SCFM vs. CFM, also air flow/pressure across a regulator
Grant Erwin
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
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
> 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
Kathy and Erich Coiner
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" <grant...@kirkland.net> wrote in message
news:vue6fh9...@corp.supernews.com...
Roy J
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.
Richard J Kinch
> 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.
Richard J Kinch
> 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:
Gary Coffman
>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
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
jim rozen
>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
==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================
Gary Coffman
>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
jim rozen
>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.
Grant Erwin
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
jim rozen
>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 rozen
>Boy, I wish I had 2 flowmeters.
You don't need them! You understand
the physics behind it instead, which
is better.
Tim Williams
news:bsa0h...@drn.newsguy.com...
> >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
> 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
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
Roger Head
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
Ned Simmons
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 <vuhhvjg...@corp.supernews.com>,
grant...@kirkland.net says...
jim rozen
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 rozen
>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.
jim rozen
>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
Grant Erwin
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
Jeff Wisnia
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."
Ned Simmons
grant...@kirkland.net 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
Gary Hallenbeck
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
ATP
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.
Tim Williams
news:Xns945AB31CFB3...@216.196.97.131...
> 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.
ATP
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.
ATP
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.
ATP
> In article <vuh92pc...@corp.supernews.com>, 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
>
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.
jim rozen
> 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 rozen
>... 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 rozen
>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 rozen
==================================================
Richard J Kinch
> 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.
Richard J Kinch
> 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.
Richard J Kinch
> 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).
Richard J Kinch
> 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.
Richard J Kinch
> http://www.cleandryair.com/scfm_vs__icfm_vs__acfm.htm
This page correctly explains that CFM, SCFM, ACFM, etc., ALL refer to "the
volume of air that is compressed each minute and it is measured on the
_inlet_ side of the compressor." The "S" or "A" prefixes simply further
specify the temp and humidity of this inlet air.
Richard J Kinch
> CFM X psi for a compressible gas is not analogous to volts
> X amps.
Actually, it is analogous:
CFM:DC amps
pressure:DC volts
power = CFM * pressure : volts * amps.
Grant Erwin
> 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.
I'm obviously not the only one who's "wrong", Richard. Actually, I'm relieved
because now that I understand what you mean by CFM it makes everything you
say make sense. I took it to just be a three letter acronym for "cubic feet
per minute" which could be at any pressure. I did this because my whole adult
life I've seen compressors specified this way. If you try a quick google for
compressor cfm psi
you will instantly see what I mean - everyone specifies their compressors
at a cfm @ a psi.
So it's all a matter of a simple misunderstanding. In the circles you travel
in, CFM has some specialized meaning. To the rest of the world, it simply
means cubic feet per minute.
So what do you call simple cubic feet per minute as a volumetric gas flow?
Grant Erwin
Grant Erwin
application engineer for I-R for a lot of years. As far as I am concerned
he speaks with the voice of infinite knowledge about the air compressor
industry.
Looks like I-R (like Quincy and the other major manufacturers) don't know
about your special meaning for CFM either.
Grant
Grant Erwin
cleandryair.com and why should I believe them over what is obvious, what
is common practice in the air compressor industry, and what is taught in
every engineering school?
I suspect this page is trying to redefine some terminology to give themselves
some kind of business advantage. I have no problem with this, but don't expect
me to believe it.
Grant Erwin
jim rozen
says...
> CFM in compressors refers to the flow rate OF THE
>INPUT AIR AT ATMOSPHERIC PRESSURE.
In your world it does. In general though,
cfm means what the dimensions imply: a
cubic foot of any material that passes by
a point in a minute. It could be water,
air, compressed gas, and it can be at any
pressure the user specifies.
The folks who inhabit this ng have a broad
knowledge base and will use *specific* terms
when suited. Otherwise a generic term like
'cfm' will be seen as an open-ended term.
Your insistance that cfm means delivery
of a specific gas at a specific pressure
makes about as much sense as saying that
a 'cupfull' ALWAYS means one cupfull of
flour in a baking recipe. Nobody is going
to agree with either of those approaches.
Besides, the original question that was posed
was about GAS REGULATORS. Not COMPRESSORS.
jim rozen
says...
>One CFM means a cubic foot of air at 1 atm pressure ("free" air), flowing
>per minute.
This may be true in some disciplines, but in general
this is NOT a true statement. Because the inhabitants
of this ng have such a diverse background they will
not want to agree with that *specific* definition.
Roger Head
Roger
Roger Head
Now my head is spinning, and I don't think it's the brandy sauce on the
Xmas pudding!
Initially I just nit-picked on a statement that CFM is a measure of mass
flow, so I posted a site ref to demonstrate the difference in the
implications of SCFM and just CFM.
Richard came back with his knickers in a twist, then again a few minutes
later acknowledging the validity of the ref site. I assumed that he had
only just got around to reading it, and now understood things.
In another post that you made a few minutes before this one that I am
replying to, you also appeared to agree with me that Richard had the
wrong end of the stick. But now you are upset about the ref site, and I
don't understand why - I'm not arguing with you, just puzzled.
What they're saying is that the measure of a compressor system is the
number of SCFM (mass of air) that it can deliver at some specified
outlet pressure (with the assumption that the outlet temperature is at
the agreed standard value, etc etc), because that is a measure of the
work that can be done by that air. Now unfortunately most compressors
aren't operating in a STP enviroment, so what they are saying is that if
you want a certain amount of work to be performed by your compressed air
then you need to look at the inlet conditions (ambient temp, pressure,
humidity, effects of inlet filters and manifolds etc etc) to determine
what capacity compressor you will need to look for. Remember that the
capacity written on the data sheet (probably as CFM @ xxx psi)
implicitly assumes that the ambient conditions are at STP.
So if we want a compressed air system to do a certain amount of work,
and we will be using it in STP ambient conditions, then (ignoring sales
hype, and a bunch of other inefficiencies, etc) we should be able to
select a suitable unit by looking at the straight data-sheet specs. But
if we want to do that same amount of work on top of a mountain, then we
will have to select a unit that has a greater specified capacity. And so
on, if other conditions vary from STP.
Please tell me what it is that you don't like about cleandryair's page.
Regards,
Roger
Ned Simmons
nob...@nowhere.com says...
Ya think? If that's the case there should be a general
conversion factor such that cfm X psi X constant = watts.
What would that factor be?
Ned Simmons
jim rozen
says...
>conversion factor such that cfm X psi X constant = watts.
>What would that factor be?
Hmm. This might actually work out, the energy
stored in a pressure vessel is (IIRC) 1/2 P(sq) x V
so if you imagine an amount of air at a given
pressure passing a point in a pipe, then
that would be a time rate change of energy.
I think I would have to get ahold of my copy
of Haliday and Resnick at work, with the units
conversion tables at the back, to give the
constant though.
Grant Erwin
volume of an air compressor. Thus to them 16 cfm @ 90 psi would have no
meaning. As this is in direct conflict with the way many if not all industrial
air compressors are specified (if you don't believe me just google on:
compressor cfm psi
and see all the bazillions of specs in this format, it only takes a second)
I find their attempt to limit this terminology to be unhelpful. That page
does little to shed light and in my opinion is designed to steer people to
their way of thinking and thus their product line whatever that is.
There *are* valid thermodynamic concerns here, but they aren't relevant to
my original question. It is true that it is inefficient to compress air to
a high pressure (with lots of heat) and then remove the heat and moisture
to get clean dry air. Obviously you are spending electrical $$$ to dump heat
into some heat exchanger. This is a fact. What I would find a more helpful
fact is the information how to get clean dry air without doing the above.
I have always understood that the way to go is to get a 2-stage compressor
which can run a high volume of air at a high pressure (like 180 psi, gaged,
or 180 psig) and then pipe that air to a refrigerated air dryer and then
filter the moisture and oil drops out of the air. I think Richard Kinch's
point is that you will get a lot more energy out of that same compressor
if you use it hot and wet. Very true, but you can't spray-paint a car with
hot wet air, nor can you run it into a plasma cutter, nor is all that
moisture good for your air tools.
Way back when I first posted the question, I was wondering about using a
"regular" air compressor to generate air for HVLP painting. As I knew for
sure you couldn't get the kind of volume out of my compressor that a HVLP
gun needs, I wondered if you could "transform" it to a much lower pressure
but much higher volume. I now believe enough light has been shed on that issue
to say yes you can get higher volume by regulating to a lower pressure as
long as the temperature doesn't drop.
I try not to get embroiled in argumentative discourse on this NG but I seem
to have "fallen from grace" this time. I can only say that I believe all
parties have been respectful, and anyone who has taken the time to read all
of this material and to consider each poster's comments deliberately, will have
learned a considerable amount.
Grant Erwin
Kirkland, Washington
Grant Erwin
equals 1000 watts based on what you can get out of a 1.5 hp compressor.
I have also sadly lost my copy of Halliday and Resnick, so I am not inclined
to do all the physics calculations to get a better answer. Obviously, the
temperature and barometric pressure will play a part too.
Grant Erwin
Mark Rand
wrote:
>"jim rozen" <jim_m...@newsguy.com> wrote in message
>news:bsa0h...@drn.newsguy.com...
>> >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
It can be a tricky one to grasp at first. This is an adiabatic expansion. No
external work is being done and no heat is being added to, or taken from, the
gas. The reason it cools down is because it is doing work on _itself_ during
the expansion.
Mark Rand
RTFM
jim rozen
>
>You're obviously right, Ned. I believe it's something like 4 cfm @ 90 psi
>equals 1000 watts based on what you can get out of a 1.5 hp compressor.
That's going to be close, as 1 hp is 750 watts.
jim rozen
>
>The cleandryair page claims that the ONLY meaning of CFM is as the input
>volume of an air compressor. Thus to them 16 cfm @ 90 psi would have no
>meaning. As this is in direct conflict with the way many if not all industrial
>air compressors are specified
Ha ha, not only that, but there are about a thousand
other uses for the term 'cubic feet per minute' that
have nothing whatsoever to do, with air compressors!
jim rozen
>It can be a tricky one to grasp at first. This is an adiabatic expansion. No
>external work is being done and no heat is being added to, or taken from, the
>gas. The reason it cools down is because it is doing work on _itself_ during
>the expansion.
But in a real world regulator, it will not be quite a true
adiabatic expansion. Some heat will come in from the outside,
so it will want to be isothermal too.
ATP
> In article <vui26m8...@corp.supernews.com>,
> grant...@kirkland.net 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?
>>
>
> 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.
>
Whenever you let a confined gas expand there has to be a loss in its
potential to do work. Consider a pneumatic cylinder, it has more potential
energy in a more compressed or closed state. Is there any difference between
doubling the enclosed volume of a cylinder and expanding compressed air
through a regulator? That cylinder could do a certain amount of work
expanding, f x d. You would end up with double the amount of air at half the
psi, yet work was accomplished expanding the cylinder so there must be a
loss of potential energy in the compressed air. That's my "hand waving"
argument.
Mark Rand
wrote:
>You're obviously right, Ned. I believe it's something like 4 cfm @ 90 psi
>equals 1000 watts based on what you can get out of a 1.5 hp compressor.
>I have also sadly lost my copy of Halliday and Resnick, so I am not inclined
>to do all the physics calculations to get a better answer. Obviously, the
>temperature and barometric pressure will play a part too.
>
>Grant Erwin
Nearer 250 Watts of useful power. A _lot_ of the power going into the
compressor ends up heating the air. Think of what can be done with a 500 Watt
electric angle grinder and then think of what a fearsome beast a 1000 watt die
grinder would be!
Mark Rand
RTFM
ATP
The best bet with an HVLP gun is to use a turbine or regenerative blower
like the ones Spencer makes (vortex blowers) The moisture is generally not a
problem under reasonable humidity conditions. I have an Accuspray gun and
run it both ways. Using a compressor it will keep a 5 HP 3 phase compressor
running more than 50% of the time and you end up with a lot of moisture to
handle.
>
> Way back when I first posted the question, I was wondering about
> using a "regular" air compressor to generate air for HVLP painting.
> As I knew for sure you couldn't get the kind of volume out of my
> compressor that a HVLP gun needs, I wondered if you could "transform"
> it to a much lower pressure but much higher volume. I now believe
> enough light has been shed on that issue to say yes you can get
> higher volume by regulating to a lower pressure as long as the
> temperature doesn't drop.
>
>
I think Ned Simmons is on the right track in this thread. Yes, you can get a
higher volume of air at lower pressure but it is not efficient to do so with
a regulator. Compressing air just to decompress it without recovering any
energy is extremely wasteful. Applications using compressed air that need to
move large volumes of air use the venturi principle, not regulators. Do you
see how the transformer analogy does not apply?
Roger Head
I really didn't want to get involved in this whole discussion, so I
should just bite my tongue and leave it here, but... just a couple of
little things:
In those bazillions of specs the CFM is generally the displacement of
the pump, which we all know is a far cry from the amount of air that it
actually delivers. Free air delivery (FAD) is a spec that comes closer
to what you actually get out of it, but it very often isn't quoted
because it doesn't look nearly as good on the brochure.
I don't know what cleandryair are selling either, but I have to disagree
with you - I think that the page is useful in alerting a potential buyer
to the need to consider more than just the CFM@psi on the brochure when
they are deciding on a unit.
Roger
Ned Simmons
jim_m...@newsguy.com says...
> In article <MPG.1a54d659c...@news.suscom-maine.net>, Ned Simmons
> says...
>
> >conversion factor such that cfm X psi X constant = watts.
> >What would that factor be?
>
> Hmm. This might actually work out, the energy
> stored in a pressure vessel is (IIRC) 1/2 P(sq) x V
> so if you imagine an amount of air at a given
> pressure passing a point in a pipe, then
> that would be a time rate change of energy.
>
If your formula is right (it sounds plausible to me), take
a tank where V=1 and P=1.
(P^2 X V)/2 = 1/2
Expand the gas so P=1/2 and V=2 and plug back into the
formula and you get 1/4. In other words, no constant
factor, no analogy, and expansion thru a regulator is
irreversible and wasteful of energy.
Ned Simmons
Ned Simmons
@news4.srv.hcvlny.cv.net>, walter...@unforgiven.com
says...
> >
> > CFM X psi for a compressible gas is not analogous to volts
> > X amps.
> >
> > Ned Simmons
>
> Whenever you let a confined gas expand there has to be a loss in its
> potential to do work. Consider a pneumatic cylinder, it has more potential
> energy in a more compressed or closed state. Is there any difference between
> doubling the enclosed volume of a cylinder and expanding compressed air
> through a regulator? That cylinder could do a certain amount of work
> expanding, f x d. You would end up with double the amount of air at half the
> psi, yet work was accomplished expanding the cylinder so there must be a
> loss of potential energy in the compressed air. That's my "hand waving"
> argument.
>
Works for me. That's pretty much the same argument I made
in the old post I pointed to with the Google link. Here it
is again.
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
Gary Coffman
>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.
Power is not a conservative quantity. Energy is, and the energy not
released is retained in the tank. In other words, power = energy/time.
The regulator effectively changes (lengthens) the time over which a
given amount of energy is released from the tank. That does show
up as reduced output power. But power isn't what's conserved,
energy is, and it is retained in the tank until it is released.
>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".
Those parasitic losses are relatively negligible in a well designed system.
Gary
Richard J Kinch
> If you try a quick google for
> compressor cfm psi
> you will instantly see what I mean - everyone specifies their
> compressors at a cfm @ a psi.
Right. That's exactly what I have been saying. CFM must always have an
associated pressure to indicate performance. The mass of air is
nevertheless always referenced to the density and pressure of free air
(1 atm), not the stated psi.
For example, my compressor will deliver 6 CFM at 90 psi. That means it
will inhale 6 cubic feet of free air, and push it out at 90 psi.
> So it's all a matter of a simple misunderstanding. In the circles you
> travel in, CFM has some specialized meaning. To the rest of the world,
> it simply means cubic feet per minute.
No, there is no misunderstanding. CFM with respect to air compressors
only means what I have been describing. It has nothing to do with the
"circles" I travel in.
I would recommend _Machinery's Handbook_ on this subject if you are
still unconvinced.
> So what do you call simple cubic feet per minute as a volumetric gas
> flow?
Cubic feet per minute.
Richard J Kinch
> Looks like I-R (like Quincy and the other major manufacturers) don't know
> about your special meaning for CFM either.
The statement I was quoting, "Obviously the 10 cfm @1000 psig results in a
much larger SCFM than the 600cfm @ inches of H20." is in error, whoever
said it. The first item respresents 10 cfm of free air, the second 600 cfm
of free air, and the latter is much larger. The meaning of "SCFM" vs
"CFM" has nothing to do with the pressure differences being asserted.
Gary Coffman
>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
That's right. When you see a compressor rated at say 10 CFM @ 90 PSI,
that means 10 CFM of *inlet* air compressed to 90 PSI. It does NOT mean
10 cubic feet of 90 PSI air. This should make sense when you notice that
a compressor may have several fairly similar CFM ratings given at several
different pressures. That's because the only differences are due to
differences in the pumping *efficiency* of the compressor at different
delivery pressures. It is still inhaling roughly the same amount of air per
stroke at the same strokes per minute, so it still has roughly the same
CFM rating whether that's given at a delivery pressure of 45 PSI, 90 PSI,
or 140 PSI. It just takes longer to initially pump up the tank to the higher
delivery pressures.
Gary
Richard J Kinch
> In your world it does. In general though,
> cfm means what the dimensions imply: a
> cubic foot of any material that passes by
> a point in a minute
CFM *does* refer to a plain old cubic foot per minute, but at the INLET,
not the OUTPUT. There's no other "world" involved, or lack of generality.
Gary Hallenbeck
I DO understand what CFM and SCFM mean. I worked with compressor
specification and air system analysis ,around the worl, for 25 years.
Please explain upon what authority you base your contention about CFM.
For one thing your definition of "Standard Conditions", which is what
the S stands for, does not match anything I have ever seen anywhere in
the world and believe me I have seen about every definition of SCFM
there is. A CFM is just what it says it is. One Cubic Foot per
Minute at any temperature, pressure or humidity. It can be used to
define VOLUME flow for any gas, but does not necessarily have any
relation to MASS flow (except in an accidental context). Common
compressor terms include ACFM, ICFM, SCFM, etc,etc.. ACFM is "actual
cfm which refers to whatever the compressor produces at discharge
conditions(say 125 psi , 300 deg F and 100% rh for instance). ICFM
is "inlet cfm" which is flow at whatever conditions happen to be at
inlet (say 12.7 psi , 20 deg F and 14% rh) , SCFM is the MASS flow at
standard conditions which, in the USA are generally 14.696 psig, 60
deg F and 0% relative humidity. In Europe it is standardized as 1 bar
pressure 0 deg C and 0% relative humidity or alternately 1 bar,25 deg
C and 0% rh, the primary requirement being to use the same convention
at all times. Incidentally in the USA, 379 STANDARD cubic feet of
air weigh 28.79 lbm and contain one mole of air. SCFM seldom has any
relationship to conditions at the inlet of the compressor since 0% rh
is virtually unobtainable in nature. Inlet pressure, temperature and
rh have very large effects on system performance and can result in
changes of several hundred horsepower on medium to large compressors.
Gary Hallenbeck
Senior Technical Specialist (Ret.)
Western Region
Ingersoll-Rand Air Compressors
On Wed, 24 Dec 2003 19:32:33 -0600, Richard J Kinch
<nob...@nowhere.com> wrote:
>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.
Richard J Kinch
> I guess you just read that site, eh Richard?
I read it, yes, at least the parts about CFMs. Why?
Richard J Kinch
> Power is not a conservative quantity. Energy is, and the energy not
> released is retained in the tank. In other words, power = energy/time.
Your statement is contradictory, although I think I understand what you
mean. Power can be converted to heat and thus not "conserved". I
suppose what I should say is that the power going into the regulator
either comes out the other side, or is lost as heat.
>>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".
>
> Those parasitic losses are relatively negligible in a well designed
> system.
I don't know what you mean by "parasitic" here. Regulators are
inherently inefficient devices, and part of the input power is always
wasted (ultimately) as waste heat by them.
Think of the case where you have a "perfect" regulator lowering the
input pressure to ambient pressure (0 psig). For example, imagine a
pressurized tank that you instantly crack open in half. Or for a better
imaginary example, imagine that you have a compressed tank, and you can
magically make the tank disappear instantly, leaving the mass of
compressed air suddenly unconfined. Where did the power go? You get a
noisy bang, and a lot of turbulence, which ultimately degrades into heat
equal to all the energy previously stored in that compressed air.
Similar puzzle: shorting a perfect capacitor with a perfect conductor.
Where does the energy go?
Gary Hallenbeck
<nob...@nowhere.com> wrote:
>Roger Head writes:
>
>> http://www.cleandryair.com/scfm_vs__icfm_vs__acfm.htm
>
>This page correctly explains that CFM, SCFM, ACFM, etc., ALL refer to "the
>volume of air that is compressed each minute and it is measured on the
>_inlet_ side of the compressor." The "S" or "A" prefixes simply further
>specify the temp and humidity of this inlet air.
Contrary to what this basically informative, but somewhat flawed
website tells you, Flow is not necessarally measured at the inlet of
the compressor(in fact infrequently so except in the case of
centrifugal compressors which are usually rated in ICFM.) Every
specification I have ever seen requires quotation in SCFM. And from
your own website reference,which you have misquoted comes"By itself,
CFM tells you nothing about the weight of the air that is inside the
cube or the work that is available from the air. The first letter, "S"
or "I" or "A", ties the volume to a set of conditions that will define
the air in the cube and in doing so, define the available work."
Notice that the S ties the VOLUME to a specific set of conditions.
They then proceed to define S correctly and I and A in somewhat
misleading terms.
Gary H.
Richard J Kinch
> In those bazillions of specs the CFM is generally the displacement of
> the pump, which we all know is a far cry from the amount of air that it
> actually delivers.
That should always be called "displacement CFM", which is indeed a bogus
spec.
If a commercial product has a sticker claiming CFM at some psi, it had
better deliver that at the output fitting, unless otherwise set forth as
the theoretical displacement or some other nonsense. That the cheap
imports have misrepresented such standards of performance, doesn't nullify
the clear meaning of the specification.
Gary Hallenbeck
If you read my reply to Richard, I think you may see some of the
discomfort that Grant is expressing. The site is essentially correct,
but contains some definite technical mistakes. I spent 25 years
working for Ingersoll-Rand in the air compressor business so I have a
certain expertise in this area.
1. Compressor capacity is not determined on the inlet side,
except for Centrifugal units which are usually constant mass flow
machines and dependent on inlet conditions to a very large degree.
Most systems are rated in SCFM required. In specifying a centrifugal
machine it will be rated in delivered SCFM for specific inlet
conditions.
2. ACFM is a term I have only heard used in rating very small,
cheap compressors and is usually similar to "Free Air CFM" or "CFM
displacement" These terms are used to make the compressor look
comparable to a more expensive unit which is rated in CFM at a
specific pressure.
3 SCFM is the only way of comparing or rating compressor
delivery that gives a standard reference. In the compressor game
salesmen will use all sorts of numbers to rate efficiencies and flows
to make their machine look better on paper. SCFM or lbm or other mass
flow per horsepower eliminates the tom foolery and gives an apples to
apples comparison.
These three items lead me to believe that the website is either well
intentioned and poorly informed or that there is a particular economic
slant to the information presented.
Gary Hallenbeck
On Thu, 25 Dec 2003 14:10:44 GMT, Roger Head <rlh...@bigpond.net.au>
wrote:
Gary Hallenbeck
Gary
Sorry, but you are mistaken. Most compressors are rated in delivered
flow. Centrifugal units are frequently rated in inlet cfm. Somewhere
in this thread I gave the example of a compressor rated at 10 cfm and
a very high pressure. Specifially I will use a high pressure unit
that existed at Mare Island Naval Shipyard in Vallejo CA, It produced
15 CFM at 4500 psi, but it's inlet capacity was about 500 cfm at 14.7
psi. Think of it this way. If you have an air tool that requires 4.5
cfm at 90 psi, what size compressor do you need to run it? Remember
it needs 4.5 cfm at 90 psi. Convert 4.5 cfm at 90 psi and 100% rh to
cfm at inlet conditions and I think you will find you have a much
larger number.
Gary H
On Fri, 26 Dec 2003 01:41:01 -0500, Gary Coffman <ke...@bellsouth.net>
wrote:
jim rozen
says...
>
>Gary Coffman writes:
>
>> Power is not a conservative quantity. Energy is, and the energy not
>> released is retained in the tank. In other words, power = energy/time.
>
>Your statement is contradictory,
Actually not at all. It is exactly, and technically
completely correct. The conservations laws do indeed
talk about *energy* and not the time rate of energy
delivery.
> although I think I understand what you
>mean. Power can be converted to heat and thus not "conserved". I
>suppose what I should say is that the power going into the regulator
>either comes out the other side, or is lost as heat.
I suggest you read up an introductory thermodynamics book.
First off, gary's comments were that power (time rate
of energy delivery) is *not* conserved. What you mean to
say above is that "*energy* can be converted into heat..."
which is of course true, but in doing so it IS conserved
as heat is a form of energy. If you do all the bookkeeping
all the numbers add up. Mechanical work, heat, kinetic
energy, stored potential energy, all sum to a constant.
>Similar puzzle: shorting a perfect capacitor with a perfect conductor.
>Where does the energy go?
The only place it can go: radiation.
Don Foreman
<grant...@kirkland.net> wrote:
>I have been thinking about the issue of model equivalency between an air
>regulator and a transformer again.
I'm tuning in late here, sorry. I skimmed thru the postings and see
some confustion: the terms CFM and SCFM have been used somewhat
interchangably which may be causing some of the confusion.
SCFM, as was asserted a couple of times, is a measure of mass flow ;
whatever the actual flow rate is, it's normalized to what that amount
of air (at standtard temp) would be at atmospheric pressure.
CFM is actual flow, more analagous to current in an electric circuit.
For temperature the same, V1 CFM at P1 pressure would expand to 2V1
CFM at 1/2 P1 -- Boyle's gas law. In that sense ( constant
temperature or isothermal expansion), a regulator acts like a
transformer.
However, an accurate model would have to take thermodynamics into
account: air cools as it expands so if there is flow then there is
also temperature difference from P1 to P2 unless additional energy is
added downstream to warm the air back up. The equations describing
this for air can be found in Machinery's Handbook.
Because air on the low side is cooler than air on the high side, the
low-side air will have correspondingly less volume than it would have
had if there were no temperature change.
Air flow meters: it's not hard to kludge a fairly decent airflow
meter using hot wire aenomometry. There are probably many website
about it. One can use bead thermistors, or even small incandescant
bulbs with holes punched in the evelopes, though the latter tends to
be a bit fragile. Calibrate by holding it out a car window on a
stick long enough to get away from the car body and driving at various
speeds on a calm day -- or spinning it with a variable-speed motor in
the lab if you can conjure some sliprings.
Don Foreman
<grant...@kirkland.net> wrote:
>I have been thinking about the issue of model equivalency between an air
>regulator and a transformer again.
Hot wire aenomometers: check out
http://www.efunda.com/designstandards/sensors/hot_wires/hot_wires_theory.cfm
A simple filament driven by constant current (monitor voltage) would
be a start. I've used automotive taillight bulbs with the envelope
carefully broken away. Smaller bulbs are more sensitive, also more
fragile.
A bridge setup using two bulbs, one in the flow and one shielded from
flow, both driven by the same constant current, provides some
compensation for ambient temperature.
Both of these schemes assume constant resistance of the filament which
isn't so, but may be close enough for foolin' around -- particularly
with a calibrated device used at about the same ambient temp most of
the time. Ya don't run the filaments anywhere near red hot, just
warm. Delta -T may be small enough that delta R with delta-T may be
negligable compared to effects of air velocity.
For higher accuracy, consider small resistors (maybe surface mount)
with thermistor beads glued to them in close thermal contact. Use
fine wires or traces so as not to heat-sink them. Drive both
resistors with current controlled by opamps controlled by the
respective thermistors so both resistors are held at the same
temperature. The voltage difference on the resistors would then be a
function of air velocity -- perhaps about proportional to the square
root of velocity. This should make a pretty decent instrument,
though a bit slower than just a filament.
The mass flow meters in some fuel-injection systems work by hot-wire
aenomometry, as Bosch. (At least they used to)
Honeywell made a device called the "microbridge" which was essentially
a silicon hot-wire aenomometer bridge on a chip. Those things were so
fast they'd respond to sound waves. We used them with small
orifices to measure small pressures; if someone was talking loudly
near the duct I'd see audio on the pressure signal!
Don Foreman
<SPAMBLOC...@goldengate.net> wrote:
>Hot wire aenomometers: check out
>http://www.efunda.com/designstandards/sensors/hot_wires/hot_wires_theory.cfm
>
Some of this stuff is starting to come back to me.
A direct-heated aenomometer actually depends on the element having
some tempco of resistance w.r.t. temperature. Bead thermistors work
well but tungsten works OK also, with tempco of about 0.45%/degC.
As air movement removes heat, temp drops, resistance drops and
voltage drops (with current drive). This produces an error signal
which can be amplified to raise the current until it nearly makes up
the power lost to the wind with only a small temp change necessary to
produce the error signal necessary to re-establish equilibrium.
Note that such a device is a mass flow meter so it would measure SCFM
rather than CFM Given velocity (hence CFM) of air at higer
pressure would produce higher reading on the anomometer.
Don Foreman
<compre...@earthlink.net> wrote:
>
>Gary
>
>Sorry, but you are mistaken. Most compressors are rated in delivered
>flow.
Several web sites make it clear that SCFM connotes weight of air
delivered. 1 CFM is defined as a cubic foot of air if it were at
standard condx -- of which there are several definitions.
An efficient two-stage pump's delivery in what they call ACFM varies
only slightly from 100 PSI to 175 PSI delivery pressure. Further,
the rated capacity at rated pressure is not much less than piston
displacement. See
http://www.abacamerican.com/Bel%20Aire%20electric.htm
for examples.
Some say CFM, others say ACFM, it's a mess!
Richard J Kinch
> 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.
Richard J Kinch
> 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.
jim rozen
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
Gary Coffman
>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
Ned Simmons
ke...@bellsouth.net says...
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
Loren Coe
> 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
Don Foreman
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.
Richard J Kinch
> 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.
Richard J Kinch
> 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.
Richard J Kinch
>>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.
Gary Coffman
>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 here: http://air.ingersoll-rand.com/CMP/sap/scdlubvertlevel3.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
Richard J Kinch
> 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.
Don Foreman
<nob...@nowhere.com> 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.
Grant Erwin
> Yes it makes sense.
> Go here: http://air.ingersoll-rand.com/CMP/sap/scdlubvertlevel3.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
Ned Simmons
SPAMBLOC...@goldengate.net says...
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
Mark Rand
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
Loren Coe
> Gary Coffman wrote: >
>> Yes it makes sense.
>> Go here: http://air.ingersoll-rand.com/CMP/sap/scdlubvertlevel3.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
Don Foreman
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.
ATP
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.
ATP
> Gary Coffman wrote:
>
>> Yes it makes sense.
>> Go here: http://air.ingersoll-rand.com/CMP/sap/scdlubvertlevel3.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.