In article ,
says...
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.

You're still neglecting the compressibility of the medium.
Units consistency does not guarantee a correct analysis;
ft*lb work vs. lb*ft torque, for example.

As I said, in a hydraulic system the mechanical power
available is equal to pressure * flow. That power is used
(or wasted and turned into heat, as occurs in a hydraulic
regulator or relief) when there is a pressure drop across a
device.

The difference in the case of a compressible fluid is that
there is additional energy available, either as heat or
work, when the fluid expands as a result of the delta P.

Consider the difference between the amount of work that can
be done by a tank of compressed air vs. a tank of
pressurized water.


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.

The piston analogy is exactly on point. It illustrates the
loss thru a regulator, which is the whole point of this
thread, in as simple a manner as I can imagine.

Let me try one more time. You have the system I illustrated
before. Initially the reservoir is charged to 3 atm abs.
There is a 30 lb weight on the piston balancing the
pressure. You reduce the weight in small increments and
when the system comes to mechanical and thermal equilibrium
after each reduction, record the remaining weight and
position of the piston. Since the system is free to
exchange heat with the environment and we wait for
equilibrium before taking our measurements, the process is
isothermal. The red line represents the resulting data. The
area under the line equals the mechanical work done by the
expanding air.

Now repeat the experiment, except put a regulator set to 2
atm between the reservoir and the cylinder and balance the
reduced pressure with a 15 lb weight. The green line
represents the recorded values. Obviously the work done by
the expanding gas is less.

The gas was at the same pressure, temperature, and volume
at the beginning of the two processes, and the end states
are identical as well, so clearly there had to be a loss.
If it wasn't the gas expanding thru the regulator before
doing any mechanical work, then what?

Additionally, without the regulator the process is
reversible merely by replacing the weights, i.e., reversing
the direction of the work. Not so if the regulator is in
the system.

Ned Simmons