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.