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.