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  #21   Report Post  
Old December 24th 03, 02:22 AM
Roger Head
 
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Default SCFM vs. CFM, also air flow/pressure across a regulator

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

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

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

Roger




Gary Coffman wrote:

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

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

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

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

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


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

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

Gary



  #22   Report Post  
Old December 24th 03, 02:25 AM
Ned Simmons
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

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

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

This is the last post in the thread:

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

Ned Simmons

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

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

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

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

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

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:

Gary Coffman writes:


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



Impossible.

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

We know: Power = CFM * pressure.

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

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

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



  #23   Report Post  
Old December 24th 03, 02:59 AM
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Roger Head says...

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


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

Jim

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

  #24   Report Post  
Old December 24th 03, 03:34 AM
Grant Erwin
 
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Default SCFM vs. CFM, also air flow/pressure across a regulator

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

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

Grant Erwin

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

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

This is the last post in the thread:

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

Ned Simmons

In article ,
says...

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

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

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

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

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

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:


Gary Coffman writes:



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


Impossible.

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

We know: Power = CFM * pressure.

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

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

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




  #25   Report Post  
Old December 24th 03, 03:51 AM
Jeff Wisnia
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator



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."




  #27   Report Post  
Old December 24th 03, 04:35 AM
Gary Hallenbeck
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator



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

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

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

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

Grant Erwin

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

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

This is the last post in the thread:

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

Ned Simmons

In article ,
says...

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

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

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

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

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

I think it *is* like a transformer.

Grant Erwin

Richard J Kinch wrote:


Gary Coffman writes:



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


Impossible.

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

We know: Power = CFM * pressure.

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

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

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



  #28   Report Post  
Old December 24th 03, 04:46 AM
ATP
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

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


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

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

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

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

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

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

Gary


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


  #29   Report Post  
Old December 24th 03, 07:35 AM
Tim Williams
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

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


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

Tim

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


  #30   Report Post  
Old December 24th 03, 03:42 PM
ATP
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

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

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


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

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

Ned Simmons


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




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