In article ,
says...
On Thu, 1 Jan 2004 19:41:19 -0500, Ned Simmons
wrote:


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.

I think we're each and both right, wrong, and confused.

If power is defined as rate of energy flow, it does seem clear that
power into the regulator is greater than that coming out. Same number
of grams per sec flowing on both side, high-side higher-pressure air
has more potential energy per gram. I'm as yet unable to reconcile
that with the other definition of power. You're satisfied to explain
that away as irrelevant because air is compressible. Could be, but
the explaination doesn't help me understand things -- my problem, I
know.

I wouldn't say I'm satisfied. I'm very confident that it's
due to the compressibility of the gas, but would be happier
if I could give a more analytical explanation.


However, your example is flawed. I did the math. The height vs
pressure line for the case of no regulator and 30 lb weight is not a
straight line but is curved even in an isothermal case. The reservoir
is all volume on the high-pressure side of the piston and that keeps
changing. There's a displacement term in the denominator that makes
things nonlinear.

Oops. I got caught up in trying to to pick values that make
it easy to plot the end points. The correct plot still
shows the difference in the work output for the two cases,
but the difference is not as great.


I set up the integrals for 45 PSIA (30PSIG) to start with and 30 lb
supported by 1 sq inch of piston, 10 cu in initial reservoir, I let
MathCAD solve the integrals. When you remove weight, you're
harvesting potential energy: amount of weight times how far you
move it back to original height. The integrals show that in the
process of removing the first 15 lb of weight, you only harvest about
32 ft lb of energy. You're never moving the weight very far. Height
at system pressure of 15 PSIG (30 PSIA) is only 5 inches. As you
remove the rest of the weight and lower it to it's initial position,
you harvest an additional 162 ft-lb of energy. The end position when
nearly all weight has been removed is 20 inches. You're only moving
half the weight in the second 15 lbs, but you're moving it a lot
further so you're harvesting a lot of energy there.

In the regulated case, you start with 15 pounds because that's all
the air will support. Once you've removed an infinitesimal amount of
weight, the weight will be elevated until the whole system is
depleted to a hair under 15 PSI (neglecting regulator headroom).
Thereafter, the situation is the same as before because the regulator
is no longer regulating anything. You thereafter get the same 162
ft-lb of energy when removing the rest of the weight. That's less
than the 194 ft-lb you started with but not all that much less.

I just did the same in Mathcad and am getting the same
results. (But shouldn't your units be inch-pounds?) I'm
still playing with the various pressures trying to get a
feel for how this effects the loss.

Would you mind emailing me your Mathcad file so I can
compare to what I'm doing? I'm using Mathcad 2000.


Could a regulator delivering a couple of HP worth of air be
dissipating a significant fraction of an HP as heat? I guess we'd
have to do the thermo to see if the air flow and the cooling due to
expansion would keep the regulator temp from rising. The regulator
is, if anything, cooler than ambient in my system. Note that if any
energy dissipated as heat in the regulator is conveyed to the
downstream air, then that energy is conserved unless it gets away as
heat -- not likely if the reservoir is only slightly above ambient and
expansion in the regulator cools it to below ambient. That's the
case in my system. Exhaust air from my air tools is cooler than
ambient; siometimes much cooler.

The original post involved regulating shop air down to a
much lower pressure to run an HVLP spray gun. I'm pretty
sure that the losses in that case would be significant.
Figuring when it actually becomes uneconomical to do so is
another matter.


That leaves inefficiency back at the pump, where air is compressed
and then cooled. That heat energy is generally lost -- but I've
stipulated that all along. I still think most or all of the
inefficiency due to overcompression and regulation occurs at the pump.
If that heat can be managed, then the system can be efficient.


It wouldn't surprise me at all if the cost of
overcompressing is greater than the loss at the regulator.
But the question that got this all started was whether a
regulator is analogous to a transformer and whether it's
possible to equate voltage and current to pressure and flow
in a compressible medium.

Note that I'm still avoiding discussions of thermo g.

Ned Simmons