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

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

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

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

Grant Erwin

Gary Coffman wrote:

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

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


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

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

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

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

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

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

Gary