Don Foreman wrote:
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.

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.

At 100 PSI the pump pushes twice as much volume at half the pressure
per cycle as it does at 200 PSI but the energy transfer is the same.
It just pushes with less force for a longer time *while moving air*
at lower pressure. In the higher pressure case there's more
alternating exchange of energy between cylinder and flywheel.

Thermodynamics enters in if heat is lost during compression as is
generally the case. Thus Ingersoll's generally-true note that
compressing to pressure higher than needed is inefficient. The loss
is at the compressor, not at the regulator. The compressor gets hot
but the regulator does not, right?

However, if thermal energy is preserved by being removed from the
compressor and used to warm expanding air, then there is essentially
no loss in efficiency.

Take a look at the example Ned posted and consider that the piston is used
to push some mechanical device of variable resistance- the resistance is
infinitesimally less than the available force from the air side along the
entire path. At maximum extension the resistance approaches 0. I think these
conditions woud extract the maximum amount of work out of our fixed volume
of pressurized air. Under these conditions a doubling of pressure would
result in a doubling of force and distance, a quadrupling of work
accomplished, as Ned stated. In your earlier example with a fixed weight,
much of the potential of the air is squandered- consider an airtight piston
of negligible weight- it would also reach an equilibrium with 1 atm on each
side, but there would be almost no work accomplished. The amount of work
that is accomplished is dependent on the design of our device- to say that
at the end a certain weight was only lifted x amount in each case does not
prove that the unregulated case is not capable of doing more work. My
apologies to Ned if I have misunderstood his example.