air regulator is necessarily equal. The pressure drops. Power
is lost and turned into heat.

No. A *shunt* regulator would behave the way you describe, but this
is a series regulator.

The regulator is a feedback controlled valve. It is not a dissipative
device. It controls the mass flow of air from the tank such that a
particular set pressure develops against the flow resistance existing
downstream of the valve. It does not bleed excess air to atmosphere
in order to do this, it does not get perceptibly hot. So any energy not
required to achieve the set downstream pressure *remains in the tank*.
It is not lost or turned to heat.

SCFM *is* a measure of mass flow. At equilibrium flows, the mass
flow into the valve *is* the same as the mass flow out of the valve.
If we consider this mass flow analogous to current, then we can
apply Kirchhoff's laws and show that the current is everywhere
the same in a series mesh.

We can also use Kirchhoff's laws to show that voltage (pressure)
drops across the regulator valve and across the downstream flow
resistances sum to tank pressure. In other words, the sum of the
voltages around a series mesh equal zero, with the tank pressure
(analogous to a battery) treated as positive and the pressure drops
across the various flow resistances treated as negative.

But the key thing to understand here is that a resistance need not
be *dissipative*. A good electrical example is a triode tube, or *valve*
as the British called them. The control voltage on the grid changes
the current flow through the tube by modulating what we call the
plate resistance. But this isn't an actual dissipative resistance.
It is a *mathematical fiction* we use to model the plate current
valving action of the grid.

Similarly, the diaphram of the air regulator merely modulates the
flow through the valve by opening or closing the valve. Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.

Gary

I am an electrical engineer and so I view things from that perspective. I
considered the issue of the regulator and essentially my analysis agrees with
Gary's 100%. I believe 2 things to be true:

1. Mass is conserved (what goes into the regulator must come out)
2. To first order, energy is conserved through the regulator

That's why I don't see why flow x pressure wouldn't be constant across a
regulator in steady state.

Postulate a big air tank pressurized to 180 psi, with a long (long enough so
the air has time to cool to ambient) pipe to an ideal regulator which regulates
the pressure down to 90 psi. The regulator's output is a pipe of the same size
which is connected to a constant load. The cfm going into the regulator is
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at
90 psi?

Grant Erwin

Gary Coffman wrote:

On Tue, 23 Dec 2003 02:19:00 -0600, Richard J Kinch wrote:

An air regulator is not at all analogous to an electric transformer.
The proper analogy is to a three-terminal voltage regulator. Energy is
lost; that is how the regulator works. The CFM (or SCFM) on either side
of the air regulator is necessarily equal. The pressure drops. Power
is lost and turned into heat.


No. A *shunt* regulator would behave the way you describe, but this
is a series regulator.

The regulator is a feedback controlled valve. It is not a dissipative
device. It controls the mass flow of air from the tank such that a
particular set pressure develops against the flow resistance existing
downstream of the valve. It does not bleed excess air to atmosphere
in order to do this, it does not get perceptibly hot. So any energy not
required to achieve the set downstream pressure *remains in the tank*.
It is not lost or turned to heat.

SCFM *is* a measure of mass flow. At equilibrium flows, the mass
flow into the valve *is* the same as the mass flow out of the valve.
If we consider this mass flow analogous to current, then we can
apply Kirchhoff's laws and show that the current is everywhere
the same in a series mesh.

We can also use Kirchhoff's laws to show that voltage (pressure)
drops across the regulator valve and across the downstream flow
resistances sum to tank pressure. In other words, the sum of the
voltages around a series mesh equal zero, with the tank pressure
(analogous to a battery) treated as positive and the pressure drops
across the various flow resistances treated as negative.

But the key thing to understand here is that a resistance need not
be *dissipative*. A good electrical example is a triode tube, or *valve*
as the British called them. The control voltage on the grid changes
the current flow through the tube by modulating what we call the
plate resistance. But this isn't an actual dissipative resistance.
It is a *mathematical fiction* we use to model the plate current
valving action of the grid.

Similarly, the diaphram of the air regulator merely modulates the
flow through the valve by opening or closing the valve. Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.

Gary

In article , Gary Coffman says...

The regulator is a feedback controlled valve. It is not a dissipative
device. It controls the mass flow of air from the tank such that a
particular set pressure develops against the flow resistance existing
downstream of the valve. It does not bleed excess air to atmosphere
in order to do this, it does not get perceptibly hot.

Indeed because expansion is occuring inside the regulator,
under certain conditions, they may start to chill preceptably.

SCFM *is* a measure of mass flow. At equilibrium flows, the mass
flow into the valve *is* the same as the mass flow out of the valve.

Right! Otherwise, the darn thing's gonna get so heavy
it will tip over the tank.

Jim

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On Tue, 23 Dec 2003 09:41:46 -0800, Grant Erwin wrote:
I am an electrical engineer and so I view things from that perspective. I
considered the issue of the regulator and essentially my analysis agrees with
Gary's 100%. I believe 2 things to be true:

1. Mass is conserved (what goes into the regulator must come out)
2. To first order, energy is conserved through the regulator

That's why I don't see why flow x pressure wouldn't be constant across a
regulator in steady state.

Postulate a big air tank pressurized to 180 psi, with a long (long enough so
the air has time to cool to ambient) pipe to an ideal regulator which regulates
the pressure down to 90 psi. The regulator's output is a pipe of the same size
which is connected to a constant load. The cfm going into the regulator is
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at
90 psi?

10 CFM of course. As you note, mass is conserved, or as Kirchhoff's laws
tell us, current is everywhere the same in a series mesh. There is nowhere
else for the air to go. If it has a certain mass flow into the valve, it has to
have exactly the same mass flow out of it.

The thing people seem to be having trouble grasping is that CFM is a
measure of mass flow. This is pretty obvious for an incompressible liquid
like water, but for a gas, CFM has to be stated in terms of a standard
temperature and pressure. That's been defined by the standards bodies
to be 68 F and 1 atmosphere.

Gary

In article , Grant Erwin says...

Postulate a big air tank pressurized to 180 psi, with a long (long enough so
the air has time to cool to ambient) pipe to an ideal regulator which regulates
the pressure down to 90 psi. The regulator's output is a pipe of the same size
which is connected to a constant load. The cfm going into the regulator is
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at
90 psi?

Don't do the problem in cubic feet, simply convert into number
of molecules (and for simplicity, perform it with nitrogen) that
you calculate using teh universal gas law.

So many molecules of nitrogen enter the regulator, the
same number leave it. If the temperature at the inlet and
outlet are the same, then the volume will scale inversely
like the pressure.

P(1) X V(1) = P(2) X V(2)

Pressure goes down by a factor of two, the
volume will increase by two.

The pressures btw should be absolute, not gage, not
a problem to do in psig if the pressures are high.

Jim

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

In article , Gary Coffman says...

10 CFM of course. As you note, mass is conserved,

But there is twice the mass flow in the upstream
(hp) side in teh example. In one minute, 10
cu feet of air at 180 psi flows in. If you count
the number of atoms (mass) you will find that the
same number come out the downstream side at 90
psi. They just take up more room at a lower pressure.

I would say that if 10 cubic feet of atoms at
180 psi flow in, then 20 cubic feet of atoms
will flow out, at 90 psi.

PV = PV, universal gas law and all.

PV = nKT

n is not varying, same number of atoms.
KT is constant if you allow the gas to
come to thermal equibrium at the outlet of
the regulator.

So P1/P2 = V2/V1 and all.

Jim

Jim

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

Gary Coffman wrote:

Postulate a big air tank pressurized to 180 psi, with a long (long enough so
the air has time to cool to ambient) pipe to an ideal regulator which regulates
the pressure down to 90 psi. The regulator's output is a pipe of the same size
which is connected to a constant load. The cfm going into the regulator is
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at
90 psi?


10 CFM of course. As you note, mass is conserved, or as Kirchhoff's laws
tell us, current is everywhere the same in a series mesh. There is nowhere
else for the air to go. If it has a certain mass flow into the valve, it has to
have exactly the same mass flow out of it.

This doesn't sound right to me. Let's think about the amount of air molecules
that go into the regulator during one minute. It's the number of molecules
in ten cubic feet at some temperature at 180 psi (which isn't an absolute
pressure to be sure). Now that many molecules have to come out the other side
in that minute, right? (Kirchoff and all that.) The gas law is PV = nRT. If
we call the input side 1 and the output side 2, then we can write P1V1 = nRT.
Since the number of molecules, n, is the same, and the temperature is the same,
and since R is a constant, then P2V2 = nRT. So once again I do not see why
P1V1 shouldn't equal P2V2.

I can (finally!) see why you can't plug in gage pressure into this equation.
The absolute pressure is (I believe) gage pressure plus 14.7 psi.

Therefore I predict the answer V2 = (180+14.7)*10/(90+14.7) = 18.6 cfm as
long as the temperature on both sides is equal.

Boy, I wish I had 2 flowmeters.

Grant Erwin

In article , Grant Erwin says...

Boy, I wish I had 2 flowmeters.

You don't need them! You understand
the physics behind it instead, which
is better.

Jim

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"jim rozen" wrote in message
...
The regulator is a feedback controlled valve. ... it does not get
perceptibly hot.

Indeed because expansion is occuring inside the regulator,
under certain conditions, they may start to chill preceptably.

So where does the heat go? I'm going to guess that compressive systems
such as this act as heat pumps, thus all the heat goes back to the
compressor, while cooling occurs at the tools and regulator. Eh?

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

Gary Coffman writes:

Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.

Impossible.

We agree that mass flow (CFM) is conserved (equal) on either side of the
regulator.

We know: Power = CFM * pressure.

The regulator imposes: pressure (out) pressure (in).

Thus, since CFM is equal on both sides of the regulator, but pressure
decreases, there is a loss of power in the output compared to the input.
This ends up as heat as a direct consequence of the restriction that
creates turbulence and lowers the pressure. This waste heat is mostly
added to the output flow, even though the output may be at a cooler
temperature due to expansion. Some of the power lost is hissing noise
that radiates away and also eventually becomes heat. None of this is
"visible" to the source, so it is not "simply retained in the tank".

Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.

OK. CFM does not equal mass flow. CFM means simply cubic feet of air per
minute. The amount of mass in that air is proportional to both the pressure
and the temperature, as described by the gas law PV = nRT.

I agree there are second-order losses in a regulator. It might make some
noise, and it might warm up a little or cool down a little. This is also
true of a transformer.

You know, what's *really* amazing is that I know Gary Coffman and Richard
Kinch and I are all intelligent, good writers, and well educated, and yet
we get THREE different answers.

I'd just love to get to the point of agreement.

So, Richard, I DISAGREE that cfm is equal on either side of the regulator.
It cannot be because mass is conserved! (See my previous posts)

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:

Gary Coffman writes:


Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.


Impossible.

We agree that mass flow (CFM) is conserved (equal) on either side of the
regulator.

We know: Power = CFM * pressure.

The regulator imposes: pressure (out) pressure (in).

Thus, since CFM is equal on both sides of the regulator, but pressure
decreases, there is a loss of power in the output compared to the input.
This ends up as heat as a direct consequence of the restriction that
creates turbulence and lowers the pressure. This waste heat is mostly
added to the output flow, even though the output may be at a cooler
temperature due to expansion. Some of the power lost is hissing noise
that radiates away and also eventually becomes heat. None of this is
"visible" to the source, so it is not "simply retained in the tank".

Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.

In article , Richard J Kinch
says...

We agree that mass flow (CFM)

Stop right there.

Mass flow and CFM are not the same thing.
Once you realize this you can begin to
think about the issue in a different way.

Cubic feet per minute is only a time
rate of volume. So in a given time
(say one minute) a give volume of material
will pass through your system.

Depending on the pressure inside the volume,
you can have all different amounts of material.
For example, a liter of gas at atmosperic
pressure has only a few atoms. If you image
the same volume at scuba tank pressure, it
has a lot more atoms inside.

Jim

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In article , Tim Williams says...

So where does the heat go?

In the case where large amounts of gas are
expanding, the expansion valve will cool,
and the heat from the surrounding area will
flow into the region. Also the gas will
exit the expansion valve at a lower
temperature than it came in at.

This is how gas liquifers are designed,
either with direct expansion valves or
with expansion turbines, where the gas
actually performs mechanical work as it
expands, and thus cools.

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"The thing people seem to be having trouble grasping is that CFM is a
measure of mass flow."

Hmmm, well.... CFM is volume/min, and per-se is not related to mass
flow. SCFM, on the other hand, indirectly specifies a mass flow because
the air is at STP.

Have a look at http://www.cleandryair.com/scfm_vs__icfm_vs__acfm.htm

Roger




Gary Coffman wrote:

On Tue, 23 Dec 2003 09:41:46 -0800, Grant Erwin wrote:

I am an electrical engineer and so I view things from that perspective. I
considered the issue of the regulator and essentially my analysis agrees with
Gary's 100%. I believe 2 things to be true:

1. Mass is conserved (what goes into the regulator must come out)
2. To first order, energy is conserved through the regulator

That's why I don't see why flow x pressure wouldn't be constant across a
regulator in steady state.

Postulate a big air tank pressurized to 180 psi, with a long (long enough so
the air has time to cool to ambient) pipe to an ideal regulator which regulates
the pressure down to 90 psi. The regulator's output is a pipe of the same size
which is connected to a constant load. The cfm going into the regulator is
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at
90 psi?


10 CFM of course. As you note, mass is conserved, or as Kirchhoff's laws
tell us, current is everywhere the same in a series mesh. There is nowhere
else for the air to go. If it has a certain mass flow into the valve, it has to
have exactly the same mass flow out of it.

The thing people seem to be having trouble grasping is that CFM is a
measure of mass flow. This is pretty obvious for an incompressible liquid
like water, but for a gas, CFM has to be stated in terms of a standard
temperature and pressure. That's been defined by the standards bodies
to be 68 F and 1 atmosphere.

Gary

No, as Richard said, a linear regulator is a better, but
not very good, analogy. I think any analogy between
electricity and a compressible gas is doomed. You really
need to resort to thermodynamics to do this subject
justice.

But air regulators *are* lossy devices, and not just
incidental losses as in a transformer. Last time this came
around (and it looks like you started it that time too,
Grant g) I tried to avoid thermo (which I don't feel
qualified to preach after a 30 years lapse) and explain the
loss by resorting to an example of the potential of a given
mass of air do do work on a piston, starting from different
pressures.

This is the last post in the thread:

http://groups.google.com/groups?
q=g:thl672785172d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8
&selm=MPG.189e9353cdce4f36989893%40news.rcn.com

Ned Simmons

In article ,
says...
OK. CFM does not equal mass flow. CFM means simply cubic feet of air per
minute. The amount of mass in that air is proportional to both the pressure
and the temperature, as described by the gas law PV = nRT.

I agree there are second-order losses in a regulator. It might make some
noise, and it might warm up a little or cool down a little. This is also
true of a transformer.

You know, what's *really* amazing is that I know Gary Coffman and Richard
Kinch and I are all intelligent, good writers, and well educated, and yet
we get THREE different answers.

I'd just love to get to the point of agreement.

So, Richard, I DISAGREE that cfm is equal on either side of the regulator.
It cannot be because mass is conserved! (See my previous posts)

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:

Gary Coffman writes:


Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.


Impossible.

We agree that mass flow (CFM) is conserved (equal) on either side of the
regulator.

We know: Power = CFM * pressure.

The regulator imposes: pressure (out) pressure (in).

Thus, since CFM is equal on both sides of the regulator, but pressure
decreases, there is a loss of power in the output compared to the input.
This ends up as heat as a direct consequence of the restriction that
creates turbulence and lowers the pressure. This waste heat is mostly
added to the output flow, even though the output may be at a cooler
temperature due to expansion. Some of the power lost is hissing noise
that radiates away and also eventually becomes heat. None of this is
"visible" to the source, so it is not "simply retained in the tank".

Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.

In article , Roger Head says...

Hmmm, well.... CFM is volume/min, and per-se is not related to mass
flow. SCFM, on the other hand, indirectly specifies a mass flow because
the air is at STP.

Other gasses are also specified in SCFM.
It doesn't have to be air. You would have different
mass flows for one scfm of, say, hydrogen vs the
same scfm of xenon.

Jim

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

Yes, I started that thread too. The reason that this has come back is that
I was never convinced the last time. Let me ask YOU, Ned, to answer my
question? I'll repeat it:

Postulate a big air tank pressurized to 180 psi, with a long (long
enough so the air has time to cool to ambient) pipe to an ideal
regulator which regulates the pressure down to 90 psi. The regulator's
output is a pipe of the same size which is connected to a constant
load of such a size as to make the cfm going into the regulator
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator
at 90 psi?

Grant Erwin

Ned Simmons wrote:
No, as Richard said, a linear regulator is a better, but
not very good, analogy. I think any analogy between
electricity and a compressible gas is doomed. You really
need to resort to thermodynamics to do this subject
justice.

But air regulators *are* lossy devices, and not just
incidental losses as in a transformer. Last time this came
around (and it looks like you started it that time too,
Grant g) I tried to avoid thermo (which I don't feel
qualified to preach after a 30 years lapse) and explain the
loss by resorting to an example of the potential of a given
mass of air do do work on a piston, starting from different
pressures.

This is the last post in the thread:

http://groups.google.com/groups?
q=g:thl672785172d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8
&selm=MPG.189e9353cdce4f36989893%40news.rcn.com

Ned Simmons

In article ,
says...

OK. CFM does not equal mass flow. CFM means simply cubic feet of air per
minute. The amount of mass in that air is proportional to both the pressure
and the temperature, as described by the gas law PV = nRT.

I agree there are second-order losses in a regulator. It might make some
noise, and it might warm up a little or cool down a little. This is also
true of a transformer.

You know, what's *really* amazing is that I know Gary Coffman and Richard
Kinch and I are all intelligent, good writers, and well educated, and yet
we get THREE different answers.

I'd just love to get to the point of agreement.

So, Richard, I DISAGREE that cfm is equal on either side of the regulator.
It cannot be because mass is conserved! (See my previous posts)

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:


Gary Coffman writes:



Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.


Impossible.

We agree that mass flow (CFM) is conserved (equal) on either side of the
regulator.

We know: Power = CFM * pressure.

The regulator imposes: pressure (out) pressure (in).

Thus, since CFM is equal on both sides of the regulator, but pressure
decreases, there is a loss of power in the output compared to the input.
This ends up as heat as a direct consequence of the restriction that
creates turbulence and lowers the pressure. This waste heat is mostly
added to the output flow, even though the output may be at a cooler
temperature due to expansion. Some of the power lost is hissing noise
that radiates away and also eventually becomes heat. None of this is
"visible" to the source, so it is not "simply retained in the tank".

Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.

Gary Coffman wrote:



snipped


But the key thing to understand here is that a resistance need not
be *dissipative*. A good electrical example is a triode tube, or *valve*
as the British called them. The control voltage on the grid changes
the current flow through the tube by modulating what we call the
plate resistance. But this isn't an actual dissipative resistance.
It is a *mathematical fiction* we use to model the plate current
valving action of the grid.

Not really Gary:

AC plate resistance is the "fictional one" and is defined as the dynamic ratio of
plate voltage to changes in plate current at a *constant* grid voltage.

DC plate resistance is the ratio of plate voltage to plate current and it is a
*dissipative* resistance. That's what makes the plate of a triode (or a diode,
tetrode or pentode tube) glow red hot if you "push it" too much.

Don't take it too hard Gary G I had to confirm my dusty memories of this stuff
at:

http://www.lh-electric.4t.com/vt_primer4.html

Jeff (Who burned his fingers more than once on those big black metal metal 6L6s he
couldn't see had a cherry red plate.)
--

Jeff Wisnia (W1BSV + Brass Rat '57 EE)

"If you can smile when things are going wrong, you've thought of someone to blame
it on."

In article ,
says...
Yes, I started that thread too. The reason that this has come back is that
I was never convinced the last time. Let me ask YOU, Ned, to answer my
question? I'll repeat it:

Postulate a big air tank pressurized to 180 psi, with a long (long
enough so the air has time to cool to ambient) pipe to an ideal
regulator which regulates the pressure down to 90 psi. The regulator's
output is a pipe of the same size which is connected to a constant
load of such a size as to make the cfm going into the regulator
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator
at 90 psi?


If we assume that the air behaves as an ideal gas and the
pressures are absolute, then there'll be 20 CFM @ 90 psia
flowing out of the regulator. I don't think there's ever
been any question about that. But that doesn't mean that
there isn't a loss in the potential of that air to do work.

CFM X psi for a compressible gas is not analogous to volts
X amps.

Ned Simmons

P1V1/T1 = P2V2/T2 !!!! You guys need to quit trying to complicate
the simple answer! Unless you are talking SCFM in which case it will
be 10 SCFM since any loss internal to the regulator will be reflected
in the pressure at the outlet side of the regulator and temperature is
canceled out in conversion to SCFM. Mass WILL be conserved. If you
doubt me, try Compressed Air and Gas Data or any High School physics
textbook. A cubic foot VOLUME is a cubic foot VOLUME regardless of
pressure, temperature or phase of the moon. A STANDARD cubic foot, on
the other hand is a specific mass of the gas in question. Which is
why a 600 cfm compressor can have a 1/3 hp motor and a 10 cfm
compressor can require a 500 hp motor. The 600 cfm compressor being a
cooling fan producing flow @ an inch or two of water and the 10 cfm
compressor producing flow @ 1000 psig. Obviously the 10 cfm @1000 psig
results in a much larger SCFM than the 600cfm @ inches of H20. There
will be no heating of the regulator since expansion of a gas is an
endothermic sort of thing requiring enegy input. Incidentally the
numbers for "STANDARD" conditions vary somewhat from country to
country, but in the US we typically have used 14.696 psi, 60 deg. F
and 0% relative humidity

Gary Hallenbeck
On Tue, 23 Dec 2003 19:34:21 -0800, Grant Erwin
wrote:

Yes, I started that thread too. The reason that this has come back is that
I was never convinced the last time. Let me ask YOU, Ned, to answer my
question? I'll repeat it:

Postulate a big air tank pressurized to 180 psi, with a long (long
enough so the air has time to cool to ambient) pipe to an ideal
regulator which regulates the pressure down to 90 psi. The regulator's
output is a pipe of the same size which is connected to a constant
load of such a size as to make the cfm going into the regulator
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator
at 90 psi?

Grant Erwin

Ned Simmons wrote:
No, as Richard said, a linear regulator is a better, but
not very good, analogy. I think any analogy between
electricity and a compressible gas is doomed. You really
need to resort to thermodynamics to do this subject
justice.

But air regulators *are* lossy devices, and not just
incidental losses as in a transformer. Last time this came
around (and it looks like you started it that time too,
Grant g) I tried to avoid thermo (which I don't feel
qualified to preach after a 30 years lapse) and explain the
loss by resorting to an example of the potential of a given
mass of air do do work on a piston, starting from different
pressures.

This is the last post in the thread:

http://groups.google.com/groups?
q=g:thl672785172d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8
&selm=MPG.189e9353cdce4f36989893%40news.rcn.com

Ned Simmons

In article ,
says...

OK. CFM does not equal mass flow. CFM means simply cubic feet of air per
minute. The amount of mass in that air is proportional to both the pressure
and the temperature, as described by the gas law PV = nRT.

I agree there are second-order losses in a regulator. It might make some
noise, and it might warm up a little or cool down a little. This is also
true of a transformer.

You know, what's *really* amazing is that I know Gary Coffman and Richard
Kinch and I are all intelligent, good writers, and well educated, and yet
we get THREE different answers.

I'd just love to get to the point of agreement.

So, Richard, I DISAGREE that cfm is equal on either side of the regulator.
It cannot be because mass is conserved! (See my previous posts)

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:


Gary Coffman writes:



Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.


Impossible.

We agree that mass flow (CFM) is conserved (equal) on either side of the
regulator.

We know: Power = CFM * pressure.

The regulator imposes: pressure (out) pressure (in).

Thus, since CFM is equal on both sides of the regulator, but pressure
decreases, there is a loss of power in the output compared to the input.
This ends up as heat as a direct consequence of the restriction that
creates turbulence and lowers the pressure. This waste heat is mostly
added to the output flow, even though the output may be at a cooler
temperature due to expansion. Some of the power lost is hissing noise
that radiates away and also eventually becomes heat. None of this is
"visible" to the source, so it is not "simply retained in the tank".

Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.

Gary Coffman wrote:
On Tue, 23 Dec 2003 02:19:00 -0600, Richard J Kinch
wrote:
An air regulator is not at all analogous to an electric transformer.
The proper analogy is to a three-terminal voltage regulator. Energy
is lost; that is how the regulator works. The CFM (or SCFM) on
either side of the air regulator is necessarily equal. The pressure
drops. Power is lost and turned into heat.

No. A *shunt* regulator would behave the way you describe, but this
is a series regulator.

The regulator is a feedback controlled valve. It is not a dissipative
device. It controls the mass flow of air from the tank such that a
particular set pressure develops against the flow resistance existing
downstream of the valve. It does not bleed excess air to atmosphere
in order to do this, it does not get perceptibly hot. So any energy
not required to achieve the set downstream pressure *remains in the
tank*. It is not lost or turned to heat.

SCFM *is* a measure of mass flow. At equilibrium flows, the mass
flow into the valve *is* the same as the mass flow out of the valve.
If we consider this mass flow analogous to current, then we can
apply Kirchhoff's laws and show that the current is everywhere
the same in a series mesh.

We can also use Kirchhoff's laws to show that voltage (pressure)
drops across the regulator valve and across the downstream flow
resistances sum to tank pressure. In other words, the sum of the
voltages around a series mesh equal zero, with the tank pressure
(analogous to a battery) treated as positive and the pressure drops
across the various flow resistances treated as negative.

But the key thing to understand here is that a resistance need not
be *dissipative*. A good electrical example is a triode tube, or
*valve* as the British called them. The control voltage on the grid
changes the current flow through the tube by modulating what we call
the plate resistance. But this isn't an actual dissipative resistance.
It is a *mathematical fiction* we use to model the plate current
valving action of the grid.

Similarly, the diaphram of the air regulator merely modulates the
flow through the valve by opening or closing the valve.
Mathematically, this has the same appearance as a dissipative
resistance, but there is *no dissipation*. Energy not used is simply
retained in the tank.

Gary

In normal circumstances while enough air is being used. In the case of gas
regulators, and probably any critical regulator, a valve closing too quickly
on the downstream side will cause a buildup of pressure that the regulator
will not be able to prevent by modulating. The excess gas is vented to the
atmosphere. It has to be vented to some lower pressure region or there is no
way the regulator can reduce the pressure. Gas utilities have to tweak
spring rates to get the right performance and supply out of regulators. WRT
the postulated problem of X CFM going in, that's really like starting with
the problem already solved and backing out the answer, since there is no
ideal regulator. The chambers of the regulator would have to be sufficiently
large as to cause no significant resistance to flow for us to be confident
that 10 CFM at 100 PSI with no regulator would result in 20 CFM at 50 PSI on
the outlet side of the regulator. Starting the problem with what is coming
out of the regulator would make more sense to me.

"Richard J Kinch" wrote in message
. ..
Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.

Actually, that'd be a rotary phase converter type deal. An inverter would
be a whistle, and a transformer would be acoustical in nature. This makes
sense since the only way to regulate DC is a variable resistance (as a
pass or shunt regulator), likewise it is for air at pressure.

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

Ned Simmons wrote:
In article ,
says...
Yes, I started that thread too. The reason that this has come back
is that I was never convinced the last time. Let me ask YOU, Ned, to
answer my question? I'll repeat it:

Postulate a big air tank pressurized to 180 psi, with a long (long
enough so the air has time to cool to ambient) pipe to an ideal
regulator which regulates the pressure down to 90 psi. The
regulator's output is a pipe of the same size which is connected to
a constant load of such a size as to make the cfm going into the
regulator measured to be 10 cfm @ 180 psi. What cfm will come out of
the regulator at 90 psi?


If we assume that the air behaves as an ideal gas and the
pressures are absolute, then there'll be 20 CFM @ 90 psia
flowing out of the regulator. I don't think there's ever
been any question about that. But that doesn't mean that
there isn't a loss in the potential of that air to do work.

CFM X psi for a compressible gas is not analogous to volts
X amps.

Ned Simmons

You're correct, it can not do the same mechanical work, that energy has been
accounted for in the thermodynamic changes, as you stated previously. The
process is obviously not reversible without the addition of mechanical
energy, we can not use a "reverse" regulator to go to less CFM at a higher
pressure, so the transformer analogy does not apply.

Gary Coffman wrote:
On Tue, 23 Dec 2003 02:19:00 -0600, Richard J Kinch
wrote:
An air regulator is not at all analogous to an electric transformer.
The proper analogy is to a three-terminal voltage regulator. Energy
is lost; that is how the regulator works. The CFM (or SCFM) on
either side of the air regulator is necessarily equal. The pressure
drops. Power is lost and turned into heat.

No. A *shunt* regulator would behave the way you describe, but this
is a series regulator.

The regulator is a feedback controlled valve. It is not a dissipative
device. It controls the mass flow of air from the tank such that a
particular set pressure develops against the flow resistance existing
downstream of the valve. It does not bleed excess air to atmosphere
in order to do this, it does not get perceptibly hot. So any energy
not required to achieve the set downstream pressure *remains in the
tank*. It is not lost or turned to heat.

SCFM *is* a measure of mass flow. At equilibrium flows, the mass
flow into the valve *is* the same as the mass flow out of the valve.
If we consider this mass flow analogous to current, then we can
apply Kirchhoff's laws and show that the current is everywhere
the same in a series mesh.

We can also use Kirchhoff's laws to show that voltage (pressure)
drops across the regulator valve and across the downstream flow
resistances sum to tank pressure. In other words, the sum of the
voltages around a series mesh equal zero, with the tank pressure
(analogous to a battery) treated as positive and the pressure drops
across the various flow resistances treated as negative.

But the key thing to understand here is that a resistance need not
be *dissipative*. A good electrical example is a triode tube, or
*valve* as the British called them. The control voltage on the grid
changes the current flow through the tube by modulating what we call
the plate resistance. But this isn't an actual dissipative resistance.
It is a *mathematical fiction* we use to model the plate current
valving action of the grid.

Similarly, the diaphram of the air regulator merely modulates the
flow through the valve by opening or closing the valve.
Mathematically, this has the same appearance as a dissipative
resistance, but there is *no dissipation*. Energy not used is simply
retained in the tank.

Gary

The regulator may not be "losing" air, (except in the shutdown case I
mentioned in another post) but Richard is correct that the capacity to do
practical mechanical work is lost as the air expands. If this was not the
case you could get free air conditioning and still use the air to do the
same work, which would be a free lunch.

jim rozen wrote:
In article , Grant Erwin says...

Boy, I wish I had 2 flowmeters.

You don't need them! You understand
the physics behind it instead, which
is better.

Jim

You two may understand the mass flow, but if you think it about it a little
more, you will see that Richard Kinch is correct, the transformer analogy is
not applicable. This is a thermodynamic problem, and an interesting
illustration of how inefficient it is to compress air to a high pressure if
it is only going to be used at a much lower pressure.

In article , ATP says...

the transformer analogy is
not applicable.

Analogies are useful for certain things,
but I think you are correct, I would not
use 'transformer' as an analog for a
gas regulator.

Jim

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

In article , ATP says...

... we can not use a "reverse" regulator to go to less CFM at a higher
pressure, ...

Well if you figure out a way to do this, please
let me know about it right away! I want to
invest in your company then.



Jim

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

In article , Grant Erwin says...

Yes, I started that thread too. The reason that this has come back is that
I was never convinced the last time. Let me ask YOU, Ned, to answer my
question? I'll repeat it:

Postulate a big air tank pressurized to 180 psi, with a long (long
enough so the air has time to cool to ambient) pipe to an ideal
regulator which regulates the pressure down to 90 psi. The regulator's
output is a pipe of the same size which is connected to a constant
load of such a size as to make the cfm going into the regulator
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator
at 90 psi?

Grant Erwin

Ned Simmons wrote:
No, as Richard said, a linear regulator is a better, but
not very good, analogy. I think any analogy between
electricity and a compressible gas is doomed. You really
need to resort to thermodynamics to do this subject
justice.

But air regulators *are* lossy devices, and not just
incidental losses as in a transformer. Last time this came
around (and it looks like you started it that time too,
Grant g) I tried to avoid thermo (which I don't feel
qualified to preach after a 30 years lapse) and explain the
loss by resorting to an example of the potential of a given
mass of air do do work on a piston, starting from different
pressures.

This is the last post in the thread:

http://groups.google.com/groups?
q=g:thl672785172d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8
&selm=MPG.189e9353cdce4f36989893%40news.rcn.com

Ned Simmons

In article ,
says...

OK. CFM does not equal mass flow. CFM means simply cubic feet of air per
minute. The amount of mass in that air is proportional to both the pressure
and the temperature, as described by the gas law PV = nRT.

I agree there are second-order losses in a regulator. It might make some
noise, and it might warm up a little or cool down a little. This is also
true of a transformer.

You know, what's *really* amazing is that I know Gary Coffman and Richard
Kinch and I are all intelligent, good writers, and well educated, and yet
we get THREE different answers.

I'd just love to get to the point of agreement.

So, Richard, I DISAGREE that cfm is equal on either side of the regulator.
It cannot be because mass is conserved! (See my previous posts)

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:


Gary Coffman writes:



Mathematically,
this has the same appearance as a dissipative resistance, but there
is *no dissipation*. Energy not used is simply retained in the tank.


Impossible.

We agree that mass flow (CFM) is conserved (equal) on either side of the
regulator.

We know: Power = CFM * pressure.

The regulator imposes: pressure (out) pressure (in).

Thus, since CFM is equal on both sides of the regulator, but pressure
decreases, there is a loss of power in the output compared to the input.
This ends up as heat as a direct consequence of the restriction that
creates turbulence and lowers the pressure. This waste heat is mostly
added to the output flow, even though the output may be at a cooler
temperature due to expansion. Some of the power lost is hissing noise
that radiates away and also eventually becomes heat. None of this is
"visible" to the source, so it is not "simply retained in the tank".

Look, if you want an air analog to an electrical transformer, you would
have to have an air motor (not a diaphragm regulator) doing the pressure
reduction, with the motor output used to regeneratively compress more
free air. That way, power would be conserved, like a transformer,
subject to some mechanical inefficiency.




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

In article , Grant Erwin says...

Postulate a big air tank pressurized to 180 psi, with a long (long
enough so the air has time to cool to ambient) pipe to an ideal
regulator which regulates the pressure down to 90 psi. The regulator's
output is a pipe of the same size which is connected to a constant
load of such a size as to make the cfm going into the regulator
measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator
at 90 psi?

Because you posed the question as "CFM" output, not
SCFM (and of course, if you *were* specifying SCFM
then you could not say "90 psi output!") then the
answer has to be 20 psi as long as the pressures
are in absolute, not gage. Close for gage, but
not exact.

Jim

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

Grant Erwin writes:

CFM does not equal mass flow

No, this is wrong again. CFM *does* equal mass flow.

I think you don't understand what "CFM" and "SCFM" mean.

One CFM means a cubic foot of air at 1 atm pressure ("free" air), flowing
per minute. Or the equivalent mass of air at any other pressure. It does
NOT mean one cubic foot of air at any other pressure.

The "S" prefixed simply specifies the input free air is understood to be at
68 deg F and 36 percent relative humidity, to simplify the variations of
system performance (usually, but not always, slight) due to those
variables.

Gary Hallenbeck writes:

A cubic foot VOLUME is a cubic foot VOLUME regardless of
pressure, temperature or phase of the moon. A STANDARD cubic foot, on
the other hand is a specific mass of the gas in question. Which is
why a 600 cfm compressor can have a 1/3 hp motor and a 10 cfm
compressor can require a 500 hp motor. The 600 cfm compressor being a
cooling fan producing flow @ an inch or two of water and the 10 cfm
compressor producing flow @ 1000 psig. Obviously the 10 cfm @1000 psig
results in a much larger SCFM than the 600cfm @ inches of H20.

Again, you are misunderstanding.

"SCFM" and "CFM" are the SAME THING, except that the "S" prefix indicates
the input air for the system is specified to be 68 deg F and 36 percent
relative humidity, while "CFM" without the "S" prefix just *does not*
specify what the free air temperature and humidity are. Thus performance
of a system in "CFM" could be "better", "about the same", or "worse" than
in SCFM, depending on the temp and humidity of the ambient environment.

Roger Head writes:

Hmmm, well.... CFM is volume/min, and per-se is not related to mass
flow.

No, this is incorrect. See my other posts in this thread.

SCFM, on the other hand, indirectly specifies a mass flow because
the air is at STP.

Also incorrect. Units of SCFM say NOTHING about the pressure of the air
(versus 1 atm in STP). The temperature of the INPUT AIR is 68 deg F
(versus 25 deg C in STP) and the humidity of the INPUT AIR is 36 percent
relative humidity (STP does not specify humidity).

An SCFM figure tells you NOTHING about the pressure or the temperature or
the humidity of the COMPRESSED air. It merely says how much INPUT AIR is
present in whatever the system delivers in compressed form. That is why an
SCFM figure without an associated pressure is meaningless. And why, to be
complete, you need to factor in a cooler and drier as well, to improve the
temperature and humidity of the compressed air (otherwise it is hot and
wet, not good for tool performance or longevity).

jim rozen writes:

Cubic feet per minute is only a time
rate of volume.

No, no, no, no, no. CFM in compressors refers to the flow rate OF THE
INPUT AIR AT ATMOSPHERIC PRESSURE.

Depending on the pressure inside the volume,
you can have all different amounts of material.

I understand what you mean by that, but that is not what the term "CFM"
means. "CFM" does NOT mean cubic feet per minute of COMPRESSED AIR. It
means the cubic feet per minute of the INPUT AIR it took to make the
COMPRESSED AIR. Thus one CFM is always the same mass flow at any delivery
pressure.