Discussion:
SCFM vs. CFM, also air flow/pressure across a regulator
(too old to reply)
Grant Erwin
2003-12-22 15:07:55 UTC
Permalink
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
2003-12-22 15:56:34 UTC
Permalink
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.
Post by Grant Erwin
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 = 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!
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
Grant Erwin
2003-12-22 16:23:39 UTC
Permalink
Post by Roy J
SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi
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
Kathy and Erich Coiner
2003-12-22 16:48:17 UTC
Permalink
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
Post by Grant Erwin
Post by Roy J
SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi
SCFM on either side of a regulator is the same.
That's interesting. How did you come to that conclusion? Applying that to
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
Roy J
2003-12-22 21:23:23 UTC
Permalink
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.
Post by Grant Erwin
Post by Roy J
SCFM is STANDARD cfm ie one cubic foot of air at 14.7 psi
SCFM on either side of a regulator is the same.
That's interesting. How did you come to that conclusion? Applying that to
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
Richard J Kinch
2003-12-23 08:19:00 UTC
Permalink
Post by Grant Erwin
That's interesting. How did you come to that conclusion? Applying that
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
Gary Coffman
2003-12-23 17:17:22 UTC
Permalink
Post by Richard J Kinch
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
Grant Erwin
2003-12-23 17:41:46 UTC
Permalink
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
Post by Gary Coffman
Post by Richard J Kinch
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
Gary Coffman
2003-12-23 18:35:27 UTC
Permalink
Post by Grant Erwin
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
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
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
Grant Erwin
2003-12-23 20:25:30 UTC
Permalink
Post by Gary Coffman
Post by Grant Erwin
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
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
jim rozen
2003-12-23 20:30:36 UTC
Permalink
Post by Grant Erwin
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
2003-12-23 20:06:30 UTC
Permalink
Post by Gary Coffman
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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Roger Head
2003-12-24 02:22:34 UTC
Permalink
"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
Post by Gary Coffman
Post by Grant Erwin
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
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
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
jim rozen
2003-12-23 18:40:59 UTC
Permalink
Post by Grant Erwin
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
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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jim rozen
2003-12-23 18:11:55 UTC
Permalink
Post by Gary Coffman
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.
Post by Gary Coffman
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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==================================================
Tim Williams
2003-12-23 21:01:54 UTC
Permalink
Post by jim rozen
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
Richard J Kinch
2003-12-23 22:34:25 UTC
Permalink
Post by Gary Coffman
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.
Grant Erwin
2003-12-23 22:57:26 UTC
Permalink
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
Post by Richard J Kinch
Post by Gary Coffman
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.
Ned Simmons
2003-12-24 02:25:50 UTC
Permalink
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
Post by Grant Erwin
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
Post by Richard J Kinch
Post by Gary Coffman
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.
Richard J Kinch
2003-12-23 08:07:58 UTC
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Post by Grant Erwin
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
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