On Wed, 31 Dec 2003 19:37:33 -0500, Ned Simmons
wrote:

In article ,
says...
On Mon, 29 Dec 2003 23:04:25 GMT, "ATP"
wrote:



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?

Good question! I think the answer is "no" in an isothermal process.

Power flow is pressure * volume flow, volume here being actual volume
at that pressure. To avoid confusion with what "CFM" means, let's
define volume flow as actual cu in / sec at given pressure. If I
have 1 cu in/ sec at 200 PSIA on the high side I have 2 cu in/sec at
100 PSIA on the low side (if temp is the same). Power out = power in
ergo no power loss in the regulator.

No, speaking loosely thermodynamically, the power is
proportional to the pressure *squared* times flow. Pressure
* flow is proportional to power for a non-compressible
fluid, for instance in a hydraulic system, but not for a
compressible gas.

Thermodynamics are not at issue when the checkvalve opens because
pressure is essentially constant. Power flow is pressure * volume
flow -- look at the units. ( lbf/in^2) * (in^3/sec) = in * lbf /
sec which is power.

Look at the example I posted again, except now imagine a
case where the initial pressure in the reservoir is 5
atmospheres absolute (twice the gage pressure of the
original case), and no regulator. The initial force on the
piston will be twice the unregulated 3 atm case, and the
total travel of the piston will also be doubled. The total
work done by the piston will be 4x.

The piston analogy was a bad idea, apologies for that. It obscures
the issue of transferred power vs store and release of potential
energy. Work is done in raising the weighted piston (assuming the
weight doesn't become zero in the process) that can be recovered when
the piston returns to it's initial position.


But you observed that it takes more power to squeeze a given volume
of atmospheric air to a higher pressure, so where does that excess
power go?

There's a checkvalve in the compressor. The pump squeezes the air
until the pressure equals that in the tank; there's no flow until that
happens. The power up to that point has gone into potential energy in
the form of cylinder pressure. When the checkvalve opens, flow
occurs at essentially constant pressure if the reservoir is large
compared to the cylinder, and energy is transfered to the reservoir.
When the cylinder pressure drops below reservoir pressure, the
checkvalve closes, flow stops, and the potential energy in the
cylinder pressure is returned to the flywheel until the next
compression stroke.

It seems you're neglecting the fact that the compressor
needs to draw in a fresh charge of air on the down stroke.
The dead volume above the piston is very small at TDC, so
there may be a small "kick" as the crank goes over the top
and the small volume of remaining air expands, but unless
the pressure in the cylinder falls below atmospheric (or
the pressure of the previous stage) pretty quickly, the
cylinder will not refill.

Said fact agreed to and not neglected. The fact remains that the
energy transferred to the reservoir is volume flow * pressure while
checkvalve is open, while energy expended compressing the air that
did not flow is recovered (less heat loss -- I addressed that) until
the pressure in the cylinder is back to atmospheric.

Marks Handbook has a discussion of single and multistage
air compression and the thermo behind the relative
efficiencies.

I'm familiar with it. That treatment assumes that heat of
compression is lost. If it is (as it usually is) then compressing
beyond end-use pressure is indeed inefficient. My point was and is
that the loss isn't in the regulator but in the heat loss at the
compressor. If that heat energy is preserved then there is no
significant penalty to overcompressing and post regulation. There are
also some significant economies in pipe sizes to minimize flow loss
(less at higher pressure and lower volume) and reservoir sizes
required if reservoirs are used.

The original question in this thread was about loss in a regulator.
I continue to assert that there is little power or energy loss in a
regulator. I'll further note that there is value in a small shop
in minimizing reservoir size by operating at higher pressure with a
2-stage compressor and regulating down at the point of use.

I once had a similar discussion with Scott Foss (a leading expert
in the economics of compressed air usage) and a VP of Engrg at
Ingersoll in Paducah KY. We were in complete agreement both with
our analyses and with the realization that it's very difficult to get
plant engineers to think beyond rote recitation of that which is
printed in Marks with underlying assumptions unquestioningly accepted
though not fully understood. Too bad, since some plants spend
over $1mil/year in energy cost to compress air, so even minor
percentage savings can be significant.

Current practice is still to reduce leaks and stage pumps to demand.
Good ideas to be sure, but far short of what could be done with
advanced controls and energy management at the enterprise level.